A Geometric Approach to Differential Forms by David Bachman

[Tom Mattson]

Hello folks,

I found a lovely little book online called A Geometric Approach to Differential Forms by David Bachman on the LANL arXiv. I've always wanted to learn this subject, and so I did something that would force me to: I've agreed to advise 2 students as they study it in preparation for a presentation at a local mathematics conference. :eek:

Since this was such a popular topic when lethe initially posted his Differential Forms tutorial (http://www.physicsforums.com/showthread.php?t=2953), and since it is so difficult for me and my advisees to meet at mutually convenient times, I had a stroke of genius: Why not start a thread at PF? :cool:

Here is a link to the book:

http://xxx.lanl.gov/PS_cache/math/pdf/0306/0306194.pdf

As Bachman himself says, the first chapter is not necessary to learn the material, so I'd like to start with Chapter 2 (actually, we're at the end of Chapter 2, so hopefully I can stay 1 step ahead and lead the discussion!)

If anyone is interested, download the book and I'll post some of my notes tomorrow.

[mathwonk]

That seems like a gentle enough introduction to differential forms.

I do recommend though at least using them to prove the fundamental theorem of algebra, brouwers fixed point theorem, or even the non existence of vector fields on a 2 sphere. I taught all these in my advanced calculus class in ellensburg, washington in 1972.

let me sketch these:

1) by stokes theorem, if the image of a map of I x S^1 (interval cross the circle) into R^2 misses the origin, then the integral of the pullback of the angle form: dtheta = [-ydx + xdy]/(x^2+y^2), is the same over both copies of the circle {0} x S^1 and {1} x S^1.

Now it not hard to show that if f is a polynomial of degree n, and we choose the radius of our circle large enough, then the map given by H(t,z)
= z^n + tf(z) misses the origin.

But then the integral of dtheta over the image of the circle via f is 2πn.

On the other hand if there were no root of f inside the circle, then again by stokes theorem, this integral would be zero. hence there is such a root.

2) This time we integrate the solid angle form over the sphere, observing it changes sign if we pull back by the antipodal map, sending x to -x. On the iother hand, if there were a non zero tanbgent vector field on the sphere, we could use it to tell us which direction to flow around the sphere from x to -x, thus getting a homotopy as above that implies the two integrals should be the same.

Since the solid angle form integrates to something like 4π) or at least something non zero) over the sphere this is a contradiuction.

3) Brouwers fix point theorem: If some smooth map of the disk to itself has no fixed point then it enables us to write down a map of I x S^1 to S^1, which is the identity on {1}x S^1. But then the integral of dtheta around the circle would be zero and it is not.


My suggestion is that machinery should be built only for a purpose. If you are going to define and belabor the macinery of differential forms and stokes theorem, then you should use it for something.

[selfAdjoint]

Tom, I'm interested; I have the book in my Favorites and have done the excercises of Chapter two. This is very nice! Much more instructive than the MTW approach.

Just an added note; I recently bought Schroedinger's book Spacetime Structure And I'm reading that along with this. S. does a masterful intro to tensors and especially densities, so the parallels to Bachman's text are clear. Since workers in GR, etc, commonly switch back and forth, the combination is a very productive one.

[mathwonk]

i suggest re - reading my post after finishing the book of bachman. it could follow the very last section there.

[Tom Mattson]

Mathwonk, thank you for your suggestions. If you or anyone else thinks that there are some interesting applications that we can investigate before the end of the book, just give a holler.


Just an added note; I recently bought Schroedinger's book Spacetime Structure And I'm reading that along with this. S. does a masterful intro to tensors and especially densities, so the parallels to Bachman's text are clear. Since workers in GR, etc, commonly switch back and forth, the combination is a very productive one.


Sounds good, I'll order it.

I'll be posting notes over the next couple of hours. They will include section summaries, solutions to the exercises, and my own questions. I've asked my advisees to sign up at PF so they can ask questions of their own.

I've also added Bachman's name to the thread title. That way Google searches for the book will be more likely to turn up this thread. Could boost membership at PF.

[Tom Mattson]

Section 1: Coordinates for Vectors

This language of differential forms is new to me, so I think it's important to take note of and summarize the important definitions and concepts. My summary of the text is in black, my homework solutions and comments on what I think needs elucidation are in blue, and my questions are in red.

Tangent Spaces
The section begins with an example of a tangent space. The example is a tangent line to a curve C at point p. The tangent space T_pC of curve C at point p is the space in which all the tangent vectors to C exist.

Bachman also makes the point that the point p is the point at which all of the tangent vectors have their tail. This serves to distinguish T_pC from C in the event that C is a straight line.

Coordinates of Points on Curves and in Planes
Coordinates are described in terms of functions or mappings. For instance on our curve C Bachman considers the point p on a curve C whose x coordinate is 5. He explains that what is really meant is that there exists a coordinate function x : C \rightarrow \mathbb {R} such that x(p)=5. Thus the function "eats" points and "spits out" real numbers. Similary he defines coordinates in the plane P, for which we naturally need 2 functions.

Coordinates of Vectors in Tangent Spaces
Once coordinates on a curve C and in a plane P are defined, the issue of coordinates in T_pP is addressed. Since we are talking about coordinates of vectors in a vector space, the first thing we need is a basis for that space. Bachman "derives" the basis as follows:


\frac {d(x+t,y)}{dt}=<1,0>


\frac {d(x,y+t)}{dt}=<0,1>


where (\cdot , \cdot ) denotes a point in P and <\cdot , \cdot > denotes a vector in T_pP.

Here is my first question.

I say that Bachman "derives" the basis because it looks so contrived. It is obvious that T_pP is just a carbon copy of \mathbb {R}^2 with a different origin. So why not simply use the well-known fact from linear algebra that a basis for this space is {<1,0>,<0,1>}?

Now that the basis has been chosen, we write a vector \mathbf{V} \in T_pP as \mathbf{V} = dx<1,0> + dy<0,1>, dx,dy \in \mathbb{R}.

This represents a conceptual break from the manner in which many calculus books are written. dx and dy are our familiar differentials, which are typically thought of as infinitesimal quantities. Now they are regarded as real-valued coordinate functions in T_pP. The break from the "infinitesimal" conception of dx was foreshadowed on page 39 in Chapter 2.

Illustrative Example
In the example in which we are asked to consider the tangent line to the graph of y=x^2 at the point (1,1), we are given an interpretation of differentials that is not made apparent in most calculus books. He continues with the notion of differentials as coordinate functions by labeling the axes of the coordinate system based at (1,1) with dx and dy, as shown. He presses the point even further by writing down the equation of the tangent line in this coordinate system: dy=2dx, or \frac {dy}{dx}=2.

This leads to my second question.

I have always read and been taught that \frac {dy}{dx} is not to be thought of as a quotient. This point is usually made when introducing the Chain Rule. But if dx and dy are real-valued functions, then there should be no reason why the derivative could be considered a quotient. Can any of our more experienced members comment on how the two points of view may be reconciled?

Bachman also mentions that the tangent line that we are interested in is coincident with T_{(1,1)} \mathbb {R}^2.

This leads to my third question.

Why is this line referred to as a tangent space to \mathbb {R}^2? Why is it not referred to as the tangent space to the curve?



Exercise 3.1
My plan is to post all my solutions, but unfortunately I don't know how to draw vectors with LaTeX, so a verbal description will have to do. This exercise is simple enough, so that shouldn't be a problem.

(1) I have a vector whose tail is at (1,-1) with components 1 and 2.
(2) I have a vector whose tail is at (0,1) with components -3 and 1.

[mathwonk]

i have not read the book yet, but the whole point of differentials on a curve, is that the derivative IS a quotient of them.

I.e. a differential is a linear function on the tangent space. Since the tangent space to a curve is one dimensional, the space of linear functions is also one dimensional.

thus any two linear functionals are scalar multiples of each other, so their quotient is a scalar. this is not true for differentials on higher dimensional tangent spaces.

I cannot explain why this pont of view is prohibited in elementary calculus. perhaps they do not wish to do the work necessary to justify it.

[mathwonk]

well i think you are in for some trouble using this book just because it is free, and i recommend using spivak instead.

anyway, he is not very precise in describing the tangent space Tp(P). it is described more precisely in spivak as {p}xP, so that he does not use the same notation (1,0) and (0,1) for vectors in Tp(P) as for vectors in the disjoint space Tq(P). i.e. he should say {p}x(1,0), etc...

But anyway...

OK further corrections to his sloppiness:

He calls a point of Tp(P) by the name dx(1,0) + dy(0,1), where he says dx and dy are in R. This is not correct, but not too far off. the usual sloppy notation from classical calculus, but wasn't the point here to get things right?

Ok, anyway, he means if v is a vector in Tp(P) then since dx and dy are independent linear functionals on Tp(P), then dx(v) or more precisely dxp(v), is an element of R, so completely precisely, but not too neatly:

he means dxp(v)(1,0)p + dyp(v)(0,1)p is a representation of a point of Tp(P).

you see dx is certaoinly not an element of R, nor even a linear fucntional ,on Tp(P). rather dx is a fucntion whose value at each point p is a inear fucntional on Tp(P). so we need some such notation as dx(p) or dxp. but he seems not to want to introduce enough notation to be correct.

I do not know if i have the patience to correct all this, but you probably do not need me to.

I do suggest you are in for an interesting time reading this somewhat careless treatment of the subject however.

But it is not so far wrong as to be impossible, and the point of math is to have fun, so if you like this book, go for it.

i do suggest spivaks calculus on manifolds however for anyone wanting it explained correctly and precisely.

[Tom Mattson]

Sorry mathwonk, I just now accidentally hit "edit" instead of "quote", so your last post was momentarily replaced by mine. But I put everything back in order.

well i think you are in for some trouble using this book just because it is free, and i recommend using spivak instead.


That's OK. We're here to talk to each other, not do a book review. So I think we can take advantage of the incomplete or rough spots to suit our own purposes.


OK further corrections to his sloppiness:


Let's not be too ungracious. I've invited Bachman here via email to participate in the discussion. :wink:


i do suggest spivaks calculus on manifolds however for anyone wanting it explained correctly and precisely.

I've ordered Apostol and Spivak, per your recommendation.

Mathwonk, thank you for making your points. I'll look at them more thoroughly tomorrow, after I've copped some zzzzz's. :zzz:

[rdt2]

Both the book and this thread look promising - so I'll try to keep up. The fact that the text may sacrifice some rigour at this stage is a positive bonus. In many of the textbooks the wood is too obscured by the trees for them to be useful for self-tuition.

Mind you, my first problem as a stress analyst is to convince myself and my students that adopting a differential forms approach is worth the effort - there's a lot of investment in traditional tensor analysis. So if anyone can fire in some examples from fluid mechanics rather than quantum mechanics, I'd be grateful.

[Bachman]

Hello all,

My name is Dave Bachman. Tom, thanks so much for inviting me to join your thread, and for looking at my book! The version that is up on the arXiv is a little old. A more current one is available on my web page at:

http://pzacad.pitzer.edu/~dbachman

The idea of the text is that one can teach differential forms to freshmen and sophmores instead of the traditional approach to vector calc. I did not write it so that mathematicians, or even grad students, can learn differential forms. There are many good books out there targeted for this audience.

For this reason there is a lot of sacrifice of rigour for readability. The idea was not to "get it right", in the sense of presenting the material with all of its gory, technical details. Another reason I wrote the book was to present the geometric intuition behind forms, which is often lacking in more rigourous texts.

The new version that is up on my web page contains many new exercises, and a new first chapter on the basics from multivariable calculus. There is a lot of time there spent on parameterizations, sicne I had found this to be the biggest stumbling block in learning the rest of the material. Also the new version contains re-writes of several sections that were previously found to be awkward.

I am once again teaching out of my book, and every time I do this I post a new "edition". The next edition, which will be posted in about two months, will contain a new chapter on symplectic forms, as well as many new exercises that are a little more thought-provoking.

As to the comment that it is free.... I'lll try to keep a free version available on the web, but the text is currently being evaluated by a publisher.

Thanks again! I'lll try to write more when I have time....

Dave.

[Tom Mattson]

My name is Dave Bachman. Tom, thanks so much for inviting me to join your thread, and for looking at my book!


Thanks for coming! :smile:


The version that is up on the arXiv is a little old. A more current one is available on my web page at:

http://pzacad.pitzer.edu/~dbachman


I had noticed that, but only after we started. Do you recommend we switch over?


The idea of the text is that one can teach differential forms to freshmen and sophmores instead of the traditional approach to vector calc.


That's exactly why I picked it. I would like to see something like this form the basis of a "Calculus IV" course where I work. That said, I'm not trying to flesh this out to the level of the Advanced Calculus course that mathwonk mentioned. At least not for the purposes of this thread. Personally, I'd love to go through Spivak, and I will once I get it.

Thanks again! I'lll try to write more when I have time....


Great! If possible, could you (or anyone else lurking in this thread) comment on the 3 questions I put in red font in post #6?

Thanks,

[selfAdjoint]

my second question.

I have always read and been taught that is not to be thought of as a quotient. This point is usually made when introducing the Chain Rule. But if and are real-valued functions, then there should be no reason why the derivative could be considered a quotient. Can any of our more experienced members comment on how the two points of view may be reconciled?


The reason the teachers say the derivative is not a quotient is because old textbooks used to use "atomic" differentials and compute it by dividing them, which is convenient (many engineers still think that way) but that is invalid given limit concepts. The derivative is actually a limit of quotients between finite quantities. In the differential forms area the limit is sort of built in, so that when you take the tangent space you have ALREADY got the tangent, with its slope, the derivative. So then if you take a basis in the new space based on that slope, you can play differential without violating rigor.

[Tom Mattson]

OK, I think my second question is covered pretty well. I'll wait another day for anyone who would like to comment on my first and third questions. Then I'll post my notes on the next section.

[Tom Mattson]

OK, I think I've figured out the answers to my other 2 questions.

My first one was:


Here is my first question.

I say that Bachman "derives" the basis because it looks so contrived. It is obvious that T_pP is just a carbon copy of \mathbb {R}^2 with a different origin. So why not simply use the well-known fact from linear algebra that a basis for this space is {<1,0>,<0,1>}?


I plotted the points (x,y), (x+t,y), and (x,y+t) in the plane P. Then I drew vectors from (x,y) to each of the other two points. If I consider that (x,y) is the origin of the coordinate system with axes dx and dy, then I see that the vectors I drew are based in this coordinate system. Taking the derivative of the coordinates leads to the advertised unit vectors, no matter where (x,y) is located in P. So, I can sort of see why this is used as a procedure for determining the basis of T_pP.

I still don't really like it, because it does not explicitly appeal to the linear algebraic notion of a basis. I'd really like it if someone could tell me why this viewpoint is useful, but I won't complain about it again.

My third question pertained to the illustrative example on pp 18-19. It was the tangent space determined from the tangent line of the parabola y=x^2 at (1,1).


This leads to my third question.

Why is this line referred to as a tangent space to \mathbb {R}^2? Why is it not referred to as the tangent space to the curve?



The point that this question is driving at is the apparent variance with the convention from the beginning of the chapter, in which Bachman names the tangent space determined from the tangent line to a curve C as T_pC. But here he calls it T_{(1,1)}\mathbb {R}^2. I am thinking that you can replace T_pC with a tangent space to \mathbb {R}^2 provided that the points along which the tangent spaces exist are constrained to the curve C. That is, any tangent space T_{(x,x^2)}\mathbb {R}^2 is a tangent space to y=x^2.

OK, I will pause for any corrections or additions to this post before posting the next set of notes and homework solutions.

Thanks everyone, this is a real help so far.

[Bachman]

A few quick replies...

First, I do recommend switching to the most current edition, if only because there are more (and better) exercises. If you are really considering the text for Calc IV then the first chapter of the most current edition should definitely be covered, if only as a review from Calc III.

Now on to your question. There must be some confusion generated by something I wrote, but I'm not sure what it is. The tangent space to the curve C ($T_pC$) is a line made up of tangent vectors. The tangent space to $R^2$ at the point $p$ is a plane, with basis $dx$ and $dy$. The line $T_pC$ sits in the plane $T_pR^2$, but it is certainly not the whole plane. So $T_pC$ is a proper subspace of $T_pR^2$. Does this help?

Dave.

[Data]

To get LaTeX typesetting here, just use [ tex ] and [ /tex ] tags (without the spaces). You can double-click on others' math to see how as well~

[Hurkyl]

We also have [ itex ] for LaTeX in paragraphs... it's rendered smaller so it lines up with ordinary text.

The tangent space to $R^2$ at the point $p$ is a plane, with basis $dx$ and $dy$.

Aren't dx and dy supposed to be cotangent vectors, not tangent vectors?

[Tom Mattson]

First, I do recommend switching to the most current edition, if only because there are more (and better) exercises.


OK, I'll switch over.


Now on to your question. There must be some confusion generated by something I wrote, but I'm not sure what it is.

Here is why there is confusion:

On page 17 of the arXiv edition of the book (edit: that's page 47 in the newer version), you refer to the tangent space defined by the tangent line to a curve C as T_pC, not T_p\mathbb{R}^2. Then on pp18-19 (edit: that's pp 48-49 in the newer version), in what I would think is a completely analogous situation, you refer to the tangent space of y=x^2 as not the tangent space of that curve, but as the tangent space T_{(1,1)}\mathbb{R}^2.


Does this help?


Sorry, but no. :redface:

[Bachman]

Oh yes, of course. Thank you. What I meant to say was "The tangent space to \mathbb R^2 at the point p is a plane, with AXES dx and dy ."

[Bachman]

Tom,

I'm still not sure where the confusion lies. The tangent space to C is a line, denoted as T_pC . At the bottom of page 18 I say "We are no longer thinking of this tangent line (i.e. the space T_pC ) as lying in the same plane that the graph does. Rather, it lies in \mathbb T _{(1,1)} \mathbb R ^2 ."

I'm not sure how you are getting the impression, from this, that T_pC is all of \mathbb T _{(1,1)} \mathbb R ^2 .

By the way, thanks all for the latex advice.

Dave.

[Tom Mattson]

I'm not sure how you are getting the impression, from this, that T_pC is all of \mathbb T _{(1,1)} \mathbb R ^2 .


OK, I've got it. The tangent space to the parabola is a proper subspace of T_{(1,1)}\mathbb{R}^2. No problem.

[mathwonk]

my apologies dave, for the picky mathematician criticisms of a text aimed at undergrads. tom is also helping me learn which explanatins are tenable for the desired audience.

clearly you yourself know what the correct version is, and have made didactic choices based on teaching experience.

i would edit out the ungracious late night posts but cannot do so now after a certain number of days have passed.

roy

[Tom Mattson]

Section 2: 1-Forms

Once again:

My notes are in black.
My comments and homework solutions are in blue.
My questions are in red..

I'll pause 24 hours for discussion, questions, and corrections. If none are forthcoming, then I will post the next section of my notes tomorrow night at about the same time.

1-Forms
A 1-form \alpha is a linear function that maps vectors into real numbers. Since it is called "linear", we require it to satisfy:

\alpha (\mathbf{v}+\mathbf{w})=\alpha (\mathbf{v}) + \alpha (\mathbf {w})

\alpha (k \mathbf{v})=k\alpha (\mathbf{v})


Quick question:

Are "1-form" and "linear functional" synonymous?


The geometric interpretation of \omega is that of a plane whose graph passes through the origin in the dx-dy coordinate system. Fixing our attention on 1-forms on T_p\mathbb {R}^2, we see that our general 1-form is \omega = a dx +b dy. This is the equation of a plane in T_p\mathbb{R}^2 \times \mathbb{R}.


Just a note of clarification for students: "\times" denotes a Cartesian product, which makes n-tuples out of elements of sets. For instance \mathbb{R} \times \mathbb{R} is the set of all ordered pairs of real numbers. And in our case, T_p\mathbb{R}^2 \times \mathbb{R} indicates that we are forming n-tuples from ordered pairs in T_p\mathbb{R}^2 (the coordinates for dx and dy) and a member of \mathbb{R} (the value of \omega).


Illustrative Example
For \omega (<dx,dy>)=2dx+3dy, evaluate \omega (<-1,2>).

This is easily done by plugging in the components of <-1,2> into the right places in \omega:

\omega (<-1,2>)=(2)(-1)+(3)(2)=4

And we are to take note that \omega (<-1,2>) is just the dot product <-1.2> \cdot <2,3>

Note that we can make a vector out of the coefficients in \omega. We can call it < \omega >=<a,b>. This notation is not introduced until Section 2.3, but I think it would be nice to have it now for shorthand.

So a recipe for evaluating a 1-form on a given vector is:


\omega (V) = <\omega> \cdot V



This brings us to the main point of the section: the geometric interpretation of 1-forms.


Evaluating a 1-form on a vector is the same as projecting onto some line and then multiplying by some constant.


This of course has the huge advantage of being independent of coordinates. Anyone who has studied relativity can see the value of this!


So now we know how to use a given 1-form to determine the projection of a vector onto a line, and we can then determine the scaling factor. What if we want to do things the other way around? What if I am given a line L, a scaling factor k, and a vector V? Recall from vector calculus that the dot product is related to the projection of a vector onto a line:

proj_{\mathbf {u}}\mathbf {v}=\frac {\mathbf {u} \cdot \mathbf {v}}{|\mathbf {v}|}

So say I want to write down a differential form that projects vectors onto a line L: dy=c dx and scales them by a factor of k (this will be asked of us in the Exercises). Since the slope of L is c=\frac {c}{1}, it is readily seen that a vector that is parallel to L is W=<1,c>. Since we are looking for the projection of V onto a line parallel to W, we look at:


proj_WV=\frac {W \cdot V}{|W|}


proj_WV=\frac {<1,c> \cdot V}{\sqrt {1+c^2}}


Upon comparing this with our expression for \omega above, it should be clear that our vector W is nothing other than <\omega>. Furthermore, I can scale the projection by a factor of k by multiplying both sides of the above projection by that factor.


k{}proj_WV=k \frac{<1,c> \cdot V}{\sqrt {1+c^2}}


So we can now find the differential form \omega that projects V onto dy=cdx and scales by a factor of k, because we have just derived a function that does that very thing. Recognizing that:


<\omega>=<\frac{k}{\sqrt {1+c^2}},\frac{ck}{\sqrt {1+c^2}}>


we have:


\omega=\frac{k}{\sqrt {1+c^2}}dx+\frac{ck}{\sqrt {1+c^2}}dy



1-Forms in \mathbb{R}^n
All of this straightforwardly generalizes to n dimensions. There is no need for elaboration.

[Tom Mattson]

Looks like my last post was too big, so I'm splitting it up.


Exercise 3.2
(1) \omega = -dx+4dy. That means that < \omega > =<-1,4>.
\omega (<1,0>)=<-1,4> \cdot <1,0>=-1
\omega (<0,1>)=<-1,4> \cdot <0,1>=4
\omega (<2,3>)=2\omega (<1,0>) + 3\omega (<0,1>)=10

Note that I used a linear combination of \omega(<1,0>) and \omega(<0,1>) to evaluate \omega(<2,3>). This is done in the spirit of Bachman's second geometric interpretation of \omega, which is:



Evaluating a 1-form on a vector is the same as projecting onto each coordinate axis, scaling each by some constant, and adding the results.



It should not be difficult to see that this is true in general.

(2) Find the line that \omega projects onto.
Since the line is parallel to <-1,4> and it passes through the origin in T_p\mathbb{R}^2, it must be dy=-4dx.

Exercise 3.3
I will use the formula I derived in these Section notes.
(1) c=2 and k=2, so \omega=\frac{2}{\sqrt {5}}dx+\frac{4}{\sqrt {5}}dy.
(2)c=\frac {1}{3} and k=\frac{1}{5}, so \omega=\frac{3}{5 \sqrt {10}}dx+\frac{1}{5 \sqrt {10}}dy.
(3)c=0 and k=3, so \omega=3dx.
(4) Here c is undefined, but in light of (3) it shouldn't be too taxing to see that \omega=\frac{1}{2}dy.
(5) Since 1-forms are linear, we have superposition, so \omega=3dx+\frac{1}{2}dy.

[mathwonk]

answer to quick question: usually a 1 form is defined on manifold as a family of linear functionals, i.e. not as one linear function from vectors to numbers, but as an assignment of such a function to each point of the manifold.


in my usual notation dx is a 1 form, and its value at p dx(p) is a linear functional on the tangent space Tp(M).

this is analogous to the distinction between f' and f'(p). in fact the differential of f, in local coordinates x, is the 1 form f'dx whose value at p is f'(p)dx(p). more simply, if incorrectly, written as f'(p)dx.

reactions from the others? i may be out of step here, but i am trying to point out what most people in the community are going to mean by these terms.

[mathwonk]

there is some discrepancy in the literature in the use of the word "form". algebraists do indeed use the word for a linear functional. lang in his algebra book, calls an alternating k tensor, a k - form. classicists (analysts?) have long used the word "form" for linear functionals, and algebraists also used it for homogeneous polynomials of higher degrees.

differential geometers who use it as i said above, are thus left without a good short word for the value of a k form at a point, and must call it an "alternating k tensor" as spivak does in his little "calculus on manifolds".

there are two ideas though, a covector, and a field of covectors. call them what you will.

[Bachman]

I make a distinction in my book between "1-form" and "differential 1-form." A 1-form is, indeed, a linear functional. It acts on a single tangent space. So, choosing a specific point p, a 1-form is a linear functional on T _p \mathbb R^n . A "differential 1-form", on the other hand, is a (differentiable) choice of 1-form for each tangent space. You'll get to this in the next chapter.

Dave.

[mathwonk]

forgive me for not reading more closely. i have already perused the whole book quickly. since i already "know" everything in it, i am too impatient to read along in detail. so my comments should be pretty much ignored by learners.

[Gza]

I had a question after reading prof. bachman's book. On page 45 of the new edition, he shows a function denoted by ω within the integrand, to be a an n-form, based upon the n vectors that ω takes in as an input. Isn't ω none other than the jacobian? Here's the integral from page 44 with the text "Area" replacing ω, to show the purpose of ω.

/
| f(φ(r, θ))Area [∂φ/∂r(r, θ),∂φ/∂θ(r, θ)]drdθ (1)
/


Area [∂φ/∂r(r, θ),∂φ/∂θ(r, θ)] = | <∂φ/∂r(r, θ)> X <∂φ/∂θ(r, θ)> | (2)


if i'm correct, the right side of 2 is the jacobian. how does this relate to n-forms on a "bigger picture" level?

[Bachman]

The equation on page 45 is supposed to motivate the study of n-forms. The integrand there is not an n-form. But it IS a function that takes two vectors and returns a real number. The point illustrated there is that you need such a function if your answer is going to be independent of the choice of parameterization. For such an integrand to be an n-form, it must also be linear (which the "Area" function is not in \mathbb R^3 ).

Dave.

[mathwonk]

Dave, when you say an n form is "linear" do you mean what most people call "n - linear"? i.e. linear in one variable at a time?

and are they also alternating?

[Bachman]

Yes, yes. Technically, an n-form on a vector space M is a multi-linear, alternating operator on the cartesian product of n copies of M.

Dave.

[Gza]

I hate to jump off the immediate topic of the material in the book, but I just had a quick question about the application of differential forms. Would learning it simply help me to broaden my understanding of calculus, or would it also have some sort of practical (applying to physics, i'm a physics major) applications as well? I'm familiar with the concept of stating maxwell's equations in the language of differential forms, thus making them simpler, but i'm already pretty much comfortable with them in the integral and differential formulations of the laws. What other areas of physics and math would be open to me after study of differential forms?

[Tom Mattson]

What other areas of physics and math would be open to me after study of differential forms?

Anything involving vector fields, for starters. You can use them in Fluids, GR, and of course as you already noted, EM. The last chapter of Bachman's book discusses EM theory. They can also be applied to thermodynamics. But I am going to ask that this thread be reserved for a sequential discussion of the book. We can talk about all the applications you want at the end.

Since the discussion of my last set of notes has died down, I am going to post the next set later tonight.

Stay tuned...

[Tom Mattson]

Section 3: Multiplying 1-Forms

The first problem here is how to define a product of 1-forms. Why not \omega \cdot \nu (V) \equiv \omega (V) \cdot \nu (V)? Because it’s nonlinear.


To make the violation of linearity more explicit, note that superposition is violated:


\omega\cdot\nu(V_1+V_2)=\omega(V_1+V_2)\cdot\nu(V_ 1+V_2)


\omega\cdot\nu(V_1+V_2)=[\omega(V_1)+\omega(V_2)]\cdot[\nu(V_1)+\nu(V_2)]


\omega\cdot\nu(V_1+V_2)=\omega(V_1)\cdot\nu(V_1)+\ omega(V_2)\cdot \nu(V_2)+\omega(V_1)\cdot\nu(V_2)+\omega(V_2)\cdot \nu(V_1)


\omega\cdot\nu(V_1+V_2)\neq\omega\cdot\nu (V_1)+\omega\cdot\nu(V_2)


And note that the scaling property is violated:


\omega\cdot\nu(cV)=\omega(cV)\cdot\nu(cV)


\omega\cdot\nu(cV)=c^2\omega(V)\cdot\nu (V)


\omega\cdot\nu(cV)\neq c\omega\cdot\nu(V)



So instead of taking the simple product of \omega and \nu, we define the wedge product \omega \wedge \nu. Since we can use \omega and \nu to act on V_1 and V_2 to generate pairs of numbers, it stands to reason that the natural geometric setting in which we should be operating is the a plane, namely the \omega - \nu plane.

Notation
(a,b) denotes a point in the x-y plane.
<a,b> denotes a vector in the x-y plane.
[a,b] denotes a vector in the \omega - \nu plane.


Quick question:

Is there any subtle distinction between the coordinates of a vector and the components of a vector, or are they synonymous?


Geometric Interpretation of the Wedge Product
We don't want to use our product of 1-forms to generate a pair of vectors, we want to use it to generate a number. That number is defined to be the signed area of the parallelogram spanned by the vectors [\omega(V_1),\nu(V_1)] and [\omega(V_2),\nu (V_2)] in the \omega - \nu plane.


As we know from Calculus III, two vectors V_1=<a,b> and V_2=<c,d> in \mathbb {R}^2 span a parallelogram with signed area given by:


Area=(V_1\timesV_2) \cdot \hat {k}=\left |\begin{array}{cc}a&b\\c&d\end{array}\right|=ad-bc


Similarly two vectors [\omega(V_1),\nu(V_1)] and [\omega(V_2),\nu(V_2)] in T_p\mathbb{R}^2 span a parallelogram with signed area given by:


Area=\omega \wedge \nu (V_1,V_2)=\left |\begin{array}{cc}\omega(V_1)&\nu(V_1)\\\omega(V_2)&\nu(V_2)\end{array}\right|=\omega(V_1)\nu(V_2)-\omega(V_2)\nu(V_1)


Clearly the sign of the area depends on the order of the vectors in the cross product or the wedge product, as the case may be.



Just anticipating an obvious question that would be asked by an astute student:

If all we're doing here is defining the wedge product in terms of something that could just as easily be expressed in terms of a cross product, why bother defining the wedge product at all? Why not just take the cross product of vectors in the \omega - \nu plane?



We noted earlier that we did not want the simple product of 1-forms because it is nonlinear, and I showed as much in my notes. Now I want to show that the wedge product is linear.

Superposition
Checking the superposition property on \omega \wedge \nu (V_1, V_2) leads us to the following.


\omega\wedge\nu(V_1+V_2,V_3)=\left|\begin{array}{c c}\omega(V_1+V_2)&\nu(V_1+V_2)\\\omega(V_3)&\nu(V_3)\end{array}\right|



\omega\wedge\nu(V_1+V_2,V_3)=\left|\begin{array}{c c}\omega(V_1)+\omega(V_2)&\nu(V_1)+\nu(V_2)\\\omega(V_3)&\nu(V_3)\end{array}\right|



\omega\wedge\nu(V_1+V_2,V_3)=[\omega(V_1)+\omega(V_2)]\nu(V_3)-\omega(V_3)[\nu(V_1)+\nu(V_2)]



\omega\wedge\nu(V_1+V_2,V_3)=[\omega(V_1)\nu(v_3)-\omega(V_3)\nu(V_1)]+[\omega(V_2)\nu(V_3)-\omega(V_3)\nu(V_2)]



\omega\wedge\nu(V_1+V_2,V_3)=\omega\wedge\nu(V_1,V _3)+\omega \wedge\nu(V_2,V_3)


Check.

In a similar fashion it can be shown that:

\omega\wedge\nu(V_1, V_2+V_3)=\omega\wedge\nu(V_1,V_2)+\omega\wedge\nu( V_1,V_3)

[Tom Mattson]

Section 3: Multiplying 1-Forms (cont'd)


Scaling
The other property to check is scaling.


\omega\wedge\nu(cV_1,V_2)=\left|\begin{array}{cc}\ omega(cV_1)&\nu(cV_1)\\\omega(V_2)&\nu(V_2)\end{array}\right|



\omega\wedge\nu(cV_1,V_2)=\left|\begin{array}{cc}c \omega(V_1)&c\nu(V_1)\\\omega(V_2)&\nu(V_2)\end{array}\right|



\omega\wedge\nu(cV_1,V_2)=c\omega(V_1)\nu(V_2)-c\omega(V_2)\nu(V_1)



\omega\wedge\nu(cV_1,V_2)=c\omega\wedge\nu (V_1,V_2)


Check.

In a similar fashion it can be shown that:

\omega\wedge\nu(V_1,cV_2)=c\omega\wedge\nu(V_1,V_2 ).

Because \omega\wedge\nu(V_1,V_2) is linear in both variables, it is said to be bilinear. See the exchange between mathwonk and Bachman in Posts #32-33 on n-linearity.


Lastly, we address the issue of signed areas. When we defined the wedge product we defined it as the signed area of the parallelogram spanned by the vectors [\omega(V_1),\nu(V_2)] and [\omega(V_2),\nu(V_2)].

Bachman sez:


Should we have taken the absolute value? Not if we want to define a linear operator.



My next question is for the students:

Would any of you like to show this? Check my notes for how to show linearity and non-linearity (think superposition and scaling).

[Bachman]

If all we're doing here is defining the wedge product in terms of something that could just as easily be expressed in terms of a cross product, why bother defining the wedge product at all? Why not just take the cross product of vectors in the \omega-\nu plane?


Because the \omega-\nu plane is two-dimensional, and cross products are only defined for three-dimensional vectors.

Dave.

[Tom Mattson]

Because the \omega-\nu plane is two-dimensional, and cross products are only defined for three-dimensional vectors.


OK but that just changes the question. My ficticious student could then say that the same is true of the x-y plane, but that we can define cross products by defining a third axis that is orthogonal to the 2 existing axes. Why can't the same be done for the \omega-\nu plane?

[mathwonk]

the cross product is defined for n-1 vectors in n-space, and the value is a vector in that space. hence it is only defiend for 2 vectors in 3 space. [which orthogonal direction are you going to choose for a given plane in n-space?]

It also depends on a choice of determinant for the larger space, i.e. of n-form.

the wedge product is defined for two vectors in n-space, and the value is a 2-vector, an element of a space of dimension "n choose 2".

[Gza]

I like the geometric interpretation of the 2-form as the area of the parallelogram of the projection of vectors <V1> and <V2> onto the plane spanned by <ω> and <ν>, multiplied by the area of the parallelogram formed by <ω> and <v>, since it seems like a natural extension of the geometric interpretation of the one form, involving the dot product of <ω> and <V>; but it still seems difficult for me to switch back between this geometric interpretation of forms, and the idea of a 2-form for instance, as being a function ω^v:T_p\mathbb{R}^3 X T_p\mathbb{R}^3 -> \mathbb{R} . For learning purposes, how exactly should one think about forms?

[mathwonk]

klingon interpretation:
a k form is sort of like a bird of prey that hovers over the space looking for a k-cycle. when it sees one it gobbles it up and spits out a number.

[Gza]

What is a k-cycle, if i may ask? I would assume it to be a collection of k, n-vectors within T_p\mathbb{R}^n, is this close?

[Tom Mattson]

the cross product is defined for n-1 vectors in n-space, and the value is a vector in that space. hence it is only defiend for 2 vectors in 3 space.


Yep, I know all that. What I was originally asking is this:

From the point of view of a calculus student, what would be your answer to the following question at this stage in the game:

The Big Question:
"Why are we introducing the wedge product to find the area of a parellelogram, when we could just as well take a projection of a cross product, which we already know how to do?"

I already know that cross products and wedge products are two different animals, and I also know that we will eventually integrate them (actually, my advisees and I are doing that now). What I am asking is, do I tell a student who asks the question above to just sit tight and wait to see why we introduce the wedge product, or is there some reason that it's necessary now?


[which orthogonal direction are you going to choose for a given plane in n-space?]


Well, you said it yourself: the cross product is defined for n-1 vectors in n-space. I am still not seeing why taking cross products with our vectors living in the \omega - \nu plane is prohibited, as long as a 3rd axis is defined.

But if the answer to my Big Question above is, "You tell the student to sit tight and wait until the next chapter", then I'll settle for that.

By the way, my copy of Spivak's Calculus on Manifolds is due in on Saturday, and my copy of his Calculus is due in 2 weeks later. If the latter is all it's cracked up to be, then I may try to get my Department Chair to switch over. We currently use Larson, Hostetler and Edwards, which I am certain you would call a "cookbook".

More notes tomorrow...

[Tom Mattson]

Section 3: Multiplying 1-Forms (cont'd)

Here are my homework solutions for the exercises that cover the material we've done so far. In my last set of notes, I posted a question to the students on the nonlinearity of 2-forms when the area of the parallelogram is unsigned. I'll post my solution to that tomorrow, if no one takes me up on it. I'll also finish posting Section 3.3 notes tomorrow.


Exercise 3.4

(1) Evaluating the four 1-Forms:
\omega(V_1)=<2,-3> \cdot <-1,2>=-8
\nu(V_1)=<1,1> \cdot <-1,2>=1
\omega(V_2)=<2,-3> \cdot <1,1>=-1
\nu(V_2)=<1,1> \cdot <1,1>=2

(2) Evaluating the 2-Form:


\omega\wedge\nu(V_1,V_2)=\left |\begin{array}{cc}\omega(V_1)&\nu(V_1)\\\omega(V_2)&\nu(V_2)\end{array}\right|



\omega\wedge\nu(V_1,V_2)=\omega(V_1)\nu(V_2)-\omega(V_2)\nu(V_1)


\omega\wedge\nu(V_1,V_2)=(-8)(1)-(-1)(1)=-7


(3) Expressing \omega\wedge\nu as a multiple of dx\wedge dy.
Let V_1=<w,x> and V_2=<y,z>. Then \omega\wedge\nu(V_1,V_2)=5(wz-xy).

Letting dx\wedge dy act on the same two vectors yields dx\wedge dy(V_1,V_2)=wz-xy. On comparison it is readily seen that the constant of proportionality is c=5.

Exercise 3.5
Skew-symmetry of \omega\wedge\nu(V_1,V_2)


\omega\wegde\nu(V_1,V_2)=\left |\begin{array}{cc}\omega(V_1)&\nu(V_1)\\\omega(V_2)&\nu(V_2)\end{array}\right|



\omega\wegde\nu(V_1,V_2)=\omega(V_1)\nu(V_2)-\omega(V_2)\nu(V_1)



\omega\wegde\nu(V_1,V_2)=-[\omega(V_2)\nu(V_1)-\omega(V_1)\nu(V_2)]



\omega\wedge\nu(V_1,V_2)=-\left |\begin{array}{cc}\omega(V_2)&\nu(V_2)\\\omega(V_1)&\nu(V_1)\end{array}\right|



\omega\wedge\nu(V_1,V_2)=-\omega\wedge\nu(V_2,V_1)


Exercise 3.6
Using the result from the previous exercise and letting V_1=V_2=V:


\omega\wedge\nu(V,V)=-\omega\wedge\nu(V,V)


2\omega\wedge\nu(V,V)=0


\omega\wedge\nu(V,V)=0


Exercise 3.7
Done in Notes.

Exercise 3.8


\omega\wedge\nu(V_1,V_2)=\left |\begin{array}{cc}\omega(V_1)&\nu(V_1)\\\omega(V_2)&\nu(V_2)\end{array}\right|



\omega\wedge\nu(V_1,V_2)=\omega(V_1)\nu(V_2)-\omega(V_2)\nu(V_1)



\omega\wedge\nu(V_1,V_2)=-[\nu(V_1)\omega(V_2)-\nu(V_2)\omega(V_1)]



\omega\wedge\nu(V_1,V_2)=-\left |\begin{array}{cc}\nu(V_1)&\omega(V_1)\\\nu(V_2)&\omega(V_2)\end{array}\right|



\omega\wedge\nu(V_1,V_2)=-\nu\wedge\omega(V_1,V_2)


Exercise 3.9


\omega\wedge\omega(V_1,V_2)=\left |\begin{array}{cc}\omega(V_1)&\omega(V_1)\\\omega(V_2)&\omega(V_2)\end{array}\right|



\omega\wedge\omega(V_1,V_2)=\omega(V_1)\omega(V_2)-\omega(V_2)\omega(V_1)



\omega\wedge\omega(V_1,V_2)=0


Exercise 3.10
Distribution of \wedge over +.


(\omega+\nu)\wedge\psi(V_1,V_2)=\left |\begin{array}{cc}(\omega+\nu)(V_1)&\psi(V_1)\\(\omega+\nu)(V_2)&\psi(V_2)\end{array}\right|



(\omega+\nu)\wedge\psi(V_1,V_2)=\left |\begin{array}{cc}\omega(V_1)+\nu(V_1)&\psi(V_1)\\\omega(V_2)+\nu(V_2)&\psi(V_2)\end{array}\right|



(\omega+\nu)\wedge\psi(V_1,V_2)=[\omega(V_1)+\nu(V_1)]\psi(V_2)-[\omega(V_2)+\nu(V_2)]\psi(V_1)



(\omega+\nu)\wedge\psi(V_1,V_2)=[\omega(V_1)\psi(V_2)-\omega(V_2)\psi(V_1)]+[\nu(V_1)\psi(V_2)-\nu(V_2)\psi(V_1)]



(\omega+\nu)\wedge\psi(V_1,V_2)=\left |\begin{array}{cc}\omega(V_1)&\psi(V_1)\\\omega(V_2)&\psi(V_2)\end{array}\right| + \left |\begin{array}{cc}\nu(V_1)&\psi(V_1)\\\nu(V_2)&\psi(V_2)\end{array}\right|



(\omega+\nu)\wedge\psi(V_1,V_2)=\omega\wedge\psi+\ nu\wedge\psi

[mathwonk]

Tom, I assumed you were working in n space, in which case there is no natural way to choose a 3rd axis. were you actually working in 3 space?

in that case I would say to the student that there is a special definition that works in 3 space but never works again, and we are trying to learn a method that will always work.

[If the stated purpose of your course is to learn about differential forms, it seems odd that a student would say, I don't want to learn how it is done with differential forms, I'd rather do it the old way.]

but maybe he is asking what does differential forms have to offer if his old way works as well.

in that case i would appeal to the fact that the diff forms approach generalizes to higher dimensions.

[Tom Mattson]

Tom, I assumed you were working in n space, in which case there is no natural way to choose a 3rd axis. were you actually working in 3 space?


In this particular case we are working in 2-space, and taking advantage of a 3rd axis when talking about the cross product. As I said, I was wondering what to say to a student in regards to why we couldn't take the cross product in the \omega - \nu plane.


in that case I would say to the student that there is a special definition that works in 3 space but never works again, and we are trying to learn a method that will always work.


Good enough, then.


[If the stated purpose of your course is to learn about differential forms, it seems odd that a student would say, I don't want to learn how it is done with differential forms, I'd rather do it the old way.]

but maybe he is asking what does differential forms have to offer if his old way works as well.


Exactly. My advisees are making a presentation to an undergraduate math conference, and one of their points is that differential forms is superior to the old way in which vector calculus is typically presented. And before they do that, they will be giving a practice presentation to a skeptical faculty at our community college. i am just trying to anticipate the objections that they might raise.


in that case i would appeal to the fact that the diff forms approach generalizes to higher dimensions.

There we have it, then. Thanks.

[mathwonk]

Tom,

Please excuse for rattling on, but i think i can do better than my last post, in the light of day.

I am a little rusty on cross products, but it seems to me that for one thing, differential forms methods are easier.

so maybe one could work up a little demonstration of the superior ease of wedge products.

e.g. one could use the properties of wedge products to actually compute the formula for a determinant. e.g. taking the wedge of v^w =
(ae1 + be2)^(ce1+de2)

gives ac e1^e1 + bc e2^e1 + ad e1^e2 + bd e2^e2

= ac (0) - bc (e1^e2) + ad e1^e2 + bd (0) = (ad-bc) e1^e2.

the same thing works for two vectors v,w in 3 space and gives three terms, where each term is then visibly a 2 by 2 determinant, i.e. the area of a projection of the parallelogram spanned by v,w into one of the three coordinate planes.

again, excuse me if i am out of touch with skillful use of cross products, but it seems to me that in that approach one simply memorizes all the formulas, and either memorizes the explicit coefficients of a cross product, or writes it as a formal determinant, and then must already know how to expand a determinant.


so of course in the one dimension where they overlap, the two methods are equivalent, since both amount to forming a 3 by 3 determinant, but the one seems more natural to me, and easier, since it is absed on axioms instead of memorized formulas. it also generalizes better.


it also gives an algebra for geometry, as originally envisioned by grassman, i.e. he was trying to calculate with objects which represented liens, and planes and 3 spaces etc,,,in n space.


thus one thinks of a simple ("decomposable") wedge product v^w^u, as representing the span of the 3 vectors u,v,w, in n space, except it degenerates to zero if they are dependent.

so it is sort of a tool for detecting when r vectors in n spaces are dependent.


thats all i can think of.

best wishes,

roy


oih yes, the cross product method is also less natural since even in three space it replaces a vector parallelogram, spanned by v and w, with a single vector vxw perpendicualr to that parallelogram, and having length equal to the area of the parallelogram.


why does one want to replace a natural geometric object like a parallelogram by a single vector, perpendicualr to it?

Even though it seems to me unnatural, one pretty aspect of that duality, is the pythagorean theorem. i.e. there are two pythagorean theorems, one for the parallelogram, wherein the square of the area of the parallelogram equals the sum oif the squares of the areas of the three projected parallelograms. this is dual to the fact that the squared length of the cross product vector equals the sum of the squares of the lengths of its three projected vector components.


so the general phenomenon is that a sequence of r independent vectors in n space span an r dimensional parallelogram, and it is dual to another (n-r) dimensional parallelogram with presumabl;y the same area?


this duality depends on having an inner product, whereas the wedge product formulation does not. moreover in general there is no good reason to replace an r dimensiona parallelogram by an n-r dimensional one.

but in the one case of three space, it lets us replace an object possibly less intuitive, i.e. a parallelogram, by a simpler one, a vector.


so the cross product approach has many disadvantages:

1) it depends on more structure, namely that of a dot product and consequent notions of orthogonality.

2) it has less intuitive meaning. i.e. what is the point of representing a planar object by a vector object?

3) it is special to three dimesional space where 2- planes are orthogonally dual to lines.

4) it is harder to calculate with, at least for me, whereas the wedge product ahs all its rules for calculating "built in", so that computing with it is easy and mechanical.

5) wedge multiplication meshes well with (exterior) differentiation d, rendering all vector calculus formulas the same, i.e. there is no longer three versions of stokes theorem (greens theorem, gauss theorem, stokes theorem, divergence theorem) but only one.

anyone can remember it:
the integral of dP over K, equals the integral of P over the boundary of K.

[where d(fdx) = df ^ dx for example,....so curl(fdx + gdy) = d(fdx+gdy)

= [df/dx dx + df/dy dy]^ dx + [dg/dx dx + dg/dy dy] ^ dy

= dg/dx - df/dy] dx ^ dy (I have to run to class so i hope this is somewhere near right.]

i.e. integration makes d the "adjoint" of boundary.

In fact probably the nicest mechanical calculation associated to wedge products is that of grad, curl, and div.

i.e. the computation of grad f, curl (w) and div(M) becomes absolutely trivial. even i can remember them. more detail on this if desired.

i think a good demonstration of the effectiveness of wedge products would be a demonstration of how, when combined with d, it uniformizes all these classical theorems.

[mathwonk]

here is anotherr eason not to sue cross products in 2 space by choosing another orthogonal direction:

in 2 space the issue is simply to compoute a 2 by 2 determninant. it seems a big waste of energy to go to three dimensions, then compute a 3 by 3 determinant most of whose components are zero, just to get a 2 by 2 determinant.


so cross products in 2 space are even easier to dismiss as a reasonable method.

[mathwonk]

a look at the generalized stokes theorem on page 104 of dave's book, and his nice table on page 110, contrasting the different looking classical version of the theorems with the completely unified looking versions on the right side of the table, should convince most people this is the way to go.

[mathwonk]

a look at the generalized stokes theorem on page 104 of dave's book, and his nice table on page 110, contrasting the different looking classical version of the theorems with the completely unified looking versions on the right side of the table, should convince most people this is the way to go.

for me personally, this lovely synthesis made me feel i could relax about these theorems after merely understanding green's theorem for a rectangle!

[Gza]

I have a question regarding how you would find the constituent "wedged" one-forms making up a 2-form if you happened to know the 2-form. For instance, if you knew a 2-form to be α = 3dx^dy + 2dy^dz +4 dx^dz, how would you find both one-forms, and if so, would they even be unique? I tried a painful method of writing out the one forms with yet to be determined constants, and plugging in the basis vectors <1,0,0>, <0,1,0> and <0,0,1>, trying to match the constants with the "scaling factors" for each term in α. I'm sure there is a smarter way to do this, but how?

[Tom Mattson]

Gza,

The way you describe is exactly how we did it. There are exercises that ask us to do precisely this a little later, and I'll post my solutions probably tomorrow, after I've fully digested mathwonk's posts (*burp*).

BUT, this method is not all that painful. I took \alpha=a_1dx+a_2dy+a_3dz and \beta=b_1dx+b_2dy+b_3dz. Note that we have 6 constants, but only 3 constraining equations. That means that you get to pick 3 of the constants, so no the choices are not unique. Once you pick 3, finding the other 3 is easy.

My standard way of doing it is to let a_1=a_2=b_1=1.

[mathwonk]

start with the last and shortest one.

[mathwonk]

this is not much since my best intentions yesterday foundered on lack of energy, end of week binge, and ignorance. but so what, here goes: maybe someone else will fix it.

the idea of grassman was apparently to create an algebra of geometric objects. i.e. he wanted to generalize the algebra of one dimensional vectors to an algebraic technique allowing him to add also 2 dimensional objects, 3 dimensional objects etc.

so think about a vector spanning a line. there are many vectors spanning the same line, and they differ only by a scale factor, the quotient of their lengths.

to generalize we let a pair of vectors represent a parallelogram spanning a plane. two different parallelograms in that plane span the same plane and differ only by a scale factor, the quotient of their areas. so we equate two parallelograms if they span the same plane and have the same area.

given two vectors, their product is the parallelogram they span, up to this equivalence relation. hence dependent vectors have product zero.

now how do we add two such parallelograms? acn we do this so as to get another parallelogram? well we could try, in three space, in the following way: size them up [within their equivalence classes] so they have the same length side on one side of each and thus fit together as two sides of a parallelepiped. then they span a unqie parallelpiped, which thus has a third side, which might be their sum.

alternatively we could use the dot product on three space to replace each parallelogram by a single vector as follows: given an ordered parallelogram, find a vector orthogonal to the paralll... and take it to have length equal to the area of the parallelogram, and be oriented so as to obey the right hand rule, i.e. form the "cross product" of the two sides of the parallelogram.

then in the reverse order, a vector also determiens a plane orthogonal to it, as well as an equivalence class of parallelograms in that plane all having area equal to the length of the given vector.

then to add two parallelograms we could simply add their cross product vectors and then pass back to the asociated parallelogram. hopefully this gives thesame answer as the first methd but i have not thought about why it should except that life is often simple, and i am an optimist.

in particular, this seems to show that the sum of two parallelograms is always another parallelogram, up to equivalence, in three space.

but what happens in 4 space? when we try toa dd two parallelograms, the planes they spane need not meet,a nd so there is no parallelepiped, and no unique orthogonal vector. the oprthogoanjl complement now is another plane, which gives no advantage over the original object. so we must simply add the parallelograms in a more formal way.

i.e. now we allow formal sums of two or more parallelograms, and call them something like 2 - chains, or whatever. again we have an equivalence relation, and we et a vector space of these things but no longer is it true that every object in this space is a simple parallelogram, i.e. the product of just two vectors.

but anyway we do get an algebra of objects generated by parallelepipeds of various dimensions.


now there is a dual construction, which starts not from vectors, but from covectors, i.e. from linear functionals, like x and y, and so os, the coordinate functions on R^n.

we can also form products of these guys, and that is what is happening in constructing bilinear functions or tensors of form x(tensor)y.

but we are dpoing the alternating thoery, so we have things like x^y, or dx^dy.

we add them formally. and instead of being parallelograms, they are objects that assign generalized "areas" to parallelograms,........


ok i pooped out. somebody else will have totake over. please do not begin too negatively. this is obviously still in the right brain [?] fantasizing stage.

oddly though this already suggests that dually all 2 forms on 3 space are actually writable as a product of two one forms.

is that obvious? i.e. the space of 2 forms on R^3 has dimension 3, and is spanned by dx^dy, dx^dz, dy^dz.

the space of one forms is also 3 dimensional spanned by dx,dy,dz. so if we multiply we get a bilinear map oneforms x oneforms-->twoforms. surjective?

it seems to be. i.e. given two one forms mapping to a 2form, think gweometrically of two vectors mapping toa plane. in that plane there are a two dimensioonal way of ways to choose a vector hence a 4 dimensional way to choose 2 vectors spanning it. but if they must span a parallelogram with fixed area that cuts down the family to three dimensions. so tha map above has three dimensional fibaers, hence a 6 dimensional image. so it is onto.???


oh yes, i was trying to elaborate on the natural algebraic construction of wedge products in all dimensions, and note how special the cross product phenomenon is to three dimensons. yipes time flies when you are haivng fun, and i have missed the firat NCAA game!

no wonder no one is responding. its like the day italy was in the world cup and i drove through the deserted streets of rome completely unhindered by traffic.

[Tom Mattson]

no wonder no one is responding.

Doesn't mean we're not reading. I especially liked posts #48 and #51; thanks a lot for that. As I said, my advisees are doing 2 presentations: one in 2 weeks for the faculty at our school, and another in 4 weeks for the Conference. I am thinking that the first one will be more of a pitch to sell differential forms to the faculty, while at the Conference the ladies plan on talking about the generalized Stokes' theorem.

[Tom Mattson]

Section 3: Multiplying 1-Forms (cont'd)

Picking up from page 24 in the arXiv version of the book (edit: that's page 54 in the newer version) , right after Exercise 3.10, we come to the geometric interpretation of the action of \omega\wedge\nu on a pair of vectors V_1 and V_2. I think that the argument leading up to the interpretation is clear enough to not expand on, so I'm just going to present the conclusion. If any of the students reading this thread have any questions about it, go ahead and ask.


Evaluating \omega\wedge\nu onthe pair of vectors (V_1,V_2) gives the area of parellelogram spanned by V_1 and V_2 projected onto the plane containing the vectors <\omega> and <\nu>, and multiplied by the area of the parallelogram spanned by <\omega> and <\nu>.


Then there is the word of caution: This interpretation is only valid if our 2-form is the product of 1-forms. We will later see that this is always the case, at least for 2-forms on T_p\mathbb{R}^3.


Exercise 3.11
This exercise seems to be flawed. On the LHS we have a 2-form acting on a pair of vectors. This quantity is a real number. But on the RHS we have a 2-form that is not acting on anything. This quantity is, well, a 2-form! Correct me if I'm wrong, but in order for that equation to be correct then either the wedge product on the LHS should not be acting on those two vectors, or the 2-form on the RHS should be acting on the same pair of vectors. That's how I interpret the problem.

So in essence what we are asked to show is that any 2-form on T_p\mathbb{R}^3 can be expressed as the product of 1-forms. Here goes.

Let \omega=w_1dx+w_2dy+w_3dz and \nu=v_1dx+v_2dy+v_3dz be 1-forms. Now consider the wedge product \omega\wedge\nu.


\omega\wedge\nu=(w_1v_2-w_2v_1)dx \wedge dy+(w_1v_3-w_3v_1)dx \wedge dz+(w_2v_3-w_3v_2)dy \wedge dz


Now set our expression for \omega\wedge\nu equal to c_1dx \wedge dy+c_2dx \wedge dz+c_3 dy \wedge dz. Equating components yields:


w_1v_2-w_2v_1=c_1


w_1v_3-w_3v_1=c_2


w_2v_3-w_3v_2=c_3


Since there are 3 equations and 6 constants, we can choose 3 of the constants (Note: Letting all the components of a either of the 1-forms equal 1 will not work, and letting any of the components equal to 0 will not work.) A convenient choice is w_1=w_2=v_1=1. This yields:


\omega=dx+dy+\frac{c_2-c_3}{c_1}dz


\nu=dx+(c_1+1)dy+(c_2+\frac{c_2-c_3}{c_1})dz
.

This choice for 3 of the constants is only valid if c_1 \neq 0. Other choices can be found that are valid for c_2 \neq 0 and c_3 \neq 0, so that all 2-forms with either one or no constants equal to zero are covered. If two constants are equal to zero then it is trivially easy to express the 2-form as a product of 1-forms.


This exercise, together with the discussion before it, are supposed to lead us to the following conclusion.


Every 2-form projects the parallelogram spanned by V_1 and V_2 onto each of the (2-dimensional) coordinate planes, computes the resulting (signed) areas, multiplies each by some constant, and adds the results.


Note now that there is no need for the word of caution that was supplied after the first geometric interpretation. Both may now be applied to "every 2-form" because every 2-form on T_p\mathbb{R}^3 is expressible as a product of 1-forms.

Exercise 3.12
\omega\wedge\nu (<1,2,3>,<-1,4,-2>)=\left |\begin{array}{cc}\omega(<1,2,3>)&\nu(<1,2,3>)\\\omega(<-1,4,-2>)&\nu(<-1,4,-2>)\end{array}\right|


\omega\wedge\nu (<1,2,3>,<-1,4,-2>)=\left |\begin{array}{cc}8&3\\21&-8\end{array}\right|

\omega\wedge\nu (<1,2,3>,<-1,4,-2>)=-127

Exercise 3.13
Given two 1-forms, we are asked to find the 2-form that is their wedge product.


\omega\wedge\nu=-11dx \wedge dy+4dy \wedge dz+3dx \wedge dz


On comparision it is obvious that c_1=-11, c_2=4, and c_3=3.

Exercise 3.14
Now we are asked to go the other way: given four 2-forms, we are asked to express them as products of 1-forms.

(1) Use the skew-symmetry property.
3dx\wedge dy+dy\wedge dx=3dx\wedge dy-dx\wedge dy=2dx \wedge dy

(2) Use the distribuitve property.
dx\wedge dy+dx\wedge dz=dx\wedge (dy+dz)

(3) Use the results from (1) and (2).
3dx\wedge dy+dy\wedge dx +dx\wedge dz=2dx\wedge dy+dx\wedge (dy+dz)

Now use the distirbutive property again.
3dx\wedge dy+dy\wedge dx+dx\wedge dz=dx\wedge (2dy+dz)

(4) This one's more involved. Using the method I described above 2.11 (defining two 1-forms \omega and \nu and letting w_1=w_2=v_1=1), I get:

\omega=dx+dy+7dz
\nu=dx+2dy+11dz

Note that this pair of 1-forms is not unique.


That's it for now. I really don't have any questions on this section, so I will post my notes and questions on Sections 3.4 and 3.5 once any discussion on this section dies down.

Till next time...

[mathwonk]

Re: "This interpretation is only valid if our 2-form is the product of 1-forms. We will later see that this is always the case, at least for 2-forms on R^3."

I think I essentially proved this in post 55.

[mathwonk]

we say a k form is "decomposable" if it is a product of one forms. then gerometrically this is sort of dual to a k chain being simply a k plane.

now recall that 2 planes in three space also form a linear space namely the dual space, at least projectively. i.e. the dual of projective 2 space is also a projective 2 space.

the same holds in all dimensions, i.e. the dual of rpojective 3 space is a also a projective 3 space, but the elements are made up of hyperplanes in projective 3 space, i.e. projective palnes, hence spanned by triples of "points" in projective space, i.e. by triples of vectors in the underlying vector space.

so the space of projective lines in projective 3 space corresponds to the decomposable 2 forms on a 4 dimensiopna vector space like R^4. these do not form a vector space, but a quadric cone in a 6 dimensional vector space.

i.e. when we take sums of 2 planes, or 2 forms, in 4 space, we get a linear space, but not all elements are simple products, for the geometric reason that projective lines in projective three space do not form a linear space.

so the fact that any 2 form is a product of one forms in 3 space is equivalent to the fact that the dual of a projective plane is also a projective plane.

in projective 3 space however, note there are various different kinds of pairs of lines, some meet, some do not.

however the algebraic constructions above do allow us to assign coordinates to lines in projective 3 space. i.e. take any plane in a 4 diml vector space, and it will be the zeroes of a pair of linear functions f,g. then represent that plane by f^g.

when f^g is written as a linear combination of dx, dy, dz, dw, we get coordinates for our plane in R^4, i.e. our line in P^3.

since the wedge product map R^4 x R^4 still has 3 dimensional fibers as abnove, the image this time, of decomposable 2 forms, is 5 diemsnional, while the space of all 2 forms is 6 dimensional, so we get a hypersurface in a 6 dimensional vector space or in a 5 dimensional rpojective space. this hypersurface is called the grassmannian variety of all "lines in P^3".

hey this geometric approach to forms is pretty cool. I am learning something after all. thanks dave! this always seems to happen to me when a subject is being well explained, even if i think i already know it.

i never really grasped this algebra - geometry link before for k planes in R^n.

[mathwonk]

building on the previous discussions, i believe that one can characterize those 2 forms on four space, i.e. those linear combinations of products of dx0, dx1, dx2, dx3, which are products of two one forms, by the equation p01p23 - p02p13+ p03p12 = 0, where pij is the coefficient of dxi^dxj.

here is a little trick to see that in 4 dimensions not all 2 forms are products of one forms. since the product of a one form with itself is zero, if W is a 2 form which is a product of one forms, then W^W = 0. But note that [dx^dy + dz^dw] ^ [dx^dy + dz^dw] = 2 dx^dy^dz^dw is not zero. so this 2 form is not a product of one forms.


since there is only one condition on a 2 form in 4 space to be a product of one forms, this must be it.

Note if we wedge p01dx0^dx1 + p02 dx0^dx2 + p03 dx0^dx3 + p12 dx1^dx2 + p13 dx1^dx3 + p23 dx2^dx3 with itself note we get something like

2(p01p23 - p02p13+ p03p12) dx0^dx1^dx2^dx3 which must be zero, if this 2 form is going to be a product of one forms.


I just learned something else new! I had it hard wired into my brain that any form wedged with itself is zero, but this is false! it does hold for one forms, and i was just mostly in the habit of wedging one forms together, and thinking about them exclusively.

in three space of course, if you wedge two 2 forms togetehr you get a 4 form, and thsoe are all zero on 3 space, so the same confusion can arise. also another reawson is that in 3 space all 2 forms are products of one forms, so again they wedge to zero with themselves, again for special reasons that do not generalize.

[mathwonk]

Gza, the discussion reveals that the one forms having a given 2 form as product are certainly not unique. for example if N and M are any one forms at all

N^M = N^(N+M) = N^(cN+M) = (cM+N)^M, for any constant c.

geometrically if we think about representing a plane and an oriented area, by an oriented parallelogram, any parallelogram in that plane having oriented area equal to that number would do. so the wedge product of any two independent vectors in that plane oriented properly, and with fixed product for their lengths, would have the same wedge product.

thus even if you fix one vector and its length, even then the other vector is not fixed. only its projection orthognal to the first vector is fixed. even if you also fix the length of the other vector, there still seem usually to be 2 choices for it.

the abstract discussion i gave mentioned the map from pairs of one forms to their wedge product, and stated that the "fibers" of this map are three dimensional. in particular the fibers are not single points as they would be if the two one forms were determined by their product.

i.e. thinking again geometrically, given a plane, how many ways are there to pick two indepedent vectors in it? each vector can be chosen in a 2 dimensional family of ways, hence the pair can be chosen in a 4 dimensional family of ways.

even if we fix their orientation and the area of the parallelogram they span, we only lose one parameter, so it brings down the fiber dimension from 4 to three.

[mathwonk]

it would seem that geometrically, to factor a 2 form, you would just find two independent vectors both perpendicular to the vector of coefficients of the 2 form. there are lots of those. then adjust the lengths by a scalar.

this is just solving a single homogeneous linear equation in three unknowns.

[Gza]

it would seem that geometrically, to factor a 2 form, you would just find two independent vectors both perpendicular to the vector of coefficients of the 2 form.

So on what geometric basis would I be able to consider the coefficients of a two form as a vector? I'm having a hard time visualizing it.

[mathwonk]

to paraphrase some of my physicist friends on here,
if it has three numbers its a vector right?

so use the zen approach, if it looks like a vector and quacks like a vector, treat it as a vector.


see the full solution in the next post.

[mathwonk]

well here is how i thought of it: i figured the wedge product of two one forms has components which were 2by2 determinants, so they were essentiaslly the same as the components of the cross product (in 3 space). that mkeans the vector with those components should be perpendicular to the pl;ane spanned by the original two vectors, assuming they were independent.

now to perove that one would use the lagrange expansion of a determinant but i can't do thnat in my head so i just assumed it worked. then lets see, oh yes, that means that we are essentiaslly given the cross product of the two vectors and are l;ooking fopr the two vectors, which mkeans we want two vectors perpendicualr to the given vector, and spanning a parallelogram with area given by the length of the gove vbector. so i guess to be honest it was all inspired by the cross product interpretation whichw e are not using, i.e. eschewing.

but so what, if it helps, use it. just a suggestion, as it seemed easier than what i was hearing as a solution method. of course if it fails miserably i have egg on my face.
\
lets try one:


the product of oh, dx and dy is dx^dy, which has coefficients (1,0,0).

so the perp is (0,1,0) and (0,0,1). i.e. dy and dz, oops. i don't give up though but must understand what is going on.

AHA! the right way to assign coordinates is no doubt to call dx^dy dual to dz hence to (0,0,1), so in fact the coefficients of dx^dy should be (0,0,1), hence perpendicualr to (1,0,0) and (0,1,0), i.e. to dx and dy.

but of course this is cheating to make it work out. you need to give a decent explanation that works in general, but i still believe it.

why don't you give this a little shot? see if ti works for a little more complicated one like dx^dy + dx^dz. this ahs coords (0,0,1) + (0,1,0) = (0,1,1) or maybe (0,0,1) - (0,1,0) = (0,-1,1).

anyway, the perp is either (1,0,0) and (0,1,1), or (1,0,0) and (0,1,-1).

try both. multiply (1,0,0) = dx times (0,1,1) = dy + dz and get hey! dx^dy + dx^dz!!

it works!

what do you think, was i just lucky? got to go now, marge is getting implants on the simpsons.

[mathwonk]

ok: a dydz + b dzdx + c dxdy = (a,b,c)

has orthocomplement spanned by (-b,a,0), (0,-c,b), if b is not zero.

hence we try [-bdx + ady]^[-cdy+bdz]

= bcdxdy -b^2 dxdz + abdydz = bcdxdy + b^2 dzdx + abdydz

= b [a dydz + b dzdx + c dxdy].

so just divide one of the one forms by b.

if b=0, use the basis (0,1,0), (-c,0,a), for vectors orthogonal to (a,b,c).

then we get dy^(-cdx + adz) = ady^dz + c dx^dy.

what about this Gza?

[*melinda*]

hi everyone!
I’m one of the students who will be presenting this topic at a conference. It’s taken me a while to sign on, but now that I’ve jumped in I’ll hopefully be able to add to the discussion regularly.
~First, to answer Tom’s question on post #37… Why don’t we take the absolute value of the signed area? The property of superposition gives us the equality below.

\omega\wedge\nu(V_1+V_2,V_3)=\omega\wedge\nu(V_1,V _3)+\omega\wedge\nu(V_2,V_3)

If the absolute value is taken for all three wedge products, it’s pretty easy to see that the right side of the equation will not always equal the left side. This can be checked by plugging some vectors in, computing and taking note of the result. That’s what I did.
~Also, on pg. 26 of the arXiv version of the book Bachman says, “To give a 2-form in 4-dimensional Euclidian space we need to specify 6 numbers.” A question similar to this statement is asked a little further ahead in the reading. My question is, can this be treated as a combination? 4choose2 = 6. I also noticed that to give a 3-form in 3-space (3choose3 = 1), you need to specify one number

[hypermorphism]

~Also, on pg. 26 of the arXiv version of the book Bachman says, “To give a 2-form in 4-dimensional Euclidian space we need to specify 6 numbers.” A question similar to this statement is asked a little further ahead in the reading. My question is, can this be treated as a combination? 4choose2 = 6. I also noticed that to give a 3-form in 3-space (3choose3 = 1), you need to specify one number
That's the right track. To prove the general form, first note that the set of k-forms on an n-dimensional vector space is a vector space. Then find a basis for the set of k-forms (note that a one-form wedged with itself is zero, and reordering a wedge product simply changes the sign, in the same manner as the even or oddness of a permutation). Since the size of the basis determines the dimension of the vector space, which determines how many numbers are necessary to specify an element of the space, counting the size of the basis (which you will find is a combination) will tell you how many numbers you need.

[mathwonk]

*melinda* : a basis for the k forms in n variables would be all k fold wedge products of the n one forms dx1,....dxn. but note that these prodcucts are zero unless al k of the forms multiplkied are distinct. so there are exactly n choose k ways to find k distinct ones.

[mathwonk]

i know you guys skipped chapter 1, but i have learned so much reading your posts id ecided to try again reading the book. here are some tiny remarks that may be of help to dave in proofreading:

on page 15, ex 1.3 should say the area is |ad-bc|, if area is meant to be non negative. or else it should probably be called "oriented area".

in the next line the definition of determinant is also incorrect since it is defined as an area instead of an oriented area.

such obvious mistakes seem to be purposeful, but they do not make logical sense to me. i.e. it is incompatible in one line to say an area is a number that could have two values, one of them negative, and in the next line to define a determinant as an area, which can only be non negative??


what did you want to achieve here dave? are approaching the subject from the point of view that a few small inaccuracies will not matter to beginners?

if so, then please ignore all this. but if you want a proofreader, here goes.


same comment top of page 16, that "volume" formula is not always non negative.


line 2 of section on multiple variables: "these spaces a very familiar" should be "these spaces are very familiar"

a point of philosophy: it might be safer to say that picturing R^20 is very difficult for most of us. certainly some people think they can do it. in the other direction, the picture at the top of the elementary school blackboard does not allow one to picture R^1 either because it is not long enough.

but these are matters of taste. still why discourage anyone who wants to try to picture R^20? indeed you have already sketched how to do it in the introduction, as a product of 10 copies of R^2.

for example imagine 20 parallel copies of R, erected at the points 1,2....,20 on the x axis. and then imagine choosing one point on each line, perhaps connected by a zigzag line. thats a general point of R^20.

I admit these depictions do not allow one to "see" all of R^20, but no more does a line segment allow one to see all of R^1.

but this kind of thing could go on forever.


bottom page 18: it is not quite true to say we define the integral via evenly spaced subdivisions. indeed the integral is only defined for functions for which the type of spacing does not affect the outcome of the limit. if you want to say you are defining the integral of continuous functions this would be ok. but it is not too hard to define a non (riemann) integrable function such that the limit described will exist and not be equal to some other limits with other spacings.

same comment for volume integrals on page 19.

perhaps the word "compute" would be more appropriate than "define", since we do compute integrals this way when they exist.

ok on page 22 there is a caveat that technical issues are being ignored (like continuity). such caveats should probably be placed at the beginning of the discussion. even simpler is just to say at the beginning that we are discussing the case for continuous functions, since then everything said is actually true.

at the top of page 33, a parameterization for a surface is required to be one to one and onto, but in example 1.12 page 36, the parametrization given there of the unit disc is not one to one. perhaps it would be better to allow parametrizations which fail to be one to one on the boundary of the domain? (as in this standard example.)

the reader will face the same challenge in trying to solve ex 1.26 by a one to one parametrization.

[mathwonk]

chap 2: page 39, same incorrect statement about defining integrals via evenly spaced subdivisions occurs again.

problems witth the definition of parametrization raises its head again on page 40. on page 23 a parametrization of a curve was defiend as a one to one, onto, differentiable map from (all of) R^1 to the curve, (although most exampels so far have not bee defiend on all of R^1, so it might have been better to say from an interval in R^1.

more significant, the first example given on page 40 is not differentiable at the end points of its domain. so again it might be well to say the parametrization, although continuous on the whole interval may fail to be differentiable at the endpoints.

this is the beginning of another potential situation where one probably is intending to integrate this derivative even though it is not continuous or even bounded on its whole domain. this problem is often overlooked in calculus courses. i.e. when the "antiderivative" is well defined and continuous on a closed interval, it is often not noticed that the derivative is not actually riemann integrable by virtue of being unbounded.

indeed as i predicted, exercise 2.1 page 43 asks the reader to integrate the non - integrable function, derivative of (1-a^2)^(1/2), from -1 to 1.

this function is not defined at the endpoints of that interval and is also unbounded on that interval. interestingly enouhg it has a bounded continulous "antiderivative" which enables one to "integrate" it, but not by the definition given in the section, since the limit of those riemann sums does not in fact exist.

the polar parametrization of the hemisphere, on page 44, is again not one to one. and again the third coordinate function of the parametrization phi is not differentiable wrt r at r=1, hence the integral written is again not defined by a limit of riemann sums.

it seems worthwhile to face head on this problem about many natural parametrizations often not being one to one, and point out that for questions of integration, there is no harm in non one to one ness occurring on sets of lower dimension, since the integral over those sets will be zero.

Stieltjes is misspelled on page 44, both the t and one e are omitted.

the language at the bottom of page 45 describes regions parametrized by R^1, R^2, and R^n, although what is apparently meant, and what is done, is to parametrize by rectangular blocks in those spaces.

[Gza]

what about this Gza?

I understand now, thank you. :approve:

[mathwonk]

does anyone appreciate my comment about sqrt(1-x^2) not being differentiable at
x= 1?

this is the familiar fact that the tangent line to a circle at the equator is vertical.

it is rather interesting that this derivative function can be "integrated" in some sense (i.e. as an improper integral) in spite of being unbounded.

does anyone agree that the polar parametrizations given are not actually one to one? and does anyone see why that does not matter?

(but that it does call for a new definition of parametrization?)

[Haelfix]

My apologies for not having read the text so im sure its already been pointed out.

One endless source of confusion for me when I was learning this stuff is the notion of axial and polar vectors. At first glance its easy and obvious, but then terminology starts getting confused, particularly when you learn clifford algebras and some peoples pet concepts to reinvite notation via geometric algebra.

People get in endless debates about how to properly distinguish these different types of *things*. eg What constitutes active and passive transformations of the system, what is a parity change, do we take Grassman or Clifford notation blah blah blah.

Unfortunately if you want a cutesy picture of whats going on, alla MTW (forms now look like piercing planes) some of this stuff becomes relevant or else you quickly end up with ambiguities.

Most of the confusion goes away when you get into some of the more abstract and general bundle theory, but then the audience quickly starts getting pushed into late undergrad/early grad material and the point is lost.

[Tom Mattson]

does anyone appreciate my comment about sqrt(1-x^2) not being differentiable at
x= 1?

this is the familiar fact that the tangent line to a circle at the equator is vertical.


Yes, but we're not there yet. As I said in the beginning, I want to march through the book sequentially. The purpose of this thread is twofold:

1. To help my advisees for their presentation.
2. To see if a book such as Bachman's could be used as a follow-up course to what is normally called "Calculus III".

It doesn't really help to achieve my primary goal (#1) if we jump all over the place. My advisees are in Chapter 4 (on differentiation), and we are using this thread to nail down any loose ends that we left along the way in our effort to keep moving ahead.

I'll be posting the last of my Chapter 2 notes tonight and tomorrow. Once the discussion has died down I'll start posting notes on Chapter 3, which is about integration. I'll also try to pick up the pace.

Thanks mathwonk and everyone else for your useful comments, especially post #65 by mathwonk.

edit to add:

By the way mathwonk, my copy of Spivak's Calculus on Manifolds is in. Great book, thanks for the tip! One of my advisees (*melinda*) picked up Differential Forms with Applications to the Physical Sciences by Flanders. What do you think of it?

[mathwonk]

i like flanders.


i do not understanbd your reamrk about the sequential treatment, and not being up to my comment yet.

if you are talking about amrching sequentially throguh bachmann, i started on page 1, and those comments are about chapters 1 and 2. how can someone be in chapter 4 and not be sequentially up to chapters 1 and 2 yet?


are you talking about chapter 4 of some other book?

it seems to me you guys are still way ahead of me.

[mathwonk]

flanders had a little introductory article in a little MAA book, maybe Studies in Global Geometry and Analysis (ISBN:0883851040)
Chern, S.S., that first got me unafraid of differential forms, by just showing how to calculate with them.

i had been frightened off of them by an abstract introduction in college. i had only learned their axioms and flanders showed just how easy it is to multiply them. i liked the little article better than his more detailed books.

[Tom Mattson]

i do not understanbd your reamrk about the sequential treatment, and not being up to my comment yet.


Never mind my comment. I was looking at the arXiv version of Bachman's book, in which page 39 is in Chapter 3 (the chapter on integrating 1-forms).

To prevent further confusion, I am now going to burn the arXiv version and exclusively use the version from his website. I'll re-do the chapter and section numbers in my notes.

[mathwonk]

thats right, there were two versions of the book!

[Haelfix]

Flanders is sorta the defacto reference book on differential forms for US math majors. You get some treatment in Spivak, and also some good stuff in various physics books, but its not quite the same.

A modern book some people liked a lot was Darling's book on Differential forms.

Regardless I am a little bit wary of placing too much weight on intuitive pictures of the whole affair. Differential forms to me are much ore of a formal language that makes calculations tremendously simpler (not to mention the fact that they are much more natural geometric objects what with being coordinate independant and hence perfect for subjects like cohomology and algebraic geometry). Notation changes from area to area and I suspect having too rigid a 'geometric' intution might actually hurt in some cases.

I guess im just a little bit disenchanted with some of the earlier attempts to 'picture' whats happening, like the piercing plane idea from MTW (Bachmans text has a good section where they explain why that whole thing doesn't quite work out well in generality)

[Tom Mattson]

Section 4: 2-forms on T_p\mathbb{R}^3

Here is the next set of notes. As always comments, corrections, and questions are warmly invited.


Exercise 3.15

Try as you might, you will not be able to find a 2-form (edit: on T_p\mathbb{R}^3) which is not the product of 1-forms. We in this thread have already argued as much, and indeed in the ensuing text Bachman explains that he has just asked you to do something that is impossible. Nice guy, that Dave. :tongue2:


This brings us to the two Lemmas of this section. I feel that the details of the proofs are straightforward enough to omit, so I am just going to talk about what the lemmas say. If any of our students has any questions about the proofs, go right ahead and ask.

Lemma 3.1 reinforces the idea that was first brought up by Gza: The 1-forms whose wedge product make up a 2-form are not unique.

Lemma 3.2 is really what we want to see: It is the proof that any 2-form is a product of 1-forms. The lemma itself states that if you start with two 2-forms that are the product of 1-forms, then their sum is a 2-form that is the product of 1-forms. That is, any 2-form that can be written as the sum of the product of 1-forms, is itself a product of 1-forms.


Note: There is a typo in Bachman's proof (both versions of the book).

Where it says:

"In this case it must be that \alpha_1\wedge\beta_1=C\alpha_2\wedge\beta_2, and hence \alpha_1\wedge\beta_1+\alpha_2\wedge\beta_2=(1+C)\ alpha_1\wedge\beta_1",

it should say:

"In this case it must be that \alpha_1\wedge\beta_1=C\alpha_2\wedge\beta_2, and hence \alpha_1\wedge\beta_1+\alpha_2\wedge\beta_2=(1+C)\ alpha_2\wedge\beta_2".


Bachman goes from the last statement in black above to concluding that "any 2-form is the sum of products of 1-forms."


To explicitly show this, start with the most general 2-form:

\omega=c_1dx \wedge dy+c_2dz \wedge dy+c_3dz \wedge dx


Now use the distributive property:

\omega=(c_1dx+c_2dz) \wedge dy +c_3dz \wedge dx


And there we have it.


This leads us to the following conclusion:


Every 2-form on T_p\mathbb{R}^3 projects pairs of vectors onto some plane and returns the area of the resulting parallelogram, scaled by some constant.


There is thus no longer any need for the "Caution!" on page 55.

edit: That is, there is no need for it when we are dealing with 2-forms on T_p\mathbb{R}^3. See post #82.


Exercise 3.16

Now that we know that every 2-form on T_p\mathbb{R}^3 is a product of 1-forms, this is a piece of cake. Just look at the following 2-form:

\omega(V_1,V_2)=\alpha\wedge\beta(V_1,V_2)
\omega(V_1,V_2)=\alpha(V_1)\beta(V_2)-\alpha(V_1)\beta(V_2)
\omega (V_1,V_2)=(<\alpha>\cdot V_1)(<\beta>\cdot V_2)-(<\alpha>\cdot V_2)(<\beta>\cdot V_1)

This 2-form vanishes identically if either V_1 or V_2 (doesn't matter which) is orthogonal to both <\alpha> and <\beta>.

Exercise 3.17

Incorrect answer edited out:

The above argument does not extend to higher dimensions because not all 2-forms are factorable in higher dimensions.

Counterexample:

Take the following 2-form on T_p\mathbb{R}^4:

\omega=dx \wedge dy + dz \wedge dy +dz \wedge dw + 2dx \wedge dw.

Try to factor by grouping:

(dx+dz) \wedge dy + (dz+2dx) \wedge dw,

and note that we can go no further. It turns out that no grouping of terms will result in a successful factorization.



Exercise 3.18

Maybe I'm just being dense, but I do not see how to solve this one. The hint right after the exercise doesn't help. If l is in the plane spanned by V_1 and V_2, then of course the vectors that are perpendicular to V_1 and V_2 will be perpendicular to l.

Anyone want to jump in here?

[Bachman]

Hi all,

Sorry I have been silent for a few days. Busy, busy busy...

And even now I do not have time to give proper responses, but here are a quick few....

Mathwonck, please read a bit more carefully if you are going to take on a role as "proofreader":

To your comment about integrating with evenly space intervals: there is a discussion of this on page 41.
To your comment on saying that we want an "oriented area": I couldn't use the word "oriented" because at this point students have no idea what an orientation is. In fact, at that point in the text I do not even assume that the student realizes that the deterimant can give you a negative answer (although I am sure this seems obvious to you). I do, however, emphasize this by inentionally computing an example where the answer is negative, and then pointing out that we really don't want "area", but rather a "signed area". It's all there.

Next.... there is a rather long discussion here about factoring 2-forms into products. Mathwonk has a "proof" in one of his earlier posts, but this was a little bit of wasted effort, since this is the content of Section 4 of Chapter 3.

Also, Tom.... be careful! The CAUTION on page 55 is ALWAYS something to look out for. The point of Section 4 of Chap 3 is that dimension 3 is special, because there you can always factor 2-forms. The next edition of the book will have a new section about 2-forms in four dimensions, with particular interest on those that can NOT be factored.

Hopefully more tomorrow... I should give you more of a hint on Exercise 3.18.

Dave.

[mathwonk]

Dave I am sorry to see my corrections are not welcomed by you. They are accurate however.

As an expert I probably should have not gotten involved since everyone is having fun, and my corrections are invisible to the average student. But you did ask for comments in your introduction. When you do that, you should expect to get some.

I think this book is nice for a first dip into the topic, but I have a concern that a person learning the subject from this source will be left with a certain amount of confusion, due to the imprecise discussion, and non standard language, which will cause problems in trying to discuss the material with more knowledgable people.

If followed up with Spivak however it should be fine. And any source that gets people involved and allows them friendly access to a topic is good. This is the strength of Dave's book. I don't know who they sent it to for reviewing, but Dave, I think you might get some comments like mine from other reviewers.

[mathwonk]

for tom and students: you can argue that diff forms are useful in the 10 or more dimensions physicists apparently use now for space time, and they are also easily adaptable to the complex structures used there and in in string theory (Riemann surfaces, complex "Calabi Yau" manifolds).

[Tom Mattson]

Hi all,

Sorry I have been silent for a few days. Busy, busy busy...


Glad to see you back. :smile:


Also, Tom.... be careful! The CAUTION on page 55 is ALWAYS something to look out for. The point of Section 4 of Chap 3 is that dimension 3 is special, because there you can always factor 2-forms.


Whoops. I've put in an edit that corrects my remark about the Caution. I've also changed my answer to Exercise 3.17, which was evidently wrong.

[mathwonk]

another comment about selling differential forms to your audience. Dave has a nice application in chapter 7 showing that their use reduces Maxwell's equations from 4 to 2.

[AKG]

The line l = \{\vec{r}t + \vec{p} : t \in \mathbb{R}\} for some \vec{r},\ \vec{p} \in T_p\mathbb{R}^3. Suppose \vec{v},\ \vec{w} \in T_p\mathbb{R}^3 such that l \subseteq Span(\{\vec{v},\ \vec{w}\}). Then the set \{\vec{p},\ \vec{v},\ \vec{w}\} is linearly dependent, hence:

\det (\vec{p}\ \ \vec{v}\ \ \vec{w}) = 0[/itex]

Define \omega such that:

[tex]\omega (\vec{x},\ \vec{y}) = \det (\vec{p}\ \ \vec{x}\ \ \vec{y}) \ \forall \vec{x}, \vec{y} \in T_p\mathbb{R}^3

You can easily check, knowing the properties of determinants, that \omega is an alternating bilinear functional, and hence a 2-form. If you want, you can express it as a linear combination of dx \wedge dy,\ dy \wedge dz,\ dx \wedge dz, and it shouldn't be hard, but probably not necessary.

EDIT: actually, to answer the question as given, perhaps you will want to write \omega in terms of those wedge products, and determine \vec{p} from there. Then, to find l you just need to choose any line that passes through \vec{p}. Any two vectors containing that line will have to contain \vec{p}, hence those three vectors must be linearly dependent, hence their determinant will be zero, and since \omega depends only on \vec{p} and not the choice of \vec{r}, you're done.

[*melinda*]

hi
~Thanks everyone on the feedback to my question. It’s so reassuring to know when you’ve got the right idea!
~For exercise 3.17 (post 81), Tom says:

“The above argument does not extend to higher dimensions because not all 2-forms are factorable in higher dimensions”.

~I can see why this is the case in exercise 3.16, but it seems like there’s a bit more to this than a simple question of factorability. I’m probably way off, but I was thinking that it has more to do with some general property of 3-space that makes it inherently different than say, 4-space or any other space for that matter. Then again, I suppose that not being able to write a 2-form as a product of 1-forms in R^4 could very well be a general property of higher dimensions. Unfortunately these are ideas that I don’t know very much about yet, so please excuse if my questions are a bit silly or obvious.

[Haelfix]

For applications, I know of many places in physics where differential forms are useful, even to an undergrad.

First and foremost, the often quoted derivation of maxwells equations in a very neat and elegant form.

The fundamental equations of thermodynamics as well are often cast in differential form notation. You instantly get out several relations that are painful to get in other notation.

Finally general relativity/String theory etc

One thing to note though.. I really didn't see at the time the advantage of using differential forms in those situations, I often would ask 'why not just use tensor calculus instead'? And I was right in the sense that you will get very compact notation (if you suppress the irratating indices) just as quickly as with differential forms without the added hassle of learning the new, somewhat unintuitive language.

I was wrong though in the deeper meaning of these objects. It wasn't until I learned of Yang Mills theory, and principle bundles as applied to general relativity, that the full power of differential forms became instantly apparent.

Modern Physics fundamentally wants to be written down in coordinate invariant, read diffeomorphism invariant language. It doesn't necessarily want to know about metrics, and things like that. Indeed there are situations where such concepts stop you from seeing the global topology of the problem, and it is in that sense that differential forms immediately become obvious as THE god given physical language.

[mathwonk]

melinda,

pardon me if my posts have been unhelpful. I will try to explain why a 2 form is never a product of one forms in any dimension higher than 3.

Let V be the space of one forms on R^n, and let V^V be the space of 2 forms. Then since V has coordinates dx1,....dxn, and has dimension n, V^V has coordinates dxi^dxj with i <j, so has dimension = bonomial coefficient "n choose 2".


Now, just look at the product map, VxV-->V^V, taking a pair of 1 forms f,g to their product f^g. The question is when is this map surjective?

Without going into it too much, I claim that this map cannot raise dimension, much as a linear map cannot, so since the domain has dimension 2n and the range has dimension (1/2)(n)(n-1), it follows that as soon as the second number outruns the first, the map cannnot be surjective.

In particular for n > 5, the map cannot be surjective, but actually this occurs sooner than that, I claim for n > 3.

The key is to look at the dimension of the fibers of the map. Here there is a principle almost exactly the same as the "rank - dimension" theorem in linear algebra.

i.e. if we can discover the dimension of the set of domain points which map to a given point in the target of ther map, then the dimension of the actual image of the map cannot be more than the amount by which the dimension of the domain exceeds this "fiber" dimension. i.e. if (f,g) is a general point of the domain VxV, then the dimension of the set of 2 forms which are products in V^V, cannot be more than 2n - dim of the set of one forms having the same product f^g as f and g.


Now it helps to think geometrically, i.e. of f and g as vectors and f^g as the parallelogram they span. Then two other vectors have the same product if and only if they span a parallelogram in the same plane as f and g, and also ahving the same area.

So there is a 2 dimenmsional family of vectors in that plane, hence a 4 dimensional fmaily of pairs of vectors in that plane spanning it, but if choose only thos having the right area, there is noly a three dimnsional family.

Thus the inverse image of a general product f^g is 3 dimensional in VxV. Thus the dimension of the image of the rpoduct map, in V^V, i.e. the dimension of the family of factorable 2 forms, equals 2n - 3. we see this is less than (1/2)(n)(n-1) as soon as n >3.

so for n > 3, it never again happens that all 2 forms are a product of two 1 forms.

does that help?

if you look back at some of my free flying posts earlier you will probably see that these ideas are there, but not explained well.

[mathwonk]

An apology and some comments:

I apologize for making critical comments no one was interested in and which stemmed from not reading Dave's introduction well enough. He said there he was not interested in "getting it right", whereas "get it right" is my middle name (it was even chosen as the tagline under my photograph in high school, by the yearbook editor, now I know why!) I have always felt this way, even as an undergraduate, but apparently not everyone does. My happiest early moments in college came when the fog of imprecise high school explanations was rolled away by precise definitions and proofs.

On the first day of my beginning calculus class the teacher handed out axioms for the reals and we used them to prove everything. In the subsequent course the teacher began with a precise definition of the tangent space to the uncoordinatized euclidean plane as the vector space of translations on the plane.

E.g. if you are given a translation, and a point p, then you get a tangent vector based at p by letting p be the foot of the vector, then applying the translation to the point p and taking that result as the head of the vector.

This provides the isomorphism between a single vector space and all the spaces Tp(R^n) at once. Then we proceeded to do differential calculus in banach space, and derivatives were defined as (continuous) linear maps from the get go.

So I never experienced the traditional undergraduate calculus environment until trying to teach it. As a result I do not struggle with the basic concepts in this subject, but do struggle to understand attempts to "simplify" them.

I am interested in this material and will attempt to stifle the molecular imbalances which are provoked involuntarily by imprecise statements used as a technique for selling a subject to beginners.

One such point, concerning the use of "variables" will appear below, in answer to a question of hurkyl.

to post #6 from Tom, why does Dave derive the basis of Tp(R^2) the way he does? instead of merely using the fact that that space is isomorphic to R^2, hence has as basis the basis of R^2?

I think the point is that space is not equal to R^2, but only isomorphic to R^2. Hence the basis for that space should be obtained from the basis of R^2 via a given isomorphism.

Now the isomorphism from Tp(R^2) to R^2 proceeds by taking velocity vectors of curves through p, so Dave has chosen two natural curves through p, the horizontal line and the vertical line, and he has computed their velocity vectors, showing them to be <1,0> and <0,1>.

So we get not just two basis vectors for the space but we get a connection between those vectors and curves in the plane P. (Of course we have not proved directly they are a basis of Tp(P), but that is true of the velocity vectors to any two "transverse curves through p").

So if you believe it is natural to prefer those two curves through p, then you have specified a natural isomorphism of Tp(R^2) with R^2. In any case the construction shows how the formal algebraic vector <1,0> corresponds to something geometric associated to the plane and the point p.


In post #18, Hurkyl asks whether dx and dy are being used as vectors or as covectors? This is the key point that puzzled and confused me for so long. Dave has consciously chosen to extend the traditional confusion of x and y as "variables" on R^2 to an analogous confusion of dx and dy as variables on Tp(R^2).

The confusion is that the same letters (x,y) are used traditionally both as functions from R^2 to R, and as the VALUES of those functions, as in "let (x,y) be an arbitrary point of R^2."

In this sense (x,y) can mean either a pair of coordinate functions, or a point of R^2. Similarly, (dx,dy) can mean either a pair of linear functions on Tp(R^2) i.e. a pair of covectors, or as a pair of numbers in R^2, hence a tangent vector in Tp(R^2) via its isomorphism with R^2 described above.

So Dave is finessing the existence of covectors entirely.

This sort of thing is apparently successful in the standard undergraduate environment or Dave would not be using it, but it is not standard practice with mathematicians who tend to take one point of view on the use of a notation, and here it is that x and y are functions, and dx and dy are their differentials.

There is precedent for this type of attempt to popularize differentials as variables and hence render them useful earlier in college. M.E. Munroe tried it in his book, Calculus, in 1970 from Saunders publishers, but it quickly went out of print. Fortunately I think Dave's book is much more user friendly than Munroe's.

(Munroe intended his discussion as calculus I, not calculus III.)

In post #43, Gza asked what a k cycle is, after I said a k form was an animal that gobbles up k cycles and spits out numbers.

I was thinking of a k form as an integrand as Dave does in his introduction, and hence of a k cycle as the domain of integration. Hence it is some kind of k dimensional object over which one can integrate.


Now the simplest version would be a k dimensional parallelpiped, and that is spannned by k vectors in n space, exactly as Gza surmised. A more general such object would be a formal algebraic sum, or linear combination, of such things, and a non linear version would be a piece of k dimensional surface, or a sum or lin. comb. of such.


now to integrate a k form over a k diml surface. one could parametrize the surface via a map from a rectangular block, and then approximate the map by the linear map of that block using the derivative of the parameter map.

Then the k form would see the approximating parametrized parallelepiped and spit out a number approximating the integral.

By subdividing the block we get a family of smaller approximating parallelepipeds and our k form spits out numbers on these that add up to a better approximation to the integral, etc...


so k cycles of the form : "sum of parallelepipeds" do approximate non linear k cycles for the purposes of integration over them by k forms.

The whole exercise people are going through trying to "picture" differential forms, may be grounded in the denial of their nature as covectors rather than vectors. I.e. one seldom tries to picture functions on a space geometrically, except perhaps as graphs.

On the other hand I have several times used the technique of discussing parallelepipeds in stead of forms. That is because the construction of 2 forms from 1 forms is a formal one, that of taking an alternating product. the same, or analogous, construction that sends pairs of one forms to 2 forms, also sends pairs of tangent vectors to (equivalence classes of) parallelograms.

I.e. there is a concept of taking an alternating product. if applied to 1 forms it yields 2 forms, if applied to vectors it yields "alternating 2 - vectors".

In post #81, Tom asked for the proof of the lemma 3.2 that all 2 forms in R^3 are products of 1 forms. I have explicitly proved this in the most concrete way in post #66 by simply writing down the factors in the general case.

In another post in answer to a question of Gza I have written down more than one solution to every factorization, proving the factors are not unique.

Also in post #81, Tom asked about solving ex 3.18. What about something like this?
Intuitively, a 1 form measures the (scaled) length of the projection of a vector onto a line, and a 2 form measures the (scaled) area of the projection of a parallelogram onto a plane. Hence any plane containing the normal vector to that plane will project to a line in that plane. hence any parallelogram lying in such a plane will project to have area zero in that plane.

e.g. dx^dy should vanish on any pair of vectors spanning a plane containing the z axis.

Notice that when brainstorming I allow myself the luxury of being imprecise! there are two sides to the brain, the creative side and the critical side. one should not live exclusively on either one.

[Bachman]

Melinda,

You can also see that in dimensions bigger than three you will not always be able to factor 2-forms by just writing one down. If there are at least four coordinates then consider the following 2-form:

\omega=dx_1 \wedge dx_2 + dx_3 \wedge dx_4

Now, if this 2-form could be written as \alpha \wedge \beta then

\omega \wedge \omega=\alpha \wedge \beta \wedge \alpha \wedge \beta=0

But when you compute \omega \wedge \omega for the above 2-form you do not get zero. The conclusion is that this 2-form can never be factored.

Dave.

[Bachman]

Dear all,

I have been going through my book agaiin with my current students and we have found a few errors. I'll post them:

Exercise 1.6 (4) The coefficient should be \frac{2}{5} instead of \frac{5}{2}
Exercise 3.21 ... then V_{\omega}=\langle F_x, F_y, F_z \rangle .
Exercise 4.8 The form should be 2z\ dx \wedge dy + y\ dy \wedge dz -x\ dx \wedge dz . The answer should be \frac{1}{6} .
Exercise 4.13 Answer sholuld be \frac{32}{3}

If anyone finds any more please let me know!!

Dave.

[mathwonk]

Dave's example recalls post #60:

"here is a little trick to see that in 4 dimensions not all 2 forms are products of one forms. since the product of a one form with itself is zero, if W is a 2 form which is a product of one forms, then W^W = 0. But note that [dx^dy + dz^dw] ^ [dx^dy + dz^dw] = 2 dx^dy^dz^dw is not zero. so this 2 form is not a product of one forms."

Indeed if n= 4, we have argued above that the subspace of products has codimension one in the space of 2 forms, and it seems the condition w^w = 0 is then necessary and sufficient for a 2 form to be a product.

[mathwonk]

Here is another use of the constructions Dave is explaining to us: analyzing the structure of lines in 3 space.

For example what if we consider the old problem of Schubert: how many lines in (projective) 3 space meet 4 general fixed lines? This has been tackled valiantly in another thread by several people, some successfully.

I claim this can be solved using the algebriac tools we are learning.

I am going to try to wing this along the lines of the discussion so far, so Dave, feel free to jump in and correct, clarify, or augment my misstatements.

We have been seeing that a 2 form assigns a number to a pair of vectors. Since every 2 form is a linear combination of basic ones, i.e. of products of one forms, it suffices to know how those behave, and we have been seeing that e.g. the one form dx^dy seems to project our two vectors into the x, y plane and then take the oriented area of the parallelogram they span.

Now just as in linear algebra when we "mod out" a domain vector space by the kernel of a linear transformation, to make the new domain space into a space on which the transformation is one to one, we could also try to mod out the space of pairs of vectors, by equating two pairs to which every 2 form assigns the same number.

Now it suffices as remarked above, to equate two pairs of vectors if the basic two forms dxi^dxj all agree on them. From the discussion so far, it seems this means we should equate two pairs of vectors if the parallelogram they span has the same oriented area when projected into every pair of coordinate planes.

Now I claim this just means the two pairs of vectors span the same plane, and the parallelograms they span have the same area, and the same orientation. So this essentially contains the data of the plane they span, plus a real scalar.

We denote the equivalence class of all pairs equivalent in this way to v,w by the symbol v^w. Then we have taken alternating products of vectors, just as before we took alternating products of one forms, i.e. of functionals.

i.e. the same formal rules hold; v^w = - w^v, v^(u+w) = v^u + v^w, v^aw = av^w, etc....

But we again cannot add these except formally, so we consider also formal linear combinations of such guys: v^w + u^z, etc....

Now just as in 4 space and higher, not all 3 forms were products of one forms, so also not all 2-vectors are simple ones of form v^w.

E.g. in 4 space the same condition must hold as remarked above for 2 forms, i.e. that a 2 vector T is a simple product if and only if T^T = 0.

Now we have constructed a linear space of alternating 2 vectors T, in which those that satisfy the property T^T =0 correspond to products v^w. For vectors in R^4, this linear space has dimension "4 choose 2" = 6. So the space of all 2 vectors in R^4 is identifiable with R^6.

I claim this has the following interpretation:

by definition projective 3 space consists of lines through the origin of R^4, so 2 planes in R^4 correspond to lines in projective 3 space.

Now each 2 plane in R^2 is represented by a simple 2 vector, i.e. a product v^w, in fact by a "line" of such 2 vectors, since v^w and av^w represent the same plane, just accompanied by a different oriented area.

so 2 planes in R^4 are represented by the lines through the points of R^6 representing simple 2 vectors. Moreover this subset of R^6 is defined by the quadratic equation T^T = 0, hence 2 planes in R^4 are represented by a quadratic cone of lines in R^6.

If we consider the projective space of lines through the origin of R^6, we have the space of all lines in projective three space, represenetd as a quadric hypersurface of dimension 4 in the projective 5 space defined by all 2 vectors in R^4.


Now in projective 3 space we ask what it means algebraically for two lines to meet? i.e. when do the two pairs of simple 2 vectors u^v, and z^w represent planes in R^4 that have a line in common? Well it means that u^v^z^w = 0, (since this happens when the 4 diml parallelepiped they span has volume zero in 4 space).

Consequently when u^v is fixed, this is a linear equation in z^w, hence the lines in projective 3 space meeting a given line, correspond to a linear hyperplane section in 5 space, on the quadric of all lines. hence the lines meeting 4 given lines in 3 space, would be the intersection of our quadric of all lines, with 4 linear hyperplanes.

But 4 linear hyperplanes in P^5 meet in a line, so the lines in 3 space meeting 4 given lines, correspond to the points of P^5 where a quadric hypersurface meets a line, i.e. exactly 2 points.


You might ask an audience, consisting of skeptics as to the value of alternating form methods, if they can solve that little geometry problem as neatly using classical vector analysis.

[mathwonk]

I guess to make sure that quadric meets that line in 2 points, I should have chosen an algebraically closed field, like the complex numbers, to work over, instead of the reals?

[mathwonk]

It finally dawned on me what Dave is doing and why he calls this a geometric approach to differential forms.

given a vector space V, the space of linear functions on V is the dual space V*. But if we define a dot product on V we get an isomorphism between V* and V. I.e. then a linear functional f on V is represented by a vector w in V. The value of f at a vector v is given by projecting v onto the line spanned by w and multiplying the length of the projection by (plus or minus) the length of w.


Now suppose we jack that up by one degree to bilinear functions. I.e. given a dot product, a bilinear alternating functional which is an alternating product of two linear forms, is represented by a parallelogram, such that the action of the function on a pair of vectors becomes projection of those two vectors into the plane of the parallelogram, taking (plus or minus) the area of the image parallelogram, and multiplying by the area of the given parallelogram.

So this approach has more structure than strictly necessary for the concept of differential forms, but allows them to be represented as (a sum of) projection operators.

nice.

In that spirit, one is led to pose geometric versions of the factorization questions asked above in R^3:
1) given two parallelograms in R^3, find one parallelogram such that the bilinear function defined by the sum of those two given parallograms equals the one given by projection on the one resultant parallelogram.
2) give a geometric proof in R^4 that the bilinear function defined by the sum of dx^dy and dz^dw, cannot be equal to the function defined by projection on the plane spanned by any one parallogram.

In short the use of a dot product, allows one to have an isomorphism between the space V*^V* of 2 forms and the more geometric object V^V I defined above, which I said was analogous to the space of 2 forms.

Dave, you have obviously put a lot of thought into this.

[mathwonk]

another in my wildly popular series of commentaries:

towards a more fully geometric view of differential forms.

It seems after reading Dave's section on how [to and] not to picture differential one forms, he does not advocate there the use of the dot product. I.e. he suggests picturing the kernel planes of the field of one forms in R^3, a view point which depends only on the nature of a one form as functional, having a kernel, and not on its nature as a dot product.

I.e. I would have thought one might use the picture of the one form df, for example as a "gradient field", i.e. as a vector field whose vector at each point is given by the cooordinate vector of partial derivatives of f in the chosen coordinate dircetions.

I guess Dave is not doing this because he wants to give us a coordinate invariant view of forms although coordinates seem to be used in the projected area point of view introduced earlier.

If we pursue this, we have an interpretation of every one form as a vector, namely the vector perpendicular to the kernel hyperplane, with length equal to the valoue of the functional on a unit vector.

Then we truly have a geometric object representing a one form (although it depends on a dot product), and moreover we can add one forms and representing vectors interchangeably. I.e. the vector representing the sum of two one forms, is the geometric vector sum of the vectors representing each of them.

In this same vein, if we represent a 2 form on R^3 as an oriented parallelogram, as suggested above, and in R^4 as a formal sum of oriented parallelograms, then we do get a geometric representation of 2 forms, i.e. as a sum of parallelograms.

But to have a fully geometric interpretatioin we should haver a geometric view also of addition of 2 forms. so as asked before, given two parallelograms in R^3, what is a geometric construction of a parallelogram in R^3 represenmting their sum as 2 forms?

And since in R^4, we have a 6 dimensional space of 2 forms, and it is one quadratic condition to be represented by just one parallelogram, we ask what is the geometric condition on a pair of parallelograms that their sum be represented by just one parallelogram, and then what is that parallelogram?

Well, we already know part of this don't we? Because Dave's condition w^w = 0, for this says that the two parallelograms have a sum represented by just one parallelogram if ands only if they span together only a 3 space in R^4. And then surely the construction is the same as the construction in R^3, whatever that is.

If we try to avoid the choice of dot product, as Dave does in his "kernel plane" interpretation of one forms, what would be the correct interpretation?

If we restrict to factorable 2 forms, is there a geometric kernel plane interpretation?

peace.

More free flowing conjectures: We "know" that in projective 5 space the point represented by the coordinates of a 2 form on R^4 is factorable into a product of one forms if and only if satisfies w^w = 0, i.e. if and only if it lies on the 4 dimensional quadric hypersurface defiend by that degree two equation in the coordinates of the 2 form.

Now what is the geometric condition for the sum of two factorable 2 forms to still be factorable? Would it be that the line joining those two points on the quadric still lies wholly in the quadric? I.e. just as a quadric surface in P^3 is doubly ruled by lines, a quadric 4 fold in P^5 also contains a lot of lines.

Just wondering and dreaming. And urging people who want a "geometric" view of the subject to explore further what that would mean.

peace.

[Tom Mattson]

Sorry I've been away for so long. Work gets in the way of what I really want to do, sometimes. :frown:

The line l = \{\vec{r}t + \vec{p} : t \in \mathbb{R}\} for some \vec{r},\ \vec{p} \in T_p\mathbb{R}^3. Suppose \vec{v},\ \vec{w} \in T_p\mathbb{R}^3 such that l \subseteq Span(\{\vec{v},\ \vec{w}\}). Then the set \{\vec{p},\ \vec{v},\ \vec{w}\} is linearly dependent, hence:

\det (\vec{p}\ \ \vec{v}\ \ \vec{w}) = 0[/itex]

Define \omega such that:

[tex]\omega (\vec{x},\ \vec{y}) = \det (\vec{p}\ \ \vec{x}\ \ \vec{y}) \ \forall \vec{x}, \vec{y} \in T_p\mathbb{R}^3

You can easily check, knowing the properties of determinants, that \omega is an alternating bilinear functional, and hence a 2-form. If you want, you can express it as a linear combination of dx \wedge dy,\ dy \wedge dz,\ dx \wedge dz, and it shouldn't be hard, but probably not necessary.


OK thanks, but as you recognized this is answering the reverse question: Given the line, find the 2-form.


EDIT: actually, to answer the question as given, perhaps you will want to write \omega in terms of those wedge products, and determine \vec{p} from there. Then, to find l you just need to choose any line that passes through \vec{p}. Any two vectors containing that line will have to contain \vec{p}, hence those three vectors must be linearly dependent, hence their determinant will be zero, and since \omega depends only on \vec{p} and not the choice of \vec{r}, you're done.

Right, this is what I was wondering about. I think I've worked it out correctly. Here goes.

Exercise 3.18
Let \omega=w_1dx \wedge dy +w_2dy \wedge dz +w_3dz \wedge dx.
Let A=<a_1,a_2,a_3> and B=<b_1,b_2,b_3> be vectors in \mathbb{R}^3.
Let C=[c_1,c_2,c_3] be a vector in T_p\mathbb{R}^3 such that C=k_1A+k_2B. So the set {A,B,C} are dependent. That implies that det|C A B|=0.

Explicitly:


det [C A B]=\left |\begin{array}{ccc}c_1&c_2&c_3\\a_1&a_2&a_3\\b_1&b_2&b_3\end{array}\right|



det [C A B]=c_1(a_2b_3-a_3b_2)-c_2(a_1b_3-a_3b_1)+c_3(a_1b_2-a_2b_1)


Now let \omega act on A and B. We obtain the following:


\omega (A,B)=w_1(a_1b_2-a_2b_1)+w_2(a_2b_3-a_3b_2)+w_3(a_3b_1-a_1b_3)


Upon comparing the expressions for det [C A B] and \omega (A,B) we find that \omega (A,B)=0 if w_1=c_3, w_2=c_1, and w_3=c_2. So the line l is the line that is parallel to the vector [w_2,w_3,w_1]. So I can write down parametric equations for l as follows:


x=x_0+w_2t


y=y_0+w_3t


z=z_0+w_1t



I'll wait for any corrections on this before continuing. If this is all kosher, then I'll post the last of my Chapter 3 notes and we can finally get to differential forms, and the integration thereof.


Also in post #81, Tom asked about solving ex 3.18. What about something like this?
Intuitively, a 1 form measures the (scaled) length of the projection of a vector onto a line, and a 2 form measures the (scaled) area of the projection of a parallelogram onto a plane. Hence any plane containing the normal vector to that plane will project to a line in that plane. hence any parallelogram lying in such a plane will project to have area zero in that plane.


That's helpful. I have to admit I don't really like this geometric approach. But I think that I haven't warmed up to it yet because it still feels uncomfortable. I very much prefer to formalize the antecedent conditions and manipulate expressions or equations until I have my answer, as I've done with all my solutions to the exercises so far. It's my shortcoming, I'm sure.

[mathwonk]

have you read post 98?

I apologize if my comments are not of interest. I am stuck between trying to be helpful and just letting my own epiphanies flow as they will.


I appreciate your patience.

[Tom Mattson]

have you read post 98?


Not yet, but I will.


I apologize if my comments are not of interest. I am stuck between trying to be helpful and just letting my own epiphanies flow as they will.


No, your comments are very much of interest. I'm glad you're making them, and I'm glad that they will be preserved here so that we can go over them at leisure later. But right now, the clock is ticking for us. We are preparing to present some preliminary results to the faculty at our school. Basically the ladies (Melinda and Brittany, who has been silent in this thread so far, but she has been reading along) will be presenting the rules of the calculus, why it is advantageous, and a physical application (Maxwell's equations). The centerpiece of the presentation will be the same as the centerpiece of the book: the generalized Stokes theorem.

Once the presentation to the faculty is done, we will have 2 weeks until the conference. During that time we will get back to your comments.


I appreciate your patience.

That's what I should be saying to you!

[AKG]

So the line l is the line that is parallel to the vector [w_2,w_3,w_1].As I said, l is the (or rather, any) line containing [w_2,w_3,w_1], not parallel to it. Actually, since the plane spanned by two vectors passes through the origin (and since a plane is a subspace if and only if it passes through the origin), you can choose the line parallel to that vector, but this seems like more work.

\omega = \omega _1 dx \wedge dy + \omega _2 dx \wedge dz + \omega _3 dy \wedge dz

\omega(A, B) = \omega _1 (a_1b_2 - b_1a_2) + \omega _2 (a_1b_3 - b_1a_3) + \omega _3 (a_2b_3 - b_2a_3)

= p_3(a_1b_2 - b_1a_2) - p_2(a_1b_3 - b_1a_3) + p_1(a_2b_3 - b_2a_3)

= \det (P A B)

So P = (p_1, p_2, p_3) = (\omega _3, -\omega _2, \omega _1). (I believe you have the above, or something close, in your post).

If we choose a line containing P, then any pair of vectors A, B that span a plain containing that line will also have to conatin P. Then {P, A, B} is dependent, so the determinant is 0. Therefore it is sufficient (and easier) to choose a line containing P. The line parallel to P may not contain P (if the line doesn't pass through the origin), and hence the plane containing the line may not contain P, and hence the set {P, A, B} may not be dependent, so the determinant may not be zero, and so \omega (A, B) may not be zero. To claim that the plane containing the line parallel to P can be done, but requires (a very little) more proof. You know that the line parallel to P, paramterized by t, contains points for t=0 (let's call it P0) and t=1 (P1). So the plane contains these two points. Now you know that P1 - P0 = P. Since the plane in question is a subspace, it is closed under addition and scalar multiplication, and since it contains the line, it contains P1 and P0, and hence P1 - P0, and hence P.

So anyways, you have it right, and if you want to choose a line parallel to P, you may want to throw in that extra bit that allows you to claim that P is in the plane. One more remark: You have A and B in R³, and C in the tangent space. It seems as though you should have them all in R³, or all in the tangent space.

[mathwonk]

tom, thank you very much!

the one geometric thing i added recently may be too far along to be useful to your students but it addresses the geometry of whether a 2 form is or is not a product of one forms, in R^4.

the answer is that 2 forms in R^4 form a vector space of dimension 6, and in that space the ones which are products of one forms form a quadratic cone of codimension one.

I think I also have the answer to the geometric question of what it means to add two 2 forms in R^3, both of which are products of one forms. i.e. to add two paralleograms.


i.e. take the planes they span, and make them parallelograms in those planes, sharing one side.

then take the diagonal of the third side of the parallelepiped they determine, and pair it with the shared side of the two paralleograms.

maybe that is the parallelogram sum of the two parallelograms? at lea`st if the teo parallelograms are rectangles?

ok i know your students do not have time for this investigation, but i am trying to throw in more geometry.

of course i agree with you, the geometry is a little unnatural.

these suggestions are not worked out on paper but just in my head on the commute home from work, but they gave me some pleasure. and i had your students in mind, maybe at some point some will care about these comments.

best,

roy

[mathwonk]

Tom, here are a few more comments on how to possibly convince skeptics of the value of differential forms.

These are based on the extreme simplification of the variuous stokes, greens, gauss theorems as stated in dave's book.

The point is that when a result is simplified we are better able to understand it, and also to understand how to generalize it, and to understand its consequences.

I also feel that you sell the popwer of some tool more effectively if you give at elast one application of its power. I.e. not just simplifying statements but applying those simpler statements to prove something of interest. hence in spite of the demands on the reader I will sketch below how the insight provided by differential forms, leads to a proof of the fundamental theorem of algebra.

(I actually discovered these standard proofs for myself while teaching differential forms as a young pre PhD teacher over 30 years ago, and taught them in my advanced calc class.)

It is of course true that every form of stokes theorem, in 3 dimensions and fewer, has a classical statement and proof.

But I claim none of those statements clarify the simple dual relationship between forms and parametrized surfaces.

i.e. in each case there is an equation between integrals, one thing integrated over a piece of surface [or curve or threefold], equals something else integrated over the boudary of the surface [or curve or threefold].

But in each case the "something else" looks different, and has a completely different definition. i.e. grad(f) looks nothing like curl(w), nor at all like div(M).

It is only when these objects, functions, one forms, two forms, threeforms, are all expressed as differential forms, that the three operations, grad, curl, div, all look the same, i.e. simply exterior derivative "d".

then of course stokes theorem simply says <dS,w> = <S, dw>.


Now that is clear already from what is in the book. But once this is done, then forms begin to have a life of their own, as objects which mirror surfaces, i.e. which mirror geometry.

I.e. this reveals the complete duality or equality between the geometry of parametrized surfaces S, and differential forms w. There is a gain here because even though taking boundary mirrors taking exterior derivative, what mirrors exterior multiplication of forms? I.e. on the face of them, forms have a little more structure than surfaces, which enables calculation a bit better.

Eventually it turns out that multiplication of forms mirrors intersection of surfaces, but this fact only adds to the appeal of forms, since they can then be used to calculate intersections.

Moreover, who would have thought of multiplying expressions like curl(w) and grad(f)? without the formalism of forms?

Already Riemann had used parametrized curves to distinguish between surfaces, and essentially invented "homology", the duality above reveals the existence of a dual construction, of "cohomology".

I.e. if we make a "quotient space" from pieces of surfaces, or of curves, we get "kth homology", defined as the vector space of all parametrized pieces of k dimensional surfaces, modulo those which are boundaries.

this object measures the difference between the plane (where it is zero) and the punctured plane (where it is Z), because in the latter there exists a closed curve which is not the boundary of a piece of parametrized surface, namely the unit circle. Then a closed curve represents n if it wraps n times c.c. around the origin.

This difference can be used to prove the fundamental theorem of algebra, since a polynomial can be thought of as a parametrizing map. Moreover a globally defined polynomial always maps every closed curve onto a parametrized curve that IS the boundary of a piece of surface. namely, if C is the boiundary of the disc D, then the image of C bounds the image of D!.


But we know that some potential image curves, like the unit circle, are not boundaries of anything in the complement of the origin. Hence a polynomial without a zero cannot map any circle onto the unit circle one to one, nor onto any closeed curve that winds around the origin,

Hence if we could just show that some circle is mapped by our polynomial onto such a curve, a curve that winds around the origin (0,0), it would follow that our polynomial does not map entirely into the complement of (0,0). I.e. that our polynomial must "have a zero"!

So it all boils down to verifying that certain curves in the punctured plane are not boundaries, or to measuring how many times they wind around the origin. How to do this? How to do it even for the simple unit circle? How to prove it winds once around the origin?

Here is where the dual object comes in. i.e. we know from greens theorem or stokes theorem or whatever you want to call it, that if w is a one form with dw = 0, then w must have integral zero over a curve which is a boundary.

Hence the dual object, cohomology, measure the same phenomena, as a space of those differential forms w with dw = 0, modulo those forms w which themselves equal dM for some M.

Hence, how to see why the unit circle, does wind around the origin?

Answer: integrate the "angle form" "dtheta" over it. if you do not get 0, then your curve winds around the origin.

here one must must realize that "dtheta" is not d of a fucntion, because theta is not a single valued function!

so we hjave simultaneously proved that fact.

anyway, this is taking too long.

but the solid angel form, integrated =over the 2 sphere also proves that the 2 sphere wrapos around the origin in R^3, and proves after some argument, that there can be no never zero smooth vector field on the sphere, i.e. that you cannot comb the hair on a billiard ball.

[straycat]

Hey all,

I have been going through the book and following the very interesting discussion here. David, I definitely fall into the category of people who like to learn things in a visual way, so I am finding your book to be a nice introduction to the subject. (As for my math background, btw, I majored in electrical engineering as an undergrad and graduated in 1993 -- since then I have been in the medical field, so I'm a bit rusty! :smile: )

As time permits I may join in the discussion. For now I thought I'd post something on this:

for example if N and M are any one forms at all

N^M = N^(N+M) = N^(cN+M) = (cM+N)^M, for any constant c.


In keeping with the spirit of the geometric interpretation, I was inspired when I got to mathwonk's post to make a powerpoint visualization to demonstrate
N^M = N^(cN+M). You can download it from my briefcase at briefcase.yahoo.com/straycat_md in the "differential forms" folder. It's got animations so you have to view it as a "presentation" and then click on the spacebar to see things move (vectors appearing, etc.). Tell me what you think! :)

Regards,

straycat

[mathwonk]

hey! i loved that. i did not realize myself why it was true geometrically until i saw your picture! its just that the area of a patrrallelogram does not change when you translate one side parallel to itself, keeping it the same length.

cool!

[Tom Mattson]

Hey everybody!

My advisees, Melinda and Brittany, gave their practice presentation to the faculty on Friday, and they just ate it up. I was thinking that many of them would not have been exposed to forms, and I was right. After leading up to it the ladies showed how quickly the classical versions of Stokes' Theorem and the Divergence Theorem pop right out of the Generalized Stokes' Theorem. They thought it was beautiful.

I'll be returning to this thread with more notes tomorrow.

[mathwonk]

congratulations!

[straycat]

Tom, Melinda, and Brittany: let me add my congratulations as well!

I have a question for you. In your attempts to "sell" differential forms as an area of study, what are the branches of mathematics against which you are competing, or against which you would compare differential forms? I am wondering in particular whether Hestenes' geometric algebra (also called Clifford Algebra, I think) would be one of these "competitors." I guess a way to phrase the question would be: for a given typical application of differential forms, what other branches of mathematics might be used for the same application? (I hope this is not too off the topic of David's book.)

David Strayhorn

[Haelfix]

Yea this is a thorny issue of notation and the war still rages in specialist circles.

As a physicist I was very interested in Hestenes work at first, but upon further review it seems a tad rigid. It really boils down to a choice of how much structure you want to have on a manifold without losing all information. Eg the minimal amount of structure we can place such that we retrieve the good results we know about, at that point philosophy comes into play (as well as potential physics).

Hestenes basically goes with the philosophy that all manifolds are isomorphic in some sense to a vector space and starts his algebra from there, as opposed to the usual covering space method which somewhat a priori picks a notion of coordinates. The cool thing (for a physcisist) is that the dirac operator instantly is promoted to a very natural geometrical object, as fundamental as length.

The problem is tricky and I'd love to start a new thread on the subject with experts more familiar with the problem. I tried to get a category theorist explain the problems to me, but I must admit a lot of it went way over my head.

[kleinwolf]

I just wanted to ask about the non-linear forms for Area.....How can I generalized the formula in the work there for finding the area of the boundary of a 3D manifold (Agree with me that the boundary of a 3D manifold is 2-dimensional ?) :

Area=\sqrt{(dx\wedge dy)^2+(dx\wedge dz)^2+(dx\wedge dz)^2}

which gives the area of a 2Dimensional manifold, with x=x(t,p), y=y(t,p), z=z(t,p)....But what if I have the boundary of a 3D manifold (4 coordinates parametrized by 3 free variables) ??

[Tom Mattson]

Finals week is wrapping up, and the girls and I are going to get back to doing more work on this right afterwards. Their presentation at the Conference (http://www.skidmore.edu/academics/mcs/hrumc.htm) went very well. They were among the best of the day, which is pretty amazing considering that this was their first speaking engagement.

We also got to hear the keynote speaker, Ken Ribet, talk about Fermat's last theorem. But that's for another thread.

Sorry for the delay, and see you later this week.

[mathwonk]

to straycat, i confess i am a little puzzled by the question, but perhaps it is only because it has a "what is this good for" sound to me.

I.e. there are only a few natural constructions possible in mathematics, starting from a given amount of data, and one needs to know all of them.


I.e. starting from a differentiable manifold, almost the only construct possible is the tangent bundle. then what more refined constructions can be made? one can take sections of it, dualize it, and perform the various multilinear constructions on it, e.g. alternating, or symmetric.

but thats about it.


not to know about any of these, such as sections of the exterior powers of the dual bundle, (i.e. differential forms), would seem to be folly.

I.e. I cannot imagine an argument for NOT knowing about differential forms, and clifford algebras too for that matter.

Its not like theres a huge amount of constructions out there and you only need one. Theres only a few useful constructions that anyone has been able to think of, and you need them all to understand the objects of study.

Its big news when anyone thinks of a new one, like moduli spaces of manifodls or bundles on manifolds, and related iinvariants like characteristic classes or gauge theory.

but thats just a mathematician talking.

Suppose you want to understand a ring. what do you look at? well you could ask how many elements it has, quite interesting if it is finite, not at all other wise.

Then you could ask about its group of units, whether it is commutative, whether it embeds in a field, what its relation is to its "prime ring", i.e. smallest subring containing 1, (i.e. dimension as a vector space if a field, or transcendence degree); prime elements versus irreducible elements, possible uniqueness of factorability into primes, structure of its ideals; then you could ask what its various modules are like, are they all free? what resolutions do they admit? i.e. their projective dimension, representations, then their set of prime ideals and geometric structures possible on these such as spectrum, zariski topology, krull dimension, components, possible mappings to or from standard rings like polynomial rings.

what else? there is really a limited amount of interesting constructions possible. one should not have to argue in favor of learning something about them. i guess the only argument is that life is finite, but most of us have some spare time. thats why we post here on PF.

The big excitement aboiut Wiles work on FLT was not that he solved it, but that he invented some new tools that other people think they can also use to solve new problems and push matters further. That's why a whole generation of young number theorists jumped with glee on his work and began studying it eagerly.

Useful tools are all too rare. we should treasure them and contemplate them when we get the chance. Are there really people out saying, "well i know differential forms have been around for decades, they are the basic tool for defining fundamental invariants like deRham cohomology, they have a huge literature devoted to them, are part of the accepted language of manifolds by all mathematicians, and physicists like John Archibald Wheeler used them in the standard text on gravitation, but are they really important enough for me to learn about?"

[straycat]

Its not like theres a huge amount of constructions out there and you only need one.

Well the main motivation for my question is to try to understand to what extent and in what way tensor analysis, differential forms, and Clifford algebras are different, and to what extent they are minor variations on the same thing.

To make an analogy: there are multiple formulations of quantum mechanics [1], such as wave mechanics, the matrix formulation, Feynman's path integral, etc etc. You could argue that any practicing physicist should know all of them, but I think that most do not. So it's worthwhile to develop arguments for why they should spend the time to do so.


i guess the only argument is that life is finite, but most of us have some spare time. thats why we post here on PF.


Well, don't underestimate the "life is short" argument! :wink: I'm not a mathematician by trade, so most of my time is spent on other things. I could be watching Star Wars right now. :cool:

David

[1] Styer et al. "Nine Formulations of Quantum mechanics. Am J Phys 70 (3), 288.

[Doodle Bob]

I have a question for you. In your attempts to "sell" differential forms as an area of study, what are the branches of mathematics against which you are competing, or against which you would compare differential forms? I am wondering in particular whether Hestenes' geometric algebra (also called Clifford Algebra, I think) would be one of these "competitors." I guess a way to phrase the question would be: for a given typical application of differential forms, what other branches of mathematics might be used for the same application? (I hope this is not too off the topic of David's book.)

I have no idea whether it's been brought up or not. But, the example I think of when reading your question, is that of Maxwell's Equations. It is obviously entirely possible to study them without any knowledge of differential forms. However, if you do have the machinery of forms behind you, you can rewrite the equations extremely succinctly. If I recall correctly, it boils down to two: dF=0 and d^*F=0. The extra bonus of this is that then one can study Maxwell-like forms on other manifolds besides E^3.

The other example is that of symplectic and contact geometries, which of course wouldn't exist without the use of forms. Now, this is a mathematician writing here whose work lies within the realm of this geometry. So, it's important to me. And apparently to a few physicists out there too.

It's a bit dated but Harley Flanders' text (differential forms and its application to the physical sciences) gives several examples of how forms can be used in various parts of science and mathematics.

[mathwonk]

up until this morning i would not have known a clifford algebra if it spoke to me, but while doing my exercises and lying down, i perused artin's geometric algebra and read the definitions, lemmas and consequences, over about 15 minutes, since life is finite. From that fairly innocent acquaintance, it seems to me they are a tool for studying the structure of groups of linear transformations which preserve various inner products.

For instance the group of "regular" positive Clifford elements leaving the original space invariant, map onto the group of positive inner product preserving linear transformations of the original space, with kernel the non zero elements of the underlying field.

This gives them applications to understanding the Lorentz group of rotations of 4 dimensional space time in special relativity which do not interchange past and future.

From this brief perspective, I would say many physicists should know about them, but that their interest is vastly more restricted than that of the very general and flexible tool of differential forms, which everyone who does calculus can benefit from. In particular anyone who wants to study general as opposed to only special relativity seems destined to require differential forms.

[mathwonk]

ALFLAC!!! [why is this not sufficient? does the hierarchy here think us unable to communicate with a single word?]

[Haelfix]

"but that their interest is vastly more restricted than that of the very general and flexible tool of differential forms, which everyone who does calculus can benefit from. In particular anyone who wants to study general as opposed to only special relativity seems destined to require differential forms."

Thats the problem, if you ask Hestenes and the Geometric Algebra people, they will tell you Differential forms are a subset of the more general Clifford algebra construction they use.

That is not however how I learned it, and why its somewhat confusing. For instance, typically in physics Clifford algebras primarily arise when you want to stick a spin geometry (read spinor bundles) on a manifold. This is topologically restricting from the getgo, amongst other things you need a choice of complex structure and I think the other is that the second STiefel Whitney class is identically zero.

I guess it just means I don't understand Geometric Algebra, b/c not only is their definition of differential forms/manifolds different than what I learned and how I use it daily, it also seems their 'Clifford algebras' are somewhat different than I learned. For instance one second of googling gives 4 camps
http://www.ajnpx.com/html/Clifford/4CliffordCamps.html

I asked a math proffessor about this the other day, and he babbled something (he was clearly confused too) about how they are trying to generalize cross products and how their construction is really only good in dimensions 3 and 4.

[mathwonk]

I tried some of those links but they sound like crackpots to me, and I do not want to waste any more time pursuing reading their stuff. If anyone seriously believes these guys have made differential forms obsolete, fine. I cannot help further. (math professor talking here.)

[Haelfix]

Lol, I thought so too. They sound too grandiose, with huge claims etc.

However serious people, take them seriously. Hestenes is at Cambridge, and he has managed to convince quite a few physicists to write books on his approach, etc.

Go figure.

[mathwonk]

well i noticed he is at cambridge, but he still seemed to be claiming to rewrite the whole mathematical basis of physics so i figred he is most likely a nutcase anyway.

of course we could be wrong. i mean i acknowledge that i also am a pod person mascarading as a normal human being.

[dextercioby]

Is that David Hestenes,a guy who was at "Department of Physics and Astronomy,Arizona State University,Tempe,Arizona" ...?

I've got a lecture by him at the 1996 Fourth International Conference on Clifford Algebras and Their Applications to Mathematical Physics",Aachen,D called "Spinor Particle Mechanics".

Back then he didn't seem to be a crackpot.He's published in peer reviewed journals.I dunno what happened in between,there are 9 years,after all...

Daniel.

[Doodle Bob]

That is not however how I learned it, and why its somewhat confusing. For instance, typically in physics Clifford algebras primarily arise when you want to stick a spin geometry (read spinor bundles) on a manifold. This is topologically restricting from the getgo, amongst other things you need a choice of complex structure and I think the other is that the second STiefel Whitney class is identically zero.

I guess it just means I don't understand Geometric Algebra, b/c not only is their definition of differential forms/manifolds different than what I learned and how I use it daily, it also seems their 'Clifford algebras' are somewhat different than I learned. For instance one second of googling gives 4 camps
http://www.ajnpx.com/html/Clifford/4CliffordCamps.html


You all will have to forgive me for my lame answer to the posed question. I didn't realize that the questioner was asking the question from such a sophisticated point of view. I was not aware that anyone disagreed with how to define Clifford algebras, et al., having myself assumed that it was all decided already. Looking at my copy of Spin Geometry, I think that maybe the idea is to create spin-like structures on a broader class of manifolds besides those with zero 2nd Stiefel-Whitney classes.

I asked a math proffessor about this the other day, and he babbled something (he was clearly confused too) about how they are trying to generalize cross products and how their construction is really only good in dimensions 3 and 4.

Since any odd-diml. complex projective space (among other higher dimensional creatures) is spin, maybe he was referring to the Seiberg-Witten equations which use spin geometry (and hence Clifford algebras) but seem restricted to the 4-diml. case.


I really do find it hard to believe though, that differential forms will become completely obsolete. Technically, Riemannian geometry has replaced calculus, but you still need the basic 1-diml. real calculus to actually do anything.

[mathwonk]

here is my perspective on "new" algebras, as derived from the old fashioned education i received in the 60's and currently visible in books such as lang's algebra.

an (associative) "algebra" A (with identity), over a ring R, is an abelian additive group with an associative bilinear multiplication, for which an element called 1 acts as the identity, equipped with a ring map from R to A, "preserving identities".


Given any module M over R there is a universal such object T(M) called the tensor algebra of M over R. There is always a module map from M into T(M), and the image generates T(M) as an algebra.

If M is free of rank s over R, then T(M) is a non commutative polynomial ring over R generated by s "variables", which can be chosen to be any s free generators of M as a module.

The beauty of this object is, it contains in its DNA the data of all possible such algebras over R. I.e. if B is any associate R algebra with identit, equipped with a module map M-->B whose image generates B over R, then there is a unique surjective R algebra map T(M)-->B such that the composition M-->T(M)-->B equals the given map M-->B.

Hence the "new" algebra B, is merely a quotient T(M)/I, of the universal algebra T(M) by some ideal I. In this sense there are no new algbras of this type, as they are all constructed out of T(M).


For example, if S(M) is the "symmetric algebra" of M over R, which just equals the usual commutative polynomial algebra over R, with algebra generators or "variables" equal to the module generators of M, then S(M) = T(M)/I where I is the 2 - sided ideal generated by elements of form
x(tens)y - y(tens)x.

and if E(M) is the exterior algebra of M over R, (whose elements are linear combinations of wedge products of things like dx, dy, dz, when dx, dy, dz are generators of M over R), then E(M) is just the quotient of T(M) by the ideal generated by elements which contain repeated factors like x(tens)x.


Now the usual definition of a Clifford algebra is that it is an associative algebra with identity, built on a vector space M over a field R, plus a quadratic form q ("inner product"), as follows: the algebra C(M) is equipped with a module map M-->C(M) such that the image of the element x, in C(M) has square equal to q(x).1. I.e. if x is in M, and q(x) is its "squared length" under the form q, then in C(M), we have x^2 = q(x).1. And the elements of M generate C(M) as an algebra over R. morover C(M) is universal for all such algebras, i.e. every other one is a quotient of C(M).

But in particuilar C(M) is an associative algebra generated by M. Hence there is a unique surjective R algebra map T(M)-->C(M) realizing C(M) as a quotient of form T(M)/I for some unique ideal I in T(M), containing elements of form
x(tens)x - q(x).1, and presumably generated by these.

Now I fully admit to being a novice here, but i fail to see how anyone can fail to deduce from this that the key construction to understand in all of this is the tensor product.

Moreover, as the Clifford algebra involves extra structure which is not always present, namely the form q, it is clearly a more special derivative of T(M) than is the exterior algebra E(M), i.e. differential forms.

Furthermore, what "new" algebras are possible? unless they are non associative. (and mathematicians have also studied non associative algebras but i have not myself.)

Anyone claiming to construct a new associative algebra generated by elements of a module M, makes one wonder if they are unaware of the basic universal constructions that have been on the scene and even dominated it since the 1950's.

Of course this all concerns only the local, i.e. pointwise side of the story. The usefuleness of these constructs to physicists should be influenced, perhaps decidely, by their global manifestations in physics.


Notice that even if I am completely wrong, I have purposely given you enough data to decide for yourself.

If someone in a competing camp wishes to share more sophisticated and newer definitions for these concepts, I assume we will all be grateful.

Oh yes, and Riemannian geometry cannot possibly replace calculus, as Riemannian geometry also invovles an inner product which is unnecessary for intrinsic ideas of calculus.

[mathwonk]

OK, I have looked on the webpage http://modelingnts.la.asu.edu/GC_R&D.html and in particular perused the short simplified version of GA, intended for high school teachers. there is of course nothing there which is new in the mathematical sense, but some which would seem new to high school students (although I had some of this material in second grade! from a student teacher experimenting on us with trigonometry), an mr hestenes' goal there is to advocate incorporating some well known ideas of vector algebra, exterior algebra, and quadratic forms, into high school geometry, which he calls geometric algebra.

so he is not a real crackpot since he advocates something both useful and correct, and which he also seems to understand; but he is sort of a missionary, and hence comes on like a crackpot by advertising his mission in overly glowing terms, claiming he is going to revolutionize physics education and provide the universal answer to all communication problems between the two sciences, and harking back to the golden days of the 19th century, and so on.

this makes his non technical stuff sound a little fishy. but there is a similar movement by people who dress up like patch adams and try to sell calculus to reluctant students with books called "streetwise calc for dummies" and so on, and they are real mathematicians who have done some people some good, or at least some of my friends think so.

so i for one am glad mr hestenes is out there pumping for more use of vector algebra in high school and college. and although this stuff did get published in am. j. physics it seems, it would be hard for me to believe it occurs in any research math journals. but i have a finite amount of evergy and interest to devote to this type of thing. but i say in this case, more power to him.

i try to do exactly the same type of thing in my teaching, i.e. take known ideas, which are however not having the impact they should have at lower levels, and force them in there, hopefully after having understood them myself that is. i do it right here on this forum all the time. i am not talking about anything mathematical here that is not extremely well known to most practicing mathematicians. my very modest contribution to things like the discussion of clifford algebgras is just to pick up a book not everyone may have access to, read it quickly as a mathematician, and report back here to the best of my ability.

[Doodle Bob]

Oh yes, and Riemannian geometry cannot possibly replace calculus, as Riemannian geometry also invovles an inner product which is unnecessary for intrinsic ideas of calculus.

As I'm sure you know, standard calculus on the real numbers uses the Euclidean norm to define convergence of limits of sequences (among others), which can be derived from the Euclidean inner product, although I suppose one could develop most if not all of standard calculus by defining any old Hausdorff topology on R and defining convergence of sequences from there. Derivatives might end up looking a little strange, if the topology is...

Please take no offense. I was just being cheeky.

[mathwonk]

are you under the impression that all norms arise from inner products?

i.e. that all banach spaces are hilbert spaces?

[Doodle Bob]

are you under the impression that all norms arise from inner products?

i.e. that all banach spaces are hilbert spaces?

Of course not. I just know that the standard distance norm on R can be seen (if one wants to) as coming from the (albeit rather trivial) inner product on R.

[Tom Mattson]

OK school's out, and I'm back for real. Let's finish this book by the end of the summer! Hoo-rah!

I'd like to pick up from where we left off in the book: Exercise 3.18. I posted a solution, to which AKG commented. I haven't looked at his comments in a while, but I do have questions on them. Naturally anyone is free to answer.

Here's my solution to the Exercise.



Exercise 3.18
Let \omega=w_1dx \wedge dy +w_2dy \wedge dz +w_3dz \wedge dx.
Let A=<a_1,a_2,a_3> and B=<b_1,b_2,b_3> be vectors in \mathbb{R}^3.
Let C=[c_1,c_2,c_3] be a vector in T_p\mathbb{R}^3 such that C=k_1A+k_2B. So the set {A,B,C} are dependent. That implies that det|C A B|=0.

Explicitly:


det [C A B]=\left |\begin{array}{ccc}c_1&c_2&c_3\\a_1&a_2&a_3\\b_1&b_2&b_3\end{array}\right|



det [C A B]=c_1(a_2b_3-a_3b_2)-c_2(a_1b_3-a_3b_1)+c_3(a_1b_2-a_2b_1)


Now let \omega act on A and B. We obtain the following:


\omega (A,B)=w_1(a_1b_2-a_2b_1)+w_2(a_2b_3-a_3b_2)+w_3(a_3b_1-a_1b_3)


Upon comparing the expressions for det [C A B] and \omega (A,B) we find that \omega (A,B)=0 if w_1=c_3, w_2=c_1, and w_3=c_2. So the line l is the line that is parallel to the vector [w_2,w_3,w_1]. So I can write down parametric equations for l as follows:


x=x_0+w_2t


y=y_0+w_3t


z=z_0+w_1t




AKG responded thusly.


So anyways, you have it right, and if you want to choose a line parallel to P, you may want to throw in that extra bit that allows you to claim that P is in the plane.


I did not use your P though. I used a vector C that is in the plane spanned by A and B. I did that for the purpose of choosing a line l that is parallel to C, so that the plane spanned by A and B is guaranteed to contain l. The only thing I did not determine was the point (x_0,y_0,z_0), but this can be found easily knowing the vector parallel to l and the equation of the plane.


One more remark: You have A and B in R³, and C in the tangent space. It seems as though you should have them all in R³, or all in the tangent space.


That is not consistent with any of the reading thus far. The rest of the chapter discussed forms defined on T_p\mathbb{R}^n that act on vectors in \mathbb{R}^n. Am I misunderstanding something?

[mathwonk]

i tried to answer this exercise in post 91, or so, as follows:

"Also in post #81, Tom asked about solving ex 3.18. What about something like this?
a 1 form measures the (scaled) length of the projection of a vector onto a line, and a 2 form measures the (scaled) area of the projection of a parallelogram onto a plane. Hence any plane containing the normal vector to that plane will project to a line in that plane. hence any parallelogram lying in such a plane will project to have area zero in that plane.

e.g. dx^dy should vanish on any pair of vectors spanning a plane containing the z axis.'

does that make any sense?

i have forgotten now but it sems the point is that 2 forms on R^3 are decomposable? reducible? whatever?

[Tom Mattson]

i tried to answer this exercise in post 91, or so, as follows:

(snip)

does that make any sense?


Yes, it made sense. It's just that the next few exercise deal with the line that was to be found in 3.18, which is why I wanted an algebraic result. I'll chew on your answer a little longer and see if I can't answer the other questions with it.

[pmb_phy]

are you under the impression that all norms arise from inner products? Are you under the impression that the norm doesn't arise from a inner products?

From the context presented here the norm of a vector is the square root of the inner product of a vector. The defnition of norm of of continuous when ..... sorry but [f(x) = f(x)

[Rev Prez]

My next question is for the students:

Would any of you like to show this? Check my notes for how to show linearity and non-linearity (think superposition and scaling).


Without going through all the steps, scaling returns |\omega\wedge\nu(cV_1,V_2)| = |c||\omega\wedge\nu(V_1,V_2)| \not= c|\omega\wedge\nu(V_1,V_2)| for c < 0.

Rev Prez

[mathwonk]

pmb phy, the point of my post was that calculus for normed spaces depends only on the norm, hence makes sense in any banach space. and yes, i am under the impression, as are most people, that there exist banach spaces which are not hilbert spaces. i.e. there are norms which do not arise from dot products. (sup norm on continuous functions on [0,1].)

the message is that the derivative is a more basic concept than is the dot product, since the derivative makes pefect sense, with exactly the same definition, in situations where the dot product does not. of course people may disagree, but to me the evidence seems clear. :smile:

[Tom Mattson]

OK, back to my quandry. I feel like I'm on the cusp of finally moving past it, but there is a little nagging detail here.

AKG said to me the following:


One more remark: You have A and B in R³, and C in the tangent space. It seems as though you should have them all in R³, or all in the tangent space.


And I replied as follows:


That is not consistent with any of the reading thus far. The rest of the chapter discussed forms defined on T_p\mathbb{R}^n that act on vectors in \mathbb{R}^n. Am I misunderstanding something?


On closer inspection of the text, it seems that I was wrong. But it seems as though there is actually a contradiction in the book. The notation <\cdot , \cdot> is explicitly said on p. 48 to denote vectors in a tangent space, and 1-forms on \mathbb{R}^n are said on p 50 to act on vectors of the form <dx,dy>, which means that they are vectors in the tangent space T_p\mathbb{R}^n. But looking at the diagram at the top of page 53, he plots the vectors V_1 and V_2 at the origin of a set of axes marked with x,y,z. This denotes the space \mathbb{R}^3, no? Well, if a 1-form acts on vectors from T_p\mathbb{R}^3, then I wonder why the axes aren't labeled dx,dy,dz?

OK, so here is what I'd like to know:

Just where do the vectors which are the arguments of a 1-form on T_p\mathbb{R}^n live? Do they live in the tangent space, or in \mathbb{R}^n itself?

And mathwonk: I'm not ignoring your geometric answer to Exercise 3.18. It's just that, as I said, it looks like we need an expression for the line l to move on to the other Exercises.

Boy I can't wait to be done with this chapter.

[mathwonk]

i pointed out long ago several of the many imprecisions and errors in this book, such as you are now noticing.

in this case there, is no big worry. i.e. there is a natural isomorphism between R^n and any of its tangent spaces. so there is no real problem in identifying one with the other.

of course it is no help in understanding the authors conventions.



according to mr bachman's earlier statements to me, the arguments for what he calls a 1 form are indeed elements of the tangent space.

[quetzalcoatl9]

The Flanders book, "Differential form and applications to the physical sciences" threw me for a loop with this. In the book, he started by referring to differential forms as "vectors". since the book came highly recommended to me, I began doubting everything that I thought I knew about forms until that point...

He was referring to them as vectors within the dual space (and later making a correspondence between them with vectors in E^3), which indeed they are...but he didn't lay that out until like 50 pages later, and it was unnecessarily confusing.

[mathwonk]

look, at every point of R^n, or any manifold, there is a tangent space. a form is a linear function on the tangent space, and a field of forms is a choice of a such a linear function on every tangent space. thats all there is to it. whatever language each person uses is only a distraction. just get the idea, then deal with each author's variations in language.


a confusion then is that for R^n the space itself is naturally isomorphic to every tangent space. so what?

if you prefer a book that actually writes down everything correctly and precisely the first time, read spivak instead of bachman.

[Tom Mattson]

in this case there, is no big worry. i.e. there is a natural isomorphism between R^n and any of its tangent spaces. so there is no real problem in identifying one with the other.


I understand that one space is a carbon copy of the other. If you recall, that was the reason I was moaning about the strange way in which he introduced the basis for T_p\mathbb{R}^2. But in this case there is a problem in identifying one with the other, because different two origins of the "home space" of V_1 and V_2 in Exercise 3.18 results in two different lines, and the line is the answer to the question. And the fact that that answer has to be used in the next 2 exercises makes it even worse.

To be honest, my advisees and I left this behind long ago just to move forward. We've finished all of chapters 4 and 5, and much of chapter 6 (up to Stokes' theorem). We just had no choice but to abandon this because of the deadline of the conference.


if you prefer a book that actually writes down everything correctly and precisely the first time, read spivak instead of bachman.


I would agree with that if you're talking about a course in advanced calculus. But I don't want to give up yet on the idea of a course in forms for college sophomores. But if I were going to suggest a course like that to my Department Chair, I can see now that I would want to invest the time putting it together myself, rather than just using this book.

OK, that's enough of that. I've solved Exercises 3.18 through 3.21. Solutions forthcoming shortly.

[Tom Mattson]

Note: All symbols used in Exercises 3.18 through 3.21 have the same meaning.

Exercise 3.18
Let \omega=w_1dx \wedge dy +w_2dy \wedge dz +w_3dz \wedge dx.
Let V_1=<a_1,a_2,a_3> and V_2=<b_1,b_2,b_3> be vectors in T_p\mathbb{R}^3.
Let V_3=<c_1,c_2,c_3> be a vector in T_p\mathbb{R}^3 such that C=k_1V_1+k_2V_2. So the set {V_1,V_2,V_3} are dependent. That implies that det|V_3 V_1 V_2|=0.

Explicitly:


det [V_3 V_1 V_2]=\left |\begin{array}{ccc}c_1&c_2&c_3\\a_1&a_2&a_3\\b_1&b_2&b_3\end{array}\right|



det [V_3 V_1 V_2]=c_1(a_2b_3-a_3b_2)-c_2(a_1b_3-a_3b_1)+c_3(a_1b_2-a_2b_1)


Now let \omega act on V_1 and V_2. We obtain the following:


\omega (V_1,V_2)=w_1(a_1b_2-a_2b_1)+w_2(a_2b_3-a_3b_2)+w_3(a_3b_1-a_1b_3)


Upon comparing the expressions for det [V_3 V_2 V_1] and \omega (V_1,V_2) we find that \omega (V_1,V_2)=0 if w_1=c_3, w_2=c_1, and w_3=c_2. So the line l is the line that contains the vector V_3=<w_2,w_3,w_1>. So I can write down parametric equations for l as follows:


x=w_2t


y=w_3t


z=w_1t

[Tom Mattson]

Exercise 3.19

Let ||V_1 \times V_2|| \equiv A, the area of the parallelogram spanned by V_1 and V_2.

Now look at \omega (V_1,V_2).

\omega (V_1,V_2)= w_1(a_1b_2-a_2b_1)+w_2(a_2b_3-a_3b_2)+w_3(a_3b_1-a_1b_3)

Recalling that V_3=<w_2,w_3,w_1> we have the following.

\omega (V_1,V_2)=V_3 \cdot (V_1 \times V_2)
\omega (V_2,V_2)=||V_3||A cos( \theta ),

where \theta is the angle between V_3 (and therefore l) and both V_1 and V_2. Noting that this dot product is maximized when \theta is 90 degrees, we have our result.

Exercise 3.20

Let N \equiv V_1 \times V_2.

Recalling the action of \omega on V_1 and V_2 from the last Exercise, we have the following.

\omega (V_1,V_2)=V_3 \cdot (V_1 \times V_2)

Noting the definition of N we see that we can immediately identify V_3 with V_{\omega}, and the desired result is obtained.

Exericise 3.21

Start by manipulating the expression given in the Exercise.

\omega= F_x dy \wedge dz - F_y dx \wedge dz + F_z dx \wedge dy
\omega = F_z dx \wedge dy + F_x dy \wedge dz - F_y dx \wedge dz
\omega = F_z dx \wedge dy + F_x dy \wedge dz + F_y dz \wedge dx

I used commutativity of 2-forms under addition to get to line 2, and anticommutativity of 1-forms under the wedge product to get to line 3.

Noting that V_3=<c_1,c_2,c_3>=<w_2,w_3,w_1> (Exercise 3.18) and noting that V_3=V_{\omega} (Exercise 3.20), it can be seen that V_{\omega}=<F_x,F_y,F_z>

[mathwonk]

well on the positive side, some people actually learn more,by correcting errors of an imprecise book, than by plodding thriough one where all the i's are dotted for you. I think that may the case here. you seem to be learning a lot.

[Tom Mattson]

Too true. I sometimes hand out fallacious arguments to my students and ask them to find the errors.

Notes on Section 3.5 will be forthcoming shortly, and then we can finally get on to differential forms and integration.

Yahoo!

[phcatlantis]

Is it safe to say this thread is dead? I'm working through Bachman on my own and the discussion here has been pretty helpful.

[nrqed]

Hello folks,

I found a lovely little book online called A Geometric Approach to Differential Forms by David Bachman on the LANL arXiv. I've always wanted to learn this subject, and so I did something that would force me to: I've agreed to advise 2 students as they study it in preparation for a presentation at a local mathematics conference. :eek:

Since this was such a popular topic when lethe initially posted his Differential Forms tutorial (http://www.physicsforums.com/showthread.php?t=2953), and since it is so difficult for me and my advisees to meet at mutually convenient times, I had a stroke of genius: Why not start a thread at PF? :cool:

Here is a link to the book:

http://xxx.lanl.gov/PS_cache/math/pdf/0306/0306194.pdf

As Bachman himself says, the first chapter is not necessary to learn the material, so I'd like to start with Chapter 2 (actually, we're at the end of Chapter 2, so hopefully I can stay 1 step ahead and lead the discussion!)

If anyone is interested, download the book and I'll post some of my notes tomorrow.


I ahve a question on the example of the integral presented in Example 3.3 (pages 40-41, from the hep archive).

He seems to go from dx^dy directly to dr^dt, where r and t are parametrizations of the upper half unit sphere, x= r cost, y = r sin t, z = sqrt(1- r^2), r ranging from 0 to 1 and t from 0 to 2 pi.

I don't understand that, it seems to me that dx^dy = r dr ^ dt.

Any one can help?

Thanks


Patrick

[George Jones]

The extra r is there.

(z^2) dx^dy was transformed to (1 - r^2) r dr^dt.

Regards,
George

[nrqed]

The extra r is there.

(z^2) dx^dy was transformed to (1 - r^2) r dr^dt.

Regards,
George

Yes, of course...:redface: Thanks

(I simply made the change of variables x,y -> r,t into dx^dy and got r dr^dt. Now I see that his \omega_{\phi(x,y)} calculates the Jacobian which is included automatically in the way I did it. Now I see that he literally meant to replace dx^dy by dr^dt without taking derivatives...that confused me).

Thanks..

On a related note...I know that I will sound stupid but I still find very confusing that the symbols "dx" and "dy" are used sometimes to represent infinitesimals and sometimes to represent differential forms. :eek:

Anyway....

[Doodle Bob]

On a related note...I know that I will sound stupid but I still find very confusing that the symbols "dx" and "dy" are used sometimes to represent infinitesimals and sometimes to represent differential forms. :eek:


Umm... that's on purpose since the one forms dx and dy are defined so that one can do the calculus without all this infinitesimal nonsense.

BTW whatevre is the obsession with infinitesimals? I thought that Bishop Berkley firmly nailed the last nail of their coffin way back in the 1600s. And Cauchy showed us how to do all of analysis and hence calculus without thinking once about them. Virtually no one that I know of in the research field actually thinks in terms of these. Don't we have enough non-computable numbers to deal with (e.g. the vast majority of irrational numbers) without willfully adding more?

[Hurkyl]

I thought that Bishop Berkley firmly nailed the last nail of their coffin way back in the 1600s.
I'm not sure what you mean, but I'm afraid you mean that using infinitessimals can make no sense! But we've had nonstandard analysis since the 1950s, which can be used to put infinitessimals on a perfectly rigorous foundation.

[George Jones]

I'm not sure what you mean, but I'm afraid you mean that using infinitessimals can make no sense! But we've had nonstandard analysis since the 1950s, which can be used to put infinitessimals on a perfectly rigorous foundation.

I'm not sure, but I think that Doodle Bob was referring to these when

Don't we have enough non-computable numbers to deal with (e.g. the vast majority of irrational numbers) without willfully adding more?

Regards,
George

[Doodle Bob]

I'm not sure what you mean, but I'm afraid you mean that using infinitessimals can make no sense! But we've had nonstandard analysis since the 1950s, which can be used to put infinitessimals on a perfectly rigorous foundation.

George's hunch was correct -- I just see nonstandard analysis as rather redundant given that it only adds structure to R that is not really needed for anything. But I am sort of contradicting myself above there. Berkley was railing against Newton's use of fluxions, which were his version of infinitesimals and which he used very much nonrigorously.

But, I am aware that nonstandard analysis has been rigorously established (I always thought it was much older than the 50s). My general feeling, though, is that it's not all that necessary. Sure, standard analysis can be a pain to learn -- at least, it is for my students right now. But eventually one can get used to it -- and even appreciate its utility and elegance.

[Hurkyl]

My general feeling, though, is that it's not all that necessary.
That's true -- a statement is true in standard analysis if and only if it's true in nonstandard analysis. So it doesn't provide any extra power. Any NSA proof could be directly translated into a standard proof, but it's messy.

But it's alledged that NSA proofs are shorter, cleaner, and more intuitive. If so, then there is a practical reason to use it. Mathworld even claims that there will eventually be NSA theorems that will never be proven in a standard way because the proof will be too long to ever write down. (I don't know if that's just sensationalism, or if there's actually been work done on complexity analysis of the two approaches)

I'm told physicists prefer to think in terms of infinitessimals. *shrug*


And who says they have to be uncomputable? :smile: You could just introduce a symbol e for a particular infinitessimal, and do computations with the resulting structure. It's no less "computable" than doing arithmetic with polynomials.

[George Jones]

I'm told physicists prefer to think in terms of infinitessimals. *shrug*

Try to separate physicists from their infinitesimals, and they get downright hostile.

Regards,
George

[nrqed]

Try to separate physicists from their infinitesimals, and they get downright hostile.

Regards,
George


I admit not having the mathematical sophistication of all you guys and I am obviously too stupid to understand advanced maths to start with but here is what I mean.

When I see an integral \int dx dy F(x,y), I think of the Riemann sum lim_{\Delta x, \Delta y \rightarrow 0} \sum F(x,y) \Delta x \Delta y}. That`s what I mean by infinitesimals. As far as I know, this is the onlyway to actually compute an integral (if anyone knows how to get from the expression with differential forms to an actual number without using the above formula, I will be happy to be enlightened from my complete lack of knowledge).

Now, when I see integrals in terms of differential forms defined, I always reach a point where the author will something like '' we *define* '\int dx^dy F(x,y) to be \int dx dy F(x,y) (ex: Equation 8.12 of Felsager..I don't have Frankel with me right now but he does exactly the same thing). On the left side there are differential forms, on the right side there are what I call infinitesimal.

It's this step which bothered my weak and feeble brain. I have always thought that there should be a way to get from the left side to the right side without saying ''well, we have to take this as a *definition*''. This sounds like there is nothing one can do with the left side so one makes this leap by saying that it has to be taken as a definition. I have always wondered why one cannot simply ''feed'' two infinitesimal vectors to the dx^dy to get a scalar dx dy. But that is never presented this way in books, it is always presented as a definition. So on one side, there are differential forms labeled dx, dy and on the other side there are the dx and dy that I call infinitesimals (which, in my mind, are defined throught the limiting procedure I gave at the very top, which is something one can actually work with to get a numercal value (let's say on a computer).

If there was an explicit step going from one side to the other I could understand the connection between the diff forms dx, dy and the ''infinitesimals'' (defined through the limit) dx, dy. But no explicit step is ever shown. This is why I find it confusing to use the same symbols ''dx'' and ''dy'' on both sides because (to my very unsophisticated and slow intellect), they have totally different meaning on the two sides!!


I guess that the notation ''dx'' stems from this one form being the exterior derivative of the coordinate function? And this why when we do a change of variables we can use the usual rules of derivatives to get the corresponding differential form in the new basis?
I was hoping a discussion along those lines but I am obviously too unsophisticated and dumb to discuss those things.

(as you have surely guessed by the stupidity of my comments, I am a physicist).


Regards

Patrick

PS: I hope that it`s clear that my response is not aimed at George specifically..but I had to reply to one of the posts

[Hurkyl]

Well, everything has to have a definition, right? The question is about the motivation of the definitions!


One of the basic ideas of differential geometry is that it's very easy to define things on Euclidean space, and then we extend those definitions to apply to any manifold.


Our picture of a differential form is that it acts like some sort of geometric measure, right? So, say, if we integrate a 3-form in R³, it had better look like an ordinary volume integral. But which 3-form should we take as a "standard"? This one seems most natural:


\int_R f \, dx \wedge dy \wedge dz := \int_R f dV


If I write \omega := f \, dx \wedge dy \wedge dz, then observe that we also have:


f = \omega \left( \frac{\partial}{\partial x} \otimes \frac{\partial}{\partial y} \otimes \frac{\partial}{\partial z} \right)


So we can think of it as evaluating our 3-form on a triple of tangent vectors, and thus producing a scalar function. We then integrate that scalar function with the ordinary volume measure in R³. In other words:


\int_R \omega = \int_R \omega(\mathbf{e}_1 \otimes \mathbf{e}_2 \otimes \mathbf{e}_3) \, dV



Ah, but this generalizes!

An n-dimensional surface in your manifold M is really just a map from the Euclidean n-cube [0, 1]n to your manifold M.

But if we have an n-dimensional surface S into M, we can map differential forms on M back to differential forms on the cube. Recall that for a covector:

S^*(\omega) =
\omega \left( \frac{\partial S}{\partial x_1} \right) \, dx_1 + \cdots +
\omega \left( \frac{\partial S}{\partial x_n} \right) \, dx_n

And this easily extends to an n-form:

S^*(\omega) =
\omega \left( \frac{\partial S}{\partial x_1} \otimes \cdots
\otimes \frac{\partial S}{\partial x_n} \right) dx_1 \wedge \cdots \wedge dx_n

And integration over a surface is defined in terms of an integral over the parameter space:


\int_S \omega := \int_{[0, 1]^n} S^*(\omega)
= \int_{[0, 1]^n}
\omega \left( \frac{\partial S}{\partial x_1} \otimes \cdots
\otimes \frac{\partial S}{\partial x_n} \right) \, dV


so, your intuition was right in a sense -- when we integrate our n-form, we are applying it to some vectors and producing a scalar function, to which we apply an ordinary integral. In particular, we apply it to the standard tangent vectors defined by the parametrization!

[Hurkyl]

This is why I find it confusing to use the same symbols ''dx'' and ''dy'' on both sides because (to my very unsophisticated and slow intellect), they have totally different meaning on the two sides!!
This is, in fact, something I really dislike about differnetial geometry, and has been an obstacle to my understanding... even when it was still in the context of doing ordinary multivariable calculus.

On the one hand, it's convenient to use the same terminology for things that are "essentially" the same, but on the other hand, when you're learning, it makes it awfully difficult to figure out what you're really doing! :frown:

[George Jones]

Patrick,

I apologize if my brief comment caused offence.

The comment was not directed in any way at you - in fact, quite the opposite.

Nor did I intend that anyone should draw the inference that I think physicists are stupid and slow - they aren't. But I have observed that some physicists show incredible stubborness with respect to learning new mathematical techniques. Conversely, many mathematicians are not willing to "let down their hair" enough to see the deep ideas in physics behind the informal mathematics used by physicists.

I have commented on these issues before in this forum, and yesterday's comment was meant as a lightheated continuation of these comments. As such, it was a complete failure.

I do not expect you or anybody else to agree with my views. However, my experience is that you have always been interested in the interplay between physics and mathematics.

The question of what level of mathematical rigor is apropriate for physics is a difficult and context-dependent one. All that can be said for sure is that it lies in the open interval (a , b), where a = no rigor and b = level of mathematical analysis.

Too much rigor can be a real hindrance to communicating physics, and to doing physics. The same for too little rigor. I certainly do not feel that physicists should give up their informal approach to differentials, which they have used profitably for so long.

I enjoy your well thought out posts, both your answers and your questions. You often ask questions to which I don't know the answer, and I geatly appreciate the discussions provoked by your questions and answers.

Regards,
George

[nrqed]

Patrick,

I apologize if my brief comment caused offence.

The comment was not directed in any way at you - in fact, quite the opposite.


Thank you for your thoughtful reply. Sorry if I misunderstood the tone of your reply. I know that my questions may sound very naive but I assure you that they are not meant to criticize the rigor of mathematicians but are genuine attempts to understand the concepts.

I have a PhD and a postdoc in theoretical particle physics but when it comes to differential geometry I am at the same level as a beginning undergraduate student in maths.

As a professor, I ask only one thing of my students: that they make a genuine effort to understand what I am explaining. If they do, I can forgive them for taking time to understand or for trying to offer counterarguments to explanations I am giving. As long as there is a genuine desire to learn and to understand, I am quite patient.

I cannot ask you or anyone here spending time replying to my posts, of course. But the only thing I ask , for anyone taking the time to reply,is for patience. Any question/comment/counterargument I may offer is purely with the intention of understanding, not of putting down rigor or to be argumentative.



Nor did I intend that anyone should draw the inference that I think physicists are stupid and slow - they aren't. But I have observed that some physicists show incredible stubborness with respect to learning new mathematical techniques.

I agree completely!
But the reason I am posting those questions is that I am trying to learn new mathematical techniques.

In learning a mathematical technique, I see two stages: The first stage is to learn how to *apply* it...how to get results using the technique. The second stage is to understand the deep reasons why it works, to understand the underlying concepts and foundations. (I know that mathematicians prefer to follow the opposite order....and often not to even bother with the applications :wink: but I still think that the best way to learn a new technique is to see how it works before understanding the why.)

I can do simple calculation with differential forms, and I could stop there and not bother about the "why" and deeper foundations. But that does not satisfy me. This leads me to very naive questions and I realize that.




Conversely, many mathematicians are not willing to "let down their hair" enough to see the deep ideas in physics behind the informal mathematics used by physicists.

I have commented on these issues before in this forum, and yesterday's comment was meant as a lightheated continuation of these comments. As such, it was a complete failure.

I do not expect you or anybody else to agree with my views. However, my experience is that you have always been interested in the interplay between physics and mathematics.

The question of what level of mathematical rigor is apropriate for physics is a difficult and context-dependent one. All that can be said for sure is that it lies in the open interval (a , b), where a = no rigor and b = level of mathematical analysis.

Too much rigor can be a real hindrance to communicating physics, and to doing physics. The same for too little rigor. I certainly do not feel that physicists should give up their informal approach to differentials, which they have used profitably for so long.


I agree. Except that I was not asking to "lower" the level of rigor. My goal was not to say "why bother with this definition guys when this less rigorous approach works perfectly well". That was not my intent at all and if it came out that way, I apologize!

My comment was more "I am really trying to understand this and I don't see why this step is needed. Why couldn't this other way be possible, even if rigor is to be maintained...why is it necessary to involve a definition at that step". I did not want to sacrifice rigor, but to understand why it would be non rigorous to do it another way.
I


I enjoy your well thought out posts, both your answers and your questions. You often ask questions to which I don't know the answer, and I geatly appreciate the discussions provoked by your questions and answers.

Regards,
George

Thank you. And I certainly have learned a lot from your posts, especially on GR.

Best regards,

Patrick

[rdt2]

If you think you have it bad as a physicist, nrqed, pity us poor engineers. I'm not interested in GR as such but differential forms look as though they might be useful in fluid mechanics and I've been trying to acquire the tools for A Very Long Time. I still have a problem with 'dx' and it doesn't help when the most recent book I bought (on GR) talks of the 'infinitesimal interval dx' and then the 'exterior derivative dx' two chapters later!

Interesting that you say:

In learning a mathematical technique, I see two stages: The first stage is to learn how to *apply* it...how to get results using the technique. The second stage is to understand the deep reasons why it works, to understand the underlying concepts and foundations. (I know that mathematicians prefer to follow the opposite order....and often not to even bother with the applications but I still think that the best way to learn a new technique is to see how it works before understanding the why.)

Without wishing to insult either group, I'd tended to lump physicists and mathematicians together, thus, from my own notes to students:

This isn’t how engineers approach a new subject. Their training is different from that of physicists and mathematicians and their aim is usually to get to a particular application as quickly as possible, leaving the development of the general theoretical framework until later. For example, ‘stress’ first appears on p.27 of Ashby and Jones’ ‘Engineering Materials’, in the context of simple uniaxial structures, but p.617 of Frankel’s ‘Geometry of Physics’, in the context of a general continuum. Engineers therefore tend to go from the particular to the general rather than vice versa and their ‘worked examples’, preferably taken from familiar engineering fields e.g fluid mechanics or stress analysis, rather than relativity or quantum mechanics, usually start with ‘Calculate…’ rather than with ‘Prove…’. Consequently, many of the otherwise-excellent maths and physics textbooks, including Flanders, aren’t really suitable as engineering texts.

Apologies to both, or all three?

ron.

[nrqed]

Well, everything has to have a definition, right? The question is about the motivation of the definitions!


One of the basic ideas of differential geometry is that it's very easy to define things on Euclidean space, and then we extend those definitions to apply to any manifold.


Our picture of a differential form is that it acts like some sort of geometric measure, right? So, say, if we integrate a 3-form in R³, it had better look like an ordinary volume integral. But which 3-form should we take as a "standard"? This one seems most natural:


\int_R f \, dx \wedge dy \wedge dz := \int_R f dV



Yes.

Just a quick comment: then the obvious question is: what is "dV" on the rhs? It is not a form, it is an "infinitesimal". And it is this expression that one can actually use to compute something tangible (a number) using a computer, say. And yet I get beaten up when I talk about d(something) being an infinitesimal!:wink: But they *must* be introduced at some point, differential forms notwithstanding!!! (again, by infinitesimal I mean what is defined within a Riemann sum)




If I write \omega := f \, dx \wedge dy \wedge dz, then observe that we also have:


f = \omega \left( \frac{\partial}{\partial x} \otimes \frac{\partial}{\partial y} \otimes \frac{\partial}{\partial z} \right)


So we can think of it as evaluating our 3-form on a triple of tangent vectors, and thus producing a scalar function. We then integrate that scalar function with the ordinary volume measure in R³. In other words:


\int_R \omega = \int_R \omega(\mathbf{e}_1 \otimes \mathbf{e}_2 \otimes \mathbf{e}_3) \, dV




Good But I wonder why this is not done this way in books (as opposed to *defining* the integration as you wrote in your first equation).

The way I (thought) I understood it, forms and the wedge product provide an elegant way to calculate (signed) areas or volumes (or length element or higher dimensional quantities).

I found neat the first time I saw the Jacobian arising naturally and how natural it is to go from one coordinate system to another. This is beautiful. But it is surprising that after seeing all this power for the calculation of areas and volumes, one gets to a point where it would seem the *most* natural of all to use this machinery to get an infinitesimal area or a volume element but instead of feeding vectors to the forms, one introduces a definition!

To emphasize my point, let's say that one is working with the boring case of finding the area of a surface in 2D Euclidian space. Let's say one starts with Cartesian coordinates (x,y) and one wants to go to polar coordinates. The way I used to see it, using forms provided a wonderful way to do this. But now it seems that one would have to do the following:

a) Define the integral over the area measure dx dy to be an integral over the wedge product dx^dy

b) do your change of variable to r, theta

c) Now you have an integral over dr ^ d(theta). (times r^2 but I am just looking at the differential form part) What to do with this? You have to *define* it as an integral over the measure dr d(theta) !!!
So it seems that every time one does a change of variable, one must introduce a new definition to connect with something that can be integrated over explictly (in the sense of "put in a computer").
That does not sound very useful!


What I have always hoped to hear is the following (I know this is wrong, I ma just trying to show my reasoning):

One is integrating over the usual measure dx dy. But this really comes from the wedge product dx^dy to which two basis vectors of the Cartesian coordinate system have been fed.This implies that this expression is already in a specific coordinate system. As such, it is not in a suitable form to change basis. The "real" expression (before having fed any basis vectors) is the integral over dx^dy. *Now* one can do a change of variables which leads to r^2 dr^ d(theta). Now, to finally get an expression that is useful numerically, one must feed two basis vectors of the coordinate system one is working with. Feeding the basis vectors of the polar coordinate system turns dr^d(theta) into the usual dr d(theta) and voila! We have made a change of coordinate system in which the Jacobian very naturally arises and without having to *define* something new.



Ah, but this generalizes!

An n-dimensional surface in your manifold M is really just a map from the Euclidean n-cube [0, 1]n to your manifold M.

But if we have an n-dimensional surface S into M, we can map differential forms on M back to differential forms on the cube. Recall that for a covector:

S^*(\omega) =
\omega \left( \frac{\partial S}{\partial x_1} \right) \, dx_1 + \cdots +
\omega \left( \frac{\partial S}{\partial x_n} \right) \, dx_n

And this easily extends to an n-form:

S^*(\omega) =
\omega \left( \frac{\partial S}{\partial x_1} \otimes \cdots
\otimes \frac{\partial S}{\partial x_n} \right) dx_1 \wedge \cdots \wedge dx_n

And integration over a surface is defined in terms of an integral over the parameter space:


\int_S \omega := \int_{[0, 1]^n} S^*(\omega)
= \int_{[0, 1]^n}
\omega \left( \frac{\partial S}{\partial x_1} \otimes \cdots
\otimes \frac{\partial S}{\partial x_n} \right) \, dV


so, your intuition was right in a sense -- when we integrate our n-form, we are applying it to some vectors and producing a scalar function, to which we apply an ordinary integral. In particular, we apply it to the standard tangent vectors defined by the parametrization!

Exactly what I was hoping for! But why then the need to define an integral over forms in terms of an integral over an ordinary measure (which is what I can an infinitesimal)?? It would seem to me that the definition should be at a different level, it should be that the integral over the differential forms is defined to be the same as the integral over the forms to which the tangent vectors have been fed! *That* would make sense to me. But I have never seen in a book the connection between integration over forms and integration over infinitesimals described that way. They never seem to mention *feeding vectors to the forms* in going through that step. It is always defined as going directly from integration over forms to integration over infinitesimals, period. Why?

Thank you for your help!

[ObsessiveMathsFreak]

I'm bumping this in the hope that discussions on the book are still ongoing.

I'm finding the book very good so far. It's explaining the meaning and application of differential forms far better than most introductions, which just present rather abstract definitions.

However, I'm having trouble with the presentation of the derivative of a differential form. This is presented in chapter 5 on page 89 in the updated version of the book. Here's the extract.

The goal of this section is to figure out what we mean by the derivative of a
differential form. One way to think about a derivative is as a function which measures
the variation of some other function. Suppose \omega is a 1-form on R2 . What do we mean
by the “variation” of \omega ? One thing we can try is to plug in a vector field V . The result is a function from R^2 to R. We can then think about how this function varies near a point p of R^2 . But p can vary in lots of ways, so we need to pick one. In Section 6 of Chapter 1 we learned how to take another vector, W , and use it to vary p. Hence, the derivative of \omega , which we shall denote d\omega, is a function that acts on both V and W . In other words, it must be a 2-form!

Let’s recall how to vary a function f (x, y ) in the direction of a vector W at a
point p. This was precisely the definition of the directional derivative:

\nabla_W f(p) = \nabla f(p) \cdot W


This all seems fine. The derivative of the one-form d\omega (V,W)will be the rate of change of in the direction of W, of the one-form applied to a fixed V. I would thus be inclined to think that d\omega(V,W) = \nabla_W \omega (V) . However the book goes on to say that;

d\omega(V,W) = \nabla_W \omega (V) - \nabla_V \omega(W)

What is the reason for this second term, and why is W, a change in R^2, being used as input to a one-form on tangent space V?

[Hurkyl]

k-forms are antisymmetric in its arguments. So, if we intend the derivative of a 1-form to be a 2-form, we need to define the derivative so that it is antisymmetric. (I'm sure there are other reasons, but this one jumps to mind the quickest)

The derivative you mention is simply \nabla \omega, which is not a 2-form, and thus not a candidate to be the exterior derivative.

[ObsessiveMathsFreak]

But then what is d\omega(V,W)? It is clearly not the rate of change of the form with respect to the evaluation point. Are we just defining things for convienience here?

[mathwonk]

i guess the reason you want an alternating object is so the area of a parallelogram spanned by two copies of the same vector, i.e. a flat parallelgram, will be zero?

[ObsessiveMathsFreak]

OK. It's become clear to me that in general d\omega(V_1, \ldots V_n, W) is not the derivative of \omega(V_1, \ldots V_n) in the direction of W. It appears to be some kind of derivative in multiple directions, but the exact interpretation of what the derivative of a general form is escapes me.

However, having read further on in the text, I think the best I can really say is that the differential of a form is something that satisfies the generalised Stokes equation. This seems a little circular though. I'd like to be able to say what the differential of a form is in its own right.

[mathwonk]

in mathematics you always get to choose where to begin. i.e. what your definitions are. then the 0therv ersions of the same thing become theorems. so if you define a derivative of a form as something that satisfies the stokes theorem, then you are defining it in the same way as the dual of a linear operator os defiend in vector space theory.

then you have to prove that its matrix is the transpose of the other one in appropriate corrdinates, or that it behaves a certain way wrt the inner product.

if you define a derivative more directly, them you have to prove stokes theorem.

what a form is directly, is a measure of volume as i said in my last post. then to geta measure of 4 volume from a measure of three volume, you first take the boiundary of the 4 block, and then apply the measure of three volume to that.

that is the derivative of the three form, and also a local version of the adjoint property mentioned before.

alwys think of what you want your object to measure. then read the definition.

[nrqed]

Try to separate physicists from their infinitesimals, and they get downright hostile.

Regards,
George
You know, I have been thinking about this lately and the following came to mind.

It's hard to give up entorely on infinitesimals when they have been used for years and have faithfully led to correct results.

Just a very simple example. Let's say a student asks to find the electriv field at a point produced by an infinite line of charge. Then you set up the integral of k dq / r^2 {\hat r} and so on. Now, how is one to think about dq here? As a differential form?:eek: Or as an infinitesimal (again in the sense of writing the integral as a sum and taking the limit etc etc)?

I am probably too simple minded to see it but I have a hard time seeing how this can be seen in terms of differential forms!!

Patrick

[mathwonk]

patrick, what is a field? i.e. what does it do? how do you detect its presence? say, given the charge?

do you put a particle in there anbd see if it accelerates? or let something move through there and see how it is diverted?

i am trying to see if the field acts on vectors, i.e. tangent vectors, or moving particles.

if it acts on vectors then it should be represented by a differential form.

i.e. it would be a covector field, as opposed to a vector field.

[nrqed]

patrick, what is a field? i.e. what does it do? how do you detect its presence? say, given the charge?

do you put a particle in there anbd see if it accelerates? or let something move through there and see how it is diverted?

i am trying to see if the field acts on vectors, i.e. tangent vectors, or moving particles.

if it acts on vectors then it should be represented by a differential form.

i.e. it would be a covector field, as opposed to a vector field.

That's a good question.

well, the answer that any physicist would give is the first one you mentioned: you place a "test charge" (i.e. a physicist's idealization of an "infinitesimal" charge) charge at a point and see if it accelerates.

On the other hand, in mathematical physics books, people usually say that E is a one-form. They say that one should think at grabbing the charge, moving it through the E field and measuring the work done by the electric "field" on the charge. So that the E field, integrated over the path gives a number, the work.

It's very confusing to start from the conventional physics equations and to figure out what quantities are truly differential forms, what quantities are vector fields.

I actually have a few questions about this which I will post soon.

Regards,

Patrick

[Bachman]

Just a quick note to officially announce the release of my book, "A Geometric Approach to Differential Forms." It has been published by Birkhauser, and is available via their webisite, Amazon.com, Barnes & Noble, etc.

I have done what I can to keep the purchase price low. I think its in the $35-40 range. There have been many significant additions/corrections since the versions that were put up on the web, such as the one on the arxiv.

So please support the author and buy yourself a copy! To make it easy for you, here's a link to the book at Amazon:
http://www.amazon.com/Geometric-Approach-Differential-Forms/dp/0817644997/sr=8-6/qid=1158608172/ref=pd_bbs_6/104-3264928-9327100?ie=UTF8&s=books

Thanks!
Dave Bachman

[cliowa]

patrick, what is a field? i.e. what does it do? how do you detect its presence? say, given the charge?

do you put a particle in there anbd see if it accelerates? or let something move through there and see how it is diverted?

i am trying to see if the field acts on vectors, i.e. tangent vectors, or moving particles.

if it acts on vectors then it should be represented by a differential form.

i.e. it would be a covector field, as opposed to a vector field.

In general a field may vary according to some physical properties of the object you're studying. Depending on the environment you're working in (classical mechanics, electromagnetism) one describes a physical object by certain quantities. Examples: mass, charge, spin,... and the state of motion, which - in a classical context - is described by the momentum, forces and position. In most classical mechanics situations the mass of the object you're observing is constant. That makes it a lot easier.
An electromagnetic field for example depends upon tangent vectors, as the Lorentz force depends on the velocity of an object.

But I feel that the issue of whether to take a field or a 1-form isn't really deep, as you're dealing with settings in \mathbb{R}^n and you can "convert" quite easily between the two.

[cliowa]

@patrick:
I think the trouble you're having goes much deeper than differential geometry. Your problem is integration in \mathbb{R}^n, in my eyes. Once you have understood the notion of integration in higher dimensions, the integrating part of differential geometry will not be so hard to understand. One way of looking at manifolds is as objects that are locally isomorphic to a euclidean space, and it is exactly this property one is using to integrate, differentiate and so forth. If you know how to integrate in euclidean space, you'll simply "pull back" your form on the manifold to the euclidean space and integrate there. And that, of course, is a definition, but I feel it's quite reasonable.

[gts87]

Sorry, hate to dig up an old thread but I found it relevant to my question. I'm using this book to teach myself about differential forms and the generalized Stokes' theorem.

On page 32 (2006 edition) Bachman asks to evaluate the following integral:

\int_{R} f(\phi(r,\theta))Area(\partial\phi/\partial r, \partial\phi/\partial\theta)drd\theta

Where Area(.,.) is the area of the parallelogram spanned by the vectors \partial\phi/\partial r, \partial\phi/\partial\theta. The function is f(x,y,z) = z^2 and the region R is the top half of the unit sphere, parametrized by \phi(r,\theta) = (rcos\theta, rsin\theta, \sqrt{1 - r^2}). This was part of a discussion of how to generalize the integral (if anyone has read/is reading this book). It would be simple enough to just evaluate the integral \int_{R} (1 - r^2)r dr d\theta, using the standard change of variables, but for the sake of the development I took this approach. However, it did not give me the desired answer (\pi/2, which is also the answer he got when using differential forms to integrate later in the book, p. 56). This is how I did it:

\int^{2\pi}_{0}\int^{1}_{0}(1-r^2)|\partial\phi/\partial r \times \partial\phi/\partial\theta| dr d\theta

\int^{2\pi}_{0}\int^{1}_{0}(1-r^2)|<r^2cos\theta/\sqrt{1 - r^2}, r^2sin\theta/\sqrt{1 - r^2}, r>| dr d\theta


\int^{2\pi}_{0}\int^{1}_{0}(1-r^2)\sqrt{(r^4cos^2\theta + r^4sin^2\theta)/(1 - r^2) + r^2} dr d\theta


\int^{2\pi}_{0}\int^{1}_{0}(1-r^2)\sqrt{(r^2/(1 - r^2)} dr d\theta


\int^{2\pi}_{0}d\theta\int^{1}_{0}r\sqrt{1 - r^2} dr

\int^{2\pi}_{0}d\theta\int^{1}_{0}r\sqrt{1 - r^2} dr

(\int^{2\pi}_{0}d\theta)(-1/2)\int^{0}_{1}\sqrt{u} du Letting u = 1 - r^2

\pi\int^{1}_{0}\sqrt{u} du = 2\pi/3.

There may be something really obvious I'm missing here. I used the magnitude of the cross product of \partial\phi/\partial r, \partial\phi/\partial\theta for the Area function, but it's not giving the right answer. Appreciate any help!

[hanskuo]

Hi Tom Mattson,
I'm reading this book now.

[n!kofeyn]

This is slightly off topic and I know you have a book you're already using, but I thought I'd share this reference. There is a book by William Burke (http://www.ucolick.org/~burke/home.html) called Div, Grad, Curl Are Dead (http://www.ucolick.org/~burke/home.html), and it seems to be a bit of a treasure in this subject. Although I haven't read the book, I read the introduction and it seems at least relevant to what you are studying.