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?



Bertini theorem for big divisors and klt pairs
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!)


