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12:14 AM
It’s not E[X] as angle, it’s E[XY] as inner product and E[X(1)] as a special case of that
With E[X] having the interpretation, I guess, of the component of X in the direction of 1 (as a random variable)
And X-E[X] therefore being orthogonal to 1
 
12:40 AM
Hey chat
Could anyone link me a good proof of the fundamental theorem of Galois theory?
The injection part for $F \mapsto \operatorname{Gal}(K/F)$ is pretty okay
But Lang’s proof in Undergraduate Algebra uses the primitive root theorem (since he works over fields of characteristic 0)
And his Algebra proves the much more general case for possibly infinite algebraic extensions
I’m searching for a proof that works for every finite extension
(and is readable for a first-timer)
 
I think any conventional proof invokes the primitive element theorem
The primitive element theorem is also important in its own right, so maybe you should read up on that instead
 
1:15 AM
In other news, I just found an inversion-free proof of the injectivity of the Fourier transform that only requires tools from complex analysis and integration theory and it makes me very happy
 
Constant real part --> a holomorphic function is constant, right?
I think it is just Liouville.
 
That's just Cauchy-Riemann
 
That's true, it forces the imaginary part to be constant right away.
Bounded would need something more like Liouville affords.
Or maybe you can do it with CR.
 
That's essentially equivalent to Liouville
Since bounded implies bounded real part
 
1:33 AM
@Thorgott do you enjoy complex analysis?
 
I do, but I don't understand it
 
Do you loath applying the residue theorem to find improper integrals
 
lol I just had to do that in a tutorial today
A lot of applications of the residue theorem are pretty slick, but quite a few are also just plain ugly
 
I hate contour integrals with a passion and I have to do two tonight for an assignment.
Did you see the residue question Ted helped me with yesterday?
#5
You seem much better at complex than me so you probably can do it without hints, but if you scroll back in the chat history you can see Ted help me work it out.
 
1:50 AM
Looks like a variant of the argument principle
 
2:09 AM
Hmm, I think I'm not seeing something
I feel like that integral should add up the zeros of $f(z)=a$ (with multiplicity) and substract the poles of $f$ (with multiplicity), but it seems I'm mistaken about that latter term
Two example calculations seem to agree with this
 
Agree with the mistaken part?
 
Agree with the poles turning up in the result
 
Hmmm, I didn't check about the poles.
 
Take $f(z)=\frac{1}{z-\frac{1}{2}}$, $a=0$ and the unit circle. Clearly, $\frac{1}{z-\frac{1}{2}}=0$ has no solutions. Now , $\frac{1}{2\pi i}\int_{|z|=1}\frac{-z\frac{1}{(z-1/2)^2}}{\frac{1}{z-1/2}}dz=-\frac{1}{2\pi i}\int_{|z|=1}\frac{z}{z-\frac{1}{2}}dz=-\frac{1}{2}$, which is minus the only pole of $f$ (counted with its multiplicity 1).
 
 
2 hours later…
4:01 AM
 
4:24 AM
What's that, Akiva?
 
4:35 AM
looks fractal-ish
 
What's a good proposal distribution to sample from a multi-modal distribution with support in $[0,1]^n$?
The peaks of the distribution are very narrow, so I have to use importance sampling.
I was thinking something along the lines of a mixture of products of beta distributions, but I don't know the number of components to use, since I don't know the number of modes.
 
5:27 AM
@Thorgott just sat down and checked, and you are right about the poles.
Same proof effectively as the roots
 
Cute @AkivaWeinberger
 
@Rithaniel Fractal I found somewhere
Not sure how it's made
 
6:08 AM
(b) Prove that if A and B be two square invertible matrices, then AB and BA have same charac-
teristic roots.
Any hint on this question ^?
 
 
1 hour later…
7:18 AM
@TedShifrin ^^^^^^^ that was an amazing explanation. Thank you for letting upload your classes to YouTube, they are amazing!!
 
8:13 AM
Morning
 
8:57 AM
@EdwardEvans guten Morgen
 
9:07 AM
Grüß Gott
 
 
2 hours later…
10:55 AM
0
Q: Set theory and relation

maths studentLet $\mathcal{C}$ be a collection of subsets of $[n]$ with the property that if $A, B \in \mathcal{C},$ then $A \cap B \neq \varnothing .$ (For example, $\mathcal{C}=\{\{1,2\},\{1,3\},\{2,3\},\{1,2,3\}\}$ has this property.) What is the largest that $\mathcal{C}$ can be? Hint: You will have to pr...

 
11:47 AM
Question: $A$ is a Dedekind ring, $\mathfrak{p}$ a non-zero prime ideal in $A$ and suppose $\mathfrak{a} \subset A_\mathfrak{p}$ is a non-zero ideal in the localisation. Since Dedekind rings are stable under localisation we have the factorisation $\mathfrak{a} = (\mathfrak{p}A_\mathfrak{p})^n$ for some $n \in \Bbb N$.
Take an $a \in (\mathfrak{p}A_\mathfrak{p})^n \setminus (\mathfrak{p}A_\mathfrak{p})^{n+1}$. Why does it follow that $(a) = (\mathfrak{p}A_\mathfrak{p})^n$
(that factorisation exists because Dedekind rings have prime ideal factorisation and that guy is the only prime of the localisation)
 
Does $E[X|Y=y] = E[X]$ mean X, Y are independent random variables??
 
No, take $X=Y\sim\delta_1$
 
Obviously $(a) \subseteq (\mathfrak{p}A_\mathfrak{p})^n$ but what's stopping smth like $(a) = (\mathfrak{p}A_\mathfrak{p})^{n-1}$
 
How we see mathjax in mobile @Thorgott
 
All I'm saying is to take X and Y to be equal and dirac-distributed
 
12:05 PM
durr lower powers of primes contain the larger powers, not the other way
no wait
yeah
that's right hahaha
 
12:24 PM
Looking for a nice geometry problem? There is one. It already has an answer, but finding a (mostly) coordinate-free reasoning would be great.
 
12:44 PM
@EdwardEvans is that even true? I'm looking at things like $(2) = (2,1+\sqrt{-5})^2$ in $\Bbb Z[\sqrt{-5}]$
oh what am I thinking about
yeah maybe just use the $\mathfrak p$-adic valuation
oh wait you're looking at the localization
then isn't it trivial
the localization is a DVR
 
@Leaky the proof is to show that $A_\mathfrak{p}$ is a DVR lol
 
if you still want a proof, let $a = \varpi^n u$; for any $\varpi^m v \in (\mathfrak p A_\mathfrak p)^n$, $m \ge n$, then $\varpi^m v = (\varpi^n u) (\varpi^{m-n}) (vu^{-1}) \in (a)$
aha
 
It's fine I got it hehe
thanks tho
 
ok
 
Currently defining $S$-integers, units, ideals, ideal classes
 
12:49 PM
\varpi looks so weird
 
lol
varpi sounds like a Finnish name
also hi @Leaky and @Alessandro
 
how can I let the browser dissplay the latex here?
ops got it
 
@quallenjäger: Look at the link near "LaTeX in chat" on this page
 
what a horrid definition: let $S$ be a set of primes of a number field $K$. Then $\mathcal{O}_{K,S} := \lbrace \frac{a}{b} \in K : a, b \in \mathcal{O}_K,\ b\notin \mathfrak{p}\ \text{for all}\ \mathfrak{p} \notin S\rbrace$
 
12:52 PM
just swa it
 
I think the definition gets better when phrased in terms of $\mathfrak{p}$-valuations
 
The definition gets better when you study something else instead :P
 
wait there's other maths?
 
Just come to some set theory
 
12:59 PM
Want to hear something weird I learned recently? I swear it's algebra, totally not set theory
 
hahaha go for it
 
So if you have a group $G$ there is the dual group $G^\ast=\mathrm{Hom}(G,\Bbb Z)$ and the usual canonical map $G\to G^{\ast\ast}$ given by $g\mapsto(h\mapsto h(g))$
 
I think I know what you're gonna say rofl
 
And a group is reflexive if this map is an iso
@EdwardEvans I won't say anything then mwahahaha
 
hahaha
nooo carry on
 
1:03 PM
"every free Abelian group is reflexive" is equivalent to "there are no measurable cardinals"
 
alright I didn't know you were gonna say that
 
lol what were thinking about?
 
"totally not set theory"
 
It's clearly group theory come on
 
lol
hang on lemme see if the thing I'm thinking about is even a thing
 
1:04 PM
Whitehead's problem maybe?
 
So you mentioned Pontryagin duality right?
 
Yes, I have some vague idea of how it works from some stuff I read on operator algebras mostly
 
and the canonical map you mentioned blah blah blah but there's some other construction where the map is totally unnatural and dependent on a basis etc.
but I've forgotten it and now I'm sad
 
Hm, I'm afraid I don't know what you had in mind but let me know if you find it again because it sounds interesting
 
yeah I'm just looking for it in my notes lol
Oh it's there
$G \cong \hat{\hat{G}}$ canonically but $G \cong \hat{G}$ non-canonically
if
$G$ is a finite abelian group
@Lukas explained it well in the construction of $L$-functions lecture
@Alessandro I get the feeling that lots of statements that seem to have nothing to do with set theory end up implying stuff in set theory
set theoretic theorem vampirism
 
1:13 PM
@EdwardEvans just like for vector spaces, makes sense
 
1:51 PM
Hey chat.
How do I prove that $(a,b) = R \implies (a^m, b^n) = R$ for $(a,b)$ denoting the left ideal generated by those elements?
 
@Lucas what happens if you assume $(a^m, b^n) \neq R$ ?
 
I’m not sure. Could you elaborate?
 
Well if $(a^m, b^n) \neq R$ then there's a maximal ideal $\mathfrak{m}$ that contains $(a^m, b^n)$ right?
 
I’m thinking about the $1$, obviously
Yes.
I don’t think I should use maximal ideals, since the book just defined what an ideal is.
(I’m reading the first chapters again)
 
Hmm that's how I'd prove it
what kind of ring is $R$?
 
2:03 PM
None specified.
I’m almost sure that there are missing hypotheses
 
probably not, but in the scenario $R$ a commutative ring one can do the following: $(a^m, b^n) \subset \mathfrak{m}$ means that $(a)^m \subset \mathfrak{m}$ and $(b)^n \subset \mathfrak{m}$, and a maximal ideal is in particular also a prime ideal, so $(a) \subset \mathfrak{m}$ and $(b) \subset \mathfrak{m}$, i.e. $(a,b) \neq R$, contradiction
in fact that works for any ideals $\mathfrak{a}, \mathfrak{b}$, not just principal ideals
 
Hmm, interesting question
 
This is the first problem on ideals from the book “A Course in Galois Theory”, by D. H. Garling
 
lol I have that book here
 
To moderators: Consider math.meta.stackexchange.com/questions/31167/… . Could I ask you to ask the SE developers to fix the problem mentioned there?
 
2:28 PM
How can we say just by looking at it that the number of pivot columns is 2?
I know that Column 3 and 4 are dependent on 1 and 2...But how does that affect the number of pivot columns?
 
The number of pivot columns is the rank
 
@Thorgott Assume that the concept of rank isn't introduced in the course yet....
 
If you perform Gaussian elimination, the last two columns will get eliminated precisely because you can obtain them as linear combinations of the first two
 
2:44 PM
@Just_A_Man I am not sure whether the mods can influence this in some way - but still, if you want the mods to see this, posting a message in Math Mods' Office might be more reasonable that posting here.
You could also improve the post on meta a bit. For example, from the post it's not clear whether it's about or . You could mention what you used to view the post when the problem appeared.
 
@Thorgott they didn't get eliminated...
 
@Archer Gaussian elimination in the columns
(Autocorrect)
 
@MartinSleziak Thx; will do in a moment.
 
 
1 hour later…
4:12 PM
@manooooh Not sure what you're referring to, but I'm glad you find it helpful.
 
@Archer You look at the echelon form (or rref) and pick out those columns in which the pivots appear. You can't tell just by looking at the original matrix. (Well, of course, that's a bit of a lie, but ...)
Hi, demonic @Alessandro
 
Hi.
Oh, Ted: we forgot (or at least I did) to check what happens with poles in that integral $\int \frac{zf'(z)}{f(z)-a}\,dz$. Thorgott pointed it out, but the calculation is basically the same just with $f(z)-a = \frac{g(z)}{(z-z_0)}$.
 
4:48 PM
@anakhro: I don' t know what you're talking about. How can $f(z)=a$ if $z$ is a pole? This is about holomorphic functions.
 
The question had $f$ meromorphic.
So $f$ might have poles.
 
Hmm, OK, so we're looking at zeroes of $f(z)-a$ and poles of $f$. Weird.
Better just to stick to the argument principle and look at zeroes and poles. That's the way to understand this, anyhow.
 
Same method yields that it subtracts the poles (with multiplicity).
 
Yeah.
Without the $z$ in there, this is the proof of the argument principle.
 
Is that how it is proved? I don't remember.
I realized it's been over 4 years since I took a course in complex analysis.
 
4:53 PM
BTW, this leads to a beautiful result. If you do $\dfrac{z^k f'(z)}{f(z)-w}$ analogously, you will prove that any symmetric function of the roots of $f(z)=w$ is a holomorphic function of $w$.
 
I will check that!
But later since I have to do function analysis :((
 
What do you mean :((, functional analysis is beautiful
 
This functional analysis is less than beautiful.
Gelfand Naimark theory is beautiful.
Marcinkiewicz interpolation is not. :(
 
I'm not familiar with that
And it doesn't sound like something I want to be familiar with
 
In mathematics, the Marcinkiewicz interpolation theorem, discovered by Józef Marcinkiewicz (1939), is a result bounding the norms of non-linear operators acting on Lp spaces. Marcinkiewicz' theorem is similar to the Riesz–Thorin theorem about linear operators, but also applies to non-linear operators. == Preliminaries == Let f be a measurable function with real or complex values, defined on a measure space (X, F, ω). The distribution function of f is defined by λ f ( t ) = ω...
 
4:58 PM
It says "non-linear"
This is forbidden in nice functional analysis
 
Someone needs to write a good book on functional analysis for once.
And measure theory.
 
Sheldon Axler has just written a graduate real analysis book. A few people mentioned it here.
 
I think there's nice books for both already
 
I thought Axler's was an undergraduate measure theory book.
 
It's GTM.
 
5:02 PM
@AlessandroCodenotti by defeating Alireza Firouzja, Magnus Carlsen is now #3 in Tata Steel
 
There's a GTM course being offered here next semester, I tried to write an email to the professor to ask about the syllabus but discovered that the mail on his website isn't working...
 
In any case, I did look at Axler's but I did not find it to be anything special. It wasn't a piece of garbage, though. And he typesets nice.
@AlessandroCodenotti any suggestions on functional analysis books?
Something to make me love the subject.
 
@TedShifrin the part where you are explaining in 4 dimensions with two different pictures
 
@TedShifrin You got flagged.
 
@anakhro At which level? I like Brezis but you might be looking for something more advanced instead
 
5:06 PM
@AlessandroCodenotti first course in functional analysis. Something with a proof of the Hahn-Banach theorem in it.
I did like these notes: imsc.res.in/~sunder/fa.pdf
 
I'm assuming that flag is nothing!? I see no animosity between @TedShifrin and @AlessandroCodenotti, both of whom are room owners. If whoever's flagging wants to express an objection, please try using words first. Or appealing to local ROs/mods.
Or maybe it was just a misclick....
[shuffles back to RPG-land]
 
@anakhro Brezis starts with the Hahn-Banach theorem on the very first page
 
Functional Analysis, Sobolev Spaces and Partial Differential Equations?
 
Can you guys see who flagged? Because if it was me, it was definitely a misclick.
 
5:09 PM
First five or six chapters are functional analysis, then it does PDEs (I've read very little of the PDEs part of the book)
 
@AlessandroCodenotti I will take a look, and hopefully it will be a good read. :)
What do you like about it?
 
@nitsua60 A quick estimate from the transcript search shows that Ted called me demonic as a greeting almost 300 times, so I think it's safe to assume that the flag was either trolling or a missclick
 
Public announcement: Alessandro and Ted are good friends who have a sense of humor. Whoever is flagging needs to stop.
12
(P.S. I remember exactly when that appellation started, and I'm sure Alessandro does too.)
 
@anakhro IIRC a local mod can see who cast a flag while it's still active, and SE staff can dig back to find out at any time. But that's definite overkill at this point =)
 
@anakhro I think it's very readable overall. I also like how he does the general theory before the special case of $L^p$ spaces (and as a consequence how early he introduces and emphasizes the weak and weak* topology)
@TedShifrin I'm not sure actually. Was there a driving licence involved?
 
5:14 PM
Yes ... you were endangering my life.
 
@AlessandroCodenotti I look forward to the weak/weak* topology emphasis. Definitely not a concept I have worked with a lot.
Thanks for the suggestion!
 
You're welcome
That's the book I studied from in my first functional analysis course
 
What sort of stuff are you studying these days?
 
I have exams soon so it's mostly a combination of set theory and differential geometry
And I'll have another exam in March on semigroups of operators on Hilbert spaces
(I should also resume working on my master thesis at some point after the exams...)
 
What are you doing in your set theory and differential geometry courses?
 
5:42 PM
the diffgeo one was an introductory course, so smooth manifolds, bundles, Stokes, de Rham cohomology, flows, a little Riemannian manifolds and so on
The set theory course is called Models of Set Theory II which is a follow up to a course I did last semester in which forcing was introduced. Here we saw class forcing, product forcing, iterated forcing, forcing axioms and applications of all of those
 
Forcing sounds so cool. I read a little paper by Timothy Chow which was an intro to forcing.
He never actually did do forcing technically, but he explained a little of the background and the idea that Cohen had.
Cohen was a genius.
 
@anakhro I think I know which paper you mean
The basic idea of forcing is very intuitive, we want to add a new object, let's say a new subset of $\Bbb N$ to our model, which is a function $\Bbb N\to 2$. We just approximate it with finite functions that are in our model
 
But from this to getting a working tool there's a lot of technical issues to deal with
 
There was once upon a time I considered doing research in foundations. In particular, I enjoyed type theory.
 
5:48 PM
I've been told that Cohen's original presentation skipped some of them which had to be patched up later, but I haven't read it myself
 
Alas! the allure of geometry sucked me in and now I find myself distanced from that old love.
Would be interesting to read Cohen's original paper.
 
I took a course on homotopy type theory last semester, I thought it's cool but not really something I'd like to work with
 
A professor of mine told me that Cohen never actually did much after forcing. Was working on a difficult problem of a sort with a student or something? Just rumours.
I know nothing of HoTT, but I enjoyed working with CCCs and doing lambda calculus stuff.
A friend later did her M.Sc. in that stuff, but I think she is in CS now instead of math.
 
@anakhro I don't know
@anakhro I think the modern presentation is much more polished
I did read a paper by Cohen once, but it is not too related (the construction of two Baire spaces whose product is not Baire, Cohen's construction does use forcing though)
 
Polished is usually a characteristic of the modern presentation. :)
"Gödel himself wrote a letter to Cohen in 1963, a draft of which stated, "Let me repeat that it is really a delight to read your proof of the ind[ependence] of the cont[inuum] hyp[othesis]. I think that in all essential respects you have given the best possible proof & this does not happen frequently. Reading your proof had a similarly pleasant effect on me as seeing a really good play.""
 
6:22 PM
I've noticed I no longer have a fear of liminf and limsup. On one hand, this is a good thing. On the other hand, I still can't use them well and it's a bad that I've reached such a position of apathy towards them.
 
6:50 PM
Hey @Ted
 
Hi @Lukas
 
Ciao @Alessandro come stai?
 
Bene grazie, tu?
 
The lecture schedules for next semester have been announced today and it turns out that there are three lectures I really want to attend, but whose dates happen to intersect pairwise each. What a shame :/
 
Benissimo, grazie. Ho appena avuto un appuntamento con una ragazza bella :)
 
6:55 PM
@Thorgott :(
$\left|\int|f_n| - \int|f-f_n| - \int|f|\right| \leqq 2\int|f|$
If $f_n\to f$ pointwise, and $|f_n|<\infty$, how does this inequality follow?
 
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