Why are homogeneous polynomials sections of $\mathcal{O}(k)$

Question

Trying to understand the following statements in Huybrecht's complex geometry:

Why does the restriction to $\mathcal{O}(-k)$ provide a holomorphic section of $\mathcal{O}(k)$?

Answer 1

So the first question is what is a holomorphic section of $\mathcal O(k)$? It is a map $\mathbb P^n \longrightarrow \mathcal O(k)$ such that $s(l)$ is in the fiber over $l$ for any $l \in \mathbb P^n$. Going off of what you said in your comment, the fiber over $l$ of $\mathcal O(k)$ is the dual of the fiber of $l$ of $\mathcal O(-k)$. This fiber is just $l^{\otimes k}$. Thus, given a homogeneous polynomial of degree $k$ we need to associate to each $l$ a linear map $l^{\otimes k} \longrightarrow \mathbb C$.

To explain this, let's start with the case $k = 1$. In fact, let's just look at the polynomial $z_i$. This should describe a linear function $l \longrightarrow \mathbb C$. $z_i$ is the $i^{th}$ coordinate function, so the obvious choice here is to let it be projection onto the $i^{th}$ coordinate. Every degree 1 homogeneous polynomial is a linear combination of the $z_i$, so we've defined this for all degree 1 homogeneous polynomials.

Now let's go back to degree $k$. As discussed, given some $f \in \mathbb C[z_0, \dots, z_n]_k$, we need a linear map $l^{\otimes k} \longrightarrow \mathbb C$. By universal property of the tensor product, this is the same as a multilinear map $l^k \longrightarrow \mathbb C$. Well a good way to get such a multilinear map is to take $k$ linear maps $g_1, \dots, g_k: l \longrightarrow \mathbb C$ and define $(x_1, \dots, x_k) \mapsto \prod g_i(x_i)$. Since we already showed how each linear homogeneous polynomials gives a linear map, this allows us to define such a map for any $k$-fold product of the $z_i$. Indeed, each $z_i$ in this product yields a linear map $g_i$ and we multiply them together as just discussed. This is independent of which coordinate each $z_i$ in the product is associated to, as multiplication in $\mathbb C$ is commutative. Now take linear combinations to get the general case. Of course, I haven't checked that these sections are holomorphic, but it's clear for the $z_i$ and the rest come from combinations of these. A computation in local coordinates shouldn't be too hard to show holomorphicity if necessary.