# 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.

• Although the answer shows that the construction from first paragraph can't be generalized in most straightforward way, perhaps there still is some kind of generalization. Would be glad to hear any ideas. – Grigory M Apr 28 '11 at 13:17

## 1 Answer

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).

• It's certainly true (and thank you for spelling this out!), but not quite... explicit. I see a contradiction, but don't understand what's going on: where does the construction (sketched in the last paragraph of the OP) breaks? — the map is not well-defined? or it is not a fibration? – Grigory M Apr 26 '11 at 13:45
• I'm not sure, and, unfortunately, I'm under a huge time crunch until 10:30 tomorrow morning (I shouldn't even be on this site now!!). I'll think about it and get back to you, but I suspect the problem is well-definedness. – Jason DeVito Apr 26 '11 at 13:56
• Sorry to distract you :). Actually, it's almost obvious that there is no reason for the map to be well-defined (I'll update OP, probably; the problem is, one have to be more careful with definition of octonionic Hopf fibration already). – Grigory M Apr 26 '11 at 16:48
• (related: mathoverflow.net/questions/14698) – Grigory M Nov 28 '11 at 23:12
• @Grigory: If I had to guess, I learned my entire argument from Torcuato's post. He is certainly the one who caused me to notice it. Thanks for finding that! – Jason DeVito Nov 29 '11 at 2:48