# How can I get a small resolution for the binomial fourfold $x_1 x_2 x_3- y_1 y_2=0$ in $\mathbb{C}^5$?

I consider the singular fourfold $X$ defined as follows:

$$X: \quad x_1 x_2 x_3 -y_1 y_2=0\quad \text{in}\quad \mathbb{C}^5.$$

Its singular locus is a bouquet of three planes meeting at the origin:

$$Sing(X):\quad y_1=y_2=x_1 x_2=x_2 x_3=x_1 x_3=0.$$

How can I described the small resolution of this space (if any)?

As Alex Woo says, this is a toric example, and hence can be solved with toric methods. Your variety is $\mathrm{Spec} \ \mathbb{C}[S]$ where $S$ is the semigroup ring generated by $(1,0,0,1)$, $(0,1,0,1)$, $(0,0,1,1)$, $(0,0,0,1)$ and $(1,1,1,2)$. (These correspond to the variables $x_1$, $x_2$, $x_3$, $y_1$ and $y_2$ respectively.) This is a saturated semigroup, so toric methods apply with no subtleties. The dual cone is generated by $(1,0,0,0)$, $(0,1,0,0)$, $(0,0,1,0)$, $(-1,-1,0,1)$, $(-1,0,-1,1)$, $(0,-1,-1,1)$. Toric resolutions of this singularity correspond to triangulations of this cone.

Notice that the six generators of the dual cone all lie in the plane $w+x+y+3z=1$. In this plane, they form a triangular prism. We can draw our pictures in three coordinates by discarding the final coordinate. However, we need to recognize that, if we do this, a lattice point means a point $(x,y,z)$ such that $x+y+z \equiv 1 \mod 3$. (This is the point I screwed up earlier.)

So we want to understand triangulations of the prism with vertices $(1,0,0)$, $(0,1,0)$, $(0,0,1)$, $(0,-1,-1)$, $(-1,0,-1)$, $(-1,-1,0)$. Now, suppose that our subdivision has a face $d$ which is contained in a face of the triangular prism of dimension $e$. The cones on these faces have dimension $d+1$ and $e+1$; the corresponding torus orbits have dimension $3-d$ and $3-e$, so the fibers here have dimension $e-d$. So smallness means that $2(e-d) < e+1$. For $e=0$, $1$, $2$, $3$ this gives $d \geq 0$, $1$, $1$ and $2$, respectively. In other words, we must add no new vertices to the triangular prism, and we may only add new edges within $2$-faces.

Fortunately, the standard triangulation of the triangular prism has this property. There are three tetrahedra: $$\mathrm{Hull} {\Large (} (1,0,0), \ (0,1,0), \ (0,0,1), \ (0,-1,-1) {\Large )}$$ $$\mathrm{Hull} {\Large (} (0,1,0), \ (0,0,1), \ (0,-1,-1), \ (-1,0,-1) {\Large )}$$ $$\mathrm{Hull} {\Large (} (0,0,1), \ (0,-1,-1), \ (-1,0,-1), \ (-1,-1,0) {\Large )}$$ In my previous update, I worried that these are not unimodular, because the "lattice point" $(0,0,0)$ lay on the $2$-faces of some of them. However, that is actually not a lattice point. (It corresponds to $(0,0,0,1/3)$ back in $4$-space.) Sorry about the confusion.

• Note to original poster: Please see UPDATE! And I won't be insulted if you withdraw the acceptance. – David E Speyer Jul 13 '10 at 20:27
• Sorry, just noticed another error. Editing again, more slowly this time. – David E Speyer Jul 13 '10 at 20:29
• All fixed now, I think. – David E Speyer Jul 13 '10 at 20:34
• It's small by the explicit computation of fiber dimensions in the third paragraph. Yup, I agree that there should be some flops. There are 6 symmetric versions of this triangulation, by permuting the three coordinates. Many of them are linked by flops. For example, the first two tetrahedra above form a square pyramid, with vertex $(0,0,1)$. Triangulating this square pyramid the other way gives a flop. – David E Speyer Jul 13 '10 at 21:23
• My final comment answered a now-deleted question. – David E Speyer Jul 13 '10 at 21:23

Warning: This seems to be a really bad way of answering this question (but it at least tells you there is one).

The intersection of the opposite Schubert cell $X_\circ^{13425}$ with the Schubert variety $X_{34512}$ is defined by that equation in the appropriate coordinates. This tells you that locally around the Schubert point $e_{13425}$, the Schubert variety is isomorphic to the product of that fourfold with $\mathbb{C}^{4}$.

Since the permutation $34512$ is 321 and hexagon avoiding, the Bott--Samelson resolution is small for that Schubert variety.

As you have a binomial, I would expect there to be some toric answer that is at least a little more general than this one.

• This is probably not the most direct way but it is an interesting way to look at it. I am not familiar with the subject of Schubert varieties and the Bott-Samelson resolution. Do you have a reference? – JME Jul 13 '10 at 19:49
• For quite some time the standard introduction to Schubert varieties has been the last two chapters of Fulton's book [i]Young Tableaux[/i]. I quite like Brion's lecture notes from a mini-course he gave in Warsaw in 2003: arXiV:math/0410210 (and there is also a book version somewhere). Only the first 2 chapters are general background (including explanation of the Bott-Samelson resolution) and the last 2 give his proof of the Buch-Fulton conjecture. – Alexander Woo Jul 13 '10 at 22:35
• And more specific references: the kind of calculation giving this equation in my paper with Alex Yong: <i>Governing Singularities of Schubert Varieties</i>, J. Algebra 320 (2008), 495-520, and the fact that 321 hexagon avoiding Schubert varieties have small Bott--Samelson resolutions is proven by Sara Billey and Greg Warrington in <i>Kazhdan--Lusztig polynomials for 321 hexagon avoiding permutations</i> J. Algebraic Combin. 13 (2001), 111--136. – Alexander Woo Jul 13 '10 at 22:40