When are fiber bundles reversible?

Question

My question, in its most general form is this:

Given a fiber bundle $F\rightarrow E\rightarrow B$, when is there a fiber bundle $B\rightarrow E\rightarrow F$?

Here, F,E, and B can lie in whichever category you wish, but I'm mostly interested in the case where all 3 are smooth closed manifolds.

Now, I realize that the initial answer is "unless E is a product, essentially never", so here is a more focused question (with background).

I've been studying a certain class of free actions of the 3-torus $T^3$ on $S^3\times S^3\times S^3 = (S^3)^3$. For each of these actions, by quotienting out by various subtori, I can show that the orbit space $E=(S^3)^3/T^3$ simultaneously fits into 2 fiber bundles:

$$S^2\rightarrow E \rightarrow S^2\times S^2$$ and $$S^2\times S^2\rightarrow E\rightarrow S^2$$ where the structure group for both bundles is $S^1$.

(In fact, the class of actions also gives rise to examples where either $S^2\times S^2$ can independently be replaced with $\mathbb{C}P^2\sharp -\mathbb{C}P^2$, the unique nontrivial $S^2$ bundle over $S^2$.)

By computing characteristic classes for (the tangent bundle to) E, I know that for an infinite sublcass of the actions I'm looking at, E is not homotopy equivalent to $S^2\times S^2\times S^2$, and each of the E are pairwise nondiffeomorphic.

I suspect the reason I could find so many E which fit into "reversible" fiber bundles is strongly related with the fact that the fiber and base are so closely related.

And so, I ask

For fixed manifold M, what is the relationship between bundles $X\rightarrow E\rightarrow M$ and $M\rightarrow E'\rightarrow X$ where $X$ is some $M$ bundle over $M$?

And just in case there is no general relationship,

Is there a reason I should have expected there to be a relationship in my examples, even though in general there isn't?

Answer 1

In terms of Seifert fiber spaces, there are two examples when you consider torus bundles over $S^1$ among the ones which use periodic mapping classes: These are $(No,1|(1,0))$ and $(No,1|(1,1))$ which are respectively $K\times S^1$ and $K\times_{\tau}S^1$, where $\tau$ is the unique Dehn's twist on the Klein bottle.

Those fibrations are not unique because also $K\times S^1=(NnI,2|(1,0))$ and $K\times_{\tau}S^1=(NnI,2|(1,1))$. Curiously, if $T=S^1\times S^1$ is the 2-torus, then $T\hookrightarrow K\times S^1\to S^1$ is a "non-trivial" fibering.