How does one compute the Hurewicz homomorphism for a (symplectic) nilmanifold?

Question

I have a symplectic six-dimensional nilmanifold $X:=G/\Gamma$ in hand, characterized by the sextuple $(0,0,12,13,14+23,24+15)$, which records the exterior derivatives of a basis of $\Gamma$-invariant $1$-forms on the Lie algebra $\mathfrak{g}$ of $G$. I have computed basis elements for all of the cohomology groups $H^{k}(X;\mathbf{R})$. I am seeking to compute the Hurewicz homomorphism for this manifold. Theoretically, of course, this should be easy if I can find generators for $\pi_{2}(X)$ and $H_{2}(X;\mathbf{Z})$. Is there a way to get generators for these groups, just using knowledge of the cohomology and of the Lie algebra $\mathfrak{g}$?

Answer 1

We have $\pi_2(X) \cong \pi_2(G)$, and $\pi_2$ of a connected Lie group vanishes. So in this case the Hurewicz homomorphism is zero.

If $G$ is a simply connected nilpotent Lie group, then it is diffeomorphic to $\mathbb{R}^6$ and in particular contractible. This means that $G/\Gamma$ is an Eilenberg-MacLane space $B \Gamma$, so its homology and cohomology are the group homology and cohomology of $\Gamma$ (and its homotopy groups are $\pi_1 \cong \Gamma$, all higher homotopy vanishing). You can try to compute $H_2(\Gamma, \mathbb{Z})$ using Hopf's formula.