Adams and Atiyah give a wonderfully simple proof of the Hopf invariant 1 problem that uses the Adams operations on K-theory to reduce the Hopf Invariant 1 question to an elementary number theory question. In this theme, I think we should also be able to reduce the Hopf Invariant 1 problem to a number theoretical question about the L-polynomials.

Recall the Hopf Invariant 1 problem asks for which $n$ there is $f: S^{2n-1} \rightarrow S^n$, $\operatorname{cofiber}(f)=X$ has its middle dimensional cohomology generator square to a generator of its top dimension cohomology. Of course, this implies that $X$ has Poincare duality, i.e. it is a Poincare duality space.

It is not difficult to show that for $f$ to have Hopf invariant 1, $n$ must be a power of 2, so let us assume such an $n$ and that $n>2$. Thus $X$ has cohomology concentrated in even dimensions. We might ask when is $X$ actually the homotopy type of a manifold. The first obstruction is a lift of the Spivak normal fibration (that $X$ has as a result of being a PD space) to $BTop$. Recall $G$ is used to denote the space classifying stable spherical fibrations

Since the homotopy groups of $G/Top$ are the surgery obstruction groups of the trivial group, the homotopy groups are trivial in odd dimensions. So all obstructions to lifting must be trivial because the cohomology of $X$ is in even dimensions. Hence, we have a lift from $BG$ to $BTop$.

This means we have a surgery problem in dimension divisible by four, so if the surgery obstruction vanishes $X$ has the homotopy type of a $8k$-manifold that is $4k-1$ connected with signature 1. Perhaps it is helfpul to mention here that the surgery obstruction will just be the difference of signatures.

$\bf Question:$ Suppose I have a 8k-manifold $M$ so (1) the rank of $H^{4k}(M)$ is 1, (2) Hirzebruch L-polynomials for $M$ have contributions only from $p_{2k}$ and $p_{k}^2$, (3) $p_k$ is some multiple $n$ of the generator $x$ of $H^{4k}(M)$, and (4) that $x^2$ generates $H^{8k}(M)$, can we deduce a contradiction?

Of course, the solution of the Hopf invariant 1 problem implies no such manifold can exist (in dimensions greater than 16, but I am wondering if this can be proven only from the $L$ polynomials.

  • $\begingroup$ When you state your question, what are you already assuming about $X$? Just that it is a manifold, or something more? $\endgroup$ Commented Oct 12, 2020 at 17:50
  • $\begingroup$ @MichaelAlbanese X is the cofiber of a map of spheres. In the body of the question I show that (under the relevant conditions) if this map has Hopf invariant 1, the primary obstruction vanishes. So if we were able to somehow show the secondary surgery obstruction vanished, this would imply that $X$ is in fact a manifold. So for the purposes of the bold faced question, $X$ should be assumed a topological manifold, I will edit to make more clear. $\endgroup$ Commented Oct 12, 2020 at 18:00

1 Answer 1


If I have understood your question correctly, the answer is no.

A rational projective plane is a closed $2n$-dimensional manifold $M$ with $H^*(M; \mathbb{Q}) \cong \mathbb{Q}[\alpha]/(\alpha^3)$ where $\deg\alpha = n$. Such manifolds were studied by Su in her paper Rational Analogs of Projective Planes. Note that if $n = 4k$, a rational projective plane satisfies the criteria (1) - (4). However, in that same paper, Su used the Barge-Sullivan theorem to show the existence of a 32-dimensional rational projective plane. In particular, the conditions you state are insufficient to obtain a solution of the Hopf invariant one problem.

  • $\begingroup$ Yes I do think this is a counterexample! The paper is very interesting. Thank you, I will wait a while before accepting. $\endgroup$ Commented Oct 12, 2020 at 18:28
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    $\begingroup$ There is also the later paper "On dimensions supporting a rational projective plane" by Kennard and Su where they show rational projective planes exist in dimensions 128 and 256. They also show 4,8,16,32, 128, 256 are the only dimensions <=512 supporting such a manifold. I think this was independently obtained and announced by Zagier in work with Kreck, see the talk "What are the Betti numbers of a manifold?" by Zagier, Nov 30 2017, available on youtube; the 512 bound in their case is ~1000000. In the talk he expresses doubt that there are any further dimensions supporting such a manifold. $\endgroup$ Commented Oct 12, 2020 at 19:40

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