TITLE: Why are the holomorphic line bundle sections finite dimensional?
QUESTION [16 upvotes]: I'm trying to understand the Borel--Weil theorem  at the moment (not the whole Bott--Borel--Weil theorem as has been asked elsewhere). However, I am having a little difficulty finding a direct proof, so I started to try and reconstruct it for myself. The question I can't seem to find an answer for is why should the space of holomorphic sections should be a finite dimensional space in the first place? Is there any easy way to see why this should be the case? It seems to me that if one can answer this question, then the theorem follows directly from the classification of the finite dimensional reps of $G$.
As I mention in a comment below I am more (but not exclusively!) interested in algebraic ways of proving finite dimensionality.

REPLY [10 votes]: Here is an algebraic version of Margaret's answer that gives an explicit bound on the dimension of the space of section. This reasoning can be also adapted to the case of complex (non-algebraic) manifolds, but without an explicit bound.
Claim. Suppose $X^k\subset \mathbb CP^n$ is a $k$-dimensional projective variety. Let $D$ be the zero divisor of a section of a line bundle $L$ on $X^k$. Then the dimension  of the space of sections of $L$ is at most the binomial coefficient: 
$$\binom{k+ deg(D)\cdot deg(X)}{deg(D)\cdot deg(X)}$$
The proof of this statement uses just Bezout theorem. Indeed from Bezout it follows that 
at a fixed (say smooth) point $x\in X^k$ a section of $L$ can vanish at most to order $deg(D)\cdot deg(X)$ (to see this cut $X^k$ through $x$ by a generic plane $\mathbb CP^{n-k+1}$ and intersect the obtained curve with $D$). Now, the binomial coefficient is the dimension of the space of all $deg(D)\cdot deg(X)$-jets at $x$.
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