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Grojnowski constructs a $S^1$-equivariant cohomology theory over a complex elliptic curve $E$, designed to trivially satisfy: $$E^*_{S^1}(pt) = E$$

The functor $E^*_{S^1}(-)$ takes in a space $X$ with an $S^1$ action, and is stalkwise defined over each point $e \simeq (e_1, e_2)$ in the elliptic curve $E \simeq S^1 \times S^1$, where $X^e := X^{e_1, e_2} = \{ x \in X \text{ s.t. } xe_1 = xe_2 = x \}$.

$$(E^*_{S^1}(X))_{(e)} := H_{S^1}(X^{e})$$

Grojnowski proves that $E^*_{S^1}(X)$ is a sheaf over $E$. More specifically, he proves that $E^*_{S^1}(X)$ is a $\mathbb{Z}/2$-graded sheaf of super-commutative $\mathcal{O}_E$-algebras.

How can we similarly specify such a sheaf on an elliptic curve over $\mathbb{F}_p$?

I understand that over a curve $E$ in positive characteristic, if the curve is disconnected, we just use empty gluing conditions to construct a sheaf over $E$ that is stalkwise $H^*_{S^1}(X^e)$.

However, with such a naive approach it seems like we can't expect to vary the elliptic curve $E$ over [a substack of $M_{ell}$ which has both height 1 and height 2 curves] and have the corresponding cohomology theories $E^*_{S^1}(-)$ which are defined stalkwise over this substack satisfy a sheaf condition.

Here's my question: When we continuously vary the elliptic curve $E$, how does it affect Grojnowksi's construction?

Technical aside and a minor confusion:

It is not possible for $E^*_{S^1}(pt)$ to give us the elliptic curve $E$ in the usual way (as a ring of global functions) for elliptic curves (as examples of projective varieties) don’t have interesting global functions. Presumably to remedy this, Grojnowski builds the sheaf over $E'$, a mild modification of the elliptic curve $E$.

$$E^*_{S^1}(-): \{S^1\text{-spaces}\} \to \{\text{Sheaves over } E'\}$$

where $E'$ is defined as $E$ tensored with the lattice of cocharacters $\text{Hom}_{\text{Grp}}(S^1, T)$ of the compact torus $T = (S^1)^{\times n}$.

$$E' := E \otimes_{\mathbb{Z}} \text{Hom}_{\text{Grp}}(S^1, T)$$

The lattice of cocharacters of $T$ is identified with the $\mathbb{R}$-linear dual of the Lie algebra of $T$. In our case, $T = S^1$, and $\text{Hom}_{\text{Grp}}(S^1, S^1) = \mathbb{Z}$, so $E' := E \otimes_{\mathbb{Z}} \mathbb{Z} = E$. The construction degenerates quite a bit in the $S^1$ case, so how are we still able to recover $E$ from the ring $E^*_{S^1}$?

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    $\begingroup$ Why do you say "$E$ is disconnected" in positive characteristic? That is false. And at the start you write that an elliptic curve is Spec of a ring, which is also false. $\endgroup$
    – grghxy
    Jun 27, 2015 at 15:13
  • $\begingroup$ Isn't this what the derived algebraic geometry theory of elliptic cohomology is for? I'm not an expert, but I am talking about this sort of thing: math.harvard.edu/~lurie/papers/survey.pdf $\endgroup$
    – Will Sawin
    Jun 27, 2015 at 17:21
  • $\begingroup$ @grghxy You're right. It’s not possible for $\text{Spec } E_{S^1}^*(pt)$ to give us the elliptic curve in the usual way (as a ring of global functions), rather, it's my understanding that we get $\text{Spec } E_{S^1}^*(pt) = E \otimes Hom(S^1, T)$, where we are identifying $Hom(S^1, T)$ with the lattices on the dual of the Lie algebra (of E), $S^1$ is the multiplicative circle group, and $T$ is a compact torus. $\endgroup$
    – Catherine Ray
    Jun 27, 2015 at 17:45
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    $\begingroup$ The case where $G = S^1$ collapses Grojnowski’s construction quite a bit. Here, $G = T, \mathfrak{g} = \mathfrak{t}$, and $Hom(T, S^1) = Hom(S^1, S^1) = \mathbb{Z}$. So $\text{Spec } E_T = \mathbb{Z} \otimes_{\mathbb{Z}} E$, which gives us $\text{Spec } E_T = E$. $\endgroup$
    – Catherine Ray
    Jun 27, 2015 at 17:46
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    $\begingroup$ Does not the Ginzburg-Kapranov-Vasserot version work everywhere? $\endgroup$
    – მამუკა ჯიბლაძე
    Jul 3, 2015 at 5:38

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