Mathematics - 2019-07-24 (page 1 of 2)

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Mike Miller

03:13

@BalarkaSen kiwisbybeat.netlify.com/verse3.html

@BalarkaSen What this argument shows me is that it's equivalent to $(S^n \times D^n)/(S^n \times S^{n-1})$ but I'm not sure I see that this is what you think it is

But maybe this is just a generalization of the torus case as you say

Balarka Sen

03:41

@MikeMiller this is A+

@MikeMiller yeah this is easier to argue; it's the Thom construction on the trivial bundle on S^n

I do not see how to spell out the torus argument in this case but I'm too sleepy and on the airport so not in the best shape

Mike Miller

04:11

@BalarkaSen Oh this is a nice observation. The Thom space you get when you add a rank one trivial bundle just is the suspension of the previous Thom space

So this is $\Sigma^n(S^n \vee S^0)$

You can see the two factors in the torus case

Secret

04:50

ams.org/journals/notices/201907/rnoti-p1034.pdf

Leaky Nun

05:02

@BalarkaSen $(S^n \times S^n)/(\ast \times S^n) = (\Bbb R^n \times S^n + \ast \times S^n)/(\ast \times S^n) = \Bbb R^n \times S^n + \ast = \Bbb R^n + \ast + \Bbb R^{2n} = (\Bbb R^n + \ast) \vee (\ast + \Bbb R^{2n}) = S^n \vee S^{2n}$

lol no

ok suppose $H_n = L_n H_0$, which is the case for nice enough spaces, then we are interested in $H_n(-;G) = L_n(H_0 - \otimes G)$

Grothendieck says that $L_p (- \otimes G) L_q H_0 - \implies H_{p+q}(-;G)$

i.e. $\operatorname{Tor}_p(H_q-,G) \implies H_{p+q}(-;G)$

as you noted, over PID we have $\operatorname{Tor}_{\ge 2} = 0$

7 hours ago, by Balarka Sen

$\cdots \to E^2_{n, 0} \to E^2_{n-2, 1} \to H_{n-1} \to E^2_{n-1, 0} \to E^2_{n-3, 1} \to \cdots$

how does this work

no this is wrong, the LES above is for concentrated in q=0,1

here we have p=0,1

so $E^2 = E^\infty$

so $0 \to E^2_{0,n} \to H_n \to E^2_{1,n-1} \to 0$

i.e. $0 \to H_n(-) \otimes G \to H_n(-;G) \to \operatorname{Tor}(H_{n-1}(-),G) \to 0$

let's check the UCT...

(aha, this is why you said UCT!)

2 days ago, by Leaky Nun

$0 \to H_i(X; \mathbf{Z})\otimes A \, \overset{\mu}\to \, H_i(X;A) \to \operatorname{Tor}(H_{i-1}(X; \mathbf{Z}),A)\to 0$

praise Grothendieck!

wait, if in general $H_n \ne L_n H_0$ then what is $L_n H_0$?

oh wait this makes no sense in general, because $\mathsf{Top}$ isn't an abelian category!

so above I should replace $L_n H_0$ with $L_n$ of the global section of the constant sheaf

the correct statement is that for nice enough spaces, $H_n$ is a left derived functor

since it coincides with the sheaf cohomology of the constant sheaf $\Bbb Z$

3 hours later…

Balarka Sen

08:28

@MikeMiller oh good point

Mike Miller

08:53

@BalarkaSen I think the distinction between geometry and homotopy theory is that this is the unreduced suspension.

2 hours later…

MatheinBoulomenos

10:40

@Balarka is there a nice abstract reason why $\mathrm{vol}(\Bbb{H}/\mathrm{SL}_2(\Bbb Z)) < \infty$ or do you need to use a fundamental domain?

loch

11:23

- sheaf cohomology gives you singular cohomology of your space (not homology)
- I don't think anyone uses the notation L_nH_0, usually L_nF or R_nF, for whatever your functor F is
- I don't think singular homology is realised as left derived functors for some functor on sheaves (at least ive never heard of such thing) .. you can get something called borel moore homology using sheaves though

hm based on this link mathoverflow.net/questions/66401/… maybe you can

MatheinBoulomenos

11:38

@Leaky exercise: X is a compact Hausdorff space, then C(X) is Jacobson iff X is totally disconnected

Perturbative

12:33

Suppose I have a fibration $p : E \to B$ with fiber $F$, let $G$ be an abelian group, what is the action of $\pi_1(B)$ on $H_{\bullet}(F; G)$?

Balarka Sen

13:07

@MatheinBoulomenos Yeah just use the fundamental domain. It's a $(2, 3, \infty)$-triangle, so has area $\pi - (\pi/2 + \pi/3 + \pi/\infty) = \pi/6$.

By Gauss-Bonnet if you wish

@Perturbative $\pi_1 B$ acts on $F$ by monodromy. Namely, given a loop on $B$ based at a point $x \in B$, you can lift it to a path starting at any point $f \in F$ and ending at $\tilde{f} \in F$.

If the homotopy class of this loop is $\alpha$, we define the action by $\alpha \cdot f = \tilde{f}$. This is a continuous action of $\pi_1 B$ on $F$.

Look at the induced action on $H_*(F)$

@loch The sheafification of the singular cochains presheaf gives an acyclic resolution of the constant sheaf, so you can prove that the singular cohomology agrees with the Cech cohomology. That shows it's left derived functors for the global section functor on the constant sheaf, right?

@loch The sheafification of the singular cochains presheaf gives an acyclic resolution of the constant sheaf, so you can prove that the singular cohomology agrees with the Cech cohomology. That shows it's left derived functors for the global section functor on the constant sheaf, right?

loch

13:31

@BalarkaSen I think that sounds about right - but it's right derived functors not left derived functors since the global sections functor is left exact :p

Ryan Unger

@ÉricoMeloSilva so is Fernando gonna be your first year advisor

I got an email from Alice Chang

s.harp

14:13

oof its so hot today

Sila

14:33

can someone help me to find an example of : an example of $ψ:C∙→D∙$ co-chain map that exists :
$ψi:C^i→D^i$ is an surjective map (for i>=0) but $ψ∗:H^k(C∙)→H^k(D∙)$ is not a surjective map (for k>=0)? I want to check if an identity map or $f: R→S1$ can be an example of surjective cochain map like in a question? if not , how I can find like this map? The difintion of cochain map in a booklet is: cochain map between cochain complex ⟨D∙,δ⟩ and ⟨C∙,d⟩ is a homomorphisim chains $ϕn:Cn→Dn$ (for n>=0) such as δn◦ϕn=ϕn+1◦dn

MatheinBoulomenos

@Sila you can use the fact that a short exact sequences of cochain complexes induces a long exact sequence on the cohomology.

Sila

@MatheinBoulomenos, do yoy mean foe example :

@MatheinBoulomenos, 0→A→iB→jC→0 induces a long exact sequence of homology groups ...→δHn(A)−→i∗Hn(B)−→j∗Hn(C)→δHn−1(A)−→i∗Hn−1(B)−→j∗Hn−1(C)→δ

MatheinBoulomenos

14:57

yes, except for cochain complexes, you get cohomology and an increasing long exact sequence. Anyway, here's a simple example Let $C^0=C^1=\Bbb Z$ and $C^i=0$ else and let $C^0 \to C^1$ be the identity. Let $D^0=\Bbb Z$ and $D^i=0$ else. Let $\phi_0:C^0 \to D^0$ be the identity and $\phi_i:C^i \to D^i$ zero else. Then $\phi_i:C^i\to D^i$ is surjective for all $i$. But $H^0(C^\bullet)=0$ and $H^0(D^\bullet)=\Bbb Z$, so $\phi_*:H^0(C^\bullet) \to H^0(D^\bullet)$ can't be surjective

Ryan Unger

@MatheinBoulomenos chains are just cochains but indexed incorrectly

MatheinBoulomenos

I agree

Secret

0

Q: Bijection of proper classes

I have two proper classes which intuitively are like bijectively equivalent, i.e. for every element of one of these two classes we can define an expression for the corresponding element of the other class, and these behave nicely (like a bijection). I wonder, is the notion of bijection extended ...

lo.logic set-theory

$Mathematics$ The Star Wars Room :--O--:

A^T ATx + BB8^2y+|----|C3PO

I hate this auto room freezing system

Sila

15:16

@RyanUnger, what do you mean by incorrectly ?

@MatheinBoulomenos, what the example being if :

Ultradark

Hey everyone

Sila

$ϕ^i: A^i → B^i$ is a bijective map for (i>=0) but $ϕ∗ : H^k(A•) → H^k(B•)$ is not bijective for $(k>=0)$ in the two examples I mean by : ϕ∗:= dϕ that is meaning: if the map $ϕ: C to C'$ then $ϕ∗:H(C) to H(C')$

Thomas Nicholson

15:33

Hi all, wondering where I might be able to get another set of eyes on a proof I've put together for properties of Farey sequences, of course already well proven in a number of sources, just a practice of doing it myself for a specific purpose in an article.

MatheinBoulomenos

@Sila there's no example like that

If $\phi^i:A^i \to B^i$ is bijective for all $i$, then $\phi$ is an isomorphism of chain complexes and hence it induces an isomorphism on cohomologies

Mike Miller

@ThomasNicholson I don't have any reason to believe that one of us will or will not be able to help, but the usual advice is to ask, don't ask to ask --- most people just don't respond to the request to ask :)

Ultradark

@MikeMiller I have a question for you. What advice could you give to an individual like myself starting a Master's program in math in one month?

Thomas Nicholson

15:49

@MikeMiller Right, was just trying to be polite, considering I haven't got a clue if outsourcing something like that is appropriate in the chat :D it is about 2 1/2 pages long, relevant for a music article I have written. I am a hobby mathematician at best, so just wanted to get a verification that there are no glaring holes in it...

Secret

So, I was thinking about some irreflexive but transitive order:

where a<b<c and for all x, x<x

It is easy to see a<a, but how to prove a<a+b?

Semiclassical

16:05

Is that enough to show a<a+a?

I'm not sure it is.

s.harp

if you have 0<a, a<b, as well as compatability with + (meaning out of x<y follows x+c<y+c for all c) then 0+a<a+a<b+a

Secret

right

Akiva Weinberger

16:27

The 25th of April is Tau Approximation Day

It's not as close to tau as 22/7 is to pi, but it's the closest we can get with calendar dates

Sila

@MatheinBoulomenos, In the booklet which I read in, rememberd that there is an examples but I do not know how to find like these cochain maps. in the booklet is wrote that there is an example for a cochain map $ψ: A• → B•$ that exists :
$ψi:A^i→B^i$ is an bijective map (for i>=0) but $ψ∗ :H^k(A∙)→H^k(B∙)$ is not a bijective map (for k>=0)?

Akiva Weinberger

November 14th is pi/4 Approximation Day, which is basically the same approximation as 22/7≈pi, but better because it uses the American date order

(jk)

Sila

@MatheinBoulomenos,$phi$ and $psi$ is are different cochain maps.

1 hour later…

Secret

18:03

Can you have set theories where the powerset operation has fixed points?

Rithaniel

Define "fixed point." Are you taking about an element which is in both a set and it's powerset?

Secret

Basically, can you have a set theory T such that given A is a set, and |A| is a cardinality then |P(A)|=|A|?

I knew ZF cannot do that, nor do most common varieties of set thoery I heard of

because they all rely on some von neumann universe thing that will end up proving cantor theorem

Rithaniel

Well, the best chance you might have with that is some infinite set, because the power set of a finite set you're screwed on, unless you change the definition of power set.

Secret

yeah make sense

Rithaniel

Is ZF + "All infinite sets have equal cardinalities" a consistent axiom system? If so, then there you go.

Secret

18:12

it should be consistent, since you are basically taking a equivalence class over all infinite sets

Rithaniel

Well, it's not gonna be safe to just assume that it's consistent. You gotta prove it is, and that might take a little effort.

Perturbative

Thanks! @BalarkaSen

Secret

18:29

ok, so by dry run the proof with that extra axiom on cantor theorem, the result will be quite bizarre:

Cantor's theorem

In elementary set theory, Cantor's theorem is a fundamental result that states that, for any set A {\displaystyle A} , the set of all subsets of A {\displaystyle A} (the power set of A {\displaystyle A} , denoted by P ( A ) {\displaystyle {\mathcal {P}}(A)} ) has a strictly greater cardinality than A {\displaystyle A} itself...

It basically means the diagonal set has to be empty in order to avoid the contradiction

which also means all functions $f$ are bijective on infinite sets

Rithaniel

I believe Cantor's theorem actually shows that that axiom system I suggested is inconsistent.

Secret

won't having B empty block the first line of the iff (at the cost of the bizarre consequence that all subsets of A are empty)?

Ted Shifrin

18:44

Oh oh. I've stumbled into the logic room.

Secret

lol

Rithaniel

Heya Ted.

MatheinBoulomenos

Hi @Ted

Ted Shifrin

Hi @Rithaniel @Mathein @Secret

And hi @loch

MatheinBoulomenos

@Ted I'm cutting a lot of geometry from my thesis to keep it at a reasonable length

Ted Shifrin

18:47

Why do you cut geometry, of all things??!!

Hrumph.

MatheinBoulomenos

because it's not that essential for what I'm doing. Jacobians only yield the construction for weight 2

Ted Shifrin

But you need PPAV in general?

Ryan Unger

@TedShifrin yo

MatheinBoulomenos

You need étale cohomology for the general construction

Ted Shifrin

yo @Ryan

Ryan Unger

18:48

I found the page in that paper with the theorem

Ted Shifrin

LOL, @Mathein, so much for geometry :P

Oh, cool. What page?

Ryan Unger

Second to last page

66, I think

it's not demarcated as a theorem, it's just italicized near the bottom of a paragraph

MatheinBoulomenos

@Ted I thought up a proof on how to directly reconstruct a compact Riemann surface from its field of functions, though I guess it's standard

Rithaniel

Well, Secret, if you require that any subset of an infinite set is empty, I'm pretty sure that you can prove that those "infinite" sets are actually the empty set, which contradicts the ZF axiom of inifinity.

Ryan Unger

PPAV?

MatheinBoulomenos

18:50

principally-polarized abelian varieties

Ted Shifrin

Oh, the thing he credits to Franz?

principally polarized abelian variety

oops, late as usual

MatheinBoulomenos

damn

Ryan Unger

I did some digging and Franz proved it for n=3 lens spaces

Ted Shifrin

@ryan: But he's doing only the sphere here. You were talking about all space forms.

Ryan Unger

I don't think this proof is written in English anywhere

@TedShifrin Spherical space forms

It's not true for flat ones and it's true for hyperbolic by Mostow

Ted Shifrin

18:52

Oh, so it's not true for constant $K\le 0$? !!

Aha.

I admit I've not thought about it.

Ryan Unger

for flat you can just take two flat tori with different radii

diffeomorphic but not isometric

Ted Shifrin

I thought we said simply connected?

Ryan Unger

for simply connected this is well known

Ted Shifrin

Yeah, that's what I thought.

Ryan Unger

the amazing this is that it's true for any space forms (with finite volume) in the non-flat cases

Ted Shifrin

18:54

I guess Minding's Theorem says that for surfaces if you have a map preserving Gaussian curvature, then it is an isometry.

Have you looked carefully at Wolf? Surely this result is in there.

Ryan Unger

I googled that and it's a local result for surfaces

I looked very carefully at Wolf

Ted Shifrin

I sure wish I hadn't gotten rid of that book.

Ryan Unger

The proof uses techniques completely different from his book

Ted Shifrin

Fair enough.

Ryan Unger

Though you might be able to brute force it from the complete classification of spherical space forms

MatheinBoulomenos

18:55

The idea is this: for a Riemann surface $X$, take the field of meromorphic functions $\mathcal{M}(X)=K$. Then one can consider the set $ZR(K)$ of discrete valuations on $K$ that are trivial on $\Bbb C \subset K$. For $v \in ZR(K)$, we have a valuation ring $K_v$ inside $K$ with residue field $\Bbb C$ and the quotient map $K_v \to \Bbb C$ recovers the evaluation of functions at the point corresponding to $v$

Ted Shifrin

@Mathein: I admit this is the sort of thing I've never thought through, but it's well known.

MatheinBoulomenos

(Using the theory of extension of valuations on finite extensions, one can prove that points of $ZR(K)$ correspond to points in $X$, by reducing it to $\Bbb{P}^1(\Bbb C)$)

Ryan Unger

@TedShifrin So the real claim is that if a manifold diffeomorphic to a spherical space form has two constant curvature metrics, they are in fact isometric

Ted Shifrin

Surely it's in Hartshorne :)

Ryan Unger

(same constant)

Ted Shifrin

18:56

@Ryan: Wait. Aha.

So that is a generalization of Minding, of course.

Ryan Unger

Wikipedia tells me that Minding says constant Gauss curvature determines the isometry type of balls on a surface

MatheinBoulomenos

@Ted yeah but the punchline is then you can consider every $f \in K$ as a map $ZR(K)
\to \Bbb{P}^1(\Bbb C)$. Now you take the initial topology with respect to all these maps for the analytic topology on $\Bbb{P}^1(\Bbb C)$

Ryan Unger

This is of course true for any Riemannian manifold for the sectional curvature

MatheinBoulomenos

It took me a while to figure out why this actually recovers the topology of $X$

Ryan Unger

And then there's the Cartan lemma that lets you extend this to complete simply connected manifolds globally

At least, do Carmo credits Cartan with this

I imagine Hopf and Killing knew about it

Ted Shifrin

18:59

Yeah, Minding is proved using geodesic normal coordinates, so it's local.

Cartan knew everything.

Ryan Unger

So de Rham's theorem is proved using Reidemeister torsion, which I know nothing about

I can't find any evidence that this proof was written up anywhere else

Ted Shifrin

I thought Reidemeister torsion came along way later with Atiyah-Singer ...

Ultradark

Toponogov's theorem affords a characterization of sectional curvature in terms of how "fat" geodesic triangles appear when compared to their Euclidean counterparts. The basic intuition is that, if a space is positively curved, then the edge of a triangle opposite some given vertex will tend to bend away from that vertex, whereas if a space is negatively curved, then the opposite edge of the triangle will tend to bend towards the vertex.

Ryan Unger

Are you thinking about Ray-Singer torsion

Ultradark

I assume there's a similar thing for quadrilaterals?

Ted Shifrin

19:01

Not really, @Ultra. You're talking about geodesics spreading out of a fixed vertex.

Ryan Unger

The two share a wiki article for what it's worth

I wonder if one can prove this using some kind of index theory

Ted Shifrin

I used to know a little about Reidemeister torsion, but I remember nothing now.

Ryan Unger

@ÉricoMeloSilva Transcript is all good. They didn't email me so I emailed them...

Ultradark

Yeah true.

Ted Shifrin

OK, I have to run some errands. BBL.

Ryan Unger

19:03

ciao

Ultradark

see ya

MatheinBoulomenos

The proof goes like this: as all meromorphic functions on $X$ are continuous, the "identity" $X \to ZR(K)$ is continuous by the universal property of an initial topology. Now as $X$ is compact, we only need to show that $ZR(X)$ is Hausdorff. First, I claim that $\mathcal{M}(X)$ separates points:
Let $x \in X$, then consider the divisor $D=(g+1) \cdot x$.
By Riemann-Roch $L(D) \geq 2$, so there is a non-constant function $f$ in $L(D)$. $f$ has only a pole at $x$, so it separates $x$ from every other points.

see ya @Ted

ÍgjøgnumMeg

Evening all

MatheinBoulomenos

evening @ÍgjøgnumMeg

Rithaniel

Sup, Meg

ÍgjøgnumMeg

19:05

Hey @Mathein and @Rithaniel

Ultradark

Hey @ÍgjøgnumMeg

ÍgjøgnumMeg

Hey @Ultradark :)

Secret

@Rithaniel ah right because union of empty sets are empty

s.harp

19:21

@Secret whats the intersection over an empty index set? 8)

Rithaniel

So, define an equivalence relation on the integers $\equiv_p$ such that, if $x=np^a$ and $y=mp^b$ with $\text{gcd}(n,p)=\text{gcd}(m,p)=1$, then $x\equiv_p y\iff n=m$. Then does it seem correct to say that applying standard multiplication to the set of all these equivalence classes provides us with a monoid?

Secret

cap(A)={x: x in A} but there is no such x since A will be empty

sorry brain fart:

Since A is empty, the condition is vacuously true, thus all x applies and you get the whole set

s.harp

thats true :) and I think its horrible

for example in bourbaki in the definition of a topology they only ask that an arbitrary union and a finite intersection over opens is open

as the empty intersection is a finite intersection and so the whole set is open, the empty union is then the empty set and the empty set is open

making life difficult by making it simpler

Ryan Unger

the one that got me is disjoint submanifolds are transversal

Secret

well... the whole set and empty set are always clopen in any topology, thus it is not much of a deal

Ryan Unger

19:31

Hirsch has a comment about this

Rithaniel

So many comments on here use words I don't recognize.

Such as "transversal"

Ultradark

clopen

Rithaniel

clopen means simultaneously closed and open. That one I'm lucky enough to know.

Ultradark

haha nice

transversal means something like "going across"

I think

I'm thinking of transversal lines from 9th grade geometry

Rithaniel

Well, yeah, but in math in general, words have precise meanings. I would predict that transversal has something to do with reaching across an entire manifold in some way, but until I look up the definition, it's just a guess.

Also, definitions often rely on other terms, which must then be looked up independently.

Ultradark

19:40

everything I say on here is wrong :/

Ryan Unger

transversal is a local thing

submanifolds are transversal at a point of intersection if the tangent spaces add up to give the tangent space of the ambient manifold

Rithaniel

Ah, and disjoint submanifolds have no point of intersection, so they're vacuously transversal.

Ryan Unger

yeah

user image

Rithaniel

Does transveral-ness always have to occur at a particular point? Or can you say "these two submanifolds are tranversal" if they're transveral at all points of intersection

Ryan Unger

yeah they're transversal if each point of intersection is transversal

Rithaniel

19:46

Is there a sense of measuring transversal-ness? Such as "60% of points of intersection are transversal, so these two submanifolds are 60% transversal."

s.harp

two sub-manifolds can also be "in general position" to one another

Ryan Unger

@Rithaniel don't think so

s.harp

i dont recall preceisly what that means, but i think it is a synonym for transverse

Ultradark

what is a submanifold

Ryan Unger

general position is the term for PL transversality

Rithaniel

19:47

A subset of a manifold that is itself a manifold.

and a manifold is a space that approximates euclidean space locally.

Ultradark

@Rithaniel why do you sometimes put periods after your sentences and sometimes don't?

Ryan Unger

lol

Ultradark

I always just do this

Rithaniel

Heh, a lack of paying attention, most likely.

s.harp

you are exposed and thus banished from the chat for a period of 24 days

Rithaniel

19:51

I do it when I write notes by hand, too.

Nooooooo! That's a day for every hour in a day!

Ultradark

I think I'm going to switch to this now.

It makes me sound too short though.

Idk if I like it.

Rithaniel

I don't pay attention to periods, and no one's ever called me out for my touch-and-go use of them until now.

Ultradark

Yoooooooo

washingtonpost.com/news/speaking-of-science/wp/2015/12/08/…

Leaky Nun

.

Rithaniel

Huh.....

Ultradark

20:04

It's honestly not that big of a deal just food for thought

ÍgjøgnumMeg

@Rithaniel elements of $\Bbb Z/\equiv_p$ just look like $\frac{x}{p^{v_p(x)}}$ I guess

clearly $1$ is in there (the class of $1$ is just the set of all powers of $p$) and associativity of multiplication is from that of $\Bbb Z$

so yeah I'd say that seems right

Ryan Unger

tf is $\equiv_p$

ÍgjøgnumMeg

@Ryan Rithaniel defined it abve

above*

@Rithaniel Nb: I'm a total noob and have been away from mathematics for a year so a lot of the shit I say is to be taken with a fistful of salt

Leaky Nun

20:23

$\pi_1(\Omega S^2) = \pi_2(S^2) = H_2(S^2) = \Bbb Z$

can we construct a universal cover of $\Omega S^2$

let's compute $H_n(\Omega S^2)$

$\Omega S^2 \to PS^2 \to S^2$

$\begin{array}{c} H_0(\Omega S^2) & 0 & H_0(\Omega S^2) \\ H_1(\Omega S^2) & 0 & H_1(\Omega S^2) \\ \vdots \end{array}$

ok so $H_n(\Omega S^2) = \Bbb Z$

oh no this means I can't deform $\Omega S^2$ to something finite dimensional

this is sad

@loch help

MatheinBoulomenos

It seems crazy that there's a classification of injective modules over a commutative noetherian ring. Injective modules seem "harder" than projective modules, but I don't think there's any classification of projective modules over a noetherian commutative ring

Leaky Nun

20:38

@MatheinBoulomenos i'm sad

Ryan Unger

oh god why would you try to classify modules

MatheinBoulomenos

why not?

Ryan Unger

how many are there

Leaky Nun

@RyanUnger that's... like the entirety of representation theory?

ÍgjøgnumMeg

There are exactly 13

MatheinBoulomenos

20:39

rofl

Leaky Nun

exactly as many as the prime ideals in the polynomial ring over a principal ideal domain

Ryan Unger

groups seem marginally more interesting than just random modules

Leaky Nun

classification of finite simple groups uses classification of Lie groups which uses classification of Lie algebra

and Lie algebra somewhat feels like modules lol

Ryan Unger

classification of finite simple groups is an insane problem

Leaky Nun

glad to know it's done

MatheinBoulomenos

20:42

over a left-Noetherian all injective left modules decompose uniquely into injective indecomposable modules. The indecomposable injective modules are in canonical bijection to prime ideals if the ring is commutative Noetherian

Leaky Nun

@MatheinBoulomenos what does $\Bbb Q/\Bbb Z$ correspond to

and $\Bbb Q$

MatheinBoulomenos

$\Bbb Q$ is indecomposable, it corresponds to $(0)$

Ryan Unger

latex z looks so awful

MatheinBoulomenos

$\Bbb Q/\Bbb Z$ is not indecomposable, we have $\Bbb{Q}/\Bbb{Z}=\bigoplus_p \Bbb{Q}_p/\Bbb{Z}_p$ and each factor corresponds to $(p)$

ÍgjøgnumMeg

nice

Leaky Nun

20:47

you what

loch

@LeakyNun whats up

Leaky Nun

19 mins ago, by Leaky Nun

oh no this means I can't deform $\Omega S^2$ to something finite dimensional

loch

is that bad

Leaky Nun

that's sad

loch

lol
why is it sad

Leaky Nun

20:49

because then I can't visualize it

MatheinBoulomenos

I have just stopped trying to visualize things in topology

too hard

loch

same

hopefully balarka will save you

MatheinBoulomenos

lol

Leaky Nun

how does the homology of a covering space relate to the homology of a space?

loch

hmm im not sure what you can say -- eg if you look S^n -> \R \P^n

you can say eg that the euler characteristic of the covering space = degree of covering* your base though

Leaky Nun

20:58

hmm

yeah I should have looked at examples lol

Ryan Unger

@loch can one prove this in the de Rham setting

I used this in a Ricci flow seminar but wasn't able to give an "analytic" proof

Mike Miller

@LeakyNun I'm pretty sure you can visualize $\Omega S^2$ strictly better than some 5-manifold

Leaky Nun

Sep 20 '17 at 17:54, by Balarka Sen

I think the only proof of $\pi_4(S^2) \cong \Bbb Z/2$ I know is by Pontryagin-Thom isomorphism

@MikeMiller what's P-T isomorphism and why does it prove this?

Mike Miller

I'm not a search engine

2

Leaky Nun

I googled it and I couldn't find anything useful

Wiki says something about cohomology groups

MatheinBoulomenos

21:10

It's a known fact that $\Bbb{A}_{\Bbb Q}$ is self-dual. This implies that the dual of $\Bbb Q$ is $\Bbb{A}_{\Bbb Q}/\Bbb Q$.
Consider $0 \to \Bbb Z \to \Bbb Q \to \Bbb Q/\Bbb Z \to 0$. Taking Pontryagin duals, we obtain $0 \to \widehat{\Bbb Z} \to \Bbb{A}_{\Bbb Q}/\Bbb{Q} \to S^1 \to 0$ Obviously $\Bbb{A}_{\Bbb Q}/\Bbb{Q} \to S^1$ is not a covering, but it should be an inverse limit of coverings, so like a "procovering". If one takes the inverse limit of $z \to z^n$, then one obtains another "procovering" with "fiber" $\widehat{\Bbb Z}$, but I can't see why that is isomorphic to $\Bbb{A}_{\…

Mike Miller

@RyanUnger Let me try this. Let $\tilde M \to M$ be a finite covering space, with fiber $F$. One has a corresponding flat bundle $\Bbb R[F]$ over $M$. First you should see that for any bundle with connection $E$ you have $\chi(H^*_{dR}(M;E)) = (\dim E) \cdot \chi(M)$, which is probably best done via homotopy invariance of index. Now argue that $\Omega^*_{dR}(M;\Bbb R[F]) = \Omega^*_{dR}(\tilde M)$.

I haven't checked all the details but this should work fine.

Ryan Unger

@MikeMiller So you just lift $d$ to $\Bbb R[F]$ (never seen that notation before) in the sensible way?

What is $\Bbb R[F]$ called

Leaky Nun

7

A: Visualize Fourth Homotopy Group of $S^2$

For me, the most satisfactory explanation of the isomorphism $\pi_4(S^2)\cong \mathbb{Z}/2$ is in Pontryagin's beautiful survey Smooth manifolds and their applications in homotopy theory. 1959 American Mathematical Society Translations, Ser. 2, Vol. 11 pp. 1–114 American Mathematical Soci...

this looks promising

MatheinBoulomenos

I thought that standard proof for the Euler characteristic thing is to lift a triangulation (or a CW decomposition I guess)

Mike Miller

@RyanUnger the covering map gives you a representation $\pi_1 M \to S_n$, where $n$ is the cardinality of the fiber. a representation of the fundamental group into $O(n)$ gives a flat bundle with fiber $\Bbb R^n$. a flat bundle comes equipped with a flat connection, by definition depending on your definition, but in any case in a local trivialization it's just $d$.

@MatheinBoulomenos he wanted an analytic proof

Ryan Unger

21:20

@MatheinBoulomenos Sure, and that's the reason I gave during the talk

Mike Miller

I named it $\Bbb R[F]$ because $F$ is the fiber and this is a bundle of rank $|F|$

The rest is up to you

Leaky Nun

6

A: Visualize Fourth Homotopy Group of $S^2$

One way I see is the following. To start with, $\pi_{4}(S^{3}) \simeq \pi_{4}(S^{2})$ because of the existence of the Hopf fibre sequence $S^{1} \rightarrow S^{3} \rightarrow S^{2}$ and moreover the isomorphism is given by composition with the Hopf map. Hence, as was observed in the comments, yo...

7

A: Not null homotopic map from $S^3$ to $S^2 \vee S^2$

Consider the attaching map $S^3 \to S^2 \vee S^2$ of the $4$-cell in the standard CW-structure of $S^2 \times S^2$. If this is nullhmotopic, $S^2 \times S^2$ would be homotopic to $S^2 \vee S^2 \vee S^4$ as homotopic attaching map implies homotopy equivalent spaces. But cup square $\alpha \smile...

@BalarkaSen lol

Leaky Nun

21:57

@MatheinBoulomenos where can I learn about [X,Y]?

nlab perhaps

it seems to be a group when we have a "group structure" on X

this sounds similar to some Yoneda construction

ok it's the morphism in the naive homotopy category

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