Mathematics - 2020-07-10 (page 1 of 2)

So, the projection map is $p : (X\times Y)/G \to X/G$, $[x, y] \mapsto [x]$. Let $q : X \to X/G$ be the quotient map. Any point in the fiber $q^{-1}([x])$ has a neighborhood $U$ which is disjoint from other $gU$'s (other meaning g \ne 1). I claim that $q(U)$ is the trivializing neighborhood for $[x]$.

The map $\phi : p^{-1}(q(U)) \to q(U)\times Y$, $[x, y] \mapsto ([x], y)$ should be a homeo.

It sure looks bijective. I mean, if $([x], y) = ([x'], y)$ where $[x], [x'] \in q(U)$ then $x = x'$, because $[x] = [x']$ means that $x = gx'$ for some $g$, but this means $U \cap gU \ne \varnothing$ so $g = 1$.

Actually, I even need to check well-definedness.

oh of course it is. By the same argument: if $[x,y] = [x', y']$, then $g(x, y) = (x', y')$ for some $g$. But $gx$ 's are all distinct so $g = 1$.

(I should have checked this while defining the projection map).

feynhat

1:50 AM

$\phi$ is continuous and open because, on $p^{-1}(q(U))$, its nothing but $(x, y) \mapsto (x, y)$ as, inside the set $p^{-1}(q(U))$, the class $[x, y]$ is just a singleton.

Looks good?

feynhat

3:15 AM

Yeah he ded.

Balarka Sen

who's dead

feynhat

:O

Balarka Sen

im not verifying point set topology in the crack of dawn man

this is too hard for me

$Y \to (X \times Y)/G \to X/G$ is obviously a bundle

you just crush the $Y$ copy

feynhat

oh okay.

Balarka Sen

i bet what you said is right

i trust your topology

feynhat

3:20 AM

Another question, if all the fibers defret to the base point, then does the bundle defrets to the base space?

@BalarkaSen Big mistake.

I guess it depends on what the structure group is, right?

Balarka Sen

@feynhat As in? What's the basepoint?

There must be a canonical choice of basepoint in the first place

feynhat

(oh wait)

Balarka Sen

I mean there's no canonical section of a bundle $E \to B$

in general

you want these "fiberwise basepoints" to form a section

feynhat

Okay. Maybe take Y to be a pointed space, and your structure group is basepoint preserving homeos.

Balarka Sen

$Y \to (X \times Y)/G \to X/G$ has a canonical section if the $G$-action on $Y$ has a fixed point $y$

$(X \times y)/G$ is the image of the section

@feynhat Yeah then it's alright

Balarka Sen

6:33 AM

Four player chess is insane

every move is a blunder so you don't have to worry about anything

Edward Evans

10:10 AM

If S is an R-Algebra and M a projective R-module, how y'all showing that M o_R S is a projective S-module?

Thorgott

projective iff direct summand of a free module, so if M is a summand of R^n, then Mo_RS is a summand of S^n cause tensor commutes with direct sums

this holds for any extensions of scalars (so has nothing to do with S being an R-algebra)

Edward Evans

Right, i have a different proof and was suspicious because I didn't really use the fact that $S$ is an $R$-algebra

probably a bit overkill though

$M$ is projective so $\operatorname{Hom}_R(M, -)$ is exact, $S$ is a projective $S$-module so $\operatorname{Hom}_S(S, -)$ is exact, compose the hom-functors to get another exact functor, then tensor-hom adjunction to get $M\otimes_R S$ projective $S$-module

rofl

Thorgott

ah yeah, that's nice

@Thorgott should be R^(X), projective modules don't need to be f.g.

Edward Evans

sure

maybe $R$-module was meant instead of $R$-algebra idk

Leaky Nun

10:29 AM

@EdwardEvans well the operator o_R S preserves properties of modules so this is obviously true :P

Balarka Sen

unital ring homomorphism R -> S is literally what an R-algebra is, no?

Leaky Nun

proof by philosophy

@BalarkaSen yes

Balarka Sen

so whats the fuss about

Thorgott

you don't need a ring structure on S for this to be true

ah wait, you do

Balarka Sen

what does M o_R S being an S-module mean

what are you smoking dude

Thorgott

10:31 AM

yeah, you're right

don't trust anything I say

Edward Evans

it's a pset so I can'T really write "o_R S preserves properties of modules so this is obv true"

Thorgott

first I can't do computations, now I can't even algebra anymore

my existence has lost all pretension of worth

Edward Evans

ah well

hahahaha

Thorgott

I mean, o_RS isn't even exact

Balarka Sen

sneak edit

just to switch to the grothendieck point of view

Edward Evans

10:35 AM

FlAt

Thorgott

my brain is working in weird ways rn

and by weird ways I mean not at all

Edward Evans

nah $S$ isn't assumed to be flat

Balarka Sen

finitely presented flat implies projective i believe

and projective is locally free

completely unrelated

Edward Evans

a flat module is isomorphic to $\Bbb R^2$ by geometry

Thorgott

projective iff locally free only holds for finitely presented modules

Balarka Sen

10:37 AM

finitely generated is enough

Thorgott

in one direction yes, cause a projective module is f.p. iff f.g., but I don't believe it works in the other

Balarka Sen

you just want Nakayama to go through

and for that you only need fg

Thorgott

I mean, f.g. projective modules over local rings are free

but projectivity is local only for f.p. modules, not for f.g. modules, I think

Balarka Sen

ah you mean the other direction yeah

that is true

Thorgott

this stuff's way too confusing

Balarka Sen

10:39 AM

Q is locally free over Z

i bet you modify this to be fg somehow

Edward Evans

Q = <1/1>_Z

Thorgott

why is Q locally free over Z

Edward Evans

I can't actually do algebra, I just make shit jokes

Thorgott

you already inverted everything

Balarka Sen

Yeah that's not true

You're right

wikipedia has an example en.wikipedia.org/wiki/…

boolean rings jesus christ

Ah I see the point. If it's finitely presented you have a presentation $0 \to K \to F \to M \to 0$ where $F$ is free, $K$ is fg

You need Nakayama on $K$ and globalize

That is why fp and fg isn't enough

Thorgott

10:49 AM

I think the point is that for f.p. modules, localization commutes with Hom

that's why projectivity is local for f.p. modules

but this fails for f.g. modules

ah, I see what you're saying too

that's the point of why f.g. projective modules over local rings are free

Balarka Sen

Hold on I'm confused. Localizing the above exact sequence I get $0 \to K_m \to F_m \to M_m \to 0$ on every maximal ideal, and $M_m$ is free so the sequence splits and $F_m \cong (K_m \oplus M_m) \cong (K \oplus M)_m$

local global principle works only for fg modules, right?

locally isomorphic implies globally isomorphic

Thorgott

nah, works always

Balarka Sen

So I am confused

Thorgott

if you're localizing the same iso, of course

Balarka Sen

aha

Thorgott

10:54 AM

cause exactness of localization + being 0 is local

Balarka Sen

i dont get a canonical map $F \to K \oplus M$, correct.

yeah this proof approach doesn't work and you have to do some explicit shit with Hom(-, P)

ok whatever

Thorgott

no, I think it does

your SES splits

Balarka Sen

why?

Thorgott

because M is projective

Balarka Sen

No M is only locally free

I was going the other direction

fg projective implies locally free is obvious Nakayama

Thorgott

10:57 AM

ah

I wouldn't say obvious, but yeah, that direction does follow from Nakayama

if your module is f.g., then locally free is the same as locally projective for that reason

Balarka Sen

sure

Thorgott

so the crux is why projectivity isn't local for f.g., but only for f.p.

and that's because localization commutes with Hom for f.p., but not in general for f.g.

Balarka Sen

sure that makes sense

Thorgott

and indeed, Hom(P,-) is exact iff Hom(P,-)_p is exact for all prime ideals p, but this is Hom(P_p,-_p) for f.p. modules and that's exact because P_p projective

Balarka Sen

i buy it tho i have not checked it

Thorgott

11:04 AM

that's the entire argument, there's nothing to check

Balarka Sen

yeah i mean im not paying attention

Thorgott

except for why localization commutes with Hom for f.p., but not f.g.

that's a tedious thing tho

Balarka Sen

i believe it

Thorgott

I actually don't know a f.g., but not f.p. counterexample, but I was told there is one

Balarka Sen

its in wikipedia

Thorgott

11:05 AM

the thing is that f.g., but not f.p. requires non-Noetherian rings and that's ugly

does that also work as counterexample for localization commuting with Hom?

oh, it has to

otherwise it wouldn't be a counterexample for projectivity vs locally free either

Balarka Sen

mhm

Thorgott

precisely because that's all the above argument needs in terms of being f.p.

well, it was good to think this through again, I now understand it better than before

Balarka Sen

fp flat implies projective is not trivial iirc

but i forget details

if M is flat over local ring (A, m) then Tor(A/m, M) = Tor(M, A/m) = 0 (commutativity of Tor is nontrivial), i.e., for any SES 0 -> P -> Q -> M -> 0 if you tensor with A/m it's still exact. Then just choose a presentation 0 -> K -> A^n -> M -> 0 where n is a minimal generating set for M and tensor with A/m and observe the last map (A/m)^n -> (M/m)^n becomes an isomorphism so K/mK = 0 so K = 0 by Nakayama as K is fg

so fp flat is locally free and fp locally free is projective by your argument

Thorgott

11:21 AM

I see

I don't know the Tor business yet

Balarka Sen

you should be able to write down a concrete proof that if M, N are modules over A such that for any exact sequence 0 -> P -> Q -> M -> 0, tensoring with N leaves it exact => for any exact sequence 0 -> P' -> Q' -> N -> 0, tensoring with M leaves it exact

thats what im using

Thorgott

weird

I don't buy that

the latter would in particular imply that M is flat over A

Balarka Sen

no why would it mean that

N is a fixed module

the latter isn't saying (-) o M is exact

Thorgott

oh duh, not all modules P,Q fit in such a sequence

Balarka Sen

its true because Tor is commutative :)

Thorgott

11:29 AM

sure, I'm gonna believe

Balarka Sen

it might be fun to try to write an explicit argument though

should be a baby spectral sequence argument

Thorgott

I'm not seeing it immediately and am still occupied with figuring out why two isn't equal to infinity

maybe I'll try later

Balarka Sen

yeah let me know if you try

i see the idea and its not easier than writing a proof of Tor(M, N) = Tor(N, M) in general

oh well

Mary Star

12:15 PM

Hello!!

How is $\mathbb{Z}_8\times \mathbb{Z}_8/\langle (4,1)\rangle $ defined?

It is the set $\{(x,y)+(4,1)\}$ with $(x,y)\in \mathbb{Z}_8\times \mathbb{Z}_8$, right?

I want to calculate the order of $(6,2)\in \mathbb{Z}_8\times \mathbb{Z}_8/\langle (4,1)\rangle$.

We have that $2\cdot (6,2)=(12,2)$ which is equivalent to $(4,2)$. Then $3\cdot (6,2)=(12,6)=(4,6)$. Then $4\cdot (6,2)=(0,0)$.

So is the order equal to $4$ ?

Or do we have to consider also the defintion of $\mathbb{Z}_8\times \mathbb{Z}_8/\langle (4,1)\rangle $ ?

Thorgott

oh my god, I have finally uncovered the mistake in my computations

I'm so ineffably stupid

I need a grad student to teach me the art of crying in a corner

Edward Evans

12:32 PM

pick a corner, done that?

Now cry in it

you wanna stop crying right?

Don't, keep with it

Thorgott

ah, that's a good tip

which corners are best?

Edward Evans

Always pick an Inside corner, never an outside corner, you cant sit in those

Thorgott

ok, I feel encouraged now

gonna commit cry

Edward Evans

Immer in einer Ecke, denn in einer Kante kann man nicht sitzen - Immanuel Kant

Thorgott

lol

Balarka Sen

12:40 PM

@LeakyNun I just played Qxj13+#

Just4PlayerThings

Thorgott

I'm doing this against my will, but out of obligation, so that makes it morally right to the Kantian

For the record, it turned out that $2=2$ after all

Edward Evans

Rofl

turns out LCFT is hard and I have to present on it in like 2 weeks

Thorgott

The local coordinate Fourier transform?

Edward Evans

right

..

Balarka Sen

large conformal field theory

obviously

Edward Evans

12:46 PM

obvi

Thorgott

Lie comonoid functor transversality

Balarka Sen

Langlands-Connes-Fomin theory

you know how physicsts have those weird names

BBGKY hierarchy

Edward Evans

wot

Thorgott

The Lipschitz compact-final topology

Balarka Sen

BBGKY hierarchy

In statistical physics, the BBGKY hierarchy (Bogoliubov–Born–Green–Kirkwood–Yvon hierarchy, sometimes called Bogoliubov hierarchy) is a set of equations describing the dynamics of a system of a large number of interacting particles. The equation for an s-particle distribution function (probability density function) in the BBGKY hierarchy includes the (s + 1)-particle distribution function, thus forming a coupled chain of equations. This formal theoretic result is named after Nikolay Bogolyubov, Max Born, Herbert S. Green, John Gamble Kirkwood, and Jacques Yvon. == Formulation == The evolution of...

look at this abomination

Edward Evans

12:49 PM

loool

Balarka Sen

BBGKY man

so catchy

Edward Evans

nah I can't take that seriously

Balarka Sen

lmfao

that was left field

Edward Evans

hahahaha

Thorgott

LMAO

Balarka Sen

12:51 PM

Bogoliubov sounds like a Pokemon

wtf

Edward Evans

sounds like a londoner starting a fight

Thorgott

@MaryStar No, it's the set $\{(x,y)+\langle(4,1)\rangle\colon(x,y)\in\mathbb{Z}_8\times\mathbb{Z}_8\}$, i.e. the set of cosets of $\mathbb{Z}_8\times\mathbb{Z}_8$ under the equivalence relation $(x,y)\sim(x^{\prime},y^{\prime})\colon\Leftrightarrow(x-x^{\prime},y-y^{\prime})\in\langle(4,1)\rangle$. See en.wikipedia.org/wiki/Quotient_group

Edward Evans

$\operatorname{Hom}_{S^{-1}R}(S^{-1}(-), S^{-1}N)$ is exact on sequences of finitely generated modules over a noetherian ring $R$

and $N$ injective

again by composition of exact functors

confirm

lol

Thorgott

localization commutes with Hom for f.p. modules and f.p. and f.g. are the same over noetherian rings

Edward Evans

okey nise

and Hom(- , N) is exact cuz N injective

Balarka Sen

12:56 PM

ye

Edward Evans

yeboi

Thorgott

and localization is exact, ofc

Edward Evans

yeah haha

Balarka Sen

actually you can also just argue localization of injective modules over Noetherian rings are injective

and then you dont need the finite generation stuff anymore

Edward Evans

Yeah that's the next exercise

Balarka Sen

12:58 PM

nice

Edward Evans

else that'd have been my approach

Balarka Sen

actually maybe thats a bit circular

i need something like what u wrote to prove that

so it makes sense thats the next exercise :)

Edward Evans

That $\operatorname{Hom}_{S^{-1}R}(- , S^{-1}N)$ is exact?

Balarka Sen

yeah heh

Edward Evans

lol the hint for the exercise is to use the Baer criterion

Balarka Sen

1:01 PM

right that makes sense

Thorgott

if $M\subset\mathbb{R}^n$ is a $k$-dimensional Riemannian submanifold, then surely the volume of $M$ as a Riemannian manifold and the $k$-dimensional Hausdorff measure of $M$ agree (maybe up to some pi factor), right?

Balarka Sen

Yes, but as I recall you need to work to prove this

If $M \subset \Bbb R^n$ is compact, cover the Riemannian manifold by geodesic $\delta$-balls and use a volume comparison theorem, like Bishop-Gromov inequality

For noncompacts you exhaust by compact things

It's just some technical bullshit

Federer has it

Thorgott

it should suffice to check this locally, so we can take $M$ to be the graph of some function or sth, then explicitly write down the volume form and it should just be a computation, no?

Balarka Sen

I don't know what this computation is

You need a (sequence of) cover(s) to compute Hausdorff dimension

What are your covers

geodesic $\delta$-balls are the correct thing to want, and then you use this:

19 hours ago, by Balarka Sen

I think the function $\text{vol}_g/dx_1 \wedge \cdots \wedge dx_n$ is $1 - \frac16 \sum_{ij} \text{Ric}_p(\partial_i, \partial_j) x_i x_j + O(|x|^3)$ in normal coordinates at a point $p$

Thorgott

I don't know what geodesic $\delta$-balls are

can't we just use coarea formula in some form

surely that should be the point

Balarka Sen

1:14 PM

No man the point is what I said

Anyway this conversation is a waste of time. Look it up in Federer

Thorgott

actually, this should be precisely what the coarea formula says with the right setup

Balarka Sen

OK dude

Just do the computation instead of waffling philosophy :P

Thorgott

the Gramian determinant of a parametrization should be the volume form

Balarka Sen

What is the fucking cover

Thorgott

no cover man

I'm not gonna compute the Hausdorff dimension of something by definition

who does that

Balarka Sen

1:17 PM

Hausdorff dimension is computed by a cover you absolute retard

This is so frustrating

Thorgott

dude, we have theorems for a reason

that's like telling me to compute a Riemann integral as limit of Riemann sums when I could use the FTC instead

Edward Evans

I love your arguments

Thorgott

I think he actually ragequit lol

Edward Evans

top banter

Balarka Sen

Ah look here is a simple argument without any of your big theorem shit. Cover the manifold by $1 + \delta$-Lipschitz charts. Simply just that.

On $\Bbb R^k$, Hausdorff measure = volume form. So this says (volume form) $\leq$ $(1 + \delta)$(Hausdorff measure)

That's it

Thorgott

1:29 PM

So take $M\subset\mathbb{R}^n$ to be globally parametrized by $\varphi\colon U\rightarrow M$, $U\subseteq\mathbb{R}^k$ open. The area formula (coarea formula was a misthink) states $\int_M\sqrt{|\det(J\varphi^TJ_\varphi)|}=\int_{\mathbb{R}^k}\mathcal{H}^0(M\cap\varphi^{-1}(x))d\mathcal{H}^k(x)$. The Gramian determinant on the LHS should be the volume form, so LHS=vol(M), but the cardinality on the RHS is one for x in M and 0 else, so the RHS is $\mathcal{H}^k(M)$

Balarka Sen

This is completely useless

Look at above proof

It's literally 1 line. No coarea nonsense

Leaky Nun

@BalarkaSen brilliant

Thorgott

Why can I cover my manifold with Lipschitz charts? $C^1$ implies locally Lipschitz, so I get sth Lipschitz after possibly restricting, but how do I control the Lipschitz constant?

Balarka Sen

What do you mean by controlling the Lipschitz constant? Around any point you can draw a $(1 +\delta)$-Lipschitz chart by $C^1$-ness. That gives you a cover. Pass to a countable subcover by Lindelof property

Thorgott

$C^1$ means I can control the Lipschitz constant in terms of the norm of the derivative, but why should I be able to make that arbitrarily close to $1$?

Michael

1:38 PM

Is there a mathematical notation to signify that a number has been rounded?

Balarka Sen

Because you can find a chart $\Phi : U \subset M \to B(0, r)$ centered at any point $x \in M$ such that $d\Phi_x = I$, so $\|d\Phi\|$ is close to $1$...?

Super rigorously, use the exponential map I mentioned before. $\exp_p : B(0, r) \subset T_x M \to M$ is a chart with $d\exp_p(0) = I$.

For $r$ small enough

You're living in $\Bbb R^n$, this is equivalent to projecting to the tangent space $T_x M \subset \Bbb R^n$.

Anyway if this is convincing just note that this implies in a coordinate system $(x_1, \cdots, x_n)$ around $x$ given by $\exp_x$, $\int_{A} dx_1 \wedge \cdots \wedge dx_n \leq (1 + \delta)^k \mathcal{H}(A)$

And then you still have to compare $dx_1 \wedge \cdots \wedge dx_n$ with $d\text{vol}_g$. This is given by the formula I mentioned above, $d\text{vol}_g = f dx_1 \wedge \cdots \wedge dx_n$ where $f(x) = 1 + O(|x|^2)$

So you can choose $r$ to be small enough so that $\|f\| \leq 1 + \delta$

Putting it all togather, $\int_A d\text{vol}_g \leq (1 + \delta)^{k+1} \mathcal{H}(A)$

True for any $\delta$. So one inequality is proved. For the other half, proceed similarly, but use $1 - \delta$-Lipschitzness of the chart $\Phi^{-1}$ this time. You get $\int_A d\text{vol}_g \geq (1 - \delta)^{k+1} \mathcal{H}(A)$

This is a simple, down to earth proof. Quoting theorems is journalism, not mathematics.

Anyway I'm off

299792458

2:09 PM

Hello Math Stack people. I had a question, which perhaps isn't a good fit for the main site, therefore asking in chat.

I am evaluating some pFq hypergeometric functions at some specific points, and seem to be getting different answers from MATLAB and Wolfram Alpha evaluation of the same function.

Does anyone know any reliable way of computing them?

Case in point: 3F2([2,-48.-49],[-46.5,-47.5],1

feynhat

Sanity check: If G is a finite group that acts freely on a contractible space X. Then X is an EG?

Thorgott

2:31 PM

@Balarka i dont buy the reverse direction

Mike Miller

3:47 PM

@feynhat That's the definition of EG

I guess formally one would define for a topological group $G$ that an $EG$ is a contractible space with $G$-action so that $p: EG \to EG/G = BG$ is a principal $G$-bundle

For a finite (or compact) group that's just asking that G act freely

For a discrete group they call that 'free and properly discontinuous' or 'covering space action'

Balarka Sen

4:56 PM

@Thorgott Here's the relevant blackboxes. Given any Riemannian manifold $(M, g)$, for any point $p \in M$ $\DeclareMathOperator{\vol}{vol}$

(1) There's $r > 0$ and a diffeomorphism $\exp_p : B(0, r) \subset T_p M \to M$ such that $d\exp_p(0) = I$. The image is called geodesic ball of radius $r$ around $p$. For a reference, check do Carmo.
(2) In local coordinates $(x_1, \cdots, x_n)$ given by $\exp_p$ around $p$, the Riemannian volume form is related to the form $dx_1 \wedge \cdots \wedge dx_n$ by $d\vol_g = f(x) dx_1 \wedge \cdots \wedge dx_n$ where $f(x) = 1 + O(\|x\|^2)$. This follows f…

As far as I agree there is nothing wrong with this argument. Sorry for being a prick earlier, I shouldn't be having heated math discussions when I haven't slept for like the whole day. It is just atrocious to me that you'd think we need to invoke big theorems like coarea formula for simple results like this.

To be 100% honest I do not know the coarea formula beyond it's name, but from what you said I'd be surprised if it's real prowess really lies in proving theorems about good subsets in $\Bbb R^n$ like manifolds, I'd imagine it gives you analogous results for $k$-currents instead of $k$-manifolds. This seems to be just soft analysis.

There is one point where you might object I am using $M$ is more regular a manifold than $C^1$, in point (2) where the Ricci curvature pops up, because curvature lives in the $C^2$ world. But actually we didn't need that anywhere; all we needed was $f$ is $C^1$, $f(0) = 1$, $f'(0) = 0$ - the 1st order Taylor's formula with reminders.

Anyway apologies for being rude earlier

that was uncalled for

Balarka Sen

5:17 PM

And also when I say it is globally true because locally true because basis I look at the countable subcover by these geodesic balls $\exp_p(B(0, r_p))$ where $r_p$ is my chosen radius at $p$ after all the shrinking. This exists by Lindelofness of manifolds. Then I can argue by breaking a set up into disjoint pieces each fitting inside one of these balls, and countable additivity of measures

just to leave no stone unturned