8 Geometry of the quotient surface

Let \(e_1,e_2,e_3\) be the elementary symmetric polynomials in \(X,Y,Z\), and put \(\delta =X^2Y+Y^2Z+Z^2X\). The integral quotient is \(\mathscr {S}=\operatorname {Proj}(\mathbb {Z}[X,Y,Z]^{C_3})\), with generic fiber \(S\). We describe its cubic embedding and its minimal resolution.

Proposition 9 The invariant cubic model

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The invariant cubics

\[ A=e_1^3,\qquad B=e_1e_2,\qquad C=e_3,\qquad D=\delta \]

identify \(\mathscr {S}\) with the cubic hypersurface in \(\mathbb {P}^3_\mathbb {Z}\) defined by

\begin{equation*} F(A,B,C,D):=B^3+AD^2-ABD+3ACD+A^2C-6ABC+9AC^2=0. \end{equation*}

The divisor \(\mathscr {D}_0=\{ e_1=0\} \) corresponds to the line \(\{ A=B=0\} \) in these coordinates. On the generic fiber,

\begin{equation*} 3D_0\sim -K,\qquad \mathcal{O}(3D_0)\cong \mathcal{O}(1),\qquad (-K)^2=3. \end{equation*}

In particular, this cubic embedding is anticanonical.

Proof in Lean ▼

Compute the cyclic invariant ring and its third Veronese. The resulting cubic relation gives the embedding. Adjunction gives the anticanonical class, and the hyperplane intersection has degree three.

Equation 40

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\begin{equation*} F(A,B,C,D):=B^3+AD^2-ABD+3ACD+A^2C-6ABC+9AC^2=0. \end{equation*}

Proof in Lean ▼

The linked Lean proof establishes this statement.

Equation 41

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\begin{equation*} 3D_0\sim -K,\qquad \mathcal{O}(3D_0)\cong \mathcal{O}(1),\qquad (-K)^2=3. \end{equation*}

Proof in Lean ▼

The linked Lean proof establishes this statement.

Equation 42

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\begin{equation*} \delta ^2-(e_1e_2-3e_3)\delta +e_2^3+e_1^3e_3-6e_1e_2e_3+9e_3^2=0. \end{equation*}

Proof in Lean ▼

The linked Lean proof establishes this statement.

Proposition 10 The standard rational form

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Over \(\mathbb {Z}[1/6]\), \(\mathscr {S}\) is isomorphic to the cubic hypersurface

\begin{equation*} x_0(x_1^2+3x_2^2)=x_3^3 \end{equation*}

in \(\mathbb {P}_{\mathbb {Z}[1/6]}^3\). Its generic fiber \(S\) is a del Pezzo surface of degree \(3\) and geometric Picard rank \(1\). Its three geometric singularities are of type \(\mathbf A_2\): one is rational, and the other two are conjugate over \(E=\mathbb {Q}(\sqrt{-3})\). Over \(E\), the coordinates \(x_1'=x_1+\sqrt{-3}x_2\) and \(x_2'=x_1-\sqrt{-3}x_2\) give the split toric cubic \(x_0x_1'x_2'=x_3^3\). In these coordinates, \(D_0=\{ x_0=x_3=0\} \).

Proof in Lean ▼

Apply the explicit change of coordinates over \(\mathbb {Z}[1/6]\). Split the norm over \(E\), and compute the singular points and their local equations. The toric lattice gives the Picard rank.

Equation 43

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\begin{equation*} x_0(x_1^2+3x_2^2)=x_3^3 \end{equation*}

Proof in Lean ▼

The linked Lean proof establishes this statement.

Equation 44

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\begin{equation*} \begin{aligned} x_0& =A,& x_3& =A-3B,\\ x_1& =A-\tfrac 92B+\tfrac {27}{2}C,& x_2& =3D-\tfrac 32B+\tfrac 92C. \end{aligned}\end{equation*}

Proof in Lean ▼

The linked Lean proof establishes this statement.

Proposition 11 The resolution and boundary

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The surface \(S\) admits a minimal resolution \(\pi \colon X\to S\) over \(\mathbb {Q}\). Put \(D_{\mathrm{red}}=(\pi ^{-1}D_0)_{\mathrm{red}}\). Then \(X\) is a weak del Pezzo surface of degree \(3\), of geometric Picard rank \(7\). The geometric components of \(D_{\mathrm{red}}\) form a chain with self-intersections

\begin{equation*} (-2)-(-2)-(-1)-(-2)-(-2). \end{equation*}

The divisor \(L=-K_X-D_{\mathrm{red}}\) is nef and not numerically zero, but is not big. More precisely,

\begin{equation*} K_X\cdot D_{\mathrm{red}}=-1,\qquad D_{\mathrm{red}}^2=-1,\qquad L^2=0. \end{equation*}

Proof in Lean ▼

Subdivide the three singular cones to obtain the minimal smooth fan. Compute the exceptional chain, divisor classes, and intersections in that fan. The intersection numbers establish the assertions about \(L\).

Equation 45

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\begin{equation*} (-2)-(-2)-(-1)-(-2)-(-2). \end{equation*}

Proof in Lean ▼

The linked Lean proof establishes this statement.

Equation 46

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\begin{equation*} K_X\cdot D_{\mathrm{red}}=-1,\qquad D_{\mathrm{red}}^2=-1,\qquad L^2=0. \end{equation*}

Proof in Lean ▼

The linked Lean proof establishes this statement.