Basic question: $H^1$ and $H^{0,1}$

Question

Please could you explain why for a smooth projective variety over $\mathbb{C}$ (or - if you prefer the analytic world - compact complex manifold) $T$ we have $H^1(\mathcal{O}_T)\simeq H^{0,1}(T)$ as in the bottom of page 13 in this set of notes:

http://www.cgtp.duke.edu/ITP99/morrison/cortona.pdf

Thank you very much.

Answer 1

This is a special case of Dolbeault's theorem.

(By the way, a compact complex manifold is not the same thing as a smooth projective variety. Not all compact complex manifolds are projective.)