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14 votes
Accepted

Non-Abelian Hodge theory

I personally like the notes by Eugene Xia: Abelian and Non-Abelian Cohomology to build intuition. But for a definitive source, I would read Simpson: Moduli of representations of the fundamental group ...
Sean Lawton's user avatar
  • 8,529
12 votes

primary decomposition for nonabelian cohomology of finite groups

Yes, this is true. Let $F$ be the free product of all the Sylow subgroups of $G$: $$ F = \mathop\ast_{P\text{ Sylow}} P $$ The inclusions $P \to G$ together give a group homomorphism $F \to G$. On ...
Tyler Lawson's user avatar
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9 votes
Accepted

Long exact sequence of cohomology from 2-groups

$c$ is your crossed module, or 2-group, in a sense. Anything more concrete will depend on a choice of a cocycle description of the pointed set $[BH,B^3A]$. For example, $[BH,B^3A]$ classifies central ...
Konrad Waldorf's user avatar
9 votes
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Is there a notion of Čech groupoid of a cover of an object in a Grothendieck site?

Take $U=\coprod_{i∈I}Y(U_i)$, where $Y\colon C\to\mathop{\rm Presh}(C,{\rm Set})$ is the Yoneda embedding. We have a canonical morphism $U→Y(X)$. The Čech groupoid of $J_c$ can now be defined as the ...
Dmitri Pavlov's user avatar
8 votes
Accepted

primary decomposition for nonabelian cohomology of finite groups

It turns out the answer is no. Here I'll sketch a counter-example in which $H^1(G;M)$ is non-trivial (in fact infinite), while $H^1(H;M)$ is trivial for all proper subgroups $H<G$. Let $G=A_5$, ...
Mark Grant's user avatar
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6 votes
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Non-abelian Ext functor and non-abelian $H^2$

EDITED, taking into account the comments of Donu Arapura. As JLA wrote, a homomorphism $f\colon G\to N$ gives an extension \begin{equation}\label{e:E} 1\to K\to E\to G\to 1.\tag{E} \end{equation} This ...
Mikhail Borovoi's user avatar
5 votes

Second nonabelian group cohomology: cocycles vs. gerbes

Nonabelian $H^2$ in Galois cohomology can be defined in terms of: (1) cocycles, (2) extensions, (3) gerbes. The relations between these three definitions are described in Section 2.2 of Le principe de ...
Mikhail Borovoi's user avatar
5 votes

Second nonabelian group cohomology: cocycles vs. gerbes

I am not sure if this is the most general definition, but my proposal for 2nd nonabelian cohomology is combinatorial descent data in a cosimplicial crossed gropoid, modulo gauge equivalence. This is a ...
Amnon Yekutieli's user avatar
5 votes

Non-Abelian Hodge theory

I recently attented a nice online talk by Pengfei Huang and he indicated two sources: the first chapter of his own phd Non-abelian Hodge theory and some specializations - TEL - Thèses en ligne ...
Niels's user avatar
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4 votes
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What extra structure does the group of automorphisms of a torsor carry?

I am not sure what kind of criterion you are after. You can give an answer in terms of cocycles or an answer in terms of killing an obstruction. Briefly, if $\mathcal{G}$ is a bundle of groups on $Y$ ...
Tony Pantev's user avatar
  • 6,239
3 votes

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

For simplicity we write $G$ for $G^{\rm der}$ and $M_0$ for $M_0^{\rm der}$, then $G$ is a connected reductive group, and $M_0$ is a Levi subgroup of a minimal parabolic $P_0$ of $G$. Let $$\xi\in{\rm ...
Mikhail Borovoi's user avatar
1 vote
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Connecting homomorphism in non-abelian cohomology

$\newcommand{\diag}{{\rm diag}} \newcommand{\sH}{{\mathcal H}} \newcommand{\R}{{\mathbb R}} \newcommand{\HH}{\sf H} \newcommand{\V}{{\sf V}} \newcommand{\B}{{\sf B}} \newcommand{\C}{{\Bbb C}} $No, ...
Mikhail Borovoi's user avatar
1 vote
Accepted

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

In the meantime I found the right nlab pages to read... The answer seems to be 2-groups! Specifically automorphism 2-groups. I will write up what I have come to understand so far (though I have not ...
Johannes Hahn's user avatar
1 vote

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

If you have a morphism $f:G\to N\,,$ then you get a $K$-extension of $G$ by pulling back the $K$-extension of $N\,.$ The morphism $f$ lifts to a morphism into $M$ if and only if this extension is ...
Josh Lackman's user avatar
  • 1,198

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