9
$\begingroup$

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?
$\endgroup$
  • $\begingroup$ **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

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.