$S^6$ as the total space of bundle

Question

Does there exist a 4-manifold $M$ such that $S^6$ is the total space of an $S^2$-bundle over $M$?

If the bundle is orientable, then according to Bott and Tu: $H^{\ast}(S^6)=H^{\ast}(M)\otimes H^{\ast}(S^2),$ so it doesn't seem to be possible in this case.

Answer 1

There is no manifold $M$ with this property. First of all, note that $M$ must be closed.

Associated to the fibration $S^2 \to S^6 \to M$ we have the long exact sequence in homotopy

$$\dots \to \pi_{n+1}(M) \to \pi_n(S^2) \to \pi_n(S^6) \to \pi_n(M) \to \pi_{n-1}(S^2) \to \dots$$

It follows that $\pi_1(M) = 0$ (so $M$ is orientable), $\pi_2(M) = 0$, and $\pi_3(M) = \mathbb{Z}$. By the Hurewicz Theorem, $H_1(M; \mathbb{Z}) = H_2(M; \mathbb{Z}) = 0$, and $H_3(M; \mathbb{Z}) = \mathbb{Z}$. But this is impossible as $M$ must satisfy Poincaré duality.

In fact, once we know $\pi_1(M) = \pi_2(M) = 0$, it follows from Freedman's Theorem that $M$ is homeomorphic to $S^4$, while $\pi_3(S^4) = 0 \neq \mathbb{Z}$.

Answer 2

If the bundle is not oriented, there is a double covering $p:M'\rightarrow M$ such that the pullback of $f:S^6\rightarrow M$ by $p$ is oriented. $p^*f:N\rightarrow M'$ induces a covering $N\rightarrow S^6$, you deduce that $N=S^6$ and apply your argument again.