What is the right definition of equivariant Cech cohomology is so that given a $G$-space $X$, $H^1_G(X;H)$ classifies $G$-equivariant principal $H$-bundles on $X$?

H is abelian, right? If H is not abelian, you will fail to construct the long exact sequence past the first cohomology group (cf. Giraud's Cohomologie non-abelienne).
– Harry GindiSep 28 '10 at 15:08

2

At this point, it really makes most sense to define the cohomology of the orbifold X/G. That orbifold is best encoded by the groupoid whose objects are X and whose arrows are XxG. In that way, you avoid the possibility of doing various non-sensical constructions (such as keeping track of different G-space structures on H - no offence).
– André HenriquesSep 28 '10 at 16:56

What is a $G$-equivariant principal $H$-bundle?
– Martin BrandenburgSep 28 '10 at 17:22

Andre is right to point out that the way I first went at it doesn't make sense. There's some extra data that I need to encode in the definition of cochains in order to capture that a G-equivariant principal H-bundle E has isomorphisms $g:E_x\rightarrow E_{gx}$ for any $g\in G$ and $x\in X$ and I′d confused this data with a notion of $G$−action on $H$. Thanks for the replies.
– Jesse WolfsonSep 28 '10 at 21:05

You say: right definition but usually that requires a meaning for 'right'. For example, working with $G$-spaces the whole time look at what a group object in that category looks like, here you may need to replace $G$-spaces by spaces over $BG$. Now look what torsor / principal bundle objects over a G-space in that category would be and so on. That is very general and may be way too general for what you want but is one interpretation. André's Orbifold approach (and orbifold cohomology à lå Moerdijk and Pronk?) may answer your query by interpreting things differently.
– Tim PorterJun 21 '11 at 16:34

groupoidwhose objects areXand whose arrows are XxG. In that way, you avoid the possibility of doing various non-sensical constructions (such as keeping track of different G-space structures on H - no offence). – André Henriques Sep 28 '10 at 16:56