random cohomology for quantum nerds

Leaky Nun

01:08

$\newcommand{\R}{\Bbb R}\newcommand{\H}{\Bbb H}$what on earth is $\H \otimes_\R \H$

shouldn't it be 16 dimensional

I guess it is $\R^{4 \times 4}$

ah you make $K^{n \times n}$ equivalent in the Brauer group

Leaky Nun

01:43

in Mathematics, 5 hours ago, by MatheinBoulomenos

@LeakyNun let $L/K$ a Galois extension with $G=\mathrm{Gal}(L/K)$ and let $f \in H^2(G,L^\times)$ a 2-cocycle, then we can construct a central simple algebra over $K$ like this. Take as a set the group algebra $L[G]$ and define the multiplication based on the relations $\lambda \cdot \sigma = \sigma \cdot \sigma(\lambda)$ and $\sigma \cdot \tau = (\sigma \circ \tau) f(\sigma,\tau)$

let's do this with the trivial 2-cocycle

then we have $\langle 1, \sigma, i, i\sigma \rangle$

$i \sigma = -\sigma i$

$\begin{array}{c|c} \times & 1 & c & i & ic \\\hline 1 & 1 & c & i & ic \\\hline c & c & 1 & -ic & -i \\\hline i & i & ic & -1 & -c \\\hline ic & ic & i & c & 1 \end{array}$

hmm my other source didn't mention the $\lambda \cdot \sigma = \sigma \cdot \sigma(\lambda)$ thing

Leaky Nun

02:06

but it can't be just $\lambda \cdot \sigma = \sigma \cdot \lambda$ because $\Bbb R^{2 \times 2}$ isn't commutative

ok I found another source containing that rule

now how am I supposed to show that that thing is $\Bbb R^{2 \times 2}$

well we know that $\Bbb C$ corresponds to $1 = \begin{pmatrix}1&0\\0&1\end{pmatrix}$ and $i = \begin{pmatrix}0&-1\\1&0\end{pmatrix}$

how about $c = \begin{pmatrix}0&1\\1&0\end{pmatrix}$

then $ci = \begin{pmatrix}1&0\\0&-1\end{pmatrix}$ and $ic = \begin{pmatrix}-1&0\\0&1\end{pmatrix}$

this must be it

so now how do we generalize to $L/K$

wait, if we interpret $\R^{2 \times 2}$ as $\operatorname{End}_\R(\Bbb C)$ then $c = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$

ok this looks perfect

claim: the set $L[G]$ under the multiplication rule $\lambda \cdot \sigma = \sigma \cdot \sigma(\lambda)$ and $\sigma \cdot \tau = (\sigma \circ \tau)$ is isomorphic to $\operatorname{End}_{K-\mathsf{Vec}}(L)$

where we send $\lambda \in L$ to $\ell \mapsto \lambda \ell$ and $\sigma \in G$ to $\sigma$

$\lambda \ell$ looks like $\mathcal{M}$

so we only need to check that for $x \in L$ we have $\sigma(\lambda x) = \sigma(\lambda) \sigma(x)$

maybe we should have $\sigma \cdot \lambda = \sigma(\lambda) \cdot \sigma$ instead

but it's ok since matrix algebras are isomorphic to their opposites

@MatheinBoulomenos so this is how the trivial element corresponds to $\Bbb R$

Leaky Nun

03:16

btw it is Dedekind's theorem that $G$ is $L$-linearly independent, which gives us that the map $L[G] \to \operatorname{End}_{K-\mathsf{Vec}}(L)$ is an isomorphism

Leaky Nun

03:37

claim: $\newcommand{Hom}{\operatorname{Hom}}$$\Hom(\widehat{\Bbb Z},\Bbb Q/\Bbb Z) = \Bbb Q/\Bbb Z$

$\varphi \mapsto \varphi(1)$

is this obvious?

I can't even come up with a single map

or is there a continuity assumption

Leaky Nun

03:53

we can write every $x \in \widehat{\Bbb Z}$ as $\sum_{n=1} a_n n!$ where $0 \le a_n \le n$

no matter what $\varphi(1)$ is, $\varphi(n!) = 0$ for sufficiently large $n$

so if we have continuity then indeed there is a unique extension

so $\varphi \mapsto \varphi(1)$ is injective

MatheinBoulomenos

04:30

$\mathrm{Hom}(\widehat{\Bbb Z}, \Bbb Q/ \Bbb Z) = H^1(\widehat{\Bbb Z}, \Bbb Q/\Bbb Z) = \varinjlim H^1(\Bbb Z/n\Bbb Z, \Bbb Q/Bbb Z) = \varinjlim Hom(\Bbb Z/n\Bbb Z, \Bbb Q/\Bbb Z) = \varinjlim \Bbb Z/n\Bbb Z = \Bbb Q/\Bbb Z$

Leaky Nun

@MatheinBoulomenos there's no continuity assumption?

and in general do we have $\operatorname{Hom}(\varprojlim A_\bullet, B) = \varinjlim \operatorname{Hom}(A_\bullet, B)$?

Hom isn't right exact...

or is it because $\Bbb Q/\Bbb Z$ is divisible

where's my elementary proof using $\sum_{n=1} a_n n!$

I think Hom(-,Q/Z) is right exact because Q/Z is injective

MatheinBoulomenos

There is a continuity assumption

Leaky Nun

cool

MatheinBoulomenos

@LeakyNun not in general no

Leaky Nun

can you prove it?

MatheinBoulomenos

04:38

Can I prove what?

Leaky Nun

that there is a continuity assumption

MatheinBoulomenos

Nobody works with $\widehat{\Bbb Z}$ as a discrete group, lol

Leaky Nun

why $H^2(G,L^\times) = H^1(G,T_1(L))$?

so we need $1 \to T_1(L) \to ? \to L^\times \to 1$

what is $T_1(L)$ even

and what do acyclic $G$-modules look like

they look injective

MatheinBoulomenos

What is $T_1(L)$?

Leaky Nun

numdam.org/article/AST_1992__209__215_0.pdf

where T1(K) is the algebraic k-torus associated to the augmentation ideal I(G) of G.

theoremoftheday.org/Algebra/Main/TotDMainTheorem.pdf

gotta love the $\tau$ that appears out of nowhere

MatheinBoulomenos

04:49

@LeakyNun I don't know

Leaky Nun

ok thanks

in Mathematics, Oct 17 at 12:32, by MatheinBoulomenos

The Albert-Brauer-Hasse-Noether theorem says that for any global field $K$ with sets of places $\Sigma$, there's a short exact sequence $0 \to \mathrm{Br}(K) \to \bigoplus_{v \in \Sigma} \mathrm{Br}(K_v) \to \Bbb Q/\Bbb Z \to 0$. That splits non-canonically.
By LCFT we have a canonical isomorphism, called invariant map $\mathrm{Br}(K_v)=\Bbb Q/\Bbb Z$ if $v$ is non-archimedean and $\mathrm{Br}(K_v)=(\frac{1}{2}\Bbb Z)/\Bbb Z$ if $v$ is real and obviously $\mathrm{Br}(K_v)=0$ if $v$ is complex. The map $\bigoplus_{v \in \Sigma} K_v \to \Bbb Q/\Bbb Z$ is given by summing up those local invari…

where did you see this short exact sequence?

it's not on wiki etc

MatheinBoulomenos

In ANT 3

Leaky Nun

Neukirch?

MatheinBoulomenos

No I meant my lecture

Leaky Nun

oh

oh and did you read my solution above

MatheinBoulomenos

04:52

But I'm sure it is in standard texts on GCFT

Leaky Nun

to why the trivial element corresponds to $K$

MatheinBoulomenos

Yeah

That's how you can prove Galois descent from Dedekind's theorem on linear independence of characters

Leaky Nun

cool

MatheinBoulomenos

@LeakyNun okay now I get it.

$I_G$ is a $\Bbb Z[G]$-Module that is finite free over $\Bbb Z$

The category of those is equivalent to Algebraic k-tori that split over K

Leaky Nun

h- h- how..

oh because Z = Z[Gal(K/K)]

and Z[G] = Z[Gal(K/k)]

like I can sort of see why they have the same "colour"

but I can't see the equivalence

I remember tori having characters

maybe that's the equivalence

the Galois group acts on the characters

MatheinBoulomenos

05:03

Yes

Leaky Nun

(I only know this because of the project I did in year 1, if you still remember)

that was the best and worst month in my life

but what's the opposite direction

ok lemme dig up my project

MatheinBoulomenos

The functor is given by sending a torus T over k that splits over K to the base change $K \otimes_k T$ and then taking the character group of that, remembering the Galois action

Leaky Nun

and the opposite functor is?

MatheinBoulomenos

uhm

Not sure if you can write that down in a simple way

Leaky Nun

because that's what we need now

or we can treat I_G as the kernel of Z[G] -> Z

Z[G] and Z are both in the category right

MatheinBoulomenos

05:09

Yes

Leaky Nun

what tori correspond to them?

one of them must be $\mathsf{GL}_{1,k}$

MatheinBoulomenos

Clearly Z corresponds to that

Leaky Nun

the character of $T$ is $\operatorname{Hom}(T_K, \mathsf{GL}_{1,K})$

so we're looking at $\operatorname{Hom}(\mathsf{GL}_{1,K}, \mathsf{GL}_{1,K})$

it does look like $\Bbb Z$ doesn't it

the self natural transformations of the functor $A \mapsto A^\times$

MatheinBoulomenos

We only need the K-points with the Galois action here, I think you can get those by looking at $\mathrm{Hom}(A, K^\times) $

Leaky Nun

it's a subset of $\operatorname{Hom}_K(K[x,x^{-1}],K[x,x^{-1}])$

it's $\{ p \in K[x,x^{-1}]^\times \mid m(p) = ?? \}$

where $m: K[x,x^{-1}] \to K[x,y,x^{-1},y^{-1}]$

I think we need $m(p) = p \otimes p$

$p = \sum a_n x^n$, $m(p) = \sum a_n x^n y^n$, $p \otimes p = \sum \sum a_m a_n x^m y^n$

$N = N^2 \implies N \in \{0,1\}$ so $p = a x^n$

$a = a^2 \implies a \in \{0,1\}$ so $p = x^n$ or $0$

but $P$ is invertible so $p = x^n$

@MatheinBoulomenos there you go, the character group is Z

now what corresponds to Z[G]...

might it be $\mathsf{GL}_{1,K}$

look Z[G] is the... generator of Z[G]-Mod?

so we need a cogenerator of the group schemes

we need A with Hom(-,A) = -

how about a cogenerator of group

I don't think it exists

Z[G] still generates your category right

Leaky Nun

05:44

@MatheinBoulomenos how does H^q(G,M) look like w.r.t H^q(G,M[i]/M[i+1]) if M = lim M/M[i]?

MatheinBoulomenos

@LeakyNun Z[G] correspond to the Weil restriction of $K^\times$ to $k$

Leaky Nun

the what restriction?

MatheinBoulomenos

the Weil restriction

Leaky Nun

thanks, didn't hear you the first time

Weil restriction

In mathematics, restriction of scalars (also known as "Weil restriction") is a functor which, for any finite extension of fields L/k and any algebraic variety X over L, produces another variety ResL/kX, defined over k. It is useful for reducing questions about varieties over large fields to questions about more complicated varieties over smaller fields. == Definition == Let L/k be a finite extension of fields, and X a variety defined over L. The functor Res L / k...

MatheinBoulomenos

Weil restriction

In mathematics, restriction of scalars (also known as "Weil restriction") is a functor which, for any finite extension of fields L/k and any algebraic variety X over L, produces another variety ResL/kX, defined over k. It is useful for reducing questions about varieties over large fields to questions about more complicated varieties over smaller fields. == Definition == Let L/k be a finite extension of fields, and X a variety defined over L. The functor Res L / k...

lol

Leaky Nun

05:50

yeah so it is as I conjectured

28 mins ago, by Leaky Nun

might it be $\mathsf{GL}_{1,K}$

MatheinBoulomenos

well, almost

Leaky Nun

so now we need to look at $\mathsf{GL}_{1,k} \to \mathsf{GL}_{1,K}$

MatheinBoulomenos

$\mathrm{GL}_{1,K}$ is not a torus over $k$

Leaky Nun

what do you mean by almost

oh then what is?

MatheinBoulomenos

$\mathrm{Res}_{K/k}(\mathrm{GL}_{1,K})$

Leaky Nun

05:52

they're really the same thing right

MatheinBoulomenos

well, no

Leaky Nun

you just compose the structural map

MatheinBoulomenos

no

Leaky Nun

then how does it work?

MatheinBoulomenos

it's described in the Wiki article

Leaky Nun

05:57

ok we're in the 2nd case where X = Spec(K[x,y]/(xy=1))

this is too complicated lol

x=x1e1+x2e2+...+xnen, y=y1e1+y2e2+...+ynen

xy-1 = ???

MatheinBoulomenos

for example $\Bbb C^\times=\mathrm{Spec}(\Bbb C[x_1,x_2]/(x_1x_2-1))$ we have a $\Bbb R$-basis of $\Bbb C$ given by $1,i$. Write $x_1=y_{1,1}+iy_{1,2}$ and $x_2=y_{2,1}+iy_{2,2}$. Then $0=x_1x_2-1=(y_{1,1}+iy_{1,2})(y_{2,1}+iy_{2,2})-1=(y_{1,1}y_{2,1}-y_{1,2}y_{2,2}-1)+i(y_{1,1}y_{2,2}+y_{1,2}y_{2,1})$

Thus $\mathrm{Res}_{\Bbb C/\Bbb R}(\Bbb C^\times)=\mathrm{Spec}(\Bbb R[y_{1,1},y_{1,2},y_{2,1},y_{2,2}]/(y_{1,1}y_{2,1}-y_{1,2}y_{2,2}-1,y_{1,1}y_{2,2}+y_{1,2}y_{2,1}))$

Leaky Nun

brilliant

that's pretty complicated

@MatheinBoulomenos so which finite central simple algebra corresponds to 1/2 for Q_2?

actually I should know this

MatheinBoulomenos

06:12

Quaternions with Q_2 coefficieints

Leaky Nun

wait what

ok what about 1/3

MatheinBoulomenos

uhm

Leaky Nun

$\frac m n$ is found in $\newcommand{\Gal}{\operatorname{Gal}}$$H^2(\Gal(K_n/K), K_n^\times) = H^2(\Gal(K_n/K), \Bbb Z) = H^1(\Bbb Z/n\Bbb Z, \Bbb Q/\Bbb Z) = \Hom(\Bbb Z/n\Bbb Z,\Bbb Q/\Bbb Z) = \Bbb Z/n\Bbb Z$

so it corresponds to a matrix algebra of $K$-dimension $n \times n$

MatheinBoulomenos

@LeakyNun no

the matrix algebra is the trivial element of the Brauer group

Leaky Nun

I mean a matrix algebra over some algebra

the K-dimension is $n \times n$

MatheinBoulomenos

06:18

well, sure $M_{1 \times 1}(A)=A$ for any algebra $A$...

Leaky Nun

$M_{n/p \times n/p}(A)$ for some $p^2$-dimensional algebra $A$

that is what I mean

MatheinBoulomenos

elements of the Brauer group are only well-defined up to Morita equivalence

so $M_{k \times k}(A)$ and $A$ are the same

Leaky Nun

yeah but I'm using your $L[G]$ construction

MatheinBoulomenos

what is $p$?

Leaky Nun

a divisor of $n$

unknown

not necessarily prime

I ran out of letters

MatheinBoulomenos

06:21

okay

Leaky Nun

so for 1/3 we need to take $\Bbb Q_2(\zeta_7)$

MatheinBoulomenos

yes

Leaky Nun

I am not ready to deal with 9-dimensional objects so

MatheinBoulomenos

For an unramified extension, the norm group is really simple

choosing a uniformizer in the base field, say $p$ in $\Bbb Q_p$ gives you canonical generators for all $H^2(\Bbb Q_p(\zeta_{p^n-1})/\Bbb Q_p,\Bbb Q_p(\zeta_{p^n-1})^\times)$

Leaky Nun

what is the generator?

so we need to specify $f(\theta^i, \theta^j)$

MatheinBoulomenos

06:25

If $L/K$ is an unramified extension of local fields and $\pi \in K$ is a uniformizer, then $N_{L/K}(L^\times)=\pi^{[L:K]\Bbb Z} \mathcal{O}_K^\times$

Leaky Nun

s.t. $f(\theta^i, \theta^j \theta^k) = f(\theta^i \theta^j, \theta^k) f(\theta^i, \theta^j) / \theta^i(f(\theta^j, \theta^k))$

why are we looking at norm groups?

MatheinBoulomenos

because that's the 0th Tate cohomology

and we have a cyclic Galois group

Leaky Nun

oh

that doesn't look like Z/nZ to me

or do you take the quotient

MatheinBoulomenos

well, we haven't taken a quotient yet

Leaky Nun

yeah you take the quotient

MatheinBoulomenos

06:27

then you get $\Bbb Z/n\Bbb Z$, canonically generated by the class of $\pi$

Leaky Nun

so what's the 2-cocycle now

MatheinBoulomenos

just apply the isomorphism between $H^2$ and $\widehat{H}^0$ :P

Leaky Nun

16:00

$\chi(g^2) = \sum \lambda^2 = 2 \sum \lambda_i \lambda_j - (\sum \lambda)^2$

Leaky Nun

16:29

no

Transcript for

♦ $Mathematics$ random cohomology for quantum nerds

Transcript for

♦ random cohomology for quantum nerds

♦ $Mathematics$ random cohomology for quantum nerds