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Let $s \geq 0$ be fixed. The $J$-homomorphism includes $\pi_{8s+1}(SO) = \mathbb Z/2$ in $\pi_{8s+1}^s$, the $(8s+1)$-th stable homotopy group of spheres.

Now regard $\pi_{8s+1}^s = \pi_{8s+1} ((B\Sigma_{\infty})^+)$ , where $\Sigma_{\infty}$ is the group of compactly supported permutations of $\mathbb N$.

The obvious map (i.e. write permutations as permutation matrices) $\Sigma_{\infty} \to O$ induces a map $\pi_{8s+1}^s \to \pi_{8s+1}(BO) = \pi_{8s}(O)= \mathbb Z/2$.

Is the composition $\mathbb Z/2 = \pi_{8s+1} (SO) \to \pi_{8s+1}^s \to \pi_{8s} (O) = \mathbb Z/2$ trivial or nontrivial?

For $s = 0$ this map is nontrivial, since the two elements of $\pi_1^s$ correspond to odd and even permutations in $\Sigma_{\infty}$, whereas the two elements in $\pi_1(BO) = \pi_0(O)$ correspond to determinant $\pm1$ matrices and the map $\Sigma_{\infty} \to O$ preserves this structure.

So what happens for positive $s$? [I have a strong feeling that the map is then trivial, but maybe I am wrong...]

Remark: This map factors through $K_{8s+1}(\mathbb Z)$.

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The composition is trivial. Composition with $\eta$ acts trivially on the source and nontrivially on the target, for positive $s$, which implies the claim. This argument does not apply for $s=0$, because $J : SO \to QS^0$ does not deloop.

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  • $\begingroup$ Today, I was reading parts of Akhil Mathew's notes of a course Michael Hopkins taught in 2012, see math.berkeley.edu/~amathew/256y.pdf. On page 126, there is an exercise asking basically the same question as I did and the answer is the opposite of what you claimed. (There is no proof.) So what is the right answer, at the end of the day?? $\endgroup$
    – Jens Reinhold
    Commented Jun 25, 2015 at 1:11
  • $\begingroup$ Proposition 7.19 on page 46 of J. Frank Adams' paper "On the groups J(X). IV", Topology 5 1966 21–71, states that the composite in question ($d_R J : \pi_r(SO) \to Z/2$ for $r \equiv 1 \mod 8$) is an isomorphism for $r=1$ and zero for $r>1$. I trust Adams in this (especially in view of the simple argument given above). $\endgroup$
    – John Rognes
    Commented Sep 11, 2015 at 14:17

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