A Geometric Approach to Differential Forms by David Bachman

1. Mar 14, 2005

Tom Mattson

Staff Emeritus
Hello folks,

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

Since this was such a popular topic when lethe initially posted his Differential Forms tutorial, and since it is so difficult for me and my advisees to meet at mutually convenient times, I had a stroke of genius: Why not start a thread at PF?

Here is a link to the book:

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

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

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

Last edited by a moderator: Apr 21, 2017
2. Mar 14, 2005

mathwonk

That seems like a gentle enough introduction to differential forms.

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

let me sketch these:

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

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

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

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

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

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

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

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

Last edited: Mar 15, 2005
3. Mar 15, 2005

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

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

Last edited: Mar 15, 2005
4. Mar 15, 2005

mathwonk

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

5. Mar 15, 2005

Tom Mattson

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

Sounds good, I'll order it.

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

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

6. Mar 15, 2005

Tom Mattson

Staff Emeritus
Chapter 3: Forms

Section 1: Coordinates for Vectors

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

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

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

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

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

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

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

Here is my first question.

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

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

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

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

This leads to my second question.

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

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

This leads to my third question.

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

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

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

Last edited: Mar 31, 2005
7. Mar 15, 2005

mathwonk

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

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

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

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

8. Mar 16, 2005

mathwonk

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

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

But anyway...

OK further corrections to his sloppiness:

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

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

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

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

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

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

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

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

Last edited by a moderator: Mar 16, 2005
9. Mar 16, 2005

Tom Mattson

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

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

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

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

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

10. Mar 16, 2005

rdt2

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

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

11. Mar 16, 2005

Bachman

Hello all,

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

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

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

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

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

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

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

Dave.

12. Mar 16, 2005

Tom Mattson

Staff Emeritus
Thanks for coming!

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

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

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

Thanks,

13. Mar 16, 2005

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

Last edited: Mar 16, 2005
14. Mar 16, 2005

Tom Mattson

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

15. Mar 16, 2005

Tom Mattson

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

My first one was:

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

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

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

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

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

Thanks everyone, this is a real help so far.

16. Mar 17, 2005

Bachman

A few quick replies...

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

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

Dave.

17. Mar 17, 2005

Data

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

18. Mar 17, 2005

Hurkyl

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

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

19. Mar 17, 2005

Tom Mattson

Staff Emeritus

OK, I'll switch over.

Here is why there is confusion:

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

Sorry, but no.

Last edited: Mar 31, 2005
20. Mar 17, 2005

Bachman

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

21. Mar 17, 2005

Bachman

Tom,

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

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

By the way, thanks all for the latex advice.

Dave.

22. Mar 17, 2005

Tom Mattson

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

Last edited: Mar 17, 2005
23. Mar 17, 2005

mathwonk

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

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

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

roy

Last edited: Mar 17, 2005
24. Mar 17, 2005

Tom Mattson

Staff Emeritus
Chapter 3: Forms

Section 2: 1-Forms

Once again:

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

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

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

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

Quick question:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

we have:

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

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

Last edited: Mar 31, 2005
25. Mar 17, 2005

Tom Mattson

Staff Emeritus
Chapter 3: Forms

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

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

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

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

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

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

Last edited: Mar 31, 2005