Let $V$ be a simply connected smooth projective complex variety defined over the rationals. Then for any integer $n\geq 2$ the group $\pi_n(V)$ is finitely generated abelian so profinite completion preserves its torsion. Thus we get an action of the absolute Galois group of $\mathbb{Q}$ on the torsion subgroup of $\pi_n(V)$. Is there an example where (for fixed $V$ and $n$) it does not factor through the group of homotopy classes of homotopy equivalences $V\to V$?

  • 1
    $\begingroup$ That should be true for practically any example. For example $\pi_n\mathbb P^k=\pi_n S^{2k-1}$ so the action of homotopy equivalences can only be $\pm1$. But $\pi_{2k-1}\mathbb P^k=\mathbb Z(-k)$ which probably propagates through to the higher homotopy groups: I don't know the Galois action on $\pi_{2k+2}\mathbb P^k\cong \mathbb Z/24$ is, but I imagine that $\pi_{2k+4}\mathbb P^{k+1}=\pi_{2k+2}\mathbb P^k(-1)$, so infinitely many choices have bigger action than $\pm1$. If you just fix $V=\mathbb P^1$, the groups are harder to compute, but I imagine that they detect the cyclotomic character. $\endgroup$ – Ben Wieland Jun 29 at 19:22
  • $\begingroup$ Oh, yeah, I meant to say that this question sounds the like the title of Sullivan's MIT notes, so, uh, check that out, ok? . . . I now claim that in the stable range $\pi_{2n+k-1}\mathbb P^n=\pi_k^S(-n)$. The Galois group acts on the space $S^{2n-1}_p$, which has automorphism group $\mathbb Z_p^\times$. Everything is determined by the action on $\mathbb Z_p$. But what's special in the stable range is that the action is the same on all the homotopy groups. (Outside the stable range, that's not true. eg, the Hopf invariant is quadratic.) $\endgroup$ – Ben Wieland Jun 29 at 23:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.