An Introduction to Riemann Surfaces and Algebraic Curves: Complex 1-Tori and Elliptic Curves
IIT Madras, , Prof. T.E. Venkata Balaji
Updated On 02 Feb, 19
IIT Madras, , Prof. T.E. Venkata Balaji
Updated On 02 Feb, 19
4.1 ( 11 )
An Introduction to Riemann Surfaces and Algebraic Curves Complex 1-Tori and Elliptic Curves by Dr. T.E. Venkata Balaji, Department of Mathematics, IIT Madras. For more details on NPTEL visit httpwww.nptel.iitm.ac.insyllabus111106044Goals To see how the topological quotient of the universal covering of a space by the deck transformation group (which is isomorphic to the fundamental group of the space) gives back the space In particular, the topological universal covering of a Riemann surface (which inherits a unique Riemann surface structure as shown in the previous lecture) modulo (or quotiented by or divided by) the fundamental group gives back the Riemann surface To see that nontrivial deck transformations are fixed-point free To see why any Riemann surface with universal covering the Riemann sphere is isomorphic to the Riemann sphere itself To get a characterisation of discrete subgroups of the additive group of complex numbers To use the above characterisation to deduce that a Riemann surface with universal covering the plane has to be isomorphic to either the plane itself, or to a complex cylinder, or to a complex torusKeywords Holomorphic covering, holomorphic universal covering, group action on a topological space, orbit of a group action, equivalence relation defined by a group action, quotient by a group, topological quotient, quotient topology, quotient map, transitive action, deck transformation, open map, Riemann sphere, one-point compactification, stereographic projection, Moebius transformation, unique lifting property, group of translations, admissible neighborhood, module, submodule, subgroup, discrete submodule, discrete subgroup
Sam
Sep 12, 2018
Excellent course helped me understand topic that i couldn't while attendinfg my college.
Dembe
March 29, 2019
Great course. Thank you very much.