Euler characteristic and genus

We now want to give the precise definition of genus. We can start with the famous formula of Euler. Given a polyhedron

with V vertices, E edges and F faces

V - E + F = 2

The unstated assumption is that the surface of the polyhedron is homeomorphic to the sphere. However, we can form polyhedra homeomorphic to other surfaces. The following is homeomorphic to a torus, Graph

In such cases, we get

V - E + F = 0

The point is that this expression called the Euler characteristic χ depends only on the surface. If S is compact oriented surface (e.g. a complex curves with points of infinity included) then χ(S) is known to even, and the genus g can be defined by

χ(S) = 2 - 2g

The genus can be understood intuitively as the “number of holes”. For example, a genus two surface or “two holed doughnut” is depicted below. Graph

Consider the hyperelliptic curve X defined by an equation y² = p(x) where p(x) has an odd degree, say n, with distinct roots. We give a formula (due to Riemann and Hurwitz) for the genus g of X. We triangulate C ∪{∞} so that the roots of p(x) and ∞ are included among the vertices. Let’s say that there are V verticies, E edges and F faces. So that V - E + F = 2. The preimage of this triangulation gives a triangulation of X with V ′ vertices etcetera. We have V ′- E′ + F′ = 2 - 2g. The map from X → C ∪{∞} is two to one except at vertices lying over the roots and ∞ where it one to one. Thus V ′ = 2V - n - 1, E′ = 2E, F′ = 2F, so that

g = 1 - 1(V′ - E ′ +F ′) = n-- 1 2 2

Although the definition is topological, it turns out that the genus is strongly connected to the complex function theory of a curve. The relationship is given by the Riemann-Roch theorem. We will explain a special case. First observe that a holomorphic function on a compact connected complex curve X is necessarily constant.

Proof: Since X is compact, the modulus of the function must attain a maximum at some point p. By the maximum modulus principle the function must be constant in a neighbourhood of p. This implies that f = f(p) everywhere since the difference has nonisolated zero.

Therefore nonconstant meromorphic functions on X must have singularities. Which raises the question, what are simplest possible singularities? Choose points p₁,…p_k X and positive integers n₁,…n_k. Then the Riemann-Roch theorem tells us that there is a nonconstant function with poles only at p_i of order at most n_i, provided that n₁ + … + n_k > g. A standard proof uses a refined form of the Euler characteristic.

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