Oh, I'll try to remember.

I said "hi chat", I don't think I greeted you specifically

Hi @Ted :P

Much nicer :)

Suppose $F\subset K \subset X$, $F$ is closed (relative to $X$) and $K$ is compact. Let $\{V_{\alpha}\}$ be an open cover of $F$. If $F^c$ is adjoined to $\{V_{\alpha}\}$, we obtain an open cover $\omega$ of $K$. Since $K$ is compact, there is a finite subcollection $\phi$ of $\omega$ which covers $K$ and hence $F$. If $F^c$ is a member of $\phi$ we may remove it from $\phi$ and still retain an open cover of $F$. We have thus shown that a finite subcollection of $\{V_{\alpha}\}$ covers $F$.

How do we know $F^c$ is a member of $\phi$?

What if it's not? Then we can't remove it.

Then you have no problem

$F^c$ has nothing to do with the covering of $F$

It's just that it wasn't in the original $V_{\alpha}$

Oh, I think I see. Where do we use the fact that $F$ is closed?

So if we extract a finite subcover of $V_{\alpha} \cup F^c$, it may contain $F^c$, so we need to get rid of it

We use that $F$ is closed because $F^c$ is open. Otherwise, $V_{\alpha} \cup \{F^c\}$ wouldn't be an /open/ cover of $K$

@Daminark ah got it, thanks

@Alessandro Neato.

Let me modify the proof and see if it still works

Is $|E:F|<\infty$ needed or is $E/F$ algebraic sufficent? We used it to justify the last isomorphism of groups because $|E:F|<\infty$ implies that $|\text{Gal}(E/F)|=|E:F|$, but I don't think that's actually needed to show the correspondence between normal subgroups and intermediate Galois extensions

Suppose $F\subset K \subset X$, $F$ is closed (relative to $K$) and $K$ is compact. Let $\{V_{\alpha}\}$ be an open cover of $F$. If $F^c$ ($F^c = K \cap F^c$) is adjoined to $\{V_{\alpha}\}$, we obtain an open cover $\omega$ of $K$. Since $K$ is compact, there is a finite subcollection $\phi$ of $\omega$ which covers $K$ and hence $F$.We have thus shown that a finite subcollection of $\{V_{\alpha}\}$ covers $F$.

Back later. Everybody's got everything well in hand. :)

Bye @Ted

Bye @Ted

Okay now let's talk about transfer...

Lol see you @Ted!

@Alessandro Infinite Galois theory is kind of tricky.

see you @TedShifrin

be well Monsieur @ted

nvm doesnt work

@BalarkaSen So we're back at finite groups again :P

I think there is some issue about subgroups which are closed in the profinite topology

I'm trying to recall the story

Argh. I just spent 30 minutes trying to use "high-tech" methods to solve something which is basically just AM-GM... LOL

math.berkeley.edu/~gbergman/245/3.3.pdf

This seems cool

profinite topology?

@orbit-stabilizer My friend took that class, and I sat in once or twice. Bergman is fantastic

How did he find it?

In what sense?

Just looked it up, that's a weird topology

@Alessandro I think only profinite-closed normal subgroups of $Gal(E/K)$ are coming from algebraic subextensions $E/F$ or something

Was it an interesting course? A useful course?

Where's @MatheinBoulomenos when you need him?

@orbit-stabilizer More interesting than useful for my friend at the time, but well worth it.

@Alessandro On the contrary, it's the most natural topology on infinite Galois groups! :)

That stuff is probably going to go over my head atm anyway, I'll stick to the nice part of Galois theory :P

I don't know this story

We did the classical stuff about degree $\ge5$ polynomials and roots expressible via radicals today, cool stuff

I told you to learn wild Galois theory with me, but you refused. Top 10 anime refusals

oh wow

I wondered if this would be true, but i'm still amazed that it is

@BalarkaSen I have way too much topics to learn for uni already this semester to do any more maths, I'm taking 6 courses

@Alessandro Yikes

When does your semester end?

the lectures end in December, I'll have a couple of exams in December and more between January and February

@TedShifrin @BalarkaSen $F(x,y,z,w)=0$ is equivalent to Cayley's nodel cubic surface $G(x,y,z,w)=wxy+xyw+yzw+zwy$ up to a projective transformation(!!)

It's dinner time, I'll be back later!

@Semiclassical Oh

Oh wow

You're damn right

@Alessandro Got it

specifically it's $(x,y,z,w)\mapsto (x+w,y+w,z+w,w+x+y+z)$

with $F(x+w,y+w,z+w,w+x+y+z)=4 G(x,y,z,w)$

Ok, now it's pretty easy to verify that it's irreducible over $\Bbb C$. That more or less finishes the argument I had in mind.

yeah

that is really really cool

Truly

...oh ffs. equation (6): mathworld.wolfram.com/CayleyCubic.html

oh, wait

no, that's not quite the right one

equation (7) is closer but probably still requires a change of variables

Right

I wonder what you get when you blow those 4 double points up

sup semi

hi @GFauxPas

Not 14, that's for it's Hessian

It's just 4, like the real locus

Hello everyone! Is saying "What do perfect squares is 11 between?" Its construction is different to me to judge whether it is correct. I heard the phrase while watching a video and it sounded to me as mentioned above. Thanks in advance!

No, it doesn't sound right

What perfect squares is 11 between

that sounds fine

remove the 'do'

"What two perfect squares is 11 between"

Oh, I think you meant 'two' inetad of di

do

@orbit-stabilizer @Semiclassical Got it!

@BalarkaSen now I wonder what this translates to for the one algebraic identity I came up with

Of course the correct answer is $|C_9|$ and $|C_{16}|$.

excuse me to everybody. Hi

Can I have an help?

sure, what's your question?

At this moment I have not ask a question why?

I have received the message below.

Sorry, I don't know...

@BalarkaSen that is correct. If you take any subgroup $H \leq \operatorname{Gal}(E/K)$, you can of course look at the fixed field $E^H$, but if you look at $\operatorname{Gal}(E,E^H)$, this will generally not be $H$ again, but the closure of $H$ in the profinite topology, so if you restrict it to closed subgroups, you get a bijection

read that meta question/answer

@Semiclassical How I do?

click the link in the question title

@Semiclassical My english is ugly and at this moment i am into my car and my laptop have the empty battery

Then probably you shouldn't be worrying about posting more questions at the moment.

Do you guys study using pomodoro?

Hey again everyone

I don't see how tomatoes are helpful in stuying maths

@Semiclassical thanks a lot for your help. hi

pomodoro technique*

@MatheinBoulomenos Ah, I see, yes. Cool

also lmao @ the tomato comment

@Perturbative do you take 5 minute breaks?

you*

@MatheinBoulomenos ah, nice

that's when you tell someone to throw tomatoes at you every time you make a mistake

I'd like to think I take 5 minute breaks, but in reality it's closer to 50 mins

I have come to the conclusion that I can't into time management, so I have given up completely

My study pattern is pure randomness

But it works

@Perturbative oh wow. Haha - so 25 minutes work, 50 minute break?

Really it's not too surprising that not every subgroup of $\operatorname{Gal}(E/K)$ corresponds to a field extension, because the Galois group of every extension is profinite, but there are plenty of examples of subgroups of profinite groups which are not profinite

@BalarkaSen I'm sure it's not pure randomness. Otherwise that would be a very big deal

Nah, it totally is

I do whatever at a given interval of time

It's totally topologically mixing my man

Being a profinite group sounds like fancy and difficult stuff, but for Galois groups, this basically boils down to the fact that every Galois extension may be written as a union of all finite subextension, so the action of a Galois automorphism is completely determined by how it acts on the finite Galois subextension

Is there true randomness beyond quantum phenomenon?

@BalarkaSen hmm. apparently Cayley's cubic is (up to isomorphism) the only cubic with four double points.

@Semiclassical That's quite cool

so that makes this much less surprising

yep

i feel time management very difficult for me, too.

11 is between a lot of perfect squares

Ugh my night lamp is busted

mine too; burnt out bulb

i just like to do things which by chance interest me at that moment disregarding how much time is elapsed therein.

I'll just max out my monitor intensity and work with that

@BalarkaSen hmm, here's an interesting statement I just found

my watch and clock both don't walk due to the batteries are out of electricity.

@BalarkaSen this may delight you, the easiest proofs I know that closed subgroups of profinite groups and quotients by normal closed subgroups are again profinite are purely topological. (You can also contruct a Galois extension with any given profinite group as Galois group and use the Galois correspondence, but that feels more like a hack)

Hah

"Even before the discovery of the 27 lines, in a paper of 1844 Cayley studied what we now call Cayley's 4-nodal cubic surface. He finds its equations and describes its plane sections which amounts to describe its realization as the image of the plane under the map given by the linear system of cubic curves passing through the vertices of a complete quadrilateral." (emph. added)

@MatheinBoulomenos Still sad that you like all the bad topologies and not the actually nice topologies

I don't dislike the nice topologies, I'm just bad at working with them

now I want to make a pun on path-ology

      haha

@Mathei Yeah that's fair

I find schemes easier than manifolds. What is wrong with me?

Very glad to see you realize there is something wrong with you

You are a mutant, like the rest of the algebraists

Not actually a human

If I'm given the choice between being human and being an algebraist, I'll choose algebraist without hesitation

@BalarkaSen what I still find crazy is that this is originating from trigonometry

ya now that I know what manifolds are it seems totally natural

takes up very little new mental effort

@MatheinBoulomenos I knew you'd take that as a compliment

I'd call SE badly moderated.

@KevinDriscoll Right, it's very natural to the eyes

Tons of non-sourced questions directly from assigned work.

Yeah of course it is super badly moderated

I was making a Trump meme

Smooth manifolds are equivalent to a full subcategory of locally ringed spaces, so there's that

I dont know if this gets me any cred here, but Im gonna try it out just to see

I hung out with the Smarter Every Day guy last night while he made a video with a colleague

Oh cool!

at least you didn't hang out with Vsauce Michael

I did not

I can't wait for next semester

Taking a course on graph theory

Should be fun times

@KevinDriscoll That's when you know you are acquiring actual culture

your ability to find youtube meme videos is scary

@orbit-stabilizer oh that'll be sick

Also woo just finished bio!

@Daminark How did it go?

My fellow mutant arrived. Hey @Daminark

Actually it doesn't matter - finishing bio is the only thing that matters

It went alright, we are allowed to bring in a sheet of notes, and I wrote mine last night based on the review PowerPoint

That's pretty cool

The writing was ridiculously compact but it worked, got most of the questions. Enough that I probably did as well as I really need to since I don't care much about that class. So I'm done

Yo @Mathein!

So it's only Complex now?

And algebra

Those are both next Friday so I can breathe a bit

Right, that

Mutant subject

(I've also got my paper on the Dold-Thom theorem, that homotopy groups of the infinite symmetric power of a space are isomorphic to its integral homology groups, though I've got time to do that)

u have to teach me as u write it boi

Will do, will do

@TedShifrin Heya! I'd been looking for you!

@BalarkaSen hmm hmm hmm

there's some deep s*** here

Ooh nice pictures

@BalarkaSen what i'm interested right now is the suggestion that my curve is equivalent to some 3-by-3 matrix (whose entries are linear in x,y,z,w) being singular

Right, I need you geometry people that know about bundles and stuff :O

Am I right when I say that a section of a rank-1 vector bundle over some nice manifold M can be interpreted as a smooth function $M \to \mathbb{R}$ ?

@TastyRomeo Well, no.

That's only if the bundle is trivial

@Semiclassical Interesting algebraic geometry

yeeeeep

It seems like a much more complicated version of Siegel domains

In general the fibers of a rank 1 vector bundle does not have any canonical identification to each other, so that you can call them as a map to $\Bbb R$

I heard about those for the first time today

hi can someone help me with stochastic processes?

cool

Ah, right... hmmm

@TastyRomeo If the section is globally nonvanishing then sure

(because it gives a trivialization)

Right. If you have a globally nonvanishing section, your line bundle is trivial.

@Semiclassical And those are in turn related to some spaces that I will study

This happens iff the global sections are a free $C^{\infty}(M)$ module

Right... I should check if I'm allowed to assume that

@Mathein Truly

@MatheinBoulomenos That's just a complicated way of saying that the vector bundle is trivial though.

Free modules <=> Trivial bundles in the Serre-Swan correspondence

Basically, I need to find a one-to-one correspondence between covariant derivatives (defined as maps from sections of $E \otimes TM$ to sections of $E$) and one-forms on $M$

for rank 1 vector bundles, mind.

that looks basically like a version of the Hom-Tensor adjunction

Such a covariant derivative is an element of $\Gamma(TM; End(E))$

Definition: The anti-derivative of $x\mapsto 1\over x on $(0,\infty)$ and which maps $1$ to $0$ is called the natural logarithm of $x$ and noted $\ln$

So it's an $End(E)$-valued $1$-form on $M$, I believe

Why is it necessary to add the assertion $\ln(1)=0$

(@Danu is this correct?)

@BalarkaSen I really want to find how to write $4xyzw-w(x+y+z-w)^2$ as $\det(Ax+By+Cz+Dw)$ for symmetric 3-by-3 matrices

aaaand End(E) doesn't happen to be magically isomorphic to R?

Hmm, for a line bundle it may be.

Yeah it should be

It's $E^* \otimes E$

Ahh, of course it is. The identity endomorphism is a nowhere zero section of $End(E)$

It's trivial

if ln(1) is not 0 then e^p = 1 for p = ln(1) != 0 right?

What's a simple example of a group whose derived series doesn't stabilize (at a finite ordinal)?

A free group?

I don't actually know, it just seems plausible :)

Free group of rank $r \geq 2$ definitely doesn't stabilise in a finite number of terms. I have no idea about ordinals though

You could argue in a tiny bit complicated way that if $\varphi_{ij}$ are the transition functions of $E$, then the transition functions of $E^*$ are $(\varphi_{ij})^{-1}$ because you're a line bundle. So the transition functions of $E^* \otimes E$ is $\varphi_{ij} \varphi_{ij}^{-1}$ because you get multiplied when you tensor, so it's just identity

@TastyRomeo I was interested in a finite number of steps, but why does this example work?

That's more or less the intuition. $E^*$ twists backwards corresponding to $E$

The forward/backwards twists cancel in tensor

@BalarkaSen Okay... I'll need some time to think about this :P

@Semiclassical I'm afraid I don't know the answer

@BalarkaSen Hmm. Some thinking out loud. Suppose $f(x,y,z)=\det(Ax+By+Cz+D)$ for 3-by-3 matrices A,B,C,D

stuff I know: $f(x,y,z)=f(z,x,y)=f(y,x,z)=f(1-x,1-y,z)$

free groups are actually hypoabelian, so the transfinite central series stabilizes at the identity

I also know that $f(x,x,1)=0$

Doesn't the transfinite series always stabilize (not necessarily at the identity)?

The simplest way for that permutation symmetry to hold is if $A=B=C$

it does always stabilize, in fact it stabilizes at the largest perfect subgroup

in which case I'd also need $\det(A(x+y+z)+B)=\det(A(2-x-y+z)+B)$ and $\det(A(2x+1)+B)=0$ for all $x$

free groups on $\geq 2$ elements don't stabilize at a finite number of steps, as the commutator subgroup of any free group is again free (like every subgroup of a free group) and the commutator for free non-abelian groups is a always a proper nontrivial subgroup

should have $\det(Ax+By+Cz)=4xyz$ if I'm interpreting some stuff right

which eliminates $A=B=C$ as an option

@MatheinBoulomenos Ah, I see, thanks! So a free group on $2$ generators contains an infinite descending chain of subgroups that are all free on $2$ generators? That's weird

No, it's even weirder

@BalarkaSen Yes, that's always the case

the derived subgroup of a free group on 2 generators is free on countably infinite generators

No, wait

Covariant derivatives are an affine space over endomorphism-valued 1-forms---but they're not tensorial (they have a Leibniz rule)

@MatheinBoulomenos uhm... I can see why it's free and a proper subgroup but I have no idea about that

@Danu Oh. So difference of two covariant derivatives is an End(E)-valued 1-form?

@BalarkaSen That's more or less what I found on wikipedia, I think: en.wikipedia.org/wiki/Connection_(vector_bundle)

Yeah OK I misremembered.

Not really good at this stuff

exactly

$$(\nabla_1 - \nabla_2) \in \Omega^1(M;End(E))$$

@AlessandroCodenotti that's not that easily to prove actually. I can't think of purely algebraic proof atm

cause the part where the vector field acts on the function gets canceled

Right

that's the non-tensorial part

Thanks

Hrmf. But that ruins the idea of the one-to-one correspondence, or not? :/

Not that I'm really following with the whole End(E)-valued stuff yet...

Not if you already have a canonically chosen covariant derivative on your bundle