Mathematics - 2018-10-12 (page 5 of 5)

Perturbative

21:00

@ÍgjøgnumMeg Let me try and prove that now

ÍgjøgnumMeg

@Perturbative the fundamental theorem of arithmetic is your friend

Leaky Nun

@TobiasKildetoft what is the signifcance of the T thing?

I don't know what twisting means

Tobias Kildetoft

@LeakyNun Ahh, we have a homomorphism $f: G\to G$ and a representation $\varphi: G\to GL_n$, so we can get a new rep by composing the two to get the twisted representation $\varphi\circ f: G\to GL_n$.

Leaky Nun

ok

Tobias Kildetoft

So now each $g$ acts as $f(g)$ instead.

Leaky Nun

21:03

when you say "twisting" I think of "conjugation"

go on then

Alucard

@ÍgjøgnumMeg what means the $\to$ here?

ÍgjøgnumMeg

@Alucard ignore me it was a shi**y joke

lol

Tobias Kildetoft

Now, it is not obvious from this, but all of these $L(n)$ are irreducible, and all irreducible $SL_2$ representations have this form

Alucard

ah ok^^ @ÍgjøgnumMeg

ÍgjøgnumMeg

@Alucard are you aware of the notation $f : X \to Y$ though?

Tobias Kildetoft

21:05

So the main thing left is to try to understand these better

Leaky Nun

@TobiasKildetoft wait, if you generate an invariant subspace with one vector, isn't it guaranteed to be irreducible?

Alucard

@ÍgjøgnumMeg yeah

a mapping

Tobias Kildetoft

@LeakyNun Not necessarily. Think of the subspace generated by the neutral element inside the group algebra of a finite group.

ÍgjøgnumMeg

@Alucard that's what I meant then

Leaky Nun

that's too many words

ok I get it

why not though

ah ok

because we're generating it as a vector space

not just considering its orbit

Tobias Kildetoft

21:07

Right, we might have some linear combination of the generated vectors itself be invariant

Leaky Nun

ok

Tobias Kildetoft

But in this case, it works out with this specific choice of vector

Alucard

a 3d object with zero height is no 3d object i guess...

Leaky Nun

now what

Tobias Kildetoft

Now, we also know that for $n < p$ we have $L(n) = \nabla(n)$, and we might feel like we understand these $\nabla(n)$ better (it is for example very easy to find their dimension, whereas it is right now not so obvious what the dimension of $L(n)$ should be for large $n$)

And this is where Steinberg's tensor product formula comes in and saves us. It states that if $n = a_mp^m + a_{m-1}p^{m-1} + \cdots + a_1p + a_0$ is the $p$-adic expansion of $n$, then $L(n)\cong L(a_0)\otimes L(a_1)^{(1)}\otimes\cdots\otimes L(a_m)^{(m)}$.

ÍgjøgnumMeg

21:11

@Alucard $(x, y, 0)$ is an element of $\Bbb R^3$

Alucard

@ÍgjøgnumMeg ah, i see

Tobias Kildetoft

And this is again isomorphic to $\nabla(a_0)\otimes \nabla(a_1)^{(1)}\otimes\cdots\otimes\nabla(a_m)^{(m)}$

Leaky Nun

did you say p-adic

ÍgjøgnumMeg

triggerd

Tobias Kildetoft

@LeakyNun I just mean base $p$

Leaky Nun

21:12

:P

to me, the p-adic expansion has Teichmuller coefficients

(should have)

ÍgjøgnumMeg

lol

Tobias Kildetoft

So this immediately gives us the dimension of $L(n)$

Leaky Nun

as the product of the coefficients? this doesn't seem right

Tobias Kildetoft

Steinberg's tensor product theorem is btw now specifically for $SL_2$. It just needs some more setup to explain what it means for arbitrary semisimple groups

Leaky Nun

no, as the sum

Tobias Kildetoft

21:14

@LeakyNun Product of one more than each coefficient

Leaky Nun

oh

ok

Tobias Kildetoft

(which means that it is ok to add $0$s to the expansion)

Leaky Nun

that sounds rather unnatural

but ok

Tobias Kildetoft

You can work out the case $L(p+1)$ to see it in action if you want.

Leaky Nun

so L(p+1) is supposed to be L(1) x L(1)

Tobias Kildetoft

21:16

Right, except with the second factor twisted

ohh, and tensor product, not direct product

Leaky Nun

I guess I don't know how to work it out...

the natural thing to do is to send X otimes X^p to X^(p+1)

this should be it

lol

Tobias Kildetoft

So check if that works out with $Y$ as well and the multiplication

Leaky Nun

I mean, I can see why it works

a00a^-1 sends X^(p+1) = X * X^p to a^(p+1) X^(p+1) = aX * a^p X^p

just like the one with the tensor product

Tobias Kildetoft

Right, but you need to use that $p$ is the characteristic of the field at some point to see that this works out

(otherwise it will not)

Leaky Nun

right

can we continue

Tobias Kildetoft

21:27

So now we have a complete description of all irreps of $SL_2$ as certain tensor products of twists of these representations on homogeneous polynomials

I need to be off to bed soon though, so we should probably stop here for now. Next step would probably be something like tilting modules, which I am less sure of a good description of in these terms

Leaky Nun

@TobiasKildetoft then how do we get GL2

Tobias Kildetoft

Right, let $Det$ be the $1$-dimensional rep where a matrix in $GL_2$ acts by its determinant

All the $L(n)$ we defined before lift to representations of $GL_2$ (define them the same way).

And now any irrep of $GL_2$ will be of the form $L(n)\otimes Det^m$ for some integer $m$

And these are still all distinct

Leaky Nun

what is det^m?

Tobias Kildetoft

acting by $m$'th power of the determinant

Leaky Nun

do we really need tensor?

Tobias Kildetoft

21:34

Not sure what you mean by "need"

Leaky Nun

can you write it without tensor

Tobias Kildetoft

sure, this is essentially twisting by that power of the determinant, like we did with a hom from $G$ to itself

or rather, not really twisting, but multiplying by that scalar in addition to the usual action

Leaky Nun

hmm

Tobias Kildetoft

So if $g$ would act as the linear map $f$ without the $Det^m$ there, it now acts by $Det(g)^m f$

Leaky Nun

ok

Tobias Kildetoft

21:38

Ohh, and if we want representations of $GL_2(\mathbb{F}_{p^r})$ (still reps over the algebraically closure of that finite field), then we just take the $L(n)$ with $n < p^r$ and the $m$ with $0\leq m < p^r-1$

@LeakyNun Anyway, now I really do need to be off to bed. Cya

Leaky Nun

thank you very much

see you

so can someone tell me what kind of rep is this

I don't think this is just group rep

but "homogeneous polynomials with deg n" isn't really a scheme

ÍgjøgnumMeg

22:00

@Perturbative how's it going? :P

Perturbative

@ÍgjøgnumMeg I got the trivial direction, didn't manage to get the other one yet :p

ÍgjøgnumMeg

@Perturbative the trivial direction being if $p^2 \mid n$ then... ?

Perturbative

I'm not sure if this is correct, but $p^2 | n$ implies $p^2 k =n$ for some $k$, then $p^2k^2 = nk$, then $(pk)^2 \equiv 0 (mod \ n)$ so $\overline{pk}$ is nilpotent

If that is hogwash I'm sorry

I'm way too sleepy to be doing any sort of math now :p

ÍgjøgnumMeg

Right, and also $pk \neq 0$ in $\Bbb Z/(n)$

(so this is a non-zero nilpotent element)

Perturbative

Thanks for the help @ÍgjøgnumMeg

I'll take a look at the reverse direction tomorrow

ÍgjøgnumMeg

22:08

Alright, no problem :P

Perturbative

I'm off to bed now, night everyone!

ÍgjøgnumMeg

Goodnight!

loch

@LeakyNun i didn't read the above convo so i could be wrong - but is $L(n)$ just hte vector space of homog. poly of degree $n$ (and $0$)?

Leaky Nun

(also, the answer is that, if $n = p_1^{a_1} \cdots p_k^{a_k}$ then the nilpotent elements of $\Bbb Z/n\Bbb Z$ is the ideal generated by $(p_1 \cdots p_k)$ @ÍgjøgnumMeg @Perturbative)

@loch it's the invariant subspace generated by $X^n$

ÍgjøgnumMeg

@Leaky claps slowly

Leaky Nun

22:11

@ÍgjøgnumMeg I mean, that's what he initially asked, isn't it

ÍgjøgnumMeg

@Leaky sure but "the answer is ... " is not particularly helpful :P

loch

22:23

@LeakyNun ?

Leaky Nun

@loch $\nabla(n)$ is the vector space of homog. poly of degree $n$ (and $0$) equipped with the action of SL2

$L(n)$ is the invariant subspace generated by $X^n$

Alucard

22:39

maybe we are just a product of a bufferoverflow, drinks water

loch

oh ok i skimmed through it now - it's some characteristic $p$ stuff!

Leaky Nun

@loch it's rep of GL2

loch

@LeakyNun anyway i suspect the reason why Tobias mentioned AG is because $\Lambda(n)$ is the global sections of the line bundle $\mathcal{O}(n)$ over $\mathbb{P}^1$

Leaky Nun

@loch I asked KB, and he says that in this context, rep of GL2 means a group scheme morphism GL2 -> GLn

here K is alg. closed so we can recover this from the map of K-points, GL2(K) -> GLn(K)

i.e. from reps of GL2(K)

loch

sure

Leaky Nun

22:51

@loch anyway what the hell is this

loch

it's a fancy language of saying "homogeneous polynomials of degree $d$"

essentially when you work in projective space it's quite natural that people consider things like zero locus of homogeneous polynomials right --- analogous to the affine case

but unlike the affine case where polynomials are really "functions to $k$"

you cannot think of homogeneous polynomials as "functions to $k$" (because it's not well-defined if you define it the naive way) -- in fact it turns out the only regular functions on a projective variety are constants (say / k algebraically closed)

in a sense sections of bundles give you "generalised functions"

but anyway i'm not 100% confident if that's what he said related to AG - probably it's better for you to talk to him when he's back haha

en.wikipedia.org/wiki/… if you look at the example it's basically what i said

Sebastian

23:08

How rapidly is the field of statistics changing? I just started a statistics course (lagunita.stanford.edu/courses/HumanitiesSciences/StatLearning/…) and I wonder how different the content would've been, say, ten years ago

Alucard

is this a simple algebraic problem? suppose a benevolent dictator wants to build a house for every human in the world, would there be enough space?

to me it is

Leaky Nun

@Alucard there's no such thing as a benevolent dictator

Alucard

23:32

@LeakyNun it is hard to think about a 1 government world

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