MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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)?

share|cite|improve this question
up vote 7 down vote accepted

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.

share|cite|improve this answer
Note to original poster: Please see UPDATE! And I won't be insulted if you withdraw the acceptance. – David Speyer Jul 13 '10 at 20:27
Sorry, just noticed another error. Editing again, more slowly this time. – David Speyer Jul 13 '10 at 20:29
All fixed now, I think. – David 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 Speyer Jul 13 '10 at 21:23
My final comment answered a now-deleted question. – David 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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.