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.

  • 1
    $\begingroup$ 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. $\endgroup$ – 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.

  • 1
    $\begingroup$ Would sage do it? $\endgroup$ – Julien Puydt Mar 11 '11 at 8:24
  • $\begingroup$ Thank you for this precise answer. I will try to do this and post again if there is a problem. $\endgroup$ – 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 $2$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.