Posted by: Steven Sam | February 1, 2010

## Resolving du Val singularities

A while back, Sam wrote a post about du Val singularities and gave explicit calculations for the blowups of the $A_1$ and $D_4$ singularities. One neat characterization of du Val singularities is the concept of absolutely isolated double point: they can be resolved by successive blowups, where each blowup is at an isolated double point. In particular, this means that when we blowup a du Val singularity, the result is either nonsingular, or the singularities are also of du Val type. In this post I will discuss this in further detail. These are mainly rough notes for myself, so apologies in advance if it’s not so enlightening to anyone else. We work over the complex numbers.

These are related to Dynkin diagrams in the following way. Let X be a du Val singularity, and let X’ be its minimal resolution, so we have a proper birational map $X' \to X$ and X’ is nonsingular. The preimage of the origin will be a tree of projective lines, and any two either intersect transversely in a single point, or are disjoint. Furthermore, any point of intersection contains only two lines. We draw a graph whose nodes are the projective lines, and connect two nodes with an edge if the corresponding lines intersect. Then this graph will be a Dynkin diagram of type ADE. Furthermore, this configuration determines the singularity up to analytic isomorphism, so we can give them names using Dynkin diagram language. Due to the recursive nature of these resolutions, we should be able to see how these trees are “built up” from successive blowups, and that’s what I want to investigate. For the E case, I’ll just give some remarks.

One fact we’ll use about (analytic) isomorphism types of singularities is the following. Let S be the polynomial ring in n variables and let R be the ring of power series in n variables. Let f and g be polynomials in S with no constant term. Let I(f) be the ideal generated by f and its partial derivatives. Then f and g are isomorphic singularities at the origin if and only if R/I(f) is isomorphic to R/I(g). Intuitively, if S is the polynomial ring in n variables, then S/I(f) is the coordinate ring for the singular locus of the variety defined by f, and R/I(f) is the completion at the origin. This is the Mather-Yau theorem, which is stated as Theorem I.2.26 of the book Introduction to Singularities and Deformations by G.-M. Greuel, C. Lossen, and E. Shustin.

Blowing up $A_n$: This is the affine variety defined by the equation $x^2 + y^2 + z^{n+1} = 0$. We blow up the origin of X, and call the birational transform Y. Let’s describe this explicitly. Choose projective coordinates $[s:t:u]$ on ${\bf P}^2$. The blowup $\tilde{\bf C}^3$ of ${\bf C}^3$ is defined by the equations xt = sy, xu = sz, tz = uy, and it comes with a projection p to ${\bf C}^3$. Then Y is the closure of $p^{-1}(X \setminus \{0\})$ inside of $\tilde{\bf C}^3$.

It is covered by 3 open affine varieties, given by the equations s=1, t=1 and u=1, respectively. If we look at the case when u=1, we get x = sz and y = tz. So our original equation becomes $z^2(s^2 + t^2 + z^{n-1})$. This has two irreducible components, one where $z^2 = 0$, corresponding to $p^{-1}(0)$, and the other $s^2 + t^2 + z^{n-1} = 0$ which defines the closure of $p^{-1}(X \setminus \{0\})$ inside of this open set.

The exceptional divisor E is the intersection of Y with $p^{-1}(0)$. When n=1, E is defined by the equation $s^2 + t^2 + u^2 = 0$, which is isomorphic to ${\bf P}^1$. This can be verified on each open affine set.

When n>1, its intersection with the open set defined by u=1 is the set of points $((0,0,0), [a:\pm i a:1])$ where a is an arbitrary complex number, and $i^2 = -1$. Working in the open set defined by s=1, we have the equation $x^2(1 + t^2 + x^{n-1}u^{n+1}) = 0$. Again we throw out the $x^2$ factor, and the intersection with $p^{-1}(0)$ is affine line $((0,0,0), [1:\pm i: b])$ where b is an arbitrary complex number. We will get the same thing if we work in the open set defined by t=1. Note that $((0,0,0), [a:\pm i a:1]) = ((0,0,0), [1: \pm i: b])$ when a and b are both nonzero and $b = a^{-1}$. Let’s look at the limit $b \to \infty$. First we rewrite the points as $((0,0,0), [b^{-1}: \pm i b^{-1}: 1])$. Then we set $b^{-1} = 0$, so the limit point is $((0,0,0), [0: 0: 1])$ which corresponds to the case a=0 in the paragraph above. We conclude that E consists of two projective lines intersecting at the point $((0,0,0), [0:0:1])$.

When n>2, this point of intersection is a singular point of Y, and the singularity is of type $A_{n-2}$ (as can be seen from the equation $s^2 + t^2 + z^{n-1}$ above). Again we can blow this up, and by what we have just said, the exceptional divisor is either a single ${\bf P}^1$ or two copies F, F’ intersecting in a point Q. In the first case, the inverse images of E and E’ intersect this copy in different points, and in the second case we will have E intersect F in a point distinct from Q, and E’ intersect F’ in a point distinct from Q. I omit the calculations. If we continue blowing up singular points until we get something nonsingular, ultimately, we will get n projective lines $L_1, \dots, L_n$ above the origin such that $L_i$ intersects $L_{i+1}$, and all of these points of intersection are distinct.

Blowing up $D_n$: For this part, $n \ge 4$. This is the singularity defined by the equation $x^2 + y^2z + z^{n-1} = 0$. The exceptional divisor E of the blowup Y is a projective line. We use the same notation as the $A_n$ case. Then the open set given by s=1 doesn’t intersect E, while the intersection of Y with the open set defined by t=1 is given by the equation $s^2 + yu + y^{n-3}u^{n-1}$. This singularity is analytically isomorphic to a type $A_1$ singularity because the ideal generated by this equation and its partial derivatives is the maximal ideal $(s,u,y)$.

Now we examine the intersection of Y with the open set defined by u=1. The defining equation is $s^2 + zt^2 + z^{n-3}$. If n=4, it contains two singular points corresponding to $s=0, t = \pm i, z = 0$, both of type $A_1$: for example to deal with the case t=-i, we translate coordinates by $t \to t+i$ to get $s^2 + t^2z + 2izt$. Its partial derivatives are $2s, z(t+i), t(t+2i)$. Since we’re working in the power series ring, $t+i$ and $t+2i$ are units, so the partial derivatives generate the maximal ideal. Otherwise, if n>4, it contains 1 singular point $s=0,t=0,z=0$ of type $D_{n-2}$ (calculation omitted). When we blow it up enough times, this singular point will become a tree of projective lines which intersect in the pattern of $D_{n-2}$. Then the preimage of E will intersect in a single point, but it will also contain another projective line hanging off it (coming from the $A_1$ singularity in the previous paragraph). This shape is $D_n$.

Blowing up $E_6$: This is the variety X defined by the equation $x^2 + y^3 + z^4 = 0$. The exceptional divisor E of the blowup $Y_1$ is a projective line $L_1$, which contains a singularity of $Y_1$ of type $A_5$: using the notation from above, this is in the open set defined by u=1, and the defining equation is $s^2 + t^3z + z^2$. The quotient ring ${\bf C}[[s,t,z]] / (s^2 + t^3z + z^2, 2s, t^3 + 2z, t^2z)$ is isomorphic to ${\bf C}[t] / (t^5)$, which is what one gets in type $A_5$.

Blowing up that singularity, we get $Y_2 \to Y_1 \to X$, and the inverse image of the origin contains three projective lines $L_1, L_2, L_3$, all of which intersect at the same point. This is a singularity of type $A_3$. We blow it up again and get $Y_3 \to Y_2 \to Y_1 \to X$. Now the inverse image of the origin is 5 projective lines $L_1, L_2, L_3, L_4, L_5$ which all intersect at a single point. This point of intersection is still singular. Blow it up one more time to get a nonsingular surface, and the inverse image of the origin is 6 projective lines which intersect each other like in the $E_6$ Dynkin diagram. This article does the calculations out in gory detail.

Blowing up $E_7$: This is defined by the equation $x^2 + y^3 + yz^3 = 0$. The exceptional divisor E of the blowup contains a singular point of type $D_6$ in the open set defined by u=1 where the defining equation is $s^2 + t^3z + tz^2$. The quotient ${\bf C}[[s,t,z]]/I(s^2 + t^3z + tz^2)$, after some manipulations, becomes the ring ${\bf C}[t,z] / (t^5, 2tz + t^3, z^2 - 3/2 t^4)$. Replacing z by $z' = 1/2 t^2 + z$, we have $tz' = 0$ and that $z'^2$ is a scalar multiple of $t^4$, so we can do a scalar change of coordinates to get it isomorphic to ${\bf C}[t,z] / (t^5, tz, t^4 - z^2)$, which is what we get from $D_6$.

Blowing up $E_8$: This is defined by the equation $x^2 + y^3 + z^5 = 0$. The exceptional divisor E of the blowup contains a singular point in the open set defined by $u=1$ of type $E_7$: the defining equation is $s^2 + t^3z + z^3 = 0$.

-Steven