I've been reading Huybrecht's Complex Geometry, and ran across this question in section 2.5:
Let $\hat X\rightarrow X$ be the blow-up of a surface $X$ in a point $x\in X$.Show that the pull-back of sections defines an isomorphism: $$H^0(X,K_X)=H^0(\hat X,K_\hat X).$$
Does anyone have some ideas to solve this exercise?
Thanks a lot.
I write the solution here in case someone feel confused about the problem.
In fact, the problem can be generalized.$X$ can be a complex manifold of arbitrary dimension.For $dim=1$,the situation is trivial,since now,blowing-up along a point(divisor) does not change $X$,i.e.,$\hat X=X$.
For $dim\ge 2$,for one direction:$H^0(X,K_x)\rightarrow H^0(\hat X,K_\hat X)$,we make it possible by pulling up forms.For another direction,we first have restriction $H^0(\hat X,K_\hat X)\rightarrow H^0(\hat X\setminus E,K_{\hat X\setminus E})$.In the same time,$H^0(\hat X\setminus E,K_{\hat X\setminus E})\cong H^0( X\setminus\left\{ x \right\},K_{X\setminus\left\{ x \right\}})$.Using Hartog's extension theorem,since $dim(X\setminus\left\{x\right\})\ge 2$,we have $H^0( X\setminus\left\{ x \right\},K_{X\setminus\left\{ x \right\}})=H^0(X,K_x)$.
We just need to show $H^0(\hat X,K_\hat X)=H^0(\hat X\setminus E,K_{\hat X\setminus E})$,since $$H^0(E,K_E)=H^0(E,\sigma ^* {K_X}\otimes\mathcal O(E)|_E),$$ $\sigma ^* {K_X}|E$ is trivial,$\mathcal O(E)|_E=\mathcal O(-1)$ has no global section.We get $H^0(E,K_E)=0$.
$Q.E.D.$