random cohomology for quantum nerds

Leaky Nun

04:00

in Mathematics, 8 hours ago, by MatheinBoulomenos

the version for finite Galois extension is: if $L/K$ is a finite Galois extension with Galois group $G$, then if we consider the category $L^*[G]$-Mod which consists of $L$-vector spaces together with a semilinear $G$-action, then $L^*[G]$-Mod is equivalent to k-Mod via thefunctors $V \mapsto V^G$ and $W \mapsto L\otimes_K W$

in Mathematics, 8 hours ago, by MatheinBoulomenos

that's actually an equivalence of symmetric monoidal categories if we equip $L^*[G]$-Mod with the tensor product over $L$ and $K$-Mod with the tensor product over $K$

in Mathematics, 8 hours ago, by MatheinBoulomenos

thus by abstract nonsense, you also get an equivalence of the corresponding categories of monoid objects and commutative monoid objects

in Mathematics, 8 hours ago, by MatheinBoulomenos

which are $K$-algebras and $L$-algebras with a semilinear $G$-action by ring automorphisms (respectively the commutative ones for commutative monoid objects)

in Mathematics, 8 hours ago, by MatheinBoulomenos

and then continuing by abstract nonsense, you get a category equivalence of the corresponding cogroup objects in commutative $K$-algebras which are dual to affine algebraic groups

in Mathematics, 8 hours ago, by MatheinBoulomenos

there's also some Galois descent for projective varieties, but that's harder

let $V$ be a $L^\ast[G]$-module. We need to show that there is an isomorphism $V \to L \otimes_K V^G$ of $L^\ast[G]$-modules

Leaky Nun

04:19

maybe the inverse direction is easier

$L \otimes_K V^G \to V$

does $b \otimes a \mapsto ba$ work?

Conrad claims that this works

and the proof is interesting

let $v_1, \cdots, v_n$ be $K$-linearly independent vectors in $V^G$

suppose $a_1 v_1 + \cdots a_n v_n = 0$

so for each $\sigma \in G$ we have $(a_1 - \sigma(a_1)) v_1 + \cdots + (a_n - \sigma(a_n)) v_n = 0$

WLOG $a_n = 1$, so actually we have $(a_1 - \sigma(a_1)) v_1 + \cdots + (a_{n-1} - \sigma(a_{n-1})) v_{n-1} = 0$

this is a shorter relation, so by induction we have $a_i = \sigma(a_i)$

so each $a_i \in L^G = K$ to start with

so $a_i = 0$

hey I never knew Galois descent's proof is this short

wait it isn't done, we need to show that the action is preserved

so we need to show that $\sigma(a)v = \sigma(av)$

this is clear since $v = \sigma(v)$

ok so this is short

oh there's a slick proof to show that $[L:K](\dim V^G) = \dim V$

consider the projection map $V \to V^G$ sending $v$ to $\sum_\sigma \sigma(v)$

actually it's not a projection map

ok if we're in char. coprime to $[L:K] =: n$ then we can send $v$ to $\frac1n \sum_\sigma \sigma(v)$

I'll come back to this argument later (Conrad didn't use this)

Leaky Nun

06:15

there is something to complete in this proof, I don't have the time

Galois descent is a special case of Morita equivalence of matrix algebra!

$L^\ast[G] = \operatorname{End}_K(L)$ as rings

Leaky Nun

08:46

in Mathematics, yesterday, 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 $\sigma \cdot \tau = (\sigma \circ \tau) f(\sigma,\tau)$

I still don't understand why it is $\lambda \cdot \sigma = \sigma \cdot \sigma(\lambda)$ instead of the other way round

given a $L$-vec with $G$-semilinear action we define $(b \sigma)(v) = b \sigma(v)$

then $(a\sigma)((b\tau)(v)) = (a\sigma)(b\tau(v)) = a \sigma(b) \sigma(\tau(v)) = (a\sigma(b)\cdot(\sigma\circ\tau))(v)$

so $(a \sigma) \cdot (b \tau) = a \sigma(b) \cdot \sigma \cdot \tau$

so $\sigma \cdot b = \sigma(b) \cdot \sigma$?

we want this to be an $L$-algebra

which it is

I don't see why this fails

also this is isomorphic to $\operatorname{End}_K(L)$ as rings so

Leaky Nun

09:52

@MatheinBoulomenos

MatheinBoulomenos

10:02

@LeakyNun oh

maybe I misremembered

I just wanted the natural map to $\mathrm{End}_K(L)$ to be an isomorphism

but I think for that you need $\sigma \cdot \lambda =\sigma(\lambda) \sigma$ lol

I'm stupid

@LeakyNun I'm also not sure if we want $\sigma \cdot \tau = (\sigma \circ \tau) f(\sigma,\tau)$ or with the $f(\sigma,\tau)$ on the other side

one of those works

Let's see: we need associativity, so we want

@LeakyNun btw: this construction does not produce an $L$-algebra

the center is $K$, not $L$

okay I think what I gave just gives a construction of the opposite algebra

so I think it should be $\sigma \cdot \tau= f(\sigma,\tau) (\sigma \circ \tau)$ if you want $\sigma \cdot \lambda = \sigma(\lambda) \sigma$.

let's check that this satisfies associativity:

$(\sigma \cdot \tau) \cdot \theta = (f(\sigma,\tau) (\sigma \circ \tau)) \cdot \theta = f(\sigma,\tau) f(\sigma \circ \tau, \theta) (\sigma \circ \tau \circ \theta)$

on the other hand $\sigma \cdot (\tau \cdot \theta)=\sigma \cdot (f(\tau,\theta) \cdot (\tau \circ \theta))= \sigma(f(\tau,\theta)) \sigma \cdot (\tau \circ \theta)= \sigma(f(\tau,\theta)) f(\sigma,\tau \circ \theta) (\sigma \circ \tau \circ \theta)$

so we want $f(\sigma,\tau)f(\sigma \circ \tau,\theta)=\sigma(f(\tau,\theta))f(\sigma,\tau \circ \theta)$

$d^2(f)(\sigma,\tau,\theta)=\sigma(f(\tau,\theta)) \cdot f(\sigma \circ \tau,\theta)^{-1} \cdot f(\sigma,\tau \circ \theta) \cdot f(\sigma,\tau)^{-1}$

so it all works out

Leaky Nun

10:48

@MatheinBoulomenos oh and which book can I find this in?

some books don't give explicit formulas

and proceed via PGL

MatheinBoulomenos

11:11

idk

just work it out yourself

I'm not sure if you can totally avoid PGL

Leaky Nun

well I want to cite it...

MatheinBoulomenos

I think there are some explicit formulas in Jacobson - Basic Algebra 2

iirc

I've never seen the generalized form of Galois descent written down anywhere

Leaky Nun

@MatheinBoulomenos je n'ai pas le tems

MatheinBoulomenos

do you mean "temps"?

Leaky Nun

that's how Galois wrote it

MatheinBoulomenos

11:14

okay

I'm pretty sure it's "temps" in modern French

maybe Galois made a typo

Leaky Nun

yes, he misspelled it

guess why

because il n'avait pas le temps

MatheinBoulomenos

parce qu'il n'avait pas le temps?

damn it

Leaky Nun

ok P.472 contains the formulas $\sigma \cdot \tau = f(\sigma, \tau) (\sigma \circ \tau)$ and $\sigma \cdot \lambda = \sigma(\lambda) \cdot \sigma$

thank you

MatheinBoulomenos

np

Leaky Nun

19:34

so we have two resolutions of $\Bbb Z$ as a $\Bbb Z[G]$-module where $G=C_n$

MatheinBoulomenos

yes

Leaky Nun

$\cdots \to \Bbb Z[G] \xrightarrow N \Bbb Z[G] \xrightarrow{g-1} \Bbb Z[G] \xrightarrow N \Bbb Z[G] \xrightarrow{g-1} \Bbb Z[G] \xrightarrow \varepsilon \Bbb Z \to 0$

MatheinBoulomenos

jup

Leaky Nun

$\cdots \to \Bbb Z[G^5] \to \Bbb Z[G^4] \to \Bbb Z[G^3] \to \Bbb Z[G^2] \to \Bbb Z[G^1] \to \Bbb Z \to 0$

MatheinBoulomenos

if you want Tate cohomology, you can just take $\dots \to \Bbb Z[G] \xrightarrow{N} \Bbb Z[G] \xrightarrow{g-1} \Bbb Z[G] \xrightarrow{N} \Bbb Z[G] \to \dots$

Leaky Nun

19:41

I've forgotten the maps

you map $(1, g_1, g_1 g_2, \cdots)$ to $(1, g_1, g_1 g_2, \cdots)$?

that doesn't have $d^2=0$

ok the last map is still $\varepsilon$

MatheinBoulomenos

you have some alternating sum

$(g_1, \dots, g_n) \mapsto \sum_{j=1}^n (-1)^{j+1}(g_1,g_2, \dots, \widehat{g_j}, \dots, g_n)$

Leaky Nun

oh, that one

nice

now I want maps downwards

MatheinBoulomenos

they exist by general nonsense about projective objects

why would you want to write them down?

Leaky Nun

because I'm an explicit guy

since $\Bbb Z[G]$ is rank 1 it suffices to specify where $1$ gets sent to

MatheinBoulomenos

there's a more useful explicit isomorphism than the one you get by a construction like that

Leaky Nun

19:54

what is it?

$d(g,1) = g-1$ btw

MatheinBoulomenos

as a special case of Tate-Nakayama, if you fix a generator of $\alpha \in H^2(C_n,\Bbb Z)$ (which is just cyclic of order $n$ again), then for all $C_n$-modules $M$ and all $k \in \Bbb Z$, the map $\beta \mapsto \beta \cup \alpha:\widehat{H}^k(C_n,M) \to \widehat{H}^{k+2}(C_n,M)$ is an isomorphism

this is nice and useful because cup products satisfy various compatibilities with restriction/corestriction/induction/coinduction

Leaky Nun

but I want a cocycle $C_n^k \to M$

MatheinBoulomenos

the cup product is defined on the level of cocycles

Leaky Nun

oh nice

@MatheinBoulomenos what's the formula?

CF writes: let $\varphi: A \otimes B \to C$, then $\widehat{H}^p(G,A) \otimes \widehat{H}^p(G,B) \to \widehat{H}^p(G,C)$ is given by $a \otimes b \mapsto \varphi^\ast(a.b)$

this is tautological, this isn't the formula I want

MatheinBoulomenos

$\cup:C^n(G,A) \times C^k(G,B) \to C^{n+k}(G,A \otimes B)$
$(f \cup f')(g_1, \dots, g_n,g_{n+1}, \dots, g_{n+k})=f(g_1, \dots, g_n) \otimes g_1g_2 \dots g_n f'(g_{n+1}, \dots, g_{n+k})$

Transcript for

♦ $Mathematics$ random cohomology for quantum nerds

Transcript for

♦ random cohomology for quantum nerds

♦ $Mathematics$ random cohomology for quantum nerds