Proof. The proof goes along the same line as [Reference FujitaFuj15, Theorem 2.3].

(i) The equality $((\unicode[STIX]{x1D70E}^{\ast }L-xF)^{n})=(L^{n})-\operatorname{mult}_{Z}X\cdot x^{n}$ follows from the fact that $\operatorname{mult}_{Z}X=(-1)^{n-1}(F^{n})$ (see [Reference RamanujamRam73]).

For the inequality part, we can assume that $x\in \mathbb{Q}_{{>}0}$ since the function $\operatorname{vol}_{\hat{X}}(\unicode[STIX]{x1D70E}^{\ast }L-xF)$ is continuous. Take any sufficiently large $k\in \mathbb{Z}_{{>}0}$ with $kx\in \mathbb{Z}_{{>}0}$ and $kL$ Cartier. Notice that we have the long exact sequence

(2.1) $$\begin{eqnarray}\displaystyle 0\rightarrow H^{0}(\hat{X},\unicode[STIX]{x1D70E}^{\ast }(kL)-kxF)\rightarrow H^{0}(\hat{X},\unicode[STIX]{x1D70E}^{\ast }(kL))\rightarrow H^{0}(kxF,\unicode[STIX]{x1D70E}^{\ast }(kL)|_{kxF})\rightarrow \cdots & & \displaystyle\end{eqnarray}$$

Thus

$$\begin{eqnarray}\displaystyle h^{0}(\hat{X},\unicode[STIX]{x1D70E}^{\ast }(kL)-kxF) & {\geqslant} & \displaystyle h^{0}(\hat{X},\unicode[STIX]{x1D70E}^{\ast }(kL))-h^{0}(kxF,\unicode[STIX]{x1D70E}^{\ast }(kL)|_{kxF})\nonumber\\ \displaystyle & = & \displaystyle h^{0}(X,kL)-h^{0}(kxF,{\mathcal{O}}_{kxF}).\nonumber\end{eqnarray}$$

Consequently,

$$\begin{eqnarray}\operatorname{vol}_{\hat{X}}(\unicode[STIX]{x1D70E}^{\ast }L-xF)\geqslant \operatorname{vol}_{X}(L)-\lim _{k\rightarrow \infty }\frac{h^{0}(kxF,{\mathcal{O}}_{kxF})}{(kx)^{n}/n!}x^{n}.\end{eqnarray}$$

Hence it suffices to show that

$$\begin{eqnarray}\lim _{j\rightarrow \infty }\frac{h^{0}(jF,{\mathcal{O}}_{jF})}{j^{n}/n!}=\operatorname{mult}_{Z}X.\end{eqnarray}$$

We look at the exact sequence

(2.2) $$\begin{eqnarray}\displaystyle 0\rightarrow {\mathcal{O}}_{\hat{X}}(-jF)|_{F}\rightarrow {\mathcal{O}}_{(j+1)F}\rightarrow {\mathcal{O}}_{jF}\rightarrow 0. & & \displaystyle\end{eqnarray}$$

The isomorphism $F\cong \mathbf{Proj}\bigoplus _{m\geqslant 0}I_{Z}^{m}/I_{Z}^{m+1}$ yields ${\mathcal{O}}_{\hat{X}}(-jF)|_{F}\cong {\mathcal{O}}_{F}(j)$ . Since $H^{1}(F,{\mathcal{O}}_{F}(j))=0$ for $j\gg 0$ , we have an exact sequence

$$\begin{eqnarray}0\rightarrow H^{0}(F,{\mathcal{O}}_{F}(j))\rightarrow H^{0}((j+1)F,{\mathcal{O}}_{(j+1)F})\rightarrow H^{0}(jF,{\mathcal{O}}_{jF})\rightarrow 0.\end{eqnarray}$$

Hence $h^{0}((j+1)F,{\mathcal{O}}_{(j+1)F})-h^{0}(jF,{\mathcal{O}}_{jF})=h^{0}(F,{\mathcal{O}}_{F}(j))=\ell (I_{Z}^{j}/I_{Z}^{j+1})$ for $j\gg 0$ . As a result,

$$\begin{eqnarray}h^{0}(jF,{\mathcal{O}}_{jF})=\ell ({\mathcal{O}}_{X}/I_{Z}^{j})+\text{const}.\end{eqnarray}$$

for $j\gg 0$ . Hence we prove (i).

(ii) Assume $\operatorname{vol}_{\hat{X}}(\unicode[STIX]{x1D70E}^{\ast }L-xF)=(L^{n})-\operatorname{mult}_{Z}X\cdot x^{n}$ for some $x\in \mathbb{Q}_{{>}0}$ . We will show that

(2.3) $$\begin{eqnarray}h^{i}(\hat{X},\unicode[STIX]{x1D70E}^{\ast }(kL)-kxF)=o(k^{n})\quad \text{for }i\geqslant 1,k\gg 0\text{ with }kx\in \mathbb{Z}_{{>}0}.\end{eqnarray}$$

For $i\geqslant 2$ , the long exact sequence (2.1) yields

$$\begin{eqnarray}h^{i}(\hat{X},\unicode[STIX]{x1D70E}^{\ast }(kL)-kxF)\leqslant h^{i-1}(kxF,{\mathcal{O}}_{kxF})+h^{i}(\hat{X},\unicode[STIX]{x1D70E}^{\ast }(kL)).\end{eqnarray}$$

Since $\unicode[STIX]{x1D70E}^{\ast }(kL)$ is nef, we know that $h^{i}(\hat{X},\unicode[STIX]{x1D70E}^{\ast }(kL))=o(k^{n})$ . Serre vanishing theorem implies that $H^{i-1}(F,{\mathcal{O}}_{F}(j))=0$ for $i\geqslant 2$ , $j\gg 0$ . Then the exact sequence (2.2) implies that

$$\begin{eqnarray}H^{i-1}((j+1)F,{\mathcal{O}}_{(j+1)F})\cong H^{i-1}(jF,{\mathcal{O}}_{jF})\end{eqnarray}$$

for $i\geqslant 2$ , $j\gg 0$ . As a result, $h^{i-1}(kxF,{\mathcal{O}}_{kxF})$ is constant for $k\gg 0$ . Therefore, $h^{i}(\hat{X},\unicode[STIX]{x1D70E}^{\ast }(kL)-kxF)=o(k^{n})$ for any $i\geqslant 2$ .

By the asymptotic Riemann–Roch theorem we know that

$$\begin{eqnarray}\displaystyle \lim _{k\rightarrow \infty }\frac{\unicode[STIX]{x1D712}(\hat{X},\unicode[STIX]{x1D70E}^{\ast }(kL)-kxF)}{k^{n}/n!} & = & \displaystyle ((\unicode[STIX]{x1D70E}^{\ast }L-xF)^{n})\nonumber\\ \displaystyle & = & \displaystyle (L^{n})-\operatorname{mult}_{Z}X\cdot x^{n}\nonumber\\ \displaystyle & = & \displaystyle \operatorname{vol}_{\hat{X}}(\unicode[STIX]{x1D70E}^{\ast }L-xF).\nonumber\end{eqnarray}$$

Hence we have

$$\begin{eqnarray}\lim _{k\rightarrow \infty }\frac{1}{k^{n}}\mathop{\sum }_{i=1}^{n}(-1)^{i}h^{i}(\hat{X},\unicode[STIX]{x1D70E}^{\ast }(kL)-kxF)=0.\end{eqnarray}$$

Since $h^{i}(\hat{X},\unicode[STIX]{x1D70E}^{\ast }(kL)-kxF)=o(k^{n})$ for $i\geqslant 2$ , we conclude that $h^{1}(\hat{X},\unicode[STIX]{x1D70E}^{\ast }(kL)-kxF)=o(k^{n})$ as well.

Now let us prove (ii). Denote by $a$ the right-hand side of the equation in (ii). For any nef divisor, its volume is equal to its self intersection number. Hence it is obvious that $\unicode[STIX]{x1D716}_{Z}(L)\leqslant a$ . Then it suffices to show that $a\leqslant \unicode[STIX]{x1D716}_{Z}(L)$ . Equivalently, we only need to show that for $\unicode[STIX]{x1D716}>0$ sufficiently small with $a-\unicode[STIX]{x1D716}\in \mathbb{Q}_{{>}0}$ , $\unicode[STIX]{x1D70E}^{\ast }L-(a-\unicode[STIX]{x1D716})F$ is ample. Fix $\unicode[STIX]{x1D6FF}\in \mathbb{Q}_{{>}0}$ such that $\unicode[STIX]{x1D6FF}<\unicode[STIX]{x1D716}_{Z}(L)$ , that is, $\unicode[STIX]{x1D70E}^{\ast }L-\unicode[STIX]{x1D6FF}F$ is ample. By [Reference de Fernex, Küronya and LazarsfelddFKL07, Theorem A], we only need to show that

(2.4) $$\begin{eqnarray}h^{i}(\hat{X},m((\unicode[STIX]{x1D70E}^{\ast }L-(a-\unicode[STIX]{x1D716})F)-t(\unicode[STIX]{x1D70E}^{\ast }L-\unicode[STIX]{x1D6FF}F)))=o(m^{n})\end{eqnarray}$$

for any $i>0$ , any sufficiently small $t\in \mathbb{Q}_{{>}0}$ and suitably divisible $m$ . Denote $b:=(a-\unicode[STIX]{x1D716}+t\unicode[STIX]{x1D6FF})/(1-t)$ . Notice that

$$\begin{eqnarray}\frac{1}{1-t}((\unicode[STIX]{x1D70E}^{\ast }L-(a-\unicode[STIX]{x1D716})F)-t(\unicode[STIX]{x1D70E}^{\ast }L-\unicode[STIX]{x1D6FF}F))=\unicode[STIX]{x1D70E}^{\ast }L-\frac{a-\unicode[STIX]{x1D716}+t\unicode[STIX]{x1D6FF}}{1-t}F=\unicode[STIX]{x1D70E}^{\ast }L-bF.\end{eqnarray}$$

It is clear that $b<a$ for $t\in \mathbb{Q}_{{>}0}$ sufficiently small. Hence from the description of $a$ we get:

$$\begin{eqnarray}\operatorname{vol}_{\hat{X}}(\unicode[STIX]{x1D70E}^{\ast }L-bF)=(L^{n})-\operatorname{mult}_{Z}X\cdot b^{n}.\end{eqnarray}$$

Then applying (2.3) yields that $h^{i}(\hat{X},\unicode[STIX]{x1D70E}^{\ast }(kL)-kbF)=o(k^{n})$ for $i>0$ , sufficiently small $t\in \mathbb{Q}_{{>}0}$ and suitably divisible $k$ . In other words, (2.4) is true under the same condition for $i$ , $t$ and $m=(1-t)k$ . Hence we prove the lemma.◻

Proof. Part (i) is proved by Berman in [Reference BermanBer16] (see also [Reference FujitaFuj15, Theorem 3.2]).

In (ii), the direction that Ding-semistability implies K-semistability is proved by Berman in [Reference BermanBer16] using the inequality that $\text{Ding}({\mathcal{X}},{\mathcal{L}})\geqslant \text{CM}({\mathcal{X}},{\mathcal{L}})$ for any normal test configuration $({\mathcal{X}},{\mathcal{L}})$ (see also [Reference FujitaFuj15, Theorem 3.2.2]). The direction that K-semistability implies Ding-semistability is proved in [Reference Berman, Boucksom and JonssonBBJ15] based on [Reference Li and XuLX14] (see also [Reference FujitaFuj16, § 3]).◻

Proof. Let $\unicode[STIX]{x1D70E}:\hat{X}\rightarrow X$ be the blow up along $Z$ , and let $F\subset \hat{X}$ be the Cartier divisor defined by the equation ${\mathcal{O}}_{\hat{X}}(-F)=\unicode[STIX]{x1D70E}^{-1}I_{Z}\cdot {\mathcal{O}}_{\hat{X}}$ . By Theorem 15, we have

$$\begin{eqnarray}\operatorname{lct}(X;I_{Z})\cdot ((-K_{X})^{n})\geqslant \int _{0}^{+\infty }\operatorname{vol}_{\hat{X}}(\unicode[STIX]{x1D70E}^{\ast }(-K_{X})-xF)\,dx.\end{eqnarray}$$

On the other hand, by Lemma 10, we have

$$\begin{eqnarray}\displaystyle \int _{0}^{+\infty }\operatorname{vol}_{\hat{X}}(\unicode[STIX]{x1D70E}^{\ast }(-K_{X})-xF)\,dx & {\geqslant} & \displaystyle \int _{0}^{\sqrt[n]{((-K_{X})^{n})/\operatorname{mult}_{Z}X}}(((-K_{X})^{n})-\operatorname{mult}_{Z}X\cdot x^{n})\,dx\nonumber\\ \displaystyle & = & \displaystyle \frac{n}{n+1}((-K_{X}))^{n}\cdot \sqrt[n]{((-K_{X})^{n})/\operatorname{mult}_{Z}X}.\nonumber\end{eqnarray}$$

Hence we get the desired inequality

$$\begin{eqnarray}((-K_{X})^{n})\leqslant \biggl(1+\frac{1}{n}\biggr)^{n}\operatorname{lct}(X;I_{Z})^{n}\operatorname{mult}_{Z}X.\Box\end{eqnarray}$$

Proof. The decreasing, left-continuous and multiplicative properties of ${\mathcal{F}}_{v}$ follows from the corresponding properties of $\{\mathfrak{a}_{x}(v)\}_{x\in \mathbb{R}}$ . To prove that ${\mathcal{F}}_{v}$ is saturated, notice that the homomorphism

$$\begin{eqnarray}{\mathcal{F}}_{v}^{x}S_{m}\otimes _{\mathbb{C}}L^{-m}{\twoheadrightarrow}I_{(m,x)}^{{\mathcal{F}}_{v}}\end{eqnarray}$$

induces the inclusion $I_{(m,x)}^{{\mathcal{F}}_{v}}\subset \mathfrak{a}_{x}(v)$ for any $x\in \mathbb{R}$ . Thus $\overline{{\mathcal{F}}}^{x}S_{m}=H^{0}(X,L^{m}\cdot I_{(m,x)}^{{\mathcal{F}}_{v}})\subset {\mathcal{F}}^{x}S_{m}$ , which implies that ${\mathcal{F}}_{v}$ is saturated.

If $A_{X}(v)<+\infty$ , then by [Reference LiLi15, Theorem 3.1] there exists a constant $c=c(X,p)$ such that $v\leqslant cA_{X}(v)\text{ord}_{p}$ . So it is easy to see that

$$\begin{eqnarray}e_{\max }(S_{\bullet },{\mathcal{F}}_{v})\leqslant cA_{X}(v)\cdot e_{\max }(S_{\bullet },{\mathcal{F}}_{\text{ord}_{p}})<+\infty .\end{eqnarray}$$

The second inequality follows from the work in [Reference Boucksom, Küronya, Maclean and SzembergBKMS15]. Next, it is clear that $\mathfrak{a}_{x}(v)={\mathcal{O}}_{X}$ for $x\leqslant 0$ . Hence ${\mathcal{F}}_{v}^{x}S_{m}=S_{m}$ for $x\leqslant 0$ , which implies $e_{\min }(S_{\bullet },{\mathcal{F}}_{v})\geqslant 0$ . Hence ${\mathcal{F}}_{v}$ is linearly bounded.◻

Proof. We may assume that $A_{X}(v)<+\infty$ , since otherwise the inequality holds automatically. Hence Lemma 19 implies that ${\mathcal{F}}_{v}$ is good and saturated.

By Theorem 18, we know that $(X\times \mathbb{A}^{1},{\mathcal{I}}_{\bullet }^{\cdot (1/r)}\cdot (t)^{\cdot d_{\infty }})$ is sub log canonical. Since $X\times \mathbb{A}^{1}$ has klt singularities, for any $0<\unicode[STIX]{x1D716}\ll 1$ there exists $m=m(\unicode[STIX]{x1D716})$ such that

$$\begin{eqnarray}{\mathcal{O}}_{X\times \mathbb{A}^{1}}\subset {\mathcal{J}}(X\times \mathbb{A}^{1},{\mathcal{I}}_{m}^{\cdot (1-\unicode[STIX]{x1D716})/(rm)}\cdot (t)^{\cdot (1-\unicode[STIX]{x1D716})d_{\infty }}),\end{eqnarray}$$

where the right-hand side is the multiplier ideal sheaf. By [Reference Boucksom, de Fernex, Favre and UrbinatiBdFFU15, Theorem 1.2] we know that the following inequality holds for any real valuation $u$ on $X\times \mathbb{A}^{1}$ :

(3.4) $$\begin{eqnarray}A_{X\times \mathbb{A}^{1}}(u)>\frac{(1-\unicode[STIX]{x1D716})}{rm}u({\mathcal{I}}_{m})+(1-\unicode[STIX]{x1D716})d_{\infty }u(t).\end{eqnarray}$$

Let us choose a sequence of quasi-monomial real valuations $\{v_{n}\}$ on $X$ such that $v_{n}\rightarrow v$ and $A_{X}(v_{n})\rightarrow A_{X}(v)$ when $n\rightarrow \infty$ . It is easy to see that $\{\bar{v}_{n}\}$ is a sequence of quasi-monomial valuations on $X\times \mathbb{A}^{1}$ satisfying $\bar{v}_{n}(t)=1$ and

$$\begin{eqnarray}A_{X\times \mathbb{A}^{1}}(\bar{v}_{n})=A_{X}(v_{n})+1.\end{eqnarray}$$

Hence we have

$$\begin{eqnarray}\displaystyle A_{X}(v)+1 & = & \displaystyle \lim _{n\rightarrow \infty }A_{X}(\bar{v}_{n})\nonumber\\ \displaystyle & {\geqslant} & \displaystyle \frac{(1-\unicode[STIX]{x1D716})}{rm}\lim _{n\rightarrow \infty }\bar{v}_{n}({\mathcal{I}}_{m})+(1-\unicode[STIX]{x1D716})d_{\infty }.\nonumber\end{eqnarray}$$

From the definition of ${\mathcal{F}}_{v}^{x}S_{m}$ we get:

$$\begin{eqnarray}v(I_{(m,x)}^{{\mathcal{F}}_{v}})\geqslant x.\end{eqnarray}$$

Therefore,

$$\begin{eqnarray}\displaystyle \lim _{n\rightarrow \infty }\bar{v}_{n}({\mathcal{I}}_{m}) & = & \displaystyle \lim _{n\rightarrow \infty }\min _{0\leqslant j\leqslant re_{+}}\{v_{n}(I_{(m,j)}^{{\mathcal{F}}_{v}})+me_{+}-j\}\nonumber\\ \displaystyle & = & \displaystyle \min _{0\leqslant j\leqslant re_{+}}\{\lim _{n\rightarrow \infty }v_{n}(I_{(m,j)}^{{\mathcal{F}}_{v}})+me_{+}-j\}\nonumber\\ \displaystyle & = & \displaystyle \min _{0\leqslant j\leqslant re_{+}}\{v(I_{(m,j)}^{{\mathcal{F}}_{v}})+me_{+}-j\}\nonumber\\ \displaystyle & {\geqslant} & \displaystyle me_{+}.\nonumber\end{eqnarray}$$

Hence when $\unicode[STIX]{x1D716}\rightarrow 0+$ we get:

(3.5) $$\begin{eqnarray}A_{X}(v)+1\geqslant \frac{e_{+}}{r}+d_{\infty }.\end{eqnarray}$$

Therefore,

$$\begin{eqnarray}A_{X}(v)\geqslant -1+\frac{e_{+}}{r}+d_{\infty }=\frac{1}{r^{n+1}((-K_{X})^{n})}\int _{0}^{+\infty }\operatorname{vol}({\mathcal{F}}_{v}S^{(t)})\,dt.\end{eqnarray}$$

Hence we get the desired inequality. ◻

Proof. We may assume $A_{X}(v)<+\infty$ , since otherwise the inequality holds automatically. Since ${\mathcal{F}}_{v}^{mx}S_{m}=H^{0}(X,L^{m}\cdot \mathfrak{a}_{mx}(v))$ , we have the exact sequence

$$\begin{eqnarray}0\rightarrow {\mathcal{F}}_{v}^{mx}S_{m}\rightarrow H^{0}(X,L^{m})\rightarrow H^{0}(X,L^{m}\otimes ({\mathcal{O}}_{X}/\mathfrak{a}_{mx}(v))).\end{eqnarray}$$

Hence we have

$$\begin{eqnarray}\dim {\mathcal{F}}_{v}^{mx}S_{m}\geqslant h^{0}(X,L^{m})-\ell ({\mathcal{O}}_{X}/\mathfrak{a}_{mx}(v)).\end{eqnarray}$$

This implies

(3.6) $$\begin{eqnarray}\operatorname{vol}({\mathcal{F}}_{v}S^{(x)})\geqslant (L^{n})-\operatorname{vol}(v)x^{n}.\end{eqnarray}$$

So we have the estimate:

$$\begin{eqnarray}\displaystyle \int _{0}^{+\infty }\operatorname{vol}({\mathcal{F}}_{v}S^{(x)})\,dx & {\geqslant} & \displaystyle \int _{0}^{\sqrt[n]{(L^{n})/\operatorname{vol}(v)}}((L^{n})-\operatorname{vol}(v)x^{n})\,dx\nonumber\\ \displaystyle & = & \displaystyle \frac{n}{n+1}(L^{n})\cdot \sqrt[n]{(L^{n})/\operatorname{vol}(v)}.\nonumber\end{eqnarray}$$

Applying (3.3) we get the inequality:

$$\begin{eqnarray}\displaystyle A_{X}(v) & {\geqslant} & \displaystyle \frac{1}{r(L^{n})}\int _{0}^{+\infty }\operatorname{vol}({\mathcal{F}}_{v}S^{(t)})\,dt\nonumber\\ \displaystyle & {\geqslant} & \displaystyle \frac{1}{r(L^{n})}\frac{n}{n+1}(L^{n})\sqrt[n]{(L^{n})/\operatorname{vol}(v)}=\frac{n}{n+1}\sqrt[n]{((-K_{X})^{n})/\operatorname{vol}(v)}.\nonumber\end{eqnarray}$$

This is exactly equivalent to

$$\begin{eqnarray}((-K_{X})^{n})\leqslant \biggl(1+\frac{1}{n}\biggr)^{n}A_{X}(v)^{n}\operatorname{vol}(v)=\biggl(1+\frac{1}{n}\biggr)^{n}\,\widehat{\operatorname{vol}}(v).\Box\end{eqnarray}$$

Proof. It is clear that the multiplicity is preserved under completion. Next, [Reference de Fernex, Ein and MustaţădFEM11, Proposition 2.11] implies that $\operatorname{lct}$ is also preserved under completion. Hence it suffices to prove the theorem for $X=\mathbb{P}^{n}$ . This follows directly by applying Theorem 1 to the case $X=\mathbb{P}^{n}$ .◻

Proof. Let $(Y,o):=(\mathbb{C}^{n}/G,0)$ be the local analytic model of the quotient singularity $(X,p)$ . Let $H:=G\cap \mathbb{G}_{m}$ with $d:=|H|$ . Define the real valuation $u_{0}$ on $Y$ to be the pushforward of the valuation $\text{ord}_{0}$ on $\mathbb{C}^{n}$ under the quotient map $\mathbb{C}^{n}\rightarrow Y$ .

We first show that $\widehat{\operatorname{vol}}(u_{0})=n^{n}/|G|$ . Let $\widehat{\mathbb{C}^{n}}$ be the blow up of $\mathbb{C}^{n}$ at the origin $0$ with exceptional divisor $E$ . Denote by $\unicode[STIX]{x1D70B}:\mathbb{C}^{n}\rightarrow Y$ the quotient map. Then $\unicode[STIX]{x1D70B}$ lifts to $\widehat{\mathbb{C}^{n}}$ as $\hat{\unicode[STIX]{x1D70B}}:\widehat{\mathbb{C}^{n}}\rightarrow {\hat{Y}}$ , where ${\hat{Y}}:=\widehat{\mathbb{C}^{n}}/G$ . We have the following commutative diagram:

Let $F\subset {\hat{Y}}$ be the exceptional divisor of $h$ . For a general point on $F$ , its stabilizer under the $G$ -action is exactly $H$ . So $\hat{\unicode[STIX]{x1D70B}}^{\ast }F=dE$ , which implies that $u_{0}=\unicode[STIX]{x1D70B}_{\ast }(\text{ord}_{E})=d~\text{ord}_{F}$ . It is clear that

$$\begin{eqnarray}K_{\widehat{\mathbb{C}^{n}}}=\hat{\unicode[STIX]{x1D70B}}^{\ast }\biggl(K_{{\hat{Y}}}+\biggl(1-\frac{1}{d}\biggr)F\biggr).\end{eqnarray}$$

Then combining these equalities with $K_{\widehat{\mathbb{C}^{n}}}=g^{\ast }K_{\mathbb{C}^{n}}+(n-1)E=g^{\ast }(\unicode[STIX]{x1D70B}^{\ast }K_{Y})+(n-1)E$ , we get

$$\begin{eqnarray}K_{{\hat{Y}}}=h^{\ast }K_{Y}+\biggl(\frac{n}{d}-1\biggr)F.\end{eqnarray}$$

Hence $A_{X}(\text{ord}_{F})=n/d$ and $A_{X}(u_{0})=d\cdot A_{X}(\text{ord}_{F})=n$ . Lemma 24 implies that $\operatorname{vol}(u_{0})=1/|G|$ . Therefore, $\widehat{\operatorname{vol}}(u_{0})=n^{n}/|G|$ .

Since $A_{Y}(u_{0})<+\infty$ , a singular version of [Reference Jonsson and MustaţăJM12, Corollary 5.11] for klt singularities (such a statement is true thanks to the Izumi inequality for klt singularities [Reference LiLi15, Proposition 3.1]) implies that $u_{0}$ has a unique extension, say $\hat{u} _{0}$ , as a valuation on $\mathbf{Spec}\,\widehat{{\mathcal{O}}_{Y,o}}$ that centered at the closed point. By assumption we have an isomorphism $\unicode[STIX]{x1D719}:\widehat{{\mathcal{O}}_{X,p}}\rightarrow \widehat{{\mathcal{O}}_{Y,o}}$ . Let $v_{0}:=(\unicode[STIX]{x1D719}^{\ast }\hat{u} _{0})|_{{\mathcal{O}}_{X,p}}$ , then $v_{0}\in \text{Val}_{X,p}$ . By a singular version of [Reference Jonsson and MustaţăJM12, Proposition 5.13], we know that $A_{X}(v_{0})=A_{Y}(u_{0})$ . On the other hand, since the colength of any $\mathfrak{m}$ -primary ideal does not change under completion, we have $\operatorname{vol}(v_{0})=\operatorname{vol}(\hat{u} _{0})=\operatorname{vol}(u_{0})$ . As a result, we have

$$\begin{eqnarray}\widehat{\operatorname{vol}}(v_{0})=\widehat{\operatorname{vol}}(u_{0})=\frac{n^{n}}{|G|}.\end{eqnarray}$$

Hence the theorem follows as a direct application of Theorem 2 to $v=v_{0}$ .◻

Proof. Denote $W:=\mathbb{C}[x_{1},\ldots ,x_{n}]=\mathbb{C}x_{1}\oplus \cdots \oplus \mathbb{C}x_{n}$ . Then we have that

$$\begin{eqnarray}\mathbb{C}[x_{1},\ldots ,x_{n}]\cong \bigoplus _{m\geqslant 0}\text{Sym}^{m}W.\end{eqnarray}$$

Denote by $\unicode[STIX]{x1D70C}_{m}:G\rightarrow \text{GL}(\text{Sym}^{m}W)$ the induced representation of $G$ on $\text{Sym}^{m}W$ . Since $G$ is finite, $\unicode[STIX]{x1D70C}_{m}(g)$ is diagonalizable for any $m\geqslant 0$ and $g\in G$ . Denote the $n$ eigenvalues of $\unicode[STIX]{x1D70C}_{1}(g)$ by $\unicode[STIX]{x1D706}_{g,1},\ldots ,\unicode[STIX]{x1D706}_{g,n}$ . Then the eigenvalues of $\unicode[STIX]{x1D70C}_{m}(g)$ are exactly all monomials in $\unicode[STIX]{x1D706}_{g,i}$ of degree $m$ . Let $c_{m}:=\dim _{\mathbb{C}}(\text{Sym}^{m}W)^{G}$ and $d_{m}:=\dim _{\mathbb{C}}\mathbb{C}[x_{1},\ldots ,x_{n}]_{{\leqslant}m}^{G}$ . From representation theory we know that

$$\begin{eqnarray}c_{m}=\frac{1}{|G|}\mathop{\sum }_{g\in G}\text{tr}(\unicode[STIX]{x1D70C}_{m}(g)).\end{eqnarray}$$

Hence

$$\begin{eqnarray}c(t):=\mathop{\sum }_{m=0}^{\infty }c_{m}t^{m}=\frac{1}{|G|}\mathop{\sum }_{g\in G}\frac{1}{(1-\unicode[STIX]{x1D706}_{g,1}t)\cdots (1-\unicode[STIX]{x1D706}_{g,n}t)}.\end{eqnarray}$$

This is also known as Molien’s theorem [Reference MolienMol97]. Since $d_{m}=\sum _{i=0}^{m}c_{i}$ , we have

$$\begin{eqnarray}\displaystyle d(t) & := & \displaystyle \mathop{\sum }_{m=0}^{\infty }d_{m}t^{m}=\frac{c(t)}{1-t}\nonumber\\ \displaystyle & = & \displaystyle \frac{1}{|G|}\mathop{\sum }_{g\in G}\frac{1}{(1-t)(1-\unicode[STIX]{x1D706}_{g,1}t)\cdots (1-\unicode[STIX]{x1D706}_{g,n}t)}.\nonumber\end{eqnarray}$$

Here all eigenvalues $\unicode[STIX]{x1D706}_{g,i}$ are roots of unity. Using partial fraction decomposition, for any $g\neq id$ we have

$$\begin{eqnarray}\frac{1}{(1-t)(1-\unicode[STIX]{x1D706}_{g,1}t)\cdots (1-\unicode[STIX]{x1D706}_{g,n}t)}=\mathop{\sum }_{m=0}^{\infty }o(m^{n})t^{m}.\end{eqnarray}$$

Hence

$$\begin{eqnarray}\displaystyle d(t) & = & \displaystyle \frac{1}{|G|}\cdot \frac{1}{(1-t)^{n+1}}+\mathop{\sum }_{m=0}^{\infty }o(m^{n})t^{m}\nonumber\\ \displaystyle & = & \displaystyle \mathop{\sum }_{m=0}^{\infty }\biggl(\frac{1}{|G|}\binom{n+m}{n}+o(m^{n})\biggr)t^{m}.\nonumber\end{eqnarray}$$

As a result,

$$\begin{eqnarray}\lim _{m\rightarrow \infty }\frac{d_{m-1}}{m^{n}/n!}=\lim _{m\rightarrow \infty }\frac{1/|G|\binom{n+m-1}{n}}{m^{n}/n!}=\frac{1}{|G|}.\Box\end{eqnarray}$$

Proof. Denote by $\mathfrak{m}_{p}$ the maximal ideal at $p$ . We first show that $\operatorname{lct}(X;\mathfrak{m}_{p})\leqslant 1$ . Since $p$ is an isolated singularity, we may take a log resolution of $(X,\mathfrak{m}_{p})$ , namely $\unicode[STIX]{x1D70B}:Y\rightarrow X$ such that $\unicode[STIX]{x1D70B}$ is an isomorphism away from $p$ and $\unicode[STIX]{x1D70B}^{-1}\mathfrak{m}_{p}\cdot {\mathcal{O}}_{Y}$ is an invertible ideal sheaf on $Y$ . Let $E_{i}$ be the exceptional divisors of $\unicode[STIX]{x1D70B}$ . We define the numbers $a_{i}$ and $b_{i}$ by

$$\begin{eqnarray}K_{Y}=\unicode[STIX]{x1D70B}^{\ast }K_{X}+\mathop{\sum }_{i}a_{i}E_{i}\quad \text{and}\quad \unicode[STIX]{x1D70B}^{-1}\mathfrak{m}_{p}\cdot {\mathcal{O}}_{Y}={\mathcal{O}}_{Y}\biggl(\!-\!\mathop{\sum }_{i}b_{i}E_{i}\biggr).\end{eqnarray}$$

It is clear that $\operatorname{lct}(X;\mathfrak{m}_{p})=\min _{i}((1+a_{i})/b_{i})$ . Since $\unicode[STIX]{x1D70B}$ is an isomorphism away from $p$ , we have $b_{i}\geqslant 1$ for any $i$ . Since $X$ is not terminal at $p$ , there exists an index $i_{0}$ such that $a_{i_{0}}\leqslant 0$ . Hence

$$\begin{eqnarray}\operatorname{lct}(X;\mathfrak{m}_{p})\leqslant \frac{1+a_{i_{0}}}{b_{i_{0}}}\leqslant 1.\end{eqnarray}$$

So we finish the proof by applying Theorem 1 to $Z=p$ .◻

Proof. Since $v_{0}$ computes $\operatorname{lct}(\mathfrak{a})$ , we have that $\operatorname{lct}(\mathfrak{a})=A_{X}(v_{0})/v_{0}(\mathfrak{a})$ . Denote $\unicode[STIX]{x1D6FC}:=v_{0}(\mathfrak{a})$ . Then $\mathfrak{a}^{m}\subset \mathfrak{a}_{m\unicode[STIX]{x1D6FC}}(v_{0})$ since $v_{0}(\mathfrak{a}^{m})=m\unicode[STIX]{x1D6FC}$ . Therefore,

$$\begin{eqnarray}\displaystyle \operatorname{mult}(\mathfrak{a}) & = & \displaystyle \lim _{m\rightarrow \infty }\frac{\ell ({\mathcal{O}}_{X}/\mathfrak{a}^{m})}{m^{n}/n!}\nonumber\\ \displaystyle & {\geqslant} & \displaystyle \lim _{m\rightarrow \infty }\frac{\ell ({\mathcal{O}}_{X}/\mathfrak{a}_{m\unicode[STIX]{x1D6FC}}(v_{0}))}{m^{n}/n!}\nonumber\\ \displaystyle & = & \displaystyle \unicode[STIX]{x1D6FC}\,\operatorname{vol}(v_{0}).\nonumber\end{eqnarray}$$

Hence we prove the lemma. ◻

Proof. Let $\mathfrak{a}_{\bullet }$ be a graded sequence of ideal sheaves supported at $p$ . By [Reference Jonsson and MustaţăJM12, Reference Boucksom, de Fernex, Favre and UrbinatiBdFFU15], we have

$$\begin{eqnarray}\lim _{m\rightarrow \infty }m~\operatorname{lct}(\mathfrak{a}_{m})=\operatorname{lct}(\mathfrak{a}_{\bullet }).\end{eqnarray}$$

Since $\operatorname{mult}(\mathfrak{a}_{\bullet })=\operatorname{vol}(\mathfrak{a}_{\bullet })$ , if $\operatorname{lct}(\mathfrak{a}_{\bullet })<+\infty$ then

$$\begin{eqnarray}\lim _{m\rightarrow \infty }\operatorname{lct}(\mathfrak{a}_{m})^{n}\cdot \operatorname{mult}(\mathfrak{a}_{m})=\operatorname{lct}(\mathfrak{a}_{\bullet })^{n}\cdot \operatorname{vol}(\mathfrak{a}_{\bullet }).\end{eqnarray}$$

Thus we have

(4.2) $$\begin{eqnarray}\inf _{\mathfrak{a}}(\operatorname{lct}(\mathfrak{a})^{n}\cdot \operatorname{mult}(\mathfrak{a}))\leqslant \inf _{\mathfrak{ a}_{\bullet }}(\operatorname{lct}(\mathfrak{a}_{\bullet })^{n}\cdot \operatorname{vol}(\mathfrak{a}_{\bullet })).\end{eqnarray}$$

Next, Lemma 26 implies that

(4.3) $$\begin{eqnarray}\inf _{\mathfrak{a}}(\operatorname{lct}(\mathfrak{a})^{n}\cdot \operatorname{mult}(\mathfrak{a}))\geqslant \inf _{v}\widehat{\operatorname{vol}}(v).\end{eqnarray}$$

Finally, for any real valuation $v\in \text{Val}_{X,p}$ with $A_{X}(v)<+\infty$ , by [Reference Boucksom, de Fernex, Favre and UrbinatiBdFFU15, Theorem 1.2] we have

$$\begin{eqnarray}\operatorname{lct}(\mathfrak{a}_{\bullet }(v))\leqslant \frac{A_{X}(v)}{v(\mathfrak{a}_{\bullet }(v))}=A_{X}(v)<+\infty .\end{eqnarray}$$

By definition we have $\operatorname{vol}(\mathfrak{a}_{\bullet }(v))=\operatorname{vol}(v)$ . Therefore,

(4.4) $$\begin{eqnarray}\operatorname{lct}(\mathfrak{a}_{\bullet }(v))^{n}\cdot \operatorname{vol}(\mathfrak{a}_{\bullet }(v))\leqslant A_{X}(v)^{n}\cdot \operatorname{vol}(v)=\widehat{\operatorname{vol}}(v).\end{eqnarray}$$

Hence we have

(4.5) $$\begin{eqnarray}\inf _{\mathfrak{a}_{\bullet }}(\operatorname{lct}(\mathfrak{a}_{\bullet })^{n}\cdot \operatorname{vol}(\mathfrak{a}_{\bullet }))\leqslant \inf _{v}\widehat{\operatorname{vol}}(v).\end{eqnarray}$$

Combining (4.2), (4.3) and (4.5) together, we finish the proof. ◻

Proof. It is clear that $A_{X}(\text{ord}_{V})=r^{-1}$ and $\operatorname{vol}(\text{ord}_{V})=r^{n-1}((-K_{V})^{n-1})$ . Hence the right-hand side of (4.6) is equal to $\widehat{\operatorname{vol}}(\text{ord}_{V})$ . From Theorem 27 we know that (4.6) is equivalent to saying that the normalized volume $\widehat{\operatorname{vol}}$ is minimized at $\text{ord}_{V}$ over $(X,o)$ . This is equivalent to $V$ being K-semistable by [Reference Li and LiuLL16, Corollary 1.5].◻

Proof. By Izumi’s theorem we know that the filtration $\mathfrak{a}_{1}(v_{\ast })\supset \mathfrak{a}_{2}(v_{\ast })\supset \cdots \,$ defines the same topology on ${\mathcal{O}}_{X,p}$ as the $\mathfrak{m}_{p}$ -adic topology on ${\mathcal{O}}_{X,p}$ . Then Lemma 32 implies $\bigoplus _{m\geqslant 0}\mathfrak{a}_{m}(v_{\ast })$ is a finitely generated ${\mathcal{O}}_{X,p}$ -algebra. Therefore, the degree $k$ Veronese subalgebra $\bigoplus _{m\geqslant 0}\mathfrak{a}_{km}(v_{\ast })$ is generated in degree $1$ for $k\in \mathbb{Z}_{{>}0}$ sufficiently divisible. This means $\mathfrak{a}_{km}=\mathfrak{a}_{k}^{m}$ for any $m\geqslant 0$ . From inequality (4.4) we have

$$\begin{eqnarray}\displaystyle \widehat{\operatorname{vol}}(v_{\ast }) & {\geqslant} & \displaystyle \operatorname{lct}(\mathfrak{a}_{\bullet }(v_{\ast }))^{n}\cdot \operatorname{vol}(\mathfrak{a}_{\bullet }(v_{\ast }))\nonumber\\ \displaystyle & = & \displaystyle \operatorname{lct}(\mathfrak{a}_{k\bullet }(v_{\ast }))^{n}\cdot \operatorname{vol}(\mathfrak{a}_{k\bullet }(v_{\ast }))\nonumber\\ \displaystyle & = & \displaystyle \operatorname{lct}(\mathfrak{a}_{k}(v_{\ast }))^{n}\cdot \operatorname{mult}(\mathfrak{a}_{k}(v_{\ast })).\nonumber\end{eqnarray}$$

On the other hand, Theorem 27 also implies that $\operatorname{lct}(\mathfrak{a}_{k}(v_{\ast }))^{n}\cdot \operatorname{mult}(\mathfrak{a}_{k}(v_{\ast }))\geqslant \widehat{\operatorname{vol}}(v_{\ast })$ because $v_{\ast }$ is a minimizer of $\widehat{\operatorname{vol}}(\cdot )$ . So we prove the proposition.◻

Proof. We take a set of homogeneous generators of $\bigoplus _{m\geqslant 0}\mathfrak{a}_{m}/\mathfrak{a}_{m+1}$ , say $x_{1},\ldots ,x_{r}$ . Let $d_{i}$ be the degree of $x_{i}$ , i.e. $x_{i}\in \mathfrak{a}_{d_{i}}/\mathfrak{a}_{d_{i}+1}$ . Take $y_{i}\in \mathfrak{a}_{d_{i}}$ such that $x_{i}=y_{i}+\mathfrak{a}_{d_{i}+1}$ . Denote by $\bigoplus _{m\geqslant 0}\mathfrak{b}_{m}$ the graded $R$ -subalgebra of $\bigoplus _{m\geqslant 0}\mathfrak{a}_{m}$ generated by $y_{1},\ldots ,y_{r}$ .

Since $\{x_{i}\}$ generates $\bigoplus _{m\geqslant 0}\mathfrak{a}_{m}/\mathfrak{a}_{m+1}$ , we have $\mathfrak{a}_{m}=\mathfrak{b}_{m}+\mathfrak{a}_{m+1}$ . Hence for any $l>0$ the induction yields

$$\begin{eqnarray}\mathfrak{a}_{m}=(\mathfrak{b}_{m}+\mathfrak{b}_{m+1}+\cdots +\mathfrak{b}_{m+l-1})+\mathfrak{a}_{m+l}.\end{eqnarray}$$

Because $R$ is Noetherian, the ideals $(\mathfrak{b}_{m}+\mathfrak{b}_{m+1}+\cdots +\mathfrak{b}_{m+l-1})$ stabilize for $l$ sufficiently large. Denote the stabilized ideal by $\mathfrak{c}_{m}$ , then we have $\mathfrak{a}_{m}=\mathfrak{c}_{m}+\mathfrak{a}_{m+l}$ for $l$ sufficiently large. Since the filtration $\mathfrak{a}_{1}\supset \mathfrak{a}_{2}\supset \cdots \,$ defines the same topology as the $\mathfrak{m}$ -adic topology, we may take $l$ sufficiently large such that $\mathfrak{a}_{m+l}\subset \mathfrak{m}\mathfrak{a}_{m}$ . As a result, we have $\mathfrak{a}_{m}=\mathfrak{c}_{m}+\mathfrak{m}\mathfrak{a}_{m}$ which implies that $\mathfrak{a}_{m}=\mathfrak{c}_{m}$ by Nakayama’s lemma.

Now it suffices to show that $\bigoplus _{m\geqslant 0}\mathfrak{c}_{m}$ is a finitely generated graded $R$ -algebra. By definition, $\mathfrak{c}_{m}$ is generated by monomials in $\{y_{i}\}$ of weighted degree at least $m$ . For any $0\leqslant j\leqslant d_{i}$ , denote by $y_{i,j}$ the homogeneous element $y_{i}\in \mathfrak{a}_{j}$ of degree $j$ . Then any monomial in $\{y_{i}\}$ of weighted degree at least $m$ is equal to a monomial in $\{y_{i,j}\}$ of weighted degree $m$ . Hence $\bigoplus _{m\geqslant 0}\mathfrak{c}_{m}$ is generated by $\{y_{i,j}\}$ as a graded $R$ -algebra, which finishes the proof.◻

Postscript remark.

Since the first version of this article was posted on the arXiv, there has been a lot of progresses studying properties of the minimizer of $\widehat{\operatorname{vol}}$ . Here we mention two of them which are useful in our presentation.

1. – Blum [Reference BlumBlu18] proved that there always exists a minimizer of $\widehat{\operatorname{vol}}$ for any klt singularity $x\in X$ ;

2. – If a divisorial valuation $v_{\ast }=\text{ord}_{S}$ minimizes $\widehat{\operatorname{vol}}$ in $\text{Val}_{X,x}$ , then $S$ is necessarily a Kollár component (see [Reference Li and XuLX16] for a definition). In particular, $\text{gr}_{v_{\ast }}{\mathcal{O}}_{X,x}$ is finitely generated. This was proved by Blum [Reference BlumBlu18] and Li and Xu [Reference Li and XuLX16] independently.

Proof. Part (i) is a direct consequence of Theorem 2.

We need some preparation to prove part (ii). Since $v_{\ast }$ is a divisorial minimizer of $\widehat{\operatorname{vol}}$ by (i), it is induced by a Kollár component by [Reference BlumBlu18, Proposition 4.9] and [Reference Li and XuLX16, Theorem 1.2]. In particular, $\text{gr}_{v_{\ast }}{\mathcal{O}}_{X,p}$ is a finitely generated $\mathbb{C}$ -algebra. Let $\bar{v}_{\ast }$ be the corresponding divisorial valuation on $X\times \mathbb{A}^{1}$ (see (3.2)). Let $t$ be the parameter of $\mathbb{A}^{1}$ . Let $\mathbf{Spec}\,R$ be an affine Zariski open subset of $X$ . By Lemma 32, $\bigoplus _{m\geqslant 0}\mathfrak{a}_{m}(v_{\ast })$ is a finitely generated $R$ -algebra. Then

$$\begin{eqnarray}\mathfrak{a}_{m}(\bar{v}_{\ast })=\mathfrak{a}_{m}(v_{\ast })+\mathfrak{a}_{m-1}(v_{\ast })t+\cdots +\mathfrak{a}_{1}(v_{\ast })t^{m-1}+(t^{m}).\end{eqnarray}$$

In order to keep track of the degree of the graded ring $\bigoplus _{m\geqslant 0}\mathfrak{a}_{m}(\bar{v}_{\ast })$ , denote by $s$ the element $t$ in $\mathfrak{a}_{1}(\bar{v}_{\ast })=\mathfrak{a}_{1}(v_{\ast })+(t)$ . Hence $\deg s=1$ , $\deg t=0$ and we have

$$\begin{eqnarray}\mathfrak{a}_{m}(\bar{v}_{\ast })=\mathfrak{a}_{m}(v_{\ast })+\mathfrak{a}_{m-1}(v_{\ast })s+\cdots +\mathfrak{a}_{1}(v_{\ast })s^{m-1}+s^{m}R[t],\end{eqnarray}$$

where $\deg \mathfrak{a}_{i}(v_{\ast })=i$ . As an $R[t]$ -algebra, $\bigoplus _{m\geqslant 0}\mathfrak{a}_{m}(\bar{v}_{\ast })$ is generated by $\bigoplus _{m\geqslant 0}\mathfrak{a}_{m}(v_{\ast })$ and $s$ . Since $\bigoplus _{m\geqslant 0}\mathfrak{a}_{m}(v_{\ast })$ is a finitely generated $R$ -algebra, we have that $\bigoplus _{m\geqslant 0}\mathfrak{a}_{m}(\bar{v}_{\ast })$ is of finite type over ${\mathcal{O}}_{X\times \mathbb{A}^{1}}$ .

Let ${\mathcal{X}}:={\mathbf{Proj}_{X\times \mathbb{A}^{1}}\oplus }_{m\geqslant 0}\mathfrak{a}_{m}(\bar{v}_{\ast })$ , hence ${\mathcal{X}}$ is normal by Lemma 34. Denote by $g:{\mathcal{X}}\rightarrow X\times \mathbb{A}^{1}$ the projection. Denote the composite map by $\unicode[STIX]{x1D70B}:{\mathcal{X}}\rightarrow \mathbb{A}^{1}$ . The $\mathbb{Q}$ -line bundle ${\mathcal{L}}$ on ${\mathcal{X}}$ is defined as

$$\begin{eqnarray}{\mathcal{L}}:=g^{\ast }(-K_{X\times \mathbb{A}^{1}/\mathbb{A}^{1}})+\biggl(1+\frac{1}{n}\biggr)A_{X}(v_{\ast }){\mathcal{O}}_{{\mathcal{X}}}(1),\end{eqnarray}$$

where we treat ${\mathcal{O}}_{{\mathcal{X}}}(1)=(1/k){\mathcal{O}}_{{\mathcal{X}}}(k)$ as a $\mathbb{Q}$ -line bundle for sufficiently divisible $k\in \mathbb{Z}_{{>}0}$ . Let $\hat{X}:=g_{\ast }^{-1}(X\times \{0\})$ and $\unicode[STIX]{x1D70E}=g|_{\hat{X}}:\hat{X}\rightarrow X$ . Let $E$ be the exceptional divisor of $g$ given by $I_{E}:={\mathcal{O}}_{{\mathcal{X}}}(1)$ . Hence

$$\begin{eqnarray}E\cong \mathbf{Proj}~\bigoplus _{m\geqslant 0}\mathfrak{a}_{m}(\bar{v}_{\ast })/\mathfrak{a}_{m+1}(\bar{v}_{\ast })=\mathbf{Proj}(\text{gr}_{v_{\ast }}{\mathcal{O}}_{X,p})[s],\end{eqnarray}$$

where $s$ is a homogeneous element of degree $1$ . Since $\text{gr}_{v_{\ast }}{\mathcal{O}}_{X,p}$ is an integral domain, $E$ is irreducible and reduced.

We will show the following statements:

1. (a) ${\mathcal{L}}$ is $\unicode[STIX]{x1D70B}$ -nef with ${\mathcal{L}}^{\bot }=\hat{X}$ ;

2. (b) ${\mathcal{L}}$ is $\unicode[STIX]{x1D70B}$ -semiample, hence $({\mathcal{X}},{\mathcal{L}})$ is a semi $\mathbb{Q}$ -test configuration of $(X,-K_{X})$ ;

3. (c) $\text{CM}({\mathcal{X}},{\mathcal{L}})=0$ .

For (a), we see that ${\mathcal{L}}_{t}\cong -K_{{\mathcal{X}}_{t}}$ is ample whenever $t\neq 0$ . From the definition we know that ${\mathcal{L}}$ is $g$ -ample, hence ${\mathcal{L}}|_{E}$ is ample. On $\hat{X}$ , we have that

$$\begin{eqnarray}{\mathcal{L}}|_{\hat{X}}\cong \unicode[STIX]{x1D70E}^{\ast }(-K_{X})+\biggl(1+\frac{1}{n}\biggr)A_{X}(v_{\ast }){\mathcal{O}}_{\hat{X}}(1).\end{eqnarray}$$

For $k\in \mathbb{Z}_{{>}0}$ sufficiently divisible, let $Z$ be the thickening of $p$ in $X$ such that $I_{Z}=\mathfrak{a}_{k}(v_{\ast })$ . From (i) we know that $v_{\ast }$ minimizes $\widehat{\operatorname{vol}}$ , hence $\operatorname{lct}(X;I_{Z})^{n}\cdot \operatorname{mult}_{Z}X=\widehat{\operatorname{vol}}(v_{\ast })$ by Proposition 31. Since

$$\begin{eqnarray}((-K_{X})^{n})=\biggl(1+\frac{1}{n}\biggr)^{n}\,\widehat{\operatorname{vol}}(v_{\ast })=\biggl(1+\frac{1}{n}\biggr)^{n}\operatorname{lct}(X;I_{Z})^{n}\cdot \operatorname{mult}_{Z}X,\end{eqnarray}$$

by Lemma 10 and Remark 17 we have that ${\mathcal{L}}|_{\hat{X}}$ is nef with zero top self intersection number. In addition, $\unicode[STIX]{x1D716}_{Z}(-K_{X})=A_{X}(v_{\ast })/m$ . As a result, ${\mathcal{L}}^{\bot }=\hat{X}$ .

For (b), we first show that ${\mathcal{X}}$ is normal $\mathbb{Q}$ -Gorenstein with klt singularities. By Lemma 34 we have that $E=\text{Ex}(g)$ is a $\mathbb{Q}$ -Cartier prime divisor on ${\mathcal{X}}$ . Hence

$$\begin{eqnarray}K_{{\mathcal{X}}}=g^{\ast }K_{X\times \mathbb{A}^{1}}+(A_{{\mathcal{X}}}(\text{ord}_{E})-1)E\end{eqnarray}$$

is $\mathbb{Q}$ -Cartier. To show that ${\mathcal{X}}$ has klt singularities, it suffices to show that $({\mathcal{X}},\hat{X})$ is plt. By Lemma 34, we have $\bar{v}_{\ast }=\text{ord}_{E}$ . Hence ${\mathcal{X}}_{0}=\hat{X}+E$ as Weil divisors since $\bar{v}_{\ast }(t)=1$ . In particular, $K_{{\mathcal{X}}}+\hat{X}$ is $\mathbb{Q}$ -Cartier.

It is clear that

$$\begin{eqnarray}I_{E}/(I_{E}\cdot I_{\hat{X}})=I_{E}|_{\hat{X}}={\mathcal{O}}_{{\mathcal{X}}}(1)|_{\hat{X}}={\mathcal{O}}_{\hat{X}}(1)\subset {\mathcal{O}}_{\hat{X}}.\end{eqnarray}$$

Since the kernel of $I_{E}\rightarrow {\mathcal{O}}_{\hat{X}}$ is $I_{E}\cap I_{\hat{X}}$ , we have that $I_{E}\cdot I_{\hat{X}}=I_{E}\cap I_{\hat{X}}$ . On the other hand, by computing graded ideals we know that $I_{E}+I_{\hat{X}}=I_{F}$ . Denote by $\unicode[STIX]{x1D702}$ the generic point of $F$ , then ${\mathcal{O}}_{\hat{X},\unicode[STIX]{x1D702}}={\mathcal{O}}_{{\mathcal{X}},\unicode[STIX]{x1D702}}/I_{\hat{X},\unicode[STIX]{x1D702}}$ is a DVR since $\hat{X}$ is normal. Applying Lemma 35 to $(R,\mathfrak{p},\mathfrak{q})=({\mathcal{O}}_{{\mathcal{X}},\unicode[STIX]{x1D702}},I_{E,\unicode[STIX]{x1D702}},I_{\hat{X},\unicode[STIX]{x1D702}})$ yields that $E$ is Cartier at $\unicode[STIX]{x1D702}$ . Since ${\mathcal{X}}_{0}=\hat{X}+E$ is Cartier and $\hat{X}\cap E=F$ , we have that $\hat{X}={\mathcal{X}}_{0}-E$ is Cartier in codimension $2$ . Next we notice that $\hat{X}\rightarrow X$ produces a Kollár component, hence $\hat{X}$ has klt singularities. Thus $({\mathcal{X}},\hat{X})$ is plt by inverse of adjunction [Reference Kollár and MoriKM98, Theorem 5.50].

By Shokurov’s base-point-free theorem, to show ${\mathcal{L}}$ is $\unicode[STIX]{x1D70B}$ -semiample we only need to show that ${\mathcal{L}}-K_{{\mathcal{X}}/\mathbb{A}^{1}}$ is $\unicode[STIX]{x1D70B}$ -ample. It follows from Lemma 34 that $\bar{v}_{\ast }=\text{ord}_{E}$ and $E{\sim}_{\mathbb{Q}}{\mathcal{O}}_{{\mathcal{X}}}(-1)$ . Hence we have

$$\begin{eqnarray}\displaystyle -K_{{\mathcal{X}}/\mathbb{A}^{1}} & = & \displaystyle g^{\ast }(-K_{X\times \mathbb{A}^{1}/\mathbb{A}^{1}})-(A_{X\times \mathbb{A}^{1}}(\bar{v}_{\ast })-1)E\nonumber\\ \displaystyle & = & \displaystyle g^{\ast }(-K_{X\times \mathbb{A}^{1}/\mathbb{A}^{1}})+A_{X}(v_{\ast }){\mathcal{O}}_{{\mathcal{X}}}(1).\nonumber\end{eqnarray}$$

Since $0<A_{X}(v_{\ast })<(1+1/n)A_{X}(v_{\ast })$ , we see that $-K_{{\mathcal{X}}/\mathbb{A}^{1}}$ is $\unicode[STIX]{x1D70B}$ -ample. Hence ${\mathcal{L}}-K_{{\mathcal{X}}/\mathbb{A}^{1}}$ is $\unicode[STIX]{x1D70B}$ -ample.

For (c), we know that

$$\begin{eqnarray}\text{CM}({\mathcal{X}},{\mathcal{L}})=\frac{1}{(n+1)((-K_{X})^{n})}(n(\bar{{\mathcal{L}}}^{n+1})+(n+1)(\bar{{\mathcal{L}}}^{n}\cdot K_{\bar{{\mathcal{X}}}/\mathbb{P}^{1}})).\end{eqnarray}$$

By definition of ${\mathcal{L}}$ we know that

$$\begin{eqnarray}\displaystyle & \displaystyle \bar{{\mathcal{L}}}=\bar{g}^{\ast }(-K_{X\times \mathbb{P}^{1}/\mathbb{P}^{1}})+\biggl(1+\frac{1}{n}\biggr)A_{X}(v_{\ast }){\mathcal{O}}_{\bar{{\mathcal{X}}}}(1), & \displaystyle \nonumber\\ \displaystyle & \displaystyle K_{\bar{{\mathcal{X}}}/\mathbb{P}^{1}}=\bar{g}^{\ast }K_{X\times \mathbb{P}^{1}/\mathbb{P}^{1}}-A_{X}(v_{\ast }){\mathcal{O}}_{\bar{{\mathcal{X}}}}(1). & \displaystyle \nonumber\end{eqnarray}$$

Since ${\mathcal{O}}_{\bar{{\mathcal{X}}}}(1)$ is supported in $\bar{g}^{-1}((p,0))$ , we have $(\bar{g}^{\ast }K_{X\times \mathbb{P}^{1}/\mathbb{P}^{1}}\cdot {\mathcal{O}}_{\bar{{\mathcal{X}}}}(1))=0$ as a cycle. Next, it is clear that $((\bar{g}^{\ast }K_{X\times \mathbb{P}^{1}/\mathbb{P}^{1}})^{n+1})=(K_{X\times \mathbb{P}^{1}/\mathbb{P}^{1}}^{n+1})=0.$ Hence we have

$$\begin{eqnarray}\displaystyle & \displaystyle (\bar{{\mathcal{L}}}^{n+1})=\biggl(1+\frac{1}{n}\biggr)^{n+1}A_{X}(v_{\ast })^{n+1}({\mathcal{O}}_{\bar{{\mathcal{X}}}}(1)^{n+1}), & \displaystyle \nonumber\\ \displaystyle & \displaystyle (\bar{{\mathcal{L}}}^{n}\cdot K_{\bar{{\mathcal{X}}}/\mathbb{P}^{1}})=-\biggl(1+\frac{1}{n}\biggr)^{n}A_{X}(v_{\ast })^{n+1}({\mathcal{O}}_{\bar{{\mathcal{X}}}}(1)^{n+1}). & \displaystyle \nonumber\end{eqnarray}$$

Thus $\text{CM}({\mathcal{X}},{\mathcal{L}})=0$ .

Now we are ready to prove part (ii). By (b) we know that ${\mathcal{L}}$ is semiample. Denote the ample model of $({\mathcal{X}},{\mathcal{L}})/\mathbb{A}^{1}$ by $({\mathcal{Y}},{\mathcal{M}})$ , with $h:{\mathcal{X}}\rightarrow {\mathcal{Y}}$ and ${\mathcal{L}}=h^{\ast }{\mathcal{M}}$ . Then $({\mathcal{Y}},{\mathcal{M}})$ is a normal $\mathbb{Q}$ -test configuration of $(X,-K_{X})$ . Since $\bar{h}_{\ast }K_{\bar{{\mathcal{X}}}/\mathbb{P}^{1}}=K_{\bar{{\mathcal{Y}}}/\mathbb{P}^{1}}$ , we have $\text{CM}({\mathcal{Y}},{\mathcal{M}})=\text{CM}({\mathcal{X}},{\mathcal{L}})=0$ . From (a) we know that ${\mathcal{L}}^{\bot }=\hat{X}$ , so $\text{Ex}(h)=\hat{X}$ . Hence ${\mathcal{Y}}_{0}=h_{\ast }{\mathcal{X}}_{0}=h_{\ast }(\hat{X}+E)=h_{\ast }E$ . In particular, ${\mathcal{Y}}_{0}$ is a prime divisor on ${\mathcal{Y}}$ . It is clear that ${\mathcal{L}}|_{E}=(1+1/n)A_{X}(v_{\ast }){\mathcal{O}}_{E}(1)$ is ample, hence $h|_{E}:E\rightarrow {\mathcal{Y}}_{0}$ is a finite birational morphism. Since $X$ is K-polystable by assumption, hence $\text{CM}({\mathcal{Y}},{\mathcal{M}})=0$ yields that $X\cong {\mathcal{Y}}_{0}$ . Then $h|_{E}:E\rightarrow {\mathcal{Y}}_{0}$ has to be an isomorphism because ${\mathcal{Y}}_{0}\cong X$ is normal. Hence $X\cong E\cong \mathbf{Proj}~(\text{gr}_{v_{\ast }}{\mathcal{O}}_{X,p})[s]$ .◻

Proof. By Lemma 32, we know that $\bigoplus _{m\geqslant 0}\mathfrak{a}_{m}(v_{\ast })$ is a finitely generated ${\mathcal{O}}_{X}$ -algebra. Since the valuation ideals $\mathfrak{a}_{m}(v_{\ast })$ are always integrally closed, $\hat{X}$ is normal. By definition of $F$ we know that the ideal sheaf $I_{F}$ is the same as the coherent sheaf ${\mathcal{O}}_{\hat{X}}(1)$ . Let $k\in \mathbb{Z}_{{>}0}$ be sufficiently divisible so that $\mathfrak{a}_{mk}(v_{\ast })=\mathfrak{a}_{k}(v_{\ast })^{m}$ for any positive integer $m$ . Thus $\hat{X}$ is naturally isomorphic to the blow up of $X$ along the ideal sheaf $\mathfrak{a}_{k}(v_{\ast })$ . In particular, ${\mathcal{O}}_{\hat{X}}(k)=\unicode[STIX]{x1D70E}^{-1}\mathfrak{a}_{k}(v_{\ast })\cdot {\mathcal{O}}_{\hat{X}}$ is an invertible ideal sheaf. Next, ${\mathcal{O}}_{\hat{X}}/{\mathcal{O}}_{\hat{X}}(k)$ is supported at the exceptional locus of $\unicode[STIX]{x1D70E}$ . As ideal sheaves on $\hat{X}$ , we know that ${\mathcal{O}}_{\hat{X}}(1)^{k}\subset {\mathcal{O}}_{\hat{X}}(k)\subset {\mathcal{O}}_{\hat{X}}(1)$ . Hence ${\mathcal{O}}_{\hat{X}}(1)$ and ${\mathcal{O}}_{\hat{X}}(k)$ have the same nilradical as ideal sheaves, and

$$\begin{eqnarray}F_{\text{red}}=\text{Supp}({\mathcal{O}}_{\hat{X}}/{\mathcal{O}}_{\hat{X}}(1))=\text{Supp}({\mathcal{O}}_{\hat{X}}/{\mathcal{O}}_{\hat{X}}(k))\end{eqnarray}$$

which is the reduced exceptional locus of $\unicode[STIX]{x1D70E}$ . On the other hand, it is easy to see that $\text{gr}_{v_{\ast }}{\mathcal{O}}_{X,p}$ is an integral domain, so $F$ is an integral scheme.

For $k$ sufficiently divisible, we have that ${\mathcal{O}}_{\hat{X}}(k)$ is invertible and $\unicode[STIX]{x1D70E}_{\ast }{\mathcal{O}}_{\hat{X}}(mk)=\mathfrak{a}_{mk}(v_{\ast })$ for any $m\in \mathbb{Z}_{{\geqslant}0}$ . Thus ${\mathcal{O}}_{\hat{X}}(k)={\mathcal{O}}_{\hat{X}}(-lF)$ for some positive integer $l$ (so $F$ is $\mathbb{Q}$ -Cartier). Then for any $m\in \mathbb{Z}_{{\geqslant}0}$ we have the following equivalences:

$$\begin{eqnarray}\displaystyle v_{\ast }(f)\geqslant m & \Leftrightarrow & \displaystyle v_{\ast }(f^{k})\geqslant mk\Leftrightarrow f^{k}\in \mathfrak{a}_{mk}(v_{\ast })\Leftrightarrow \unicode[STIX]{x1D70E}^{\ast }f^{k}\in {\mathcal{O}}_{\hat{X}}(mk)\nonumber\\ \displaystyle & \Leftrightarrow & \displaystyle \text{ord}_{F}(\unicode[STIX]{x1D70E}^{\ast }f^{k})\geqslant ml\Leftrightarrow \text{ord}_{F}(f)\geqslant \frac{ml}{k}.\nonumber\end{eqnarray}$$

Assume $v_{\ast }(f)=m$ . If $\text{ord}_{F}(f)>ml/k$ , then we have $\text{ord}_{F}(f)\geqslant (ml+1)/k$ . Hence $\text{ord}_{F}(f^{l})\geqslant (ml+1)l/k$ which implies that $v_{\ast }(f^{l})\geqslant ml+1$ , a contradiction! Hence we have $\text{ord}_{F}=(l/k)v_{\ast }$ . Therefore, $v_{\ast }=\text{ord}_{F}$ since both $v_{\ast }$ and $\text{ord}_{F}$ are divisorial.◻

Proof. Let $x+\mathfrak{q}$ be a uniformizer of $R/\mathfrak{q}$ . Since $\mathfrak{m}=\mathfrak{p}+\mathfrak{q}$ , we may choose $x$ so that $x\in \mathfrak{p}$ . Hence we have $(x)+\mathfrak{q}=\mathfrak{m}$ . As a result,

$$\begin{eqnarray}\mathfrak{p}=(x)+\mathfrak{p}\cap \mathfrak{q}=(x)+\mathfrak{p}\mathfrak{q}.\end{eqnarray}$$

So $\mathfrak{p}=(x)$ by Nakayama lemma.◻

Proof 1.

Notice that $((-K_{X})^{n})\leqslant (n+1)^{n}$ by [Reference FujitaFuj15, Corollary 1.3]. Thus we have $((-K_{X})^{n})=(n+1)^{n}$ . Let $p\in X$ be a smooth point. From [Reference FujitaFuj15, Proof of Theorem 5.1] or Remark 17, we see that $\unicode[STIX]{x1D716}_{p}(-K_{X})=n+1$ . Then the theorem is a consequence of the forthcoming paper [Reference Liu and ZhuangLZ16] (joint with Ziquan Zhuang), where we show that if an $n$ -dimensional $\mathbb{Q}$ -Fano variety $X$ satisfies $\unicode[STIX]{x1D716}_{p}(-K_{X})>n$ for some smooth point $p\in X$ , then $X\cong \mathbb{P}^{n}$ .◻

Proof 2.

We follow the strategy and notation of the proof of Lemma 33. Let $v_{\ast }:=\text{ord}_{p}$ for a smooth point $p\in X$ . As argued in the first proof, the assumptions of Lemma 33 are fulfilled except that $X$ is only Ding-semistable rather than K-polystable. Nevertheless, we still have a semi $\mathbb{Q}$ -test configuration $({\mathcal{X}},{\mathcal{L}})$ of $(X,-K_{X})$ such that ${\mathcal{L}}^{\bot }=\hat{X}$ and $\text{CM}({\mathcal{X}},{\mathcal{L}})=0$ . Let $({\mathcal{Y}},{\mathcal{M}})$ be the ample model of $({\mathcal{X}},{\mathcal{L}})/\mathbb{A}^{1}$ , with $h:{\mathcal{X}}\rightarrow {\mathcal{Y}}$ and ${\mathcal{L}}=h^{\ast }{\mathcal{M}}$ . We will show that ${\mathcal{Y}}_{0}\cong \mathbb{P}^{n}$ .

Following the proof of Lemma 33, we know that $h|_{E}:E\rightarrow {\mathcal{Y}}_{0}$ is a finite birational morphism. Since $v_{\ast }=\text{ord}_{p}$ and $p\in X$ is a smooth point, we have that $E\cong \mathbb{P}^{n}$ . Therefore, $E$ is the normalization of ${\mathcal{Y}}_{0}$ . Consider the short exact sequence

$$\begin{eqnarray}0\rightarrow {\mathcal{O}}_{{\mathcal{X}}}(-E)\rightarrow {\mathcal{O}}_{{\mathcal{X}}}\rightarrow {\mathcal{O}}_{E}\rightarrow 0.\end{eqnarray}$$

By taking $h_{\ast }$ , we get a long exact sequence

(5.1) $$\begin{eqnarray}0\rightarrow h_{\ast }{\mathcal{O}}_{{\mathcal{X}}}(-E)\rightarrow {\mathcal{O}}_{{\mathcal{Y}}}\rightarrow h_{\ast }{\mathcal{O}}_{E}\rightarrow R^{1}h_{\ast }{\mathcal{O}}_{{\mathcal{X}}}(-E)\rightarrow \cdots \,.\end{eqnarray}$$

Since ${\mathcal{X}}_{0}=\hat{X}+E$ , we have that $h_{\ast }{\mathcal{O}}_{{\mathcal{X}}}(-E)={\mathcal{O}}_{{\mathcal{Y}}}(-{\mathcal{Y}}_{0})\otimes h_{\ast }{\mathcal{O}}_{{\mathcal{X}}}(\hat{X})$ . Notice that $\hat{X}$ is $h$ -exceptional, hence $h_{\ast }{\mathcal{O}}_{{\mathcal{X}}}(\hat{X})={\mathcal{O}}_{{\mathcal{Y}}}$ . As a result, $h_{\ast }{\mathcal{O}}_{{\mathcal{X}}}(-E)\cong {\mathcal{O}}_{{\mathcal{Y}}}(-{\mathcal{Y}}_{0})$ . On the other hand, we have

$$\begin{eqnarray}h^{\ast }{\mathcal{M}}={\mathcal{L}}=g^{\ast }(-K_{X\times \mathbb{A}^{1}/\mathbb{A}^{1}})-(n+1)E=-K_{{\mathcal{X}}/\mathbb{A}^{1}}-E.\end{eqnarray}$$

So $-E=K_{{\mathcal{X}}/\mathbb{A}^{1}}+h^{\ast }{\mathcal{M}}{\sim}_{\mathbb{Q},h}K_{{\mathcal{X}}}$ . Since ${\mathcal{X}}$ has klt singularities, we have that $R^{1}h_{\ast }{\mathcal{O}}_{{\mathcal{X}}}(-E)=0$ by the generalized Kodaira vanishing theorem [Reference KollárKol95, Theorem 10.19.4]. Thus the exact sequence (5.1) yields that $h_{\ast }{\mathcal{O}}_{E}={\mathcal{O}}_{{\mathcal{Y}}_{0}}$ , which implies ${\mathcal{Y}}_{0}\cong E\cong \mathbb{P}^{n}$ .

Since $\mathbb{P}^{n}$ is rigid under smooth deformation (cf. [Reference KollárKol96, Exercise V.1.11.12.2]), we conclude that ${\mathcal{Y}}_{t}\cong \mathbb{P}^{n}$ for general $t\in \mathbb{A}^{1}\setminus \{0\}$ , hence $X\cong {\mathcal{X}}_{t}\cong {\mathcal{Y}}_{t}\cong \mathbb{P}^{n}$ .◻

Proof. For the ‘if’ part, we may assume that $G\subset U(n)$ . Hence $G$ preserves the Fubini–Study metric $\unicode[STIX]{x1D714}_{FS}$ on $\mathbb{P}^{n}$ . Since $|G\cap \mathbb{G}_{m}|=1$ , the $G$ -action on $\mathbb{P}^{n}$ is free in codimension $1$ . This implies that the quotient metric of $\unicode[STIX]{x1D714}_{FS}$ on $\mathbb{P}^{n}/G$ is Kähler–Einstein and $((-K_{\mathbb{P}^{n}/G})^{n})=((-K_{\mathbb{P}^{n}})^{n})/|G|=(n+1)^{n}/|G|$ .

For the ‘only if’ part, let $(Y,o):=(\mathbb{C}^{n}/G,0)$ be the local analytic model of $(X,p)$ . Let $d:=|G\cap \mathbb{G}_{m}|$ . Following the proof of Theorem 23, we have a partial resolution $h:{\hat{Y}}\rightarrow Y$ of $Y$ that is the quotient of the blow up $g:\widehat{\mathbb{C}^{n}}\rightarrow \mathbb{C}^{n}$ . Since ${\hat{Y}}$ is the quotient of $\widehat{\mathbb{C}^{n}}$ , it has klt singularities. Denote by $F$ the exceptional divisor of $h$ , then we have $F\cong \mathbb{P}^{n-1}/G$ . Let $u_{\ast }:=\text{ord}_{F}$ be the divisorial on $Y$ centered at $o$ , then we have $\text{gr}_{u_{\ast }}{\mathcal{O}}_{Y,o}\cong \mathbb{C}[x_{1},\ldots ,x_{n}]^{G}$ is a finitely generated $\mathbb{C}$ -algebra. (Here we assume $\deg x_{i}=1/d$ so that the grading is preserved under the isomorphism.) Theorem 23 also implies $\widehat{\operatorname{vol}}(u_{\ast })=n^{n}/|G|$ .

Since $\widehat{{\mathcal{O}}_{X,p}}\cong \widehat{{\mathcal{O}}_{Y,o}}$ , the divisorial valuation $u_{\ast }$ induces a divisorial valuation $v_{\ast }$ on $X$ centered at $p$ as in the proof of Theorem 23. Thus $\widehat{\operatorname{vol}}(v_{\ast })=\widehat{\operatorname{vol}}(u_{\ast })$ , $\text{gr}_{v_{\ast }}({\mathcal{O}}_{X,p})\cong \text{gr}_{u_{\ast }}({\mathcal{O}}_{Y,o})$ is a finitely generated $\mathbb{C}$ -algebra, $\hat{X}:={\mathbf{Proj}_{X}\bigoplus }_{m\geqslant 0}\mathfrak{a}_{m}(v_{\ast })$ has klt singularities, and

$$\begin{eqnarray}((-K_{X})^{n})=\frac{(n+1)^{n}}{|G|}=\biggl(1+\frac{1}{n}\biggr)^{n}\,\widehat{\operatorname{vol}}(v_{\ast }).\end{eqnarray}$$

By [Reference BermanBer16] we have that $X$ is K-polystable. Hence applying Lemma 33 yields

$$\begin{eqnarray}X\cong \mathbf{Proj}~\text{gr}_{v_{\ast }}{\mathcal{O}}_{X,p}[x]\cong \mathbf{Proj}~\mathbb{C}[x_{1},\ldots ,x_{n}]^{G}[x],\end{eqnarray}$$

where $\deg x_{i}=1/d$ and $\deg x=1$ . Denote $y:=x^{1/d}$ , then $\mathbb{C}[x_{1},\ldots ,x_{n}]^{G}[x]$ is the Veronese subalgebra of $\mathbb{C}[x_{1},\ldots ,x_{n}]^{G}[y]$ where $\deg x_{i}=\deg y=1/d$ . As a result,

$$\begin{eqnarray}X\cong \mathbf{Proj}~\mathbb{C}[x_{1},\ldots ,x_{n}]^{G}[y]=\mathbb{P}^{n}/G.\end{eqnarray}$$

Denote by $H_{\infty }$ the hyperplane at infinity in $\mathbb{P}^{n}$ . Let $D$ be the prime divisor in $X$ corresponding to $H_{\infty }/G$ in $\mathbb{P}^{n}/G$ . It is clear that $(X,(1-1/d)D)$ is an orbifold as a global quotient of $\mathbb{P}^{n}$ . Denote by $\unicode[STIX]{x1D70B}:\mathbb{P}^{n}\rightarrow X$ the quotient map, then we have

$$\begin{eqnarray}\unicode[STIX]{x1D70B}^{\ast }\biggl(K_{X}+\biggl(1-\frac{1}{d}\biggr)D\biggr)=K_{\mathbb{P}^{n}},\quad \unicode[STIX]{x1D70B}^{\ast }D=dH_{\infty }.\end{eqnarray}$$

Hence $\unicode[STIX]{x1D70B}^{\ast }(-K_{X})=-K_{\mathbb{P}^{n}}+(d-1)H_{\infty }={\mathcal{O}}_{\mathbb{P}^{n}}(n+d)$ . This implies $((-K_{X})^{n})=(n+d)^{n}/|G|$ , so $d=1$ .◻