Cohomology of $\mathcal O(k)$

I am reading a paper in which it is claimed that $H^1(\mathcal O(-k),\mathcal O)=0$, where $k\geqslant 1$. Moreover, the argument also requires that $H^2(\mathcal O(-k),\mathcal O)=0$.

Here $\mathcal O(-k)$ refers to the line bundle over $\mathbb CP^1$ with first Chern class $-k$ and I am assuming that $\mathcal O$ refers to the structure sheaf on the total space of $\mathcal O(-k)$, $\operatorname{Tot}(\mathcal O(-k))$.

Is there an easy way to see this claim?

Edit. I think that $\mathcal O(-k)$ can be covered by two open sets with acyclic intersection, so that $H^2(\mathcal O(-k),\mathcal F)=0$ for any coordinates. My question about $H^1$ remains.

• What is $\textrm{Tot}(\mathcal O(k))$? – Brenin Sep 7 '13 at 20:10
• @atricolf The total space of the bundle $\mathcal O(k)$. – Earthliŋ Sep 7 '13 at 21:18

First I want to show the principle how to compute $H^1(X, \mathscr O)$ with X the total space of $\mathscr O_{\mathbb P^1}(n)$: The principle is to apply the Laurent expansion.

Hence I consider the case $n=0$. Here $X = \mathbb P^1 \times \mathbb C$. Consider the standard covering of $\mathbb P^1$ by $U_0 := \{(z_0:z_1) \in \mathbb P^1: z_0 \neq 0 \}$ and $U_1 := \{(z_0:z_1) \in \mathbb P^1: z_1 \neq 0 \}$. Denote by $t$ the complex coordinate of the fibre $\mathbb C$.

We use $\check C$ech-Cohomology and employ the a-cyclic covering $(U_0 \times \mathbb C, U_1 \times \mathbb C )$. Any cocycle $f_{01}((z_0: z_1),t)$ on $(U_0 \cap U_1) \times \mathbb C$ has the Laurent expansion with respect to $\frac {z_1}{z_0}$

$$f_{01}((z_0: z_1),t) = \sum_{k=-\infty}^{\infty}a_k(t)(\frac {z_1}{z_0})^k$$

It decomposes as coboundary $f_{01} = f_0 - f_1$ with

$$f_0((z_0: z_1),t) := \sum_{k=0}^{\infty}a_k(t)(\frac {z_1}{z_0})^k$$

$$f_1((z_0: z_1),t) := -\sum_{k=-\infty}^{-1}a_k(t)(\frac {z_1}{z_0})^k = -\sum_{k=1}^{\infty}a_k(t)(\frac {z_0}{z_1})^k.$$

Hence $H^1(X, \mathscr O) = 0$.

Secondly, I consider the general case $\mathscr O_{\mathbb P^1}(n), -n \in \mathbb N,$ from your question. The line bundle trivializes on $U_0$ with fibre coordinate $t \in \mathbb C$ and on $U_1$ with fibre coordinate $\tau \in \mathbb C$. Furthermore, it has the transition function $g_{01}(z_0:z_1) = (\frac {z_1}{z_0})^n$. Any cocycle $f_{01}((z_0: z_1),t)$ on $(U_0 \cap U_1) \times \mathbb C$ has the Laurent expansion with respect to $\frac {z_1}{z_0}$

$$f_{01}((z_0: z_1),t) = \sum_{k=-\infty}^{\infty}a_k(t)(\frac {z_1}{z_0})^k = \sum_{k=-\infty}^{\infty}a_k(g_{01}(z_0:z_1)*\tau )(\frac {z_1}{z_0})^k.$$

It decomposes as coboundary $f_{01} = f_0 - f_1$ with

$$f_0((z_0: z_1),t) := \sum_{k=0}^{\infty}a_k(t)(\frac {z_1}{z_0})^k$$

$$f_1((z_0: z_1),\tau) := -\sum_{k=-\infty}^{-1}a_k(g_{01}(z_1:z_0)*\tau )(\frac {z_1}{z_0})^k = -\sum_{k=1}^{\infty}a_k((\frac {z_1}{z_0})^n*\tau )(\frac {z_0}{z_1})^k, n \leq 0.$$

Hence $H^1(X, \mathscr O) = 0, n \leq 0$, q.e.d.