Analytification of a smooth projective variety is a compact Kähler manifold.

Question

I am reading “Fourier-Mukai transforms in algebraic geometry” by Daniel Huybrecht. On page 130 it is written that by Hodge theory there is a natural direct sum decomposition $$H^n(X,\mathbb{C})=\bigoplus_{p+q=n}H^{p,q}$$ with $\overline{H^{p,q}}=H^{q,p}$ and moreover $H^{p,q}\cong H^q(X, \Omega^p)$ where here $X$ is the analytification of a smooth projective variety over $\mathbb{C}$. I am agree that by Hodge decomposition, we have this for a compact Kähler manifold. So, if the analytification of a smooth projective variety over $\mathbb{C}$ is a compact Kähler manifold, we should have this decomposition. My question is that is this true and if not why do we have this decomposition?

Answer 1

Yes, this boils down to two facts, which you should be able to find in e.g. Huybrecht's Complex Geometry or Voisin's Hodge Theory and Complex Algebraic Geometry: I.

1. $\mathbb{P}^n(\mathbb{C})$ is a Kähler manifold. This is true, since we can endow it with the Fubini-Study metric.

2. If $M$ is a Kähler manifold and $N\subset M$ is a complex submanifold, then $(N,\omega|_N)$ is Kähler, where $\omega$ is the Kähler form of $M$.

Now every smooth projective variety $X$ over $\mathbb{C}$ admits a closed embedding $X\hookrightarrow \mathbb{P}^n$ (as schemes over $\mathbb{C}$). Applying the analytification functor realizes $X(\mathbb{C})$ as a complex submanifold of $\mathbb{P}^n(\mathbb{C})$ and combining the two above facts you are done.