Let $G$ be a compact group acting freely and properly on a spin manifold $X$. Fix a spin-structure on $X$ and suppose it is $G$-equivariant.

How does the $G$-equivariant spin-structure on $X$ induce a spin-structure on the quotient manifold $X/G$?

  • 3
    $\begingroup$ What is $G$, and how does it act on $X$ (freely, properly...)? $\endgroup$ – abx Oct 16 at 14:52
  • $\begingroup$ @abx I thought it was clear since I assumed that $X/G$ was a manifold... But I edited so that there can be no confusion. $\endgroup$ – Kafka91 Oct 16 at 15:03
  • 2
    $\begingroup$ Write $F(M)$ for the frame bundle of $M$, a principal $SO(n)$-bundle. A spin structure is a principal $\text{Spin}(n)$-bundle $F_S(M)$ with an isomorphism $$F_S(M) \times_{\text{Spin}(n)} SO(n) \cong F(M).$$ An oriented $G$-manifold (with $G$-invariant metric) comes equipped with a $G$-action on $F(M)$ (take the derivative). So the right definition of a spin $G$-manifold is a left $G$-action on $F_S(M)$ with an isomorphism as $(G, SO(n))$-spaces $$F_S(M) \times_{\text{Spin}(n)} SO(n) \cong F(M).$$ $\endgroup$ – Mike Miller Oct 16 at 15:41
  • 2
    $\begingroup$ Then when $G$ acts with the right adjectives on $M$, the space $G\backslash F_S(M)$ is a perfectly good principal $\text{Spin}(n)$-bundle over $G \backslash M$, and the above isomorphism is $G$-equivariant so descends to an isomorphism $F_S(G \backslash M) \times_{\text{Spin}(n)} SO(n) \cong F(G \backslash M)$. $\endgroup$ – Mike Miller Oct 16 at 15:41
  • $\begingroup$ @MikeMiller thanks a lot! $\endgroup$ – Kafka91 Oct 20 at 7:23

Your Answer

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

Browse other questions tagged or ask your own question.