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Cartan's theorem A says that on for a coherent sheaf ${\mathcal{F}}$ on a Stein manifold X, the fibres ${\mathcal{F}}_x$ over each point x in X are generated by global sections.

I'm wondering if there are compact analogues of these theorem. Here I consider holomorphic line bundles over a compact complex manifold X. Consider the fibre stalk ${\mathcal{O}}_{X,x}$ of holomorphic germs over some point x. Is this fibre stalk generated by quotients of global holomorphic line bundle sections? That is, given two global sections $s$ and $t$ of the same line bundle $L\to X$. The quotient $s/t$ is a global meromorphic function. Given a holomoprhic function germ $f_x\in {\mathcal{O}}_{X,x}$, we can always find such $s$ and $t$ such that $(s/t)_x=f_x$?

An analogue for Cartan's theorem B would be nice too. But I can't phrase this precisely.

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You're mixing up stalks and fibers, so what you ask is not really an analogue of Thm. A. It's false for every projective $X$ and non-algebraic germ $f$. By GAGA, $X \simeq Y^{\rm{an}}$ for smooth projective alg. variety $Y$ (disconnected if $X$), and any $L$ satisfies $L \simeq N^{\rm{an}}$ for unique alg. line bundle $N$ on $Y$, with $N(Y) \rightarrow L(X)$ an isom. Hence, ratios $s/t$ come from function field $M(Y)$ of $Y$ inside the meromorphic function field $M(X)$ of $X$, and $O_ {X,x} \cap M(Y) = O_ {Y,x}$ inside $M(X)$ by faithful flatness of $O_ {Y,x} \rightarrow O_ {X,x}$. – BCnrd Sep 4 '10 at 9:33
Another way to see obstruction is that $f_x=(s/t)_x$ forces $f_x$ to be stalk of global mero. function. "Conversely", for any connected complex manifold $X$ (no compactness!) and global meromorphic $f$ on $X$, there's a line bundle $L$ and $s,t \in L(X)$ with $t \ne 0$ so $s/t=f$. This is a good exercise in coherence: prove coherence of the ideal sheaf $J$ of "denominators" of $f$, then invertibility of $J$ (using that each $O_ {X,x}$ is a UFD), and take $L=J^{-1} = \mathcal{H}om(J,O_X)$ with $t:J \rightarrow O_X$ the inclusion and $s:J \rightarrow O_X$ mult. by $f$. – BCnrd Sep 4 '10 at 10:27
BCnrd, are you saying that every algebraic holomorphic germ is generated by quotients of holomorphic line bundle sections? (assuming X compact) – user2529 Sep 4 '10 at 14:30
Dear Colin: Sure, though in the algebraic case it's all much easier, since algebraic germs $f$ are (special) global rational functions and the theory of "divisors" is easier to set up, so one finds $L$ exactly as on an algebraic curve in the Riemann-Roch theorem, namely $L = O_X(D)$ where $D$ is the divisor of poles of that global rational function $f$ (treating the silly case $f = 0$ separately). None of this should be considered to be an analogue of Theorem A. – BCnrd Sep 4 '10 at 15:48
Dear Colin: Just to clarify, "algebraic" germ only makes sense when $X$ has an alg. structure; i.e., it is the analytification of a smooth proper variety (or more generally, alg. space) over $\mathbf{C}$. Then such an alg. structure on $X$ is unique and functorial in such $X$ (by GAGA, which is also valid for proper alg. spaces), and by work of Moishezon and Artin such an alg. structure exists if and only if the field of global mero. functions (which is always finitely generated over $\mathbf{C}$ with transcendence degree at most dim($X$), by a theorem of Siegel) has trdeg equal to dim($X$). – BCnrd Sep 4 '10 at 17:00
up vote 2 down vote accepted

Colin, I think you have answered your own question in your response to Brian Conrad. The fraction field of $\mathcal O_{X,x}$ has infinite transcendence degree over $\mathbb C$, while $\mathcal M(X)$, in Elencwajg's notation, has finite transcendence degree. For any global sections $s,t$ of a holomorphic line bundle, the ratio $s/t$ lies in $\mathcal M(X)$. So most $f$ cannot be written as $s/t$.

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i would love to give a big tick to brian's remarks. – user2529 Sep 5 '10 at 6:45

Dear Colin , for $X$ a holomorphic connected manifold, denote by $\mathcal M (X)$ its field of meromorphic functions.

A) It is not true that a germ of holomorphic function $f_x\in \mathcal O_{X,x}$ is induced by a global meromorphic function : many compact complex manifolds only have $\mathbb C$ as meromorphic functions: $\mathcal M (X)=\mathbb C$. There is an example with $X$ a surface in Shafarevich's Basic Algebraic Geometry, volume 2, page 164.

B) The best analogon to Theorem B is probably Cartan-Serre's result that for any coherent sheaf $\mathcal F$ on the compact manifold $X$, the cohomology vector spaces $H^q(X,\mathcal F), q\geq 1$ are finite-dimensional over $\mathbb C$.

(Original article: Cartan-Serre, C.R.Acad.Sci. Paris 237 (1953), 128-130)

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thank you for giving a negative answer to A. I am still hoping someone could give an example/theorem for a positive answer to A. – user2529 Sep 4 '10 at 7:40

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