# Octonionic Hopf fibration and $\mathbb HP^3$

Real, complex and quaternionic versions of Hopf fibration ($S^0\to S^n\to\mathbb RP^n$, $S^1\to S^{2n+1}\to\mathbb CP^n$ and $S^3\to S^{4n+3}\to\mathbb HP^n$) give rise to spherical fibrations $S^1\to\mathbb RP^{2n+1}\to\mathbb CP^n$ and $S^2\to\mathbb CP^{2n+1}\to\mathbb HP^n$.

Question. Does octonionic Hopf fibration $S^7\to S^{15}\to\mathbb OP^1$ give rise to a fibration $\mathbb HP^3\to\mathbb OP^1$?

(On one hand, constructing a map $\mathbb HP^3\to\mathbb OP^1$ seems(upd-1) to be easy: take preimage under $S^{15}\to\mathbb HP^3$ and then use octonionic Hopf map. On the other hand, $\mathbb HP^n$ doesn’t admit free involution for $n\ge 2$…)

(upd-1) Wrong! Let $\pi$ be the map $(a,b)\mapsto ab^{-1}$. For octonions $\pi(a,b)$ and $\pi(a\lambda,b\lambda)$ needn’t coincide (because of non-associativity). So this map is not well-defined.

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It can’t work in the usual way.

Assuming everything works just as in the previous 2 cases, the fiber would be $S^4$. It follows that the bundle $\mathbb{H}P^3\rightarrow\mathbb{O}P^1$ would be orientable since all the pieces are simply connected. Now, one would be able to “fill in the fibers” to get a bundle $D^4\rightarrow X\rightarrow \mathbb{O}P^1$, with $X$ an orientable 13-manifold whose boundary is $\mathbb{H}P^3$. But there is no such manifold.

To see there is no such manifold, recall that the usual proof that there are no free involutions comes from the fact that the first Pontryagin class $p_1(\mathbb{H}P^n) = [2(n+1)-4]u$ where $u$ is a generator of $H^4(\mathbb{H}P^n)$. Thus $p_1\neq 0$ unless $n=1$. Then then using the fact that an involution (or any diffeomorphism) must preserve $p_1$ and applying the Lefshetz fixed point theorem, we find every involution (or diffeomorphism) has a fixed point.

All we need is that $p_1\neq 0$. This, together with the ring structure, implies that the Pontryagin number $\langle p_1^n, u^n\rangle \neq 0$. It follows from a theorem of Thom, that no $\mathbb{H}P^{n}$ is the boundary of an oriented manifold. (In fact, if $n$ is even, a similar argument with Stiefel-Whitney classes shows that these manifolds are not the boundary of any manifold, orientable or not).