Mathematics - 2020-09-22

1:29 AM

@BalarkaSen I don't know if that is necessarily true. German wikipedia says that that the random variable that determines the number of dirac measures in the simple representation of a point process can take infinity as a value, contradicting english wikipedia

user2103480

1:51 AM

Yup, we need the index-variable to be possibly infinite, see Corollary 6.5 in "Lectures on the Poisson Process" by Last & Penrose

robjohn

5:16 AM

@orientablesurface the complement of the Cantor Set is open, and, in $\mathbb{R}$, an open set is the union of countably many open intervals. The Cantor Set is not the union of countably many intervals. Which is pretty much what you've said.

Balarka Sen

@user2103480 I'm reading Last & Penrose. Note carefully the measure I started with is a probability measure.

The index variable you can take to be just Poisson

ShankRam

5:29 AM

When proving theorems about linear codes, can we always assume that the generator matrix is in standard form?

S. Das

7:11 AM

0

Q: Understanding the proof of Joy of Cats Theorem 12.13

This question is regarding a step in the proof of Theorem 12.13 as stated in Joy of Cats. The theorem says, Every co-wellpowered cocomplete category with a separator is wellpowered (see this) and complete. In the course of the proof, it is shown that if $\bf{A}$ be a category which is both co-w...

category-theory

Any help is appreciated.

Anonymous

8:18 AM

In the context of the extended real number line, does it make sense to say that the limit of the sequence $\infty, \infty, \infty, \ldots$ is $\infty$?

Leaky Nun

8:32 AM

yes

Anonymous

@LeakyNun Oh, but how does the $\epsilon-\delta$ definition work? $\infty - \infty$ is undefined I guess

Anonymous

Things like $|\infty - \infty| < \epsilon$ for all $n > N$ won't work

Leaky Nun

@S.D. if you want to put a metric on the extended real number line, you can firstly contract the original metric on the real number line so that $d(x,y) < 1$ for all real $x$ and $y$

and then declare $d(x, \infty) = 1$

and then you can use the epsilon-delta definition

also, the extended real number line (if you add in both infinities) is homeomorphic to $[0,1]$

Anonymous

@LeakyNun Ummm, actually in the context in which I'm working, I need to preserve the standard metric on $\mathbb R$

Leaky Nun

then you can't use the epsilon-delta definition I guess

Anonymous

8:45 AM

I guess I need to approach the problem differently. A sequence of $\infty$'s won't work

Leaky Nun

just use the open sets

neighbourhoods of $x \in \Bbb R$ are $(x-\varepsilon, x+\varepsilon)$, hence the epsilon part

but neighbourhoods of $\infty$ are $(M, \infty]$, so you need to use the $M$-$N$ definition of limit

Anonymous

Yeah, I will think about it. Thanks a lot for the help!

Leaky Nun

i.e. $(\lim_{i \to \infty} x_i = \infty) \equiv (\forall M \exists N \forall i : i \ge N \implies x_i \ge M)$

Balarka Sen

@Alessandro I found a post-rock band called "Lakes of Wada". First album is "Star-finite Complexes"

Has to be a troll

Seems like great music

Anonymous

@LeakyNun Oh, I see your point. Yes, if we use the M-N definition of limit, it makes sense that $\infty$ is the limit of $\infty, \infty, \infty, \ldots$. :)

Leaky Nun

8:54 AM

just use filters

Anonymous

@LeakyNun Filters?

Anonymous

Oh there seems to be a Wikipedia page: en.wikipedia.org/wiki/Filters_in_topology

Anonymous

I don't know about this, interesting

Alessandro Codenotti

9:12 AM

@BalarkaSen lol I'll listen to them while doing some topology later

Balarka Sen

Actually they don't have an album just some tracks

I tried to spook them in their youtube channel

New EP coming in Oct 9 tho

Astyx

10:33 AM

Why are the $p\Bbb Z\subset \Bbb Z$'s for prime $p$ not irreducible components ?

Leaky Nun

what do you mean irreducible components

what's the topology

Zariski

so you mean Spec Z

Yes

irreducible components correspond to minimal primes right

Astyx

10:34 AM

The definition I had is maximal closed subsets that cannot be the union of two proper closed subsets

Leaky Nun

yes, and what I said is a theorem

anyway if you want to use the definition directly, consider the bigger closed subset Spec Z

Astyx

So I can't write $\operatorname{Spec} \Bbb Z = \{p\Bbb Z\} \cup \{q\Bbb Z | q\ne p\}$ ?

Leaky Nun

the second one isn't closed

Astyx

I mean, why is$\{q\Bbb Z | q\ne p\}$ not closed ?

Leaky Nun

because the generic point (0Z) is dense (closure is the whole space, i.e. every nonempty open contains it)

Astyx

10:41 AM

Oh ok

Thank you !

Secret

11:08 AM

blogs.adelaide.edu.au/maths-learning/2015/08/07/…

There was once upon a time back in high school I did made that $\sin (x+y) = \sin x + \sin y$ mistake. Good times

Alessandro Codenotti

this one? @Balarka

Secret

11

A: Does the zeroth root exist?

The $n^\text{th}$ root of a real number $x$ is $$x^{1/n}$$ If $n=0$ then $1/0$ is undefined, so there is no such thing as the $0^{\text{th}}$ root.

not defined for the same reason why you cannot divide by zero in any semirings

The map $f : x \to y$ is not injective

I think this undefineness holds for any noninjective maps $f$ whose kernel is the same as its domain

The function $y=x^0$ is also less behaved than $y=0x$ since the later it is continuous for all $x$ while $y=x^0$ is discontinuous at $x=0$

user2103480

11:58 AM

@BalarkaSen Ah yes true. But I'm not sure whether your example of throwing countably many darts is an accurate description. For the whole space, the probability that n darts are in the space is given by the poisson distribution, so almost surely we only throw finitely many darts onto X, or am I missing something

Balarka Sen

12:16 PM

@user2103480 Valid point. But I think this is OK if the measure on X is $\sigma$-finite, no?

Secret

Poisson random measure

Let ( E , A , μ ) {\displaystyle (E,{\mathcal {A}},\mu )} be some measure space with σ {\displaystyle \sigma } -finite measure μ {\displaystyle \mu } . The Poisson random measure with intensity measure μ {\displaystyle \mu } is a family of random variables { N A...

hmm interesting

so the measure used in the sum is of a special form

so they are not things with compact support as I originally thought

But the one Balarka mentioned seemed to be a sum of dirac deltas, which has compact support

tatan

1

Q: Is $([0, \sqrt 2] \cap \mathbb Q) \subset \mathbb Q$ closed, bounded, compact?

As far as I can tell it is bounded, as it's within $[0, \sqrt 2]$, and is closed as there cannot be an open neighbourhood about 0, and as it's closed and bounded it is therefore compact. However I'm not sure if closed and bounded imply compact in this situation, as I've only ever used this proper...

general-topology metric-spaces

Regarding the second comment on the accepted answer by Stephan... Shouldn't $I$ be $[0,\sqrt2]$ instead of $[0,\sqrt2)$?

Alessandro Codenotti

It makes no difference when you intersect it with $\Bbb Q$

tatan

But see this line - "But then q∈I as I is closed".. how do we directly say $I$ is closed?

Thorgott

12:33 PM

yeah, should be $[0,\sqrt{2}]$

the entire point of the example, though, is that $[0,\sqrt{2}]\cap\mathbb{Q}=[0,\sqrt{2})\cap\mathbb{Q}$, which ensures on one hand that the set is closed (and obviously bounded), but on the other that it is not compact

Sayan Chattopadhyay

@MikeMiller The Poisson question was a mistatement.

tatan

@Thorgott When he says, I is closed, he means in R, right?

Secret

numdam.org/article/SPS_1974__8__344_0.pdf

This basically elaborate what Balarka said about the counting measure relationship with the poisson measure

It holds within some compact set of positive measures and is only locally finite

That means, most of the stuff probably happened in the tail of the poisson variable

so the summands isn't a finite object even though you only need a finite number of them

Mike Miller

@SayanChattopadhyay I assume they meant what we proved

Thorgott

yes

Sayan Chattopadhyay

12:47 PM

No they actually changed the question. Now by an infinitesimal symmetry $X$ of the triple $(M, \{ \cdot, \cdot \}, H)$ where $H \in C^{infty}(M) $ now they mean to say a vector field $X$ such that $L_X(H) = 0$ and $L_X(\Lambda) = 0$, where $\Lambda$ is the bivectorfield defined by the poisson bracket

user2103480

@BalarkaSen I'd think so, for example if we take an infinite strip in R^2

Mike Miller

meh

Sayan Chattopadhyay

Yeah, I don't understand what's happening anymore

Mike Miller

You're presumably supposed to calculate $L_X(\Lambda)$ explicitly

Sayan Chattopadhyay

I dont even understand how I should think about a bivectorfield

user2103480

12:49 PM

I'm just assuming that an infinite strip actually works without any probability glitches, btw

Mike Miller

Sums of tensor products of vector fields. You give it two functions it gives an output function

Sayan Chattopadhyay

@MikeMiller Yeah, that's what I thought

Mike Miller

"Differential operators" which take 2 inputs and spit out 1 output function, which satisfy Leibniz in each coordinate

Sayan Chattopadhyay

Why do I care about them? Do they have anything interesting about them?

Mike Miller

Something like that

It's just the geometric way of expressing what a Poisson bracket is

Possion brackets are operations on pairs of functions

Bivector fields give you precisely that

Sayan Chattopadhyay

12:51 PM

Hmm okay

Mike Miller

Nobody cares about them outside the context of Poisson

Sayan Chattopadhyay

I see

These poisson guys seem very boring as of now, I thought they would say cool stuff about symplectic manifolds

Mike Miller

I don't like that stuff lol

Sayan Chattopadhyay

Maybe this is an interesting question. Does this bivectorfield have some kind of a global flow associated to it?

Balarka Sen

I know too few questions about symplectic manifolds

What can you ask about them

Sayan Chattopadhyay

12:57 PM

Classify them in $n=4$ maybe ?

Balarka Sen

What's special about "symplectic" in that question

Sayan Chattopadhyay

I have heard its harder. Or maybe that's just a plot to convince me to do symplectic geometry

Balarka Sen

Idgi

Sayan Chattopadhyay

I don't know but $\Bbb{C}P^n$ is a very nice example of a symplectic manifold. Maybe one can ask how its algebro-geometric properties are influenced because of a symplectic structure

Balarka Sen

The symplectic form on, Fubini-Study, CP^n can be interpreted as some average right

How does that go

Ah yes $\omega$ integrated over a curve is average intersection number of the curve with CP^1s in CP^n

I think

Can you do this in general

Sayan Chattopadhyay

1:02 PM

Oh hmm

General as in?

Balarka Sen

Interpret the symplectic form as average intersection number with some family of curves

symplectic guys have lots of curves

Who knows

I don't know simp geo

Sayan Chattopadhyay

That's a nice question though

Balarka Sen

i dunno lol

Mike Miller

Even the claim that there's a holomorphic sphere passing through any two given points is usually hard to prove

it's some statement about GW invariants

Balarka Sen

yeah lol

nuts

Mike Miller

1:07 PM

@SayanChattopadhyay I mean every function f you feed it gives you a vector field which you can flow. But past that idt you can say much

Balarka Sen

im convinced its impossible to learn symplectic geometry

Sayan Chattopadhyay

@MikeMiller Wow crazy

Balarka Sen

after this gromov pseudoholomorphic crap there's garbage about moduli spaces

not my cup of deal

Thorgott

is there any reason to care about symplectic geometry if you're not a physicist

Balarka Sen

symplectic structure is everywhere

i dont know the right questions to ask

Mike Miller

1:10 PM

Is there any reason to care about anything?

Balarka Sen

@Mike Some guy asked me about some signs in Morse homology

Lmao

Sayan Chattopadhyay

lol

Mike Miller

Those are notoriously irritating

Balarka Sen

yeah

Sayan Chattopadhyay

Just do it in characteristic 2 :p

Balarka Sen

1:12 PM

I got it tho, he was trying to compute Morse homology of Klein bottle

Mike Miller

RIP

Balarka Sen

lolol

Mike Miller

Idk what Morse function I would even use off the top of my head

Balarka Sen

We did it with a tilted height function like torus

think of it as immersed in R^3 I guess

Anyway my point is my only motivation behind this was to tell him a pun

which I did, at the end

Mike Miller

Ok I guess I do

Balarka Sen

1:14 PM

"the sign convention in most Morse homology books are quite badly Witten"

Mike Miller

Meh not gonna do

Balarka Sen

Yeah man forget about it who remembers $TU(a) = TM(a, b) \oplus \Bbb R \oplus TS(b)$

or whatever

you switch something and you'll forever be stuck

They needed to come up with some mnemonic

Sayan Chattopadhyay

Oh @Balarka, what text are you following for Representation Theory?

Balarka Sen

Steinberg

I have to read rep theory, I took it as audit and then forgot about it

Ok gotta run

Sayan Chattopadhyay

Ah we seem to doing that too, but the guy is overtly focused on Burnside stuff and is going to skip all the fourier analysis, which is annoying

user193319

1:50 PM

Let $$0 \rightarrow A \rightarrow C \rightarrow B \rightarrow 0$$ be an exact sequence of modules. Call the map from $A \to B$ $f$ and from $C \to B$ $g$. Suppose there is a map $k : C \to A$ such that $fk = id_{A}$. Show that $C \cong A \oplus B$.

I tried showing that the sequence is split by defining $j : B \to C$ via the following way. Let $b \in B$. By surjectivity of $g$, there is a some $c_0 \in C$ such that $g(c_0) = b$. Let $j(b) := c_0$. One obvious problem is determining whether it is well-defined. But showing $j$ is well-defined seems equivalent to showing that $g$ is invertible, in which $C \cong B$, which seems weird.

I could use some help.

Thorgott

You mean $f\colon A\rightarrow C$ and $fk=\operatorname{id}_C$?

user193319

Yes, sorry.

Oh wait. The problem statement says $f : A \to C$, $k : C \to A$ and $kf = id_{A}$

Do those compositions make sense?

Thorgott

2:07 PM

yeah, if you know that the existence of a right splitting implies a direct sum decomposition, then you can just mimic the same proof in this scenario

user193319

So, a right splitting is when there exists a map $j : B \to C$ such that $g \circ j = id_{B}$?

Thorgott

yeah

not sure if that's standard terminology for the record, but as long as you know what I mean

the point is that these scenarios are symmetrical in a sense

user193319

I do understand it, and I like the term actually. In my case I have a left splitting.

MathStudent

3:03 PM

Taylor's theorem is usually stated for an approximation around a given/fixed point. Can I still use it for an approximation around a point that is not given/fixed? For example, say I have a function $f$ that is dependant on $(x_1,x_2) \in \mathbb{R}^2$, can I use Taylor's theorem for multivariate functions to approximate $f(x_1,x_2)$ around $(x_1,x_2 - y)$ for a $y \in \mathbb{R}$?

I guess this is a pretty weird question. Not sure if I would need to give more context to properly ask.

Thorgott

3:15 PM

I mean, you can just do the usual approximatioon for each $y$ separately, what else do you want?

MathStudent

Hm maybe. It's a bit hard to explain what I want without giving overwhelmingly much context.

Here's an attempt: What I'm trying to do is show an upper bound by using Taylor. It's about testing the null hypothesis that $\tilde{\beta}_1} = 0$ where $\tilde{\beta} \in \mathbb{R}^2$ is an unknown vector of parameters. The null hypothesis can be rewritten by saying that $\tilde{\beta} = \nu$ where $\nu_1 = 0$ and $\nu_2 \in \mathbb{R}$ is arbitrary.

Now, I would like to upper bound something like $|\log(1+e^{\beta}) - \log(1+e^{\tilde{\beta}})|$ by approximating $\log(1+ e^{\beta})$ around $\beta = \beta - \tilde{\beta} = (\beta_1, \beta_2 - \nu_2)$. Would that be legitimate?

Secret

4:17 PM

Just realised something interesting in Wheels

The identity $0x+0y = 0xy$ has the same form as the logarithm identity $\ln (ab) = \ln a + \ln b$

AMDG

4:58 PM

Hello! Is there anything I can do to find the error function depicted here (orange plot) given what I have? desmos.com/calculator/wbpxs2jyjp

geocalc33

5:09 PM

continuously varying group

AMDG

What is that?

AMDG

5:58 PM

Just need a bit of a half turn to translate those peaks over a bit... I think I'll start, however, by normalizing everything to ln and exp.

This'll surely incorporate metallic means in some way as the circular functions contain all unique polygons and all unique circles.

xcodeking

6:16 PM

The residue theorem is in some sense a limit, right? Cause you can’t really define f(a) if f has a pole at a

Mike Miller

@xcodeking Where does f(a) appear in the statement of the residue theorem?

xcodeking

I guess I should have been clear. I'm talking about $\oint f(z) dz = 2\pi i Res(f,a)$ where a is the only pole of f enclosed by the integration contour

Thorgott

So what is the question?

Mike Miller

That doesn't require a notion of f(a) though

The contour encloses a; a is not on the contour

Ted Shifrin

The residue theorem is not truly a limit. It is a consequence of Green's Theorem (Stokes's Theorem) and the wonderful fact that $\int_{|z|=c}\frac{dz}z = 2\pi i$ for any $c>0$.

But, yeah, what Thor said. What is the question?

EM4

6:50 PM

hello

AMDG

Nice, I shaved off some more error from my approximation: desmos.com/calculator/xhha0bse0c

EM4

7:04 PM

I have a question on root of unity

1

Q: Roots of Unity problem

Let m and n be positive integers have that have no common factor. Prove that the set of numbers $(z^\frac{1}{n})^m$ is the same as the set of numbers $(z^m)^\frac{1}{n}$.We denote this common set of numbers by $z^\frac{m}{n}$. Show that $$z^\frac{m}{n} = \sqrt[n]{|z|^m}\left(\cos\left(\frac{m}{n}...

complex-analysis solution-verification

the book transformed 𝛽=(𝑚𝜃+2𝑘𝜋)/𝑛 into 𝛽=(𝑚𝜃+2𝑚𝑘𝜋)/𝑛

I don't know why

user76284

8:20 PM

Is the composition of flows (of vector fields) always a flow (of a third vector field)?

Thorgott

How do you compose the flows?

Balarka Sen

In an hour I will be an expert in character theory

Going to power through Ch. 4 Steinberg

Thorgott

I'm eagerly awaiting your proof of the odd order theorem

nbro

8:40 PM

Energy and Policy Considerations for Deep Learning in NLP (2019) by Emma Strubell et al.

Edward Evans

9:30 PM

@Balarka @Alessandro Rivers of Nihil - Where Owls Know My Name

Very good album

Balarka Sen

Gonna check it out

Fanks

Edward Evans

npnp

Pre-seminar talk tomorrow on Langlands correspondence for GL_2

wanna get a hard talk

cuz the prof is the guy I wanna do my master thesis with rofl

Balarka Sen

Cool

Let us know how it goes

Edward Evans

will do :D

Alessandro Codenotti

@EdwardEvans thanks, what genre is that?

The Lake of Wada EP was pretty cool @Balarka

Edward Evans

9:33 PM

errr

it's like

Balarka Sen

@Alessandro Yeah

Edward Evans

tech death

with weird prog elements

Alessandro Codenotti

now that's interesting

Edward Evans

Subtle Change is the weirdest song on the album

user2103480

10:22 PM

@BalarkaSen Not only that! Polish spaces behave well with respect to countable products (the borel sigma algebra of the topological product is the product of the borel algebras of the spaces); for weak convergence to behave well, we need the space to be polish; kolmogorov's extension theorem works up to polish spaces (maybe standard borel spaces) and for polish spaces, regular conditional distributions exist

Rando Hinn

o/

I need some help

Need to take this imgur.com/a/dOjTHGg to only contain negation and conjunction

by the formulas, not using a truth table

am a bit swamped

All I get is a permanently false result

mick

hi all

0

Q: Examples of weakly algebraicly closed commutative non-associative rings ??

Consider a commutative non-associative ring $A$ of finite dimension, that is power-associative. Now define weakly algebraicly closed in $T$ as : $$ x^2 + a_1 x + a_2 = 0 $$ always has at least one solution $x$ that belongs to $T$. ( the coefficients $a_1,a_2$ are in $T$ ) So I wonder : When is $...

abstract-algebra algebraic-geometry polynomials nilpotence solvable-groups

any ideas ??

user2103480

Ah nah it's not just weak convergence, it's crucial for measurability of the map d(X,Y), where d is a metric on a space X. So to handle the differen types of convergence we define, polish spaces are at the least very convenient to work in

Rando Hinn

10:41 PM

Ive tried every possible combination of laws

only to end up with sth like X & ~X

user2103480

I have found out the reason that it's so opaque to me: it is apparently a different approach to SPDE via martingale measures and rough paths. The trees and forests are there to abbreviate some kinds of iterated integrals.

What I'll be learning about is the so-called variational approach via some functional analysis hokus pokus that doesn't look much less disturbing; see chapter 5 in http://page.math.tu-berlin.de/~scheutzow/SPDEmain.pdf, and especially 5.2, for the definition of a solution of an SPDE, in that approach

note that polish spaces pop up in the very first theorem/remark, lol

Damn, apparently classical probability theory really lives in polish spaces

Mike Miller

I'll stick with my particular area of inscrutable math

Rando Hinn

I give up, this is hopeless to solve

Ted Shifrin

LOL, are we having an inscrutability contest?

user2103480

nah, everything's inscrutable

haven't yet seen an area that isn't

Ted Shifrin

10:53 PM

I think it really depends on the person and the person's training, strengths, etc.

Thorgott

11:04 PM

it's all hogwash

user2103480

@MikeMiller if I convince myself, I'll have a small taste of your inscrutable area though. I'm not yet sure if I will

It's topology 2; the syllabus would approximately be:

homotopy groups (homotopy groups of spheres, relative homotopy groups and the long exact sequence of a pair, hopf fibration and the long exact sequence of a fibration), then singular homology (excision, jordan-brouwer splitting theorem), CW complexes (hurewicz, whitehead), homology with coefficients, cohomology (cellular/singular, künneth's formula), poincare duality and then some

I haven't done algebra in a while and also never seriously worked through a topology 1 course so yeah dunno if it's too much work. I'd just hope some intuition rubbed off from my exposure to simplicial sets and kan complexes

I'm a bit lazy to seriously work through simplicial homology, is it necessary to do that?

For computations it probably is, but I'd hope that it's not necessary to grasp singular homology and work with it

Alessandro Codenotti

@user2103480 Apart from trivial cases the countable product is Borel isomorphic to any of its factors with Polish spaces though

I'm not sure what my point is

user2103480

@AlessandroCodenotti I'm slightly confused either way

Alessandro Codenotti

Why?

user2103480

Because I don't know whether that converts to a weaking of the conditions for good behaviour™ of the probability

Alessandro Codenotti

11:14 PM

Dunno

Anyway separable already implies that weak convergence is equivalent to convergence in the Prokhorov metric, so maybe you don't need Polish for weak convergence to beheave well

user2103480

@AlessandroCodenotti maybe I count that as good behaviour? bigthink

Alessandro Codenotti

Polish gives that the Prokhorov metric is also complete which is quite nice I suppose?

user2103480

Everything's nice in polish space

especially the beer prices

Alessandro Codenotti

@user2103480 That's why I do descriptive set theory :P

Thorgott

my comment about hogwash applies

Alessandro Codenotti

11:21 PM

Pfff

user2103480

merriam webster says hogwash's initial sense was "a semiliquid food for animals (such as swine) composed of edible refuse mixed with water or skimmed or sour milk"

accurate description of what it feels like

Alessandro Codenotti

If that isn't probability I don't know what is

user2103480

amen

Alessandro Codenotti

@user2103480 So is that what you're up to in Berlin now?

user2103480

Hell, I don't know. Next semester will be very probability-heavy

Alessandro Codenotti

11:32 PM

I'm sorry?

user2103480

Stochastic processes in neuroscience and fluids (two seperate courses), SPDE and one course which I'll decide when that extremely slow TU releases their course catalogue

Alessandro Codenotti

Aren't classes starting in like 2 weeks?

user2103480

Haha no, that would be insane. They start 2nd november

Thanks to the rona

Alessandro Codenotti

Oh ok

user2103480

@AlessandroCodenotti yeah I know

Alessandro Codenotti

11:35 PM

I actually have no idea when classes begin in Muenster

Just realized

user2103480

let's see if I lose my taste after that dose of probability, but it can't be worse than higher category theory :D

(which is a beautiful subject that I am just horrendously underqualified for in terms of prereqs)

@AlessandroCodenotti you plan to take some?

or just participate in seminars

Alessandro Codenotti

Not sure, there are a few classes that seem interesting but we'll have to see how strong my will to get out of bed is

user2103480

get out of bed?

I literally didnt for some online classes hahaha

Alessandro Codenotti

They want to have classes in person

(which I think is pretty crazy but anyway)

user2103480

@AlessandroCodenotti like, the whole uni, or just the math department?

Alessandro Codenotti

11:39 PM

At least that's what they told me this summer, maybe they changed idea in the meantime

@user2103480 I don't know

user2103480

thank god my uni is not in NRW anymore, classes are almost exclusively online in berlin

I think the regional rules are different

Alessandro Codenotti

They really didn't tell me anything, all they keep repeating is that I absolutely must be there before the 1st of October to sign my contract because otherwise ~~the ground will open beneath my feet and swallow me~~ the bureaucracy people will be mad

user2103480

@AlessandroCodenotti in what areas?

@AlessandroCodenotti be there one time or permanently?

Alessandro Codenotti

There's a descriptive set theory class (thaught my advisor so maybe I should go lol) which is very relevant to the stuff I'll be doing, as well as a geometric group theory class that looks cool (they're doing small cancellation stuff which I never learned about properly)

@user2103480 Just to sign the contract, I asked and there's no issue if I travel back to Italy in October for a while (which I'll do)

user2103480

I always forget. Thats the model theory GGT right?

And you're trained in the analysis GGT?

Alessandro Codenotti

11:43 PM

trained is a very big word

I've seen mostly the geometric side of things

user2103480

fair enough

Alessandro Codenotti

Small cancellation is something that both geometers and model theorists think about for some reason. Balarka should know more than me about what that is though

Mike Miller

@user2103480 No, you'll use cellular or something for computations. Simplicial is rarely computationally tractable

Sounds like a standard course, maybe a little fast

My personal area is low-dim topology / gauge theory. Very different flavor

user2103480

The lecturer actually works in differential & geometric topology (e.g. lectures about kirby calculus & 4-manifolds in the past). But I guess that is approximately the standard syllabus that lecturers have to teach in that topology cycle

@AlessandroCodenotti damn, münster has a lot of classes for one math department. that's surely comparable to bonn

Mike Miller

You comfortable giving me a name?

Alessandro Codenotti

11:54 PM

@user2103480 Probably not as many as Bonn, but still quite a few

Edward Evans

Lokale Galois-Darstellungen - Prof. Dr. Schneider oof

Schneider works a lot with the prof running the Langlands seminar

Alessandro Codenotti

Schneider is a big name in his field according to an arithmetic geometry guy I met in Bonn

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$Mathematics$ Mathematics