# Locally toric resolutions of compactifications

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?

• 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. – Jason Starr Nov 5 '17 at 17:00
• @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. – Dmitry Vaintrob Nov 5 '17 at 17:25
• 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. – Jason Starr Nov 5 '17 at 19:45
• 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 – Jason Starr Nov 5 '17 at 20:44

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."