Cech cohomology with respect to different sheaf.

Question

I am just wondering

Can I compute $H^1(\mathbb P^2,\mathcal O_{\mathbb P^1}(1))$? if yes, how? Note that the sheaf is not on $\mathbb P^2$, but on $\mathbb P^1$

Also is there a sheaf with this notation? $T_{1,2}$

Thanks

Answer 1

1) If $j:\mathbb P^1\hookrightarrow \mathbb P^2$ is a linear embedding one has the following formula, which might answer your question: $$H^1(\mathbb P^2,j_*\mathcal O_{\mathbb P^1}(1))=H^1(\mathbb P^1,\mathcal O_{\mathbb P^1}(1))=0 \quad (\bigstar)$$ Note that $j_*\mathcal O_{\mathbb P^1}(1)$ is a coherent sheaf on $\mathbb P^2$ of which it is perfectly legitimate to compute the cohomology.

2) The relevant context is probably the exact sequence of sheaves on $\mathbb P^2$: $$0\to \mathcal O_{\mathbb P^2}(-1)\to \mathcal O_{\mathbb P^2}\to j_*\mathcal O_{\mathbb P^1}\to 0$$ which after twisting by $\mathcal O_{\mathbb P^2}(1)$ yields $$0\to \mathcal O_{\mathbb P^2}\to \mathcal O_{\mathbb P^2}(1)\to j_*\mathcal O_{\mathbb P^1}(1)\to 0$$

3) In general, if a closed subvariety (or subscheme) $j:Y\hookrightarrow X$ is given by the quasi-coherent ideal sheaf $\mathcal I_Y$, the following exact sequence obtains: $$ 0\to \mathcal I_Y \to \mathcal O_X\to j_*\mathcal O_Y\to 0 $$ However algebraic geometers usually write just $$ 0\to \mathcal I_Y \to \mathcal O_X\to \mathcal O_Y\to 0 $$ omitting the $j_*$. This abuse of language can, unfortunately, be a source of confusion.