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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.
2
votes
$2$-cohomology group of semi-direct products
From the spectral sequence mentioned by Anton one can derive an exact sequence
in low degrees
$$ 0\to E_2^{1,0}\to E^1\to E_2^{0,1}\to E_2^{2,0}
\to \mathrm{ker}\left[E^2\to E_2^{0,2}\right]\to E …
3
votes
1
answer
421
views
Conjugation of group extensions
Let $H$ be a finite group. We write ${{\mathbb{C}}}^{*n}$ for the $n$-dimensional complex torus $({{\mathbb{C}}}^*)^n$.
We have a short exact sequence
$$ 0\to {{\mathbb{Z}}}^n\to {{\mathbb{C}}}^n\to{{ …
4
votes
Galois cohomology of linear groups over local fields
@Brian Conrad: The sequence $$1 \rightarrow \mu \rightarrow Z' \times \mathcal{G} \rightarrow G' \rightarrow 1$$ is not exact in general. Namely, the kernel $\nu$ of the map
$Z' \times \mathcal{G} \ri …
3
votes
2
answers
333
views
Quasi-isomorphism preserves group hypercohomology
I am looking for a reference for the assertion in the title.
In more detail, let $\Gamma=\{1,\gamma\}$ be a group of order 2.
Let $A$ be a $\Gamma$-module (an abelian group on which $\Gamma$ acts).
T …
1
vote
Accepted
gluing gerbes over a spectrum of a field
I don't think so. I think a gerbe bound by $A$ over the spectrum of a field $k$ gives a cohomology class $\eta\in H^2(k,A)$, and the gerbe trivializes over an extension $k'/k$ if and only if this c …
2
votes
Accepted
Regarding extensions of finite groups by Tori
(I write an answer rather than a comment in order to accommodate exact sequences.)
Let
$$0\to T\to E\to\Gamma\to 1\tag{$E_1$}$$
be your first group extension, where $T$ is the torus ${\Bbb R}^n/{\Bbb …
5
votes
1
answer
274
views
What is the meaning of this coboundary homomorphism for group hypercohomology?
$\require{AMScd}$
Let $\Gamma=\{1,\gamma\}$ be a group of order 2.
In my problem from Galois cohomology of real reductive groups I came to a commutative diagram of $\Gamma$-modules
(abelian groups wit …
6
votes
1
answer
143
views
Finite $\Gamma$-modules with trivial $H^2$, where $\Gamma$ is a group of order 2
Let $\Gamma=\{1,\gamma\}$ be a group of order 2. Let $A$ be a finite $\Gamma$-module,
that is, a finite abelian group on which $\Gamma$ acts.
It is a hopeless problem to classify finite $\Gamma$-modul …
0
votes
Quasi-isomorphism preserves group hypercohomology
I give an elementary proof of the fact that a quasi-isomorphism of short complexes (complexes of length 2) of $\Gamma$-modules induces an isomorphism on hypercohomology.
Actually, it is very close to …
2
votes
Compute corestriction map on group cohomology in Magma
Since OP found my comments helpful, I post them as an answer to have an editable text.
Let $G$ be a group, $H\subset G$ be a subgroup of finite index,
and $M$ be a right $H$-module.
I know an explicit …
8
votes
Accepted
Group cohomology question, trivial Galois action on discrete Galois module means we can say ...
If the $G_K$-action on $M$ is trivial, then
$$H^1(K,M)=\mathrm{Hom}(G_K,M),$$
and by Chebotarev's density theorem
$$ F^1(K,M)=0.$$
For details see Lemma 1.1(i) of Sansuc, J.-J. Groupe de Brauer et ari …
7
votes
1
answer
440
views
Imperfect Tate (cup product) pairing in Galois cohomology?
Let $k$ be a field of characteristic 0 with a fixed algebraic closure $\bar k$
and absolute Galois group $\Gamma={\rm Gal}(\bar k/k)$.
Let $M$ be a finite $\Gamma$-module, that is, a finite abelian g …
7
votes
0
answers
303
views
Albrecht Fröhlich's text `Groupoids, groupoid spaces and cohomology' (1965)
I am looking for Albrecht Fröhlich's unpublished text `Groupoids, groupoid spaces and cohomology' (1965). In this text Fröhlich defines cohomology of a group with coefficients in a groupoid (this was …
2
votes
Accepted
Imperfect Tate (cup product) pairing in Galois cohomology?
Let us take $k=\mathbf{Q}$ and $M=\mu_3$.
Then $H^0(\Gamma,M)=0$.
On the other hand, $M^D\simeq \mathbf{Z}/3\mathbf{Z}$, and one can show that $H^2(\Gamma,M^D)\neq 0$.
It follows that there exists an …
4
votes
0
answers
117
views
Neutral cohomology classes and restriction maps for $H^2$ in group cohomology
$\DeclareMathOperator\res{res}$
Let $G$ be a profinite group.
Let $A$ be a finite $G$-group (a finite group on which $G$ acts continuously), not necessarily abelian,
with center $Z=Z(A)$.
We say that …