Are the chern classes of the sphere bundle from a complex line bundle the same?

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 $L$?

1 Answer

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.

• 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. Aug 24, 2017 at 17:12
• Then this is the bundle $P_L\times_{S^1}S^1\cong P_L$ and it has the same Chern classes. Aug 24, 2017 at 17:24
• Ah, okay. You are taking the identity morphism $S^1 \to S^1$ and gluing $P_L$ along that. Aug 24, 2017 at 18:49
• 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. Aug 24, 2017 at 20:17
• @Tyrone, thanks +1, nice answer, I voted you up again. Jul 10, 2018 at 16:29