Compact Riemann Surfaces are Projective Varieties.

Question

We know that every compact Riemann surface is a complex compact manifold of dimension one. But why is every compact Riemann surface a projective variety?

Answer 1

As a general process this may be done in two steps.

1. Fine a line bundle with "enough sections". This is called a very ample line bundle. The criterion for this can be deduced via the classical Riemann-Roch theorem. The details of how to do this embedding are as given in this answer.

This embeds the Riemann surface as a complex sub-manifold of the projective variety.

2. The next step is to show that this complex sub-manifold of the projective space is actually algebraic. This is a theorem of Chow.

This is the way you generally do it for complex manifolds. For example, for abelian varieties, the appropriate sections can be given by theta functions. Seem Mumford's lectures for details.

Answer 2

Consider reading this senior thesis which is entirely devoted to proving (in two ways) projectivity of compact Riemann surfaces. All the proofs (that I know) are along the following lines: You construct a very ample line bundle $L$ over a Riemann surface $X$. Then you embed $X$ in ${\mathbb C}P^N$ by sending $x\in X$ to $[s_0(x): s_1(x): \ ...\ :s_N(x)]\in {\mathbb C}P^N$ where $s_i$'s are (holomorphic) sections of $L$ which "separate points".

Answer 3

As far as I know, Riemann's original proof (the "Riemann existence theorem") proceeded by constructing a meromorphic function on the compact Riemann surface $X$. (This function is actually constructed by a slightly roundabout process, which is something like first constructing its real part, which will be a harmonic function on $X$. Ultimately this harmonic function is constructed by applying the Dirichlet principle on the simply connected domain obtained by cutting $X$ open into a polygon, with boundary conditions that show that the harmonic function obtained actually gives a well-defined function when we glue the cuts back together to recover $X$.)

Such a meromorphic function realizes $X$ as a branched cover of $\mathbb P^1$, and one then argues that all branched covers of $\mathbb P^1$ are actually algebraic curves.