When are fiber bundles reversible? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-26T03:04:44Z http://mathoverflow.net/feeds/question/20971 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/20971/when-are-fiber-bundles-reversible When are fiber bundles reversible? Jason DeVito 2010-04-11T01:32:12Z 2010-04-11T01:32:12Z <p>My question, in its most general form is this:</p> <blockquote> <p>Given a fiber bundle $F\rightarrow E\rightarrow B$, when is there a fiber bundle $B\rightarrow E\rightarrow F$?</p> </blockquote> <p>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.</p> <p>Now, I realize that the initial answer is "unless E is a product, essentially never", so here is a more focused question (with background).</p> <p>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:</p> <p>$$S^2\rightarrow E \rightarrow S^2\times S^2$$ and $$S^2\times S^2\rightarrow E\rightarrow S^2$$ where the structure group for both bundles is $S^1$.</p> <p>(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 $S^2$.)</p> <p>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.</p> <p>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.</p> <p>And so, I ask</p> <blockquote> <p>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 $M$?</p> </blockquote> <p>And just in case there is no general relationship,</p> <blockquote> <p>Is there a reason I should have expected there to be a relationship in my examples, even though in general there isn't?</p> </blockquote>