I am reading Class field theory - Bonn Lectures by Neukirch.

Given a $G$ module $A$ he defines Cohomology groups $H^i(G,A) : i\in \mathbb{Z}$ by considering some complete resolution of $G$ modules and applying hom functor.

He then calculated $H^{-1}(G,A),H^0(G,A),H^1(G,A),H^2(G,A)$

It says $H^2(G,A)$ is related to group extensions. Given an abelian group $A$ written multiplicatively and an arbitrary subgroup $G$, we want to find all group extensions $\hat{G}$ of $A$ such that $A$ is normal subgroup pf $\hat{G}$ and $\hat{G}/A\cong G$. He says this is related to $H^2(G,A)$.

It says if the group $G$ acts trivially on $A$ then $H^1(G,A)=\rm{Hom}(G,A)$, in particular for $A=\mathbb{Q}/\mathbb{Z}$ we have character group $H^1(G,A)=\chi(G)$.

It says $H^0(G,A)=A^G/N_GA$ the norm residue group of $G$ module $A$ with notation $A^G=\{a\in A : \sigma a=a~\forall \sigma\in G\}$ and $N_GA=\{\sum_{g\in G}ga:a\in A\}$. It says this is useful in class field theory.

It says $H^{-1}(G,A)$ has some other concrete form.

It says $H^{-2}(G,\mathbb{Z})=G^{ab}$ abelianization of $G$.

My question is are there any higher cohomology groups that are of interest at least in some special cases?

  • $\begingroup$ Why there is a downvote? Thanks for the upvote.. $\endgroup$ – user37663 Aug 23 '16 at 12:19
  • $\begingroup$ Perhaps you should look in a textbook on cohomology of groups, e.g Ken Brown's book? $\endgroup$ – Mark Grant Aug 23 '16 at 12:21
  • $\begingroup$ @MarkGrant : I do have that book, Can you please specify what to look for,, $\endgroup$ – user37663 Aug 23 '16 at 12:26
  • $\begingroup$ Maybe. Could you specify what you mean by "of interest"? ;) $\endgroup$ – Mark Grant Aug 23 '16 at 12:29
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    $\begingroup$ Then this might not be the right forum for you question. I suggest you read the first few chapters of Brown (or at least the introduction) and then try to ask a more focussed question, either here or at MathStackexchange. $\endgroup$ – Mark Grant Aug 23 '16 at 12:40

As Serre seems to be fond of saying, life begins at $H^3$.

The first appearance of a $3$-cocycle seems to be in Teichmüller's article Über die sogenannte nichkommutative Galoissche Theorie... in Deutsche Mathematik 5. He was trying to extend Galois Theory to extensions of a commutative field which are not themselves commutative. Teichmüller's formula for the $3$-cocycle helped Eilenberg and MacLane come up with the right definition of group cohomology in

http://www.jstor.org/stable/1968966 (Group Extensions and Homology, Samuel Eilenberg and Saunders MacLane, Annals of Mathematics Second Series, Vol. 43, No. 4 (Oct., 1942), pp. 757--831).

Let me just give one example where $H^3$ is useful. Let $p$ be a prime number, and consider a finite extension $K$ of the field $\mathbf{F}_p((t))$ (or of $\mathbf{Q}_p$, which you seem to know). Let $\bar K$ be a separable algebraic closure of $K$, and $G_K=\mathrm{Gal}(\bar K|K)$. It is an absolutely crucial result that $H^3(G_K,\mathbf{Z})=0$. That's what allows you to lift projective representations $G_K\to\mathrm{PGL}_n(\mathbf{C})$ to linear representations $G_K\to\mathrm{GL}_n(\mathbf{C})$. See for example

https://eudml.org/doc/142305 (André Weil, Exercises dyadiques, Inventiones mathematicae (1974) Volume: 27, pp. 1--22).

  • $\begingroup$ Thanks for your answer. As of now, I think I can make use of your suggestion Group extensions and homology article.. I will come back again and try to understand the other part. Thanks $\endgroup$ – user37663 Aug 23 '16 at 15:12

Here's a number field example, analogous to Dalawat's local field example. Let $K$ be a number field. One can prove that $H^3(G_{\overline K/K},\overline K^*)=0$. (Although interestingly, there may be finite Galois extensions $L/K$ with $H^3(G_{L/K},L^*)\ne0$.) Anyway, Tate's cohomological construction of the pairing on the Tate-Shafarevich group $III(E/K)$ of an elliptic curve $E/K$ proceeds by first constructing a 2-cochain and then using the fact that $H^3(G_{\overline K/K},\overline K^*)=0$ to find a "correction term" that turns it into a 2-cocycle, thereby giving the desired element of $\text{Br}(K)$. So the whole construction depends on the calculation of an $H^3$.

  • $\begingroup$ I am happy that my question has got your interest.. I am not familiar with elliptic curves brauer groups... I will come back and rtry to understand this once I am ready... Thank you $\endgroup$ – user37663 Aug 23 '16 at 19:08
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    $\begingroup$ @PraphullaKoushik Sorry for the ambiguity, it's the usual Brauer group $\text{Br}(K)=H^2(G_{\overline K}/K,\overline K^*)$. One gets a pairing from on $III(E/K)$ that takes values in $\text{Br}(K)$. Hmmm...,no actually, that's not quite right, either, one gets values in $\mathbb Q/\mathbb Z$ by getting an element of the local Brauer groups and adding up their invariants. Anyway, it's not an elliptic curve Brauer group. $\endgroup$ – Joe Silverman Aug 23 '16 at 21:18
  • $\begingroup$ Thanks for the clarification. I just need some time to understand. I will try one more time... :) $\endgroup$ – user37663 Aug 24 '16 at 6:42

The cohomology of braid groups is non-trivial in every dimension and gives one a way of detecting families of orbits of disjoint points in the plane.

In a sense this isn't "cohomology of groups" for the purpose of studying cohomology of groups. The classifying space of braid groups turns out to be a configuration space of points in the plane. So the cohomology of the braid groups is the cohomology of this configuration space.

The high-dimensional cohomology (as a ring) tells one how you can construct high-dimensional families of such disjoint points in the plane, and how they intersect.

Lots of groups have very physical classifying spaces, like this. So many group cohomologies have similarly physical interpretations.

  • $\begingroup$ Are you talking about the same cohomology theory than the OP ? In class field theory one usually considers the cohomology of profinite groups. I thought Braid groups are discrete groups and their cohomology is the cohomology of abstract groups ? $\endgroup$ – tj_ Aug 23 '16 at 17:41
  • $\begingroup$ I woke up at 4am today. A little groggy, and I certainly missed that. $\endgroup$ – Ryan Budney Aug 23 '16 at 17:56

All of the higher cohomology groups naturally occur in algebraic topology. For $n \ge 3$, the cohomology group $H^n(G, A)$ describes the classification of pointed connected spaces $X$ with $\pi_1(X) \cong G, \pi_{n-1}(X) \cong A$, and fixed action of $G$ on $A$. Said another way, it classifies extensions, or more precisely fiber sequences, of the form

$$B_{n-1} A \to X \to BG.$$

This second description has the pleasant property that it continues to make sense when $n = 2$.

For something more concrete, if $k$ is a field then $H^2(G, k^{\times})$ has something to do with projective representations of $G$ over $k$. One way to describe this relationship is that a $2$-cocycle $c \in Z^2(G, k^{\times})$ is exactly what is needed to define a "twisted group algebra" where the multiplication has the form

$$g \ast h = c(g, h) gh.$$

Modules over this twisted group algebra then correspond to certain projective representations of $G$.

There is an analogue of this when $2$ is replaced by $3$: a $3$-cocycle in $Z^3(G, k^{\times})$ is exactly what is needed to define a twisted version of the monoidal category of $G$-graded $k$-vector spaces, where the $3$-cocycle now twists the associator. This construction can be used, among other things, to describe 3d Dijkgraaf-Witten theory. Similar statements are true for higher values of $3$.

  • $\begingroup$ This went totally over my head. Thanks for giving me another word Tate cohomology group.. I will try to learn about that... $\endgroup$ – user37663 Aug 23 '16 at 19:20

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