0
$\begingroup$

I feel the following fact has been used in many argument in algebraic geometry, but I was not be able to prove it or find the precise reference:

Let $X$ be a $\mathbb{Q}$-factorial variety with log canonical singularities, and $Z \subseteq X$ be a subvariety. Then the set $$\{E \mid E {\rm{~is~ an~ exceptional~ divisor~ of~ some~ resolution~ }}Y \to X, {\rm ~such ~ that~} f(E)=Z {\rm~and~} discrepancy(K_X, E) \leq 1 \}$$ is finite.

Any suggestion for the proof or references is welcome!!

Improve this question
$\endgroup$
7
  • 2
    $\begingroup$ Since these $E$ aren't in $X$, but in various possibilities for $Y$, I'd say there's a proper class of such $E$. Are you really looking for something like the set of valuations on the structure sheaf of $X$, centered at $Z$? $\endgroup$
    – Allen Knutson
    Commented Mar 31, 2015 at 2:06
  • $\begingroup$ But those aren't finite either, as iterated blowing up of a point in $\mathbb P^2$ shows. $\endgroup$
    – Will Sawin
    Commented Mar 31, 2015 at 3:24
  • $\begingroup$ Will Sawin is certainly right, there are infinitely many such $E$. But if you put stronger conditions on the singularities of $X$ (like log terminal and $\mathbb{Q}$-Gorenstein) and stronger conditions on the $E$ (like log discrepancy $\leq 1$) then the set is indeed finite... Or do you mean to consider $E$ that appear on every resolution (not just some resolution?) $\endgroup$
    – Karl Schwede
    Commented Mar 31, 2015 at 4:02
  • $\begingroup$ Dear@KarlSchwede, I had edited the condition: I do mean a variety $X$ with reasonable singularities, and the exceptional divisor with discrepancy $\leq 1$. Why in this case such $E$ is finite? $\endgroup$
    – Li Yutong
    Commented Mar 31, 2015 at 13:58
  • $\begingroup$ You really want log discrepancy $\leq 1$ (log discrepancy is discrepancy + 1). There are only finitely many because all such divisors must be extracted on any resolution of $X$. $\endgroup$
    – user47305
    Commented Mar 31, 2015 at 14:25

1 Answer 1

Reset to default
3
$\begingroup$

Ok, let me show the the following which is well known to experts (although I'm not sure where the right reference is):

Proposition: If $(X, \Delta)$ is KLT (or PLT), then there are at most finitely many divisorial valuations corresponding to divisors $E$ on birational models of $X$ such that the discrepancy of $K_X + \Delta$ along $E$ is $\leq 0$.

(Recall that if $X' \to X$ is birational then the coefficient of $K_{X'} - \pi^*(K_X + \Delta)$ along a prime divisor $E$ is called the discrepancy of $K_X + \Delta$ along $E$).

Proof: First let $\pi : Y \to X$ be a log resolution of $(X, \Delta)$. Consider then the simple normal crossings pair $(Y, -K_Y + \pi^*(K_X + \Delta))$. This pair is probably non-effective, but it is SNC.

Let $Y' \to Y$ be the composition of a sequence of blowups along the strata of $(Y, -K_Y + \pi^*(K_X + \Delta))$ such that in $(Y', -K_{Y'} + \pi'^*(K_X + \Delta) )$ every divisor on $Y'$ with discrepancy $\leq 0$ is disjoint from every other such divisor.

(This is not so hard to see how to do, note if I have two divisors with discrepancy $-1 < a \leq 0$ that intersect, then the blowup of their intersection separates the two divisors and creates a new divisor with discrepancy strictly bigger than that of the two original divisors. Keep doing this until separation is achieved).

At this point, we have reduced to the case of a pair $(Y', \Delta' = -K_{Y'} + \pi'^*(K_X + \Delta) )$ where $Y'$ is smooth, $\Delta'$ is SNC (not necessarily effective) and $\Delta'$ has finitely many prime divisors with coefficient $\geq 0$, all of which are disjoint from each other. In fact, it is easy to see that any exceptional divisor over $(Y', \Delta')$ has discrepancy $> 0$. Hence we are done. $\blacksquare$

Remark: This is false for log canonical singularities as mentioned in the comments (ie, consider $(\mathbb{A}^2, \text{Div}(xy))$).

Improve this answer
$\endgroup$
1
  • $\begingroup$ Dear Karl, I understand it now. Thank you very very much for your help!!! $\endgroup$
    – Li Yutong
    Commented Mar 31, 2015 at 18:08

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .