When are these base spaces isomorphic?

Question

Given a smooth manifold $\mathcal{M}$ that is a fibre bundle over two different base spaces, i.e., there are $\Pi_1:\mathcal{M}\rightarrow B_1$ and $\Pi_2:\mathcal{M}\rightarrow B_2$, if I can prove that the fibers are isomorphic, then necessarily the base spaces are also isomorphic?

Answer 1

Here's a perhaps more serious example: There is a compact $4$-manifold that fibers over both $\mathbb{RP}^3$ and $(S^1\times S^2)/\mathbb{Z}_2$ where, in each case, the fibers are circles. (The $\mathbb{Z}_2$-action on $S^1\times S^2$ is the 'diagonal-antipodal' action.)

To see the example, consider a $4$-dimensional real vector space $V$ endowed with a symplectic structure, i.e., a non-degenerate, skew-symmetric pairing $\omega:V\times V\to \mathbb{R}$.

The $4$-manifold $M$ will be the space of pairs $(L,E)$ where $E\subset V$ is an $\omega$-Lagrangian $2$-dimensional subspace and $L\subset E$ is a $1$-dimensional subspace. Let $\pi_1:M\to \mathbb{P}(V)\simeq\mathbb{RP}^3$ be the map $\pi_1(L,E) = L$, and let $\pi_2:M\to \operatorname{Gr}_2(V)$ be $\pi_2(L,E) = E$. The image of $\pi_2$ in $\operatorname{Gr}_2(V)$ is the space of $\omega$-Lagrangian subspaces in $V$, which is known to be diffeomorphic to $(S^1\times S^2)/\mathbb{Z}_2$.

The fiber $\pi_2^{-1}(E)$ is the set of $1$-dimensional subspaces of $E$, i.e., $\mathbb{P}(E)\simeq\mathbb{RP}^1\simeq S^1$.

The fiber $\pi_1^{-1}(L)$ is the set of $\omega$-Lagrangian spaces that contain $L$, but this is $\mathbb{P}(L^\perp/L)\simeq \mathbb{RP}^1\simeq S^1$, where $L^\perp\subset V$ is the $3$-dimensional space that is the $\omega$-annihilator of $L$ (so $L^\perp/L$ is a $2$-dimensional vector space).

Answer 2

I guess that isomorphic means diffeomorphic. Let me provide an example that seems conceptually different from the others given so far.

Take $M= \mathbb{R}^4 \times \mathbb{R}^4_{\operatorname{ex}} \times \mathbb{R}$, where $\mathbb{R}^4_{\operatorname{ex}}$ is any exotic $\mathbb{R}^4$.

You have two projections

$$ p_1 \colon M \longrightarrow \mathbb{R}^4, \quad p_2 \colon M \longrightarrow \mathbb{R}^4_{\operatorname{ex}},$$

whose fibres are diffeomorphic to $\mathbb{R}^4 \times \mathbb{R}_{\operatorname{ex}}$ and $\mathbb{R}^4 \times \mathbb{R}$, respectively.

Since there exists no exotic $\mathbb{R}^5$, these fibres are both diffeomorphic to $\mathbb{R}^4 \times \mathbb{R}$, but the bases are not diffeomorphic.

Answer 3

Let me add some more examples. It is quite challenging to classify all free actions of a group $G$ on a given space (or even decide if there is one). Classify could mean here up to $G$-equivariant diffeomorphism, homeomorphism, homotopy equivalence, and so on. One invariant of such an action is the (homotopy type of the) quotient by the group action, which then can be viewed as the base of a fiber bundle with fiber $G$.

There are plenty of group actions that can be distinguished by looking at the quotient. A classical and particularly interesting example are the actions of $\mathbb{Z}/p$ on $S^{2n-1}\subset \mathbb{C}^n$, where the groups acts on each coordinate by multiplication with a primitive $p$-root of unity. We can choose any such root at every coordinate and thus we obtain a list of actions. We would like to decide whether the two actions are the same.

The quotient by this action is called a lens space, and the classification of lens spaces up to homeomorphism differs from the classification up to homotopy equivalence. So there we obtain more examples as above, where the base space is an odd-dimensional sphere (starting with $S^3$, $S^1$ behaves differently).