A Geometric Approach to Differential Forms by David Bachman

[quantumdude]

Chapter 3: Forms

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:

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

 

[quantumdude]

Chapter 3: Forms

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.

 

[quantumdude]

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?

 

[quantumdude]

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...

 

[quantumdude]

Chapter 3: Forms

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.

 

[quantumdude]

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 let's 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?

 

[quantumdude]

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 theory, 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.

 

[quantumdude]

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.

 

[quantumdude]

Chapter 3: Forms

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 anyone 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 let's 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. that's 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.