Mathematics - 2021-02-11 (page 1 of 2)

Thorgott

00:00

but ok, with the previous computations, it's clear that this is gonna be transversal to the zero section iff that bottom half is non-degenerate, so that works out

Ted Shifrin

I've talked about this before here and on main (you can search for that phrase). The derivative of a section of a vector bundle is well-defined at a zero of the section without any connection needed on the vector bundle.

Thorgott

00:16

that's the "in general, we need a Riemannian metric to define the Hessian, but at a critical point, we can always do it" point, yes?

the pushforward of the tangent space through the zero section provides a canonical splitting of the tangent spaces of a vector bundle along the zero section

so you have an intrinsic canonical splitting into horizontal and vertical component there

and in this particular case and in coordinates, the vertical projection precisely gives me the part of that matrix that consists of the mixed partials

explicitly in this case, we have $T_{(p,0)}T^{\ast}M\cong T_pM\oplus T_p^{\ast}M$ (image of pushforward of zero section and kernel of pushforward of projection) and the composition $T_pM\rightarrow T_{(p,0)}T^{\ast}M\rightarrow T_p^{\ast}M$, the first map being the differential of $df$ and the second being vertical projection describes a bilinear map $T_pM\times T_pM\rightarrow\mathbb{R}$, which is the Hessian at the critical point

great, I never got this point before, but I think now I do

thanks @Ted

Ted Shifrin

I didn't do much.

Thorgott

00:33

you told me all I needed to figure out the rest

Thorgott

00:47

So apparently an open smooth $n$-manifold $M$ can be embedded in $\mathbb{R}^{2n-1}$. Roughly, I know that $M$ deformation retracts onto an $n-1$-dimensional submanifold $N$, strong Whitney tells me that $N$ embeds in $\mathbb{R}^{2n-2}$ and I would guess/hope that if we then take a tubular neighborhood of $N$ in $\mathbb{R}^{2n-1}$, this will be diffeomorphic to $M$. Is anything remotely close to that true?

Ted Shifrin

Sounds good to me.

user2103480

01:24

@MikeMiller @Thorgott do you have some easy introductory exercises for getting a feel for local homology

Thorgott

01:37

I don't know if there is all that much to say about local homology. Understand in which sense it is local. Compute it for a smooth manifold relative any point and for any cone relative the cone point.

user2103480

It came up mainly for justifying degree computations by local degrees and that's been meh for me. My basic intuition atm is via excision, i.e. $H_n(M, M \setminus \{x\}) \simeq H_k(D^n, D^n \setminus \{ x \}$

In a manifold. And it now seems simpler than before lol, maybe the knowledge starts to solidify a little bit

About time though, exam is on the first of march and I genuinely thought it was exactly three weeks away. Then I realized there is no 30th february

Your brains: ugly and wrinkly

My brain: smooth and beautiful

Clarinetist

What is the analogue of Lebesgue measure in the complex plane?

Thorgott

I don't think you will ever have to realistically calculate local degrees for a non-differentiable map

Clarinetist

Is it just the same one as the one in $\mathbb{R}^2$?

Thorgott

for a differentiable map, the local degree at a point is the sign of its determinant

@Clarinetist yes, the analogue of the Lebesgue measure is in fact the Lebesgue measure itself

user2103480

01:50

And incidentally, the fourier transform is also the same, to revisit what we went over yesterday

At least the part in the exponential is just x_1*Im(z) + x_2*Re(z)

Clarinetist

So here's a theorem I was presented:

Let $f$ be a Borel measurable function on $[a, b]$. Then there exist continuous functions $f_n$ on $[a, b]$ with $\lim f_n(x) = f(x)$ for all $x \in [a, b] \setminus E$, where $E$ has Lebesgue measure zero.

Especially with the interval, I'm struggling to understand what the complex-function analogue of this theorem would be.

user2103480

Uhh fourier transform only if density. I meant characteristic function, but should be the same by viewing Im(X) and Re(X) as random variables and then substituting to get a double integral, if these things have a joint density (which is the same as a density for X itself)

Thorgott

you don't actually need to exclude a measure 0 set

anyway, the natural analogue of an interval is a rectangle

but I'm kinda sure the domain can be any measurable subset and the result will still be true

probably true for the Borel sigma-algebra on any paracompact topological space, but I'm not thinking much

Mike

Hi, I have a problem about smoothness of moment generating functions on arbitrary open sets. Someone suggested an answer, but it does not have generality and I am still stuck at it. Can anyone take a look? Thanks. math.stackexchange.com/q/4018712/792125

1

Q: Prove that MGF defined on an open set is infinitely differentiable

I am doing an exercise in the book "Applied Stochastic Analysis" by E-Li-Vanden-Eijnden, and I meet this problem (on Page 26): (Exercise 1.19) Prove that if the moment generating function $M_X(t)$ can be defined on an open set $U$, then $M_X(t)\in C^{\infty}(U)$. Here is my approach: Use Taylor...

Thorgott

02:09

it follows from en.wikipedia.org/wiki/…

user2103480

@Mike in the taylor expansion, I think this should be o(t)/t

Mike

@user2103480 Personally, I find the Taylor expansion argument not a rigoruous one. Do you have other approaches to prove that? Note the open set is arbitrary.

BigSocks

Say we have a ring $A$ and its field of fractions $K$. Say we have a guy, $a/b \in K$ such that $a/b \notin A$. $A$ is integrally closed.

Does it follow that $a/b$ is *not* integral over $A$? Kind of fumbling definitions here I think

Because if it were integral, then $A$ being integrally closed would mean that $a/b$ had to be in $A$?

Thorgott

that's what integrally closed means, isn't it?

BigSocks

yeah I think so- I just haven't internalized the definition.

Really dumb because it's not that complicated. thanks though

user2103480

02:35

@Mike Hm I think I just used the wrong taylor formula. But now I get a not so nice expression e^tX = 1 + tX + o(tX)(tX) which doesnt align with your expression either

Thorgott

why not just differentiate under the integral?

user2103480

It's maybe not the best approach since you have to control the tX terms there

Thorgott

what's there to control

though tbf doing Taylor is just like manually re-proving differentiation under the integral, so it's not like this is a conceptually all too different approach

user2103480

Oh I didnt mean yours

I meant the taylor expansion

Differentiating unter the integral should be fine, but you gotta show that the derivatives have a common integrable majorant

One might also try the ol' switcheroo to get a power series locally, but gotta be careful about the necessary conditions being fulfilled

Thorgott

yeah, but it's monotonous in $t$, so you just bound by the max. value of $t$

(we can always restrict to a bounded domain since smoothness is local of course)

Ted Shifrin

02:49

@user2103480 Nah, you have uniform convergence on compacts. Sometimes you don’t need Lebesgue s****.

Hmm, I guess I didn't look. I assumed this was Taylor/Laurent.

user2103480

@TedShifrin this is a random variable, no? What exactly converges here

@Thorgott still gotta argue that X exp(t_0X) is integrable

And so on

I think using fubini locally to make this into a power series with coefficients being the moments and radius of convergence given by openness should be a good approach

Thorgott

ok, you got me there

but, like, small values of X don't make an issue, but for large values of X, you can bound X above by exp(t_0*epsilon*X) for epsilon as small as you like (by making X larger) and then the product is bounded above by exp(t_0(1+epsilon)X), which is integrable by hypothesis if we chose epsilon small enough

user2103480

Hahaha there surely is a way to use that approach

Thorgott

of course I first choose epsilon appropriately and then a cut-off for X to get a uniform bound, but this should work

user2103480

Might be tedious, might even have an elegant, quick solution

But I will better go to bed instead of thinking about completing the argumenr

Thorgott

02:59

I think this is complete enough

details are left to the reader as an exercise

user2103480

Amen

Thorgott

ez

user2103480

Good night!

Thorgott

gn

Mike

Can we move the discussion to my post? I think your epsilon-approach is interesting, can you go to my post and write an answer if you are willing to?@Thorgott

Clarinetist