# meromorphic function on projective space must be rational

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

###### Theorem.

Let $f$ be a meromorphic function on ${\mathbb{P}}^{n}$. Then $f$ is rational.

###### Proof.

Note that the zero set of $f$ and the pole set are analytic subvarieties of ${\mathbb{P}}^{n}$ and hence algebraic by Chow’s theorem. $f$ induces a meromorphic function $\tilde{f}$ on ${\mathbb{C}}^{n+1}\setminus\{0\}$. Let $p$ and $q$ be two homogeneous polynomials such that $q=0$ are the poles and $p=0$ are the zeros of $\tilde{f}.$ We can assume we can take $p$ and $q$ such that if we multiply $\tilde{f}$ by $q/p$ we have a holomorphic function outside the origin. Hence $(q/p)\tilde{f}$ extends through the origin by Hartogs’ theorem. Further since $\tilde{f}$ was constant on complex lines through the origin, it is not hard to see that $(q/p)\tilde{f}$ is homogeneous 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 $(q/p)\tilde{f}$ must be a constant, and the proof is finished. ∎

Title meromorphic function on projective space must be rational MeromorphicFunctionOnProjectiveSpaceMustBeRational 2013-03-22 17:52:13 2013-03-22 17:52:13 jirka (4157) jirka (4157) 5 jirka (4157) Theorem msc 51N15 msc 32A20 ChowsTheorem