Let $X$ be a complex variety/ manifold with one singular point $x_0\in X$. If we blow up $X$ at $x_0$, we obtain a smoot variety/manifold with exceptional divisor $Y$. How can we calculate the canonical line bundle $\omega_{\tilde X}$ of $\tilde X:=Bl_{x_0}X$?
If $X$ was smooth, then the calculation of $\omega_{\tilde X}$ is done in severalsteps:
- We know that the blow down map $\pi:\tilde X\to X$ when restricted to $\tilde X\setminus Y$ is an isomorphism with image $X\setminus x_0$. Hence $\omega_{\tilde X}=\pi^* \omega_X \otimes \mathcal O_{\tilde X}(Y)^{\otimes a}$ for some $a\in \mathbb Z$.
- Adjunction for $i:Y\hookrightarrow \tilde X$ implies $\omega_Y=i^*\omega_{\tilde X} \otimes N_{Y/\tilde X}$
- Using that $Y=\mathbb P^{n-1}$ and inserting the first equation in the second one we get $$\mathcal O_Y(-n)= i^* \pi^* \omega_X \otimes \mathcal O_Y(Y)^{\otimes a+1} $$
- Since $\pi\circ i$ is constant and the normal bundle $\mathcal O_Y(Y)=\mathcal O_Y(-1)$, this implies $n=a+1$. Hence $$\omega_{\tilde X}=\pi^* \omega_X \otimes \mathcal O_{\tilde X}(Y)^{\otimes n-1} $$
But how can I generalise this argument for the case of $x_0$ being a singular point?
One thing to change is, that the normal bundle might not be $\mathcal O_Y(-1)$ anymore, but this is no problem.
What bothers me more is the ansatz $$\omega_{\tilde X}=\pi^* \omega_X \otimes \mathcal O_{\tilde X}(Y)^{\otimes a}.$$
Can we even write this down? Is the canonical bundle $\omega_X$ well-defined? Can we instead work with $\omega_{X\setminus x_0}$. (But then the pullback by $\pi$ is not a bundle on $\tilde X$). And if there is a line bundle $L$ on $X$ such that $\omega_{\tilde X}=\pi^* L \otimes \mathcal O_{\tilde X}(Y)^{\otimes a}$, how do we need to modify the rest of the argument?
Update: In the comments it was pointed out, that singular varieties still have a canonical sheaf. Does the proof above generalise simply by replacing canonical bundle with canonical sheaf?