Isomorphism between $H^1(\mathbb{P}^n,\mathcal{O}^*)$ and $H^2(\mathbb{P}^n,\mathbb{Z})$ In Griffiths and Harris "Principles of Algebraic Geometry" in page 145 they write $$Pic(\mathbb{P}^n)\simeq H^2(\mathbb{P}^n,\mathbb{Z})=\mathbb{Z}$$ and they justify it by saying $H^1(\mathbb{P}^n,\mathcal{O})=0$. I have two questions: * They are sending to a nonexistent reference such as Chapter 1 section 7. Where is the real one? * Second, for the isomorphism to be true we need also the second cohomology group to vanish, that is $H^2(\mathbb{P}^n,\mathcal{O})=0$ as well. Is that correct? If yes where could I find a reference for that? Thanks is advance.

I don't know if it worth as an answer but...

Both questions are solved with the so called _Bott-formulae_ A particular case can be found at the top of page 108 (section 7 of **Chap 0** ) of Griffiths & Harris. This is as follows:

> **Bott Formulae:** $$ \dim {\rm H}^q(\mathbb{P}^n, \Omega^p_{\mathbb{P}^n}\otimes \mathcal{O}_{\mathbb{P}^n}(k)) = \begin{cases} \binom{k+n-p}{k}\binom{k-1}{p} & {\rm for }\quad q=0,\, 0\leq p \leq n,\, k>p\\\ 1 & {\rm for }\quad k=0,\, 0\leq p =q\leq n\\\ \binom{p-k}{-k}\binom{-k-1}{n-p} & {\rm for }\quad q=n,\, 0\leq p \leq n,\, k
This result can be found in Okonek, Schneide, Splinder - Vector Bundles on Complex Projective Spaces page 8.

The result in Griffiths & Harris is the particular case where $k=0$.