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 ...
- 3,635
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.
- 52.2k
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 $...
- 2,160
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 ...
- 59.4k
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 ...
- 59k
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 ...
- 3,644
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 \...
- 42.8k
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.
- 17.5k
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
- 196
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 ...
- 704
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]...
- 5,520
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'') ...
- 20.5k
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 ...
16
votes
Accepted
Generators for the first cohomology of free groups
They are a generating set. In fact, even more is true. Recall that for a group $G$ and a $G$-module $M$, the first cohomology group $H^1(G;M)$ is the abelian group $Der(G,M)$ of derivations $G \...
- 42.8k
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)...
- 3,031
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 "...
- 379
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 ...
- 33.2k
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
$H_3(SL_\infty(\mathbb Z);\mathbb Z) = \mathbb Z/24$
$H_3(SL_\infty(\mathbb F_q);\mathbb Z) = \mathbb Z/(q^2-1)$
...
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 ...
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)...
- 50.6k
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_{\...
- 4,659
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 $\...
- 8,194
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: ...
- 59.4k
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 $n$-...
- 150k
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
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
Accepted
Yoga of six functors for group representations?
The accepted answer here is on a rather negative note -- I don't think that's fair! In fact, I think the correct answer is that all of this works, except that it is $\pi_\ast$ that gives group ...
- 18.4k
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 $...
- 34.9k
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.
- 2,030
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 ...
- 17.1k
Only top scored, non community-wiki answers of a minimum length are eligible