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1:48 AM
a convergent series whose sub series doesnot converge?

@BAYMAX $1-\frac12+\frac13-\frac14+\dotsb$
(which equals $\ln2$)
Or, more simply, $1-1+\frac12-\frac12+\frac13-\frac13+\frac14-\frac14+\dotsb$, which equals $0$

yaaa\

They can't all be positive
It can't be absolutely convergent

ya a pinch of absolutenes
yess
absolute convergent iff every sub series converges
thanks Akiva!

6 hours later…
7:38 AM
@Mathein yey Vorlesungsverzeichnis ist veroeffentlicht worden

8:03 AM
So what are you going to take?

Algebraic Number Theory 1
Modular Forms
and maybe one other
there is a course on Adic Spaces
or differential topology 1

Cool shit

Sounds cool

yeeee
Idk how many lectures I'm supposed to take per semester
don't wanna overload myself, especially as I've been out of the game for a year
idek what adic spaces are

@ÍgjøgnumMeg enough to finish in 4 semester, but I'd be very careful if you want to take more

8:08 AM
ANT 1 + MF + DF 1 sounds like a good combination to me tbh

Yeah that seems like a good idea
@Alessandro right, I have to take some applied courses too :(

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

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

Meh
Those are actual math

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

8:11 AM
I have 4 statistics courses in my college

I hope they are not terribad

@Balarka I think I had 8 during my undergrad hahaha

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

and I know literally nothing about it

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

I had a tiny bit of physics in undergrad, masquerading as vector calculus
we did some celestial mechanics which was cool

ah yeah thats good
classical stuff
Kepler's laws

yeah, and how they discovered Uranus
or Neptune
I can~t remember which

classical mechanics is basically ODE theory done badly

ye
that was it

8:13 AM
Just read Arnold or something if you want to learn properly, would be my guess

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!

Nice!!

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

Yeah I did PDEs in undergrad, found it quite fun

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

8:14 AM
blergh

I'm excited to finally be doing pure maths at uni
instead of applied lol

Geometry is only a false prophet, logic is our saviour

laughs in HoTT

NT is Queen

8:16 AM
HoTT is quite neat to be honest

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

Let me recall quickly

brb fast, am at work lol
tho I will listen in when I return

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 $\lambda$.
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.

8:35 AM
tf

yup
The proof, as you see, is measure theory

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.

aha

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

but it wasn't theoretical
"Proof" was a bit of a taboo word among my coursemates rofl

8:38 AM
Oh it's actually the example on wiki!

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

There's a totally ad hoc proof using big matrices
It doesn't explain anything lol

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

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

8:40 AM
But by looking at their definitions it seems magical that they are independent

Oo there's a seminar on Quadratic Forms too

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

ahh that was explained to us in the stats course
but I don't remember it

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

that'll be it
we had to prove it was biased with $n$

8:42 AM
Aha

here's the list of seminars lol

Looks nice
I'd attend hyperbolic geometry lol

riemann surfaces would be cool
and affine algebraic groups

oh that's what flachen means

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

8:47 AM
Oh

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

Sniped

too late

And I got sniped while looking for an umlaut

hahaha
ae my man
ae

8:49 AM
I guess Flasche is like
Aight I gotta go
Cya all

Cya @Balarka
So 3 lectures or 2 lectures and a seminar
idk I gotta decide

1 hour later…
10:11 AM
Hi, is someone knowledgeable about orders in associative algebras?
0

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...

2 hours later…
12:11 PM
hi @BalarkaSen

is EG def. retr. to a pt.?

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

id homotopic to constant rel pt

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

12:13 PM
the strong one I guess

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

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

your model of EG is that delta-complex thing?

right

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

12:18 PM
that's not good

who cares my man

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?

ye
you can make them all 2 if you want

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

sure

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

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

topology is just a bunch of approximations

I thought that's analysis
@BalarkaSen there's a couple exercises in Hatcher about it iirc

@AlessandroCodenotti how about cellular approximation
@BalarkaSen is $S^{\omega_1}$ contractible?

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?

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

Yes.

@BalarkaSen surely $\Bbb R^{\omega_1}$ is one?
or maybe take countable support

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

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

12:41 PM
@LeakyNun why do you want to know

haha

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?

@Perturbative just Aut(A)
Z means that A is a Z-module

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

Well that notation is kinda redundant then

12:42 PM
yeah no

Thanks @LeakyNun
Why not?

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

I don't know what I meant

Oooh

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

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

@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

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

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

oh right, modular representation theory

Homology with twisted coefficients specifically

12:45 PM
Think of a $\Bbb C$-vector space. By restricting scalars, it's also an $\Bbb R$-vector space.

@LeakyNun this came out of nowhere

Ahh I see @BalarkaSen

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

lol

aha
or just GL

12:46 PM
modular representation theory is about representations of finite groups $G$ over fields $\Bbb F_p$ with $p \mid |G|$

and we use End_R(M) there
which is... not the automorphism group

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

Ah.

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

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

12:55 PM
i assume you're going to need choice to get a schauder basis

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

Is there no correct analogue for non-separable spaces
I don't really care if it's named Schauder

shudder
thats what i am doing looking at this convo
so you might say i have a basis for shuddering
brb gonna kms

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

I only want Hilbert!
Banach spaces don't even all have bases

1:01 PM
Oh ok, the definition of Schauder basis just works the same in Hilbert and Banach spaces

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

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)

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 $H$?

Yes
closure of their finite linear combinations

and do you have any uniqueness statement about sum representations?

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 $\ell^2(\kappa)$

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

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

Mm I see.

@MikeMiller I don't know :P

Okay, I will resign to being confused

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

Well that's certainly true already in the separable case

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.

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

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

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

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 $\Bbb Z[G]$.

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

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

Haar system seems to be rescaled box-looking functions

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.

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.

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?

2:07 PM
just use complex numbers

What do you mean?
Complex numbers are the problem.

just evaluate it formally

So, use the binomial theorem?

no, change the bounds to complex numbers

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

2:13 PM
just evaluate it formally

you can get a reduction formula for it!
I think
@Alessandro what's Geometric group theory like?

2:32 PM
Neat
Is there a GGT course being offered in Heidelberg?

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.

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

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

Do you also know how the word metric works?

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

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

I see
so the Cayley graph is determined by the generating set?
rather than by the group itself

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

sure

@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

nise

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

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

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

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?

Nope, I have no idea what an adic space even is
Mathei knows most likely

Me neither lol

2:48 PM
oh lord

Yeah definitely
or does @Ryan know?

no
number theory is for nerds

@RyanUnger lol

rofl, googling it leads to rigid analytic spaces

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

2:50 PM
Lol I think I'll need to take some AG before I take that so I'll leave it for now

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

Oh that's unexpected

let me do some sleuthing one moment

Hey @Mathein :P

HI @ÍgjøgnumMeg

2:53 PM
Henlo

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

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

@Mathein did you see that the Vorlesungsverzeichnis is available?

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

gah I don't have access to that journal

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

yeah $\lambda_0(\tilde M)$ iff $\pi_1M$ is amenable
@BalarkaSen might like that one

Sounds fun
I don't know much about amenability though

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

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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