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.



Computations in group cohomology
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.


