Does an oriented $S^3$ fiber bundle admit the structure of a principal $SU(2)$-bundle?

Let $S \to X$ be an $S^3$-fiber bundle over a smooth manifold $X$. If $S$ is an oriented manifold does this fiber bundle admit the structure of an $SU(2)$-principal bundle?

There is a similar theorem for the case of circle bundles and is proved in Morita's book on differential forms. Unfortunately, I do not see a way to extend his argument to this case and he does not discuss $SU(2)$-principal bundles in detail.

• A nontrivial result of A. Hatcher says that the diffeomorphism group of $S^3$ is homotopic with $O(4)$. If the bundle is orientable, then the bundle has an $SO(4)$-structure, i.e., it is the unit sphere bundle of an oriented rank $4$ real vector bundle. – Liviu Nicolaescu Sep 29 '17 at 8:47
• (I'm assuming you mean that the bundle is orientable not just the total space). From there maybe use that SO(4)/SU(2) \cong SO(3). This is topologically split, and SO(3) has nontrivial pi_4. Using the splitting gets you an oriented vector bundle on S^5 that doesn't have an SU(2) structure. Of course, with the counterexample in hand, we didn't need the Smale conjecture :) – Dylan Wilson Sep 29 '17 at 9:07
• Dylan, maybe you meant $\pi_3$, so a bundle over $S^4$? – Tom Goodwillie Sep 29 '17 at 23:22

No. (The main idea here is present in Dylan Wilson's comment.)

Every principal $SU(2)$-bundle over $S^2$ is trivial, because $\pi_1 SU(2)$ is trivial. But there is a nontrivial oriented bundle over $S^2$ with fiber $S^3$, namely the unit sphere bundle of the nontrivial rank $4$ vector bundle. (There are precisely two rank $r$ oriented vector bundles over $S^2$ if $r\ge 3$, because $\pi_1 SO(r)$ has order two.)

I should explain why the nontrivial vector bundle has nontrivial unit sphere bundle, that is, why $\pi_1 SO(4)$ injects into $\pi_1 SDiff(S^3)$.

You can use the big theorem of Hatcher ("Smale Conjecture"), which says that $\pi_k SO(4)$ maps isomorphically to $\pi_k SDiff(S^3)$ for all $k$.

Alternatively, you can use that the bundle is detected by the Stiefel-Whitney class $w_2$, and that Stiefel-Whitney classes of vector bundles are invariant under fiber homotopy equivalence of unit sphere bundle.

Or you can use $\pi_3$ instead of $\pi_1$ as suggested by Dylan Wilson, making a bundle over $S^4$; there are elements of $\pi_3 SO(4)\cong \mathbb Z\times \mathbb Z$ not coming from $\pi_3 SU(2)\cong\mathbb Z$. Again, the resulting sphere bundle is nontrivial either by Hatcher's theorem or by using characteristic classes. I'm not sure what the most elementary version of the characteristic-class argument would be.

One example of a rank $4$ real vector bundle over $S^4$ that does not admit a complex structure is the tangent bundle!