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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.

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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, $4\pi(E_2)^2=\pi^*(\pi(2E_2))^2=(2E_2+E_1)^2=-4+4-2=-2$.

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  • $\begingroup$ 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)? $\endgroup$ – PP123 Jun 9 at 15:06
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    $\begingroup$ @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 $\pi^*(K_{X_3})=K_{X_2}=2E_2+E_1$. $\endgroup$ – Mohan Jun 9 at 15:39

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