# Computing the canonical divisor and intersection numbers after contracting an exceptional divisor

I would like to get a better understanding of the following situation.

Consider the affine plane $$\mathbb{A}^2$$ and blow up in a point. Denote the blown up space by $$X_1$$ and the exceptional divisor by $$E_1$$. Blow up again in a point on $$E_1$$, yielding $$X_2$$ and $$E_2$$. The strict transform of $$E_1$$ lives on $$X_2$$. Denote it again by $$E_1$$. Next, we contract ("blow down") $$E_1$$. Denote the new space by $$X_3$$. The point to which $$E_1$$ is contracted will be a singularity of $$X_3$$.

The canonical divisor of $$X_2$$ is given by $$E_1+2\cdot E_2$$. Apparently, the canonical divisor of $$X_3$$ will be given by $$2\cdot E_2$$ but I don't understand why. Furthermore, I would like to know what the self-intersection number of $$E_2$$ (or rather, the image of $$E_2$$ under this contraction map) will be in $$X_3$$. I know how to compute these things in the smooth case.

Finally, is the contraction of $$E_1$$ really the inverse of a blow up? Is there a way to know what the intersection number of the exceptional divisor of this blow up should be? In this case this number should equal -2 (and not -1 which would be true if $$X_3$$ was smooth).

I would be grateful if someone could help me understand this or point me to some place where I can learn this.

In your case notice that $$E_1\subset X_2$$ has self-intersection number $$-2$$ and thus $$X_3$$ has a rational double point, which is Gorenstein. Thus, if $$\pi:X_2\to X_3$$ denotes the blowing down map, $$\pi^*K_{X_3}=K_{X_2}$$ and thus you get $$K_{X_3}$$ to be the image of $$2E_2$$. Self intersection of $$\pi(E_2)$$ is slightly problematic, since it is not a Cartier divisor, only $$\mathbb{Q}$$-Cartier. So, to calculate, $$4\pi(E_2)^2=\pi^*(\pi(2E_2))^2=(2E_2+E_1)^2=-4+4-2=-2$$.
• Thank you for your answer. Some things are not entirely clear to me yet. How do you know what kind of singularity you will get from the self-intersection number of the curve you contract? And how do you see that $\pi^*(\pi(2E_2))=2E_2+E_1$ (why do you need $E_1$ exactly once)? – PP123 Jun 9 at 15:06
• @PP123 If and when you can contract a smooth rational curve of self-intersection $-2$, you get a rational double point. (You could read this and much more in Artin's classic paper on rational singularities). As I said, $\pi(2E_2)=K_{X_3}$ and thus $\pi^*(K_{X_3})=K_{X_2}=2E_2+E_1$. – Mohan Jun 9 at 15:39