31
votes

Accepted

### Where should I search for computations of group cohomology rings of not-too-complicated finite groups?

Simon King and David Green maintain a computer calculated computation of the mod p cohmology of many finite $p$-groups ('order at most 128, of all but 6 groups of order 243, and of some sporadic ...

26
votes

### $H^4$ of the Monster

In arXiv:1707.08388, I calculate that the cohomology class you described has order 24 and that it is not a characteristic class in the ordinary sense.

21
votes

Accepted

### Example of group cohomology not annihilated by exponent of $G$?

For each finite group $G$ there is a $G$-module $M$ that is a free abelian group of finite rank such that $H^2(G,M)=\mathbb{Z}/|G|$.
Proof: Let $I$ be the augmentation ideal of $\mathbb{Z}G$. Then $...

21
votes

Accepted

### Abelianization of general linear group of a polynomial ring

There is a discussion on whether $K$ has two elements or is larger, which strongly affects the conclusion.
One has the determinant map $\mathrm{GL}_2(K[X])\to K^*$. To show that it's the ...

21
votes

Accepted

### Is Lie group cohomology determined by restriction to finite subgroups?

After the heavy lifting done by people on MSE and in the comments, I think it's not too bad to finish off the proof that the answer is yes.
As argued by Ben Wieland in the comments, we reduce to ...

20
votes

### Group cohomology and condensed matter

The geometric interpretation for $1$-cocyles.
Recall the following construction due to Bisson and Joyal.
Let $p:P\rightarrow B$ be a covering space over the connected manifold $B$. Suppose that the ...

19
votes

Accepted

### Is the moduli space of graphs simply connected?

Yes, it is. If $G$ is a discrete group acting on a simply-connected simplicial complex $X$, then a theorem of M. A. Armstrong says that there is a short exact sequence
$$1 \longrightarrow H \...

19
votes

Accepted

### Is the cohomology ring of a finite group computable?

As I understand it this follows from Benson's Regularity Conjecture, proved by Symonds fairly recently. It says that $b_p = 2(|G|-1)$ will do.

18
votes

Accepted

### Computations in modular cohomology of finite groups

I'll discuss $H^*(GL_n\mathbb{F}_q; k)$ first, because that is my current area of research.
1) When $\mathbb{F}_q$ and $k$ have different characteristics (although $p$ typically still divides the ...

18
votes

Accepted

### What is the cohomological dimension of the commutator subgroup of the pure braid group?

Here is the answer: for $n\geq 2$ we have $\mathrm{cd}([P_n,P_n])=n-2$.
https://arxiv.org/abs/1905.05099

18
votes

Accepted

### Second Betti number of lattices in $\mathrm{SL}_3(\mathbf{R})$

The arithmetic cocompact lattices constructed in (6.7.1) of Witte-Morris' book all have torsion-free finite index subgroups with arbitrarily large second Betti number.
I will briefly recall the ...

17
votes

Accepted

### Unifying "cohomology groups classify extensions" theorems

$\newcommand{\cA}{\mathcal{A}}\newcommand{\Ext}{\mathrm{Ext}}\newcommand{\Hom}{\mathrm{Hom}}$Let $\cA$ be an abelian category; then, $\Ext_\cA^i(A,B)$ is literally $\Hom_{D(\cA)}(A, B[i])$, where $B[i]...

16
votes

Accepted

### Acyclic aspherical spaces with acyclic fundamental groups

Yes, such things exist.
Take any finitely presented infinite acyclic group $G$, for example, Higman's group.
It is a theorem by Kervaire (''Smooth homology spheres and their fundamental groups'') ...

16
votes

### Do there exist acyclic simple groups of arbitrarily large cardinality?

I just realized this is indeed, as Neil Strickland and Tom Goodwillie predicted, not hard, thanks to the fact that a directed union of simple groups is simple. Since homology commutes with direct ...

Community wiki

15
votes

### Abstract proof that $\lvert H^2(G,A)\rvert$ counts group extensions

Here is a simple way. The extension $A \to E \to G$ induces a map of classifying spaces $BA\to BE \to BG$, which is a principal fibration, so classified by (homotopy class of) a map $BG \to BBA=K(A,2)...

15
votes

Accepted

### An intuitive explanation for group cohomology via cochains?

What I'm going to say is pretty much the same that JK34 has written in their answer, but in a more elementary approach that is hopefully adding some insight.
Suppose that you want to look at the "...

14
votes

### Solvable irreducible subgroups of the $\mathbf{GL}_n$ of $\mathbf{F}_p$ ($p$ prime)

Here's a slightly different answer, less group-theoretic and more representation-theoretic than Geoff's.
Rephrasing your question in terms of $\mathbb{F}_pG$-modules, you are asking about a faithful ...

14
votes

Accepted

### H_3 of SL(n,Z) and SL(n,F_p)

Summarizing the comments, the stable ($n\geq 3$) values of $H_3(SL_n;\mathbb Z)$ are
$H_3(SL_\infty(\mathbb Z);\mathbb Z) = \mathbb Z/24$
$H_3(SL_\infty(\mathbb F_q);\mathbb Z) = \mathbb Z/(q^2-1)$
...

Community wiki

14
votes

Accepted

### A reductive group has a quasi-split inner form

Nothing is "better-suited to using the classical language"; if you cannot express things clearly via schemes then think harder about it until you can. Also, any connected reductive group over a field ...

Community wiki

14
votes

Accepted

### Computing an explicit homotopy inverse for $B(*,H,*) \hookrightarrow B(*,G,G/H)$

Yes, there is an explicit algorithm for doing this. Pick a set of representatives $a_i \in G$ for the left cosets of $G/H$. Then the inverse map is as follows.
Given any element $(g_n,\dots,g_1, g_0H)...

14
votes

Accepted

### Trivial homology with local system

For $X = BG$ local systems on $X$ can be identified with $G$-modules, and homology with the derived tensor product $-\otimes^L_{\mathbb ZG}\mathbb Z$, i.e. $H_i(X;M) \cong \operatorname{Tor}^i_{\...

14
votes

Accepted

### loop space of a finite CW-complex

This is true for finite $\pi_1$ and false for infinite $\pi_1$: Let $\widetilde{X}$ denote the universal cover of $X$, then $\Omega\widetilde{X}$ is the unit connected component of $\Omega X$, and $\...

14
votes

### Groups all of whose extensions are split

Proposition. Given a group $G$, this happens (every exact sequence $1\to G\to H\to H/G\to 1$ splits) iff $G$ has a trivial center and $1\to G\to \mathrm{Aut}(G)\to\mathrm{Out}(G)\to 1$ splits.
Lemma: ...

14
votes

Accepted

### Relation between the cohomology group of a curve and the cohomology group of its jacobian

$\def\Alb{\text{Alb}}\def\Pic{\text{Pic}}\def\CC{\mathbb{C}}\def\ZZ{\mathbb{Z}}\def\RR{\mathbb{R}}\def\cO{\mathcal{O}}$There are two abelian varieties associated to a smooth projective connected $n$-...

14
votes

Accepted

### Explanation for $\chi(\operatorname{SL}_2(\mathbb{Z})) = -1/12$ with zeta function

(Expanding my comment into an answer)
It is not a coincidence. Relating the Euler characteristic of certain arithmetic groups to the Zeta function is a theorem due to Harder [1] from 1971. It is ...

13
votes

Accepted

### The term $H^1(N,A)^{G/N}$ in the inflation-restriction exact sequence

You do not need LHS spectral sequence for this action.
The functors $H^*(N,-)$ are the derived functors of $(-)^N:G\text{-mod}\rightarrow G/N\text{-mod}$,
so they will carry a structure of $G/N$-...

13
votes

### First Galois cohomology of Weil restriction of $\mathbb{G}_m$

One can do much better: it is not necessary to assume $L/K$ is Galois (merely separable is sufficient). And in fact one can formulate the result in a manner which works beyond that of fields, working ...

13
votes

### Cohomological dimension of $G \times G$

One fairly general class of groups $G$ for which $\operatorname{cd}(G\times G)=2\operatorname{cd} (G)$ are the duality groups. A group $G$ is a duality group of dimension $n$ if there is a dualizing $...

13
votes

Accepted

### Cohomological dimension of torsion-free groups and its subgroups

This is Theorem 3.1, p. 190, in Brown, "Cohomology of groups". He also attributes it to Serre.
As a remark, this is the reason that the virtual cohomological dimension (vcd) is well-defined.

12
votes

### Yoga of six functors for group representations?

Allow me to answer not in stacks but rather in spaces (which may be appropriate if the answer is about getting intuition; e.g. for $B\mathbb{Z}/2\mathbb{Z}=\mathbb{RP}^\infty$ there is probably ...

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