Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

It is well known that for a compact topological group $G$ acts (say, from the right) freely on a compact space $X$. Then the category of equivariant complex vector bundles on $X$, $\text{Vect}_G(X)$, is equivalent to the category $\text{Vect}(X/G)$ consists of complex vector bundles on the quotient space $X/G$.

On the other hand, we have the correspondence between vector bundles and finitely generated projective modules, i.e Serre-Swan theorem. In more details, let $A:=C(X)$ be the ring of continuous complex functions on $X$ and for $E$ a vector bundle on $X$, the global sections $M:=\Gamma(E)$ is a finitely generated projective $A$-module. Since $G$ acts on $X$, $G$ also acts on $A$ and if we denote $A^G$ for the invariant subring then we have $A^G=C(X/G)$. Moreover $G$ acts on the category $A$-Mod by twist. Explicitly, if $M$ is an $A$-module and $g\in G$, then the twisted module $M^g$ is the same as $M$ as sets but with different $A$-module structure, i.e for $a\in A$, $m\in M$ we have $a\cdot_g m:=g(a)\cdot m$. The category of equivariant $A$-modules consists of $M$ such that $M\cong M^g$ as $A$-modules for any $g\in G$ and the morphisms are $G$- equivariant $A$-module maps.

The statement in the first paragraph now becomes: the category of equivariant finitely generated projective modules of $A$ is equivalent to the category of finitely generated projective modules of $A^G$, given that the action of $G$ is free. The functor $A^G$-mod to equivariant $A$-mod is given by $$ N \mapsto A\otimes_{A^G}N. $$

My question is: is there a proof purely in terms of $A$-modules and $A^G$-modules and Barr-Beck theorem? (We have an algebraic version of all above, say in Section 4.4 of Vistoli's "Notes on Grothendieck topologies, fibered categories and descent theory". Notice that free $G$ action on space $X$ is replaced by $G$-torsor $X$ there. It is essentially Barr-Beck theorem.)

In particular

  1. Is the condition that the action of $G$ on $X$ is free is the same as $A^G\rightarrow A$ is faithfully flat or something similar to faithfully flat?

  2. How to apply Barr-Beck theorem in this case?

share|improve this question

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.