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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 : )

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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. $\blacksquare$

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