A Geometric Approach to Differential Forms by David Bachman

[mathwonk]

An apology and some comments:

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

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

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

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

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

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

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

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

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

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

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

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


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

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

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

So Dave is finessing the existence of covectors entirely.

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

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

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

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

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


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


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

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

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


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

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

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

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

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

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

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

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

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

 

[Bachman]

Melinda,

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

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

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

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

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

Dave.

 

[Bachman]

Dear all,

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

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

If anyone finds any more please let me know!

Dave.

 

[mathwonk]

Dave's example recalls post #60:

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

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

 

[mathwonk]

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

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

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

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

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

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

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

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

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

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

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

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

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

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

I claim this has the following interpretation:

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

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

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

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


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

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

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


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

 

[mathwonk]

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

 

[mathwonk]

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

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


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

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

nice.

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

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

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

 

[mathwonk]

another in my wildly popular series of commentaries:

towards a more fully geometric view of differential forms.

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

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

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

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

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

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

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

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

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

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

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

peace.

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

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

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

peace.

 

[quantumdude]

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

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]<br /> <br /> Define \omega such that:<br /> <br /> \omega (\vec{x},\ \vec{y}) = \det (\vec{p}\ \ \vec{x}\ \ \vec{y}) \ \forall \vec{x}, \vec{y} \in T_p\mathbb{R}^3<br /> <br /> 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&#039;t be hard, but probably not necessary.<br />

<br /> <br /> OK thanks, but as you recognized this is answering the reverse question: Given the line, find the 2-form.<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> 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 <i>any</i> 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&#039;re done. </div> </div> </blockquote><br /> Right, this is what I was wondering about. I think I&#039;ve worked it out correctly. Here goes.<br /> <br /> <b>Exercise 3.18</b><br /> Let \omega=w_1dx \wedge dy +w_2dy \wedge dz +w_3dz \wedge dx.<br /> Let A=&amp;lt;a_1,a_2,a_3&amp;gt; and B=&amp;lt;b_1,b_2,b_3&amp;gt; be vectors in \mathbb{R}^3.<br /> Let C=[c_1,c_2,c_3] be a vector in T_p\mathbb{R}^3 such that C=k_1A+k_2B. So the set {A,B,C} are dependent. That implies that det|C A B|=0.<br /> <br /> Explicitly:<br /> <br /> &lt;br /&gt; det [C A B]=\left |\begin{array}{ccc}c_1&amp;amp;c_2&amp;amp;c_3\\a_1&amp;amp;a_2&amp;amp;a_3\\b_1&amp;amp;b_2&amp;amp;b_3\end{array}\right|&lt;br /&gt;<br /> <br /> &lt;br /&gt; det [C A B]=c_1(a_2b_3-a_3b_2)-c_2(a_1b_3-a_3b_1)+c_3(a_1b_2-a_2b_1)&lt;br /&gt;<br /> <br /> Now let \omega act on A and B. We obtain the following:<br /> <br /> &lt;br /&gt; \omega (A,B)=w_1(a_1b_2-a_2b_1)+w_2(a_2b_3-a_3b_2)+w_3(a_3b_1-a_1b_3)&lt;br /&gt;<br /> <br /> Upon comparing the expressions for det [C A B] and \omega (A,B) we find that \omega (A,B)=0 if w_1=c_3, w_2=c_1, and w_3=c_2. So the line l is the line that is parallel to the vector [w_2,w_3,w_1]. So I can write down parametric equations for l as follows:<br /> <br /> &lt;br /&gt; x=x_0+w_2t&lt;br /&gt;<br /> &lt;br /&gt; y=y_0+w_3t&lt;br /&gt;<br /> &lt;br /&gt; z=z_0+w_1t&lt;br /&gt;<br /> [/color]<br /> <br /> I&#039;ll wait for any corrections on this before continuing. If this is all kosher, then I&#039;ll post the last of my Chapter 3 notes and we can finally get to differential forms, and the integration thereof.<br /> <br /> <blockquote data-attributes="" data-quote="mathwonk" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-title"> mathwonk said: </div> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> Also in post #81, Tom asked about solving ex 3.18. What about something like this?<br /> Intuitively, a 1 form measures the (scaled) length of the projection of a vector onto a line, and a 2 form measures the (scaled) area of the projection of a parallelogram onto a plane. Hence any plane containing the normal vector to that plane will project to a line in that plane. hence any parallelogram lying in such a plane will project to have area zero in that plane. </div> </div> </blockquote><br /> That&#039;s helpful. I have to admit I don&#039;t really like this geometric approach. But I think that I haven&#039;t warmed up to it yet because it still feels uncomfortable. I very much prefer to formalize the antecedent conditions and manipulate expressions or equations until I have my answer, as I&#039;ve done with all my solutions to the exercises so far. It&#039;s my shortcoming, I&#039;m sure.

 

[mathwonk]

have you read post 98?

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


I appreciate your patience.

 

[quantumdude]

have you read post 98?

Not yet, but I will.

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

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

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

I appreciate your patience.

That's what I should be saying to you!

 

[AKG]

So the line l is the line that is parallel to the vector [w_2,w_3,w_1].

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

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

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

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

= \det (P A B)

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

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

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

 

[mathwonk]

tom, thank you very much!

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

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

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


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

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

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

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

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

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

best,

roy

 

[mathwonk]

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

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

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

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

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

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

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

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

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

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

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


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

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

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

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

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

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

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

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


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

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

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

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

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

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

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

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

so we hjave simultaneously proved that fact.

anyway, this is taking too long.

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

 

[straycat]

Hey all,

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

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

for example if N and M are anyone forms at all

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

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

Regards,

straycat

 

[mathwonk]

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

cool!

 

[quantumdude]

Hey everybody!

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

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

 

[mathwonk]

congratulations!

 

[straycat]

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

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

David Strayhorn

 

[Haelfix]

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

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

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

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

 

[kleinwolf]

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

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

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

 

[quantumdude]

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

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

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

 

[mathwonk]

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

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


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

but that's about it.


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

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

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

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

but that's just a mathematician talking.

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

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

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

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

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

 

[straycat]

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

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

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

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

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

David

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

 

[Doodle Bob]

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

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

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

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

 

[mathwonk]

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

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

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

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

 

[mathwonk]

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

 

[Haelfix]

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

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

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

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

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

 

[mathwonk]

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

 

[Haelfix]

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

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

Go figure.