Holomorphic function from $\mathbb{C}\mathbb{P}^n$ to $\mathbb{C}$ is constant

Question

Let $f:\mathbb{C}\mathbb{P}^n\longrightarrow\mathbb{C}$ be some complex-differentiable function, where $\mathbb{C}\mathbb{P}^n$ is the complex projective space. I want to show that $f$ is constant. My idea is to use compactness of $\mathbb{C}\mathbb{P}^n$ together with the Maximum Modulus Principle, but I don’t know how to do it exactly. Can someone help me with this proof ? Thanks for your help and stay safe !

Answer 1

Your statement is a particular case of something more general.

If $X,Y$ are two Riemann Surfaces, $X$ compact and $f:X\to Y$ holomorphic non-constant map, then $Y$ is compact and $f$ is surjective.

In fact, by open-mapping theorem, $f(X)$ is open. But $X$ is compact, so $f(X)$ is such, so it's in particular closed.

Thus $f(X)\subseteq Y$ is both open and closed, and since it's not empty, it has to be the whole $Y$. That is $f$ is surjective.

Now $X=\Bbb C\Bbb P^n$ is clearly compact and since $Y=\Bbb C$ it's NOT compact, by the above statement, $f$ must be constant.

EDIT

The MMP is the following:

Let $X$ be a Riemann Surface and $f:X\to\Bbb C$ a non constant holomorphic function. Then $|f|$ does not attain its maximum.

Now if $X=\Bbb C\Bbb P^n$, then $X$ is compact thus by Weierstrass $|f|$ must attain its maximum on $X$. Thus by MMP it must be constant.