# The problem

Let $$U_j$$ for $$0\leq j\leq n$$ denote the standard coordinate charts of the complex manifold $$\mathbb{P}^n$$. Fix $$d\geq 1$$ and assume we are given holomorphic functions $$f_j:U_j\to \mathbb{C}$$ for each $$j$$ with the additionally property that $$f_i=g_{ij}f_j \enspace \text{ on } \enspace U_i\cap U_j$$ For all $$0\leq i,j\leq n$$, where $$g_{ij}=\left(\frac{z_j}{z_i} \right)^d$$. Show that there exists a homogeneous polynomial $$F$$ of degree $$d$$ such that $$F/z_j^d=f_j \enspace \text{ for all }j.$$

# Discussion

The question is one of many problems I am working through to prepare for a graduate level complex analysis exam (no knowledge of vector bundles or the more advanced machinery of manifolds is assumed). We are given a hint to proceed as follows:

1. Let $$G(z_0,...,z_n)$$ be a (nonzero) homogeneous polynomial of degree $$d$$, and set $$g_j:=G/z_j^d$$ on $$U_j$$ (for all $$j$$). Define a map $$h:\mathbb{P}^n\to \mathbb{C}, \enspace h:=\frac{f_j}{g_j} \text{ on } U_j$$ What can be said about $$h$$?

2. Let $$\pi:\mathbb{C}^{n+1}\setminus \{0\}\to \mathbb{P}^n$$ be the natural quotient map. What can be said about $$G\cdot (h\circ \pi)$$ on $$\mathbb{C}^{n+1}\setminus \{0\}$$? Why does it extend holomorphically to all of $$\mathbb{C}^{n+1}$$?

Attempt: Following step $$1$$, it is easy to argue $$h$$ is well defined on all of $$\mathbb{P}^n$$ (if we view it as a map into $$(\mathbb{C}\cup \infty)$$ )- and it is a meromorphic function on $$\mathbb{P}^n$$. Let us assume that $$h$$ could be extended to a holomorphic function on all of $$\mathbb{P}^n$$ $$(*)$$.

Then one can show using the maximum modulus principle (and the compactness +connectedness of $$\mathbb{P}^n$$) that $$h$$ must be constant. It follows that $$F:=G\cdot (h\circ \pi)$$ is holomorphic on $$\mathbb{C}^{n+1}\setminus \{0\}$$, and by Hartog's Theorem can be extended to a holomorphic map on all of $$\mathbb{C}^{n+1}$$. Since $$h$$ is constant $$F$$ is a homogeneous polynomial of degree $$d$$ (since $$G$$ is). Moreover, we have (on $$U_j$$):

$$F([z_0,...,z_n])=G([z_0,...,z_n])\cdot \left(\frac{f_j([z_0,..,z_n])}{G([z_0,...,z_n])} z_j^d\right)=f_j([z_0,...,z_n])z_j^d$$

as desired.

However, this argument relies heavily on the statement $$(*)$$. Either $$h$$ can be extended (why would this be true?) or one could choose $$G$$ more carefully so that $$h$$ is already holomorphic. Any ideas would be greatly appreciated : )

For completeness sake, I will post my solution to this problem. I was a little bit off originally, we need not extend $h$ to a holomorphic function on $\mathbb{P}^n$. We will need the following lemma:

## Lemma

If $F:\mathbb{C}^N\to \mathbb{C}$ is holomorphic of degree $d$ (i.e. $F(\alpha z)=\alpha^dF(z)$ for all $\alpha\in \mathbb{C}$ and $z\in \mathbb{C}^n$), then $F$ is a homogeneous polynomial of degree $d$.

Proof. Since $F$ is holomorphic it has a power series expansion about $z=0$:

$$F(z)=\sum_{|\alpha|\geq 0}A_\alpha z^{\alpha}$$ where $\alpha=(a_1,...,a_N)$. Set $F_i(z)=\sum_{|\alpha|=i}A_\alpha z^\alpha$. This is a finite sum (for each $i\geq 0$), and so $F_i$ is a homogeneous polynomial of degree $i$. Clearly we have $$F(z)=\sum_{i=0}^\infty F_i(z).$$ For all $\lambda\in \mathbb{C}$ and $z\in \mathbb{C}^N$ we have: $$F(z\lambda)=F(z)\lambda^d=\sum_{i=0}^{\infty}F_i(z)\lambda^i.$$ Treating $\lambda$ as a formal variable, we see that we must have have $F(z)=F_d(z)$ for each $z\in \mathbb{C}^N$. Thus $F$ is a homogeneous polynomial of degree $d$. $\blacksquare$

## Solution

Let $G(z_0,..,z_n)$ be a nonzero homogeneous polynomial of degree $d$ and set $g_j:=G/z_j^d$ on $U_j$. Next define a function $h$ on $\mathbb{P}^n$ by the assignment $$h([z]):=f_j([z])/g_j(z) \text{ if }[z]\in U_j$$ It is an easy exercise using $f_i=g_{ij}f_j$ and the definition of $g_j$ to show that $h$ is well-defined on $\mathbb{P}^n$ (it may however have poles, which come from the zeros of $G$).

Let $\pi:\mathbb{C}^{n+1}\setminus \{0\}\to \mathbb{P}^n$ be the natural quotient map. Then it is easy to see $(h\circ \pi)(\lambda z)=\lambda^0 (h\circ \pi)(z) \enspace (*)$ for all $z\in \mathbb{C}^{n+1}\setminus \{0\}$ and $\lambda \neq 0$. Define $F:=G\cdot (h\circ f)$. Since all the poles of $h\circ \pi$ come from the zeros of $G$, $F$ is holomorphic on $\mathbb{C}^{n+1}\setminus \{0\}$. By Hartog's theorem, it extends to a holomorphic function on $\mathbb{C}^{n+1}$, which necessarily has degree $d$ since $G$ has degree $d$ and $h\circ \pi$ has degree $0$ by $(*)$. Thus by the lemma, $F$ is a homogeneous polynomial of degree $d$. Additionally, by construction we have $$F(z)=f_j([z])z_j^d$$ for all $z\in U_j$ (and all $0\leq j\leq n$), which completes the proof. $\blacksquare$