Mathematics - 2019-07-22 (page 1 of 2)

Here we have to pick courses from three areas (out of the six all courses are divided into), so you need some variety but you can still avoid all of the applied courses

ÍgjøgnumMeg

I think in Heidelberg you pick your specialisation and then just plough into it

but you have to take 2 applied courses

although probability theory counts as applied and according to Mathein it's just measure theory

and functional analysis counts as applied too I believe

Balarka Sen

Meh

Those are actual math

ÍgjøgnumMeg

yeah

As long as I don't have to do lots of statistics or physics rofl

I have to wait until next winter to take AG tho

Balarka Sen

8:11 AM

I have 4 statistics courses in my college

ÍgjøgnumMeg

which is sad

Balarka Sen

I hope they are not terribad

ÍgjøgnumMeg

@Balarka I think I had 8 during my undergrad hahaha

Balarka Sen

Because theoretical statistics is fantastic, people just butcher it when they teach

ÍgjøgnumMeg

and I know literally nothing about it

Balarka Sen

8:11 AM

@ÍgjøgnumMeg Shit

I have 4 physics courses as well, and the first of them was horrible

I still picked up some Lagrangian mechanics anyway

ÍgjøgnumMeg

I had a tiny bit of physics in undergrad, masquerading as vector calculus

we did some celestial mechanics which was cool

Balarka Sen

ah yeah thats good

classical stuff

Kepler's laws

ÍgjøgnumMeg

yeah, and how they discovered Uranus

or Neptune

I can~t remember which

Balarka Sen

classical mechanics is basically ODE theory done badly

ÍgjøgnumMeg

ye

that was it

Balarka Sen

8:13 AM

Just read Arnold or something if you want to learn properly, would be my guess

ÍgjøgnumMeg

Ooo, diff. top. 1 runs from 9am to 11am Mondays and Thursdays and ANT1 is 11am to 1pm Mondays and Thursdays

it's a perfect fit!

Balarka Sen

Nice!!

Alessandro Codenotti

@ÍgjøgnumMeg if you do PDE or something like that it can be quite applied to be fair

ÍgjøgnumMeg

Yeah I did PDEs in undergrad, found it quite fun

Balarka Sen

So you get to face the angel of geometry and the devil of algebra back to back

ÍgjøgnumMeg

8:14 AM

blergh

Balarka Sen

:p

ÍgjøgnumMeg

I'm excited to finally be doing pure maths at uni

instead of applied lol

Alessandro Codenotti

Geometry is only a false prophet, logic is our saviour

Balarka Sen

laughs in HoTT

ÍgjøgnumMeg

NT is Queen

Alessandro Codenotti

8:16 AM

HoTT is quite neat to be honest

Balarka Sen

@ÍgjøgnumMeg Here's a small demonstration regarding how theoretical statistics can be fantastic. Do you know what the Gamma distribution is?

ÍgjøgnumMeg

err just about

Balarka Sen

Let me recall quickly

ÍgjøgnumMeg

brb fast, am at work lol

tho I will listen in when I return

Balarka Sen

A random variable $X$ is said to be exponentially distributed with parameter $\lambda$ if it's pdf is $f(x) = \lambda e^{-\lambda x}$ supported on $x \in [0, \infty)$. Recall that $f(x)$ is the "probability density" that $X = x$; the probability a continuous random variable takes a point as value is 0, but we can define the "proportion" $f(x) = \lim_{\varepsilon \to 0} \frac{1}{2\varepsilon}\Bbb P(X \in (x-\varepsilon, x+\varepsilon))$.

Exponential with parameter $\lambda$ says this probability density is exponentially decaying towards infinity. A very natural random variable; they can arise as stopping times of processes (eg, think of tossing a coin with a massive bias on head very fast and waiting for a head - $X$ gives you the amount of time you need to wait to get the first head)

$\text{Gamma}(n, \lambda)$ is the distribution you get by taking a bunch of independently distributed random variables $X_1, \cdots, X_n$, each $\text{Exp}(\lambda)$, and looking at $X = X_1 + \cdots + X_n$. This is the amount of time you need to wait to get $n$ heads in the earlier interpretation, say.

You can define $\text{Gamma}(\alpha, \lambda)$ for any $\alpha > 0$ real. It's a small work. Their pdf is $f(x) = \lambda^\alpha/\Gamma(\alpha) x^{\alpha - 1} e^{-\lambda x}$.

There's this odd fact which you can check but makes no sense: Let $X \sim \text{Gamma}(\alpha, \lambda)$ and $Y \sim \text{Gamma}(\beta, \lambda)$ be independently distributed. Then $X + Y$ and $X/(X + Y)$ are independently distributed.

One would be inclined to say they are "obviously dependent" because algebraically the expressions seem intertwined

I cannot give an interpretation of this fact without theoretical statistics. One way to explain this: If $Z \sim \text{Exp}(\lambda)$ then $\lambda Z \sim \text{Exp}(1)$. Therefore if $Z \sim \text{Gamma}(\alpha, \lambda)$ then $\lambda Z \sim \text{Gamma}(\alpha, 1)$. Note that $X + Y$ above is $\text{Gamma}(\alpha+\beta, \lambda)$ (think $\alpha$, $\beta$ natural numbers and use the definition for natural numbers I provided)

Then $\lambda X \sim \text{Gamma}(\alpha, 1)$ and $\lambda(X + Y) \sim \text{Gamma}(\alpha+\beta, 1)$.

This means $X/(X + Y) = \lambda X/\lambda(X + Y)$ is independent of the parameter "$\lambda$" in the story. Whereas $X+Y$ is dependent on the parameter $\lambda$.

Basu's theorem

In statistics, Basu's theorem states that any boundedly complete minimal sufficient statistic is independent of any ancillary statistic. This is a 1955 result of Debabrata Basu.It is often used in statistics as a tool to prove independence of two statistics, by first demonstrating one is complete sufficient and the other is ancillary, then appealing to the theorem. An example of this is to show that the sample mean and sample variance of a normal distribution are independent statistics, which is done in the Example section below. This property (independence of sample mean and sample varianc...

Indeed, $X+Y$ is a complete sufficient statistic for $\lambda$ and $X/(X + Y)$ is an ancillary statistic for $\lambda$. That's why they are independent.

ÍgjøgnumMeg

8:35 AM

tf

Balarka Sen

yup

The proof, as you see, is measure theory

ÍgjøgnumMeg

Yeah, this seems like something I'd like to learn about

all of my undergrad was applied stats

well some of it

we did some statistical inference

maximum likelihood estimation etc.

Balarka Sen

aha

Alessandro Codenotti

There was some similar black magic involved in the independence of estimators for the variance of a normal distribution iirc

ÍgjøgnumMeg

but it wasn't theoretical

"Proof" was a bit of a taboo word among my coursemates rofl

Alessandro Codenotti

8:38 AM

Oh it's actually the example on wiki!

Balarka Sen

@Alessandro The sample mean and sample variance thing, yeah?

Alessandro Codenotti

Yes

Balarka Sen

There's a totally ad hoc proof using big matrices

It doesn't explain anything lol

Alessandro Codenotti

I don't remember if we even proved it in the course I took

Balarka Sen

All these non-obvious independence results can be derived as a corollary from Basu's theorem

Alessandro Codenotti

8:40 AM

But by looking at their definitions it seems magical that they are independent

ÍgjøgnumMeg

Oo there's a seminar on Quadratic Forms too

Balarka Sen

Also, nobody will explain to you why the sample variance has an $n-1$ in the denominator and not an $n$

ÍgjøgnumMeg

ahh that was explained to us in the stats course

but I don't remember it

Balarka Sen

The point is that if you do it with $n$, it'll be a biased estimator. This can be checked

ÍgjøgnumMeg

that'll be it

we had to prove it was biased with $n$

Balarka Sen

8:42 AM

Aha

ÍgjøgnumMeg

here's the list of seminars lol

Balarka Sen

Looks nice

I'd attend hyperbolic geometry lol

ÍgjøgnumMeg

riemann surfaces would be cool

and affine algebraic groups

Balarka Sen

oh that's what flachen means

ÍgjøgnumMeg

Yeah

Funny thing: a Klein bottle was originally a Klein surface and was mistranslated as bottle into English

because bottle is "Flasche" and surface is "Flaeche"

which sound very similar

Balarka Sen

8:47 AM

Oh

Alessandro Codenotti

There's a story that the Klein bottle is called a bottle because someone translated Fläche as Flasche

ÍgjøgnumMeg

Sniped

Balarka Sen

too late

Alessandro Codenotti

And I got sniped while looking for an umlaut

ÍgjøgnumMeg

hahaha

ae my man

ae

Balarka Sen

8:49 AM

I guess Flasche is like

flask

Aight I gotta go

Cya all

Alessandro Codenotti

Bye

ÍgjøgnumMeg

Cya @Balarka

So 3 lectures or 2 lectures and a seminar

idk I gotta decide

abenthy

10:11 AM

Hi, is someone knowledgeable about orders in associative algebras?

0

Q: Commensurability of groups associated to orders in $\mathbb{Q}$-algebras

Let $A$ be a finite-dimensional $\mathbb{Q}$-algebra and let there be given two $\mathbb{Z}$-orders $O_1, O_2 \subset A$. It is known that the unit groups $(O_1)^\times$ and $(O_2)^\times$ are commensurable, that is, their intersection has finite index in both of them. Does this also hold when I...

abstract-algebra

Leaky Nun

12:11 PM

hi @BalarkaSen

Balarka Sen

\o

Leaky Nun

is EG def. retr. to a pt.?

Balarka Sen

what is your definition of a deformation retract

there is a strong notion, which is keeping the subspace you're retracting to fixed for all times

Leaky Nun

id homotopic to constant rel pt

Balarka Sen

there is a weak notion, which doesn't demand that

Leaky Nun

12:13 PM

the strong one I guess

Balarka Sen

gotcha

it should be but i haven't put thought to it.

Leaky Nun

because in my notes it is noted that the given homotopy is not a def. ret.

Balarka Sen

your model of EG is that delta-complex thing?

Leaky Nun

right

Balarka Sen

right, the natural thing to do is to slide everything along the line joining it with the identity

that's not a strong deformation retract

Leaky Nun

12:18 PM

that's not good

Balarka Sen

who cares my man

Leaky Nun

I feel like these infinite-dimensional CW complexes are cheating

ok so $S^\infty$

one 0-cell, one 1-cell, two 2-cells, two 3-cells, two 4-cells, two 5-cells, etc?

Balarka Sen

ye

you can make them all 2 if you want

Leaky Nun

one 0-cell, one N-cell, two (N+1)-cells, two (N+2)-cells, ...

for any N

Balarka Sen

sure

Mike Miller

12:24 PM

Every contractible CW complex is contractible relative to a point

You can subdivide if necessary so the point is a 0-cell at which point this is the relative Whitehead theorem

Balarka Sen

i felt it should be true

the only example i know where it's contractible but not def rets to a point is

like

the zig zag infinite comb lol

Leaky Nun

topology is just a bunch of approximations

Alessandro Codenotti

I thought that's analysis

@BalarkaSen there's a couple exercises in Hatcher about it iirc

Balarka Sen

ya

Leaky Nun

@AlessandroCodenotti how about cellular approximation

@BalarkaSen is $S^{\omega_1}$ contractible?

Balarka Sen

12:38 PM

I don't give a shit man lmao

first you have to tell me what it means. is there are $\omega_1$-dimensional Hilbert space?

Leaky Nun

how about the unit sphere in $L^\infty(\Bbb R)$

Balarka Sen

Yes.

Leaky Nun

@BalarkaSen surely $\Bbb R^{\omega_1}$ is one?

or maybe take countable support

Balarka Sen

unit sphere in any separable Hilbert space is homeomorphic to the cellular $S^\infty$

IIRC

you can take a countable flag of closed subspaces to get the filtration by $S^n$'s

Also just use the shift map

Leaky Nun

how about the bounded functions $\Bbb R \to \Bbb R$

Mike Miller

12:41 PM

@LeakyNun why do you want to know

Leaky Nun

haha

Perturbative

If $A$ is an abelian group what does the notation $\operatorname{Aut}_{\mathbb{Z}}(A)$ usually denote? Some modification to the automorphism group of $A$ perhaps?

Leaky Nun

@Perturbative just Aut(A)

Z means that A is a Z-module

Mike Miller

balarka and i should ask that before answering any of your technical questions imo

Perturbative

Well that notation is kinda redundant then

Leaky Nun

12:42 PM

yeah no

Perturbative

Thanks @LeakyNun

Why not?

Mike Miller

"yeah no" I think means "yeah, it's terrible"

Leaky Nun

I don't know what I meant

Perturbative

Oooh

Mike Miller

it's the opposite of the california convention, in which "no yeah" means "no, it's terrible, you're right"

and yeah no means "haha, that's a no from me"

the only reason they would use that notation is if the groups $A$ are more naturally modules over some other ring, like $\Bbb R$ or something

Leaky Nun

12:44 PM

I don't think we ever care about the automorphism group of a module

at least I've never heard of it... wait...

Galois theory

oh no that's the aut group of an algebra

Balarka Sen

@Perturbative The notation is useful when you have two rings flying around, one is an algebra over the other, and $A$ is a module over one of 'em

Leaky Nun

@MikeMiller have you heard about automorphism group of a module?

Perturbative

@LeakyNun They're popping up in the background material to some spectral sequences I was looking at

Leaky Nun

oh right, modular representation theory

Perturbative

Homology with twisted coefficients specifically

Balarka Sen

12:45 PM

Think of a $\Bbb C$-vector space. By restricting scalars, it's also an $\Bbb R$-vector space.

Mike Miller

@LeakyNun this came out of nowhere

Perturbative

Ahh I see @BalarkaSen

Mike Miller

certainly I've heard of automorphism groups of modules, for instance famously there's $GL_n$

Balarka Sen

lol

Leaky Nun

aha

or just GL

Mike Miller

12:46 PM

modular representation theory is about representations of finite groups $G$ over fields $\Bbb F_p$ with $p \mid |G|$

Leaky Nun

and we use End_R(M) there

which is... not the automorphism group

Alessandro Codenotti

@BalarkaSen $\ell^2(\omega_1)$ (this works for any cardinal)

Balarka Sen

Ah.

Strange business

Leaky Nun

@AlessandroCodenotti so just $l^2(\Bbb R)$?

Alessandro Codenotti

It's the same as $\ell^2(\Bbb N)=\ell^2(\omega)$, everything needs to have countable support to be square summable, the inner product is defined in the same way, $\langle x,y\rangle=\sum x_i\overline{y_i}$, there's just more basis vectors

@LeakyNun If you assume $\mathsf{CH}$, sure

Mike Miller

12:55 PM

i assume you're going to need choice to get a schauder basis

Alessandro Codenotti

I don't think there can be any Schauder basis, this space is not separable

Mike Miller

Is there no correct analogue for non-separable spaces

I don't really care if it's named Schauder

Balarka Sen

shudder

thats what i am doing looking at this convo

so you might say i have a basis for shuddering

brb gonna kms

Alessandro Codenotti

Not that I know of, but I don't know anything concerning non separable Banach spaces apart from occasional encounters with $L^\infty$

Mike Miller

I only want Hilbert!

Banach spaces don't even all have bases

Alessandro Codenotti

1:01 PM

Oh ok, the definition of Schauder basis just works the same in Hilbert and Banach spaces

Mike Miller

All I'm asking is "Does every Hilbert space have a basis in an appropriate sense"

Alessandro Codenotti

Then yes, they all have an orthonormal basis

(and an Hamel one, just like every vector space, but those are pretty much useless)

(both of this statements use choice)

Mike Miller

ok, in what sense, since I an now to understand schauder basis is the wrong notion?

sure they're all orthonormal, but beyond that you just mean linearly independent and closure is $H$?

Alessandro Codenotti

Yes

closure of their finite linear combinations

Mike Miller

and do you have any uniqueness statement about sum representations?

Alessandro Codenotti

1:07 PM

Hilbert spaces are classified by dimension, if $\kappa$ is the cardinality of an orthonormal basis of $H$ then $H$ is isometrically isomorphic to $\ell^2(\kappa)$

Balarka Sen

It feels like if you demand closure of finite lin combos is full $H$ then it's separable, but maybe I am wrong? Can't you pick a countably many of these guys which would span a dense subspace

Mike Miller

so you're saying yes, there is a uniqueness statement. why is this not what's already called a schauder basis

@BalarkaSen Finite lin combos of an uncountable set though

Balarka Sen

Mm I see.

Alessandro Codenotti

@MikeMiller I don't know :P

Mike Miller

Okay, I will resign to being confused

Alessandro Codenotti

1:11 PM

Maybe that's why having a Schauder basis is way more interesting for Banach spaces, everything works well anyway in Hilbert spaces

Mike Miller

Well that's certainly true already in the separable case

Perturbative

Lets say I have a group $G$ and an abelian group $A$ and a representation $\rho : G \to \operatorname{Aut}_{\mathbb{Z}}(A)$ It's stated that $\rho$ endows $A$ with the structure of a left $\mathbb{Z}G$ module by taking the action

$$\left( \sum_{g \in G}m_gg\right)\cdot a = \sum_{g \in G}m_g\rho(g)(a)$$
But this action isn't clear to me, for example it isn't stated what is meant by $m_g$ and it seems like we have a possibly infinite (depending on the order of $G$) sum in the brackets whereas $\mathbb{Z}G$ has only finite sums.

Alessandro Codenotti

Yes, any orthonormal basis of a separable Hilbert space is also a Schauder basis

Mike Miller

I don't even know a Schauder basis for $L^p$, $1 < p \neq 2 < \infty$

Alessandro Codenotti

So I guess that for Banach spaces having a Schauder basis is somehow behaving "like an Hilbert space"

@MikeMiller I don't even know if one exists

Mike Miller

1:13 PM

@Perturbative Think this through. Your reference is telling you that there is an action of $\Bbb Z[G]$. Therefore whatever they write down to prove that... will be using an element of $\Bbb Z[G]$. So you conclude that $m_g$ is an assignment of an integer to each element of $G$, with only finitely many nonzero. That is, you conclude that the term in parentheses is your author writing down a generic element of $\Bbb Z[G]$.

Perturbative

Ahh sorry I was implicitly assuming the $m_g$ was non-zero

That makes sense

Alessandro Codenotti

wiki says that there are Schauder basis for $L^p([0,1])$ en.wikipedia.org/wiki/Schauder_basis#Examples

Mike Miller

Haar system seems to be rescaled box-looking functions

Martin Sleziak

BTW I have collected some references for "every Hilbert space is isomorphic to $\ell_2(A)$ for some set $A$ in this answer: Does uncountable summation, with a finite sum, ever occur in mathematics?

Although I see you've already moved away from Hilbert spaces.

@MikeMiller I have asked about this kind of basis here: Are uncountable “Schauder-like” bases studied/used?

But I won't be able to tell you much about the uncountable case, either.

Mike Miller

I saw that earlier. I didn't link it because there wasn't much conclusive in the 0 answers :)

I guess what I missed when reading wikipedia is the statement that the basis elements form a sequence, by which they mean are indexed by $\Bbb N$. This is I think a very unnatural restriction.

This is more a notational complaint than anything else.

user193319

1:56 PM

Where $a,b \in \Bbb{C}$ and $t \in [0,1]$, how does one compute $\int_{0}^{1} [tb + (1-t)a]^ndt$? My thought was to do a u-sub., say $u = tb + (1-t)a$. But the problem is that the bounds will change to complex numbers, and I haven't learned how to deal with that case.

Am I supposed to use the binomial theorem?

Leaky Nun

2:07 PM

just use complex numbers

user193319

What do you mean?

Complex numbers are the problem.

Leaky Nun

just evaluate it formally

user193319

So, use the binomial theorem?

Leaky Nun

no, change the bounds to complex numbers

user193319

But I haven't learned how to compute such integrals yet.

Leaky Nun

2:13 PM

just evaluate it formally

ÍgjøgnumMeg

you can get a reduction formula for it!

I think

@Alessandro what's Geometric group theory like?

Alessandro Codenotti

2:32 PM

Neat

Is there a GGT course being offered in Heidelberg?

ÍgjøgnumMeg

Yeah

Die geometrische Gruppentheorie beschäftigt sich mit dem Zusammenhang zwischen Gruppenwirkungen auf geometrischen Objekten und algebraischen Eigenschaften der Gruppe. In der Vorlesung werden endlich erzeugte Gruppen betrachtet. Einer endlich erzeugten Gruppe kann man einen Graphen, den sogenannten Cayley-Graphen zuordnen, auf dem die Gruppe wirkt. Aus dem Studium dieses Graphen lassen sich interessante Eigenschaften der Gruppe zeigen.

is the course description

Im Zentrum der Vorlesung werden freie Gruppen und hyperbolische Gruppen stehen. Die Methoden sind sowohl geometrischer als auch algebraischer Natur.

Alessandro Codenotti

So do you know how the Cayley graph of a group is defined?

ÍgjøgnumMeg

Yeah I believe so

we looked at the Cayley graph of the free group on two generators at a sumemr school I attended lol

Alessandro Codenotti

Do you also know how the word metric works?

ÍgjøgnumMeg

Nooope

$\lvert g \rvert$ is the shortest length of a word which evaluates to $g$

which is the word norm

and then the distance between $g$ and $h$ is the norm of $gh^{-1}$

okay

Alessandro Codenotti

2:39 PM

Anyway it looks like you're going to look at Cayley graphs and hyperbolic groups, so you'll probably talk about ends of groups and see results like "a group can have either $0,1,2$ or infinitely many ends" or "a group with two ends is virtually $\Bbb Z$" (both by Freudenthal-Hopf) and Stallings theorem (a group with infinitely many ends splits as either a free product with amalgamation or as an HNN extension)

@ÍgjøgnumMeg Right so that's the starting point, you give the Cayley graph a metric, so that now you have a geometric object associated to every group

However if you look for example at $\Bbb Z$ with generating set $\{1\}$ the Cayley graph looks like a line, while if you look at $\Bbb Z$ with generating set $\{2,3\}$ the Cayley graph looks like a braid

ÍgjøgnumMeg

I see

so the Cayley graph is determined by the generating set?

rather than by the group itself

Alessandro Codenotti

This is because we're looking at the fine structure, while GGT is concerned with the coarse/large scale structure, for example even without knowing the definition it should be clear that both those Cayley graphs have two ends

ÍgjøgnumMeg

sure

Alessandro Codenotti

@ÍgjøgnumMeg Yes, to fix this one introduces a notion of "quasi-isometry" which is a very coarse notion, turns out that all Cayley graphs of the same finitely generated group are quasi isometric

ÍgjøgnumMeg

nise

Alessandro Codenotti

2:45 PM

So we can speak about the Cayley graph as long as we're interested in properties which are invariant under quasi-isometries

ÍgjøgnumMeg

well, sounds interesting, but I think I should take Diff Top instead of that

Alessandro Codenotti

GGT is somewhat niche, diff top will surely be more useful generally speaking

ÍgjøgnumMeg

yeah sure, looks like something interesting though

maybe I'll take it in later years or smth

Do you know what adic spaces is about?

Alessandro Codenotti

Nope, I have no idea what an adic space even is

Mathei knows most likely

ÍgjøgnumMeg

Me neither lol

Ryan Unger

2:48 PM

oh lord

ÍgjøgnumMeg

Yeah definitely

or does @Ryan know?

Ryan Unger

no

number theory is for nerds

Alessandro Codenotti

@RyanUnger lol

ÍgjøgnumMeg

rofl, googling it leads to rigid analytic spaces

Alessandro Codenotti

Do you know about amenable groups? People call it geometric group theory but it's all functional analysis in disguise @Ryan

ÍgjøgnumMeg

2:50 PM

Lol I think I'll need to take some AG before I take that so I'll leave it for now

Ryan Unger

there were ggt people at my undergrad so I've heard about it

amenability shows in up in geometric analysis for some reason

I don't remember the paper rn

it was some gmt paper lmao

Alessandro Codenotti

Oh that's unexpected

Ryan Unger

let me do some sleuthing one moment

ÍgjøgnumMeg

Hey @Mathein :P

MatheinBoulomenos

HI @ÍgjøgnumMeg

ÍgjøgnumMeg

2:53 PM

Henlo

Alessandro Codenotti

I've been reading about group C*-algebras lately, turns out that a (discrete) group is amenable iff its group C*-algebra is nuclear

Ryan Unger

ok so the first result that I remember is if $M$ is compact, locally conformally flat and has negative scalar curvature, then $\pi_1M$ is not amentable

very random

that's not the cool theorem

but its proof uses the cool theorem

ÍgjøgnumMeg

@Mathein did you see that the Vorlesungsverzeichnis is available?

MatheinBoulomenos

yeah

Ryan Unger

link.springer.com/article/10.1007%2FBF02566228

there

Tobias Kildetoft

2:55 PM

@AlessandroCodenotti I remember trying to learn about hyperbolic groups at one point, since it seemed like there might be a way to generalize some pseudo-Anosov stuff to them. But I got lost early on, so I never got anywhere

Ryan Unger

gah I don't have access to that journal

Alessandro Codenotti

@TobiasKildetoft We talked about hyperbolic groups in the GGT course, but I have no idea what pseudo Anosov means :P

Ryan Unger

yeah $\lambda_0(\tilde M)$ iff $\pi_1M$ is amenable

@BalarkaSen might like that one

Balarka Sen

Sounds fun

I don't know much about amenability though

Alessandro Codenotti

What's $\lambda_0(\tilde{M})$?

Tobias Kildetoft

2:58 PM

@AlessandroCodenotti It is something completely unrelated actually (that I could not give even a vague definition of any longer. I attended a master class that happened to be all about it)

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$Mathematics$ Mathematics