3
$\begingroup$

Here by algebraic stack I mean an algebraic stack over the etale site $\textbf{Sch}/\mathbb{C}$.

So I've read from various nonrigorous sources that the upper half plane $\mathcal{H}$ is a fine moduli space for framed elliptic curves. If the data of a framing can be expressed in algebraic geometry (maybe as some choice of basis for the first etale cohomology), then it ought to be able to be seen as an algebraic stack. However, in this case, it should be covered by schemes $U_i$ etale over $\mathcal{H}$, and I don't really see what these schemes would be.

Improve this question
$\endgroup$
4
  • 2
    $\begingroup$ It is a fine moduli space in the category of complex-analytic spaces; there is no need to bring in schemes or stacks. $\endgroup$
    – user27920
    Commented Oct 18, 2014 at 2:21
  • 2
    $\begingroup$ ... but it is not an algebraic stack. The framing uses the standard complex topology, it cannot be expressed in algebraic terms. $\endgroup$
    – abx
    Commented Oct 18, 2014 at 5:48
  • 2
    $\begingroup$ By the way, a semi-rigorous source is Proposition 2.2 in Deligne's Formes modulaires et représentations $\ell$-adiques (Séminaire Bourbaki, 21 1968/69, no. 355). If you can find an unburned copy of Conrad's book on the Ramanujan conjecture, that has a rigorous development (which disagrees with Deligne by a sign). $\endgroup$
    – S. Carnahan ♦
    Commented Oct 18, 2014 at 7:58
  • $\begingroup$ @oxeimon Perhaps you might consider $[\mathbb{A}^1_{\mathbb{C}}/\langle\sigma\rangle]$ where $\sigma$ denotes complex conjugation. Would this be an algebraic model of the upper half plane? $\endgroup$
    – Leo Alonso
    Commented Apr 28, 2017 at 17:43

1 Answer 1

Reset to default
4
$\begingroup$

The upper half plane is not an algebraic stack over ${\mathbb C}$.

Suppose it were, then it would be in fact an algebraic space, since its points do not have any isotropy groups. But it is also smooth and one dimensional. It is easy to show that algebraic spaces over ${\mathbb C}$ which are smooth, separated, one dimensional are in fact schemes.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .