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 $H_i(N(r, k), \Bbb Z)$? Is there anything besides 3?

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 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 $\leq 3$ and there's numerical evidence that it's true always).

  • Should we expect any kind of stability in free case — a variation of Church-Ellenberg-Farb representation stability, for example?
  • $\begingroup$ **Church**—Ellenberg—Farb $\endgroup$ – Andy Putman Dec 21 '17 at 15:32
  • $\begingroup$ I think I can fix it. $\endgroup$ – Denis T. Dec 21 '17 at 16:21

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