32
$\begingroup$

(First posted on math.SE, with no answers.)

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...

$\endgroup$

2 Answers 2

47
$\begingroup$

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 $n=k-1$.

$\endgroup$
1
  • 4
    $\begingroup$ Thanks! That's more straightforward than I thought; I guess the appearance of $n = k - 1$ was a clue I should've taken more seriously. $\endgroup$ May 26, 2015 at 4:02
12
$\begingroup$

I want to add that having a fiber bundle with total space a sphere is very restrictive even if you don't assume that the base is a sphere too. This has been studied. I will exclude the obvious cases of the fiber or the base being points. The starting point is a fairly elementary observation due to Whitehead and Spanier that for a fiber bundle $F\to E\to B$ where all spaces are connected finite dimensional CW complexes, if the fiber $F$ is contractible in $E$ then $F$ must be an $H$-space. This of course restricts things a great deal and by playing with rational cohomology it's easy to see that if $E$ is a sphere (which can only be odd dimensional) the fiber must rationally be an odd dimensional sphere too. Browder later showed that $F$ can only be homotopy $S^1, S^3, S^7$. This implies that the base is a homotopy $\mathbb{CP}^n$, a homology $\mathbb{HP}^n$ or a homotopy $S^8$ respectively. This yields the result when one assumes that the base is a sphere as a special case.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.