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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?

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By "algebraic function" do you mean "regular"? –  Qiaochu Yuan Jan 5 '10 at 0:55
    
I mean one that can be represented as a quotient of polynomials. –  Dyke Acland Jan 7 '10 at 18:19
    
so, yes, I mean regular. –  Dyke Acland Jan 7 '10 at 19:28

2 Answers 2

up vote 5 down vote accepted

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.

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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 '10 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 '10 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 '10 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 '10 at 19:03
    
Maybe not a complex Lie group, but still a complex manifold. –  Dyke Acland Jan 7 '10 at 19:08

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 !

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Ahh, you just barely beat me to it! –  Charles Siegel Jan 5 '10 at 1:01
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Sorry... Our answers are astonishingly close in time and content ! –  Olivier Benoist Jan 5 '10 at 1:07
    
Clearly, we're both completely, 100% correct, then. Except for my typo, corrected by Mariano. –  Charles Siegel Jan 5 '10 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 '10 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 '10 at 12:59

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