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 $L$?


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.

| cite | improve this answer | |
  • $\begingroup$ 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 '17 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 '17 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 '17 at 18:49
  • 1
    $\begingroup$ 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 '17 at 20:17
  • $\begingroup$ @Tyrone, thanks +1, nice answer, I voted you up again. $\endgroup$ – wonderich Jul 10 '18 at 16:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.