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The sphere bundle of a complex line bundle $$ L \to M $$ is an $S^1$-bundle over $M$. Moreover, since complex vector bundles are always orientable, we have that the induced $S^1$-bundle is principal.

Since Chern-Weil theory gives us a way to construct chern classes for principal bundles, does the chern class of this $U(1)$-bundle agree with the chern class for $L$?

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Yes. If $P_L$ is the principal $S^1$-bundle associated to $L$ then there is an isomorphism of line bundles $P_L\times_{S^1}\mathbb{C}\cong L$ over $X$ by definition. The Chern classes are natural under pullback so both the bundles have the same Chern classes.

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  • $\begingroup$ No, this is the sphere bundle constructed by taking all norm 1 vectors in the fibers. I do not know if they are the same. $\endgroup$
    – 54321user
    Aug 24, 2017 at 17:12
  • $\begingroup$ Then this is the bundle $P_L\times_{S^1}S^1\cong P_L$ and it has the same Chern classes. $\endgroup$
    – Tyrone
    Aug 24, 2017 at 17:24
  • $\begingroup$ Ah, okay. You are taking the identity morphism $S^1 \to S^1$ and gluing $P_L$ along that. $\endgroup$
    – 54321user
    Aug 24, 2017 at 18:49
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    $\begingroup$ We know $P_L\times_{S^1} \mathbb{C}\cong L$. Therefore we have a bundle map $P_L\times_{S^1}(\mathbb{C}-0)\rightarrow P_L\times_{S^1}S^1$ induced by the $S^1$-equivariant map $\mathbb{C}-0\ni v\mapsto v/|v|\in S^1$ which identifies the later bundle as the sphere bundle of $L$. The map $(e,z)\mapsto e\cdot z$ is then a bundle-isomorphism $P_L\times_{S^1}S^1\cong P_L$ so they have the same characterisic classes. $\endgroup$
    – Tyrone
    Aug 24, 2017 at 20:17
  • $\begingroup$ @Tyrone, thanks +1, nice answer, I voted you up again. $\endgroup$
    – wonderich
    Jul 10, 2018 at 16:29

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