35 votes
Accepted

$H^4(BG,\mathbb Z)$ torsion free for $G$ a connected Lie group

I try to give an argument without spectral sequences, not sure if this can be considered non-computational though. At least, there is a non-computational syllabus: torsion classes in $H^4(BG,\mathbb{Z}...
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34 votes

What is the status of the Friedlander-Milnor conjecture today?

Here is what I know about the status of the question (as of today): At the moment, there is no field of characteristic 0 for which the Friedlander-Milnor conjecture is known for a semisimple ...
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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 ...
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  • 3,525
25 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.
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22 votes
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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 $...
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  • 2,102
21 votes
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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 ...
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  • 51.8k
20 votes
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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 ...
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  • 49.7k
19 votes

(co)homology of symmetric groups

The paradoxical answer is that it is annoying but straightforward to determine $H_k(S_n)$ for particular values of $k$ and $n$, but it is very easy to write down $H_*(\coprod BS_n)$, encompassing all $...
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  • 29.1k
19 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 ...
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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.
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18 votes
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Is there a finitely presented group with infinite homology over $\mathbb{Q}$?

Thompson's group F is an example. It's finitely presented and, according to this paper of Ken Brown, the integral homology is free abelian of rank 2 in every positive dimension.
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  • 22.5k
18 votes
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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 ...
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18 votes
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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 \...
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  • 39.9k
18 votes
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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
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16 votes
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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'') ...
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16 votes
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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]...
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  • 4,773
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 ...
15 votes

(co)homology of symmetric groups

The answer to the question 1) is "yes". The classifying space of the symmetric group is of finite type, so its integral homology is determined by its rational homology and $p$-local homology for all $...
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  • 2,978
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)...
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  • 2,978
14 votes

Why is the standard definition of cocycle the one that _always_ comes up??

Late to the party as usual, but: the goal of this answer is to convince you that the standard convention for $2$-cocycles is so natural that you should consider it perverse to consider any other ...
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14 votes

Non-smooth function with all differences of translates smooth?

At least under the axiom of choice the answer is negative: The are discontinuous additive functions $f;\mathbb R \to \mathbb R$ (as far as I remember, that's why Hamel inveted Hamel basis) and for ...
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14 votes

Universal coefficient theorem for group homology and cohomology

I do not know a good reference for this, but if $k$ is a field and $M$ is a $k[G]$-module and $M^{\ast} = Hom(M,k)$ is its dual, then there is a natural isomorphism from $H^k(G;M^{\ast})$ to $Hom(H_k(...
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  • 39.9k
14 votes
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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
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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)...
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14 votes
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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 ...
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14 votes
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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_{\...
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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 $\...
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  • 6,049
14 votes
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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 ...
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13 votes
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Fundamental theorem of K-theory for loop groups over $\mathbb{F}_1$?

Let $G_n := W(\tilde{A}_{n-1})$. If I understand your description correctly, there is an extension $$1 \to G_n \to S_{n} \wr \mathbb{Z} \overset{sum}\to \mathbb{Z} \to 1$$ and so a $\mathbb{Z}$-Galois ...
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13 votes
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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 $G/N$-...
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  • 4,597

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