Homogeneous polynomials of degree $k$ provide holomorphic sections of $\mathcal{O}(k)$

Question

Trying to understand a description of how homogeneous polynomials of degree $k$ gives us a holomorphic section of $\mathcal{O}(k)$ for any complex manifold from Huybrechts's "Complex Geometry".

I can't understand how to get a linear map $(\mathbb{C}^{n+1})^{\otimes k}\rightarrow\mathbb{C}$ from a homogeneous polynomial $s$ of degree $k$. For $k=1$ simply the evaluation map works. But it shouldn't work for $k>1$.

Answer 1

The keyword is polarization.

Recall that a linear map $(\mathbb{C}^n)^{\otimes k} \rightarrow \mathbb{C}$ is defined by a multilinear form $T(v_1, \dots, v_k)$ and setting all equal entries gives us a homogeneous polynomial of degree $k$ in $n$ variables: $p(u) = T(u,\dots, u)$.

Conversely, if we have a degree $k$ homogeneous polynomial $p(x)$ in $n$ variables we make a multilinear form setting $$ T(v_1, \dots, v_k) = \left.\frac{1}{k!}\frac{\partial^k}{\partial t_1 \dots \partial t_k}p(t_1v_1 + \dots + t_kv_k)\right|_{t=0} $$

Note that if the degree is two this is the relation between quadratic and bilinear forms. One can readly check that these processes are inverses of each other if $T$ is symmetric.

Answer 2

Indeed, we can get a map: $\mathrm{Sym}^k(\mathbb{C^{n+1}})\rightarrow\mathbb{C}$.

Actually, we can identify $(\mathrm{Sym}^d(\mathbb{C^{n+1}}))^*\cong \mathrm{Sym}^k(\mathbb{C^{n+1}})$ with $\mathbb{C}[z_0, z_1,...,z_n]_k$.