It is known that for free $ k$ -step nilpotent group on $ r$ generators $ N(r, k)$ its integral homology is torsion-free in degrees $ \leq 3$ (obvious for 1 and 2, Igusa&Orr computations for 3). However, there’s a nontrivial 3-torsion in $ H_4(N(\geq 4, 2), \Bbb Z)$ — which can be computed more or less directly using Kuz’min theorem (that integral homology coincide for two-step free nilpotent Lie algebras and groups) and equals $ \Lambda^4(N(r, 2)_{ab} \otimes \Bbb Z/3)$ .

• Is there any 2-torsion ever in $ H_i(N(r, k), \Bbb Z)$ ?

Also we can assemble all $ N(r, k)$ ‘s in one compound via canonical projections $ $ \phi_{k}: N(r, k+1) \to N(r, k)$ $ and obvious simplicial maps

$ $ \delta^{r}: N(r+1, k) \to N(r, k), \,\sigma^{r}: N(r, k) \to N(r+1, k)$ $

Can we see some interesting structure in this bigraded thing (some sort of “nilpotent integral Steenrod algebra”)? For example:

• $ H_1(\phi)$ is always identity, $ H_2(\phi)$ is always zero, $ H_3(\phi)$ go nontrivially from $ 2k$ to $ k$ and then vanish — what about higher homology?
• simplicial groups $ H_1(N(\cdot, k))$ are contractible; computation of homotopy type of $ H_2$ and $ H_3$ seems interesting and doable

Further, there’s a conjecture of Millionscshikov that for integral jet groups $ J_k := t + \sum_{i = 2}^{k+1}a_it^i, a_i \in \Bbb Z$ with composition modulo $ t^{k+2}$ as operation have stable homology with $ \textrm{rk} \, H_s(J_k) = s+2$ th Fibonacci number for $ k$ sufficiently large (I think that it’s now proven for degree $ \leq 3$ ).

• Should we expect any kind of stability in free case — a variation of Farb-Ellenberg representation stability, for example?