K-stability of Fano threefolds of rank $2$ and degree $14$ as double covers

We follow the computation of stability thresholds for projective bundles from [ZZ22] (see Lemma 2.2). Let $c\in(0,1)$ be a rational number. Let $\phi:Y\to\mathbb{P}^{2}$ be the $\mathbb{P}^{1}$-bundle induced by central projection from $p$. Denote by $(V,\Delta):=(\mathbb{P}^{2},cC_{0})$ and $\Delta_{Y}:=\phi^{*}\Delta$. Then $Y\cong\mathbb{P}_{V}(L^{-1}\oplus\mathcal{O}_{V})$ where $L=\mathcal{O}_{\mathbb{P}^{2}}(1)$. Let $r=3-2c$ so that $L\sim_{\mathbb{Q}}-r^{-1}(K_{V}+\Delta)$. Let $V_{\infty}$ be the section of $Y$ at infinity. Thus we have $c\overline{S}_{0}=\Delta_{Y}+2cV_{\infty}$. By Lemma 2.2 with $a=0$ and $b=2c$, we have

$$\delta(Y,c\overline{S}_{0})=\min\left\{\frac{(3-2c)\delta(V,\Delta)}{\frac{3}{4}\frac{B^{4}-A^{4}}{B^{3}-A^{3}}},\frac{1}{\frac{3}{4}\frac{B^{4}-A^{4}}{B^{3}-A^{3}}-A},\frac{1-2c}{B-\frac{3}{4}\frac{B^{4}-A^{4}}{B^{3}-A^{3}}}\right\},$$

(2.1)

where $A=r-(1-a)=2-2c$ and $B=r+(1-b)=4-4c$. Thus we have

$$\frac{3}{4}\frac{B^{4}-A^{4}}{B^{3}-A^{3}}=\frac{15}{7}A=\frac{15}{7}(2-2c).$$

(2.2)

Combining (2.1) and (2.2) yields

$$\delta(Y,c\overline{S}_{0})=\min\left\{\frac{28(3-2c)}{45(2-2c)}\delta(V,\Delta),\frac{28}{17(2-2c)},\frac{28(1-2c)}{11(2-2c)}\right\}.$$

(2.3)

Suppose we take $c=\frac{3}{17}$, then it is easy to see that

$$\frac{28(3-2c)}{45(2-2c)}=\frac{28}{17(2-2c)}=\frac{28(1-2c)}{11(2-2c)}=1.$$

Since $c<\frac{3}{4}$, by [LS14, Theorem 1.5] (see also [Fuj20]) the log Fano pair $(V,\Delta)=(\mathbb{P}^{2},cC_{0})$ is K-polystable, which implies $\delta(V,\Delta)=1$. Thus (2.3) becomes $\delta(Y,\frac{3}{17}\overline{S}_{0})=1$ which implies K-semistability of $(Y,\frac{3}{17}\overline{S}_{0})$ by [FO18, BJ20]. ∎

The proof is the same as [ZZ22, Section 3] after replacing $V$ and $Y$ therein by $(V,\Delta)$ and $(Y,\Delta_{Y})$ respectively. ∎

Since $\pi^{*}\lambda$ induces a special degeneration $(Y,\frac{3}{17}\overline{S})\rightsquigarrow(Y,\frac{3}{17}\overline{S}_{0})$, the statement follows from Proposition 2.1 and the openness of K-semistability [BLX22, Xu20]. ∎

Let $\mu$ be a $1$-PS in $\mathrm{PGL}(4)$ given by $\mu(t)\cdot[x,y,z,w]=[tx,t^{-1}y,z,w]$. Then clearly $\lambda^{\prime}$ and $\mu$ generates a $\mathbb{T}=\mathbb{G}_{m}^{2}$-action on $(\mathbb{P}^{3},S_{0}^{\prime})$ which lifts through $\pi^{*}$ to a $\mathbb{T}$-action on $(Y,\overline{S}_{0}^{\prime})$. Since the $\mathbb{T}$-action on $Y$ is of complexity $1$, by Theorem 3.2 we only need to check $\beta_{(Y,\frac{2}{9}\overline{S}_{0}^{\prime})}(F)$ and $\operatorname{Fut}_{(Y,\frac{2}{9}\overline{S}_{0}^{\prime})}$.

We first show that $\beta_{(Y,\frac{2}{9}\overline{S}_{0}^{\prime})}(F)>0$ for every $\mathbb{T}$-invariant prime divisor $F$ on $Y$. In fact, all $\mathbb{T}$-invariant prime divisors on $Y$ are vertical from the following classification. Denote by $\overline{H}_{x},\overline{H}_{y},\overline{H}_{z},\overline{H}_{w}$ the strict transform of the coordinate hyperplanes of $\mathbb{P}^{3}$ to $Y$. A straightforward analysis of the $\mathbb{T}$-action on $\mathbb{P}^{3}$ shows that every $\mathbb{T}$-invariant prime divisor on $Y$ belongs to one of the following classes:

1. (i)

   $E$;

2. (ii)

   $\overline{H}_{w}$;

3. (iii)

   $\overline{H}_{x}$, $\overline{H}_{y}$, $\overline{H}_{z}$;

4. (iv)

   $\overline{T}_{s}:=\pi_{*}^{-1}T_{s}$ where $T_{s}:=(xyw+sz^{3}=0)\subset\mathbb{P}^{3}$ for $s\neq 0$.

Next, we split into four cases according to the above classes. We will frequently use the equality

$$S_{(Y,cD)}(F)=(1-c)S_{Y}(F)$$

for an anti-canonical surface $D\subset Y$ and $c\in(0,1)$ which follows from $-K_{Y}-cD\sim_{\mathbb{Q}}(1-c)(-K_{Y})$ and [BJ20, Lemma 3.7(i)].

(i) It is clear that $A_{(Y,\frac{2}{9}\overline{S}_{0}^{\prime})}(E)=A_{Y}(E)=1$. Next we compute $S_{Y}(E)$. We know that $-K_{Y}\sim 4\overline{H}_{w}-2E$ where $\overline{H}_{w}\sim\pi^{*}\mathcal{O}_{\mathbb{P}^{3}}(1)$. The pseudoeffective cone of $Y$ is generated by $\overline{H}_{x}\sim\overline{H}_{w}-E$ and $E$. The nef cone of $Y$ is generated by $\overline{H}_{w}$ and $\overline{H}_{x}$. Thus we have $-K_{Y}-tE\sim 4\overline{H}_{w}-(2+t)E$ is nef if $0\leq t\leq 2$, and not big if $t\geq 2$. Thus

	$\displaystyle S_{Y}(E)$	$\displaystyle=\frac{1}{(-K_{Y})^{3}}\int_{0}^{\infty}\operatorname{vol}(-K_{Y}-tE)dt$
		$\displaystyle=\frac{1}{56}\int_{0}^{2}(4\overline{H}_{w}-(2+t)E)^{3}dt$
		$\displaystyle=\frac{1}{56}\int_{0}^{2}(4^{3}-(2+t)^{3})dt=\frac{17}{14}.$

As a consequence,

$$\beta_{(Y,\frac{2}{9}\overline{S}_{0}^{\prime})}(E)=1-\frac{7}{9}\cdot\frac{17}{14}=\frac{1}{18}>0.$$

(ii) We have $A_{(Y,\frac{2}{9}\overline{S}_{0}^{\prime})}(\overline{H}_{w})=A_{Y}(\overline{H}_{w})-\frac{2}{9}\operatorname{ord}_{\overline{H}_{w}}(\overline{S}_{0}^{\prime})=\frac{7}{9}$. Next we compute $S_{Y}(\overline{H}_{w})$. It is clear that $-K_{Y}-t\overline{H}_{w}=(4-t)\overline{H}_{w}-2E$ is nef if $0\leq t\leq 2$, and not big if $t\geq 2$. Thus

	$\displaystyle S_{Y}(\overline{H}_{w})$	$\displaystyle=\frac{1}{(-K_{Y})^{3}}\int_{0}^{\infty}\operatorname{vol}(-K_{Y}-t\overline{H}_{w})dt$
		$\displaystyle=\frac{1}{56}\int_{0}^{2}((4-t)\overline{H}_{w}-2E)^{3}dt$
		$\displaystyle=\frac{1}{56}\int_{0}^{2}((4-t)^{3}-2^{3})dt=\frac{11}{14}.$

As a consequence,

$$\beta_{(Y,\frac{2}{9}\overline{S}_{0}^{\prime})}(\overline{H}_{w})=\frac{7}{9}-\frac{7}{9}\cdot\frac{11}{14}=\frac{1}{6}>0.$$

(iii) Since $\overline{H}_{x}\sim\overline{H}_{y}\sim\overline{H}_{z}$, their $S$-invariants are the same. As neither of these three divisors is contained in $\operatorname{Supp}(\overline{S}_{0}^{\prime})$, their log discrepancies with respect to $(Y,\frac{2}{9}\overline{S}_{0}^{\prime})$ are the same which is $1$. It suffices to show that $\beta(\overline{H}_{x})>0$ as these three divisors have the same $\beta$-invariant. Next we compute $S_{Y}(\overline{H}_{x})$. The divisor $-K_{Y}-t\overline{H}_{x}=(4-t)\overline{H}_{w}-(2-t)E$ is nef if $0\leq t\leq 2$ and not big if $t\geq 4$. When $2<t<4$, it admits a Zariski decomposition

$$-K_{Y}-t\overline{H}_{x}=(4-t)\overline{H}_{w}+(t-2)E,$$

which implies that $\operatorname{vol}(-K_{Y}-t\overline{H}_{x})=((4-t)\overline{H}_{w})^{3}$ when $2\leq t\leq 4$. Thus we have

	$\displaystyle S_{Y}(\overline{H}_{x})$	$\displaystyle=\frac{1}{(-K_{Y})^{3}}\int_{0}^{\infty}\operatorname{vol}(-K_{Y}-t\overline{H}_{x})dt$
		$\displaystyle=\frac{1}{56}\left(\int_{0}^{2}((4-t)\overline{H}_{w}-(2-t)E)^{3}dt+\int_{2}^{4}((4-t)\overline{H}_{w})^{3}dt\right)$
		$\displaystyle=\frac{1}{56}\left(\int_{0}^{2}((4-t)^{3}-(2-t)^{3})dt+\int_{2}^{4}(4-t)^{3}dt\right)=\frac{15}{14}.$

As a consequence,

$$\beta_{(Y,\frac{2}{9}\overline{S}_{0}^{\prime})}(\overline{H}_{x})=1-\frac{7}{9}\cdot\frac{15}{14}=\frac{1}{6}>0.$$

(iv) Since $\overline{T}_{s}\sim 3\overline{H}_{w}-2E$ for every $s\neq 0$, their $S$-invariant are the same. We also have $\operatorname{ord}_{\overline{T}_{s}}(\overline{S}_{0}^{\prime})\leq 1$, which implies $A_{(Y,\frac{2}{9}\overline{S}_{0}^{\prime})}(\overline{T}_{s})\geq\frac{7}{9}$. Next we compute $S_{Y}(\overline{T}_{s})$. The divisor $-K_{Y}-t\overline{T}_{s}=(4-3t)\overline{H}_{w}-(2-2t)E$ is nef if $0\leq t\leq 1$ and not big if $t\geq\frac{3}{4}$. When $1<t<\frac{3}{4}$, it admits a Zariski decomposition

$$-K_{Y}-t\overline{T}_{s}=(4-3t)\overline{H}_{w}+(2t-2)E,$$

which implies that $\operatorname{vol}(-K_{Y}-t\overline{H}_{x})=((4-3t)\overline{H}_{w})^{3}$ when $1\leq t\leq\frac{4}{3}$. Thus we have

	$\displaystyle S_{Y}(\overline{T}_{s})$	$\displaystyle=\frac{1}{(-K_{Y})^{3}}\int_{0}^{\infty}\operatorname{vol}(-K_{Y}-t\overline{T}_{s})dt$
		$\displaystyle=\frac{1}{56}\left(\int_{0}^{1}((4-3t)\overline{H}_{w}-(2-2t)E)^{3}dt+\int_{1}^{\frac{4}{3}}((4-3t)\overline{H}_{w})^{3}dt\right)$
		$\displaystyle=\frac{1}{56}\left(\int_{0}^{1}((4-3t)^{3}-(2-2t)^{3})dt+\int_{1}^{\frac{4}{3}}(4-3t)^{3}dt\right)=\frac{29}{84}.$

As a consequence,

$$\beta_{(Y,\frac{2}{9}\overline{S}_{0}^{\prime})}(\overline{T}_{s})\geq\frac{7}{9}-\frac{7}{9}\cdot\frac{29}{84}=\frac{55}{108}>0.$$

So far we verified that $\beta_{(Y,\frac{2}{9}\overline{S}_{0}^{\prime})}(F)>0$ for every $\mathbb{T}$-invariant prime divisor $F$ on $Y$. It remains to show that $\operatorname{Fut}_{(Y,\frac{2}{9}\overline{S}_{0}^{\prime})}=0$ on the cocharacter lattice $N=\mathrm{Hom}(\mathbb{G}_{m},\mathbb{T})$ of $\mathbb{T}$. Let $\lambda_{0}$ and $\lambda_{1}$ be two $1$-PS in $\mathrm{PGL}(4)$ given by

$$\lambda_{0}(t)\cdot[x,y,z,w]=[t^{3}x,y,tz,w]\quad\textrm{and}\quad\lambda_{1}(t)\cdot[x,y,z,w]=[x,t^{3}y,tz,w].$$

Then it is not hard to see that $\lambda_{0}$ and $\lambda_{1}$ form a basis of $N_{\mathbb{Q}}:=N\otimes_{\mathbb{Z}}\mathbb{Q}$. Meanwhile, the involution $\tau:\mathbb{P}^{3}\to\mathbb{P}^{3}$ defined by $[x,y,z,w]\mapsto[y,x,z,w]$ induces an involution $\pi^{*}\tau\in\operatorname{Aut}(Y,\overline{S}_{0}^{\prime})$ such that $\tau\lambda_{0}\tau^{-1}=\lambda_{1}$. Thus it suffices to show that $\operatorname{Fut}_{(Y,\frac{2}{9}\overline{S}_{0}^{\prime})}(\pi^{*}\lambda_{0})=0$ as this implies $\operatorname{Fut}_{(Y,\frac{2}{9}\overline{S}_{0}^{\prime})}(\pi^{*}\lambda_{1})=0$ and hence the vanishing of $\operatorname{Fut}_{(Y,\frac{2}{9}\overline{S}_{0}^{\prime})}$ on the entire $N_{\mathbb{Q}}$.

Let $v$ be the monomial divisorial valuation on $\mathbb{P}^{3}$ centered at $(x=z=0)$ such that $v(x)=3$ and $v(z)=1$. Then it is clear that $\lambda_{0}$ is the $1$-PS induced by $v$. As abuse of notation we also denote by $v$ the lifting valuation $\pi^{*}v$ on $Y$. According to [Fuj19, Theorem 5.1], we have $\operatorname{Fut}_{(Y,\frac{2}{9}\overline{S}_{0}^{\prime})}(\lambda_{0})=\beta_{(Y,\frac{2}{9}\overline{S}_{0}^{\prime})}(v)$. Thus it suffices to show $\beta(v)=0$. Since $\pi:Y\to\mathbb{P}^{3}$ is isomorphic at the generic point of the center of $v$, we know that

$$A_{(Y,\frac{2}{9}\overline{S}_{0}^{\prime})}(v)=A_{(\mathbb{P}^{3},\frac{2}{9}S_{0}^{\prime})}(v)=A_{\mathbb{P}^{3}}(v)-\frac{2}{9}\cdot v(S_{0}^{\prime})=4-\frac{2}{9}\cdot 3=\frac{10}{3}.$$

(3.1)

Next, we compute $S_{Y}(v)$. Let $F_{0}$ be the exceptional divisor of the $(3,0,1)$-weighted blow up in the affine $(x,y,z)$ with $w=1$. Then it is clear that $v=\operatorname{ord}_{F_{0}}$. Thus we have $\operatorname{vol}(-K_{Y}-tv)=\operatorname{vol}(\mathcal{O}_{\mathbb{P}^{3}}(4)-2E-tF_{0})$. As both $E$ and $F_{0}$ are toric divisors over $\mathbb{P}^{3}$, we have

$$\operatorname{vol}(\mathcal{O}_{\mathbb{P}^{3}}(4)-2E-tF_{0})=3!\cdot\operatorname{vol}(P_{t}),$$

where

$$P_{t}:=\{(u_{0},u_{1},u_{2})\in\mathbb{R}_{\geq 0}^{3}\mid 2\leq u_{0}+u_{1}+u_{2}\leq 4\textrm{ and }3u_{0}+u_{2}\geq t\}.$$

Let

$$Q_{t}:=\{(u_{0},u_{1},u_{2})\in\mathbb{R}_{\geq 0}^{3}\mid u_{0}+u_{1}+u_{2}\leq 1\textrm{ and }3u_{0}+u_{2}\geq t\}.$$

Then it is clear that $\operatorname{vol}(P_{t})=4^{3}\operatorname{vol}(Q_{\frac{t}{4}})-2^{3}\operatorname{vol}(Q_{\frac{t}{2}})$. Using convex geometry it is not hard to show that

$$\operatorname{vol}(Q_{t})=\begin{cases}\frac{1}{6}-\frac{1}{6}t^{2}+\frac{2}{27}t^{3}&\textrm{if }0\leq t\leq 1;\\ \frac{1}{108}(3-t)^{3}&\textrm{if }1\leq t\leq 3;\\ 0&\textrm{if }t\geq 3.\end{cases}$$

Computation shows that $\int_{0}^{3}\operatorname{vol}(Q_{t})dt=\frac{1}{6}$. Thus we have

	$\displaystyle\int_{0}^{\infty}\operatorname{vol}(-K_{Y}-tv)dt$	$\displaystyle=\int_{0}^{\infty}6\operatorname{vol}(P_{t})dt$
		$\displaystyle=\int_{0}^{12}6\cdot 4^{3}\operatorname{vol}(Q_{\frac{t}{4}})dt-\int_{0}^{6}6\cdot 2^{3}\operatorname{vol}(Q_{\frac{t}{2}})dt$
		$\displaystyle=6\cdot(4^{4}-2^{4})\int_{0}^{3}\operatorname{vol}(Q_{t})dt=240.$

As a result, we have

$$S_{(Y,\frac{2}{9}\overline{S}_{0}^{\prime})}(v)=\frac{7}{9}S_{Y}(v)=\frac{7}{9}\cdot\frac{1}{56}\int_{0}^{\infty}\operatorname{vol}(-K_{Y}-tv)dt=\frac{7}{9}\cdot\frac{1}{56}\cdot 240=\frac{10}{3}.$$

(3.2)

Combining (3.1) and (3.2), we get $\beta_{(Y,\frac{2}{9}\overline{S}_{0}^{\prime})}(v)=\frac{10}{3}-\frac{10}{3}=0$. Thus the proof is finished. ∎

Since $\pi^{*}\lambda^{\prime}$ induces a special degeneration $(Y,\frac{2}{9}\overline{S})\rightsquigarrow(Y,\frac{2}{9}\overline{S}_{0}^{\prime})$, the statement follows from Proposition 3.1 and the openness of K-semistability [BLX22, Xu20]. ∎

Let $\sigma:X\to Y$ be the double cover branched along a smooth anti-canonical surface $\overline{S}\subset Y$. By [Der16, Theorem 1.3], [LZ22, Theorem 1.2], and [Zhu21, Corollary 4.13], it suffices to show K-stability of $(Y,\frac{1}{2}\overline{S})$ as $\operatorname{Aut}(X)$ is finite according to [CPS19, Lemma 12.4]. By [ADL21, Theorem 2.10] (see also [JMR16, Corollary 1]), we know that $(Y,(1-\epsilon)\overline{S})$ is K-stable for $0<\epsilon\ll 1$. Combining Corollaries 2.3 and 3.3, we know that $(Y,c\overline{S})$ is K-semistable for some $c\in(0,\frac{1}{2})$ (more precisely, $c=\frac{3}{17}$ in family №2.8(a) or $c=\frac{2}{9}$ in family №2.8(b)). Thus the interpolation of K-stability [ADL19, Proposition 2.13] implies that $(Y,\frac{1}{2}\overline{S})$ is K-stable. The existence of Kähler-Einstein metrics follows from [CDS15, Tia15]. Thus the proof is finished. ∎