Proof. By inversion of adjunction, we know that $\mathcal X$ has klt singularities. Since $\mathcal D$ and $\mathcal D_b$ are $\mathbb {Q}$ -Cartier Weil divisors on $\mathcal X$ and $\mathcal X_b$ , respectively, we know that both $\mathcal {O}_{\mathcal X}(m\mathcal D)$ and $\mathcal {O}_{\mathcal X_b}(m\mathcal D_b)$ are Cohen–Macaulay for any $m\in \mathbb {Z}$ by [Reference Kollár and Mori40, Corollary 5.25]. Hence $\mathcal {O}_{\mathcal X_b}(m\mathcal D_b)\cong \mathcal {O}_{\mathcal X}(m\mathcal D)\otimes \mathcal {O}_{\mathcal X_b}$ for any $m\in \mathbb {Z}$ . By Kawamata–Viehweg vanishing, we know that $H^i\left(\mathcal X_b,\mathcal {O}_{\mathcal X_b}(m\mathcal D_b)\right)=0$ for any $b\in B$ and $m,i\geq 1$ . Hence the statement for $m>0$ follows from the semicontinuity theorem and flatness of $\mathcal {O}_{\mathcal X}(m\mathcal D)$ over B, and it is obvious for $m\leq 0$ .

Proof. For simplicity, we assume that T is connected. By [Reference Kollár37, Theorem 4.33], there exists a locally closed decomposition $T'\to T$ such that for any morphism $q:W\to T$ , the divisorial pullback $f_W: (\mathcal X_W, \mathcal D_W)\to W$ of $f:(\mathcal X,\mathcal D)\to T$ satisfies that $\mathcal D_W$ is flat over W if and only if q factors as $q: W\to T'\to T$ . It is clear that $\mathcal D$ is flat over T if and only if $T'=T$ . Thus it suffices to show that for any morphism $B\to T$ from a smooth curve B, the divisorial pullback $f_B:(\mathcal X_B,\mathcal D_B)\to B$ of f satisfies the demand that $\mathcal D_B$ be flat over B. It is clear that $f_B$ is also a $\mathbb {Q}$ -Gorenstein smoothable log Fano family. By the proof of Lemma 2.11, we know that $\mathcal {O}_{\mathcal X_B}(-\mathcal D_B)$ is flat over B, and its fiber over $b\in B$ is isomorphic to $\mathcal {O}_{\mathcal X_{B,b}}\left(-\mathcal D_{B,b}\right)$ , which is $S_2$ . Hence [Reference Kollár37, Definition-Lemma 4.19] implies that $\mathcal D_B\to B$ is flat. This finishes the proof.

Proof. Fix a sufficiently divisible $m\in \mathbb {Z}_{>0}$ . Denote

$$ \begin{align*} \chi(k):=\chi_0(mk),\quad \tilde{\chi}(k)=\chi_0(mk)-\chi_0(mk-r), \quad \text{and}\quad N_m:=\chi_0(m)-1. \end{align*} $$

Recall that in [Reference Ascher, DeVleming and Liu6, Section 3.1], we construct a locally closed subscheme $Z^{\mathrm {klt}}$ of the relative Hilbert scheme $\mathrm {Hilb}_{\chi }\left(\mathbb {P}^{N_m}\right)\times \mathrm {Hilb}_{\tilde {\chi }}\left(\mathbb {P}^{N_m}\right)$ which parametrizes $\mathbb {Q}$ -Gorenstein smoothable log Fano pairs $(X,cD)$ such that they are embedded into $\mathbb {P}^{N_m}$ by $\lvert -mK_X\rvert $ and X is klt. Denote by Z the dense open subscheme of $Z^{\mathrm {klt}}$ parametrizing $(X,D)$ , where both X and D are smooth. Let $Z_c^\circ $ be the Zariski open subset of $Z^{\mathrm {klt}}$ parametrizing K-semistable log Fano pairs $(X,cD)$ . Denote by $Z_c^{\mathrm {red}}$ the reduced scheme supported on $Z_c^\circ $ . Then $\mathcal {K}\mathcal M_{\chi _0,r,c}$ is defined as the quotient stack $\left[Z_c^{\mathrm {red}}/\mathrm {PGL}(N_m+1)\right]$ . Hence it suffices to show that $Z^{\mathrm {klt}}$ is smooth, which would then imply that $Z_c^{\mathrm {red}}$ is smooth. The following argument is inspired by [Reference Ascher, DeVleming and Liu6, Lemma 9.7].

Denote by $Z^{\mathbb {Q}\mathrm {F}}$ the locally closed subscheme of $\mathrm {Hilb}_{\chi }\left(\mathbb {P}^{N_m}\right)$ parametrizing $\mathbb {Q}$ -Gorenstein smoothable $\mathbb {Q}$ -Fano varieties X that are embedded into $\mathbb {P}^{N_m}$ by $\lvert -mK_X\rvert $ . Since we are in dimension $2$ , any point $\mathrm {Hilb}(X)\in Z^{\mathbb {Q}\mathrm {F}}$ corresponds to a log del Pezzo surface X with only T-singularities. Hence X has unobstructed $\mathbb {Q}$ -Gorenstein deformations by [Reference Hacking29, Theorem 8.2], [Reference Hacking and Prokhorov31, Proposition 3.1], and [Reference Akhtar, Coates, Corti, Heuberger, Kasprzyk, Oneto, Petracci, Prince and Tveiten1, Lemma 6]. Thus $Z^{\mathbb {Q}\mathrm {F}}$ is a smooth scheme. Denote by $Z^{\textrm {sm}}$ the Zariski open subset of $Z^{\mathbb {Q}\mathrm {F}}$ parametrizing smooth Fano manifolds X such that there exists a smooth divisor $D\sim _{\mathbb {Q}}-rK_X$ . The openness of $Z^{\textrm {sm}}$ follows from the openness of smoothness, $H^0(X,\mathcal {O}_X(D))$ being constant since $H^i(X,\mathcal {O}_X(D))=0$ for $i \geq 1$ by Kodaira vanishing, and the fact that smooth families of Fano manifolds have locally constant Picard groups. Denote $Z^{\mathrm {bs}}:=\overline {Z^{\textrm {sm}}}\cap Z^{\mathbb {Q}\mathrm {F}}$ . Hence $Z^{\mathrm {bs}}$ is the disjoint union of some connected components of $Z^{\mathbb {Q}\mathrm {F}}$ . Denote the first projection by $\mathrm {pr}_1: Z^{\mathrm {klt}}\to \mathrm {Hilb}_{\chi }\left(\mathbb {P}^{N_m}\right)$ . Clearly $\mathrm {pr}_1\left(Z^{\mathrm {klt}}\right)$ is contained in $Z^{\mathbb {Q}\mathrm {F}}$ . We claim that $\mathrm {pr}_1\left(Z^{\mathrm {klt}}\right)=Z^{\mathrm {bs}}$ , and that the restriction morphism $\mathrm {pr}_1:Z^{\mathrm {klt}}\to Z^{\mathrm {bs}}$ is proper and smooth.

We first show that $\mathrm {pr}_1\left(Z^{\mathrm {klt}}\right)=Z^{\mathrm {bs}}$ and $\mathrm {pr}_1:Z^{\mathrm {klt}}\to Z^{\mathrm {bs}}$ is proper. Since Z is a dense open subset of $Z^{\mathrm {klt}}$ , we know that

$$ \begin{align*} Z^{\textrm{sm}}=\mathrm{pr}_1(Z)\subset\mathrm{pr}_1\left(Z^{\mathrm{klt}}\right)\subset \overline{\mathrm{pr}_1(Z)}\cap Z^{\mathbb{Q}\mathrm{F}}=\overline{Z^{\textrm{sm}}}\cap Z^{\mathbb{Q}\mathrm{F}}=Z^{\mathrm{bs}}. \end{align*} $$

Hence the surjectivity of $\mathrm {pr}_1:Z^{\mathrm {klt}}\to Z^{\mathrm {bs}}$ would follow from its properness. We will verify properness by checking the existence part of the valuative criterion. Let $0\in B$ be a pointed curve with $B^\circ :=B\setminus \{0\}$ . Consider two morphisms $f^\circ : B^\circ \to Z^{\mathrm {klt}}$ and $g:B\to Z^{\mathbb {Q}\mathrm {F}}$ such that $g\rvert _{B^\circ }=\mathrm {pr}_1\circ f^\circ $ . It suffices to show that $f^\circ $ extends to $f:B\to Z^{\mathrm {klt}}$ such that $g=\mathrm {pr}_1\circ f$ . We have a $\mathbb {Q}$ -Gorenstein smoothable family $p:\mathcal X\to B$ induced by g, and a $\mathbb {Q}$ -Cartier Weil divisor $\mathcal D^\circ $ on $\mathcal X^\circ :=p^{-1}(B^\circ )$ induced by $f^\circ $ whose support does not contain any fiber $\mathcal X_b$ , and $\mathcal D^\circ \sim _{\mathbb {Q},B^\circ }-rK_{\mathcal X^\circ /B^\circ }$ . We define $\mathcal D:=\overline {\mathcal D^\circ }$ . Then, by taking Zariski closure, it is clear that $\mathcal D\sim _{\mathbb {Q}, B}-rK_{\mathcal X/B}$ , since $\mathcal X_0$ is a $\pi $ -linearly trivial Cartier prime divisor on $\mathcal X$ . Thus $(\mathcal X,\mathcal D)\to B$ is a $\mathbb {Q}$ -Gorenstein smoothable log Fano family. This finishes proving the properness and surjectivity of $\mathrm {pr}_1:Z^{\mathrm {klt}}\to Z^{\mathrm {bs}}$ .

Finally, we will show that $\mathrm {pr}_1:Z^{\mathrm {klt}}\to Z^{\mathrm {bs}}$ is a smooth morphism. Indeed, we will show that it is a smooth $\mathbb {P}^{N_r}$ -fibration, where $N_r:=\chi _0(r)-1$ . If $\mathrm {Hilb}(X,D)\in \mathrm {pr}_1^{-1}(Z^{\textrm {sm}})$ , then we know that $h^0(X,\mathcal {O}_X(D))=\chi (X,\mathcal {O}_X(D))=\chi _0(r)$ , since $H^i(X,\mathcal {O}_X(D))=0$ for any $i\geq 1$ by Kodaira vanishing. Hence the fiber over $\mathrm {Hilb}(X)\in Z^{\textrm {sm}}$ is isomorphic to $\mathbb {P}(H^0(X,\mathcal {O}_X(D)))\cong \mathbb {P}^{N_r}$ . Hence we may restrict to the case when $\mathrm {Hilb}(X)\in Z^{\mathrm {bs}}\setminus Z^{\textrm {sm}}$ . Assume that $(X,D)\in Z^{\mathrm {klt}}$ is $\mathbb {Q}$ -Gorenstein smoothable where $\pi : (\mathcal X,\mathcal D)\to B$ is a $\mathbb {Q}$ -Gorenstein smoothing over a pointed curve $0\in B$ with $(\mathcal X_0,\mathcal D_0)\cong (X,D)$ . Then by Lemma 2.11, we know that $\pi _*\mathcal {O}_{\mathcal X}(\mathcal D)$ is locally free with fiber over $b\in B$ isomorphic to $H^0\left(\mathcal X_b,\mathcal {O}_{\mathcal X_b}(\mathcal D_b)\right)$ . Hence it is easy to conclude that for any effective Weil divisor $D'\sim D$ , the pair $(X,D')$ is also $\mathbb {Q}$ -Gorenstein smoothable. Since the Weil divisor class group $\mathrm {Cl}(X)$ of X is finitely generated, we know that there are only finitely many Weil divisor classes $[D]$ such that $[D]= -r[K_X]$ in $\mathrm {Cl}(X)\otimes _{\mathbb {Z}} \mathbb {Q}$ . Hence the fiber $\mathrm {pr}_1^{-1}(\mathrm {Hilb}(X))$ is isomorphic to a disjoint union of finitely many copies of $\mathbb {P}^{N_r}$ . However, since $\mathrm {pr}_1:Z^{\mathrm {klt}}\to Z^{\mathrm {bs}}$ is proper with connected fibers over a dense open subset $Z^{\textrm {sm}}$ and $Z^{\mathrm {bs}}$ is normal, taking Stein factorization yields that $\mathrm {pr}_1$ has connected fibers everywhere. Hence $\mathrm {pr}_1^{-1}(\mathrm {Hilb}(X))\cong \mathbb {P}^{N_r}$ for any $\mathrm {Hilb}(X)\in Z^{\mathrm {bs}}$ . Therefore, $\mathrm {pr}_1$ has smooth fibers and smooth base, which implies that $Z^{\mathrm {klt}}$ is Cohen–Macaulay. Hence, miracle flatness implies that $\mathrm {pr}_1$ is flat and hence smooth. The proof is finished.

Proof. This follows from interpolation (see [Reference Ascher, DeVleming and Liu6, Proposition 2.13] or [Reference Dervan18, Lemma 2.6]), since the pair $\left(\mathbb {P}^1 \times \mathbb {P}^1, \frac {2}{d}C\right)$ is klt (resp., lc) and $\mathbb {P}^1\times \mathbb {P}^1$ is K-polystable.

Proof. The normality of $\mathcal {K}\mathcal M_{\chi _0, d/2, c}$ and $KM_{\chi _0, d/2, c}$ is a direct consequence of Theorem 2.21. For the rest, notice that there are only two smooth del Pezzo surfaces of degree $8$ up to isomorphism: $\mathbb {P}^1\times \mathbb {P}^1$ and $\mathbb {F}_1$ . In addition, they are not homeomorphic, since their intersection pairings on $H^2(\cdot , \mathbb {Z})$ are not isomorphic. By Proposition 4.1 we know that $\left(\mathbb {P}^1\times \mathbb {P}^1,cC\right)$ , where C is a smooth $(d,d)$ -curve, is always parametrized by $\mathcal {K}\mathcal M_{\chi _0, d/2, c}$ . If d is odd, then $-\frac {d}{2}K_{\mathbb {F}_1}$ is not represented by any Weil divisor, since it has fractional intersection with the $(-1)$ -curve on $\mathbb {F}_1$ . Hence $\mathbb {F}_1$ will not appear in $\mathcal {K}\mathcal M_{\chi _0, d/2, c}$ when d is odd. The proof is finished.

Proof. Let $\mathcal X\to T$ be a $\mathbb {Q}$ -Gorenstein smoothing of X – that is, $0\in T$ is a smooth germ of pointed curve, $\mathcal X_0\cong X$ , and $\mathcal X_t\cong \mathbb {P}^1\times \mathbb {P}^1$ for $t\in T\setminus \{0\}$ . By passing to a finite cover of $0\in T$ , we may assume that $\mathcal X^\circ \cong \left(\mathbb {P}^1\times \mathbb {P}^1\right)\times T^\circ $ , where $\mathcal X^\circ :=\mathcal X\setminus \mathcal X_0$ and $T^\circ := T \setminus \{0\}$ . First, using [Reference Hacking29, Lemma 2.11], we show that $\mathrm {Cl}(\mathcal {X}) \cong \mathbb {Z}^{2}$ . Indeed, consider the exact sequence

$$ \begin{align*} 0 \to \mathbb{Z}X \to \mathbb{Z} X \to \mathrm{Cl}(\mathcal{X})\to \mathrm{Cl}(\mathcal{X}^\circ) \to 0,\end{align*} $$

which gives $\mathrm {Cl}(\mathcal {X}) \cong \mathrm {Cl}(\mathcal {X}^\circ )\cong \mathbb {Z}^{2}$ .

Now we follow the proof of [Reference Hacking29, Proposition 6.3]. First note that there is an isomorphism $\mathrm {Pic}(\mathcal {X}) \to \mathrm {Pic}(X)$ , and so we obtain the inequality

$$ \begin{align*} \rho(X) = \dim \mathrm{Pic}(X) \otimes \mathbb{Q} = \dim \mathrm{Pic}(\mathcal{X}) \otimes \mathbb{Q} \leq \dim \mathrm{Cl}(\mathcal{X}) \otimes \mathbb{Q} = 2,\end{align*} $$

with equality if and only if $\mathcal {X}$ is $\mathbb {Q}$ -factorial.

Proof. Define $\beta :=1-cd/2\in (0,1)$ . We know that an index n point $x\in X$ is a cyclic quotient singularity of type $\frac {1}{n^2}(1,na-1)$ or $\frac {1}{2n^2}(1, 2na-1)$ , where $\gcd (a,n)=1$ . If $\mu _x=0$ , then the orbifold group of $x\in X$ has order $n^2$ , which implies that $\widehat {\mathrm {vol}}(x,X)=\frac {4}{n^2}$ by [Reference Li and Liu51, Proposition 4.10]. Hence Theorem 2.6 implies that

$$ \begin{align*} 8\beta^2 =(-K_X-cD)^2 \leq \frac{9}{4}\widehat{\mathrm{vol}}(x,X)=\frac{9}{n^2}. \end{align*} $$

This shows that $n\leq \frac {3}{2\sqrt {2} \beta }=\frac {3}{\sqrt {2}(2-cd)}$ . Similarly, if $\mu _x=1$ , then $x\in X$ has orbifold group of order $2n^2$ , which implies that $n\leq \frac {3}{4 \beta }=\frac {3}{2(2-cd)}$ . Hence the first terms in the index upper bounds are verified.

The rest of this proof is devoted to verifying the second terms in the index upper bounds. We know that $dK_X+2D\sim 0$ when d is odd and $\frac {d}{2}K_X+D\sim 0$ when d is even. If $x\not \in D$ , then $n\mid d$ , hence $n\leq d$ (in fact, $n\leq \frac d{2}$ if d is even). Hence the second terms are verified for $x\not \in D$ .

From now on, let us assume $x\in D$ . Let $\left(\tilde {x}\in \widetilde {X}\right)$ be the smooth cover of $(x\in X)$ , with $\widetilde {D}$ being the preimage of D. Assume $\tilde {x}\in \widetilde {X}$ has local coordinates $(u,v)$ where the cyclic group action is scaling on each coordinate. Let $u^i v^j$ be a monomial appearing in the equation on $\widetilde {D}$ with minimum $i+j=\mathrm {ord}_{\tilde {x}}\widetilde {D}$ .

Case 1. Assume d is even and $\mu _x=0$ . Then the orbifold group of $x\in X$ has order $n^2$ . Since the finite-degree formula is true in dimension $2$ by [Reference Li, Liu and Xu52, Theorem 4.15], we have

$$ \begin{align*} \widehat{\mathrm{vol}}\left(\tilde{x},\widetilde{X},c\widetilde{D}\right) =n^2\cdot\widehat{\mathrm{vol}}(x,X,c D). \end{align*} $$

On the other hand, Theorem 2.6 implies that

$$ \begin{align*} 8\beta^2=(-K_X-cD)^2\leq \frac{9}{4}\widehat{\mathrm{vol}}(x, X, cD)=\frac{9}{4n^2}\widehat{\mathrm{vol}}\left(\tilde{x},\widetilde{X},c\widetilde{D}\right). \end{align*} $$

So we have

(4.1) $$ \begin{align} n\leq \frac{3\sqrt{ \widehat{\mathrm{vol}}\left(\tilde{x},\widetilde{X},c\widetilde{D}\right)}}{4\sqrt{2}\beta} \leq \frac{3\left(2-c~\mathrm{ord}_{\tilde{x}}\widetilde{D}\right)}{4\sqrt{2}\beta}. \end{align} $$

In particular, we have $n<\frac {3}{2\sqrt {2}\beta } $ . We know that $\mathrm {lct}_{\tilde {x}}\left(\widetilde {X};\widetilde {D}\right)> c$ , and Skoda [Reference Skoda70] implies $\mathrm {lct}_{\tilde {x}}\left(\widetilde {X};\widetilde {D}\right)\leq \frac {2}{\mathrm {ord}_{\tilde {x}}\widetilde {D}}$ , so we have $ \mathrm {ord}_{\tilde {x}}\widetilde {D}<\frac {2}{c}$ . Since $\frac {d}{2}K_X+D\sim 0$ , we have $i+(na-1)j\equiv \frac {d}{2}na\mod n^2$ , which implies $i\equiv j\mod n$ .

If $\beta \geq \frac {3}{2\sqrt {2}d+3}$ , then $n<\frac {3}{2\sqrt {2}\beta }\leq d+\frac {3}{2\sqrt {2}}$ , which implies $n\leq d+1$ . Thus we may assume $\beta <\frac {3}{2\sqrt {2}d+3}$ . Then

$$ \begin{align*} i+j= \mathrm{ord}_{\tilde{x}}\widetilde{D}<\frac{2}{c}<d+\frac{3}{2\sqrt{2}}. \end{align*} $$

Hence $i+j\leq d+1$ . Assume to the contrary that $n\geq d+2$ . Then $i\equiv j\mod n$ and $i+j<n$ implies that $i=j$ . Hence $i+(na-1)j\equiv \frac {d}{2}na\mod n^2$ implies $i\equiv \frac {d}{2}\mod n$ . But since $i\leq \frac {d+1}{2}<n$ , we know that $i=j=\frac {d}{2}$ . Then formula (4.1) implies that

$$ \begin{align*} n\leq \frac{3(2-c(i+j))}{4\sqrt{2}\beta}=\frac{6\beta}{4\sqrt{2}\beta}<2. \end{align*} $$

We reach a contradiction.

Case 2. Assume d is even and $\mu _x=1$ . Then the orbifold group of $x\in X$ has order $n^2$ . By a similar argument as in case 1, we know that

$$ \begin{align*} 8\beta^2=(-K_X-cD)^2\leq \frac{9}{4}\widehat{\mathrm{vol}}(x, X, cD)=\frac{9}{8n^2}\widehat{\mathrm{vol}}\left(\tilde{x},\widetilde{X},c\widetilde{D}\right). \end{align*} $$

Hence

(4.2) $$ \begin{align} n\leq \frac{3\sqrt{ \widehat{\mathrm{vol}}\left(\tilde{x},\widetilde{X},c\widetilde{D}\right)}}{8\beta} \leq \frac{3\left(2-c~\mathrm{ord}_{\tilde{x}}\widetilde{D}\right)}{8\beta}. \end{align} $$

In particular, we have $n<\frac {3}{4\beta }$ .

If $\beta \geq \frac {3}{4d+3}$ , then $n<\frac {3}{4\beta }\leq d+\frac {3}{4}$ , which implies $n\leq d$ . Thus we may assume $\beta <\frac {3}{4d+3}$ . Then

$$ \begin{align*} i+j= \mathrm{ord}_{\tilde{x}}\widetilde{D}<\frac{2}{c} <d+\frac{3}{4}. \end{align*} $$

Hence $i+j\leq d$ . Assume to the contrary that $n\geq d+1$ . Then $i\equiv j\mod n$ and $i+j<n$ implies $i=j$ . Hence $i+(na-1)j\equiv \frac {d}{2}na\mod n^2$ implies $i\equiv \frac {d}{2}\mod n$ . But since $i\leq \frac {d}{2}<n$ , we know that $i=j=\frac {d}{2}$ . Then formula (4.2) implies that

$$ \begin{align*} n\leq \frac{3(2-c(i+j))}{8\beta}=\frac{6\beta}{8\beta}<1. \end{align*} $$

We reach a contradiction.

Case 3. Assume d is odd and $\mu _x=0$ . In this case we have $dK_X+2D\sim 0$ , which implies $2(i+(na-1)j)\equiv dna \mod n^2$ . If n is odd, then clearly $i\equiv j\mod n$ . By the same argument as case 1, we know $i=j=\frac {d}{2}$ if $n\geq d+2$ , hence a contradiction.

If n is even, then we do a finer analysis. Since both d and a are odd, from $2(i+(na-1)j)\equiv dna \mod n^2$ we know that $i-j\equiv \frac {n}{2}\mod n$ . Thus $n\leq 2(i+j)<\frac {4}{c}=\frac {2d}{1-\beta }$ . Besides, formula (4.1) implies that $n<\frac {3}{2\sqrt {2}\beta }$ . Hence

$$ \begin{align*} n<\min\left\{\frac{2d}{1-\beta},\frac{3}{2\sqrt{2}\beta}\right\}\leq \frac{2\sqrt{2}(2d)+3}{2\sqrt{2}(1-\beta)+2\sqrt{2}\beta}=2d+\frac{3}{2\sqrt{2}}. \end{align*} $$

Thus $n\leq 2d$ . Assume to the contrary that $n=2d$ ; then $i+j\geq \frac {n}{2}=d$ . Hence formula (4.1) implies that

$$ \begin{align*} 2d=n\leq \frac{3(2-c(i+j))}{4\sqrt{2}\beta}\leq \frac{3(2-cd)}{4\sqrt{2}\beta}=\frac{3}{2\sqrt{2}}. \end{align*} $$

We reach a contradiction. Thus we have $n\leq 2d-2$ .

Case 4. Assume d is odd and $\mu _x=1$ . Then by [Reference Hacking and Prokhorov31, Proposition 2.6], we know that $\rho (X)=1$ . So n is odd by Proposition 4.6. Hence $2(i+(2na-1)j)\equiv dna \mod n^2$ implies $i\equiv j\mod n$ . By a similar argument as in case 2, we know $i=j=\frac {d}{2}$ if $n\geq d+1$ , hence a contradiction.

Proof. By Proposition 4.5, we know that $\rho (X) \leq 2$ . We start with $\rho (X) = 1$ . In this case, by Proposition 4.6 we know that X is a weighted projective space of the form $\mathbb {P}\left(a^2, b^2, 2c^2\right)$ , where $a^2 + b^2 + 2c^2 = 4abc$ , or a partial smoothing. We begin enumerating the possible integer solutions and see that the first few are

$$ \begin{align*}( a,b,c) = (1,1,1), (1,3, 1), (1, 3, 5), (11, 3, 5), \dots.\end{align*} $$

We can exclude the last two (and any with higher index) by the index bound of Theorem 4.7. The first gives $\mathbb {P}(1,1,2)$ and the second gives $\mathbb {P}(1,2,9)$ . We now show that the singularity $\frac {1}{9}(1,2)$ cannot appear.

Assume to the contrary that $x\in X$ is of type $\frac {1}{9}(1,2)$ . Suppose $D \sim -2K_X$ and consider a smooth covering $\left(\tilde {x} \in \widetilde {X}\right) \to (x \in X)$ . Note that we may assume $x \in D$ , because otherwise $\mathrm {ind}(x,K_X) \leq 2$ , and we obtain a contradiction. Consider local coordinates of $\tilde {x} \in \widetilde {X}$ , namely $(u,v)$ . Let $u^i v^j$ be a monomial appearing in the equation on $\widetilde {D}$ with minimum $i+j=\mathrm {ord}_{\tilde {x}}\widetilde {D}$ . Then $i + 2j \equiv 6 \mod 9$ . Since we know that $(X, cD)$ is klt at x, we have that

$$ \begin{align*} \frac{2}{i+j} \geq \mathrm{lct}\left(\widetilde{D}\right)> c,\end{align*} $$

and so in particular $i + j < \frac {2}{c}$ . By formula (4.1) with $n=3$ and $\beta =1-2c$ , we have

$$ \begin{align*}2-(i+j)c \geq 4\sqrt{2}(1-2c).\end{align*} $$

Since this inequality holds for some $0 < c < \frac {1}{2}$ , we have $i+j\leq 3$ , because otherwise

$$ \begin{align*} 2-(i+j)c\leq 2-4c<4\sqrt{2}(1-2c), \end{align*} $$

which contradicts the previous inequality. Putting this together with $i + 2j = 6 \mod 9$ , we see that $(i, j) = (0,3)$ .

Consider the valuation w on $\widetilde {X}$ , which is the monomial valuation in the coordinates $(u,v)$ of weights $(1,2)$ . In particular, $w\left(\widetilde {D}\right) = 6$ . Moreover, $A_{\widetilde {X}}(w) = 3$ and $\mathrm {vol}(w) = \frac {1}{2}$ . Then we note that

$$ \begin{align*} \widehat{\mathrm{vol}}\left(\tilde{x}, \widetilde{X}, c\widetilde{D}\right) \leq \left(A_{\widetilde{X}}(w) - c~w\left(\widetilde{D}\right)\right)^2\mathrm{vol}(w) = \frac{(3-6c)^2}{2}. \end{align*} $$

By formula (4.1) we have

$$ \begin{align*} 4\sqrt{2}(1-2c)\leq \sqrt{\widehat{\mathrm{vol}}\left(\tilde{x}, \widetilde{X}, c\widetilde{D}\right)}\leq \frac{3-6c}{\sqrt{2}}, \end{align*} $$

which gives $4\sqrt {2} \leq \frac {3}{\sqrt {2}}$ , a contradiction. Thus the surface X with a $\frac {1}{9}(1,2)$ singularity cannot appear. In particular, the only surface with $\rho (X) = 1$ is $X \cong \mathbb {P}(1,1,2)$ .

Now we consider $\rho (X) = 2$ . By Proposition 4.6, we know that the only singular points of X are of the form $\frac {1}{n^2}(1, na-1)$ , with $n\leq 5$ . We already excluded $\frac {1}{9}(1,2)$ , so we only need to consider $n = 2, 4, 5$ .

Let us consider $n=4$ , namely a singularity of type $\frac {1}{16}(1, 3)$ . We show that this singularity cannot occur. As before, consider a smooth covering $\left(\tilde {x} \in \widetilde {X}\right) \to (x \in X)$ and suppose $D \sim -2K_X$ . Note that we may assume $x \in D$ , because otherwise $\mathrm {ind}(x, K_X) \leq 2$ , and we obtain a contradiction. Consider local coordinates of $\tilde {x} \in \widetilde {X}$ , namely $(u,v)$ . Let $u^i v^j$ be a monomial appearing in the equation on $\widetilde {D}$ with minimum $i+j=\mathrm {ord}_{\tilde {x}}\widetilde {D}$ . Then $i + 3j \equiv 8 \mod 16$ , and $i + j < \frac {2}{c}$ . By formula (4.1) with $n=4$ and $\beta =1-2c$ , we have

$$ \begin{align*} 4 \sqrt{2} (1-2c) \leq (2-c(i+j)).\end{align*} $$

Since this inequality holds for some $0 < c < \frac {1}{2}$ , we have $i+j\leq 3$ by the same reason in $n=3$ . This contradicts $i+3j\equiv 8\mod 16$ . In particular, a singularity of type $\frac {1}{16}(1,3)$ cannot occur.

Next let us consider $n=5$ , namely a singularity of type $\frac {1}{25}(1,4)$ or $\frac {1}{25}(1,9)$ . We again show that these singularities cannot occur. With the same setup as the previous paragraph, we have either $i+4j\equiv 10 \mod 25$ or $i+9j\equiv 20 \mod 25$ . Moreover, we again have $i+j\leq 3$ by the same reason in $n=3,4$ , but this contradicts the congruence equations. Therefore, a singularity of type $\frac {1}{25}(1,4)$ or $\frac {1}{25}(1,9)$ cannot occur.

After these discussions, the only case left to study is $\rho (X)=2$ where X has only singularities of type $\frac {1}{4}(1,1)$ . If X is singular, then by [Reference Nakayama62, Table 6 and Theorem 7.15] (see also [Reference Alexeev and Nikulin2]), we know that X is isomorphic to a blowup of $\mathbb {P}(1,1,4)$ at a smooth point. However, in this case X admits a $\mathbb {Q}$ -Gorenstein smoothing to the Hirzebruch surface $\mathbb {F}_1$ , which is not homeomorphic to $\mathbb {P}^1\times \mathbb {P}^1$ . This is a contradiction. Hence X is smooth and isomorphic to $\mathbb {P}^1\times \mathbb {P}^1$ .

Proof. By Proposition 4.5, $\rho (X) \leq 2$ . By the index bound of Theorem 4.7, for $ c < \frac {4 - \sqrt {2}}{2d}$ we know that $\mathrm {ind}(x, K_X) < 3$ . If $\rho (X)=1$ , then by Proposition 4.6 we know that X is Gorenstein, which implies that $X\cong \mathbb {P}(1,1,2)$ . If $\rho (X)=2$ , then by Proposition 4.6 we know that either X is smooth, hence isomorphic to $\mathbb {P}^1\times \mathbb {P}^1$ , or X has only singularities of type $\frac {1}{4}(1,1)$ . The latter case cannot happen, by the end of the proof of Theorem 4.8. Therefore, the only surfaces appearing are $\mathbb {P}^1 \times \mathbb {P}^1$ and $\mathbb {P}(1,1,2)$ .

Proof. We first show that the K-(poly/semi)stability of $\left(\mathbb {P}^1\times \mathbb {P}^1, cC\right)$ implies GIT-(poly/semi)stability of C for any $c\in \left(0,\frac {1}{2}\right)$ . Consider the universal family $\pi : \left(\mathbb {P}^1 \times \mathbb {P}^1 \times \mathbf {P}_{\left(4,4\right)}, c\mathcal C\right) \to \mathbf {P}_{\left(4,4\right)}$ over the parameter space of $(4,4)$ curves on $\mathbb {P}^1\times \mathbb {P}^1$ . It is clear that $\mathcal C \in \lvert \mathcal {O}(4,4,1)\rvert $ . Hence by Proposition 2.17 we have

$$ \begin{align*} \lambda_{\mathrm{CM}, \pi, c\mathcal C}& = -\pi_* \left(-K_{\mathbb{P}^1 \times \mathbb{P}^1 \times \mathbf{P}_{\left(4,4\right)}/\mathbf{P}_{\left(4,4\right)}}-c\mathcal C\right)^3 = -\pi_* (\mathcal{O}(2-4c,2-4c,-c))^3 \\ & = -3\left(\mathcal{O}_{\mathbb{P}^1\times\mathbb{P}^1}(2-4c,2-4c)^2\right) \mathcal{O}_{\mathbf{P}_{\left(4,4\right)}}(-c) = \mathcal{O}_{\mathbf{P}_{\left(4,4\right)}}\left(3(2-4c)^2 c\right). \end{align*} $$

Hence the CM line bundle $\lambda _{\mathrm {CM}, \pi , c\mathcal C}$ is ample whenever $c\in \left(0, \frac {1}{2}\right)$ . Hence the statement of K implying GIT directly follows from Theorem 2.16.

Next we show the converse – that is, that the GIT-(poly/semi)stability of C implies K-(poly/semi)stability of $\left(\mathbb {P}^1\times \mathbb {P}^1,cC\right)$ for $c< \frac {1}{8}$ . Indeed, using a similar argument as the proof of [Reference Ascher, DeVleming and Liu6, Theorem 5.2], with a key ingredient from properness of K-moduli spaces, it suffices to show that any pair $(X,D)$ appearing in the K-moduli stack $\overline {\mathcal {K}}_c$ for $c < \frac {1}{8}$ satisfies the conditions that $X\cong \mathbb {P}^1\times \mathbb {P}^1$ and D is a $(4,4)$ curve. Since $\mathbb {P}^1\times \mathbb {P}^1$ has no nontrivial smooth degeneration, it suffices to show that X is smooth. Assume to the contrary that X is singular at a point $x\in X$ . Then by [Reference Li and Liu51] we know that

$$ \begin{align*} 8(1-2c)^2=(-K_X-cD)^2\leq \frac{9}{4}\widehat{\mathrm{vol}}(x,X,cD)\leq \frac{9}{4}\widehat{\mathrm{vol}}(x,X)\leq \frac{9}{2}. \end{align*} $$

This implies that $c\geq \frac {1}{8}$ , which is a contradiction. Hence for $c < \frac {1}{8}$ , a K-semistable pair $(X,cD)$ must be isomorphic to $\left(\mathbb {P}^1 \times \mathbb {P}^1, cC\right)$ , where C is a $(4,4)$ curve.

Summing up, the equivalence of K-(poly/semi)stability with GIT-(poly/semi)stability yields a morphism $\phi : \mathscr {M}\to \overline {\mathcal {K}}_{c}$ which descends to an isomorphism $\mathfrak {M}\xrightarrow {\cong }\overline {K}_c$ . To conclude, it suffices to show that $\phi $ is an isomorphism between Artin stacks. The proof is similar to [Reference Ascher, DeVleming and Liu6, Theorem 3.24]. Denote $T:=\mathbf {P}_{4,4}^{\textrm {ss}}$ . Let $\pi :(\mathcal X,\mathcal D)\to T$ be the universal family. Recall from [Reference Ascher, DeVleming and Liu6, Section 3.1] and Theorem 2.21 that $\overline {\mathcal {K}}_c\cong \left[Z_{c}^\circ /\mathrm {PGL}(N_m+1)\right]$ , where $Z_{c}^\circ $ is the K-semistable locus in the Hilbert scheme of embedded by mth multiple of anticanonical divisors. Denote by $\pi ': (\mathcal X',\mathcal D')\to T'$ the universal family over $T':=Z_c^\circ $ . Let P be the $\mathrm {PGL}(N_m+1)$ -torsor over T induced from the vector bundle $\pi _*\mathcal {O}_{\mathcal X}\left(-mK_{\mathcal X/T}\right)$ . Then from [Reference Ascher, DeVleming and Liu6, Proof of Theorem 3.24] we see that there is an $\mathrm {Aut}\left(\mathbb {P}^1\times \mathbb {P}^1\right)$ -equivariant morphism $\psi : P\to T'$ whose descent is precisely $\phi $ . Hence, to show that $\phi $ is isomorphic it suffices to show that $\psi $ provides an $\mathrm {Aut}\left(\mathbb {P}^1\times \mathbb {P}^1\right)$ -torsor. Indeed, since $\pi ':\mathcal X'\to T'$ is isotrivial where all fibers are isomorphic to $\mathbb {P}^1\times \mathbb {P}^1$ , we may find an étale covering $\cup _i V_i\twoheadrightarrow T'$ such that there is an isomorphism $\rho _i: \mathcal X'\times _{T'} V_i\xrightarrow {\cong } \left(\mathbb {P}^1\times \mathbb {P}^1\right)\times V_i$ . Hence by pushing forward $(\mathcal X',\mathcal D')\times _{T'} V_i$ and its natural frame from $\mathbb {P}^{N_m+1}$ to $\left(\mathbb {P}^1\times \mathbb {P}^1\right)\times V_i$ under $\rho _i$ , we obtain a section $V_i\to P\times _{T'}V_i$ of $\psi \times _{T'}V_i$ which trivializes $\psi $ . Thus the proof is finished.

Proof. We first show that $\left(\mathbb {P}^1\times \mathbb {P}^1,\frac {1}{8}C\right)$ is K-semistable where $\left(\mathbb {P}(1,1,2), \frac {1}{8}C_0\right)$ is its K-polystable degeneration. Choose an embedding $\mathbb {P}^1\times \mathbb {P}^1\hookrightarrow \mathbb {P}^3$ as a smooth quadric surface. Then H is a hyperplane section of $\mathbb {P}^1\times \mathbb {P}^1$ . Pick projective coordinates $[x_0,x_1,x_2,x_3]$ of $\mathbb {P}^3$ such that the hyperplane section through H is given by $x_3=0$ . Then the $1$ -PS $\sigma : \mathbb {G}_m \to \mathrm {PGL}(4)$ given by $\sigma (t)[x_0,x_1,x_2,x_3]=[tx_0,tx_1,tx_2,x_3]$ provides a special test configuration of $\left(\mathbb {P}^1\times \mathbb {P}^1, \frac {1}{2}H\right)$ whose central fiber is an ordinary quadric cone with a section at infinity of coefficient $\frac {1}{2}$ – that is, isomorphic to $\left(\mathbb {P}(1,1,2), \frac {1}{2}H_0\right)$ . By [Reference Li and Liu51] we know that $\left(\mathbb {P}(1,1,2), \frac {1}{2}H_0\right)$ admits a conical Kähler–Einstein metric, hence is K-polystable. The K-semistability of $\left(\mathbb {P}^1\times \mathbb {P}^1,\frac {1}{8}C\right)$ follows from the openness of K-semistability [Reference Blum, Liu and Xu12, Reference Xu15].

Next we show that $\left(\mathbb {P}^1\times \mathbb {P}^1, cC\right)$ is K-polystable for $c\in \left(0,\frac {1}{8}\right)$ . Clearly, it is K-semistable by interpolation [Reference Ascher, DeVleming and Liu6, Proposition 2.13]. Let $(X,cD)$ be its K-polystable degeneration. By Theorem 5.2, we know that $X\cong \mathbb {P}^1\times \mathbb {P}^1$ . Since $C=4H$ , we have $D=4H_0$ for some $(1,1)$ -curve $H_0$ . If $H_0$ is reducible, then $(X,cD)$ is isomorphic to the self-product of $\left(\mathbb {P}^1, c[0]\right)$ . Since $\left(\mathbb {P}^1, c[0]\right)$ is K-unstable, we know that $(X,cD)$ is also K-unstable by [Reference Zhuang78]. Thus $H_0$ must be irreducible, which implies that $\left(\mathbb {P}^1\times \mathbb {P}^1, cC\right)\cong (X,cD)$ is K-polystable. Thus the proof is finished.