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I was reading link text and these two much simpler questions occurred to me: (i) What type of algebraic functions on complex projective varieties do the holomorphic functions correspond to? The rational functions? (ii) What type of algebraic functions on complex algebraic varieties do the holomorphic functions correspond to? The affine coordinate ring?

flag	asked Jan 5 2010 at 0:50 Dyke Acland 462●3●10

	By "algebraic function" do you mean "regular"? – Qiaochu Yuan Jan 5 2010 at 0:55
	I mean one that can be represented as a quotient of polynomials. – Dyke Acland Jan 7 2010 at 18:19
	so, yes, I mean regular. – Dyke Acland Jan 7 2010 at 19:28

Mariano Suárez-Alvarez · Accepted Answer · 2010-01-05 01:02:34Z UTC

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Well, here's a few theorems that might help:

1: On a complex projective variety, a function that is meromorphic on the whole variety is a rational function. You can get this out of the embedding into projective space.

2: On a compact complex manifold, the only globally holomorphic functions are constant. This follows from the maximum principle.

Additionally, if you're looking locally, then on an affine variety, there are a LOT more holomorphic functions than algebraic functions. On $\mathbb{C}$, you have $\mathbb{C}[z]$ for the algebraic functions, and convergent power series for holomorphic, so $e^z$ is holo but not algebraic. But every algebraic function is holomorphic.

edited Jan 5 2010 at 1:02

Mariano Suárez-Alvarez

answered Jan 5 2010 at 1:00

Charles Siegel
10.8k●2●31●81

	But the two sphere $S^2 \subset \mathbb{R}^3$ is a compact complex manifold? Surely the coordinate functions $x,y,z$ (where $x(a,b,c) = a$) are holomorphic? – Dyke Acland Jan 7 2010 at 18:15
	They're real-valued functions on $S^2=\mathbb{CP}^1$, and so can't be holomorphic, they're smooth though. – Charles Siegel Jan 7 2010 at 18:31
	Then take $SU(n,\mathbb{C})$: it's a compact complex manifold, the coordinate functions $t^i_j$ are complex valued, and must be holomorphic? No? – Dyke Acland Jan 7 2010 at 18:58
	It actually is only a real Lie group; the complexification is $SL(2,\mathbb{C})$, which of course isn't compact. – Matt Noonan Jan 7 2010 at 19:03
	Maybe not a complex Lie group, but still a complex manifold. – Dyke Acland Jan 7 2010 at 19:08

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Olivier Benoist · Answer 2 · 2010-01-07 17:56:37Z UTC

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On a complex (connected, reduced...) projective variety, the only holomorphic functions are constants, by the maximum principle. The interesting comparison theorem is the following :(algebraic) rational functions are the same as (analytic) meromorphic functions. It is a consequence of GAGA, but it is a much simpler statement that was known long before Serre.

If the variety is not supposed projective anymore, the holomorphic functions have no reason at all to be algebraic : the exponential function on $\mathbb{C}$ is not !

edited Jan 7 2010 at 17:56

answered Jan 5 2010 at 0:58

Olivier Benoist
2,706●10●18

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Ahh, you just barely beat me to it! – Charles Siegel Jan 5 2010 at 1:01

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Sorry... Our answers are astonishingly close in time and content ! – Olivier Benoist Jan 5 2010 at 1:07

Clearly, we're both completely, 100% correct, then. Except for my typo, corrected by Mariano. – Charles Siegel Jan 5 2010 at 1:08

"If the variety is not supposed projective anymore, the holomorphic functions have no reason at all to be algebraic": By Noether normalization, any irreducible complex variety of positive dimension has non-algebraic meromorphic functions. What can be said about holomorphic functions on affine varieties? Incomplete varieties? – Pete L. Clark Jan 5 2010 at 12:48

I don't think I understand your question. An irreducible complex variety of positive dimension has no non-algebraic meromorphic functions if it is projective ! (Note that by meromorphic, we mean at most poles as singularities). Noether's normalization argument shows indeed that there are non-algebraic holomorphic functions on positive-dimensional affine algebraic varieties. – Olivier Benoist Jan 5 2010 at 12:59

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