Mathematics - 2020-07-06 (page 2 of 2)

« first day (3624 days earlier) ← previous day next day → last day (1401 days later) »

4:01 PM

yeah, that's the heart, but the refinement is probably useful somewhere to someone

The main take away for me, was: 'every cohomology class has a unique harmonic form'. This what we use in the few applications that I saw: finite dimensionality of de Rham cohomology and Poincaré duality.

*for compact orientable manifolds ofcourse.

user image

@satan29

yeah, same

*and Riemannian

If you use Mayer-Vietoris argument you relax the compactness condition to finite-type.

what does finite type mean here

4:05 PM

It has a finite good cover.

doesn't every manifold?

wait nvm

I don’t know why I’m unable to get the maximum of $\sin^4 x +\cos^4 x$ by derivatives method.

I was thinking every cover can be refined to a good cover and every manifold has a finite cover, but of course cardinality can increase while refining

finite type allows "nice ends"; for example cylindrical ends S^n x (0, infty)

We have for max/min $$\frac{d}{dx} (\sin^4x +\cos^4 x)=0 \\ 4\sin^3 x \cos x - 4\cos^3 x \sin x = 0 $$ oh okay I cannot divide that Eqaution by $\sin x$ beacuse it can be zero

4:08 PM

finite good cover implies finite dimensional Cech cohomology, so that's the obstruction

for example surface of infinite genus is not finite type

i c

I feel like learning Hodge theory in any amount of detail inevitably lands one into technical functional analysis so I never learnt it

Because blackboxing the functional analysis feels like cheating

Without that it's some enhanced linear algebra

It does seem like a useful result in various dualities; I think Serre duality is an immediate corollary from Hodge theory

functional analysis is just enhanced linear algebra

@user69608 okay. so it seems i have some work to do

It's hardcore analysis

Those Sobolev embedding theorems man

4:16 PM

analysis is also just enhanced linear algebra

it's all linear algebra

Also finding harmonic representatives is like some general theme. If $M, N$ are Riemannian manifolds, what's a harmonic map $f : M \to N$? I guess $E(f) = \int \|df \|^2$ is the energy functional?

And energy minimizers are harmonic, something like this

idek why we care about things being harmonic

If $M = S^1$, harmonic maps are exactly geodesics

functional analysis is btfoing me hard

huh, that's interesting

Yeah I don't know exactly what this definition has to do with Laplacians though

6

A: How to actually find a minimizing path on a manifold?

This answer is mostly an attempt at writing down an arbitrary chunk of the theory of variational calculus on the fly while learning it. The two major references this answer is based on are (1) Milnor, "Morse theory" (part III) (2) Oancea, "Morse theory, closed geodesics, and the homology of free ...

You might find this interesting

And then the Hodge theory fact for forms become "in every homotopy class of closed loops you can find a geodesic" in this context which is true for a vast arena of cases

It's true for manifolds with negative curvature, given the homotopy class is nontrivial

4:22 PM

@BalarkaSen Can I ask something to you?

In that Oancea reference you might find the general context where it's true

Ok, @Knight

@BalarkaSen You, when in 12th Grade, knew 2nd year Undergrad Maths, right?

Some things I knew but also some things I didn't

> "should be thought of as an infinite dimensional manifold"

Is there actually a smooth structure on $\Omega_{p, q}$?

>tfw you can't have linebreaks after blockquotes

@feynhat I think it's something you call a Frechet manifold, yeah

charts are modeled on infinite-dimensional Frechet spaces

4:26 PM

@BalarkaSen Were you equally (I mean exceptionally) good at high school problems which come in comptetive exams?

I was shit bad at competitive exams

still am

@BalarkaSen wow

What’s the reason for that? I don’t believe that those questions are hard for ya

Well competitive exam problem are hard for me

If in 9th Grade you could do things with Stirling’s numbers, then how come comptetiev exam questions can be tough for ya

4:28 PM

shrug

this stuff is wild

yeah, @Thorgott, i don't understand it at all

@BalarkaSen Didn’t get that ::shrug::

hey why don't you teach me some Hodge theory one day

also "A 1969 theorem of David Henderson states that every infinite-dimensional, separable, metric Fréchet manifold X can be embedded as an open subset of the infinite-dimensional, separable Hilbert space, H (up to linear isomorphism, there is only one such space).", wild too

4:29 PM

whatever you know I mean; I really don't know anything

I understand so little of Hodge theory

I should read Warner

I'll read a little if you do so

it's like the only reference I was able to find that does Hodge theory and actually has self-contained proofs for all the necessary functional analysis in the appropriate setting

ah ok

I think he presumes Hahn-Banach, but that's harmless

4:32 PM

Mike suggested Wells once.

(or so I say, but I actually never read a proof of Hahn-Banach - I'm assuming it isn't that bad tho)

@Knight @Knight competitive exams (especially indian ones) Rarely test your mathematical ability.

yeah I learnt Hahn-Banach from some notes online once

I think an REU paper

"Differential Analysis on Complex Manifolds"? @feynhat

Yeah.

that's the one.

4:34 PM

That should be the right book

I think I can just find a proof of Hahn-Banach in Folland

in fact, I could probably pick up all the necessary functional analysis I need by reading Folland

so much to read

I'd just read whatever I need to be honest

otherwise its just rabbitholing oneself down

so many things to learn, meanwhile I'm sitting here trying to compute the Laplace-Beltrami operator on the graph of a smooth function in local coordinates

What is $\Bbb Z/3 \ast \Bbb Z/3$?

Looks like the complete bipartite graph $K_{3,3}$.

$\ast$ = join of spaces. Not the free product.

Oh lol

I was like huh?

4:39 PM

Clearly it's $\Bbb Z^2/9$

Yeah it's K_{3, 3}

Good picture

also yo

$N\times G/N=NG/N=G$

Hi @Ed

I solved algebra

4:40 PM

rofl

Hey @Ba

@BalarkaSen Its because you're a dirty algebraist.

damn

cmon man free products are good

such nice geometry

Hi Ignorant Megan.

Hello Fine Hat

Apparently the same prof that is taking alg2 this semester is also teaching alggeo next semester

which means it's gonna be solid

apparently you can use Hodge theory to prove the Peter-Weyl-theorem

Alessandro Codenotti

4:46 PM

@Thorgott The proof you usually see in books is a fairly straightfoward Zorn's lemma argument

@feynhat $K(\Bbb Z_3, 1)$ is actually kind of weird space

Alessandro Codenotti

This is not optimal in the sense that Zorn is equivalent to AC, while ZF+Hahn-Banach is strictly weaker than ZFC (and strictly stronger than ZF), but who cares

You should try to think in terms of $S^\infty$ again though

yeah, I think I saw some MO question on it's strength relative to the ultrafilter lemma or sth

ultrafilter lemma implies Hahn-Banach and reverse is open?

Haha. How did you know I was doing that?

Alessandro Codenotti

4:48 PM

@Thorgott the first is true, I don't know about the latter

Because why else would you take joins of Z/3's?

:P

Anyway when I joined K_3,3 with 3 points, I went blind.

Yeah I wouldn't blame you on that

Alessandro Codenotti

I once asked a question on related topics on MSE, turns out that ZF+every Banach space has nontrivial dual is already strong enough to get Hahn-Banach

I can't find anything about the reverse implication, so it may as well be open

Alessandro Codenotti

4:53 PM

According to wiki it seems weaker

Hahn-Banach gives real valued finitely additive measures on Boolean algebras, but the ultrafilter lemma is equivalent to such measures with values in {0,1}

Here is a better reference than wiki link.springer.com/chapter/10.1007/BFb0066014

Here is your topology fun fact of the day: if $X,Y$ are second countable Baire spaces, then $X\times Y$ is too. (It is false that products of Baire spaces are Baire, and there are about 128397219387 slight strenghtenings of Baire that are in fact productive)

For example, any product of two Baire spaces wearing a monocle and top hat is a Baire space wearing a monocle and top hat

Source: I am a topological haberdasher

smacks Fargle for good measure

5:08 PM

@satan29 That question really made me feel that I still cannot solve high-school tricky problems

Hello Ted

Hello, @Knight

I was never much good at competitive exam questions.

And hello, @Ted

@Fargle Thanks for that :-)

I did once make a 2 on the Putnam---my proudest achievement yet---but only by blind luck.

You know Fargle, the tricky question that I have posted today?

5:09 PM

Competitive exam requires cleverness. I am not clever, I just know stuff

I saw it. Can't do it

Fortunately, I just got here.

Alessandro Codenotti

@Fargle That's way above the average score

@Fargle Really? You couldn’t do it?

@BalarkaSen How was your entrance test? Wasn’t it comptetive

If someone sat down and showed me, I could follow the reasoning, but I'm not good at solving weird stuff off the cuff

5:10 PM

Yes

Quite a bit

Hi, a @Balarka and @Alessandro

@Fargle yeah same

Hi @Ted

@AlessandroCodenotti I got quite fortunate to get asked a question that gelled with my brain.

@Fargle But you’re a very respectable person here for Maths

Alessandro Codenotti

Hi Ted

5:11 PM

@Knight My strengths lie in other places than in solving that kind of problem

@BalarkaSen Oh God Why entrance tests are always competitive?

@Knight yeah I know it's pretty horrible

For example, I'm good at telling jokes that make everyone wish I could be booted from the chat :P

@Fargle I’m your comrade then, sir!

Indeed, @Fargle.

5:13 PM

@BalarkaSen How you managed it Bala? By solving previous years tests?

Yeah basically, I practiced a lot. And I barely managed to scrape it past the cut off

Was good enough for me

Being a clever problem solver has surprisingly little to do with being a quality mathematician.

user image

This is the question I'm fairly certain I got points on.

hmm, that sounds like an interesting question

I would think about this if I didn't have to think about uglier things

Try it. It's pretty fun and not all that hard to get the right answer

5:16 PM

@TedShifrin Then how about my higher education? What should I do Ted?

@Fargle You are making me very very positive, like Freddie Mercury

You’re saying that problem is a fun and I will get it, that’s very encouraging, commander

It does take time to spot it, of course---I saw it by just playing with enough numbers to see what worked and what didn't

Fargle, you have made me alive again

I’m not crazy I literally became so low by failing in that high-school question

Yesterday, I couldn't solve a problem. Today, I am still alive.

4

Alessandro Codenotti

I'm very happy Bonn has a selection process that doesn't involve an olympiad style exam

I have to give GRE if I want to apply outside, and I'm sure that will drive me nuts

5:24 PM

I'm very happy we didn't have a selection process whatsoever

The Heidelberg selection process involves being able to distinguish between squares and circles

I failed

@Balarka: GRE isn't about clever. It's about basic knowledge, especially calculus, multivariable calculus, linear algebra, then a smattering of advanced stuff.

Heidelberg hates topologists?

@TedShifrin Do I not have to be super fast or something

rofl

5:25 PM

Practice helps. Slow isn't going to work too well.

IIRC you don't have to be terribly fast

But the kinds of questions on there won't involve knowledge you don't already likely have

Especially the conceptual ones

@AlessandroCodenotti What is Bonn?

@Fargle i believe N cant have All non-zero digits?

Gotcha. I'll try a practice test one of these days

@EdwardEvans Lol :)

5:27 PM

Ok, it's 66 questions in 3 hours

That doesn't look bad

@satan29 10 doesn't have all non-zero digits, but it fails to have a unique overexpansion: $10 = 10\cdot 10^0 = 1 \cdot 10^1 + 0 \cdot 10^0$

IIRC I finished the math GRE with ample time

do orthonormal charts generally exist on Riemannian manifolds?

How does $\Bbb Z/3$ act on $S^2$?

by which I mean a chart such that the differentials form an orthonormal frame

@feynhat Ah try $S^3$

@Thorgott No

5:30 PM

S^3 in R^4 = C^2, and multiply by cube roots of unity?

Yup, this has a name

I feared as such

man how do I avoid doing computations when I'm trying to do computations

@Thorgott You're basically asking if any Riemannian manifold is locally isometric to $\Bbb R^n$ - there's a tensor which provides a restriction, called Riemann curvature tensor

If it's isometric to R^n it has "0 curvature"

ah, makes sense

Isn't the L-B operator horrible in coordinates

5:33 PM

It's not that bad, as I recall.

IIRC there's weird $1/\sqrt{|g|} g_{ij}$ terms hanging out

I don't understand it but I don't understand many things

Do it in moving frames instead :P

the codifferential in local coordaintes is $\delta\omega=\frac{-1}{\sqrt{\det g}}\sum_{i,j}\frac{\partial}{\partial x^i}(\sqrt{\det g}g_{ij}a_j)$, where $a_j$ are the coordinates of $\omega$

Yeah this is the divergence

got it

so for a function $f$, you get $\Delta f=\frac{-1}{\sqrt{\det g}}\sum_{i,j}\frac{\partial}{\partial x^i}(\sqrt{\det g}g_{ij}\frac{\partial f}{\partial x^j})$

5:37 PM

What is the intuitive meaning of $\sqrt{|g|} g_{ij}$ inside the differential btw

Clearly it's the divergence if $g$ is the Euclidean metric

because none of this stuff inside matters

uhh, the $\sqrt{\det g}$ is the correcting factor you always need to get the volume form and the $g_{ij}$ come from somewhere

I have no clue

it ultimately follows from $\ast dx^j=\sum_i(-1)^{i+1}\sqrt{\det g}g_{ij}dx^1\wedge...\wedge\widehat{dx^i}\wedge...\wedge dx^n$

Oh I see

Laplacian = div grad

grad f = g^{ij} d_i f

div(X) is just 1/sqrt(|g|) d_i(sqrt|g| X_i) always no problem

So that's all

g(grad f, X) = X(f), so that's where g^{ij} in the components of grad f comes from

Natural

ok, I'm gonna go and apply the product rule

you know what happened if I don't come back

What do you want to prove

I don't understand the Laplace de Rham operator though, why d del + del d? del d is div grad for functions

There must be a good reason why this is the correct Laplacian

Write it out in terms of $d$ and $\star$.

5:48 PM

I don't really understand * to be clear

so I don't want to do that

You have to interpret div in terms of $\star$.

Yeah I understand it's (sign) times $\ast d \ast$

Coordinates is not the way to do this.

But for forms it gets weirder for me

Alessandro Codenotti

@Knight a city in Germany, where I study

5:51 PM

@AlessandroCodenotti Are you in Germany or Italy right now?

Weitzenböck identity

In mathematics, in particular in differential geometry, mathematical physics, and representation theory a Weitzenböck identity, named after Roland Weitzenböck, expresses a relationship between two second-order elliptic operators on a manifold with the same principal symbol. (The origins of this terminology seem doubtful, however, as there does not seem to be any evidence that such identities ever appeared in Weitzenböck's work.) Usually Weitzenböck formulae are implemented for G-invariant self-adjoint operators between vector bundles associated to some principal G-bundle, although the precise...

This relates the actual Laplacian with the notion of Laplacian I would come up with, just extending del d to forms

Weird Ricci curvature terms coming in

This is all covariant derivative stuff.

Alessandro Codenotti

6:13 PM

@Knight Italy, I was on a semester break when we went into lockdown due to the virus

Hi guys

yooooo

I finished product rule

the computation actually worked out

We're proud of you, @Thor.

thanks :P

@Balarka something about wanting it to be self-adjoint

6:30 PM

doesn't that follow with no computations

it does, right?

yeah, my point is that $d\delta$ isn't self-adjoint, but $d\delta+\delta d$ is

Oh I thought you were responding to me asking what you're trying to prove

oh lol

what I proved is that if $U\subseteq\mathbb{R}^n$ is open, $0\in U$, $F\colon U\rightarrow\mathbb{R}$ is smooth, $F(0)=0$, $\nabla F(0)=0$, $h\colon\mathbb{R}^{n+1}\rightarrow\mathbb{R}$ is smooth and we interpret $G_F$ als oriented, Riemannian manifold in the canonical way, then $\Delta h(0)=\partial_{n+1}^2h(0)-\operatorname{tr}(H_F(0))\partial_{n+1}h(0)+\Delta h\vert_{G_F}(0)$

Oof

Yuck.

6:35 PM

there should be a more elegant way of solving this, because there's a hint alluding to it, but I didn't get it

but as long as the thickheaded straightforward calculation got me to the goal, I'm good

Anyone know how what the formula for the blue line might be.
I've tried playing around with it in a graphing calculator, but can't generate it using my recollection of high school math.

static.wixstatic.com/media/…

now I'm supposed to use this to show that a homogenous, harmonic polynomial restricts to an Eigenfunction of the Laplace-Beltrami operator on the sphere

@Hal looks like some kind of root function?

in terms of how it grows

It's the vertical tangent that's the clue. Otherwise it could just be logarithmic growth.

good point

Yeah I played with powers and roots. Couldn't make a form that resembled a quarter circle and looked nothing like a right angle. Also I'm not sure how to get it to start and end in the middle of the quadrant (i.e. not start at the axis and continue forever).

Sayan Chattopadhyay

6:46 PM

This seems like a dumb question to ask since the only example I have is the mobius strip but say if L is a line bundle over a manifold M. Let s be the zero section. Then is it right to say that if the self intersection number of s in L is non zero then L is non orientable?

Take sqrt(x-a) where a is the value at the left. Then limit your domain to get as much of the graph as you want.

@Sayan: What does self-intersection mean in general?

Anonymous

7:02 PM

In the proof of Schroder-Bernstein theorem, why can we assume sets $A$ and $B$ are disjoint without loss of generality?

I certainly don't do that. I reduce immediately to the case that $B$ is a subset of $A$.

Anonymous

@TedShifrin Yes I think I know a proof that way, but I'm trying to understand that statement specifically. That is, how are we preserving generality even after assuming disjointness of $A$ and $B$.

Why is it relevant?

Oh, for that proof it is, to get a unique representation.

So take a "copy" of $B$ that is disjoint from $A$. No big deal.

Anonymous

@TedShifrin Umm, what does that mean?

Anonymous

@TedShifrin Oh, I see

Anonymous

7:11 PM

@TedShifrin So basically it proves that there is a possible bijection between $B \setminus A\cap B$ and $A$. But is that enough to ensure that there is a bijection between $A$ and $B$?

Anonymous

I feel like some information and possible bijections are getting lost

Take out the intersection from them both. You can extend the bijection to be the identity on that. But it's abstract, anyhow. Just relabel all those elements as living somewhere totally different.

hi @Astyx

Anonymous

@TedShifrin Aha, that makes sense!

Anonymous

(There's another related concept I find confusing, that is the disjoint union.)

@CalvinKhor I thank you with all my heart with what I have at my disposition. My very best regards.

7:16 PM

Why do you find it confusing?

Anonymous

@TedShifrin I mean, I haven't really seen any concrete application of the disjoint union. I just keep hearing that they can be used in proofs to deal with overlapping sets

There are zillions of applications of disjoint union. For example, you might want to think of the tangent lines at various points of the unit circle, but keep track of which point we're attaching the line to. (Obviously, in $\Bbb R^2$ the lines overlap.)

Anonymous

@TedShifrin Uh, I don't get it. How is disjoint union being used in your unit circle example? I understand that points on the unit circle can be labeled by one parameter, say $ 0 \geq \theta < 2\pi$ or two parameters $(x, y)$ following a constraint.

Anonymous

Oh, perhaps you're taking the unions of the lines at each point of the unit circle

Yes, that's what I'm doing. This is an important construction in differential topology/geometry called the tangent bundle.

Anonymous

7:27 PM

So for each point $\theta$ on the unit circle we can assign an additional parameter say $\delta \in \mathbb R$ such that each $(\theta, \delta)$ represents a point on the corresponding tangent line?

Anonymous

@TedShifrin I see!

Best way to do it is to look at $\bigcup_{x\in S^1} \{x\}\times T_x S^1$, where $T_x S^1$ is the tangent line.

a perhaps more elementary usage is in defining adjunction spaces

or if we wanna stay in the realm of set theory, in defining ordinal and cardinal addition

Anonymous

@TedShifrin I think I'm getting the idea now. So for instance, in the proof of the Schroder-Bernstein theorem would it also make sense to look at bijections between $A \times \{0\}$ and $B \times \{1\}$ (where $A$ and $B$ may be overlapping)? If a bijection exists between $A \times \{0\}$ and $B \times \{1\}$ can we say that a bijection also exists between $A \setminus A\cap B$ and $B \setminus A\cap B$ ?

But then you no longer need to worry about overlaps. The sets $A' = A\times \{0\}$ and $B' = B\times \{1\}$ are necessarily disjoint.

Anonymous

7:37 PM

@TedShifrin Indeed, I get that. But my question is: if a bijection exists between $A'$ and $B'$ is it necessarily the case that a bijection exists between $A\setminus A\cap B$ and $B \setminus A\cap B$ as well?

@satan29 I get $\int_0^{2\pi}\frac{\mathrm{d}x}{(a+b\cos(x))^2}=\frac{2\pi a}{\left(a^2-b^2\right)^{3/2}}$

Anonymous

I'm trying to understand if establishing a bijection between $A'$ and $B'$ is sufficient proof for the CBS.

Yes, but the original one doesn't necessarily restrict to give it to you.

You compose bijections. $A$ is bijective to $A'$, etc.

Anonymous

@TedShifrin Oh. So because $A$ is in bijection with $A'$ and $B$ is in bijection with $B'$, it suffices to show that $A'$ is in bijection with $B'$. It's an indirect proof basically.

Not indirect at all.

Composing bijections is perfectly direct.

"indirect" suggests proof by contradiction or contrapositive.

Anonymous

7:42 PM

@TedShifrin Oh, okay :)

Anonymous

I get your point

Cool :)

Think of it this way. "in bijection with" is an equivalence relation.

Anonymous

@TedShifrin Right, got it now! By any chance do you happen to know of a good elementary set theory textbook that covers these cardinality stuff in detail? I never had a proper set theory class.

I'm not the right person to help with that.

Anonymous

Okay, no problem. Thanks for the help

7:47 PM

I suggest "Naive Set Theory" by Halmos

Anonymous

@Thorgott Ah, thanks. Does it have the ZFC formulation too?

it's entirely in ZFC

Anonymous

Ah nice, I'll get a copy then

8:03 PM

Been thoroughly bamboozled by the paper to do with Sobolev-Lorentz classes in relation to my coarea formula problem haha

I think it's time to email the author haha

I'm trying to find the equation for the 3-manifold based on the phase space of some one dimensional curves...The space of 1D curves are found from the intersections between $\log(x)\log(y)\log(z)=-a$ and $\log(1-x)\log(y)\log(z)=-b$ for parameters $a,b \in(0,\infty)$ and real variables $x,y,z \in (0,1).$

I think that it's possible that there's no equation

8:22 PM

@SayanChattopadhyay That is correct.

@satan29 if you're looking for an indefinite integral, I'd try the Weierstrass substitution $z=\tan(x/2)$.

You should phrase it in terms of the self-intersection class of the zero section, where the base is a closed manifold

In which case yes; that self-intersection class is the Poincare dual of w_1

He probably meant mod 2 anyway

Ah he said number, but the base need not be 1dim. Got it

8:58 PM

When we have a matrix in the form $J_k:=\begin{pmatrix}0 & 1 & 0 & \ldots & \ldots & 0 \\ 0 & 0 & 1 & 0 & \ldots & 0 \\ \vdots & \ddots & \ddots & \ddots & \ddots & \vdots \\ \vdots & \ddots & \ddots & \ddots & 1 & 0 \\ \vdots & \ddots & \ddots & \ddots & \ddots & 1 \\ 0 & \ldots & \ldots & \ldots & \ldots & 0\end{pmatrix}\in M_k(\mathbb{K})$ then the characteristic polynomial is $p(x)=x^k$. Is the minimal polynomial also equal to $m(x)=x^k$ ? But how can we show that?

straightforward computation is one option

see how this matrix acts on the standard basis vectors

We want to show that $k$ is the smallest integer such that $J_k^k=0$, right?
What do you mean by "how this matrix acts on the standard basis vectors" ? @Thorgott

9:20 PM

if $e_i$ is the $i$-th standard basis vector, what is $J_ke_i$?

@Thorgott The result is the $i$-th column of $J_k$, or not?

yes, can you express the $i$-th column of $J_k$ in terms of the standard basis vectors?

Let $c_i$ be the $i$-th column of $J_k$. We have that $c_1=\sum_{i=1}^k0\cdot e_i$ and $c_j=\sum_{i=1}^kd_i\cdot e_i$ where $d_{j-1}=1$ and $d_{i, i\neq j-1}=0$, right? @Thorgott

yes, so, in short, $J_ke_i=0$ for $i=1$ and $J_ke_i=e_{i-1}$ for $i>1$, agree?

9:35 PM

@Thorgott Yes!

now proceed inductively to see what $J_k^je_i$ is in general

So we have that $J_ke_i=0$ for $i=1$ and $J_ke_i=e_{i-1}$ for $i>1$. Then we get for $i=1$ that $J_k^2e_i=0$, for $i=2$ that $J_k^2e_i=J_ke_{i-1}=0$ and $J_k^2e_i=J_ke_{i-1}=e_{i-2}$ for $i>2$. So if we continue we get $J_k^ke_i=0$ for all $i\leq k$.
Is that correct? @Thorgott

yup

now, in particular, what can you say about $J_k^je_k$ for $j<k$?

@Alessandro A group with infinitely many ends decompose as a free product, right?

Alessandro Codenotti

either free product with amalgamation or HNN extension

both over a finite subgroup

This is called Stallings theorem if you want a googlable name

9:48 PM

What's a group with infinitely many ends that's not a free product?

I know Stallings

You're giving the result for e > 1

@Thorgott It is equal to $e_{k-j}$, right?

It should be true for e = infty that it's just a free product

indeed

so can $J_k^j$ be the zero matrix for $j<k$?

Alessandro Codenotti

What about dumb examples like $F_2\oplus\Bbb Z_2$?

Collapse all torsions

I was about to say

Alessandro Codenotti

9:50 PM

Ok so the question is whether a torsion free group with infinitely many ends splits as a free product

Sounds reasonable

Yeah

Also what can you say about groups of 2 ends? HNN usually gives that kinds of groups, no?

You have qi embedded copies of Z in the Cayley graph

No, if it would be then the result of the multiplication with the $k$-th standard basis vector must have been the zero vector and not $e_{k-j}$. Therefore the smallest exponent is $k$, $J_k^k=0$ and therefore the minimal polynomial is equal to the characteristic polynomial $x^k$. Is everything correct? @Thorgott

Alessandro Codenotti

@BalarkaSen yes, groups with two ends are virtually Z

Aha great

Alessandro Codenotti

I sketched a proof in the garbo room for Alex a while back, let me find it

9:53 PM

@MaryStar yeah :)

Great!! Thank you so much for your help!! :-) @Thorgott

Sorry to bother you again Thorgott, but can you just confirm something for me?

maths.ox.ac.uk/system/files/attachments/OxPDE%2016.02.pdf

Alessandro Codenotti

Ah no I was misremembering, I sketched the proof that e>2 implies e=infinity there

Bottom of page 1, they give a coarea formula and $v$ goes to $\mathbb{R}^d$

So where has this $m$ in the coarea formula and the superset of $E$ come from?

Actually I think what I should probably be asking is "whats an $m$-Jacobian" I guess

jstor.org/stable/1970577

This seems to have answers to both questions

Also seems like a nice read so I'll just read it

10:00 PM

some form of approximate Jacobian? I'm not sure tbh

Alright, actually disregard all of the above. Apparently I'm blind and missed them define it later in the paper

"denotes the m–Jacobian of v defined as the product of the m largest singular
values of the matrix ∇v(x)."

yeah ok, that coincides with the usual definition if my linear algebra isn't failing me

Alessandro Codenotti

@BalarkaSen 0.2 seems like a nice result too

yeah

Is this the original Stallings paper? Seems more algebraic

I thought the theorem on ends used purely geometric tools; maybe that's a later proof

Alessandro Codenotti

I'm not sure what the original paper is

This does look more algebraic than I remember from the ggt course though

10:05 PM

Might be better for me because I don't know enough GGT

1 hour later…

11:27 PM

is there any way I can understand stability

in differential equation

pretty confusing definition my book has.

if you can give me link to understand it better please tag me

I am off

« first day (3624 days earlier) ← previous day next day → last day (1401 days later) »

Transcript for

Jul '206

$Mathematics$ Mathematics

Associated with Math.SE; for both general discussion & math qu...

join this room
about this room

00:00

06:00

12:00

18:00

all times are UTC

$Mathematics$

site design / logo © 2024 Stack Exchange Inc; legal

mobile