Weierstrass points on rigid-analytic surfaces - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T16:34:45Z http://mathoverflow.net/feeds/question/8483 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/8483/weierstrass-points-on-rigid-analytic-surfaces Weierstrass points on rigid-analytic surfaces Christelle Vincent 2009-12-10T20:57:09Z 2009-12-11T04:40:20Z <p>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? </p> http://mathoverflow.net/questions/8483/weierstrass-points-on-rigid-analytic-surfaces/8490#8490 Answer by Felipe Voloch for Weierstrass points on rigid-analytic surfaces Felipe Voloch 2009-12-10T21:52:27Z 2009-12-10T21:52:27Z <p>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.</p> http://mathoverflow.net/questions/8483/weierstrass-points-on-rigid-analytic-surfaces/8491#8491 Answer by David Speyer for Weierstrass points on rigid-analytic surfaces David Speyer 2009-12-10T21:58:39Z 2009-12-10T21:58:39Z <p>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.</p> <p>What you want is the rigid GAGA theorem. I'm not sure what the best reference is; I refreshed my memory from <a href="http://math.berkeley.edu/~coleman/Courses/Sp08/" rel="nofollow">Coleman's lectures</a>, numbers 23-25. Rigid GAGA says: </p> <p>Let $\mathcal{X}$ be a projective rigid analytic variety. Then </p> <p>(1) $\mathcal{X}$ is the analytification of an algebraic variety $X$.</p> <p>(2) The analytificiation functor from coherent sheaves on $X$ to coherent sheaves on $\mathcal{X}$ is an equivalence of categories.</p> <p>(3) The cohomology of a coherent sheaf is naturally isomorphic to that of its analytification.</p> <p>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.</p> <p>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.</p> http://mathoverflow.net/questions/8483/weierstrass-points-on-rigid-analytic-surfaces/8529#8529 Answer by Guillermo Mantilla for Weierstrass points on rigid-analytic surfaces Guillermo Mantilla 2009-12-11T04:40:20Z 2009-12-11T04:40:20Z <p>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. <a href="http://swc.math.arizona.edu/aws/07/ConradNotes11Mar.pdf" rel="nofollow">http://swc.math.arizona.edu/aws/07/ConradNotes11Mar.pdf</a></p> <p>From Conrad's notes I remember that two references were helpful,<br /> <a href="http://wwwmath.uni-muenster.de/sfb/about/publ/heft378.pdf" rel="nofollow">http://wwwmath.uni-muenster.de/sfb/about/publ/heft378.pdf</a> and "Non-Archimedean analysis" by S. Bosch, U. Güntzer, R. Remmert </p>