Can generalization of Mumford’s theorem imply Mumford’s theorem for surface?

Question

Mumford’s theorem for surface says that for a surface $S$ with $p_g(S)\neq0$ ,$\text{CH}_0(S)$ is not representable(or infinite-dimensional). But in Voisin's LECTURES ON THE HODGE ANDGROTHENDIECK–HODGE CONJECTURES she says that the next theorem is the natural generalization of Mumford’s theorem: Let $X$ be a smooth projective variety with $\text{CH}_0(X)=\mathbb{Z} $, Then $H^{k,0}(X)=0$ for all $k> 0 $. But how can this theorem imply Munford's theorem? If not why she says it is the natural generalization.

Answer 1

I just looked at her notes. First she states the result you quote, that if $p_g(S) \neq 0$ then $\mathrm{CH}_0(S)$ is infinite dimensional. But after this she proves that if $\mathrm{CH}_0(S) \cong \mathbb Z$ then not only will $p_g = 0$ but also the irregularity vanishes. In other words $H^{2,0} = H^{1,0} = 0$. It is this statement that is then generalized to higher dimensions.