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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1 Answer
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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.
answered Nov 3, 2012 at 22:45
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3$\begingroup$ This is the correct mathematical answer; the sociological answer is that `twisted' usually just means non-trivial. As Ryan says, in this setting there's only one non-trivial bundle, so the usage is not ambiguous. $\endgroup$ Commented Nov 3, 2012 at 22:48
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1$\begingroup$ It seems to me that Ryan Budney's answer does not really answer the OP's question, although Danny Ruberman's comment does: "twisted" just means "nontrivial," and one should only really talk about "the" twisted bundle when there is only one non-trivial bundle. $\endgroup$ Commented Nov 3, 2012 at 23:25