# Computations in group cohomology

Hello,

Given a finitely presentable group $G$, I'm interested in the cup-product from $H^1$ to $H^2$ with real coefficients. I want to know if this is explicitly computable (with a computer) with a presentation of the group. More precisely, I want a program that takes the generators and relations as entries and returns the dimension of the $H^1$ and a finite generating set of linear relations between the cup-products of every couple of elements in a basis of $H^1$. (I am not really interested in all the $H^2$) Does this seem possible ?

I precise that I am not really familiar with group cohomology and I ask this question because it is certainly known if such a problem cannot be resolved with an efficient algorithm.

The problem comes in the study of Kähler groups where this cup-product plays an important role.

Thank you.

• I gave an answer below. One remark is that it is good that you aren't interested in the rest of $H^2$ -- a theorem of C. Gordon shows that this is in general not computable from a group presentation. – Andy Putman Mar 10 '11 at 18:56

EDIT: I was explaining this to a grad student today, and I realized that I didn't give any references. The result I describe below was first stated by Sullivan in

Sullivan, Dennis On the intersection ring of compact three manifolds. Topology 14 (1975), no. 3, 275-277.

He claims it is true for a 3-manifold, but all he says about the proof is that it is "a certain amount of soul searching classical algebraic topology.". In fact, the result is true for any connected CW-complex (including an Eilenberg-MacLane space, as in the group cohomology question I was answering). This whole picture was later subsumed into Sullivan's theory of 1-minimal models and rational homotopy theory in

Sullivan, Dennis Infinitesimal computations in topology. Inst. Hautes Études Sci. Publ. Math. No. 47 (1977), 269-331 (1978).

An accessible reference for this is

Griffiths, Phillip A.; Morgan, John W. Rational homotopy theory and differential forms. Progress in Mathematics, 16. Birkhäuser, Boston, Mass., 1981. xi+242 pp. ISBN: 3-7643-3041-4

The stuff on fundamental groups is in Chapter 13. I don't know where the proof I gave first appeared (I came up with it myself, but I doubt I was the first). An alternate and very pretty geometric proof is in

De Michelis, Stefano, A remark on cup products in $$H^1(X)$$, Rend. Accad. Naz. Sci. XL Mem. Mat. (5) 14 (1990), no. 1, 323-325.

This is very computable. Let $$G^{(k)}$$ be the lower central series of $$G$$, ie $$G^{(1)}=G$$ and $$G^{(k+1)} = [G^{(k)},G]$$. There are algorithmic ways to compute the quotients $$G^{(k)}/G^{(k+1)}$$ (eg using the Fox free differential calculus -- see Fox's series of papers on the free differential calculus for the details). The direct sum $$\oplus_{k=1}^{\infty} G^{(k)} / G^{(k+1)}$$ has the structure of a graded Lie algebra with the Lie bracket induced by conjugation (this is explained in many places -- I recommend the last chapter of Magnus-Karass-Solitar's book on combinatorial group theory or Serre's book "Lie Algebras and Lie Groups"). This Lie algebra is generated by the degree 1 piece, namely $$G^{(1)} / G^{(2)} \cong G^{ab}$$. The degree 2 piece is a quotient of $$\wedge^2 G^{ab}$$ by some subgroup $$R$$. I claim that understanding $$R$$ is exactly what you need to know to understand the kernel of the cup product map. Namely, we have a surjection $$\wedge^2 G^{ab} \rightarrow \wedge^2 G^{ab} / R$$ and thus a dual injection $$(\wedge^2 G^{ab} / R)^{\ast} \hookrightarrow \wedge^2 (G^{ab})^{\ast}.$$ The image of this injection is exactly the kernel of the cup product map.

Let me sketch a proof. To simplify things, let's assume that everything in sight is torsion-free (it will simplify our statements). Set $$H = H_1(G)$$ and $$H^{\ast} = H^1(G) = Hom(H,\mathbb{Z})$$. The above will allow you to compute the kernel of the cup product map $$\wedge^2 H^{\ast} \rightarrow H^2(G)$$ as follows. Consider the short exact sequence

$$1 \longrightarrow G^{(2)} \longrightarrow G \longrightarrow H \longrightarrow 1.$$

There is an associated 5-term exact sequence in group cohomology which takes the form

$$0 \longrightarrow H^1(H) \longrightarrow H^1(G) \longrightarrow (H^1(G^{(2)}))^H \longrightarrow H^2(H) \longrightarrow H^2(G).$$

Now, the map $$H^1(H) \rightarrow H^1(G)$$ is an isomorphism. Also, $$H^2(H) = \wedge^2 H^{\ast}$$, and the map $$H^2(H) \rightarrow H^2(G)$$ is easily seen to be the cup product map. What we deduce is that we have an exact sequence

$$0 \longrightarrow (H^1(G^{(2)}))^H \longrightarrow \wedge^2 H^{\ast} \longrightarrow H^2(G).$$

In other words, the kernel of the cup product map is the subgroup $$(H^1(G^{(2)}))^H$$ of $$\wedge^2 H^{\ast}$$.

Let us now interpret this subspace. It is easiest to dualize. The dual of the above inclusion is the surjection

$$H_2(H) \rightarrow (H_1(G^{(2)}))_H.$$

Now, $$H_1(G^{(2)})$$ is just $$G^{(2)} / [G^{(2)},G^{(2)}]$$, and we are killing off the action of $$H$$, which is the same as killing off the conjugation action of $$G$$. In other words, we have an isomorphism

$$(H_1(G^{(2)}))_H \cong G^{(2)} / [G,G^{(2)}] = G^{(2)} / G^{(3)}.$$

The desired claim is an immediate consequence.

• Would sage do it? – Julien Puydt Mar 11 '11 at 8:24
• Thank you for this precise answer. I will try to do this and post again if there is a problem. – mister_jones Mar 11 '11 at 12:03

Using Andy's comment and Theorem 6.1 in this paper you can easily work out a computer program to calculate what you wish from a presentation. You will only need to work with groups of nilpotency class $2$.