Skip to main content
Search type Search syntax
Tags [tag]
Exact "words here"
Author user:1234
user:me (yours)
Score score:3 (3+)
score:0 (none)
Answers answers:3 (3+)
answers:0 (none)
isaccepted:yes
hasaccepted:no
inquestion:1234
Views views:250
Code code:"if (foo != bar)"
Sections title:apples
body:"apples oranges"
URL url:"*.example.com"
Saves in:saves
Status closed:yes
duplicate:no
migrated:no
wiki:no
Types is:question
is:answer
Exclude -[tag]
-apples
For more details on advanced search visit our help page
Results tagged with
Search options not deleted user 4149

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 …
Mikhail Borovoi's user avatar
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{{ …
Mikhail Borovoi's user avatar
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 …
Mikhail Borovoi's user avatar
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 …
Mikhail Borovoi's user avatar
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 …
Mikhail Borovoi's user avatar
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 …
Mikhail Borovoi's user avatar
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 …
Mikhail Borovoi's user avatar
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 …
Mikhail Borovoi's user avatar
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 …
Mikhail Borovoi's user avatar
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 …
Mikhail Borovoi's user avatar
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 …
Mikhail Borovoi's user avatar
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 …
Mikhail Borovoi's user avatar
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 …
Mikhail Borovoi's user avatar
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 …
Mikhail Borovoi's user avatar

15 30 50 per page