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Let $W_n$ be a set of a log pair having the following property:

For any $(X, D) \in W_n$

(1)$X$ has dimensional $n$ with tirvial canonical divisor (i.e.$K_X = 0$). Moreover, $X$ is a $\mathbb{Q}$-factorial variety with canonical singularities. (one can also assume $(X, \frac{1}{2} D)$ is klt).

(2) $D$ is an integral Weil divisor which is big.

Is it true that there exists a universal $\delta > 0$ (i.e. only depends on $n$), such that we can decompose $D$ in $\mathbb{Q}$-linearly equivalent as follows: $$D \sim_\mathbb{Q} A + E,$$ where $A$ is a nef and big divisor with coefficients bigger than $ \delta$, and $E$ is an effective divisor?

Any suggestions related to the topic are welcome!!

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2 Answers 2

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I think the answer is (almost, i.e. up to birational modification) yes but non-trivial see Theorem 1.3 of arXiv:1208.4150 "ACC for log canonical thresholds" by Christopher Hacon, James McKernan, Chenyang Xu. If you assume that $D$ is integral and $(X,D/2)$ is klt, then since $K_X+D/2=D/2$ is big, the above theorem says that there exists a uniform positive integer (depending only on the dimension of $X$) such that $|m(K_X+D/2)|$ defines a birational map $\phi :X\to \mathbb P ^N=|m(K_X+D/2)|=|(m/2)D|$. If $\phi$ is a morphism (which can be arranged by replacing $X$ by a higher model), then $(m/2)D=\phi ^* \mathcal O _{\mathbb P ^N}(1)+E$ where $E$ is an effective divisor and of course $\phi ^* \mathcal O _{\mathbb P ^N}(1)$ is nef and big.

I am not sure (and skeptical about) how to do this without replacing $X$ by a higher model.

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  • $\begingroup$ Hello, thank you for your answer! But I don't understand what do you mean in the last part of your argument: if $\phi$ is a morphism, then isn't $(m/2)D = \phi^*\mathcal{O}(1)$? Moreover, could you explain more about what do you mean (as well as how to replace) $X$ be a higher model (it seems that by [BCHM], $(X,D)$ has a log terminal model). $\endgroup$
    – Li Yutong
    Commented Oct 15, 2014 at 0:44
  • $\begingroup$ What I mean is the following: suppose that $|G|$ is non-empty, then it defines a rational map $f:X\to \mathbb P ^N=|G|$. To make this in to a morphism, we take a log resolution (of $X$ and $|G|$) $g:X'\to X$ so that $g^*|G|=|M|+E$ where $|M|$ is base point free (and $E$ is a snc divisor). Then $M=f^*\mathcal O _{\mathbb P ^N}(1)$. I did not mean to consider a log terminal model (this would be "lower"). $\endgroup$
    – Hacon
    Commented Oct 15, 2014 at 18:27
  • $\begingroup$ However if you consider a log canonical model (so that $K_X+D/2$ is ample) and assume that the volume of $D$ is bounded from above, then by some results of Hacon, McKernan and Xu, it is true that a fixed multiple of $K_X+D/2$ is very ample (in particular Cartier). $\endgroup$
    – Hacon
    Commented Oct 15, 2014 at 18:29
  • $\begingroup$ Dear Prof. Hacon, thank you for your answer again, though in my case, $D$ may not be nef, I believe what you wrote is the best one can get, and the decomposition I expected may be false. $\endgroup$
    – Li Yutong
    Commented Oct 15, 2014 at 20:37
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Let's see.. Over C, if X is Q-factorial so mD is cartier for some m. mD is also big cartier, so mD ~ A+E, ample + effective. Then m'mD ~ m'A+m'E ~Q D. Does that work?

edit: nvmind, was thinking of the wrong definition of Q-equivalence :P been awake too long, maybe you can use Fujita Approximation to say something related, possibly combined with some result on boundedness of volumes of your set of (X, D)

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  • $\begingroup$ No, because I need $m$ does not depend on particular choice of $X$. Even in simpler case, where $D$ itself is Cartier, using the proof of Kodaira lemma (just as you did), I do not know if there is a universal $m$ such that $h^0(mD -A)$ is non-vanishing (where $A$ is an ample divisor). $\endgroup$
    – Li Yutong
    Commented Oct 9, 2014 at 15:37
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    $\begingroup$ Does such $m$ even exist on a fixed variety $X$? I don't see that it's even true that there exists a constant $m = m(X)$ such that $h^0(mD) > 0$ if $D$ is big and Cartier on $X$; for $h^0(mD-A)$ it's even worse. I doubt a sequence of such $D$ is going to have the decompositions you want. $\endgroup$
    – user47305
    Commented Oct 10, 2014 at 0:45

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