(Edit: Now posted to MO.)

That is:

For which positive integers $n, k \ge 1$ does there exist a submersion $S^{n+k} \to S^k$?

The discussion at this math.SE question has narrowed it down to the following two cases: either

  • $n = k-1$, in which case $k = 2, 4, 8$, realized by the complex, quaternionic, and octonionic Hopf fibrations, or
  • $n = 3k-3$, in which case $k \ge 4$ is even.

I am moderately confident that the second case doesn't occur, but don't know how to rule it out.

Here's what I can show about it: such a submersion gives rise to a smooth fiber bundle $F \to S^{4k-3} \to S^k$ where $F$ is a smooth frameable closed manifold of dimension $3k-3$. Taking homotopy fibers gives a map $\Omega S^k \to F$ whose homotopy fiber is $(4k-5)$-connected, hence which induces an isomorphism on homotopy and on cohomology up to degree $4k-5$.

This determines the cohomology of $F$ as a ring: $F$ has the cohomology of $S^{k-1} \times S^{2k-2}$. When $k = 2$ Mike Miller showed that $F$ must in fact be homeomorphic to $S^1 \times S^2$ and then gets a contradiction from looking at homotopy groups. When $k \ge 4$ we also know that $F$ is simply connected.

Aside from knowing whether it's possible to rule out the last case, I'd also be interested in a simpler argument that $k$ must be even. The argument I gave passes through both the topological Poincaré conjecture and Adams' solution to the Hopf invariant $1$ problem...


1 Answer 1


This question has been asked and answered on MathOverflow. I have replicated the accepted answer by Tom Goodwillie below.

In most cases $\pi_{n+k}(S^k)$ is a finite group, so that the homotopy fiber of any map $S^{n+k}\to S^k$ is rationally equivalent to $\Omega S^k\times S^{n+k}$ and therefore has homology in arbitrarily high dimensions and cannot be a manifold.

The only exceptions with $n>0$ have $n=k-1$.


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