MathOverflow will be down for maintenance for approximately 3 hours, starting Monday evening (06/24/2013) at approximately 9:00 PM Eastern time (UTC-4).

## Why isn’t the orbifold cohomology of $pt/G$ equal to the cohomology of $BG$?

The classifying space of a group $G$ is given by taking a contractible space $E$ equipped with a free $G$-action, and looking at the quotient, which we dub $BG$. The homotopy type of this space (and thus its cohomology) depend only on $G$, and this gives us one definition of group cohomology.

Now, we can also look at the orbifold $[pt/G]$, and compute its orbifold cohomology (as its regular cohomology is rather uninteresting). For finite groups $G$, we have the isomorphism (see Adem-Leida-Ruan) $$H^*_{orb}\big([pt/G],\mathbb{C}\big) \cong Z(\mathbb{C}G).$$

We can easily see that for finite groups the cohomology obtained is not the same as we would get from the usual group cohomology.

Why? I understand that the actual constructions are very different, but this seems very unsatisfying to me. Morally this space is a contractible one modulo a free group action, and it is a model of $BG$ if we look at it in the appropriate way, so why shouldn't their cohomologies be the same?

-
I realize that for finite groups, the torsion in H^*(BG) will go away if we use complex coefficients, but that doesn't change the fact that these will still be different. – Simon Rose Mar 3 2011 at 0:16
In Adem-Leida-Ruan I see in example 2.10, rather, that the two cohomologies agree. And see proposition 2.12 that says that over Q (so over C as well, I guess) these definitions agree with yet a third one. – Pierre Mar 3 2011 at 1:10
Oh, it has occured to me that you say "orbifold cohomology" for Chen-Ruan cohomology". Odd, considering your source is Adem-Leida-Ruan. It's also odd that someone understood you and gave an answer, well done :) – Pierre Mar 3 2011 at 1:34
Note that $[pt/G]$ is only an orbifold for finite $G$, so I'm not sure why you specify 'for finite groups'. – David Roberts Mar 3 2011 at 2:53
Yeah, I should have put the phrase "for finite groups $G$" earlier... – Simon Rose Mar 3 2011 at 15:50

Orbifold cohomology is not a model for the cohomology of the orbifold (or more generally stack) itself, but for that of its inertia stack (a.k.a. derived loop space), which parametrizes points of the stack together with automorphisms. The inertia of $BG$ is $G/G$ (which is also the correct homotopy type for the free loops in the classifying space of $G$), hence you're seeing class functions (aka the center of the group algebra). This is very natural from the point of view of topological field theory -- we're trying to capture the space of states of a topological field theory (a sigma model) on the circle, which is a measure of the space of maps from the circle into some target. In the case of a stacky target, these maps also can wrap around automorphisms of points, resulting in the inertia stack. In physics these are called "twisted sectors". The orbifold cohomology can also be described in terms of Hochschild homology of sheaves on the space itself (and Hochschild homology is an algebraic model of taking loops in great generality).
...where $G/G$ is the quotient by the adjoint action of $G$ on itself, in case it wasn't clear. – David Roberts Mar 3 2011 at 2:54