# On factorization theorem of toric birational morphisms

Let $X_{Σ′}\to X_{Σ}$ be a toric birational morphism between smooth and complete toric varieties induced by a regular subdivision $Σ′\leq Σ$, i.e. every cone in $Σ′$ is contained in a cone in $Σ$ and both fans have the same support.

Is it true that there exists a fan $Σ′′\leq Σ′$ such that $Σ″$ is constructed from $Σ$ by making a finite number of barycentric subdivisions?

Note: this follows from the following toric version of Oda's strong factorization conjecture:

Conjecture: Let $X$ and $Y$ be smooth, complete toric varieties which are birationally equivalent. Does there exist a third variety $Z$ and birational morphisms $Z\to Y$ and $Z\to X$, which are compositions of blow-ups along closed irreducible subvarieties which are the closures of torus orbits?

This is just a sketch, you may refer to Miles Reid's "Decomposition of toric morphisms" for details.

The answer is yes for surfaces (see Lemma 10.4.1 in Cox, Little, Schenck, for instance).

I would say that this is also true if $\dim X_{\Sigma'}\geq 3$.

Let us consider a wall $\omega_n\in \Sigma'(n-1)$ generated by some primitive vectors $u_1,\ldots,u_{n-1}$ in the lattice $N$. Since $X_{\Sigma'}$ is smooth all the cones will be simplicial. In particular, $\omega_n$ separates only two (smooth) cones of maximal dimension $$\sigma_n=\operatorname{cone}(u_1,\ldots,u_{n-1},u_n)$$ and $$\sigma_{n+1}=\operatorname{cone}(u_1,\ldots,u_{n-1},u_{n+1}).$$

Thus, there are integers $b_1,\ldots,b_{n+1}\in \mathbb{Z}$ and a "wall relation"

$$b_nu_n + \sum_{i=1}^{n-1}b_i u_i + b_{n+1}u_{n+1}=0$$

(Here we have that $b_n$ and $b_{n+1}$ are positive, by convexity)

Let $\alpha = \#\{ i\;|\; b_i<0\}$ and $\beta = n - \#\{ i\;|\;b_i>0 \}$.

By the Contraction Theorem in Reid's article, there exists a contraction $\varphi_R:X_{\Sigma'}\to X_R$ in the sence of Mori theory contracting all curves whose numerical class belong to the ray $R=\mathbb{R}_{\geq 0}[V(\omega_n)]$, such that $\dim \operatorname{Exc}(\varphi_R)=n-\alpha$ and that $\dim\varphi_R(\operatorname{Exc}(\varphi_R))=\beta$.

If we look at the special case where $u_2,\ldots,u_n,u_{n+1}$ are the elements of the standard basis of $\mathbb{Z}^n$ (say $u_i=e_{i-1}$) and the morphism $X_{\Sigma'}\to X_\Sigma$ is given by the contraction that sends the divisor $V(u_1)$ to a point, then we will have that

$$b_1u_1 = -\sum_{i=2}^{n+1}b_iu_i.$$

In coordinates, $b_1u_1=(-b_2,\ldots,-b_{n+1})$ and thus we can suppose $b_1=-1$, dividing the wall relation by $-b_1>0$, if necessary.

Now, it should be noticed (by computing determinants, for example) that both cones $\sigma$ and $\sigma'$ are smooth if and only if $b_n=b_{n+1}=1$.

Now, the exceptional divisor $V(u_1)$ has exactly $n$ invariant points that corresponds precisely to the cones of maximal dimension

$$\sigma_i = \operatorname{cone}(u_1,u_2,\ldots, \widehat{u_i},\ldots,u_{n+1}),$$

for $i=2,\ldots,n+1$, that contains the ray generated by $u_1$. By considering the walls $\omega_i=\operatorname{cone}(u_1,\ldots,\widehat{u_i},\ldots,u_{n})$, for $i=2,\ldots,n$, that separates $\sigma_{n+1}$ and $\sigma_i$ and that define the same contraction, we obtain by the same computations that $u_1=(1,\ldots,1)$ is the barycenter.

I think that with the same kind of arguments you can get a positive answer when the contraction sends the divisor onto a subvariety of higher dimension.

• thanks, but I still don't quite understand your argument... using your notation, why can you assume that the vectors $u_2, ..., u_{n+1}$ are the elements of the standard basis. They could not even span a full dimensional cone.. – cata Feb 2 '16 at 17:33
• I'm sorry, I think that I should add somewhere something like "Take any irreducible component of the image of the exceptional locus. This component will correspond to a cone in $\tau \in \Sigma_X$. If we choose any maximal cone $\sigma \in \Sigma$ containing $\tau$, then we can find an isomorphism $N\cong \mathbb{Z}^n$ such that $\sigma$ is generated by the elements of the standard basis of $\mathbb{Z}^n$, by smoothness." And then we work upstairs in $\Sigma'$ by considering a stellar subdivision of $\sigma$. – Pedro Montero Feb 3 '16 at 8:28
• PS: Is the same kind of argument that you can use to prove that $\mathbb{P}^n$ is the only smooth complete toric variety with Picard number one... you choose a maximal cone to be generated by the standard basis and the you determine the coordinates of the last vector by convexity (all coordinates are negative) and then computing determinants and imposing the smoothness condition (these determinants should be $\pm 1$). – Pedro Montero Feb 3 '16 at 8:35