- #1
- 3,309
- 694
What is a proof that every Riemann surface has a non-constant meromorphic function? Is it even true?
I was wondering this because if it is true then for compact Riemann surfaces without boundary
one can use the meromorphic function to produce a meromorphic 1 form whose degree - the sum of the orders of its zeros and poles - is twice its genus minus 2.
The cool thing is that a meromorphic 1 form can be combined with the metric scaling factor in isothermal coordinates to show that the total Gauss curvature is the sum of the indices of the vector field associated to the meromorphic function. this gives the Gauss Bonnet theorem without proving anything general about the sum of the indices of a vector field with isolated zeros
and without resorting to parallel translation arguments on geodesic triangles.
I was wondering this because if it is true then for compact Riemann surfaces without boundary
one can use the meromorphic function to produce a meromorphic 1 form whose degree - the sum of the orders of its zeros and poles - is twice its genus minus 2.
The cool thing is that a meromorphic 1 form can be combined with the metric scaling factor in isothermal coordinates to show that the total Gauss curvature is the sum of the indices of the vector field associated to the meromorphic function. this gives the Gauss Bonnet theorem without proving anything general about the sum of the indices of a vector field with isolated zeros
and without resorting to parallel translation arguments on geodesic triangles.