Mathematics

Dylan Moreland

12:12 AM

@Potato So $W$ is some open subset of $\mathbf R^k$?

Potato

Yes.

Dylan Moreland

Sorry, in retrospect this is obvious.

Potato

I having trouble working through the chain rule. In particular, I don't see why $D(f^{-1}\circ g)$ is the identity, nor why we can use the inverse function theorem on $f^{-1}$ when it is a map between spaces of different dimensions.

Dylan Moreland

It probably isn't the identity.

Is it?

Potato

Well, I'm having trouble working it out.

I'm actually typing my question up more nicely into a question to post, if you want to think about it a little bit and post an answer.

Dylan Moreland

12:20 AM

Psychologically it might be better to write $f\colon W \to U \cap M \subset \mathbf R^n$ where $U$ is some open set in $\mathbf R^n$ containing $x$ on which you can smoothly extend $f^{-1}$.

So it's an honest bijection and writing $f^{-1}$ feels less slimy.

Potato

Ok, just a sec. I have almost finished writing my question.

Dylan Moreland

I'll take a look.

Potato

0

Q: Why is tangent space independent of choice of coordinates?

This is a question about Spviak's Calculus on Manifolds, page 115. Let $M$ be a $k$-dimension manifold in $\mathbb{R}^n$ and let $f:W\rightarrow \mathbb{R}^n$ be a coordinate system around $x=f(a)$, where $W$ is some open subset of $\mathbb{R}^k$. We know $f'(a)$ has rank $k$ 9by definition)...

Dylan Moreland

But in the end it's the chain rule; it has to be.

Potato

Of course.

But I need to not get lazy when reading outside of class. Working through the details is a must.

Dylan Moreland

12:26 AM

So I don't think that it's the identity. I think it's an isomorphism.

Potato

Oh, right.

Dylan Moreland

So the image of $(f^{-1} \circ g)_*$ is all of $\mathbf R_a^K$. That's all.

Potato

Ah, ok.

Ok but wait.

Using the push forward of that involves taking the derivative.

Dylan Moreland

Right.

Potato

So we are actually concerned about the image of $(f^{-1}\circ g)'$

Dylan Moreland

12:29 AM

Agreed.

t.b.

Silly question: what is the model of the tangent space the two of you are using?

Potato

Did you see the question I linked?

I must go. I'll be back in a bit.

t.b.

Yes, I did. But I didn't see the definition of the tangent space there.

QED

Hello

Potato

Briefly, it's an ordered point of (point, vector)

t.b.

12:32 AM

Hi

Potato

so it's the set of tangent vectors at the point

anyway I'll return in a bit

@DylanMoreland So why is $(f^{-1}\circ g)'(b)$ an isomorphism (invertible)?

@t.b. I'm back and ready to explain how Spivak defines tangent spaces, if you want.

Dylan Moreland

12:59 AM

@Potato Because $f^{-1} \circ g$ has an inverse $g^{-1} \circ f$.

Potato

That just shows $f^{-1}\circ g$ is invertible. I'm concerned about the derivative.

Dylan Moreland

The inverse is smooth.

Do you believe that $f^{-1} \circ g$ is smooth?

Potato

Let me back up and rephrase.

Dylan Moreland

Hum. Does the inverse function theorem ever prove that something is smooth?

Potato

So for the first implication, we need to show that $g'$ = $f'\cdot(f^{-1}\circ g)'$, yes?

Dylan Moreland

1:03 AM

Looks right.

I threw up a diagram in my post. Maybe it doesn't add anything but I like diagrams.

Potato

So this is true because of the chain rule?

Dylan Moreland

Yes, because $f \circ (f^{-1} \circ g) = g$.

Maybe we should be careful about the points.

Potato

I think we should.

I think it's the points that are messing me up.

Dylan Moreland

I don't remember Spivak's notation for this. He's always pretty funny about differential notation.

Potato

Just use whatever notation you like.

Dylan Moreland

1:10 AM

Maybe that isn't the confusion here. If we use Wikipedia's convention then I'm claiming that $D_bg = D_af \circ D_b(f^{-1} \circ g)$.

All of these have domains that are open sets in Euclidean space. No worries about dimension there.

Potato

Introduction of the points is confusing me. If you use the chain rule on the right hand side, what do you get, with the subscripts included?

Dylan Moreland

The right hand side is the application of the chain rule to $D_b(f \circ (f^{-1} \circ g))$.

Potato

Oh, ok. Thanks. Now it clicks.

Dylan Moreland

Excellent.

Potato

For the second part, why is $(f^{-1}\circ g)'$ an isomorphism?

Dylan Moreland

1:18 AM

Because (and I'm shrinking $V, W$ at the beginning to ensure this makes sense) the original function $f^{-1} \circ g$ is a diffeomorphism, because it's smooth and has a smooth inverse $g^{-1} \circ f$.

Potato

I don't see how that implies the conclusion.

Dylan Moreland

Again, chain rule.

The derivative of $(f^{-1} \circ g) \circ (g^{-1} \circ f)$ is the derivative of the identity, which is the identity.

So $D_b(f^{-1} \circ g)$ is a linear map with a linear inverse, $D_a(g^{-1} \circ f)$.

Sorry, I didn't think that this was the unclear bit. Does it make sense now?

Like, if I have a diffeomorphism between open subsets of $\mathbf R^n$, then the derivative at any point is nonsingular/invertible/what have you.

This is sort of the converse to the inverse function theorem, I think.

Potato

Hmm, ok.

Ok, yes, now I get it.

Thank you!

Dylan Moreland

No trouble at all.

robjohn

@DylanMoreland nice to hear :-)

replying out of context is fun :-)

Potato

1:33 AM

How is $ds$ usually defined for one dimensional manifolds?

I want to show that if I have an oriented 1-d manifold in $R^n$, and $c:[0,1]\rightarrow M$ is orientation preserving, that the integral of $c^*(ds)$ over [0,1] is just the integral of the norm of $c'$ over [0,1] (the usual length formula).

So this is just integral of $ds$ along the manifold

Err, no.

Zhen Lin

$ds$ is not a 1-form. For one thing, $ds$ is always non-negative, but any non-trivial linear form must take negative values as well as positive values.

Potato

$ds$ here is the 1-dimensional volume element.

Zhen Lin

As you know, there is no canonical choice of a volume element...

Potato

Well, there is if the manifold is oriented and we use the standard inner product.

Zhen Lin

Sure, but then you're in the Riemannian case and everything's as you expect anyway...

Potato

1:47 AM

Basically, if I could figure out why $ds(v)=|v|$ I would be done, because then you just use the pushforward on $c$.

Well, yeah.

So if we have the usual inner product and a given orientation, why is $ds$ of a vector just the norm?

I'm a bit fuzzy on defining volume elements.

t.b.

@Potato Thanks, it's fine. I thought bringing up doubts about the very definitions you're using might help. But you appear to have figured it out by now.

Zhen Lin

@Potato: It isn't. It can be plus or minus the norm.

Potato

Ok. But we are given an orientation here so that it's plus the norm.

Ok this is all slowly making sense.

I think I need to read a serious book on differential geometry though. Spviak is very terse and has weird notation.

Zhen Lin

Yeah, but then I can take $-1$ of your vector and then $ds$ of it becomes negative.

Potato

Oh no! Don't do that!

If $f:[a,b]\rightarrow R$ is non-negative and the graph of $f$ in the x-y plane is revolved around the $x$-axis in $R^3$ to yield a solid $M$, how do I use differential forms along with the standard volume elements to derive the usual calculus formulas for these?

Zhen Lin

1:53 AM

You don't, if you don't have to!

Potato

But I want to!

Zhen Lin

There are lots of icky bookkeeping details here.

Potato

I'd be fine with a sketch, or a reference.

Zhen Lin

There are all kinds of annoying things like, "the integral of a form over a subspace of full measure is equal to the integral over the whole space".

So, suppose you're given a volume form $\omega$ for $M$. The volume of $M$ is just what you expect: $$\int_M \omega$$

To convert this to a calculus formula, choose a diffeomorphism $\phi : (0, 1)^n \to M$ with image a subspace of full measure (dense suffices!) and pull back the integral through $\phi$.

We have $$\int_M \omega = \int_{(0, 1)^n} \phi^* \omega$$

Potato

Ok.

Ick.

Do you know a good reference on this?

Zhen Lin

1:59 AM

Told you so.

Potato

Spivak is not doing it for me.

Zhen Lin

No, I don't know of any good references for differential geometry. I learned it in bits and pieces from various books / the internet / lectures.