Total spaces of orientable $2n$-sphere bundles

Question

If $\pi:E\to M$ is an orientable $S^{2n}$-bundle, then (see Bott-Tu 11.20) $$(1) \qquad \qquad H^{\ast}(E)=H^{\ast}(M) \otimes H^{\ast}(S^{2n}).$$ If $\pi:E'\to M'$ is another orientable $S^{2n}$-bundle such that $H^{\ast}(M')=H^{\ast}(M)$, then we conclude that from $(1)$ that $H^{\ast}(E)=H^{\ast}(E')$.

Question: Suppose that we further assume that $M$ and $M'$ are homeomorphic. What can we conclude about the topology of $E$ and $E'$ in this case? For example, is it true that $E$ and $E'$ are homotopy equivalent, or even homeomorphic? If this is too ambitious, what if we restrict to the case where $M$ and $M'$ are themselves homeomorphic to a sphere?

Answer 1

Consider the two $S^{2}$-bundles over $S^{2}$, $X_{2} = Bl_{p}(\mathbb{CP}^{2}) (= \mathbb{CP}^{2} \# \bar{\mathbb{CP}^{2}})$ and $X_{1} = S^{2} \times S^{2}$. Then the intersection form on $H^{2}(X_{i},\mathbb{Z})$ is even for $i=1$ and odd for $i=2$. Hence these 4-manifolds are not homotopy equivalent.

Answer 2

Any $S^k$ bundle over $S^n$ can be trivialised over both hemispheres. Using this idea one sees that topologically, such bundles are classified by the gluing map over the equator, up to homotopy, which is an element of $$\pi_{n-1}(SO(k+1))$$

So for instance, from the fact that $\pi_1(SO(3))$ is not trivial, you can derive two non-equivalent $S^2$ bundles over $S^2$.

See also Norman Steenrod's The Topology of Fibre Bundles, Section 'The fibering of spheres by spheres'.