2
$\begingroup$

I am reading Chapter 3 of the book "Singularities of the minimal model program", and I have some problems on calculating certain quantities.

Let $X$ be a projective variety, $L$ be an ample line bundle on $X$, and $K_X \sim_{\mathbb{Q}} rL$ for some $r\in \mathbb{Q}$. Then one can define the affine cone ($\S$ 3.8 loc. cit)

$$C_a(X,L):={\rm Spec}_k\sum_{m \geq 0}H^0(X, L^m),$$ and its partial resolution

$$p: BC_a(X,L):= Spec_X\sum_{m \geq 0}L^m \to C_a(X,L).$$

Let $E$ be the exceptional divisor of $p$, and $\pi: BC_a(X,L) \to X$ be the $\mathbb A^1$-bundle which identifies $E$ with $X$.

My problems are the following (it appears in Proposition 3.14, loc. cit):

(1) Why $\mathcal{O}_{BC_a(X,L)}(E)|_E = L^{-1}$?

(2) Why $\pi^*(K_X) \sim_{\mathbb Q}-rE$?

The (2) is especially confusing me: suppose $r>0$, then $K_X$ is ample, how come $\pi^*(K_X)$ is not effective (i.e.$\sim_{\mathbb Q}-rE$)?

$\endgroup$

1 Answer 1

1
$\begingroup$

Let's use the following notation $C=C_a(X,L)$ and $B=BC_a(X,L)$.

(1) This is the usual statement about the normal bundle of the exceptional divisor restricted to the exceptional. You can look at the computation at the end of II.8 in [Hartshorne] to see that this restriction is always $\mathscr O(-1)$. In this situation, $L$ is your $\mathscr O(1)$. (I know that the proof I am referring to is for "ordinary" blow-ups, but the computation is the same).

(2) I am guessing that your confusion comes from thinking projectively, but $B$ is not projective! In particular, this fact does not mean that $\pi*K_X$ is not effective. Here is an example: Let $M=-P$ denote the divisor consisting of $-1$-times a single point $P\in \mathbb A^1$. Then $M$ is not only effective, but it is actually ample.

Now in the situation with which you are struggling, $C$ is affine with trivial Picard group and $B$ is a blow-up of that, which means that all non-trivial line bundles on $B$ can be understood by their behavior once restricted to $E$.

More specifically, as explained at the start of the proof of 3.14, the projection $\pi:B\to X$ induces an isomorphism on the Picard groups. You have $\pi^*: \mathrm{Pic} X\to \mathrm{Pic} B$ giving one direction of that isomorphism and the restriction to $E\simeq X$ giving the other: $\mathrm{Pic} B\to \mathrm{Pic} E\simeq \mathrm{Pic} X$. Now, (1) implies that via this $-L$ corresponds to $E$ and hence if $K_X\sim_{\mathbb Q} rL$, then $\pi^*K_X\sim_{\mathbb Q} -rE$

$\endgroup$
2
  • $\begingroup$ Thank you very much! I got it, it boils down to the fact that the ideal sheaf of $E$ is $L+L^2+\ldots = \pi^*L$. Just one more question to make sure I understand it correctly: the $\sum_m H^0(X,L^m), \sum_m L^m$ in the definition should be viewed as the direct sum "$\oplus$" (just as in the definition of blowup), is that right? $\endgroup$
    – Li Yutong
    Dec 11, 2017 at 14:34
  • $\begingroup$ Yes, those are direct sums. $\endgroup$ Dec 12, 2017 at 1:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.