Differential Geometry

Robert Cardona

6:34 PM

This is a room to discuss anything even remotely related to differential geometry, ask questions, pose problems, etc.

Robert Cardona

6:56 PM

Chat guidelines | $\LaTeX$ in chat

Georg Lehner

8:06 PM

Ok, first easy example: Given a nondegenerate bilinear form $b$ with values in $\mathbb R^n$ on a real, finitedimensional vectorspace $V$, the set of $x \in V$ s.t. $b(x,x) = c$ for $c \neq 0$ is automatically a smooth manifold. Describe its tangent space. Do all this without ever refering to coordinate charts.
Hint: Calculate the differential of $b$. The answer is also called the "product rule".
Derive from that, that $\mathbb S^n$, $\mathbb H^n$ and $O(n)$ are manifolds and describe their tangent spaces with what you showed above.

Georg Lehner

8:31 PM

Bonus-question 1: Why is non-degenerate important? What complications can arise and can the set defined above still be a smooth manifold in special cases?

Bonus-question 2: What happens for $c=0$? Can we still construct a smooth manifold?

Georg Lehner

8:46 PM

Bonus-question 3 for Algebraic Geometers: Generalize this argument analogously for the case that $b$ is a bilinear form on $k^n$ for $k$ a commutative ring, to show that this defines a smooth affine variety in the same way. Are the conditions -non-degenerate-, $c=0$ necessary here?

Khallil

Can I pose a multivariable calculus question here?
(It concerns divergence and volume)

Robert Cardona

@Khallil, there's a calculus chat room here. Have you tried the main chat room?

Khallil

Not yet, @Robert. I'll give it a shot. Thank you. :-)

Robert Cardona

I don't mind, but you'll have better luck in the main chat room or the calculus one.

Balarka Sen

9:01 PM

@GeorgLehner Isn't that just a direct application of implicit function theorem?

Georg Lehner

@BalarkaSen: You're correct!

Balarka Sen

OK. I won't reveal anything else, then :P

Does anyone here know differential geometry of surfaces? I literally don't know anything about it and am looking for a good summary.

I am not quite sure how one uses moving frames on smooth surfaces.

Georg Lehner

@Khallil: If you just mean div, grad and volume integrals in $\mathbb R^n$, the calculus chat would be better. If you're interested in those things over riemannian manifolds, you're right here.

@BalarkaSen: Embedded surfaces or general ones?

Balarka Sen

Embedded surfaces. I am not looking for an introduction to Riemannian geometry :P

Georg Lehner

I had this book for a while ams.org/bookstore-getitem/item=GSM-61

It does surfaces via moving frames in the first few chapters and is probably exactly what you are looking for.

It's not an easy book, however.

Balarka Sen

9:11 PM

I'll have a look, thanks! It looks a bit high-brow at first glance though.

Actually, I have a good place I can read surface theory from. What I am really looking for is a summary :P

Georg Lehner

Spivaks second book in An Introduction to Differential Geometry is excellent as well and a bit more grounded. I have it in Vienna, but if I remember correctly, it contained both a chapter on moving frames and classical surface theory.

Balarka Sen

I'll check it out too. Thanks.

Khallil

I'm not quite at the level of Riemannian manifolds yet, so I'm probably in the wrong place, @Georg. :-)

Robert Cardona

9:29 PM

Question: What's the "rank theorem from multi-variable calculus"? I don't recall having learned it in the last calculus course I took, and I just opened about 4 calculus books I have on hand and didn't find it in any of them.

Question: I'm given the following theorem for the rank theorem on manifolds: Suppose $f : M \to N$ is a smooth map of constant rank $k$, then for each $p \in M$ there are coordinate neighborhoods $(U, \varphi)$ of $p$ and $(V, \psi)$ of $fp$ such that $\psi f \varphi^{-1}$ is the orthogonal projection map: $$(x^1, \ldots, x^m) \mapsto (x^1, \ldots, x^k, \underbrace{0, \ldots, 0}_{n - k}).$$

I'm uncertain why the map is keeping the first $k$ elements, as the rank of the Jacobian is the same, but we don't know which rows are linearly independent?

Georg Lehner

The rank of the map at a point is the rank of its differential. Linear Algebra says, that any matrix with rank k is equivalent to a projection matrix (that's essentially the gauss algorithm).

The rank theorem isn't from calculus but from linear algebra

Robert Cardona

I'm familiar with the "rank-nullity theorem" is this what you refer to?

Georg Lehner

Yes, exactly

The rank theorem in differential geometry says, if you know that the rank of your map is constant, than not only can the differential at a point $p$ be expressed as a projection, but this can be even extended locally for a neighbourhood of $p$.

You could also express this intuitively as: Every smooth function with the same rank looks locally the same.

Robert Cardona

What's a projection matrix? Can you refer me to the result from linear algebra you mentioned? By Gauss' algorithm, you mean row-reduction?

Georg Lehner

9:44 PM

My bad, I'm being fuzzy with my terminology. It is only a projection if the matrix is over the same vector space.

Yes, I mean row reduction.

Robert Cardona

Thanks for the intuition line, I'll keep that in mind.

What do you mean by differential?

Georg Lehner

$D_p f$

Robert Cardona

Which is the Jacobian?

Or represented by the Jacobian?

Georg Lehner

Yes, the coordinate representation of $D_p f$ is the Jacobian.

Robert Cardona

Thanks, I think you answered my first question, as for the second one, why is the map specifically keeping the first $k$ values and giving zero to the rest?

Georg Lehner

9:49 PM

Well, the differential at $p$ has rank $k$, so up to first order the image at $p$ should be a $k$-dimensial submanifold.

After all, the image of $D_p f$ is a $k$-dim subspace of $T_{f(p)} N$.

Robert Cardona

Isn't this assuming that the linearly independent rows are the first $k$ rows?

Georg Lehner

Remember that in Gauss algorithm you can swap rows to achieve that?

Since you can choose your basis however you want, you can just switch the coordinate chart such that you achieve this.

Robert Cardona

I understand the algorithm can move the linearly independent vectors up, but I don't see how that process is embedded into the definition.

Georg Lehner

Which definition do you mean?

Robert Cardona

I don't understand your second remark yet. I'm thinking about it.

The statement of the theorem doesn't mention anything about using Gauss' algorithm to get the linearly independent vectors in the Jacobian to be the first $k$ rows so that the definition of the orthogonal projection map makes sense.

I think your second remark is the reasoning I'm looking for, I just don't understand it yet.

Georg Lehner

9:55 PM

The statement was that there exist such charts. In almost any other chart the jacobian will look completely different

Robert Cardona

@GeorgLehner I think this is the key, but I'm not sure of the details about how to do this. I'm assuming that in a proof of this statement, if it had been included, would have explained this?

Georg Lehner

Remember that the jacobian transforms as $D(\psi_1 \circ \psi_2^{-1}) Df D(\phi_1 \circ \phi_2^{-1})^{-1}$ for different charts.

That's an equivalence of matrices.

In a proof of the rank theorem, you wouldn't explicitly use the gauss algorithm, but rather just the statement from the Rank-Nullity theorem and work from there.

Robert Cardona

I didn't understand the first two lines, or at least, I wasn't familiar with the result about how the Jacobian transforms or how an equivalence of matrices implies the result above.

Georg Lehner

It's essentially the functoriality of $D(-)$. But it's usually easier to understand by making a few diagrams which is hard to do here.

Robert Cardona

In any case, I was looking at a proof of it in Boothby, and it's three pages long. He does address what I was thinking about, implicitly, by mentioning that "a $k \times k$ minor of nonzero determinant in $DF(a)$..." after "suitable linear maps that permute the coordinates". I'm assuming the nonzero determinant ensures that the first $k$ rows are linearly independent.

Yeah :P

Sorry about hassling you with questions, btw :P

Georg Lehner

10:06 PM

And the gaussian algorithm transforms a matrix $A$ into one of nice form by multiplying $B A C^{-1}$ for invertible matrices $B$ and $C$.

that's why I'm mentioning it.

Is the proof of Boothby maybe just doing the statement, that a smooth map whose differential is an isomorphism is locally a diffeomorphism?

Robert Cardona

So, in this case, $A = Df$, $B = D(\psi_1 \psi_2^{-1})$ and $C = D(\phi_1\phi_2^{-1})$?

Georg Lehner

That's called the Inverse Function Theorem, and it's a bit weaker than the rank theorem.

Yes.

Robert Cardona

The theorem Boothby does is called the Rank Theorem. It looks similar to what's in Sullivan's notes, but actually proves it and says it implies the Inverse Function Theorem.

Georg Lehner

non-zero determinant of a $k \times k$ minor assures that -some- $k$ vectors are linearly independent. You can then permute basis vectors to achieve that those are the first ones.

Ok.

Because talking about subminors and that stuff seems like an overkill... I'd just quote the Rank-Nullity theorem and work from there.

Robert Cardona

In any case, I think my concern was warranted, and a proof of the statement would have to have addressed that in some form.

Georg Lehner

10:13 PM

Of course, you have to permute your basis somehow. But that has already been done in the Rank-Nullity theorem, which gives you the right transformations anyway.

Robert Cardona

I think there is a typo in Boothby :/ What's the Jacobian of $F(x^1, x^2) = \big( (x^1)^2, 2x^1x^2)$?

My calculation has a zero entry, is this correct?

Georg Lehner

Yes, since the first entry doesn't depend on $x^2$.

Robert Cardona

Boothby has $$DF(x^1, x^2) = \begin{pmatrix} 2x^1 & 2x^2 \\ 2x^2& 2x^1 \end{pmatrix}.$$

Georg Lehner

That's wrong.

in exactly one entry

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$Mathematics$ Differential Geometry