Mathematics - 2018-05-31 (page 4 of 5)

MatheinBoulomenos

21:00

but yeah complex analysis is great

we're doing modular forms and it's not clear from the definition why they are so number-theoreticy

loch

@LeakyNun around the same time - but i might work on my own stuff more

Leaky Nun

ok

GFauxPas

Hot take: complex analysis only feels cool because you're limiting yourself to such a small amd rigid class of functions

runs away

TreFox

Ok I'm scared to ask this on main because it's probably incredibly stupid, but I'm currently modeling a problem where a guy is at height $h$ and has to get over a wall of height $1$, is there any way to create a function saying how far he has to go to get over the wall without using a piecewise function? Because if he's above $1$ the function should return $0$, otherwise just $1-h$.

Ted Shifrin

@TreFox, piecewise it is, unless you use $\max(1-h,0)$.

TreFox

21:13

Dang it. Guess I'll have to, thanks.

MatheinBoulomenos

Hey @Ted

Ted Shifrin

hi @Mathein

Faust

Morning

MatheinBoulomenos

@Ted Here's a proof you'd probably like that a 2x2 matrix over $\Bbb R$ that satisfies $A^2=I$ and $A \neq \pm I$ is similar to $\begin{pmatrix}-1&0\\ 0&1\end{pmatrix}$ that avoids minimal polynomials (since they are less known then they should be imo): chat.stackexchange.com/transcript/message/44897351#44897351

Ted Shifrin

I have exercises like (and more general than) that in my linear algebra text.

MatheinBoulomenos

21:18

I see

I mean the general statement is that the image of any finite-dimensional complex represenation of a compact group is conjugate to a subgroup of $U(n)$

Ted Shifrin

Your proof is unusual. You do need to check that you have an inner product. That isn't obvious to me.

I mean, why is it positive definite?

MatheinBoulomenos

$\langle v,v\rangle + \langle Av,Av\rangle >0$

B. Mehta

Hi @TedShifrin, I've just finished a proof that if $f_n \to f$ pointwise, and $f_n$ are $L^p$ bounded then $f_n \to f$ weakly for $p \in (1,\infty)$. But in my proof I didn't seem to use that $p \neq 1$ nor that $p \neq \infty$ even though the question gave $p \in (1, \infty)$. Do you know if the statement still holds for $p = 1$ or $\infty$, or should I pick through my proof once more?

Ted Shifrin

Ah, right. My usual approach for problems like this is to show that the eigenspaces span and so it's diagonalizable, etc.

@B.Mehta: I know less of this sort of stuff (at this point in my life) than many people who are here, e.g., @Mathein, @Alessandro, @EricSilva, @Daminark.

MatheinBoulomenos

@B.Mehta I think it holds for $p=1$ but not for $p=\infty$

GFauxPas

21:23

let me see if i have this in my notes

MatheinBoulomenos

since you know that the dual space of $L^1$ is $L^\infty$ so applying a functional is just multiplying by some $L^\infty$ function and then integrating, so you can do the Hölder + dominated convergence argument I assume you did

but the dual space of $L^\infty$ is just a mess

GFauxPas

well you have to be careful because

B. Mehta

@MatheinBoulomenos Ah yeah, I did use the dual space: that explains the $p < \infty$

GFauxPas

$L^1 \subsetneq (L^\infty)^*$

so that's why, they're not duals

MatheinBoulomenos

I don't see why it shouldn't work for $p=1$

GFauxPas

21:26

take a picture and upload it to imgur?

if you dont have the proof typed out

B. Mehta

If it helps, one of the steps was to show that for $f_n \in L^p(\mathbb{R})$ for $p > 1$ and $f_n$ are $L^p$ bounded then $f \in L^p(\mathbb{R})$

That seems to be the only place where the question enforced $p > 1$

@GFauxPas I would but it's not exactly legible

GFauxPas

are you challenging me to a context of who has worse handwriting

be careful whom you pick a fight with

B. Mehta

I did that part using Fatou, but I don't see why it requires $p \neq 1$

@GFauxPas Not a challenge, just sparing you from picking through my nonsense

Leaky Nun

@MatheinBoulomenos are you interested in algebraic groups, especially tori?

Faust

おはいお ^^

MatheinBoulomenos

21:29

I'm interested in algebraic groups

Ted Shifrin

@GFauxPas: conteSt

heya @Faust

Faust

surprised chat supports japanese

Doing lots of topology :p

MatheinBoulomenos

@GFauxPas I will win that easily

Faust

well and japanese

Ted Shifrin

Is your health improving, @Faust?

Leaky Nun

21:30

@MatheinBoulomenos should I tell you what a torus is?

MatheinBoulomenos

I know what a torus is

it's a mug

Leaky Nun

so $GL_1$ is a functor $k\text{-Alg} \to \text{Ab}$ for a fixed field $k$

the easiest way to see this is via this computation

Faust

@TedShifrin small bit

Ted Shifrin

:(

Faust

i can mange 8-10hrs a day

Leaky Nun

21:32

$GL_1(A) = A^\times = \operatorname{Hom}(k[X,X^{-1}],A) = \operatorname{Hom}(\operatorname{Spec}(A), \operatorname{Spec}(k[X,X^{-1}]))$

Ted Shifrin

well, don't overdo anything, @Faust

Faust

which is alot better than it has been

Leaky Nun

contravariant, contravariant, so covariant

MatheinBoulomenos

@B.Mehta so yeah if you take the pointwise lim sup of the $|f_n|$, then you can argue either by Beppo-Levi or by Fatou that this is $L^1$ and this wil give you some dominating integrable function. Then you can apply dominated convergence. That you multiply by some $L^\infty$ function doesn't matter a bit

Leaky Nun

and you get representation for free

So now $GL_1$ is canonically identified with $\operatorname{Spec}(k[X,X^{-1}])$ as an affine $k$-scheme

Faust

21:33

got a problems textbook for topology it doesnt prove much is basically just 1600 exercises of proving relevant ideas and exercises with solutions to around half in the back of the book

Leaky Nun

For a field $k$, a $k$-scheme $A$ is a split torus if it is isomorphic to the scheme $GL_1^n$ for some $n$

if the field is ambiguous, we write $GL_{1,k}$ instead of $GL_1$

Ted Shifrin

sounds like Schaum's Outline, @Faust.

Leaky Nun

Then, for a field $k$, a $k$-scheme $A$ is a torus if there is a finite (Galois?) extension $L/k$ such that $A \times_k \operatorname{Spec}(L)$ is a split torus over $L$

i.e. $A \times_k \operatorname{Spec}(L) \cong_L GL_{1,L}^n$ for some $n$

Daminark

Hello again nerds

Leaky Nun

hi @Daminark

I should have written $T$ instead of $A$

MatheinBoulomenos

21:36

@LeakyNun I think that's equivalent to being "geometrically split", so that if you take $L$ a separable closure of $k$ then ...

Leaky Nun

indeed

what I don't understand is why we say separable closure instead of algebraic closure

are they equivalent?

MatheinBoulomenos

inseparable stuff is evil

Leaky Nun

what will happen?

MatheinBoulomenos

not sure

Ted Shifrin

work in char 0 and you won't need to worry

MatheinBoulomenos

21:38

it can happen that your schemes are no longer reduced if you base change along inseparable extensions

Leaky Nun

@TedShifrin sadly not all local fields have char 0

MatheinBoulomenos

even if they were originally reduced

Leaky Nun

oh

aha

all local fields are perfect ain't they

Daminark

@LeakyNun the horror

Leaky Nun

we don't need to care about separbility

MatheinBoulomenos

21:39

@LeakyNun no

$k((T))$ isn't perfect

Daminark

Rip

Leaky Nun

RIP

but $k$ is perfect :c

MatheinBoulomenos

if you adjoin $\sqrt[p]{T}$ where $p=\operatorname{char}(k)$

then rip

Leaky Nun

rip

GFauxPas

@B.Mehta it has something to do with $(L^1)^{**} \ne L^1$, im trying to find a page that explains why precisely

Leaky Nun

21:40

@MatheinBoulomenos can the separable closure of $k((T))$ be constructed explicitly?

MatheinBoulomenos

@GFauxPas but why do you need the double dual in the argument?

@LeakyNun yes

B. Mehta

@GFauxPas Surely the Fatou part still works?

Leaky Nun

@MatheinBoulomenos how?

Daminark

What's the question here?

Leaky Nun

who are you asking?

Daminark

21:42

GFauxPas and B. Mehta

MatheinBoulomenos

@LeakyNun ah no wait I'm not sure

Daminark

I probably wouldn't be able to participate in the algebra conversation just yet :P

MatheinBoulomenos

I know how it works for $\Bbb C_p$

B. Mehta

@Daminark chat.stackexchange.com/transcript/message/44913399#44913399

Now just wondering why $p > 1$ was necessary

Ted Shifrin

Demonark: You were on my list of people who should consider that question.

Daminark

21:44

Hmm, so right off the bat, I can say that in $\ell^1$, weak and strong covergence are the same

MatheinBoulomenos

@LeakyNun you take Hahn-Witt series over the algebraic closure of $k$ and then a subfield of that

Daminark

So that's already a bit suspect

B. Mehta

In $\ell^1$ or in $L^1$?

Daminark

The former, this isn't a proof so much as, that already feels stronger than what is true

Leaky Nun

@MatheinBoulomenos interesting

@MatheinBoulomenos i.e. separable closure of $\Bbb Q_p$?

MatheinBoulomenos

21:45

ah lol no

Hahn-Witt series are for $\Bbb C_p$

Hahn-series are for $k((T))$

Leaky Nun

hmm

MatheinBoulomenos

@LeakyNun it's a completion of the algebraic closure

Hahn series

In mathematics, Hahn series (sometimes also known as Hahn–Mal'cev–Neumann series) are a type of formal infinite series. They are a generalization of Puiseux series (themselves a generalization of formal power series) and were first introduced by Hans Hahn in 1907 (and then further generalized by Anatoly Maltsev and Bernhard Neumann to a non-commutative setting). They allow for arbitrary exponents of the indeterminate so long as the set supporting them forms a well-ordered subset of the value group (typically Q {\displaystyle \mathbb {Q...

Leaky Nun

how do you find $\Gamma$?

@AlessandroCodenotti buongiorno

Ted Shifrin

oh, is demonic Alessandro here?

Alessandro Codenotti

Buonasera, it's almost midnight!

Hi @Ted

Leaky Nun

21:47

buonasera

MatheinBoulomenos

@LeakyNun you take $\Gamma = \Bbb Q$

B. Mehta

@Daminark @GFauxPas @MatheinBoulomenos The full question is here if you're interested

Leaky Nun

how do you find the algebraic closure of $k((T))$ in the massive field $\overline k[[T^\Bbb Q]]$?

or more directly, the separable closure?

Is that even a field?

B. Mehta

a) and b) are just walking through a proof of Egorov, which didn't seem to be used for c) and d) - I'm ultimately wondering why $p > 1$ was assumed

MatheinBoulomenos

Buona quasi-mezzanotte! @AlessandroCodenotti

@LeakyNun you take the elements which are algebraic :P

@LeakyNun idk honestly

Alessandro Codenotti

21:50

@LeakyNun you write a bar over it

Leaky Nun

fair enough

MatheinBoulomenos

but note the remark "in fact, it is possible to give a somewhat analogous description of the algebraic closure of $K((T))$ in positive characteristic as a subset of $K\left[\left[T^{\Gamma }\right]\right]$" with a reference in the wiki article

Leaky Nun

@MatheinBoulomenos and for $\Bbb C_p$, what is $\Gamma$?

MatheinBoulomenos

it's still $\Bbb Q$

Leaky Nun

ok

MatheinBoulomenos

21:54

for $K$ algebraically closed of characteristic $0$, the algebraic closure of $K((T))$ inside $K\left[\left[T^{\Gamma }\right]\right]$ is given by Puiseux series, i.e. $\bigcup_{n \geq 1} K((T^{1/n}))$

that quote is saying that there is something analogous to that for finite characteristic

Leaky Nun

If I want to solve $V \otimes_k L = U$ for $V$, where $L/k$ is a Galois extension, do I do something called Galois descent?

Are you familiar with Galois descent?

MatheinBoulomenos

If you want to have the completion of the algebraic closure, you just take the topological closure of the algebraic closure inside $K\left[\left[T^{\Gamma }\right]\right]$

Leaky Nun

https://chat.stackexchange.com/search?q=Galois+descent&room=36

last time mentioned 2 years ago

MatheinBoulomenos

I know roughly what it is about, but I don't know details

Leaky Nun

RIP Galois descent

@MatheinBoulomenos can you tell me what you know about it?

Faust

21:57

@TedShifrin not sure what that is its a russian book but its pretty good theres some really hard questions in it

Ted Shifrin

ah, ok

MatheinBoulomenos

@LeakyNun there's some description in terms of non-abelian Galois cohomology, but let's not do that

@LeakyNun honestly, most of what I know is from this quite elementary exposition which I read some while ago. I couldn't explain it better from my memory than in that pdf istself

math.uconn.edu/~kconrad/blurbs/galoistheory/galoisdescent.pdf

Leaky Nun

what a coincidence

I'm reading that exact paper

Daminark

@B.Mehta so one thing that I could see as breaking it has to do with the fact that the unit ball is weakly compact in $L^p$ only for $1<p<\infty$

Like, I don't quite see how this would violate compactness

MatheinBoulomenos

okay

Daminark

22:01

Or at least not immediately

Leaky Nun

but thanks

Daminark

But that feels like the candidate for what breaks

MatheinBoulomenos

if you have questions I can try to help, but I wouldn't want to explain it from the ground up

Leaky Nun

@MatheinBoulomenos understood

verstanden

but is it what I'm looking for?

6 mins ago, by Leaky Nun

If I want to solve $V \otimes_k L = U$ for $V$, where $L/k$ is a Galois extension, do I do something called Galois descent?

I don't want to spend my time on irrelvant stuff, I don't have much time

B. Mehta

@Daminark That's reasonable - I don't think any arguments looked at a subsequence though

MatheinBoulomenos

22:03

what kind of objects are $L$ and $U$?

Leaky Nun

$L/k$ field extension

$V$ is $k$-algebra

MatheinBoulomenos

oh I meant $V$ and $U$

Leaky Nun

(so $U$ is $L$-algebra)

MatheinBoulomenos

okay, so yeah that exposition is just for vector spaces

Leaky Nun

it's actually the coordinate ring of an $k$-algebraic $k$-affine $k$-group $k$-scheme

@MatheinBoulomenos but an algebra is a vector space

MatheinBoulomenos

22:04

the thing with Galois descent for $k$-algebras is that it's just a special case of faithfully flat descent (which is a special case of mondacity theorems in category theory, but nevermind), so modern books don't tend to treat it separately

Daminark

Yeah and I don't think it's true that a bounded sequence in L^1 has a point wise covergent subsequence

MatheinBoulomenos

I haven't learned faithfuly flat descent yet

Leaky Nun

@MatheinBoulomenos so that pdf wouldn't help me?

MatheinBoulomenos

@TedShifrin knows it, but it's probably too algebraic that he wants to talk about it :P

@LeakyNun only for motivation, I guess

Leaky Nun

@TedShifrin bitteschoen

@MatheinBoulomenos so do you have any reference?

MatheinBoulomenos

22:08

@LeakyNun the one that I want to learn it from eventually is Görtz-Wedhorn "Algebraic Geometry I"

@LeakyNun maybe you could ask Tobias when he's around if the setting of (affine) algebraic groups simplifies Galois descent, that could be possible

Leaky Nun

thanks

god knows what I'm going to write in the "reference"

> many gods in math chat and in real life

MatheinBoulomenos

@LeakyNun no offence, but why are you learning so much heavy algebraic geometry all of the sudden? This takes years to learn for most, I'm not sure if you can just read it up to understand a paper

Leaky Nun

it's for a project

GFauxPas

"project"

Leaky Nun

I'm grateful that many people are willing to help me

@MatheinBoulomenos AG is just the first half of the correspondence though

the other half is the Weil group which comes from, you know, local class field theory

MatheinBoulomenos

22:15

yeah

Leaky Nun

this is crazy

@MatheinBoulomenos let's give you the whole setting

MatheinBoulomenos

I proved that the Weil group is isomorphic to the multiplicative group of a local field this tuesday

(building upon previous talks, of course)

but yeah, give me the setting, this sounds interesting

Leaky Nun

let $F$ be a field, and $K/F$ a finite Galois extension. Let $T$ be an $F$-torus (I just made this term up) that splits over $K$. Then, let $L = X^\ast(T_K) = \operatorname{Hom}(T_K, GL_{1,K})$ be the character group of $T_K$. Let $\Gamma = \operatorname{Gal}(K/F)$. Then, $\Gamma$ acts on $L$ on the right by the dictionary $l.\sigma : T_K(K) \to GL_{1,k}(K) : t \mapsto \sigma^{-1}(l(\sigma(t)))$. Then, the paper claims that it is well known that $L$ with the right-action of $\Gamma$

can uniquely determine $T$

And he goes on to say that this defines an anti-equivalence of categories between:

1. $F$-torus that splits over $K$

2. finitely-generated free abelian groups with a right $\Gamma$-action

MatheinBoulomenos

@LeakyNun this won't help you much with your concrete situtation, but I can give you some general intuition what descent is about (from a categoric perspective) if you want

Leaky Nun

ok

Kasmir Khaan

22:23

just ok leaky ?

Leaky Nun

thanks

Kasmir Khaan

show some appriciation -.-

:D :D :D

Hi yall :)

Leaky Nun

hi

Faust

Morning @KasmirKhaan you doing well?

MatheinBoulomenos

Hi @KasmirKhaan

@LeakyNun you know adjunctions, right?

Kasmir Khaan

22:23

@Faust hi thanks faust am ok and you ?

Leaky Nun

yes

Kasmir Khaan

@MatheinBoulomenos hello :D

MatheinBoulomenos

do you know the formulation with unit and counit?

Leaky Nun

I've heard of it

Faust

@KasmirKhaan sad missed the last page of my last japanese test cause it said it was two pages but there was a third so got zero on all the questions on it.

Kasmir Khaan

22:24

@Faust wait what ? you study japenese and math ? :D

Leaky Nun

Hom(L(R(-)),-) is naturally isomorphic to id, or something like that

Kasmir Khaan

@Faust and sorry to hear that

Faust

はいそおですか

Kasmir Khaan

@Faust masaka ? :O

MatheinBoulomenos

@LeakyNun not necessarily

Faust

22:25

lol i said yes, that is correct

Kasmir Khaan

:D

Faust

its good way to get extra units as the all count as two classes

Kasmir Khaan

Hmm ._. i only do math atm

and work extra

Faust

and im exceptional at rote memorization

MatheinBoulomenos

Suppose $F$ is left adjoint to $G$

Leaky Nun

22:27

my poster can rip in peace

Faust

so i cna learn vocab words alot faster than most people ^^

Kasmir Khaan

okay faust let us let mathein and leaky talk and try to lean something :D

Leaky Nun

I'll spend the entire thing just defining what a torus is

Faust

lol, i have a meet now anyway have a good night everyone.

Leaky Nun

farewell

MatheinBoulomenos

22:28

Then we can look basically at $Hom(FG-,-)$ and $Hom(-,GF-)$ to get some natural transformations $\varepsilon: FG \to id$ and $\eta: id \to GF$

you do the usual "Yoneda" trick, evaluate some map on hom sets at the identity

Leaky Nun

ok

MatheinBoulomenos

The naturality of the Hom-set bijections implies some equations for $\varepsilon$ and $\eta$, but let's not go too deep into details here

So now let's talk about something really cool in category theory: monads

Suppose you have a category with $C$, then a monad is a endofunctor $T$ of $C$ together with a natural transformation $\mu:T \circ T \to T$ "multiplication", and a natural transformation $1: \operatorname{id}_C \to T$ "unit", such that you have associativity and that $1$ is actually a unit basically

So basically like a monoid, but we have a endofunctor and our "multiplication" and "unit" maps are natural transformation

Leaky Nun

for the record I know what a group object is

(in case that will be helpful)

MatheinBoulomenos

do you know what a monoidal object is?

Leaky Nun

yes

(just group sans inverse axiom)

MatheinBoulomenos

22:35

not quite actually

Leaky Nun

is a monad a monoidal object in the 2-category?

MatheinBoulomenos

a monad is a monoidal object in the category of endofunctors

where the monoidal structure (that takes place of the product in the definition of a group object) is given by composition

Leaky Nun

so a monoidal object is an object $M \in C$ with morphisms $m : M \times M \to M$ and $e : 1 \to M$, right

MatheinBoulomenos

no

$m: M \otimes M \to M$

Leaky Nun

:o

MatheinBoulomenos

22:36

where $\otimes$ is like an additional datum that your category has

but in our case $\otimes$ is just composition of endofunctors

Leaky Nun

ok

MatheinBoulomenos

so let's look at a simple example: Take $C$ the category of sets

then we define $T(C)$ to be the set of all finite strings written in $C$ considered as an alphabet (including the empty string)

or "words" if you prefer that over strings

So let's define the unit first

it's obvious what $T$ does on morphisms I guess

Leaky Nun

free monoid generated by C?

MatheinBoulomenos

we'll get to that later yeah

but right now, it's just a set

the unit is a natural transformation from the identity functor to $T$. So for any set $X$, we define a map $X \to T(X)$ which sends $x$ to the word $'x'$

now the multiplication

remember, we have composition of functors and not products

so this is not the multiplication in the free monoid

We have $m:T(T(X)) \to T(X)$

so we take a word consisting of words $a_1, \dots, a_n$ and then we take it to the word where we concatenate $a_1a_2 \dots a_n$

it's not hard to check associativity and that the unit is actually a unit for this

Leaky Nun

what is $T(X)$?

are we identifying $X$ as a subcategory of $C$?

MatheinBoulomenos

22:43

it's the set of all words in a set $X$

lol sorry

that was dumb

$T(C)$ is nonsense

Leaky Nun

ok verstanden

MatheinBoulomenos

I defined what $T$ does on objects

okay

so one cool thing about such a monad is that we can define algebras over a monad

So when we have a monad $T$ on a category, an algebra $A$ over $T$ is an object $X$ in $T$, together with a morphism $h:T(X) \to X$

such that some diagrams commute

so basically when you go from $T^2 (X) \to X$ you could apply $Th$ first to get to $T(X)$ and then $h$ to get to $X$

or you could apply $\mu_X$ and then $h$

and these should be the same

and you can go with $\eta_X$ from $X$ to $T(X)$ and then back to $X$ with $h$ and you want that this is the identity

So can you guess what algebras over the example of a monad on $\mathbf{Set}$ are?

your observation about free monoids is very relevant for that question

okay, I'll spoil it: algebras over the $T$ we defined are just monoids

when you have a monoid $M$ then how do you define $T(M) \to M$? you just multiply your finite elements in the string together

and when you have an algebra over that monad, to recover the monoid multiplication, you only need to look at what the map $T(X) \to X$ does on words consisting of two elements

these conditions we put on $h$ translate to associativity and existence of a unit, but because of the way that our monad works. If we had taken another monad, we would get other conditions

Are you still following? @LeakyNun

Leaky Nun

yes

MatheinBoulomenos

okay so what is the connection of monads with adjunctions?

you observed correctly that $T(X)$ is just the free monoid on $X$, but considered as a set

Leaky Nun

what is $\mu_X$?

MatheinBoulomenos

22:56

@LeakyNun $\mu$ is the multiplication from the monad

that's a natural transformation

so we take the component for $X$

Leaky Nun

14 mins ago, by MatheinBoulomenos

We have $m:T(T(X)) \to T(X)$

$m$?

MatheinBoulomenos

oh sorry

I thought I used $\mu$

$\mu := m$

So when we call the forgetful functor $G:\mathbf{Mon} \to \mathbf{Set}$ and the free monoid functor $F:\mathbf{Set} \to \mathbf{Mon}$, then $T$ is just the composition of the functors $GF$

oh yeah and $\eta_X$ is $1_X$

sorry for the notational confusion

Leaky Nun

ok

MatheinBoulomenos

but the point is, this monad comes from an adjunction, at least we saw that $T$ comes from an adjunction

and as it turns out, when you describe the adjunction with unit and counit you can use these to define $\eta=1$ and $\mu=m$

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