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Apr 19, 2013 at 1:59 vote accept Will Chen
Nov 9, 2012 at 15:20 answer added stankewicz timeline score: 6
Nov 9, 2012 at 9:53 comment added Daniel Loughran I concur with Scott.
Nov 9, 2012 at 8:46 comment added S. Carnahan I don't understand why there is a vote to close. This seems to me to be a very reasonable question for someone starting in this field. Indeed, the step where you identify the upper half-plane uniformization with a map of analytic moduli problems is rarely done rigorously in the literature, if at all.
Nov 9, 2012 at 6:16 comment added Allen Knutson Jinx! $\ $
Nov 9, 2012 at 5:19 comment added Qiaochu Yuan Also, an algebraic curve over $\mathbb{C}$ doesn't have a unique $\mathbb{Q}$-structure even if one exists, e.g. the projective closures of $x^2 + y^2 = -1$ and $x^2 - y^2 = -1$ are the same over $\mathbb{C}$ but the two curves are very different over $\mathbb{R}$, let alone over $\mathbb{Q}$.
Nov 9, 2012 at 5:18 comment added Allen Knutson The plane conics $x^2 + y^2 + z^2 = 0$ and $x^2 + y^2 - z^2 = 0$ are isomorphic over $\mathbb C$, but already very different as curves over $\mathbb R$, much less $\mathbb Q$. Which is to say, their structure as compact Riemann surfaces doesn't tell you their structure over $\mathbb Q$. (This isn't an answer of why a modular interpretation might.)
Nov 9, 2012 at 5:17 comment added Qiaochu Yuan Most compact Riemann surfaces are not defined over $\mathbb{Q}$ (e.g. elliptic curves with irrational $j$-invariant). The modular interpretation tells you what the functor of points looks like, which is extra information that you don't get from the analytic structure, which only tells you what the analytification looks like.
Nov 9, 2012 at 5:03 history asked Will Chen CC BY-SA 3.0