meromorphic function on projective space must be rational


To define a rational function on complex projective space n, we just take two homogeneous polynomials of the same degree p and q on n+1, and we note that p/q induces a meromorphic function on n. In fact, every meromorphic function on n is rational.

Theorem.

Let f be a meromorphic function on Pn. Then f is rational.

Proof.

Note that the zero setPlanetmathPlanetmath of f and the pole set are analytic subvarieties of n and hence algebraic by Chow’s theorem. f induces a meromorphic function f~ on n+1{0}. Let p and q be two homogeneous polynomials such that q=0 are the poles and p=0 are the zeros of f~. We can assume we can take p and q such that if we multiply f~ by q/p we have a holomorphic functionMathworldPlanetmath outside the origin. Hence (q/p)f~ extends through the origin by Hartogs’ theorem. Further since f~ was constant on complex lines through the origin, it is not hard to see that (q/p)f~ is homogeneousPlanetmathPlanetmath and hence a homogeneous polynomial, by the same argument as in the proof of Chow’s theorem. Since it is not zero outside the origin, it can’t be zero at the origin, and hence (q/p)f~ must be a constant, and the proof is finished. ∎

Title meromorphic function on projective space must be rational
Canonical name MeromorphicFunctionOnProjectiveSpaceMustBeRational
Date of creation 2013-03-22 17:52:13
Last modified on 2013-03-22 17:52:13
Owner jirka (4157)
Last modified by jirka (4157)
Numerical id 5
Author jirka (4157)
Entry type Theorem
Classification msc 51N15
Classification msc 32A20
Related topic ChowsTheorem