Free and surface groups cohomology

What is a good reference for results on cohomology of finite rank free groups and surface groups with group ring coefficients?

I am interested in the case when the group acts on its group ring via conjugation. In particular, as for free groups I am interested in how the first cohomology of the group and its infinite index subgroups are related (how big the kernel of the restriction map is). Are there explicit ways to find these kernels (maybe in some special cases)? And also what is known about the cohomology groups of groups of compact surfaces (again with group ring coefficients and action by conjugation)? What relationships between cohomology of a free group and cohomology of a surface group can be derived from considering this surface group as a quotient group of the corresponding free group?

I don't know much about the subject but it would be very helpful to get the whole picture of these things before reading abstract proofs.

Well,I don't know why I decided to restrict myself to normal subgroups of a free group only (see comment below). Actually, it would be interesting to get any information on subgroups which are not normal too. Initially, I was interested in a question which could be formulated in terms of group theory but the question itself (or rather its infinitesimal version) led me to cohomology.

Dealing with cohomology, my first desire as a newbie was to see some good correspondence between subgroups of a free group and kernels of restrictions. Pretty soon I found that this is far from being Galois correspondence. Many subgroups have the same restriction kernel and such subgroups may even be not commensurable.

• I want to add that the particular infinite index subgroups I am nterested in are normal ones which are containg in the commutant subgroupof a free group. But I would appreciate any information on this. – N.B. Feb 3 '12 at 17:37
• The cohomology of free groups is very simple: they have cohomological dimension $1$, so cohomology is simply the invariants in degree $0$ and derivations in degree $1$—moreover, because they are free, derivations with values in a module have a very simple concrete description. – Mariano Suárez-Álvarez Feb 3 '12 at 17:46
• Thank you, Mariano. Yes, it's clear for me how to write the actual cocycles down. Mainly the question is about the kernels of restricions (their explicit generators) and the 1-cohomology of surface groups. – N.B. Feb 3 '12 at 17:52
• I have only skimmed over the question, so forgive me if this is not helpful, but what does Brown's GTM book on group cohomology say? – Yemon Choi Feb 4 '12 at 4:00
• Dear Yemon Choi, I looked at Brown's book as well as at few other books, but it considers the situation with general coefficients in a $G$-module $A$ and states just that the higher groups $H^n(G,A)=0$ for $n\geqslant 2$. Yes there are some theorems on cohomology of subgroups, but not explicitly. Why I consider infinite index subgrops: it seems to me,if I don't mistake, that if the restriction of a cocycle on a free group to a finite index subgroup is zero then the cocycle itself is zero (in the case of group ring coefficients, I did it "by hands"). – N.B. Feb 4 '12 at 6:02

Concerning your question on the relation between the cohomology of the free group and the surface group:

Let $F$ be a free group, $N \trianglelefteq F$ and $M$ an $F/N$-module. Consider $M$ as $F$-module via $F \to F/N$. Then the seven-term-exact sequence yields the exact sequence $$0 \to H^1(F/N;M) \xrightarrow[]{inf} H^1(F\;;M) \xrightarrow[]{res}H^1(N;M).$$ Thus $H^1(F/N;M)$ is a subgroup of $H^1(F\;;M)$, namely the kernel of the restriction resp. the image of the inflation.

If we interpret $H^1(F\;;M)$ in terms of derivations then $H^1(F/N;M)$ consists of those classes that are represented by derivations $F \to M$ such that $f|N = 0$. Moreover, if $N$ is the normal closure of $R \subseteq F$ then $f|N = 0$ is equivalent to $f|R=0$.

Example: $F=F_{2n}$ and $R=\lbrace [x_1,x_2] \cdots [x_{2n-1},x_{2n}]\rbrace$, i.e. $F/N$ is the fundamental group of a closed oriented surface of genus $n$. Take $\mathbb{Z}$-coefficients with trivial action. Now derivations are linear und hence vanish on commutators. Consequently $H^1(F/N; \mathbb{Z}) = H^1(F\;;\mathbb{Z}) = \mathbb{Z}^{2n}$. (One can also derive a formula for general coefficients, but it isn't very handy).

Concerning the restriction homomorphism of a free group: By the Nielsen-Schreier theorem, a subgroup of a free group is again free. This reduces the problem to compute the restriction between two free groups. This can easily been done by using the free resolutions from Brown, I (4.4) and computing a chain map between the resolutions. Since I guess that you are primarily interested in surface groups that can be treated with help of Lyndon's paper or by the method above, I leave out details.

• Thank you, Ralph. I have a question on the statement that "$H_1(F/N;M)$ consists of those classes that are represented by derivations $F\rightarrow M$ such that $f|N=0$". Generally, it seems that a derivation $f:F\rightarrow M=\mathbb{Z}[F]$ defines a derivation $\bar{f}:F/N\rightarrow M$ if $f(R)$ is a sum of terms of the form $\pm w_1R^{k_1}{w_1}^{-1}\dots w_sR^{k_s}{w_s}^{-1}, k_j\in \mathbb{Z}$ such that it is mapped to 0 under the quotient map $\mathbbZ[F] \rightarrow \mathbb{Z}[F/N]$. Why is it enough to consider onle those derivations that map $N$ exactly to zero in $\mathbb{Z}[F]$? – N.B. Feb 7 '12 at 8:44
• The inflation is induced by $Der(F/N,M) \to Der(F,M),\; f \mapsto f \circ \kappa$ with $\kappa: F\to F/N$. Hence $(f \circ \kappa)|N = 0$, since $f(\bar{1})=0$. Conversely, if $f: F \to M$ is a derivation with $f|N = 0$ you can define a derivation $\bar{f}: F/N \to M$ by $\bar{f}(\bar{x}) := f(x)$. Hence $f = \bar{f} \circ \kappa$ is in the image of $Der(F/N,M)$. Thus the image of $Der(F/N,M)$ consists exactly of the derivations that vanish on $N$. If there are futher questions please don't hesitate to ask. – Ralph Feb 7 '12 at 10:07
• Ah, now I see the problem you are faced with: $F/N$ doesn't act on $\mathbb{Z}[F]$ by conjugation, i.e. you can't take $M=\mathbb{Z}[F]$. When you consider "the cohomology groups of groups of compact surfaces (again with group ring coefficients and action by conjugation)" then the group is $F/N$ and the group ring is $M = \mathbb{Z}[F/N]$. The action of $F$ on $M$ is then given by $x \cdot \sum_i k_i \bar{y_i} = \sum_i k_i \overline{xy_ix^{-1}}$ where $x,y_i \in F,\; k_i \in \mathbb{Z}$ and the derivations for $H^1(F/N;\mathbb{Z}[F/N])$ are $f: F \to \mathbb{Z}[F/N]$ with $f|N = 0$. – Ralph Feb 7 '12 at 12:09
• If $F$ is free on $n$ elements $x_1...,x_n$, it may also be helpful to note that $Der(F,M)$ is in $1-1$ correspondence with $M^n$, given by $f \mapsto (f(x_1),...f(x_n))$. If $|R| = 1$, $$\lbrace f \in Der(F,M) \mid f|R = 0 \rbrace$$ corresponds to something like $$\lbrace (m_1,...,m_n) \mid \sum_{i=1}^n \alpha_i \cdot m_i =0 \rbrace$$ where $\alpha_i \in \mathbb{Z}[F]$ depends on $R$. – Ralph Feb 7 '12 at 12:28
• Yes, I'm dealing with $\mathbb{Z}[F/N]$ as the group ring for group of surface cohomology. So, calculating cocycles by hands becomes nontrivial. – N.B. Feb 8 '12 at 0:20