# Divisors and Riemann-Roch Intuition

1. Jun 2, 2014

### bolbteppa

Could anybody explain what divisors and the Riemann-Roch theorem are intuitively, motivating them, without any jargon or vagueries (i.e. using actual math), and preferably offering a nice example necessitating this stuff?

I'm sure there is a nice way to explain it in an absolutely natural way, that explains why it applies to classical algebraic curves, differential forms, homology etc... without ever having to use a definition like abelian group, or vector space, i.e. only using definitions Riemann had handy. Thank you.

2. Jun 9, 2014

### mathwonk

Riemann tried to describe math by the most intrinsic properties. take a polynomial. if you know the roots then you know the polynomial up to a constant multiple since if the roots are a,b,c,...d, then the polynomial is c.(x-a)(x-b)(x-c)...(x-d). for some c.

ion the same way Riemann tried tom describe meromorphic functions by their zeroes and "poles" i.e. points where the value was not zero but infinity. His ideas was that any meromorphic function on a compact surface is described up to a constant multipkle by knowing its zeroes and poles.

"divisor" is a fancy name for the zeroes and poles of a function. i.e. if a meromorphic functions has zeroes of order ri at pi and poles of order si at qi, then the "divisor" of zeroes and poles is
r1p1+...rnpn +s1q1+...+smqm.

the riemann roch theorem tells you the dimension of the space of meromorphic functions with given zeroes and poles in terms of the number of zeroes and poles, i.e. in terms of the degree of the divisor of those zeroes and poles, plus the actual location of those points.

i.e. if we specofy zeroes at p1,...,pn of orders r1,...,rn and polkes at q1,...,qm of orders s1,..,sm,

then the dimension oif meromorphic functions with zeroes at least of those orders at those points and poles at most of those orders atn those points,

is Equal to 1-g + d + i, where g = the topological genus of the surface,d = r1+...+rn -s1-...-sm, and i = the dimension of the vector space of holomorphic differential forms vanishing on the given points pi and with pokles at most on the given qi, of the given orders,

e.g. on an elliptic curve, i.e. a curve of genus 1, no differential forms have zeroes, so if we have a positive divisor of degree d, the space of meromorphic functions with poles at most at the points of our divisor has dimension d. E.g. given 2 points p1+p2, there are two independent meromorphic functions with poles at most at p1 +p2. One such is constant so we have one other defining a map S-->P^1, of degree 2, from our elliptic curve S to the Riemann sphere P^1.

probably the simplest example is the Riemann sphere, where we have all meromorphic functions are rational, and g=0, and there are no holomorphic differentials, so given any set of point p1,...,pn, there are exactly n
=1 independent rational functions with poles at those points, presumably the linear combinations of 1, 1/(x-p1),....,1/(x-pn).

does this help?

3. Jun 9, 2014

### bolbteppa

That's exactly what I was hoping for, thank you very much.

4. Jun 10, 2014

### mathwonk

If you want much more, consult:

http://www.math.uga.edu/%7Eroy/rrt.pdf [Broken]

Last edited by a moderator: May 6, 2017