Resolution of 3-fold quotient singularities

Question

This is exercise 1.10 from Reid's Young person's guide to canonical singularites.

Let $X=\mathbb{C}^3/ \mu_3$ where $\epsilon \in \mu_3$ acts by $$ (x,y,z) \to (\epsilon x, \epsilon y, \epsilon^2 z).$$ Then blowing up the origin gives us $E_1 \cup E_2$ where $E_1$ is a plane and $E_2$ is a quartic scroll.

We know $X=\text{Spec}(A)$ where $$A = \mathbb{C}[x^3, x^2 y, x y^2, y^3, xz , yz, z^3]= \mathbb{C}[u_0, u_1, u_2, u_3, v_0, v_1, w]/I$$ $$I=\langle u_0u_3-u_1 u_2, u_0u_2-u_1^2, u_1u_3-u_2^2, u_2 v_1- u_3 v_0, u_1 v_1 - u_2 v_0, u_0 v_1- u_1 v_0, \\ v_0^3 - u_0 w, v_0^2 v_1 - u_1 w, v_0 v_1^2 - u_2 w, v_1^3 - u_3 w\rangle$$

If I consider the map $$(s, t, p) \to (s, st , st^2, st^3, sp, stp, s^2p^3),$$ it seems that $s=0$ corresponds to $E_1$ and the pullback of $$s = \frac{1}{u_0^5}(du_0 \wedge du_1 \wedge du_2)^{\otimes 3 }$$ has zeros of order 1 along $E_1$.

I'm not sure how to get $E_2$. If I blow up the origin of $\mathbb{C}[u_0, u_1, u_2, u_3, v_0, v_1, w],$ how can I identify $\mathbb{P}^6$ with $E_1 \cup E_2$? Why the pullback of $s$ has has zeros of order 2 along $E_2$.

Answer 1

The quotients $\mathbf{C}^n / G$ with $G$ finite abelian group (acting linearly) are toric varieties. I present the toric description of the resolution and the discrepancies. If one needs to, one could get differential forms and coordinates from this description. The reference [CLS] is the book "Toric Varieties" by Cox, Little, Schenck.

The $\frac1{n}(1,a,b)$ singularity. Let us consider $n = 3$ and $G = \mathbf{Z}/n$ acting on $\mathbf{C}^3$ by $$(x,y,z) \mapsto (\zeta x, \zeta^a y, \zeta^b z)$$ where $\zeta$ is the primitive $n$-th root of unity and $a$, $b$ are coprime to $n$. Let $X = \mathbf{C}^3 / G$. This is the $\frac1{n}(1,a,b)$ singularity. By [CLS, Exercise 11.4.8] it is an affine toric variety with the fan generated by the cone spanned by vectors $$ v_1 = (1,0,0), \; v_2 = (0,1,0), \; v_3 = (-a,-b,n) \in \mathbf{Z}^3. $$ The determinant of the resulting $3 \times 3$ matrix equals $n$. To resolve singularities of this toric variety, one needs to perform some star-subdivisions adding new rays so that the resulting determinants are equal to one.

Resolution of $\frac13(1,1,2)$. We take $n = 3$, $a = 1$, $b = 2$, in which case we can first add the vector $ v_4 = (0,0,1) $ which subdivides the cone into three cones, with $v_1, v_2, v_4$ and $v_2, v_3, v_4$ smooth, but $ v_1, v_3, v_4 $ still singular as the corresponding determinant equals $2$. One subdivides this cone by adding $v_5 = (0,-1,2)$, and we obtain five smooth cones. This describes the toric resolution $\pi: \widetilde{X} \to X$.

The new rays $v_4$, $v_5$ give two exceptional divisors and using [CLS, Proposition 3.2.7] one checks that one of them is isomorphic to $\mathbf{P}^2$, and the other is isomorphic to $\mathbf{F}_2$ (the Hirzebruch surface); they intersect along a $\mathbf{P}^1$ (which is a line on $\mathbf{P}^2$ and the negative section on $\mathbf{F}_2$). Geometrically $\widetilde{X}$ is obtained as a composition of two blow ups of $X$: the first is a weighted blow up of the origin, and the second is the usual blow up of the resulting singular point. At this point I am confused why Miles Reid says that blowing up the singular point of $X$ gives resolution of singularities as blowing up a torus-invariant point adds just one divisor.

Discrepancies. To compute the discrepancies, one needs to express each $v_4$, $v_5$ as $$ \alpha_1 v_1 + \alpha_2 v_2 + \alpha_3 v_3, $$ and the corresponding discrepancy equals $\alpha_1 + \alpha_2 + \alpha_3 - 1$ [CLS, Lemma 11.4.10]. Using this one computes that $$ K_{\widetilde{X}} = \pi^*(K_X) + \frac13 E_4 + \frac23 E_5. $$

In particular, the singularity $\frac13(1,1,2)$ is terminal.

Terminal toric singularities. One can use this method to verify that $\frac12(1,1,1)$ is terminal, $\frac13(1,1,1)$ is canonical non-terminal, and all $\frac1{n}(1,1,1)$ are not canonical for $n > 3$; this is very easy as only one ray will be added, and its discrepancy is $\frac{3}{n} - 1$. See [CLS, Proposition 11.4.12 and Theorem 11.4.21] for a description and classification of terminal threefolds. The upshot is that toric terminal threefolds are $\frac1{n}(1,a,-a)$ and $xy = zw$. Much of this actually goes back to Miles Reid.

Credits: this question was discussed as an exercise in toric geometry in the Sheffield Algebraic Geometry learning seminar.