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}{cc}\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}{cc}\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 (