TITLE: The Hopf Invariant 1 Problem through L-polynomials
QUESTION [12 upvotes]: 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.

REPLY [13 votes]: 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.