# Resolution of 3-fold quotient singularities

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 $$E_2$$.

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.