elliptic curves and group cohomology

Question

Recently, I've been trying to understand Jacob Lurie's 2-equivariant elliptic cohomology a bit better than I had in the past.

From what I can tell, the fragment of the story that only deals with finite groups, with characteristic zero stuff, and only talks about the degree zero part of the cohomology theories is equivalent to the following claim:

CLAIM: Given the following input:
$\triangleright$ A $\mathbb Q$-algebra $R$.
$\triangleright$ An elliptic curve $E$ over $R$
$\triangleright$ A finite group $G$
$\triangleright$ A cocycle $k$ representing a class in $H^4(BG,\mathbb Z)$,
there is a way of constructing:
$\triangleright$ An $R$-module.

.

QUESTION: What is the construction which takes $R$, $E$, $G$, $k$ as input, and produces an $R$-module as output?

.

When $k=0$, the output of the construction should be ring of functions on the moduli stack $M_G$ of $G$-bundles over $E$. [Added later: this guess turned out to be wrong – see Jacob's answer]

When $R=\mathbb C$, I know how to define a line bundle $L_k$ over $M_G$, and I want the $R$-module to be the space of global sections of that line bundle.

When $R$ is some random $\mathbb Q$-algebra, I don't know how to define a line bundle over $M_G$, and in particular, I don't know how to construct its $R$-module of global sections.

Bonus questions:
$\triangleright$ What does the construction look like (in particular, does a construction exist) if one drops the assumption that $R$ is a $\mathbb Q$-algebra?
$\triangleright$ What does the construction look like if one drops the assumption that $G$ is a finite group, and one takes it to be a compact Lie group instead?
$\triangleright$ What does the construction look like if one drops the assumption that $E$ is an elliptic curve?
$\triangleright$ Any combination of the above.

Answer 1

As Charles indicates, "the moduli stack of $G$-bundles on $E$" is not quite the right thing to consider, especially if you're not working over $\mathbf{C}$. This is for two (unrelated) reasons:

1) The geometric object $M_{G}$ that you associate to a group $G$ isn't something that you can access directly (at least by the construction I know): what you can access instead is the ring of global functions on $M_{G}$, and maybe the global sections of a few other sheaves (like the line bundles you're asking about). Consequently, it's hard to tell the difference between moduli stack of $G$-bundles and its coarse moduli space. Let me ignore this in what follows. (More precisely, instead of answering the question "what geometric object and line bundle does elliptic cohomology produce?", I'll answer the question "how can I write down a geometric object and line bundle whose global sections are related to $2$-equivariant elliptic cohomology?"

2) Saying "$M_G$ is the moduli stack of $G$-bundles" doesn't really make sense, because $G$ is a compact Lie group rather than an algebro-geometric object. A more accurate statement is "$M_G$ is the moduli stack of $G(1)$-bundles", where if $G$ is connected compact Lie group, $G(1)$ denotes the split algebraic group having the same root datum. (Example: if $G = U(n)$, then $G(1) = GL_n$.)

This generally does not make sense if $G$ is not connected, for example if $G$ is a finite group. However, you can make sense of it if $G$ is a finite abelian group (for example, if $G = \mathbf{Z}/n\mathbf{Z}$, then $G(1) = \mu_{n}$; in this case, $G(1)$-bundles on $E$ correspond to $n$-torsion points on the dual abelian variety, which is just $E$ again: this recovers what Charles said. Note that we meet issue 1) here: the scheme of $n$-torsion points on $E$ is a -coarse- moduli space for $\mu_n$-bundles on $E$).

For $G$ a finite nonabelian group, you can't really make sense of $G(1)$. However, you can still make sense of a $G(1)$-bundle on $E$, at least when the order $|G|$ is invertible in $R$. To explain this, let me assume for simplicity that $R$ is an algebraically closed field of characteristic zero, and try to describe everything in way that is invariant under automorphisms of $R$. Let $\Lambda$ denote the etale fundamental group of $E$: this is a free module of rank $2$ over the ring $\widehat{ \mathbf{Z} }$ of profinite integers. The datum of a $G$-bundle on $E$ (up to isomorphism) is equivalent to the datum of map $\Lambda \rightarrow G$ (up to conjugacy). So you can define the datum of a $G(1)$-bundle on $E$ (up to isomorphism) to be the datum of a map $\Lambda(-1) \rightarrow G$ (up to conjugacy). Here $\Lambda(-1)$ denotes the twist of $\Lambda$ by the inverse of the cyclotomic character. In the language of etale cohomology, $\Lambda(-1)$ is just $H^{1}(E; \widehat{\mathbf{Z}})$.

So that's the sort of datum that $M_{G}$ is supposed to classify. You get to identify $M_{G}$ with the moduli stack of $G$-bundles on $E$ if you choose a compatible system of roots of unity in $R$, for example by taking $R = \mathbf{C}$).

Now suppose you're given a level $k$ on $G$, which we can identify with an element of $H^{4}(BG; \mathbf{Z} ) = H^3(BG; \mathbf{Q} / \mathbf{Z} )$. Let $T$ denote the profinite torus $B \Lambda(-1)$. Suitably interpeted, we have $H_2(T) = H^2(E) = \widehat{ \mathbf{Z} }(-1)$, and $M_{G}$ classifies maps $T \rightarrow BG$. You can now apply the construction that Qiaochu suggested: pull back the level to $T$ and integrate to get a class in $H^1( Spec R, (\mathbf{Q} / \mathbf{Z})(1) )$. Of course, that class will be trivial, but if you do everything at the level of cochains rather than cohomology classes you'll obtain the datum of a $( \mathbf{Q} / \mathbf{Z} )(1) = \mu_{\infty}(R)$-torsor, which determines an $R$-line.

If you say all this carefully, you'll end up with something with makes sense as long as $|G|$ is invertible in $R$ (if $|G|$ is not invertible in $R$, then I don't know a construction which avoids elliptic cohomology, although in many cases you get answers which are well behaved in pure algebraic geometry; for example, when $G$ is a symmetric group and the level is zero). I don't think it makes sense for curves of higher genus: in the discussion above, it is important that $\pi_{1} E$ is abelian.

Answer 2

This isn't an answer. André surely knows what I'm about to say already, but here is some context for guessing what this story should look like. The context is 3d Dijkgraaf-Witten theory: this is a family $Z_{G, k}$ of 3d topological field theories depending on a finite group $G$ and a cocycle $k \in Z^3(BG, \mathbb{C}^{\times})$ (note that $H^3(BG, \mathbb{C}^{\times}) \cong H^4(BG, \mathbb{Z})$).

The value $Z_{G, k}(\Sigma)$ of Dijkgraaf-Witten theory on a closed surface $\Sigma$ is a complex vector space, constructed in the following way. If $X$ and $Y$ are two spaces let $[X, Y]$ denote the mapping space. In particular, $[\Sigma, BG]$ is the space of $G$-bundles on $\Sigma$. The pullback of $k$ along the evaluation map

$$\Sigma \times [\Sigma, BG] \to BG$$

produces a cocycle in $Z^3(\Sigma \times [\Sigma, BG], \mathbb{C}^{\times})$. The transgression of this cocycle along the projection $\Sigma \times [\Sigma, BG] \to [\Sigma, BG]$ produces a cocycle in $Z^1([\Sigma, BG], \mathbb{C}^{\times})$, which one should interpret as a (flat) complex line bundle, and $Z_{G, k}(\Sigma)$ is the space of (flat) global sections of this line bundle.

In the setting where $\Sigma$ is replaced with an elliptic curve $E$ (or algebraic curve of higher genus), the transgression operation we're looking for (which I don't know how to define) should interpret cocycles in $Z^3(-, \mathbb{C}^{\times})$ as some kind of 3-line bundles, and "integrate" over $E$ to produce algebraic line bundles over $\text{Loc}_G(E)$.

In particular, if $R \neq \mathbb{C}$ then $k$ might live in something more like $Z^3(BG, R^{\times})$.