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Suppose $X$ is a variety with mild singularities (say terminal singularities), and $W$ is a center of some exceptional divisor over $X$ (i.e. there is a variety $Y$ with birational morphism $f: Y \to X$, and a divisor $E \subseteq Y$, such that $f(E) = W$). Suppose $W$ is normal (in my mind, $W$ is a minimal log canonical center).

Question: Is there a resolution $g: \tilde X \to X$ which extracts the divisor $E$ and a subset $\tilde W \subset E$, such that the natural morphism $g|_{\tilde W}: \tilde W \to W$ is birational?

If the answer is no in general, is there any special case this holds? (I cannot assume that $X$ is smooth where the claim is certainly true.)

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  • $\begingroup$ I suppose if the base field is not algebraically closed, and C is a curve with no rational points, then taking a cone over C in a projective embedding will violate this property. Indeed, the resolution is given by the single blow of the vertex of the cone, which is the total space of the O(-1) line bundle on C, and the exceptional divisor is the zero section, which has no rational points. Passing to families will probably give an example over an algebraically closed field too. $\endgroup$
    – Evgeny Shinder
    Commented Oct 23, 2018 at 14:25

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The following argument should also work when $X$ has klt singularities.

By the result of Birkar-Cascini-Hacon-M$^{\rm c}$Kernan, we can find a birational model $f:X'\rightarrow X$ extracting only one divisor $E$ with $f(E)=W$. Since $-E$ is relative ample over $X$, there is an exact sequence, $$f_*\mathcal{O}_{X'}=\mathcal{O}_X\rightarrow f_*\mathcal{O}_{E}\rightarrow R^1f_*\mathcal{O}_{X'}(-E)=0.$$ Since there is a factorization $$\mathcal{O}_X\rightarrow \mathcal{O}_W\rightarrow f_*\mathcal{O}_E,$$ the natural map $E\rightarrow W$ has connected fibers. On the other hand, result of Hacon-M$^{\rm c}$Kernan asserts that $E\rightarrow W$ must have rationally connected fibers.

The only situation that I know where one can find a birational section is when $\dim W=1$: This follows from the result of Graber-Harris-Mazur-Starr. To get a resolution of $X$, one can simply replace $X'$ by a higher model. (The argument here is wrong since we only have RCC of fibers from H-M.)

However, if $\dim W\geq2$, it seems to me the answer is related to weak approximation problem as handled in the last paper. As I remember, this is a nontrivial problem.

When $W$ is also rationally connected, to have a rational section is still not an easy question. Artin-Mumford's conic bundle over $\mathbb{P}^2$ is a unirational but non-stably-rational variety.

Edit: As remark by Kostya, I made a mistake recalling the result of H-M, where it only guarantees RCC of fibers. From Mori's list of extremal contraction on smooth threefolds, one can contract a singular irreducible reduced quadric Q in P^3 to a terminal point. The surface Q is RCC, but not RC.

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  • $\begingroup$ Thank you for your answer! Why "1. when $W$ is also rationally connected" implies the existence of a section following from GHMZ's paper? Which result did you refer to? $\endgroup$
    – Li Yutong
    Commented Sep 27, 2017 at 8:52
  • $\begingroup$ Sorry, you are right. Artin-Mumford's conic bundle over $\mathbb{P}^2$ is a unirational but non-stably-rational variety. A related survey is given by Piruka $\endgroup$
    – Ray Lai
    Commented Nov 24, 2017 at 8:49
  • $\begingroup$ Dear Ray Lai, as far as I can see Hacon and McKernan assert rational chain connectedness of E. Is E in fact rational connected? $\endgroup$
    – kostya
    Commented Dec 28, 2019 at 17:33
  • $\begingroup$ Kostya: Yes, you are right. I apologize for making another mistake. From Mori's list of extremal contraction on smooth threefolds, one can contract a singular irreducible reduced quadric Q in P^3 to a terminal point. The surface Q is RCC, but not RC. $\endgroup$
    – Ray Lai
    Commented Sep 14, 2020 at 14:16

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