Mathematics - 2019-07-27 (page 2 of 2)

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Ryan Unger

22:06

> To control error terms, we rely on a variety of norms.

this paper is a bit of a meme

Perturbative

@MikeMiller Thanks for giving this detailed correction!

Alessandro Codenotti

What is on your blog? @Mathei

Oh sorry, I didn't notice you were answering to a message

MatheinBoulomenos

22:29

@MikeMiller do you mind a silly topology question? I can visualize the isomorphism $\Bbb C/\Bbb Z \to \Bbb C^\times$ pretty well, that's easy. But if you quotient out another discrete subgroup you get that $\Bbb C^\times/q^\Bbb Z$ is a torus

Mike Miller

@BalarkaSen Let $i: M \to \Bbb R^{2n}$ be an immersion of a closed (oriented, if you like) $n$-manifold. Suppose $e(\nu(i)) = 0$. Can you homotope $i$ into an embedding? That is, does the Euler class provide the only obstruction for making an immersion into $\Bbb R^{2n}$ an embedding?

MatheinBoulomenos

I can't visualize why that is

Mike Miller

@MatheinBoulomenos Think of $\Bbb C^\times$ as a cylinder instead, replacing the radial $\Bbb R^+$ with $\Bbb R$

Then you're asking more or less why $S^1 \times \Bbb R/\Bbb Z$ is a torus, which should hopefully be visually clearer - now you are folding up the radial line into a circle

MatheinBoulomenos

yeah but if I want to think of it as a pointed palne? I guess I can't visualize a radial discrete subgroup

Mike Miller

Sure you can, $2^n$. The "point" 0 doesn't make sense anymore when you take the quotient

MatheinBoulomenos

22:34

oh I see it now

thanks

Mike Miller

With the subgroup $e^n$, which is the image of $\Bbb Z$ under the exponential homomorphism, you can take the fundamental domain $1 \leq \|z\| \leq e$ and fold together the outside

Sure

MatheinBoulomenos

I was confused because the geometry gets kinda weird

Mike Miller

Yup

MatheinBoulomenos

not all fundamental domains have the same volume anymore

visualizing things is definitely not my strength

Mike Miller

That's alright

MatheinBoulomenos

22:39

@MikeMiller I'm not sure if you like this because it's uniformization or if you shrug because it's p-adics but there is p-adic uniformization

p-adic GAGA, too

Ted Shifrin

Howdy @Mathein @MikeM

MatheinBoulomenos

Hey @Ted

Mike Miller

That's pretty cool, but yeah I already can only "see" Riemann surfaces so well. I don't think I can see them very well because they can have "special points" that are distinguished from others in interesting ways; the automorphism group rarely acts transitively

Once you go p-adic I'm lost

MatheinBoulomenos

you have p-adic GAGA, p-adic analogue of Cartan's theorems A and B, p-adic uniformization and basically analogs for most results about Lie groups

it's crazy how much carries over

the reason why I asked about this weird presentation of torus is because that's the version that carries over. Elliptic curves over $\Bbb C_p$ can be presented as $\Bbb C_p^\times/q^{\Bbb Z}$

Mike Miller

I can't draw that

Ask the pointillists

MatheinBoulomenos

22:45

@TedShifrin so can you tell me about Cartan A and B?

like over complex numbers :P

this is for Stein manifolds, right?

So Stein manifolds are like affine varities. Is this related to Serre's criterion for affinity

I mean Stein spaces sorry

we don't need smoothness

@MikeMiller I learned that PDE can be pretty cool. You have spectrum of Laplace-Beltrami and that

this seems very cool

and also really related to modular forms and generalizations

if you study the spectrum of something like hyperbolic 2-manifolds that should tell you something about Maass forms

because these are eigenfunctions for the hyperbolic Laplacian

Mike Miller

I would guess the eigenspaces are more important than the actual values but I'm no number theorists

MatheinBoulomenos

yeah that's true

Mike Miller

I guess there are reciprocity formulas and whatnot. But I don't know anything about hyperbolic geometry

MatheinBoulomenos

but I was thinking like this trace formula stuff

Mike Miller

I only do the easy stuff

MatheinBoulomenos

22:54

there are trace formulas for modular forms

Mike Miller

@MatheinBoulomenos That's what I meant instead of reciprocity

MatheinBoulomenos

and there trace formulas for Laplace-Beltrami

Mike Miller

They get much more complicated as you proceed past dim 2 but again I know nothing

MatheinBoulomenos

and I have no idea why there's a relation

Mike Miller

I mean presumably your forms are actually certain eigenspaces or something. But I can't emphasize enough I do not know this

smoking_huge_doinks

22:55

hello fellow humans

MatheinBoulomenos

@MikeMiller they live in eigenspaces at least

I actually don't know much about Maass forms which are eigenfunctions, more about the classical homorphic ones which are eigenfunctions but like for eigenvalue 0

but still is it bugging me

all these trace formulas

and I don't see how they're related

Ryan Unger

>eigenfunction with eigenvalue 0

do you mean harmonic?

MatheinBoulomenos

yes

Mike Miller

Get a book on hyperbolic geometry and number theory then

There should be at least one GTM

MatheinBoulomenos

yeah I should

maybe even some Erlangen program stuff

I know that's old, but it seems cool

@MikeMiller do you have some intuition why you can detect if a smooth manifold is simply connected by looking at flat bundles up to dimension 3, but not higher?

why should flat bundles in higher dimensions carry less information

Mike Miller

23:16

I mean because you're phrasing it in a way liable to hide the real meaning

The fundamental group is unconstrained past dimension 3

And that's just a calculation, constructing a closed oriented n-manifold with arbitrary f.p. fundamental group, when n>=4

MatheinBoulomenos

yeah that's true

I like flat bundles because I can make it seem like I'm talking about geometry but I'm actually talking about groups

Mike Miller

You can't fool me

MatheinBoulomenos