Let $BG$ denote the classifying space of a (discrete) group and $BG_+$ its disjoint union with a point.

Question: What is known about the stable homotopy groups $\pi^S_*(BG_+)$ ?

If $G$ is finite (i.e., compact) there are results in terms of the Burnside ring and completion at primes (at least for $\pi_0^S$).

I'm interested in the case of a non-compact group. Since $\mathbb{Z}$ is obvious, let's stick to the following.

Precise Question: What is known about the stable homotopy groups $\pi^S_*(BG_+)$ for $G$ a torsion-free nilpotent finitely-generated (discrete group)?

For example, take the 3-dimensional Heisenberg group $G$. What is known about $\pi^S_*(BG_+)$ for $*=0,1,2,3$ ?

In this example, the rational part is easy to calculate since $BG$ is a finite dimensional compact framed manifold, but what about the torsion information?


There is an Atiyah-Hirzebruch-Serre spectral sequence (or perhaps Atiyah-Hirzebruch-Lyndon-Hochschild-Serre SS in this context) of the form

$$E^2_{pq} = H_p(G/N, \pi_q^S(BN_+)) \implies \pi_{p+q}^S(BG_+)$$

for any normal subgroup $N \leq G$.

Take $G$ to be the Heisenberg group that you mention and $N = \mathbb{Z}$ to be the central subgroup. Then the action of $G/N$ on $N$ by conjugation is trivial, so the $E^2$ term of the above SS has trivial coefficients. Noting that $G/N = \mathbb{Z}^2$ has the homology of a torus, we have

$$E^2_{0q} = \pi_q^S(S^1_+) \mbox{, } \; E^2_{1q} = \pi_{q}^S(S^1_+)^{\oplus 2} \mbox{, and } \; E^2_{2q} = \pi_{q}^S(S^1_+),$$

since $BN = S^1$.

The only possible differential is $d_2: E^2_{2,q} \to E^2_{0,q+1}$. Now, $\pi_q^S(S^1_+) = \pi_q^S \oplus \pi_{q-1}^S$, where I'm using the notation $\pi_q^S= \pi_q^S(S^0)$. I claim that $d_2: E^2_{2,q} \to E^2_{0,q+1}$ carries $\pi_q^S$ isomorphically to itself under this map. This can be seen by comparison to the corresponding SS for computing the group homology of the Heisenberg group (which you can see done in detail in this paper by Lee and Packer; the relevant computation is Lemma 1.1) via the Hurewicz transformation $\pi_*^S \to H_*$.

The residual part of $d_2: \pi_{q-1}^S \to \pi_{q+1}^S$ must be a $\pi_*^S$-module map by general nonsense associated with the AHSS, so it is completely determined by where it sends $1 \in \pi_0^S$. This is either $0$ or $\eta^2 \in \pi_2^S$. I'm not honestly sure which, so there are two possibilities:

  • If $1 \mapsto 0$, then the associated graded of $\pi_n^S(BG_+)$ is

$$\pi_{n}^S \oplus \pi_{n-1}^S(S^1_+)^{\oplus 2} \oplus \pi_{n-3}^S = \pi_{n}^S \oplus (\pi_{n-1}^S \oplus \pi_{n-2}^S)^{\oplus 2} \oplus \pi_{n-3}^S$$

  • If $1 \mapsto \eta^2$, then the associated graded of $\pi_n^S(BG_+)$ is

$$(\pi_{n}^S/\eta^2 \pi_{n-2}^S) \oplus (\pi_{n-1}^S \oplus \pi_{n-2}^S)^{\oplus 2} \oplus (\ker \eta^2: \pi_{n-3}^S \to \pi_{n-1}^S)$$

  • 2
    $\begingroup$ Hang on, I think it has to be the first option, since $\pi_n^S$ must split off $\pi_n^S(BG_+)$. $\endgroup$ – Craig Westerland May 10 '14 at 3:34

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