Mathematics - 2020-07-08 (page 3 of 3)

10:02 PM

Oh, I understand your point now. One equivalence relation is defined on $U$ and the other on $M$

in essence, because of my observation, $[(f,V)]_U=[(f,V\cap U)]_U$ there really is no point in distinguishing $[(f,V)]_U$ and $[(f,V\cap U)]_M$

Alessandro Codenotti

Those are some weird complaint coming from someone who was complaining about having to read set theory definitions just a short while ago! :P

Balarka Sen

@EdwardEvans Within Thy Wounds, "Forests of Iniquity" seems like good DSBM

Thorgott

hey, this is just a nice exercise in elementary pedantry, no fancy-shmancy nonprincipal ultrafilters and what-not

@orientablesurface yeah, so now you can safely forget all what I've said again

Alessandro Codenotti

Speaking of which here is your random topology fact of the day! If $U$ is a nonprincipal ultrafilter on $\Bbb N$, then $U$, as a subset of the product space $2^{\Bbb N}$ cannot be Borel. In fact it doesn't even have the Baire property

Thorgott

wasn't this already yesterdays fact or was it just something similar?

nvm, it was about something completely different being not Borel and not even having the Baire property

Alessandro Codenotti

10:13 PM

I think that was about wellorderings of $\Bbb R$ as subspaces of $\Bbb R^2$

Thorgott

right

Balarka Sen

The set of continuous functions $C[0, 1]$ is obviously not measurable in the product sigma-algebra $([0, 1]^{\Bbb R}, \mathcal{B}_{[0, 1]}^{\otimes \Bbb R})$. Is it measurable in the Borel sigma-algebra of the product topology of $[0, 1]^{\Bbb R}$?

mick

hi all

0

Q: Riemann hypothesis equivalent to $\prod (\frac{4 w_i^2 + 9}{4 w_i^2 + 1}) = \prod (1 - 2/v_i) = \frac{\pi}{6}$ ??

Someone told me that the Riemann Hypothesis is equivalent to $$\prod (1 - 2/v_i) = \frac{\pi}{6}$$ where the product is over all the nontrivial zero's $v_i$. ( for the product we take conjugate pairs of nontrivial zero's ordered by size ) Is that true ? If true, how to prove it ? Can this idea be...

any ideas ?

Alessandro Codenotti

@Thorgott They both follow from the Kuratowski-Ulam theorem or its consequences

orientablesurface

Thanks for your replies @Thorgott @TedShifrin, really appreciate seeing different ways of thinking about these stuff

Balarka Sen

10:20 PM

Just think of a manifold as if you're living in it

$T_p M$ is really a small chart around $p$ for all practical purposes

Thorgott

I was like "Fubini's theorem for categories, that sounds awesome", but then I realized it's category as in "of the first/second category" and not the cool categories :/

Alessandro Codenotti

@BalarkaSen I think so because being continuous is the same as having oscillation zero at every rational

Balarka Sen

@Alessandro Yeah I buy this.

Thorgott

if your manifold lives embedded in R^n and you identify the tangent space with the appropriate affine subspace of R^n, you can actually get a chart of your manifold by orthogonally projecting to the tangent space

I don't think this is useful, but it's nice to know

Balarka Sen

@AlessandroCodenotti But "being oscillation zero at every rational" is not an event that literally depends on countably many points on $[0, 1]$, right?

Alessandro Codenotti

10:24 PM

What do you mean?

Balarka Sen

So that's why there's no paradox; events in $\mathcal{B}^{\otimes \Bbb R}$ are events which depend on countably many points in $[0, 1]$

Thorgott

yeah, oscillation is a local thing, not pointwise

Balarka Sen

Right

Hah, I am writing $\Bbb R^{[0, 1]}$ as $[0, 1]^{\Bbb R}$.

@Thorgott There's actually a canonical way to do this once you pick a Riemannian metric, it's called the exponential map

Thorgott

@Balarka I was told that a smooth structure on R^4 is diffeomorphic to the standard one iff it turns addition into a smooth map, but I can't find this with a primitive google search. Does this have a name/do you know a reference?

huh, I wasn't aware of that

Balarka Sen

I didn't know that, surprising. I guess if both addition and multiplication by $-1$ was smooth under the smooth structure on $\Bbb R^4$ it would become a Lie group and therefore not exotic

Lie groups have very rigid smooth structures; they are automatically canonically analytic for example

If anyone knows this it's @MikeM

Thorgott

10:32 PM

well, then this is my fun fact of the day, I guess

I need to learn the basics of Lie groups some day

Lie groups are automatically parallelizable, right?

can that happen for exotic R^4

Balarka Sen

yeah any topological manifold which is a topological group admits a unique analytic structure that makes it a Lie group

@Thorgott Yes, actually the tangent bundle does not recognize smooth structure

Anyway any topological bundle over R^n is trivial

because R^n is contractible

Thorgott

is the tangent bundle homeomorphism-invariant?

Ted Shifrin

what does that mean?

as a manifold or as a bundle?

We have discussed something similar before. Balarka, do you remember the example of two bundles that are diffeomorphic as manifolds but not isomorphic as bundles?

Balarka Sen

Yeah, actually in that same paper Milnor constructs two different smooth structures on a manifold M with non-isomorphic tangent bundles

IIRC

Thorgott

so in which sense does the tangent bundle not recognize smooth structure

Balarka Sen

10:38 PM

I meant the tangent bundle doesn't recognize smooth structure in the sense it can go either way. Exotic spheres always have isomorphic tangent bundles, for example, I believe

@Thorgott We should actually read some microbundle theory

Ted Shifrin

@Balarka: You remember the discussion/example to which I referred?

Balarka Sen

Yeah, I have an MSE question on that

Mike wrote an answer

Ted Shifrin

I don't berember stuff no more.

Balarka Sen

7

Q: Nonisomorphic vector bundles with diffeomorphic total spaces

What's an example of two non-isomorphic vector bundles $E,F$ over the same base such that the total spaces $Total(E), Total(F)$ are homeomorphic? Assume that rank of these bundles is the same as dimension of the base manifold. (EDIT: Mike Miller brought up the excellent point that I need to loo...

algebraic-topology differential-topology vector-bundles

Thorgott

some what

NoseBleed

10:42 PM

$|x_0 − x^∗| <δ ⇒|x(t)−x^∗|<\forall t ≥0$

can anyone explain me stability

?

Ted Shifrin

Oh, other people wrote answers. I didn't remember the answers at all.

Balarka Sen

Both people are Mike, Ted!

he baleeted account

Ted Shifrin

Oh really??!!!!

<--- stooopid

Balarka Sen

Who else would answer these fucked questions

Ted Shifrin

@NoseBleed: Stability means that if you wiggle the initial condition a tiny bit, then the solution stays close to the original one (for a while).

LOL @Balarka ... Lots of people know topology algebraic.

NoseBleed

10:45 PM

so $x^*$ is initial condition not when $f(x^*)=0$?

Balarka Sen

@Thorgott Somehow the discussion is centering on what you observed earlier. Here's the deal; let $M$ be a topological manifold. $T_p M$ makes no sense, but why can't you just declare $T_p M$ to be a small chart around $p$? Choose an atlas $(U_i, \varphi_i)$ on $M$ and try to construct an uhhhhhhh "bundle" over $M$ with fibers being these charts $U_i$ (not a vector bundle because no canonical vector space structure)

How do you do this? Well, over $U_i$ it should be the trivial vector bundle $U_i \times U_i$

So patch them togather, do $\coprod U_i \times U_i/\sim$ where $(x, v) \sim (x, \varphi_{ij}(v))$, where $\varphi_{ij}$ are the literal transition functions

Ted Shifrin

You wrote the notation, not I. I just answered your question about what stability means.

Balarka Sen

And thusly you have a bundle $E \to M$, projection being projection of $U_i \times U_i$ to first coordinate, with $\Bbb R^n$ fibers. NOT a vector bundle, and depends on the atlas

This is called the tangent microbundle; an analogue of the tangent bundle for topological manifolds

Ted Shifrin

You're asking what a stable singular point is?

Note that if you take $f(x*)=0$, then the solution with initial condition $x^*$ is the constant solution. So it agrees with what I said.

Thorgott

I guess

this is similar to the transition maps determining the bundle stuff we talked about a while ago

Balarka Sen

10:50 PM

Yeah

Thorgott

Cech cocycles or w/e

Balarka Sen

Here the transition functions are just manifold transition functions, given by the atlas

With some more thought you can convince yourself $E$ is a small neighborhood of the diagonal copy of $M$ in $M \times M$. In the smooth world, the normal bundle is indeed isomorphic to $TM$

So that's an alternative, tricky way to phrase what the tangent microbundle is. Given a top manifold $M$, take a nbhd of $M$ in $M \times M$ and declare that to be tangent bundle

microbundle equivalence is exactly like bundle equivalence but germinal around the zero section; you don't care about global things

NoseBleed

I already knew the meaning but I can't understand the inequality.

I should try drawing. Will come back if I have question.

Thorgott

ok

so it's time to ask the crucial question

why

Balarka Sen

because the problem of whether a topological manifold admits a smooth structure can be understood as an algebraic problem of when the tangent microbundle comes from an actual tangent bundle

microbundles were developed by Milnor, and was used by Kirby and Siebenmann to come up with their invariant

and that relates back to the kind of questions you were asking; how are tangent bundles of different smooth structures related?

you have an underlying topological tangent bundle so one would expect that to say something about that

This is as much as my extent of knowledge goes

Thorgott

11:06 PM

interesting

Balarka Sen

@BalarkaSen This is sort of nonsensical as written because $\varphi_{ij}$ is partially defined on $U_i$. Anyway I know how to fix this but I won't write it unless anyone's ACTUALLY curious

Thorgott

why's that an issue? you only want to identify them where they're defined either way, no?

Balarka Sen

Er I mean so $TU_i = U_i \times U_i$, right? The first copy is the base of the tangent bundle and the second copy is the fiber copy, should be identified with $\Bbb R^n$ by $\varphi_i : U_i \to \Bbb R^n$ chart homeomorphism. The idea is to build $TM$ from $TU_i$'s by patching these guys along the base copies and identifying fibers completely

Wow $\varpi$ alert

NoseBleed

@TedShifrin I understand the inequality thanks. If you start at the neighbourhood of the so called infliction point then your solution will tend to the infliction point or else it is unstable because it will diverge.

I am right right?

Balarka Sen

So if $x \in U_i \cap U_j$, I want to glue the fiber copy of $U_i \cong \Bbb R^n$ hanging out above $x$ in $TU_i$ with the fiber copy of $U_j \cong \Bbb R^n$ hanging out above $x$ in $TU_j$

NoseBleed

11:12 PM

The starting point is in neighbourhood is the initial condition.

Balarka Sen

But there's only this partially defined garbage $\varphi_{ij} : U_i \cong \Bbb R^n \dashrightarrow U_j \cong \Bbb R^n$

So, oops. Doesn't quite make sense to say $(x, v) \sim (x, \varphi_{ij}(v))$ if $v$ is not in the domain of definition of $\varphi_{ij}$, eh?

Croissant

11:24 PM

Hello! @Sebastiano asked a question regarding, kind of, non-standard integrals (meaning highschool students rarely encounter those examples). The OP, as far as I'm concerned, would like to know if there are some examples he should memorize and how to recognize integrals which require some "advanced" methods. I thought of asking here because the question might get closed due to lack of clarity and context. In the meantime, the OP included some more information.

4

Q: When to apply complex integration for integral resolutions

"Is there a criterion, a clue that makes me think that certain integrals can also be solved through complex integration and how to solve them?" When I can't solve an integral for my students of an high school, I use the numerical methods. If I have these integrals how are they solved used the com...

real-analysis integration complex-analysis proof-writing proof-explanation

Could this question remain open as a community wiki maybe?

Thorgott

ah, I see what you're saying

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$Mathematics$ Mathematics