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00:00 - 17:0017:00 - 00:00

5:00 PM
Like for a calculus class?
 
Yes.
 
That seems to be the modern trend. I didn't know about Chegg "in my day."
 
@EdwardEvans nvm I just managed to book an appointment for the 23rd of November
 
Ted had to hook up his plastic cup to a string to reach the Chegg hotline.
 
But we did catch people posting on-line precalculus test questions.
 
5:01 PM
SteezO- is back at it
 
lol
 
Oh, now someone just posted a question about a proof from my multivariable math book. I'll have to turn on my brain to think about it.
 
"$\vee$ defines an involutory contravariant autofunctor on the category of Hausdorff abelian locally compact topological groups" loool what a way to say $A \cong (A^\vee)^\vee$
woops
 
Spoken just like Thorgott would
That sentence even summoned Lukas
 
hahaha
 
5:04 PM
LIke magic?
 
say it three times into a mirror and jog on the spot for 10 seconds and Lukas turns up
 
On the bright side of my living in Germany experience it's time for Lebkuchen again so I can't complain overall
 
and Spekulatius
 
the sentence carries a lot more information than just the isomorphism
 
ALRIGHT SORRY
 
5:09 PM
you better be
bow before functoriality
 
It seems you've unsummoned Lukas?
 
@edward no reason to be scared, Galois cohomology is just étale cohomology on $\mathrm{Spec}(k)$
 
thanks @Lukas
 
No 1.0 this time?
 
you mean "ty 1,0 inc"
sniped :(
 
5:11 PM
haaaaaaa
 
Oh, hey, @Lukas :)
 
hi @Ted
@EdwardEvans the nice thing about class formations is that they provide a common framework which is purely algebraic and applies to both local and global CFT (though admittedly, it's harder to check the axioms in the global case), which means that you don't have to do the same things twice, you can just check them once in an abstract setting
 
oof nice
 
Historically global CFT was proved first via L-functions (so you had to use some complex analysis) by Artin and others. Modern algebraic proofs of global CFT actually use local CFT
 
right I heard this
I'd like to do global CFT at some point in the next year rofl
 
5:24 PM
although global CFT is harder than local CFT, I think it's not too bad once you're familiar with the group cohomology formalism
 
yeah I did LCFT via Lubin-Tate theory last semester
was cool
 
Lubin-Tate theory is nice because it is very explicit, but I think overall you learn more from the homological approach because group cohomology appears all over ANT
 
yeah right, but it's relevant for p-adic Hodge theory too so I'm happy I had that exposure already lol
something something $(\varphi, \Gamma)$
 
5:41 PM
where should I put arrow labels in a 3-dimensional commutative diagram so that it looks the least ugly
 
in the bin
 
perhaps I should succumb and planarize the cube instead
 
you should create a VR for the 3D diagram
 
@EdwardEvans Lubin-Tate theory also makes an appearance in Scholze's approch to the local Langlands correspondence for $GL_n$
 
orly
we might study perfectoid magic this semester
 
5:57 PM
we did some perfectoid stuff as exercises in the adic space course
 
Nise
Does Venjakob usually only teach really advanced courses? lol
 
mostly I think
 
I think we're going from Schneider's book "p-adic Galois reps and $(\varphi, \Gamma)$-modules" and it mentions that Fontaine's original approach to the theory can be replaced by perfectoid magic
 
Question: if $u=\sum\limits_{i=1}^\infty\langle u,e_i\rangle e_i$, then does $\overline{u}=\sum\limits_{i=1}^\infty\langle u,e_i\rangle \overline{e}_i$? ($e_i$ constitute an orthonormal basis)
I'm thinking yes, but my "geometric intuition" has been wrong before
 
what does that bar mean
 
6:12 PM
Conjugate
 
a general complex vector space doesn't have a natural complex conjugation that is independent of choosing a basis (unless it's a function space or something)
 
in which space does this take place
 
In a general complex Hilbert space
 
in a general Hilbert space, what I said above applies: there's no intrinsic notion of a complex conjugation indepedent of a basis
 
I'm trying to find the adjoint conjugate to the operator $Tx=\langle x,u\rangle v$ for fixed $u,v\in\mathbb{C}$ and that term came up
Thought that maybe this would simplify nicely, but maybe not
Also, yeah, I read, Lukas, thank you for the clarification
Ah, wait, I have been too cavalier with where I placed my conjugates in my work
Turns out that it does simplify nicely
 
6:27 PM
@Rithaniel I don't think you need to work with an orthonormal basis for that problem. Just manipulate the inner products a little bit: $\langle \langle x,u\rangle v,w\rangle=\langle x, u\rangle\langle v,w\rangle=\langle x,\overline{\langle v,w\rangle}u\rangle=\langle x,\langle w,v\rangle u\rangle$
this shows that $w \mapsto \langle w,v\rangle u$ is the adjoint
 
Ah, that is much easier than what I was doing
 
6:52 PM
Is there a functional analytic/fourier analytic version of the central limit theorem? Something along the lines of "if I convolute a pdf often enough, the corresponding sequence of integral operators defined on the bounded, continuous functions converges weakly to the integral operator corresponding to the normal distribution"
By these integral operators I just mean the expectations
 
if I'm understanding correctly, yes
there's a measure-theoretic version of the CLT stating that the if you convolve a properly normalized Borel probability measures with itself subsequently and rescale, the resulting sequence of measures weakly converges to the standard Gaussian measure
which should imply what you want
 
7:11 PM
ah nice thanks
@AlessandroCodenotti I once took a whole homotopy type theory course where the highlight was me not understanding how the equality type in the simplicial model actually works
 
I may or may not have dropped that same course :P
 
@AlessandroCodenotti algebraists taylor expansion
@AlessandroCodenotti you may or may not have been wise :D
 
how can I show the map f(x,y)=(\sqrt(1-x^2-y^2),x,y)\in R^3 is open?
 
7:26 PM
it's not when your codomain is R^3
 
sorry, the map is from U={(x,y):x^2+y^2<1} to R^3
 
then it's not open
the image lies entirely within the unit sphere
 
you are right, I am sorry I mixed it up. it is to S^2 with the subspace topology
 
the map is a homeomorphism onto its image
 
I have shown it is bijective continuous and differentiable, but I am stuck to show it is an open map
 
7:30 PM
I have a question about a binomial sum, I have a solution I just want someone to see if I am doing this right
 
you can write down the inverse easily
and see that it is continuous
 
whoops, sorry, did not realize a question was being discussed, my apologies
 
Yes of course I am so stupid :( Thank you very much!!
 
np
 
8:00 PM
@Thorgott this is of course just a restatement of CLT. you want finitely existing variance, that's absolutely crucial
everything is in $L^2$
@user2103480 you might want to check out the von Neumann ergodic theorem
 
right, I forgot to say that explicitly
that hypothesis is crucial, because it ensures that the Fourier transform of the measure is $C^2$
then you Taylor it and CLT falls out, more or less
 
pretty much. or inverse Fourier transform
there's an extension of CLT to polynomially decaying random variables for which the scaled limits converge to stable $\alpha$-symmetric distributions. $\alpha = 1/2$ is the Gaussian
 
@BalarkaSen yeah I'd prefer something that stays close to the operators. Maybe some operator algebra stuff?
 
that's why I said von Neumann
i dont know the correct second order version though
 
8:19 PM
Hi, a @Balarka.
 
Hi @Ted
cramming for a test tomorrow
 
Cram without chat :)
 
i just joined jesus christ ted
 
Did you join jesus christ ted, did you join jesus christ and told ted this, or did you join jesus and told christ ted this?
Also, did you mislay your shift key?
 
I, Just Joined Jesus Christ Ted,
 
8:22 PM
LOL @Tobias
 
hereby proclaim
 
comma before hereby
 
so there
 
Go study, silly.
OR: Go, study silly.
 
both apply, i am studying statistics
 
8:26 PM
Well, I purposely omitted the option Go, study, silly.
Because you're already way too silly and we don't need you to silly further.
 
I found a really good approximation to the prime counting function for the first 1000 numbers
 
@TedShifrn Have I told you that if $B$ is symmetric idempotent, $Q$ is symmetric positive definite such that $I - B - Q$ is positive definite, then $BQ = QB = 0$
replace "definite" by "semi-definite" if you wish
 
that's so specific lol
why does this matter
 
I'm glad you asked
Do you remember $\chi^2$ distributions
 
no I never learned that stuff
 
8:36 PM
I will give you a very natural, category theoretical introduction
If $X$ is a random variable, we can consider random variables of the form $X' A X$ where $A$ is some matrix. These are the so called "randomized quadratic forms", where you take a quadratic form and feed sample values from a specific law $X$ to it, and you get a new random variable/measure out of it.
Simplest example is $X \sim N_n(\mathbf{0}, I_n)$, multivariable standard normal, and you look at $X' X$. This is the $\chi^2_n$ distribution.
 
im out
 
Hi out, I'm dad
 
lol
 
9:24 PM
@Thorgott Still want to hear this shit?
 
@BalarkaSen I hereby introduce you to the rabbit hole of just intonation
 
Lol
 
sudden microtonal turkish guitar solo
 
 
2 hours later…
11:12 PM
@BalarkaSen Narrator: "but, in fact, the introduction was not category theoretical..."
 
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