2
$\begingroup$

Let $X$ be a smooth projective variety and let $D$ be a big $\mathbb Q$-divisor on $X$. Assume that for $m$ large $|mD|$ has no fixed components. Is there a $\mathbb Q$-divisor $D'\equiv D$ so that $(X,D)$ is klt?

For $D$ ample, this follows from the usual Bertini theorem, as we can take $D'=\frac1m H$ for $H\in |mD|$ generic, for $m$ large. Does it make sense for a similar `Bertini-type result' in the above situation?

$\endgroup$
4
$\begingroup$

If you want an example where $|mD|$ has no fixed codim-1 components, you can do the following. Let $X \dashrightarrow X^+$ be an Atiyah flop of a curve $C$, and let $H \subset X$ be a general member of a very ample linear system with $H \cdot C \geq 2$. The strict transform $\tilde{H} \subset X^+$ is still big. Let $Y$ be the resolution of the flop with exceptional divisor $E$, and $\bar{H}$ the strict transform of $H$ on $Y$.

On $Y$, we have $K_Y = g^* K_{X^+} +E$, and the strict transform is $\bar{H} = f^*H$ (since $H$ is general, it doesn't contain $C$). It's easy to check $f^*H+aE = g^*\tilde{H}$, where $a = H \cdot C$. Then $K_Y + \bar{H} = (g^*K_{X^+} + E) + f^\ast H = (g^* K_{X^+} + E) + (g^*\tilde{H} - aE)$ i.e. $K_Y + \bar{H} = g^*(K_{X^+} + \tilde{H}) + (1-a)E$. Since $a \geq 2$, the pair $(X^+,\tilde{H})$ is not numerically klt (or even numerically lc, if you take $a>2$). The problem is that $|\tilde{H}|$ is too singular along the flopped curve $C^+$; $\tilde{H}$ is already general in its linear series, so we can't get the multiplicity along $C^+$ to be any smaller, and passing to a multiple doesn't help. (Hopefully I got all my signs right!)

$\endgroup$
1
  • $\begingroup$ On the other hand, if you mean that given any $V \subset X$ of arbitrary codimension, there is some $m = m(V)$ so that $|mD|$ does not have $V$ in its base locus, then $D$ has to be nef and everything is OK. $\endgroup$ – user47305 Mar 3 '15 at 1:54
4
$\begingroup$

If $X$ is $\mathbb P ^2$ blown up at a point $O$ with exceptional divisor $E$ and $D=H+E$ where $H$ is the pull-back of the hyperplane class, then any effective divisor $D'\equiv D$ will vanish along $E$ with multiplicity $\geq 1$ (i.e. $D'\geq E$) and hence $\mathcal J(D')\subset \mathcal J(E)=\mathcal O _X(-E)$. In other words, the non-klt locus contains $E$. On the other hand, if $D$ is nef and big, then using Wilson's theorem you can show that $\mathcal J (||D||)=\mathcal O _X$ which implies that $(X,D')$ is klt for some effective $\mathbb Q$-divisor $D'\sim _{\mathbb Q}D$ (this is surely proved somewhere in Lazarsfeld positivity books).

$\endgroup$
3
  • 1
    $\begingroup$ The Wilson's theorem bit is also essentially Prop. 2.61(3) in Koll\'ar-Mori (and indeed surely in Lazarsfeld, which I don't have handy). $\endgroup$ – user47305 Mar 3 '15 at 1:10
  • 1
    $\begingroup$ Wilson's theorem is Theorem 2.3.9 in Lazarsfeld's book "Positivity in algebraic geometry I". The corollary Hacon is talking about is Proposition 11.2.18 in "Positivity in algebraic geometry II". $\endgroup$ – diverietti Mar 3 '15 at 9:53
  • $\begingroup$ @ diverietti: Thanks for finding the ref. @ Mark: Thanks for giving the correct answer (I had not noticed the "no fixed components" hypothesis). $\endgroup$ – Hacon Mar 3 '15 at 16:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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