Explain unoriented $S^2$ bundle over $S^1$.

Question

In Hamilton's classification of closed 3-d nonnegative Ricci curvature manifold, unoriented $S^2$ bundle over $S^1$ is one of the possible type. Who can give me a description of it? Many thanks!

Answer 1

Any $M$-bundle over $S^1$ is a manifold known as a mapping torus. You obtain it by picking a diffeomorphism, $\varphi: M \to M$, and letting $M_\varphi$ be the quotient manifold $M \times \Bbb R/(x,t) \sim (\varphi(x),t+1)$. The bundles $M_\varphi \to S^1$ are classified by the isotopy class of $\varphi$. For the case of $M = S^2$, there are two isotopy classes of map, one given by the identity, one given by the antipodal map. So the unoriented $S^2$ bundle is $$S^2 \times \Bbb R/(x,t) \sim (-x,t+1).$$

This is sometimes known as a twisted product and is written $S^2 \widetilde \times S^1$. Its orientation double cover is $S^2 \times S^1$, its fundamental group is $\Bbb Z$, and higher homotopy groups coincide with those of $S^2$. It (rather obviously) has $S^2 \times \Bbb R$ geometry, because we defined it as a quotient of $S^2 \times \Bbb R$ by a discrete group of isometries. If you think of $S^2 \times S^1$ as something sort of like $S^1 \times S^1$, then $S^2 \widetilde \times S^1$ is sort of like the Klein bottle. (Of course, neither $S^2 \times S^1$ nor its twisted counterpart have Euclidean geometry.)