Weierstrass points on rigid-analytic surfaces - MathOverflow

Weierstrass points on rigid-analytic surfaces

2009-12-10T20:57:09Z

Does a rigid-analytic surface defined over a nonarchimedean complete field have Weierstrass points (if its genus is big enough let's say)? Is there a good reference that (ideally) lists theorems for rigid-analytic spaces that are the analog of commonly known theorems about complex analytic spaces?

Answer by Felipe Voloch for Weierstrass points on rigid-analytic surfaces

2009-12-10T21:52:27Z

Do you mean a curve, instead of a surface? The notion of Weierstrass point is usually defined for algebraic curves (= Riemann surfaces, if the ground field is the complex numbers) and is a purely algebraic notion. For non-algebraically closed fields, there can be curves that have no Weierstrass point defined over the ground field.

Answer by David Speyer for Weierstrass points on rigid-analytic surfaces

2009-12-10T21:58:39Z

Quick note: I am going to assume you want to talk about complete curves. One can, of course, have a curve with punctures in algebraic geometry, and I'm not sure how you'd want to define a Weierstrass point on it. In rigid geometry, you have even more freedom: you can have the analogue of a Riemann surface with holes of positive area, and I think (not sure) you can also build the analogue of a Riemann surface of infinite genus. I'm going to assume you are not thinking about these issues.

What you want is the rigid GAGA theorem. I'm not sure what the best reference is; I refreshed my memory from Coleman's lectures, numbers 23-25. Rigid GAGA says:

Let $\mathcal{X}$ be a projective rigid analytic variety. Then

(1) $\mathcal{X}$ is the analytification of an algebraic variety $X$.

(2) The analytificiation functor from coherent sheaves on $X$ to coherent sheaves on $\mathcal{X}$ is an equivalence of categories.

(3) The cohomology of a coherent sheaf is naturally isomorphic to that of its analytification.

Thus, if we define Weierstrass points by the condition that the dimension of $H^0(\mathcal{O}(kp), X)$ is higher than expected, we will get the same points whether we work algebraically or analytically. It shouldn't be too hard to show that your favorite definition is equivalent to this.

Of course, all I've done is tell you how to translate between analysis and algebra. The algebra itself may be very difficult, as Felipe Voloch points out.

Answer by Guillermo Mantilla for Weierstrass points on rigid-analytic surfaces

2009-12-11T04:40:20Z

About references I think Brian Conrad's lecture notes (I'd check mainly his references) from his course in AWS07 might be a good place to look at. http://swc.math.arizona.edu/aws/07/ConradNotes11Mar.pdf

From Conrad's notes I remember that two references were helpful,
http://wwwmath.uni-muenster.de/sfb/about/publ/heft378.pdf and "Non-Archimedean analysis" by S. Bosch, U. Güntzer, R. Remmert