by the normal basis theorem $L \cong K[G]$ as $K[G]$-modules, hence $L \otimes_K L \cong K[G] \otimes_K L \cong L[G] \cong \bigoplus_{\sigma} L$

well

how about $L \otimes_K V = K[G] \otimes_K V = V[G] = \bigoplus_\sigma \sigma(V)$

yes

now let's make this explicit

first how does $K[G] \to L$ work

let $x$ be a normal element

$f(\sum a_i \sigma_i) := \sum a_i \sigma_i(x)$ looks reasonable

if only it worked like that

no it isn't K[G]-linear

suppose $f(1) = x$

then $f(\sigma) = \sigma f(1) = \sigma(x)$

wait what

ok it works

it is K[G]-linear

no no no no no

@MatheinBoulomenos we need an L-linear isomorphism $L \otimes_K V = \bigoplus_\sigma \sigma(V)$

K[G] and L are not isomorphic as rings right

no they aren't

R[Gal(C/R)] = R[Z/2Z] = R[x]/(x^2-1)

yeah so it doesn't work

why do you need a ring isomorphism?

$L \cong K[G]$ as $K[G]$-modules

because we need an L-linear isomorphism

here L is interpreted as a ring right

you don't have modules over modules

the isomorphism $K[G] \otimes_K L \cong L[G]$ is $L$-linear if you let $L$ act on the right factor of the tensor product

but that's not how $L \otimes_K V$ works

$V$ is a $L$-vector space, right?

yes

but $l_1 (l_2 \otimes v) := (l_1 l_2) \otimes v$

it's the extension of the restriction

the (co)unit

hmm, you're right

is there a functor from L-Mod to the "standard" L-vector spaces?

$L \otimes_K V \cong L \otimes_K L \otimes_L V \cong (L \otimes_K K[G]) \otimes_L V \cong L[G] \otimes_L V \cong \bigoplus_{\sigma} \sigma(V)$

@LeakyNun if you allow a sufficiently strong form of choice, yes

then you can simultanously choose a basis for all vector spaces in a compatible way

$b \otimes v \mapsto b \otimes 1 \otimes v \mapsto (b \cdot 1) \otimes v \mapsto (bv, 0, 0, 0, \cdots)$ for $b \in L$

the second isomorphism fails because again they're only isomorphic as K[G]-modules

hmm

guys

is the function f(x+iy)= $xe^y + (y + 1)e^xi$ differentiable anywhere on the complex plane?

$xe^y + (y + 1)(e^x)i$

@LeakyNun, hey, how are you? any ideas?

@topologicalmagician: You mean complex differentiable?

@MatheinBoulomenos yay I generalized it to radical extensions

hey @TedShifrin

hi Leaky

@TedShifrin yup

Did you work out whether the Cauchy Riemann equations hold?

yes, I keep getting x=\frac{-1}{2} =y as a solution

but the problem is that its written that its not differentiable anywhere

are you sure it is not written "holomorphic"?

yes, it sais to show that the function is not differentiable anywhere

You're doing something wrong.

You need both $x=y$ and $x=y+1$.

$u_x = e^y$, $u_y = xe^y$, $v_x = (y+1)e^x$, $v_y = e^x$

Oh, wait. I messed up. You need both $x=y$ and $x=-(y+1)$.

so indeed x=y=-0.5 works

So, yeah, $y=-1/2=x$ works. The only issue is, as Leaky suggests, whether they require that it be on an open set.

Depends on the definition in your course.

the question sais to show its not differentiable on any point on $\mathbb{C}$

I'm annoyed that they won't let me TA a masters course because I'm undergrad ...

challenge the director to a duel

lol

@Mathein: It would present an issue with your having authority over more advanced/older students.

@topologicalmagician: Tell me the book's definition of that phrase.

@Ted yeah I mean it kind of makes sense, but I still I think I'd be qualified

That isn't the question, @Mathein.

Sometimes policies have a rationale.

@TedShifrin A subset $U$ of $C$ is open if for every point x in U there exists an $\r>0$ such that $B(x,r)$ is contained in U

@topologicalmagician: I know what an open set is.

I'm asking for the book's definition of complex differentiable at $z=a$.

Oh, sorry I misunderstood. Hold on

"Holomorphic" requires an open set in most texts. I don't know what your text has as a convention for "complex differentiable."

@TedShifrin It just sais that a function $f: U\rightarrow \mathbb{C}$ defined on an open set U is differentiable at $c\in U$ if the lim$_{h\rightarrow 0}$ …… exists

strange

Then it would seem you're right.

challenge your professor to a duel

holomorphic if its differentiable at every point of c

And if he doesn't like that, offer a dual instead.

sorry, not c. I meant U

The map isn't holomprhic though, right?

No, if the definition is that it has to be complex diff on a neighborhood of the point ...

In my course, I said holomorphic was the same as complex differentiable (at a point).

But I distinguished between that and analytic. Whereas a lot of (older?) books confuse the issue and use analytic for differentiable. Agh.

Of course they're equivalent on open sets. That's the whole point of complex differentiability.

wait, analytic and "holomorphic" aren't equivalent?

That's a big theorem. To me the definitions are important to keep straight.

I'm just saying some books don't.

It's like saying a matrix is invertible is equivalent to saying it's non-singular. But the definitions are totally different!

Oh, I see.

@LeakyNun You've been reading too much Galois stuff

@AlessandroCodenotti correct

you should read about Galois categories

@MatheinBoulomenos This has happened in Bonn before, even though I didn't witness it first hand

do the categories engage in duels?

categories engage in duals

Ted copyrighted that pun

Maybe there are warring categories.

@MatheinBoulomenos so now what do I do... assume every extension is radical?

what if $f_\sigma(v) := \sum_\sigma \sigma(x) \otimes \sigma(x)^{-1} v$

@MatheinBoulomenos what does your abstract nonsense tell you

or maybe I just focus on $L \otimes_K L$

how exactly do you define the $L$-vector space structure on $\bigoplus_{\sigma} \sigma(L)$?

$a(b_\sigma)_\sigma := (\sigma(a) b_\sigma)_\sigma$

somehow naturality causes issues

your first objection is that $\sigma(L)$ is isomorphic to $L$

but is $\sigma(-)$ naturally isomorphic to $-$?

to be clear, $a ~ \bullet_{\sigma(V)} ~ v := \sigma(a) v$

Hi chat

$\mathrm{Hom}_K(L,K) \otimes_K L \cong \mathrm{End}_K(L) \cong L^*[G]$, where $L^*[G]$ is the twisted group algebra

Hi @Semiclassic

There’s a paper that got submitted to a journal today with me as first author

Oh, that's scary.

So that’s neat

Congratulations.

congrats @Semiclassical

$L^*[G]$ is exactly $\bigoplus \sigma(L)$

@MatheinBoulomenos so we need a basis?

and then take the dual basis?

Thanks. Now the question is if anyone actually pays attention to it :3

idk

what category are you in?

It’s a history/philosophy paper on quantum foundations stuff

I'm in L-Vec and I don't know what Hom(L,K) means

He's in the dueling category.

Did other people read it first, @Semiclassic?

With a lot of computational stuff in the middle as grist for the grind

@MatheinBoulomenos the identity map from $S \otimes_R S$ to $S \otimes_R S$ is not $S$-linear

what

@TedShifrin Not all the way through, I think. But that’s because it’s a long-ass paper

How long?

170 pages including 12 pages of bibliography, lol

@MatheinBoulomenos if you define the scalar multiplications by $s(a \otimes b) = sa \otimes b$ and $= a \otimes sb$ respectively

Holy crap. That's absurd.

well, duh

so I don't know what Hom_K(L,K) means

Lol

And you had so much trouble writing 5 pages in your thesis.

because it isn't a L-vec

it's an L vector space

A good chunk of that is solid prose contributed by the other authors

how?

define $\lambda f(v):=f(\lambda v)$

I had no idea you were involved in such a gargantuan project, @Semiclassic.

oh right

I'm a bit concerned

because I don't have an explicit formula

Keep in mind that history-type papers tend to be longer

But still, yeah

Hopefully it makes an impact :)

the morphism $L^*[G] \to \mathrm{End}_K(L)$ is given by sending $\sum_{g \in G} \lambda_g g \mapsto \sum_{g \in G} \lambda_g g$ (I know that seems tautological, but it isn't, one is a formal sum the other is a sum of k-linear endomorphisms)

It should go up on arxiv in the next day or so

I'm not concerned about this

I'm concerned about Hom(L,K)

what's concering about that?

well what's the isomorphism $L \to \operatorname{Hom}_K(L,K)$?

there's no canonical one

give me a non-canonical one

looks to the heavens and then realizes that's pointless

There’s a pretty good chance we'll need to modify the paper a bit to get it published in the journal

choose a $K$-basis for $L$, then you get a dual basis

Eg split it into 3 or something

take a bijection between those bases

well the basis is supposed to consist of only one element

so let's choose a random K-linear map L -> K and hope it works

@Semiclassic: The math journals I'm familiar with will require severe pruning. Only exceptional authors/results warrant splitting into 2 or 3 papers.

hey there is a canonical K-linear map L -> K

it's called t r a c e

I was about to say that

(The fact that you could easily split it into 3 decent-sized papers is a testament to how expansive this became)

and the trace-paring is non-degenerate for separable extension

so $b \mapsto (a \mapsto tr(ab))$

yes

@TedShifrin @LeakyNun Have a wonderful day guys, i'm going

Bye, @topologicalmagician

@topologicalmagician you too

@anakhro is it one of Littlewood's 3 principles?

if yes, that's the answer; if no, it's irrelevant to measure theory

Another option is that we publish the full account as a pamphlet and write up a short précis for the journal

But first we find out what the journal is willing to swallow :P

@LeakyNun is what one of Littlewood's 3 principles?

your question

Context:

en.m.wikipedia.org/wiki/…

I know the principles, but I don't see what they have to or don't have to do with my question or how either "yes" or "no" function as a proof of the question.

it's just a joke, I'm not a measure theorist

where did @AlessandroCodenotti go

@MatheinBoulomenos I don't feel very good

Well I found a hint for the question online and it is wacky.

I don't like when questions seem like they need some crazy guess to be done.

<--- mathematical poor sport.

@anakhro what's the hint

I've actually never seen that question before, @anakhro, and it certainly is not something I have thought about.

@TedShifrin is it because it isn't one of Littlewood's three principles? :P

smacks Leaky

Sigma-finiteness gives you an $f>0$ with $\int f\,d\mu\leqq 1$.

what

how is this relevant

Because you force $\int (1-e^{-|f_n|})f\,d\mu < 1/2^n$. Then magic happens.

Mathemagic tricks are not fun exercises.

I remember that Munkres taught me in point-set topology many decades ago that a method is a trick you use three or more times. I wonder if this trick turns into a method.

There are plenty of such tricks in real analysis.

Analysis is now officially harder for me than geometry.

is $tr$ even $K$-linear

It's depressing.

trace is not K-linear.

you need (1/n) tr right

disappears without a trace

@LeakyNun trace is K-linear

What trace are we talking

isn't tr(a) = na

yes

how does that contradict K-linearity?

wait nvm

@MatheinBoulomenos have you confirmed my calculations

it must be wrong somewhere

the first step must already be wrong

there are multiple mistakes

what are they?

for example the isomorphism of $\mathrm{End}_K(L) \cong L^*[G]$ is not given by evaluation at $1$

why did I think tr(I) = 1

You're sure working @Mathein hard for no pay!

I know that $\lambda \cdot 1 \in L[G]$ correpsonds to left multiplication by $\lambda$

@anakhro: Because all your vector spaces are $1$-dimensional.

@TedShifrin that would do it!

How are you doing by the way, Ted?

I'm still alive and kicking, thanks, @anakhro.

@LeakyNun also $\mathrm{tr}(a-) \otimes 1$ is not mapped to $b \mapsto \mathrm{tr}(a)b$

Good to hear!

I know that $v^\ast \otimes w$ is mapped to $b \mapsto v(b) w$

oh wait what

yes exactly

aha

beautiful

also you can just assume $V=L$

any $V$ is just a direct sum of $L$s

then you don't need to insert a tensor product

no.

this isn't good

$L \otimes_K L \to L^* \otimes_K L \to \mathrm{End}_K(L) \to L^*[G]$

let's start from the opposite direction

you want the $L$-vector space structure on $L^*[G]$ to come from right multiplication btw

oh they're isomorphic

it's just the transpose

I think Leaky's symbols are making me blind. Time to disappear.

@LeakyNun no it's quite different

we don't have the standard group algebra here

we have a twisted group algebra

oh no

welp I can't form an explicit formula in the opposite direction either

goodbye concreteness

$\sigma \lambda= \sigma(\lambda) \sigma$

$L \otimes_K L \to L^*\otimes_K L\to \mathrm{End}_K(L) \leftarrow L^*[G]$

$a \otimes b \mapsto (x \mapsto \mathrm{Tr}(ax) \otimes b) \mapsto (v \mapsto \mathrm{Tr}(av)b)$

$a \otimes b \mapsto tr(a-) \otimes b \mapsto (c \mapsto tr(ac) b) = (c \mapsto \sigma(c) \lambda) \leftarrow \lambda \sigma$?

would an primitive basis or a normal basis make things better?

you want $(c \mapsto \lambda \sigma(c)) \leftarrow \lambda \sigma$

ok

$L$ is commutative so it's the same

@LeakyNun not every tensor is an elementary tensor so we have no reason to expect that the image of $\lambda \sigma$ will be of that form but yeah

oh lol

I'm not sure how $tr$ reacts to a primitive/normal basis

if a primitive basis is a normal basis then are we in a cyclotomic field?

we is $L \cong L^*$ such a big deal? the first argument of the tensor product is fixed for the whole thing

@LeakyNun I think so yeah

well because that's where we define the $L$-vector space structure

$a(b \otimes v) := ab \otimes v$

actually {1,ω} isn't a normal basis

{ω,ω^2} is a normal basis

anyway I can't take trace of a primitive/normal basis

I don't see the problem, we have a sequence of $L$-linear isomorphisms

but I like formulas

well, that's not my problem

@loch should i present this crap tmr lmao

it's for 18.704

@LeakyNun what exactly are you supposed to present?

Do you even need the $\sum \sigma(V)$ thing?

@MatheinBoulomenos basically the relations between real and complex and quaternion representations of compact Lie groups

okay

people just read out of the book basically

@Ted might have a lot to say about seminar courses

I don't know if you can just work with abstract nonsense such as Galois descent for Lie groups

because you want your representations to be smooth

so they're not just functors

that's a valid point

but everything is given by formulas so they should be fine

@LeakyNun in particular, compactness is crucial for some arguments

you can't just take an inner product and average if you have a non-compact Lie group

that's true

but $\Bbb C \otimes_\Bbb R V = V \oplus \overline{V}$ works

btw what's an invariant subspace of the rep O(2) -> GL(2,C)?

that's the base change of O(2)->GL(2,R), so do the $\Bbb C \otimes_{\Bbb R} V = V \oplus \overline{V}$ thing

lol

brilliant

@LeakyNun I found a better way of doing that measure theory question btw

That isn't witchcraft.

@anakhro how

measure theory is witchcraft

@MatheinBoulomenos V isn't a complex vector space

hmm, right

Sigma finiteness gives us our $A_1\subseteq A_2\subseteq \dotsc \subseteq X$ and then we want to find a $c_n>0$ so that $B_n := \{x\in A_n \mid c_n|f_n(x)| > 1/n^2\} < 1/n^2$.

Then you use the Borel-Cantelli lemma on the sequence $B_n$.

And then it follows.

@MatheinBoulomenos I feel like given a nonzero vector $v \in \Bbb C^2$ you can rotate it by 90 degrees (using the right element in O(2)) and then you have a linearly independent vector

so O(2) -> GL(2,C) is irreducible?

but O(2) is abelian

aha (a,b) and (-b,a) being linearly independent depends on them being real right

(1,i) and (-i,1) are linearly dependent!

$\begin{pmatrix}a&b\\-b&a\end{pmatrix} \begin{pmatrix} 1 \\ i \end{pmatrix} = \begin{pmatrix} a+bi \\ -b+ai \end{pmatrix} = (a+bi) \begin{pmatrix} a \\ b \end{pmatrix}$

and that is our invariant subspace

nice

that's for SO(2) only though

it doesn't work for O(2)

hmm

$\begin{pmatrix}a&-b\\-b&-a\end{pmatrix} \begin{pmatrix} 1 \\ i \end{pmatrix} = \begin{pmatrix} a-bi \\ -b-ai \end{pmatrix} = (a-bi) \begin{pmatrix} 1 \\ -i \end{pmatrix}$

O(2) is still abelian...

unless it isn't

D_2n embeds in O(2)

haha

I have a question: How general can we obtain a simple approximation of a measurable function by simple functions if we do not need the simple function sequences to be monotonic?

@LeakyNun it's probably irreducible

yeah

any 2-d faithful representation of a nonabelian group is irreducible

where did you pull that from

since diagonal matrices commute

at least assuming it's completely reducible

@LeakyNun it just occured to me

nice

I wish theorems would just occur to me

does $\chi(g^2)$ integrate to $1$ over $O(2)$?

Heya everybody! I just have a quick question - is this proof fine? $$$$ The question is asking to prove that an increasing function is injective. $$$$ I already asked this question yesterday but didn't include a proof. So, now that I wrote one, I just wanted to confirm that it makes sense and is valid.

I feel like it integrates to $0$

since it's just 2cos(2θ)

but the book says it should integrate to 1 if it is of real type

That's my proof: $$$$ Given that a function is strictly increasing for some $a, b ∈ ℝ$, the function would always have different outputs. Thus, by the definition of injectivity, $a ≠ b$ and this implies that $f(a) ≠ f(b)$, proving that the function is indeed injective. QED

@LeakyNun are you sure you're integrating over a Haar measure? you have a $\det$-factor there

Could there be a Baire Space that is a Lindeoff space? 🤔

oh what

why?

and does that make it not integrate to 0

@LeakyNun the Haar measure for $\mathrm{GL}_n$ is given by $\int f(x) |\det(A)|^{-n} \mathrm{d}\lambda(x)$ if $\lambda$ is the n^2-dimensional Lebesgue measure

you can restrict that from $\mathrm{GL}_2$ to $O(2)$

well if you absolute then you just get 1

so it makes no difference

hmm, true

@HarryBattersby let $a,b \in \mathbb{R}$. Without loss of generality suppose $a<b$ then $f(a)<f(b)$. So the map is injective

is O(2) just D(continuum)

like instead of finitely many points you have R/Z

then do the usual construction

@topologicalmagician Do I have to state at the end the definition of an injective function as well? Or would this statement be sufficient to prove it?

@HarryBattersby Its always better to be clear in your writing.

@topologic Great. Thanks. So, would you say that the proof I provided above would be valid?

@MatheinBoulomenos ok we also care about p-adic representations right, so that L/K thing isnt pointless generalization

yeah

generality is always good

@HarryBattersby (1) "Given that a function is strictly increasing for some $a, b ∈ ℝ$, the function would always have different outputs." (What if $a=b$?) (2) "Thus, by the definition of injectivity, $a ≠ b$ and this implies that $f(a) ≠ f(b)$, proving that the function is indeed injective." ( You're supposed to assume $a\neq b$ then show $f(a)\neq f(b)$

But you have the right idea

@LeakyNun the integral of $\chi(g^2)$ over $SO(2)$ vanishes yeah, but for the integral over the $-1$-determinant matrices doesn't vanish

They say proving the contrapositive of the injection function is easier