## Why are the holomorphic line bundle sections finite dimensional?

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.

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I think that the answer to your question lies in the theory of elliptic PDEs. All harmonic functions on a compact Riemannian manifold are constant and a proof of this fact is a matter of calculation. Nevertheless, this fact can be also (at least heuristically) gleaned from the mean value property of harmonic functions. If you want more details I suggest book by Raymond Wells - Differential Analysis on Complex Manifolds. – robot Jan 27 at 18:23
You can easily reduce to the line bundle case: Given a rank $r$ vector bundle on a compact complex manifold, pull it back over the associated bundle of complete flag varieties. The new bundle has a filtration $0\subset V_1\subset V_2\subset \dots \subset V_r=V$ by subbundles such that the quotients $V_i/V_{i-1}$ are line bundles, and the conclusion for these line bundles implies it for the original bundle. – Tom Goodwillie Jan 27 at 23:09
You don't need to know in advance the finite dimensionality of the space of holomorphic sections. I think this follows from a direct algebraic calculation using Peter-Weyl, see section 6.5 in this book: "Harmonic analysis on commutative spaces" by J. Wolf. – Claudio Gorodski Jan 28 at 0:14
(Oops, I just noticed that the question was specifically about line bundles.) – Tom Goodwillie Jan 28 at 13:12

## 4 Answers

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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While the Montel theorem argument is my favorite method of proving the fact that the space $H^0(X,L)$ of sections of a holomorphic line bundle $L$ over a compact complex manifold $X$ is finitely dimensional, I should point out two more proofs of this fact, given in the monograph Holomorphic Morse Inequalities and Bergman Kernels" by Xiaonan Ma and George Marinescu (Birkh\"auser, Basel 2007).

One proof (Theorem 1.4.1) uses Hodge decomoposition theorem plus the fact that certain cohomology groups are finitely generated. (This approach also works for a holomorphic hermitian vector bundle over $X$.) This point of view is related to the above comment by @robot and @Jim Humphreys's answer.

Another proof given in this book considers spaces of $k$-jets of holomorphic sections of $L$ at $x \in X$ and norm estimates for sections (practically "by hand"). The statement the authors prove is the following Lemma 2.2.1: Let $X$ be a compact complex manifold of dimension $n$ and $L$ a holomorphic line bundle on $X$. Then for any points $x_1,...x_m$ in $X$ and $r_1,...,r_m \in \mathbb{R}_+$ such that $L$ is trivial over each polydisc $P(x_i,2r_i)$ and $X \subset \bigcup_{i=1}^m P(x_i,r_ie^{-1})$ there exists an integer $k=k(L)$ such that if $s \in H^0(X,L)$ vanishes at each point $x_i$ up to order $k$, then $s$ vanishes identically. Hence $dim H^0(X,L) \leq m \binom{n+k}{n}$.

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Since the question involves only classical ideas, it's enough to refer to work of Serre and others explained by Hartshorne in section III.5 of his text on algebraic geometry. In quite a bit of generality, one sees (starting with line bundles on the projective line) that the cohomology groups of a coherent sheaf on a projective variety (or scheme) have finite generation properties. This generalizes the earlier analytic work on manifolds, where the language of vector bundles is used.

This viewpoint was exploited in Invent. Math. papers by Demazure where he revisited Bott's theorem using more algebraic techniques. This shows how to approach the ideas of Borel-Weil and Bott algebraically, which in turn allowed Henning Andersen and others to delve more deeply into what does or doesn't work in prime characteristic for flag varieties. By now there's a lot of literature, some built into Jantzen's monograph Representations of Algebraic Groups (second edition, AMS, 2003). At the outset was Kempf's fundamental vanishing theory for dominant line bundles on flag varieties in any characteristic, which avoided Kodaira vanishing but left a lot of open questions about non-dominant line bundles and other sheaves.

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If you want the finite dimension of the space of global holomorphic sections H^{0}(E) of an holomorphic vector bundle E on a compact complex manifold, it is a very classical result. One way to prove it : choose an hermitian metric on E, it gives a natural norm (integration of the local scalar product) on H^{0}(E) which becomes a normed vector space. On the other hand, classical theory of holomorphic functions (Montel theorem) shows that H^{0}(E) is a Montel space (every closed bounded set is compact). In particular, the unit ball is compact and so H^{0}(E) is of finite dimension by a theorem of Riesz.

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if X is a projective manifold and L a holomorphic line bundle, can you use the usual Riemann surface type argument as follows? if t is a non zero section, dividing by t defines a meromorphic function, and taking its pole divisor gives a linear map into the space of divisors dominated by div(t). this space is finite dimensional since by chow's theorem supp(t) is a projective variety, and the map has one diml kernel since by the maximum principle, a holomorphic function on a compact manifold is constant. – roy smith Jan 27 at 17:21
@ unknown & Roy: Thanks for your answers. However, I was hoping that there might exist a more algebraic reason for finite dimensionality. Is it possible to show this without appealing to analysis? – John McCarthy Jan 27 at 20:43
well ask yourself the most basic question, namely why are the global sections of the structure sheaf finite dimensional, i.e. what are the global holomorphic functions? that seems inescapably analytic. So the basic reason seems to me to be that in this setting, analytic objects are essentially algebraic, and only then you can use algebraic arguments. – roy smith Jan 27 at 22:20
I really like this reasoning with Montel theorem :) – Dmitri Feb 3 at 15:54