Let $f:X\dashrightarrow Y$ be a small birational map, where $X,Y$ are normal $\mathbb{Q}$-factorial varieties. Let $\Delta_X\subset X$ be an effective $\mathbb{Q}$-divisor such that the pair $(X,\Delta_X)$ is klt. Let us take the image $\Delta_Y=f(\Delta_X)\subset Y$. Is the pair $(Y,\Delta_Y)$ still klt or not?

I am actually interested in the following case. Let $f:\mathbb{P}^n\dashrightarrow\mathbb{P}^n$ be the standard Cremona, and let $p_1,...,p_{n+3}\in \mathbb{P}^n$ be general points. We take $p_1,...,p_{n+1}$ as the center of the Cremona. Let $X$ be the blow-up of $\mathbb{P}^n$ at $p_1,...,p_{n+3}$. Then $f$ induces a small birational map $\overline{f}:X\dashrightarrow X$. If $(X,\Delta)$ is a klt pair then is $(X,\overline{f}(\Delta))$ a klt pair as well ?


2 Answers 2


Here is a very simple example and you can generalize this to get higher dimensional and more complicated ones:

Let $X=\mathbb P^2$ and $\Delta=\alpha\cdot C$ where $C$ is a smooth proper (and hence irreducible) curve of degree $>1$ and $0<\alpha<1$ a rational number. Clearly $(X,\Delta)$ is klt. Now let $p_1,p_2\in X$ be such that the line connecting $p_1$ and $p_2$ is tangent to $C$. Consider a Cremona transformation where $p_1$ and $p_2$ are blown-up.

The resulting $(Y,\Delta_Y)$ is such that $Y\simeq \mathbb P^2$ and $\Delta_Y=\alpha\cdot C_Y$ where $C_Y\subset \mathbb P^2$ is a curve with a singularity (in fact one with a non-ordinary singularity). This gives a counter-example right away if $\alpha>1/2$. Q.E.D.

Note, that you can easily deal with an arbitrary $\alpha$: If the first pair you get is still klt because $\alpha$ was sufficiently small, then do the same thing again: choose a tangent line through the singular point, two points (not on $\Delta$) on that tangent and do a Cremona with those points. In each step the singularity gets worse and no matter what $\alpha$ is, in finitely many steps you get to a pair that is not klt. (In case you wonder, the goal is to get the multiplicity of the singular point, $m$, to satisfy that $\alpha > 1/m$).


No. The point is that $\Delta_Y$ can be extremely singular along the flopped curve.

Let's use the Cremona on $\mathbb P^3$ for simplicity. Write $X$ and $X^+$ for the blow-ups on the two sides, to help us think straight. Let $W$ resolve the map, which is just a flop of six curves, and let $f : W \to X$ and $g : W \to X^+$ be the maps down. Take $H$ to be a smooth ample divisor, and $\Delta$ to be $H$ with coefficient $1/2$ to make $(X,\Delta)$ klt. You can write $f^\ast H +\sum b_i E_i = g^\ast H^+ $, where $b_i = H \cdot C_i > 0$, with $C_i$ a flopping curve (as explained here). $H$ was smooth and ample, but $H^+$ is neither: it's singular along the six flopped curves, and negative on them.

Certainly $W$ is a log resolution of $(X^+,\Delta^+)$, and to compute the discrepancies we write $K_W + \tilde{\Delta}^+ = g^\ast(K_{X^+} + \Delta^+) + \sum a_i E_i = g^\ast(K_{X^+}) + f^\ast \Delta + \sum (b_i/2) E_i + \sum a_i E_i$. The $b_i$'s can be as big as you want if you pick $H$ to be sufficiently ample. This means the log discrepancies $a_i$ have to be very negative to keep too many $E_i$'s from showing up on the left.

Since $H$ is smooth, $(X,\Delta)$ is klt. But $(X^+,\Delta^+)$ is very far from it if you picked $H$ to be sufficiently positive on some $C_i$. Of course if you arranged that the map is a $(X,\Delta)$ flip, the pair stays klt by the general results. But here the map is $(K_X+\Delta)$-positive.

  • $\begingroup$ On the other hand, it sounds like you're not blowing up very many points, so you're on a Mori Dream Space. So it is true that you can pick an $\epsilon$ such that $(Y,\epsilon \Delta_Y)$ is klt on any small $\mathbb Q$-factorial modification of your $X$, since there are only finitely many to deal with. For the blow-up at any more points this won't be the case. $\endgroup$
    – user47305
    Jul 17, 2014 at 16:50
  • $\begingroup$ Thanks for the answer. Just to better understand: if $\Delta\subset\mathbb{P}^7$ is an hypersurface of degree $7$ with points of multiplicity $6$ at $p_1,...,p_6$ and smooth in $p_7,p_8$, what happens to $f(\Delta)$. I computed that it is an hypersurface of degree $11$ with multiplicities $5$ in $p_7,p_8$ and $10$ in $p_1,...,p_6$. On the other hand $\Delta$ has positive intersection with one of the flipped lines. Can I say something in this case? $\endgroup$
    – Puzzled
    Jul 17, 2014 at 17:12
  • $\begingroup$ Hmm, something like that seems harder to say. For starters, what coefficient do you want on $\Delta$? Do you know the lc threshhold for the original guy? Assuming the divisor is movable, you might be able to compute the discrepancies by hand using the method above: start with a model where you know them, and compute what happens under each flop you need. In dim > 3 this might be annoying since there are various singular strata; I don't know those cases well. Anyway, I don't know a good general answer, but maybe somebody will suggest one. $\endgroup$
    – user47305
    Jul 17, 2014 at 17:27
  • $\begingroup$ In this case we should take $\epsilon\Delta$ with $\epsilon <\frac{6}{5}$ in order to have $(X,\epsilon\Delta)$ klt. $\endgroup$
    – Puzzled
    Jul 17, 2014 at 17:51
  • $\begingroup$ I suppose your $\Delta$ could be (but might not be -- depends whether there exist two other mult 6 singularities) the strict transform of a plane under a Cremona centered at $p_1,\ldots,p_6$ together with two points that aren't $p_7$ and $p_8$. If it is, if you write down a resolution of the standard Cremona on $\mathbb P^7$ (by successively blowing up the linear strata) you can probably explicitly compute all the discrepancies you need, by a (considerably more) painful version of the computation I sketched, applied twice: once to compute them for $\Delta$, and once for $\Delta^\prime$. $\endgroup$
    – user47305
    Jul 17, 2014 at 22:41

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