Mathematics - 2018-06-02 (page 1 of 3)

MatheinBoulomenos

00:00

or you could say coverings of $\Bbb C^\times$ since we talked about that more

Leaky Nun

@MatheinBoulomenos but their limits are different now

MatheinBoulomenos

what are you talking about? limits of categories? I don't think you're actually talking about limits in the category of categories

Leaky Nun

well I think you know what I mean

they are all canonically subcategories of some category

I'm taking limit there

MatheinBoulomenos

I'm not sure what you mean is well-defined

Ted Shifrin

wonders if Leaky forgot to cook dinner again tonight

Leaky Nun

00:02

canonically well-defined

@TedShifrin oh I ate tonight :P

thanks

MatheinBoulomenos

a category is a subcategory of many different categories

Leaky Nun

yes, but we all know which category we're in

i know it isn't well-defined

@MatheinBoulomenos don't you get canonically the universal cover?

which has galois group Z instead of Z-hat

anyway

if $K$ is a global field

choose an algebraic closure $\overline K$

then we have $\Gamma = \operatorname{Gal}(\overline{K}/K)$

does $\Gamma$ have trivial outer automorphism group?

MatheinBoulomenos

you need to take the separable closure here

Leaky Nun

right

no

let's say it's a finite extension of Q

everything works fine

MatheinBoulomenos

it definitely doesn't have trivial outer automorphism group I think

Leaky Nun

00:07

because my professor tells this story

two people constructed their own algebraic closures L and M

and then they wrote down an isomorphism from their algebraic closure to the other guy's algebraic closure

and then when you compose them together, there's a 0% chance you get the identity

you'll be conjugating by something

and then he gave the geometric analogue

that if $X$ is path-connected, then $\pi_1(X,x)$ and $\pi_1(X,y)$ are isomorphic, but not in a canonically way

it is isomorphic up to a choice of path

and then the isomorphism is a conjugation

MatheinBoulomenos

probably the absolute galois group of $\Bbb Q$ has trivial outer automorphism group, but when you have a number field of degree $>1$ over $\Bbb Q$ (say Galois), then the absolute Galois group will have non-trivial outer automorphism Galois group

this follows from the Neukirch-Uchida-Pop theorem

Leaky Nun

that's a very curious name

MatheinBoulomenos

which says that automorphisms of a number field correspond to outer automorphisms of the absolute Galois group

Leaky Nun

I see

hmm, so he's telling the story in $\Bbb Q$ then

MatheinBoulomenos

so iff it has non-trivial automorphisms it will have non-trivial outer automorphisms of the absolute Galois group

Leaky Nun

00:11

his stories are quite interesting

MatheinBoulomenos

so the only number field with that property is $\Bbb Q$

Leaky Nun

he said, there's an element of Gal(Q-bar/Q) called the complex conjugation

MatheinBoulomenos

one element? there are a lot

Leaky Nun

humour me

and then look at the three roots of X^3-2, let's call them alpha,beta,gamma

Ted Shifrin

never humor Leaky

Leaky Nun

00:12

@TedShifrin :(

which one does that element fix?

then he goes on to say that this question is nonsensical

Ted Shifrin

besides, in Britain it should be humour

Leaky Nun

everything is non-canonical

and we should look at its conjugacy class instead

i.e. we look at its one-dimensional representations

MatheinBoulomenos

one dimensional-representations throw a lot of more information away then just conjugacy

every one-dimensional representation factors over the abelianization

Leaky Nun

I see

oh, I mean, n-dimensional representations

just representations

MatheinBoulomenos

so we're essentially just looking at $\operatorname{Gal}(\overline{\Bbb Q},\Bbb Q)^{ab}$ which is $\widehat{\Bbb Z}^\times$

these are still uncanonical

Leaky Nun

00:15

why is that that?

MatheinBoulomenos

you still have conjugation in representations of higher degree

Leaky Nun

oh, it's Kronecker-Weber

MatheinBoulomenos

but you can take e.g. characteristic polynomials which will be independent of conjugation

Leaky Nun

we look at the characters

MatheinBoulomenos

okay

I'm actually taking a course on Galois representations right now

Leaky Nun

00:17

interesting

MatheinBoulomenos

we don't really work with characters, though

Leaky Nun

so in the setting of a non-archimedean local field $K$

with ring of integers $\mathcal O_K$ and residue field $k$

MatheinBoulomenos

Also one thing to note: complex representations of profinite groups are kinda boring. If $\rho: G \to \operatorname{GL}_n(\Bbb C)$ is a continuous representation where $G$ is profinite, then the image of $\rho$ is finite

Leaky Nun

@TedShifrin change my mind: $\operatorname{GL}_n$ is a scheme

Ted Shifrin

I don't do this stuff.

MatheinBoulomenos

00:20

so you just have a representation of the finite group $G/\operatorname{ker}(\rho)$

Leaky Nun

@MatheinBoulomenos so, in the setting above

@TedShifrin 2algebraic4u

@MatheinBoulomenos under the equivalence of categories above

MatheinBoulomenos

to the freshman LA class "the determinant is a homomorphism of group schemes $\mathrm{GL}_n \to \mathrm{GL}_1$"

Leaky Nun

41 mins ago, by Leaky Nun

For any non-archimedean local field $K$ (ring of integers $\mathcal O_k$ and residue field $k$), the following three categories are canonically equivalent:
1. $L/K$ finite unramified extension
2. $R/\mathcal O_K$ unramified ring extension
3. $l/k$ finite separable extension

taking the canonical non-well-defined limits of 1 and 3 respectively

MatheinBoulomenos

what

Leaky Nun

we get a separable closure K-bar of K, and an algebraic closure k-bar of k

MatheinBoulomenos

00:22

wouldn't you get unramified extension of K that may have infinite degree?

For 1

Leaky Nun

I don't know anymore

MatheinBoulomenos

well, since "taking a limit" in this sense is undefined, there's no point in arguing

Mike Miller

@Mathein I don't know about that stuff, but let's chase it through. How would you calculate colimits in a presheaf category?

MatheinBoulomenos

@MikeMiller pointwise (this works for any functor category)

and then you have to sheafify

Leaky Nun

@MatheinBoulomenos is "taking the limit of the fintie unramified extensions of K" well-defined?

MatheinBoulomenos

00:25

@LeakyNun note though that any possibly infinite unramified extension is a colimit of finite unramified extensions and every algebraic extension of $l/k$ (separable is superfluous, as we noted) is a colimit of finite algebraic extensions

both colimits are filtered actually

non-filtered probably don't exist because fields are weird categorically

Leaky Nun

is taking the algebraic closure of K a good thing to do?

MatheinBoulomenos

note that "filtered colimit" is a fancy way of taking unions here

@LeakyNun if you want to have an algebraically closed field, sure

Leaky Nun

my professor says that K-bar/K gives k-bar/k

MatheinBoulomenos

but if I should take the "limit" of that category, whatever that should be, it would be unramified extensions

gives?

in what sense

Leaky Nun

in the sense that if you have K-bar, then looking at its residue field gives k-bar

MatheinBoulomenos

00:28

There's a surjective continuous homomorphism

that's true

but you can also take $K^{ur}$

Leaky Nun

we don't need unramified for that one?

oh

MatheinBoulomenos

purely ramified extensions don't change the residue field at all

Leaky Nun

and then it turns out that every automorphism of K-bar that fixes K gives an automorphism of k-bar that fixes k

MatheinBoulomenos

sure

Leaky Nun

so Gal(K-bar/K) maps to Gal(k-bar/k)

but the latter is just Z-hat

then this map is supposed to be surjective

MatheinBoulomenos

00:30

yeah

Leaky Nun

why?

MatheinBoulomenos

that's nontrivial

Leaky Nun

ok

MatheinBoulomenos

this holds in a lot more generality

Leaky Nun

now Z lives as a dense subset of Z-hat

generated by arithmeticus Frobenius or geometricus Frobenius

MatheinBoulomenos

00:31

are you defining the Weil group now?

Leaky Nun

bingo

MatheinBoulomenos

that's isomorphic to $K^\times$

Leaky Nun

its abelianization is

MatheinBoulomenos

ah yeah

Leaky Nun

how do you define its topology?

MatheinBoulomenos

00:33

I was thinking about $W(K^{alg}/K)$ because that's what I've been working with this whole time

Leaky Nun

but that's what I said

MatheinBoulomenos

hmm, I don't think that a topology on the Weil group is used

Leaky Nun

it is

MatheinBoulomenos

I worked through the proof of local CFT and it never mattered

Leaky Nun

because the continuous 1-cocycle from W_K to ${}^LT$ is supposed to be in bijection with something

MatheinBoulomenos

00:35

I don't think the right topology on the Weil group is the subspace topology

(if you use one)

Leaky Nun

no, it isn't

my professor emphasised

you don't use the induced topology from Z-hat for Z

you use the discrete topology

now is this a pullback? do I just use the pullback topology?

MatheinBoulomenos

pullback? you mean quotient topology?

Leaky Nun

Gal(K-bar/K) surjects to Z-hat

Z with discrete topology injects to Z-hat

take the pullback and obtain $W_K$

MatheinBoulomenos

I don't think that's called a pullback

But I see what you mean

@LeakyNun the proof of that surjectivity is Thm. 9.4 in chapter 1 of Neukirch

he proves it for finite extensions, but you can lift that to infinite ones

Leaky Nun

what's the topology on ${}^LT$?

MatheinBoulomenos

00:46

what is $^LT$?

Leaky Nun

good question

it's the langlands dual

MatheinBoulomenos

no idea about that

Leaky Nun

T-hat is a complex torus equipped with a left Gal(K/F)-action

then you take the semidirect product of the two groups mentioned

and then you get ${}^LT$

@MatheinBoulomenos what should I read to learn more about the above discussion?

MatheinBoulomenos

well, for basics on local fields, Serre - local fields or Neukirch chapter 2

For the Langlands stuff, maybe Bushnell-Henniart "The local Langlands Conjecture for GL(2)"

Leaky Nun

I don't need GL(2) though

MatheinBoulomenos

00:58

this has an elementary proof of a special case

Leaky Nun

I just need GL(1)

MatheinBoulomenos

I don't know what you need

Leaky Nun

I just told you

MatheinBoulomenos

GL(1) is class field theory

Leaky Nun

I feel like ${}^LT$ is a nice group

it's easy to work with

it's just a bunch of $C^\times$ semidirectproduct a finite group

but $W_K$

here be dragons

MatheinBoulomenos

01:00

Bushnell-Henniart talks about the topology of the Weil group

Leaky Nun

I see

MatheinBoulomenos

I'd recommend Serre - local fields, as I mentioned

it doesn't have this langlands stuff

but you need CFT first anyway

Also note the answers to this MO question: mathoverflow.net/questions/66500/…

Leaky Nun