Questions tagged [group-cohomology]

In mathematics, group cohomology is a set of mathematical tools used to study groups using cohomology theory, a technique from algebraic topology. Analogous to group representations, group cohomology looks at the group actions of a group G in an associated G-module M to elucidate the properties of the group.

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Non-vanishing of group cohomology in sufficiently high degree

Atiyah in his famous paper , Characters and cohomology of finite groups, after proving completion of representation ring in augmentation ideal is the same as $ K(BG)$, gives bunch of corollaries of ...
Sam Nariman's user avatar
73 votes
10 answers
17k views

Intuition for Group Cohomology

I'm beginning to learn cohomology for cyclic groups in preparation for use in the proofs of global class field theory (using ideal-theoretic arguments). I've seen the proof of the long exact sequence ...
David Corwin's user avatar
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70 votes
1 answer
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Nontrivial finite group with trivial group homologies?

I stumbled across this question in a seminar-paper a long time ago: Does there exist a positive integer $N$ such that if $G$ is a finite group with $\bigoplus_{i=1}^NH_i(G)=0$ then $G=\lbrace 1\...
Chris Gerig's user avatar
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21 votes
1 answer
7k views

Kuenneth-formula for group cohomology with nontrivial action on the coefficient

For a trivial action on the coefficient, we have the following Kuenneth formula for group cohomology: $$ H^n(G_1 \times G_2; M) \cong [\oplus_{i= 0}^n H^i(G_1;M) \otimes_M H^{n-i}(G_2;M)] \oplus [\...
Xiao-Gang Wen's user avatar
19 votes
7 answers
3k views

Universal cover of SL2(R) admits no central extensions?

Is it true that the universal cover of $\mathrm{SL}_2(\mathbb{R})$ has no non-trivial central extensions... as an abstract group? (that's certainly true as a Lie group) Motivation: I have a projective ...
André Henriques's user avatar
19 votes
3 answers
1k views

The Teichmüller's algebraic interpretation of $H^3$ in group cohomology

In the book "Cohomology of Groups" of Kenneth S. Brown, it is told in the introduction that Teichmüller arrived to $H^3$ in an algebraic context, i.e. that Teichmüller worked with an ...
Josué Tonelli-Cueto's user avatar
14 votes
2 answers
1k views

Hilbert 90 for algebras

Let $L\diagup K$ be a Galois extension of fields satisfying $\left[L:K\right] < \infty$. Let $B$ be a finite-dimensional (as a $K$-vector space) $K$-algebra. Then, the Galois group $G$ of $L\diagup ...
darij grinberg's user avatar
7 votes
1 answer
494 views

$G$ cocycle split to a coboundary in $J$, via a group extension

Consider a generic nontrivial $d$-cocycle $\omega_d^G \in H^d(G,U(1))$ in the cohomology group of a group $G$ with $U(1)=\mathbb{R}/\mathbb{Z}$ coefficient. In otherwords, here the $d$-cocycle $\...
wonderich's user avatar
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31 votes
3 answers
6k views

(co)homology of symmetric groups

Let $S_n=\{\text{bijections }[n]\to[n]\}$ be the n-th symmetric group. Its (co)homology will be understood with trivial action. What are the $\mathbb{Z}$-modules $H_k(S_n;\mathbb{Z})$? Using GAP, we ...
Leo's user avatar
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29 votes
4 answers
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Geometric Interpretation of the Lower Central Series for the Fundamental Group?

For any group G we can form the lower central series of normal subgroups by taking $G_0 = G$, $G_1 = [G,G]$, $G_{i+1} = [G,G_i]$. We can check this gives a normal chain $G_0 > G_1 > ... > ...
Anthony Bak's user avatar
23 votes
1 answer
3k views

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

For the purposes of this question, the Friedlander-Milnor (FM) conjecture asserts an equality of the group homology for algebraic groups, and their discretizations in the following sense: Conjecture ...
user49544's user avatar
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5 votes
2 answers
633 views

Inflate a finite-group cocycle into coboundary in non-Abelian groups

Edit: In case that there is no solution for the original question, I modify to enrich the question. We like to ask a possible specific inflation a $H^3(Q, \mathbb{R} /\mathbb{Z})$ cocycle with a ...
miss-tery's user avatar
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4 votes
1 answer
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$SO(3)$ 2-cocycle trivialized to a 2-coboundary in $SU(2)$?

I was trying to understand this interesting question by example. Let me follow their previous discussion and ask: Let a generic nontrivial 2-cocycle $\omega_2^G(g_1,g_2) \in H^2(G,\mathbb{R}/\mathbb{...
miss-tery's user avatar
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70 votes
2 answers
6k views

Group cohomology and condensed matter

I am mystified by formulas that I find in the condensed matter literature (see Symmetry protected topological orders and the group cohomology of their symmetry group arXiv:1106.4772v6 (pdf) by Chen, ...
Edward Witten's user avatar
45 votes
2 answers
3k views

$H^4$ of the Monster

The Monster group $M$ acts on the moonshine vertex algebra $V^\natural$. Because $V^\natural$ is a holomorphic vertex algebra (i.e., it has a unique irreducible module), there is a corresponding ...
André Henriques's user avatar
28 votes
4 answers
4k views

Classifying Space of a Group Extension

Consider a short exact sequence of Abelian groups -- I'm happy to assume they're finite as a toy example: $$ 0 \to H \to G \to G/H \to 0\ . $$ I want to understand the classifying space of $G$. Since ...
Aaron Bergman's user avatar
26 votes
3 answers
5k views

Why is BG infinite dimensional for G finite ?

If $G \neq \lbrace 1 \rbrace$ is a finite group with classifying space $BG$ then there are infinitely many i such that $H^i(BG,\mathbb{Z}) \neq 0$. This can be found, for example, there: Non-...
tj_'s user avatar
  • 2,170
22 votes
2 answers
2k views

Proofs of the Stallings-Swan theorem

It is a well-known and deep${}^\ast$ theorem that if a group $G$ has cohomological dimension one then it must be free. This was proved in the late 60's by Stallings (for finitely generated groups) and ...
Mark Grant's user avatar
19 votes
0 answers
600 views

How is this group theoretic construct called?

Let $G$ be a finite group, $S\subset G$ a generating set, $|g| = |g|_S = $ word length with respect to $S$. Define the "defect" of $g,h$ to be $$\psi(g,h) = |g|+|h|-|gh|$$ Then $\psi:G\times G \...
user avatar
18 votes
1 answer
869 views

When do the polynomial algebra and free algebra coincide in brave new algebra?

Given an $\mathbb E_\infty$-ring (highly structured commutative ring spectrum if you want) $R$, we have the free $R$-algebra (on one generation) $R\{t\}\simeq \bigoplus_{n\ge 0} R_{\mathrm h\Sigma_n}$ ...
A Rock and a Hard Place's user avatar
16 votes
3 answers
977 views

Cohomological dimension of $G \times G$

$\DeclareMathOperator\cd{cd}$A question that I have already posted in the Mathematics section, but which seems to be too delicate for that section (see here and here): Let $\cd(G)$ denote the ...
Stephan Mescher's user avatar
16 votes
2 answers
2k views

Equivariant cohomology vs. invariant cohomology vs. cohomology of quotient space

Given a space $X$ and an action of a group $G$ on $X$, the $G$-invariant cochains with coefficients in an Abelian group $A$ define a sub-cocomplex $\mathcal{C}^{\bullet}_G$ of the cocomplex $\mathcal{...
Dominic Else's user avatar
15 votes
5 answers
2k views

Tate Cohomology via stable categories

Situation Let $G$ be a finite group and provide $G\text{-mod} := {\mathbb Z}G\text{-mod}$ with the Frobenius structure of ${\mathbb Z}$-split short exact sequences. Denote by $\underline{G\text{-mod}}...
Hanno's user avatar
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12 votes
0 answers
1k views

Non split extension isomorphic (as a group) to a split extension

$\def\Z{\mathbb{Z}}$ Let $A$ be a finite abelian group and $G$ a finite group acting on $A$. Then the extension $0 \to A \to E \to G \to 1$ is splits if and only if the corresponding $2$-cocycle is ...
Damien Robert's user avatar
8 votes
1 answer
432 views

Spin cobordism v.s. KO theory in low or in any dimensions

It seems that from this webpage, the spin cobordism is equivalent to KO theory in low dimension. If we denote the $p$-torsion part (mean $\mathbb{Z}_{p^n}$ for some $n$) $$\Omega_d(BG)_p.$$ ...
wonderich's user avatar
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7 votes
2 answers
452 views

Using Dunwoody's results on cohomological dimension to learn about a von Neumann regular group ring

Just recently I've stumbled across Warren Dicks' book Groups, trees and projective modules (1980) and I was pretty stunned. I know nothing of group cohomology, but I gather the "tree" ...
rschwieb's user avatar
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7 votes
2 answers
3k views

Conditions for the restriction $H^i(G,A)\to H^i(H,A)$ being surjective

I was wondering what the condition is for the restriction map (in group cohomology) $H^i(G,A)\to H^i(H,A)$ to be surjective. I am a little confused about when maps between cohomology groups are ...
Earthliŋ's user avatar
  • 1,181
6 votes
1 answer
311 views

Schur Multiplier of Tarski Monsters

Is it known whether the Schur Multiplier of the Tarski monsters are finitely generated?
user114539's user avatar
5 votes
1 answer
604 views

Non-vanishing of the Tate-Shafarevich kernel in group cohomology

Let $G$ be a finite group. Let $M$ be a finite $G$-module (a finite abelian group with an action of $G$). We consider a special kind of $G$-modules; in particular, our $M$ is a finite dimensional ...
Mikhail Borovoi's user avatar
2 votes
1 answer
263 views

A description of the fundamental class of the group cohomology of the fundamental group of an orientable surface of genus $g$

I asked this question on Mathematics Stack Exchange some months ago but I got no answer. Suppose one has an orientable compact surface $S$ of genus $g\ge 2$, $x\in S$, and $G=\pi(S,x)$ the fundamental ...
Nulhomologous's user avatar
52 votes
6 answers
7k views

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

This question might not have a good answer. It was something that occurred to me yesterday when I found myself in a pub, needing to do an explicit calculation with 2-cocycles but with no references ...
Kevin Buzzard's user avatar
51 votes
2 answers
2k views

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

Recently, prompted by considerations in conformal field theory, I was lead to guess that for every compact connected Lie group $G$, the fourth cohomology group of it classifying space is torsion free. ...
André Henriques's user avatar
35 votes
3 answers
1k views

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

We fix $G=\mathrm{SL}_3(\mathbf{R})$. Let $\Gamma$ be a torsion-free cocompact lattice in $G$. Is $b_2(\Gamma)=0$? Here the second Betti number $b_2(\Gamma)$ is both the dimension of the ...
YCor's user avatar
  • 60.1k
31 votes
2 answers
1k views

Is Lie group cohomology determined by restriction to finite subgroups?

Consider the restriction of the group cohomology $H^*(BG,\mathbb{Z})$, where $G$ is a compact Lie group and $BG$ is its classifying space, to finite subgroups $F \le G$. If we consider the product of ...
overcaffeinated's user avatar
25 votes
2 answers
936 views

Is super-vector spaces a "universal central extension" of vector spaces?

Is there some sense in which the category $sVect$ of super-vector spaces is the "maximal non-trivial extension" of $Vect$ as a symmetric monoidal category? Is the $\mathbb Z/2$ that shows up in the ...
André Henriques's user avatar
25 votes
3 answers
6k views

The relationship between group cohomology and topological cohomology theories

I was recently trying to learn a little bit about group cohomology, but one point has been confusing me. According to wikipedia (https://en.wikipedia.org/wiki/Group_cohomology and some other sources ...
Geoffrey's user avatar
  • 727
23 votes
1 answer
908 views

Groups whose finite index subgroups of fixed index are isomorphic

I am interested in finitely generated groups $G$ that are residually finite and have the following property: For each $d \geq 1$, $G$ has subgroups of finite index $d$, and all such subgroups are ...
Jens Reinhold's user avatar
22 votes
0 answers
2k views

Is the equivariant cohomology an equivariant cohomology?

Suppose a finite group $G$ acts piecewise linearly on a polyhedron $X$. Then there are two kinds of equivariant cohomology (or homology). $\bullet$ With coefficients in a $\Bbb Z G$-module $M$. A ...
Sergey Melikhov's user avatar
20 votes
3 answers
1k views

central extensions of Diff(S^1) and of the semigroup of annuli

$\mathit{Diff}(S^1)$ refers to the group of orientation preserving diffeomorphisms of the circle. The semigroup of annuli $\mathcal A$ is its "complexification": the elements of $\mathcal A$ are ...
André Henriques's user avatar
19 votes
2 answers
1k views

Geometric model for classifying spaces of alternating groups

The classifying space of the nth symmetric group $S_n$ is well-known to be modeled by the space of subsets of $R^\infty$ of cardinality $n$. Various subgroups of $S_n$ have related models. For ...
Dev Sinha's user avatar
  • 4,950
19 votes
3 answers
2k views

Second nonabelian group cohomology: cocycles vs. gerbes

In 1965 Jean Giraud published two Comptes Rendus notes titled "Cohomologie non abélienne", and in 1971 he published a book with the same title. In 1966 Tonny A. Springer's paper "Nonabelian $H^2$ in ...
Mikhail Borovoi's user avatar
18 votes
1 answer
1k views

Is the moduli space of graphs simply connected?

The moduli space of graphs $MG_n$ is the quotient of Culler-Vogtmann's outer space $X_n$ by the action of $\mathrm{Out}(F_n)$. It can be thought of as the space of metric graphs homotopy equivalent to ...
Jim Conant's user avatar
  • 4,838
17 votes
1 answer
560 views

Group cochains invariant under the action of the symmetric group

Let $G$ be a finite group and $A$ an abelian group. Recall the cochain groups $$ C^k = \{f: G^k \to A\} $$ and the coboundary map $$ \delta : C^k \to C^{k+1} $$ $$ (\delta f)(g_1, \ldots, ...
Kevin Walker's user avatar
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17 votes
3 answers
786 views

Does this subgroup of "even braids" have a name?

The full braid group on $n$ strands $B_n$ admits a surjective homomorphism $p\colon\thinspace B_n\to \Sigma_n$ onto the symmetric group on $n$ letters, which takes a braid to the induced permutation ...
Mark Grant's user avatar
16 votes
1 answer
2k views

The homology of the orbit space

Suppose we have an acyclic group $G$ and let $X$ be a contractible CW-complex such that $G$ acts freely on $X$ (we do not suppose that the action is proper). Is there a way to understand the homology ...
GSM's user avatar
  • 343
15 votes
1 answer
1k views

Shapiro's lemma in the language of group extensions

I am trying to understand Shapiro's lemma for $H^2$ in the concrete language of extensions of finite groups. Let $H$ be a subgroup of a finite group $G$, and let $A$ be an $H$-module. Let ${\rm Ind}...
user05811's user avatar
  • 749
15 votes
0 answers
624 views

Dijkgraaf-Witten topological invariant

We know that given a finite group $G$ and its group 4-cohomology class $w \in H^4[G;U(1)]$, we can obtain a DW topological invariant $Z_{G,w}(M^4)$ as the partition function of the DW theory on a ...
Xiao-Gang Wen's user avatar
14 votes
1 answer
654 views

What is the first Pontryagin class of the $n$-dimensional representation of $S_n$?

The symmetric group $S_n$ has an $n$-dimensional defining representation, which splits as $n = (n-1) + 1$. Although this representation exists integrally, I would like to think of this as a real ...
Theo Johnson-Freyd's user avatar
13 votes
1 answer
585 views

A cohomology theory for fusion categories

It is well known that for a finite group $G$, the associator of the fusion category of $G$-graded $k$-vector spaces is given by an element of $H^3(G,k^*)$, up to equivalence of categories. ($k^*$ is ...
Manuel Bärenz's user avatar
12 votes
3 answers
3k views

Zariski tangent spaces to representation varieties

In Bill Goldman's paper "The Symplectic Nature of the Fundamental Groups of Surfaces" (Advances, 54, 200-225, '84) it is stated that the "Zariski tangent space" to a representation space Hom$(\pi, G)/...
Dan Ramras's user avatar
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