I have often heard people talk about, say, "the" twisted $S^2$-bundle over $S^2$. My question is, what do they mean by a twisted bundle? I know that in the above example any $S^2$-bundle over $S^2$ is either $S^2 \times S^2$ or a unique non-trivial $S^2$-bundle over $S^2$. But is that how it is defined? Is a twisted bundle just a non-trivial one, or is it a specific bundle among the non-trivial ones?
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$S^2$ bundles over $S^2$ (smooth, PL, topological) are in bijective correspondence with homotopy-classes of maps: $$ S^2 \to BSO_3 $$ which up to homotopy is $$\pi_2 BSO_3 \simeq \pi_1 SO_3 \simeq \mathbb Z_2 $$ So there's precisely two non-isomorphic $S^2$-bundles over $S^2$. You can view the non-trivial $S^2$-bundle as the fibrewise one-point compactification of the vector bundle over $S^2$ whose Euler class is $1$. That vector bundle has fairly standard constructions. |
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