Skip to main content

Questions tagged [nonabelian-cohomology]

Filter by
Sorted by
Tagged with
1 vote
0 answers
42 views

When is $B^G\backslash(B/A)^G$ finite?

Let $G$ be a locally compact group, let $A,B$ be (not necessarily abelian) connected reductive complex groups equipped with continuous actions of $G$ via algebraic automorphisms. Let $\phi:A\to B$ be ...
user449595's user avatar
2 votes
1 answer
262 views

Connecting homomorphism in non-abelian cohomology

Let $G$ be a simply connected, semisimple algebraic group over $\mathbb{R}$ and let $X$ be a homogeneous space for $G$ with finite commutative stabilizer $\mu$. There is a connecting homomorphism from ...
Victor de Vries's user avatar
0 votes
0 answers
120 views

$G\cdot H$ with $G,H$ non-Abelian finite simple

Can a non-split extension of one non-Abelian finite simple group by another exist?
Daniel Sebald's user avatar
3 votes
0 answers
308 views

Non-abelian group cohomology, additional information

Let $G$ be a (profinite) group, and let $M$ be a non-abelian $G$-module. We know how to construct reasonably $H^0(G,M)$ and $H^1(G,M)$ and it turns out that $H^1(G,M)$ is just a pointed set and not ...
Or Shahar's user avatar
  • 463
4 votes
1 answer
301 views

What extra structure does the group of automorphisms of a torsor carry?

Let $Y$ be a space, and $G$ a group. For simplicity we can take $G$ to be finite, $Y$ to be a point, if this avoids technical issues. Then for $X/Y$ a $G$ torsor, so a sheaf of sets with $G$ action on ...
Chris H's user avatar
  • 1,949
6 votes
1 answer
300 views

Connection between the classifications of group extensions and group-graded algebras in terms of non-abelian cohomology

First, consider group extensions with non-abelian kernel $$1\to K\to G \to Q \to 1$$ It is well-known that these are classified by certain cohomological objects, specifically: Any such extension ...
Johannes Hahn's user avatar
6 votes
1 answer
450 views

Long exact sequence of cohomology from 2-groups

I am trying to understand the following from Principal Infinity Bundles - General by Nikolaus, Schreiber and Stevenson. So following the reference there to Nikolaus-Waldorf tells us that given any (...
Emilio Minichiello's user avatar
13 votes
2 answers
2k views

Non-Abelian Hodge theory

Let $X$ be a compact Riemann surface. I would like to find a somehow complete reference for the proof of the so called non-Abelian Hodge correspondence relating Dolbeaut, Betti and Higgs bundle moduli ...
Tommaso Scognamiglio's user avatar
6 votes
1 answer
343 views

Is there a notion of Čech groupoid of a cover of an object in a Grothendieck site?

Given a topological space $X$, and a cover $\mathcal{U} :=\cup_{\alpha \in I}U_{\alpha}$ of $X$, one can define a groupoid called Čech groupoid $C(\mathcal{U})$ of the cover $\mathcal{U}$ by $\...
Adittya Chaudhuri's user avatar
7 votes
2 answers
997 views

Non-abelian Ext functor and non-abelian $H^2$

Let $G$ be a group and $$0\rightarrow K\rightarrow M\rightarrow N\rightarrow 0$$ a short exact sequence of groups. Now these are abelian groups, if I want to show that $\text{Hom}(G,M)\rightarrow \...
curious math guy's user avatar
12 votes
0 answers
272 views

sequences in non-abelian group cohomology

In general, if we have a (pro-)finite group $G$ and a sequence of (continuous) non-abelian $G$-modules $$1\rightarrow A\rightarrow B\rightarrow C\rightarrow 0,$$ such that the image of $A$ lies in the ...
Estus's user avatar
  • 273
2 votes
1 answer
439 views

Minimal parabolic subgroups are $G(k)$-conjugate: a cohomological interpretation?

Let $G$ be a connected,reductive group over a $p$-adic field $k$. Let $M_0$ be a minimal Levi subgroup of $G$, and define $M_0^{\operatorname{der}}$ to be $M_0 \cap G_{\operatorname{der}}$. Lemma 2....
D_S's user avatar
  • 6,180
18 votes
2 answers
592 views

primary decomposition for nonabelian cohomology of finite groups

Let $G$ be a finite group, and let $M$ be a group on which $G$ acts (via a homomorphism $G\to \operatorname{Aut}(M)$). If $M$ is abelian, hence a $\mathbb{Z}G$-module, there is a primary ...
Mark Grant's user avatar
  • 35.9k
5 votes
0 answers
254 views

How can one concretely approximate a homotopy type by a scheme or (higher/derived) stack?

What methods are there to approximate an arbitrary homotopy type by an algebraic-geometric object in a concrete (read: computable) way? I know this is an ambitious question, so maybe I should ...
Patrick Elliott's user avatar
10 votes
0 answers
347 views

The non-abelian Gauss-Manin connection; non-abelian M_dR; a Grothendieck lemma for cyrstals

I'm interested in understanding the non-abelian Gauss-Manin connection on Carlos Simpson's relative de Rham Moduli space $M_{dR}(X/S,n)$ for a smooth projective morphism of schemes $X/S$. The scheme $...
Max Menzies's user avatar
3 votes
0 answers
65 views

Connection on 3-bundle given as triplet of forms

A connection on a bundle is given locally by a Lie algebra-valued 1-form. Gauge transformations act in the usual way on the forms, and form a groupoid. A connection on a 2-bundle is given locally by ...
geodude's user avatar
  • 2,129
8 votes
3 answers
2k views

Is there a homology theory that gives a *necessary and sufficient* condition for homotopy equivalence?

Is there a (non-Abelian) homology theory that realizes the following: Let $X,Y$ be manifolds with complexes $C(X),C(Y)$. Then $X$ and $Y$ are homotopy-equivalent if and only if $C(X)$ and $C(Y)$ ...
geodude's user avatar
  • 2,129
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
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