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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
 
which is sad
 
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?
 
err just about
 
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
flask
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
Q: Commensurability of groups associated to orders in $\mathbb{Q}$-algebras

abenthyLet $A$ 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.
Strange business
 
@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
 
@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
 
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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