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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
 
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
 
 
3 hours later…
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.
 
@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
 
5:29 AM
When proving theorems about linear codes, can we always assume that the generator matrix is in standard form?
 
 
2 hours later…
7:11 AM
0
Q: Understanding the proof of Joy of Cats Theorem 12.13

S. DasThis 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...

Any help is appreciated.
 
 
1 hour later…
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 $\infty$?
 
8:32 AM
yes
 
@LeakyNun Oh, but how does the $\epsilon-\delta$ definition work? $\infty - \infty$ is undefined I guess
Things like $|\infty - \infty| < \epsilon$ for all $n > N$ won't work
 
@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 $[0,1]$
 
@LeakyNun Ummm, actually in the context in which I'm working, I need to preserve the standard metric on $\mathbb R$
 
then you can't use the epsilon-delta definition I guess
 
I guess I need to approach the problem differently. A sequence of $\infty$'s won't work
 
8:46 AM
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
 
Yeah, I will think about it. Thanks a lot for the help!
 
i.e. $(\lim_{i \to \infty} x_i = \infty) \equiv (\forall M \exists N \forall i : i \ge N \implies x_i \ge M)$
 
@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
 
@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 $\infty, \infty, \infty, \ldots$. :)
 
just use filters
 
8:54 AM
@LeakyNun Filters?
Oh there seems to be a Wikipedia page: en.wikipedia.org/wiki/Filters_in_topology
I don't know about this, interesting
 
9:12 AM
@BalarkaSen lol I'll listen to them while doing some topology later
 
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
 
 
1 hour later…
10:33 AM
Why are the $p\Bbb Z\subset \Bbb Z$'s for prime $p$ not irreducible components ?
 
what do you mean irreducible components
what's the topology
 
Zariski
 
so you mean Spec Z
 
Yes
 
irreducible components correspond to minimal primes right
 
10:34 AM
The definition I had is maximal closed subsets that cannot be the union of two proper closed subsets
 
yes, and what I said is a theorem
anyway if you want to use the definition directly, consider the bigger closed subset Spec Z
 
So I can't write $\operatorname{Spec} \Bbb Z = \{p\Bbb Z\} \cup \{q\Bbb Z | q\ne p\}$ ?
 
the second one isn't closed
 
I mean, why is$\{q\Bbb Z | q\ne p\}$ not closed ?
 
because the generic point (0Z) is dense (closure is the whole space, i.e. every nonempty open contains it)
 
10:41 AM
Oh ok
Thank you !
 
11:08 AM
There was once upon a time back in high school I did made that $\sin (x+y) = \sin x + \sin y$ mistake. Good times
 
11
A: Does the zeroth root exist?

Zubin MukerjeeThe $n^\text{th}$ 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 $x=0$
 
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
 
12:16 PM
@user2103480 Valid point. But I think this is OK if the measure on X is $\sigma$-finite, no?
 
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
 
1
Q: Is $([0, \sqrt 2] \cap \mathbb Q) \subset \mathbb Q$ closed, bounded, compact?

user99777As 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...

Regarding the second comment on the accepted answer by Stephan... Shouldn't $I$ be $[0,\sqrt2]$ instead of $[0,\sqrt2)$?
 
It makes no difference when you intersect it with $\Bbb Q$
 
But see this line - "But then q∈I as I is closed".. how do we directly say $I$ is closed?
 
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
 
@MikeMiller The Poisson question was a mistatement.
 
@Thorgott When he says, I is closed, he means in R, right?
 
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
 
@SayanChattopadhyay I assume they meant what we proved
 
yes
 
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
 
@BalarkaSen I'd think so, for example if we take an infinite strip in R^2
 
meh
 
Yeah, I don't understand what's happening anymore
 
You're presumably supposed to calculate $L_X(\Lambda)$ explicitly
 
I dont even understand how I should think about a bivectorfield
 
12:49 PM
I'm just assuming that an infinite strip actually works without any probability glitches, btw
 
Sums of tensor products of vector fields. You give it two functions it gives an output function
 
@MikeMiller Yeah, that's what I thought
 
"Differential operators" which take 2 inputs and spit out 1 output function, which satisfy Leibniz in each coordinate
 
Why do I care about them? Do they have anything interesting about them?
 
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
 
12:51 PM
Hmm okay
 
Nobody cares about them outside the context of Poisson
 
I see
These poisson guys seem very boring as of now, I thought they would say cool stuff about symplectic manifolds
 
I don't like that stuff lol
 
Maybe this is an interesting question. Does this bivectorfield have some kind of a global flow associated to it?
 
I know too few questions about symplectic manifolds
What can you ask about them
 
12:57 PM
Classify them in $n=4$ maybe ?
 
What's special about "symplectic" in that question
 
I have heard its harder. Or maybe that's just a plot to convince me to do symplectic geometry
 
Idgi
 
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
 
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
 
1:02 PM
Oh hmm
General as in?
 
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
 
That's a nice question though
 
i dunno lol
 
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
 
yeah lol
nuts
 
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
 
im convinced its impossible to learn symplectic geometry
 
@MikeMiller Wow crazy
 
after this gromov pseudoholomorphic crap there's garbage about moduli spaces
not my cup of deal
 
is there any reason to care about symplectic geometry if you're not a physicist
 
symplectic structure is everywhere
i dont know the right questions to ask
 
1:10 PM
Is there any reason to care about anything?
 
@Mike Some guy asked me about some signs in Morse homology
Lmao
 
Those are notoriously irritating
 
yeah
 
Just do it in characteristic 2 :p
 
1:12 PM
I got it tho, he was trying to compute Morse homology of Klein bottle
 
RIP
 
lolol
 
Idk what Morse function I would even use off the top of my head
 
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
 
Ok I guess I do
 
1:14 PM
"the sign convention in most Morse homology books are quite badly Witten"
 
Meh not gonna do
 
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
 
Oh @Balarka, what text are you following for Representation Theory?
 
Steinberg
I have to read rep theory, I took it as audit and then forgot about it
Ok gotta run
 
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
 
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.
 
You mean $f\colon A\rightarrow C$ and $fk=\operatorname{id}_C$?
 
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?
 
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
 
So, a right splitting is when there exists a map $j : B \to C$ such that $g \circ j = id_{B}$?
 
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
 
I do understand it, and I like the term actually. In my case I have a left splitting.
 
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.
 
3:15 PM
I mean, you can just do the usual approximatioon for each $y$ separately, what else do you want?
 
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?
 
4:17 PM
Just realised something interesting in Wheels
The identity $0x+0y = 0xy$ has the same form as the logarithm identity $\ln (ab) = \ln a + \ln b$
 
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
 
5:09 PM
continuously varying group
 
What is that?
 
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.
 
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
 
@xcodeking Where does f(a) appear in the statement of the residue theorem?
 
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
 
So what is the question?
 
That doesn't require a notion of f(a) though
The contour encloses a; a is not on the contour
 
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
 
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

EM4Let 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}...

the book transformed 𝛽=(𝑚𝜃+2𝑘𝜋)/𝑛 into 𝛽=(𝑚𝜃+2𝑚𝑘𝜋)/𝑛
I don't know why
 
 
1 hour later…
8:20 PM
Is the composition of flows (of vector fields) always a flow (of a third vector field)?
 
How do you compose the flows?
 
In an hour I will be an expert in character theory
Going to power through Ch. 4 Steinberg
 
I'm eagerly awaiting your proof of the odd order theorem
 
8:40 PM
 
9:30 PM
@Balarka @Alessandro Rivers of Nihil - Where Owls Know My Name
Very good album
 
Gonna check it out
Fanks
 
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
 
Cool
Let us know how it goes
 
will do :D
 
@EdwardEvans thanks, what genre is that?
The Lake of Wada EP was pretty cool @Balarka
 
9:33 PM
errr
it's like
 
@Alessandro Yeah
 
tech death
with weird prog elements
 
now that's interesting
 
Subtle Change is the weirdest song on the album
 
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
 
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
 
hi all
0
Q: Examples of weakly algebraicly closed commutative non-associative rings ??

mickConsider 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 $...

any ideas ??
 
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
 
10:41 PM
Ive tried every possible combination of laws
only to end up with sth like X & ~X
 
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
 
I'll stick with my particular area of inscrutable math
 
I give up, this is hopeless to solve
 
LOL, are we having an inscrutability contest?
 
nah, everything's inscrutable
haven't yet seen an area that isn't
 
10:53 PM
I think it really depends on the person and the person's training, strengths, etc.
 
11:04 PM
it's all hogwash
 
@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
 
@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
 
@AlessandroCodenotti I'm slightly confused either way
 
Because I don't know whether that converts to a weaking of the conditions for good behaviour™ of the probability
 
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
 
@AlessandroCodenotti maybe I count that as good behaviour? bigthink
 
Polish gives that the Prokhorov metric is also complete which is quite nice I suppose?
 
Everything's nice in polish space
especially the beer prices
 
@user2103480 That's why I do descriptive set theory :P
 
my comment about hogwash applies
 
11:21 PM
Pfff
 
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
 
If that isn't probability I don't know what is
 
amen
 
@user2103480 So is that what you're up to in Berlin now?
 
Hell, I don't know. Next semester will be very probability-heavy
 
11:32 PM
I'm sorry?
 
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
 
Aren't classes starting in like 2 weeks?
 
Haha no, that would be insane. They start 2nd november
Thanks to the rona
 
@AlessandroCodenotti yeah I know
 
11:35 PM
I actually have no idea when classes begin in Muenster
Just realized
 
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
 
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
 
get out of bed?
I literally didnt for some online classes hahaha
 
They want to have classes in person
(which I think is pretty crazy but anyway)
 
@AlessandroCodenotti like, the whole uni, or just the math department?
 
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
 
thank god my uni is not in NRW anymore, classes are almost exclusively online in berlin
I think the regional rules are different
 
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
 
@AlessandroCodenotti in what areas?
@AlessandroCodenotti be there one time or permanently?
 
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)
 
I always forget. Thats the model theory GGT right?
And you're trained in the analysis GGT?
 
11:43 PM
trained is a very big word
I've seen mostly the geometric side of things
 
fair enough
 
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
 
@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
 
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
 
You comfortable giving me a name?
 
11:54 PM
@user2103480 Probably not as many as Bonn, but still quite a few
 
Lokale Galois-Darstellungen - Prof. Dr. Schneider oof
Schneider works a lot with the prof running the Langlands seminar
 
Schneider is a big name in his field according to an arithmetic geometry guy I met in Bonn
 

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