Suppose $U$ is a smooth, open $n$-dimensional variety over $\mathbb{C}.$ Say $X, X'$ are two proper normal-crossings compactifications of $U$. Call a map $m: X'\to X$ a modification if it is an isomorphism on $U$, i.e. $X'\times_{X}U = U$, and write in this case shorthand $X'\to^UX$. Say a modification $X'\to^UX$ is a toric modification if $U, X, X'$ are all toric varieties with compatible action by the same $n$-dimensional torus.

Definition. We say that a modification $X'\to^U X$ is of toric type if for every $x\in X$, it looks like a toric modification locally. I.e., if for $X'_x$ the fiber of $X'$ over $x$, the map on formal neighborhoods $\hat{X}'_{x}\to \hat{x}$ coincides with an analogous formal neighborhood in for a toric modification such that $U\cap \hat{x}$ is torus-equivariant (it is probably better to say that the complement to $U$ is equivariant in $\hat{x}$).

Definition. Say that two compactifications $X, X'$ of $U$ are toric-equivalent if they are related by a chain of modifications of toric type.

For example, if $X$ is a surface with local coordinatex $x, y$ at a point $x_0$ and $U$ locally looks like the complement to the line $x = 0$ or the cross $xy = 0$, then the blow-up at $x_0$ of $X$ is a toric-type modification. Using just these local models, basic considerations about birational maps of surfaces (e.g. see Tony Pantev's answer to resolution of singularities on surfaces) imply that any two normal-crossings compactifications of a smooth surface are toric-equivalent.

My question: for what higher-dimensional varieties $U$ can we say that any two normal-crossings compactifications $X$ of $U$ are toric equivalent? Is there a nice class of comapctifications $X$ that exist for nice enough $U$ which are guaranteed to be toric-equivalent?

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    $\begingroup$ That will never hold in dimensions $n\geq 3.$ Begin with a smooth point $x\in X\setminus U.$ First perform a blowing up at $x,$ say $\nu:X_1\to X,$ with exceptional divisor $E\cong \mathbb{P}^{n-1}$. Now let $C\subset E$ be a smooth curve of genus $g\geq 1$. Denote the blowing up of $X_1$ along $C$ by $\mu:X'\to X_1.$ Then $X'\to X$ has an irreducible component $F$ over $x$ that is isomorphic to a projective bundle over $C.$ Even if you replace $X'$ by a variety $X''\to X'$, that will still be true of the fiber of $X''\to X$ over $x.$ Thus, it will not be toric. $\endgroup$ Nov 5, 2017 at 17:00
  • $\begingroup$ @JasonStarr I see what you're trying to say and I agree that this point of view should lead to counter-examples. Still, in this specific example $X'$ is still "toric equivalent" to X, by the sequence of maps $\mu, \nu$. Your argument proves is that $X'$ and $X$ don't have a common locally toric refinement. $\endgroup$ Nov 5, 2017 at 17:25
  • $\begingroup$ I misunderstood your definition of "toric-equivalent". I thought that you wanted to dominate both $X$ and $X'$ by a common modification that is "toric type" over each. I see now that is now what you want. $\endgroup$ Nov 5, 2017 at 19:45
  • $\begingroup$ Now that I understand your equivalence relation, it appears to me that this follows from the Weak Factorization Theorem, Theorem 1.1 of the following report by Wlodarczyk, icm2006.org/proceedings/Vol_II/contents/ICM_Vol_2_31.pdf $\endgroup$ Nov 5, 2017 at 20:44

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I am just posting my second comment as an answer. I now understand that the equivalence relation of "toric-equivalence" is the smallest equivalence relation generated by the "toric type" relation, rather than the relation where both $X$ and $X'$ are dominated by modifications that are both "toric type."

With the definition of "toric-equivalence" as the smallest equivalence relation generated by the "toric type" relation, the conclusion appears to follow from the Weak Factorization Theorem. One formulation is Theorem 1.1 of the following report by Wlodarczyk: http://www.icm2006.org/proceedings/Vol_II/contents/ICM_Vol_2_31.pdf

  • $\begingroup$ Nice! Thank you. It looks like just blow-ups at smooth centers are enough, which is surprising to me. Do you know if there is a version of this without assuming projectiveness? $\endgroup$ Nov 6, 2017 at 21:04

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