The invariant rational function on $W$ becomes $f = (2x_1y_1-y_0-x_1^2)/(y_0-y_1^2)$ on $U \inter W$ and hence also on $X$. Under the induced $B$-quotient $\pi$ to $\proj{1}$ we see that $\pi^*([1:-1]) = \cal{Z}(x_1-y_1,y_0-y_1^2) \union \cal{Z}(x_1-y_1,z_0) = E \union \tilde{D}_{-1}$. Hence $E$ sits in the slice of hyperspace corresponding to $-1 \in \proj{1}$, and the colour $\tilde{D}_{-1}$ does not contain the minimal $G$-germ $Y$, as $D_{-1}$ did before the blow-up.

Choosing a uniformising element $\delta = (2x_1y_1-y_0-x_1^2+(y_0-y_1^2))/(y_0-y_1^2) = -(y_1-x_1)^2/(y_0-y_1^2)$ of the DVR corresponding to $-1$ and taking an affine chart $z_0 = 1$ of $X$, a simple calculation shows that $h_E = \nu_E(\delta) = 1$.

Likewise, the splitting semi-invariant $e_{2\alpha}$ from above becomes $1/(x_1-y_1)$ on $X$, giving $\ell_E = \nu_E(e_{2\alpha}) = -1$. Hence $E \mapsto (-1,-1,1) \in \cal{H}$. The positions in hyperspace of all other $G$-divisors and colours remain as always unchanged by the blow-up.

Thus the curve $Y$ has coloured data $\cal{V}_Y = \{\nu_E,\nu_{\tilde{E}_0},\nu_{\tilde{E}_{\infty}}\}$, $\cal{D}^B_Y = \{D_p \mid p \neq 0,\infty,-1\}$. It defines a supported coloured hypercone of type II in $\cal{H}$ with associated polytope $\cal{P} = \cal{P}_{-1} = \{-1\}$. Hence $X$ (and therefore the variety $\bl_Z{W}$) has the following coloured hyperfan:

[ultra thick] (-2.5,0) – (0,0);
[semithick][->] (0,0) – (2.5,0);
[ultra thick][->] (0,0) – (0,3.5);
[dashed][domain = -2.5:0] plot (, -);
[below] at (2.5,0) $\ell$;
[above] at (0,3.5) $h$;
[right] at (0.1,1) $\tilde{D}_p$;
[fill=black] (-1,0) circle [color=black,radius=0.05];
[below] at (-1,0) $-1$;
[below] at (0,-0.5) $p \neq p_f,p_v,p_e$;
(0,1) circle [radius=0.15];
[on background layer]
[pattern=my north east lines](-2.5,3.5) to (-2.5,0) to (0,0) to (0,3.5);

[ultra thick] (-2.5,0) – (0,0);
[semithick][->] (0,0) – (2.5,0);
[semithick][->] (0,0) – (0,3.5);
[ultra thick][domain = -2.5:0] plot (, -);
[fill=black] (0,2) circle [color=black,radius=0.05];
[right] at (0,2) $2$;
[fill=black] (-1,0) circle [color=black,radius=0.05];
[below] at (-1,0) $-1$;
[fill=black] (-1,0) circle [color=black,radius=0.05];
[below] at (2.5,0) $\ell$;
[above] at (0,3.5) $h$;
[below] at (0,-0.5) $p_f = -1$;
[fill=black] (-1,1) circle [color=black,radius=0.15];
[right] at (-0.9,1) $E$;
(-1,2) circle [radius=0.15];
[right] at (-0.9,2) $\tilde{D}_{-1}$;
[on background layer]
[pattern=my north east lines](-2.5,0) to (-2.5,2.5) to (0,0);

[semithick][->] (0,0) – (2.5,0);
[ultra thick] (-2.5,0) – (0,0);
[ultra thick][->] (0,0) – (0,3.5);
[below] at (2.5,0) $\ell$;
[above] at (0,3.5) $h$;
[fill=black] (1,0) circle [color=black,radius=0.05];
[below] at (1,0) $1$;
[fill=black] (0,2) circle [color=black,radius=0.05];
[right] at (0,2) $2$;
[below] at (0,-0.5) $p_e = 0$;
(1,2) circle [radius=0.15];
[right] at (1.1,2) $\tilde{D}_0$;
[fill=black] (0,1) circle [color=black,radius=0.15];
[right] at (0.1,1) $\tilde{E}_0$;
[on background layer]
[pattern=my north west lines](-2.5,0) to (-2.5,3.5) to (0,3.5) to (0,0);

[semithick][->] (0,0) – (2.5,0);
[ultra thick] (-2.5,0) – (0,0);
[ultra thick][->] (0,0) – (0,3.5);
[below] at (2.5,0) $\ell$;
[above] at (0,3.5) $h$;
[fill=black] (1,0) circle [color=black,radius=0.05];
[below] at (1,0) $1$;
[fill=black] (0,2) circle [color=black,radius=0.05];
[right] at (0,2) $2$;
[below] at (0,-0.5) $p_v = \infty$;
(1,2) circle [radius=0.15];
[right] at (1.1,2) $\tilde{D}_\infty$;
[fill=black] (0,1) circle [color=black,radius=0.15];
[right] at (0.1,1) $\tilde{E}_\infty$;
[on background layer]
[pattern=my north west lines](-2.5,0) to (-2.5,3.5) to (0,3.5) to (0,0);

figureColoured hyperfan of the blow-up of $W$ along $Z$

§.§ Blow-up of $\proj{3}$ Along the Twisted Cubic (2.27)

§.§.§ Homogeneous Space and Structure

Let $G=\sl{2}$ act on $\proj{3} = \bb{P}(S^3\bbk^2)$ as in <Ref>. The point $P = [1:0:0:1]$ has as its $G$-stabiliser the binary dihedral group $H = \tilde{D}_3$ of degree 3 and order 12. The same point has $B$-stabiliser equal to the group of sixth roots of unity, so $\dim{B\cdot P} = 2$. Hence $G\cdot P$ is the homogeneous space $G/H$, realising $\proj{3}$ (in a different way to previous examples) as a quasihomogeneous complexity-one $G$-variety.

Let $Z$ be the rational normal curve of degree 3 in $\proj{3}$, i.e. $Z = \cal{Z}(x_0x_2-x_1^2,x_0x_3-x_1x_2,x_1x_3-x_2^2)$. Then $Z$ is a closed orbit in $\proj{3}$ by rnc. We will see that $Z$ is the unique minimal $G$-germ of $\proj{3}$, and is contained in its unique $G$-divisor.

§.§.§ (Semi-) Invariant Functions

By <Ref>, there are semi-invariant functions $f_v, f_e, f_f$ in $\bbk[G]^{(B\times H)}$ of respective biweights $(3\alpha,(-1,3)), (3\alpha,(-1,1))$ and $(2\alpha,(1,2))$.

In $\proj{3}$, $x_3$ is semi-invariant of $B$-weight $3\alpha$, and $x_1x_3-x_2^2$ is semi-invariant of $B$-weight $2\alpha$ by semi. Hence the latter corresponds to $f_f$. On $G/H$, we see that, acting with $\begin{psmallmatrix} 0 & i \\ i & 0\end{psmallmatrix} \in H$ on the right, $x_3$ has $H$-weight $(-1,3)$, so $x_3$ corresponds to $f_v$ Finally, $2x_2^3-3x_1x_2x_3+x_0x_3^2$ has $B$-weight $3\alpha$ and thus by process of elimination it corresponds to $f_e$.

Now the function $f_f^3/f_e^2$ is $B$-invariant, so gives the $B$-quotient map $\pi$ to $\proj{1}$, i.e. $\pi(P) = [f_f^3(P):f_e^2(P)]$ for $P \in \proj{3}$. This defines a family of regular colours $D_p = \pi^*(p)$ in $\proj{3}$ for all $p=[\alpha:\beta] \in \proj{1}$ except:
\[p = 0: \pi^*(p) = \cal{Z}(f_e^2) = D_0,\]

\[p = \infty: \pi^*(p) = \cal{Z}(f_f^3) = D_\infty,\]

\[p = -4: \pi^*(p) = \cal{Z}(f_v^2) \union \cal{Z}(3x_1^2x_2^2-4x_1^3x_3-x_0^2x_3^2-4x_0x_2^3+6x_0x_1x_2x_3) = D_{-4} \union F\]
where $F$ is a $G$-divisor. The subregular colours $D_0, D_\infty$ and $D_{-4}$ have multiplicities 2, 3 and 2, respectively, in accordance with the pictures above.

Finally, we choose a splitting $e_{2\alpha} = f_vf_e/f_f^2 \in K^{(B)}_{2\alpha}$.

§.§.§ Coloured Data and Hyperfan of $\proj{3}$

The minimal $G$-germ is contained in $F$ and in every colour except $D_{-4}$, so has coloured data $\cal{V}_Z = \{\nu_F\}, D^B_Z = \{D_p \mid p\neq -4\}$. It therefore defines a supported coloured hypercone of type II in $\cal{H}$.

By our choice of splitting we see that $D_0 \mapsto (\ell,h) = (1,2)$ in hyperspace, $D_\infty \mapsto (-2,3)$, $D_{-4} \mapsto (1,2)$, $F \mapsto (0,1)$ and as always the regular colours $D_p \mapsto (0,1)$ for $p \neq 0,\infty,-4$.

Therefore the polytope defined by $Z$ is given by $\cal{P} = \cal{P}_0 + \cal{P}_\infty + \cal{P}_{-4} = \{1/2 - 1/3 + 0\} = \{-1/6\}$.

Hence the coloured hyperfan of $\proj{3}$ looks like:

[ultra thick](-2.5,0) – (0,0);
[semithick][->] (0,0) – (2.5,0);
[semithick][->] (0,0) – (0,3.5);
[ultra thick](0,0) – (-7/3,3.5);
[dashed](0,0) – (-2.5,2.5);
[fill=black] (0,3) circle [color=black,radius=0.05];
[right] at (0,3) $3$;
[fill=black] (-2,0) circle [color=black,radius=0.05];
[below] at (-2,0) $-2$;
[below] at (2.5,0) $\ell$;
[above] at (0,3.5) $h$;
[below] at (0,-0.5) $p_f=\infty$;
(-2,3) circle [radius=0.15];
[right] at (-2,3) $D_{\infty}$;
[on background layer]
[pattern=my north east lines](-2.5,3.5) to (-7/3,3.5) to (0,0) to (-2.5,0);

[ultra thick](-2.5,0) – (0,0);
[semithick][->] (0,0) – (2.5,0);
[ultra thick][->] (0,0) – (0,3.5);
[below] at (2.5,0) $\ell$;
[above] at (0,3.5) $h$;
[fill=black] (0,2) circle [color=black,radius=0.05];
[right] at (0,2) $2$;
[fill=black] (1,0) circle [color=black,radius=0.05];
[below] at (1,0) $1$;
[below] at (0,-0.5) $p_v=-4$;
(1,2) circle [radius=0.15];
[right] at (1.1,2) $D_{-4}$;
[fill=black] (0,1) circle [color=black,radius=0.15];
[right] at (0,1) $F$;
[on background layer]
[pattern=my north west lines](-2.5,0) to (-2.5,3.5) to (0,3.5) to (0,0);

[ultra thick](-2.5,0) – (0,0);
[semithick][->] (0,0) – (2.5,0);
[semithick][->] (0,0) – (0,3.5);
[ultra thick] (0,0) – (1.75,3.5);
[below] at (2.5,0) $\ell$;
[above] at (0,3.5) $h$;
[fill=black] (0,2) circle [color=black,radius=0.05];
[right] at (0,2) $2$;
[fill=black] (1,0) circle [color=black,radius=0.05];
[below] at (1,0) $1$;
[below] at (0,-0.5) $p_e=0$;
(1,2) circle [radius=0.15];
[right] at (1.1,2) $D_0$;
[on background layer]
[pattern=my north west lines](-2.5,0) to (-2.5,3.5) to (1.75,3.5) to (0,0);

[ultra thick] (-2.5,0) – (0,0);
[thick][->] (0,0) – (2.5,0);
[ultra thick][->] (0,0) – (0,3.5);
[dashed][domain = -2.5:0] plot (, -);
[below] at (2.5,0) $\ell$;
[above] at (0,3.5) $h$;
[right] at (0.1,1) $D_p$;
[fill=black] (-1,0) circle [color=black,radius=0.05];
[below] at (-1,0) $-1$;
[below] at (0,-0.5) $p \neq p_f,p_v,p_e$;
(0,1) circle [radius=0.15];
[on background layer]
[pattern=my north east lines](-2.5,3.5) to (0,3.5) to (0,0) to (-2.5,0);

figureColoured hyperfan of $\proj{3}$ (cubic action)

§.§.§ Blow-up

Now to get the variety we need, we blow up $Z$. To simplify what happens, we will take affine charts. Take the minimal $B$-chart for $Z$, which is $U_Z = \proj{3}\setminus{\cal{Z}(x_3)}$. By setting $x_3 =1$, we get $U_Z \inter \proj{3} = \aff{3} = \Spec{\bbk[x_0,x_1,x_2]}$. Then $Z \inter U_Z$ becomes $\cal{Z}(x_0-x_1x_2,x_1-x_2^2)$.

Now $X = \bl_{Z \inter U_Z}(\aff{3}) = \cal{Z}(z_0(x_1-x_2^2)-z_1(x_0-x_1x_2)) \sub \aff{3} \times \proj{1}$. The exceptional divisor is $E= \cal{Z}(x_1-x_2^2,x_0-x_1x_2)$. Take another affine chart $V$ defined by $z_1 =1$. Using the equation for $X$ we can eliminate $x_0$ to obtain $X \inter V = \aff{3} = \Spec{\bbk[x_1,x_2,z_0]}$. Now $E \inter V = \cal{Z}(x_1-x_2^2)$.

On $V \inter X$, the $B$-invariant above becomes $(x_2^2-x_1)/(2x_2(1+z_0))^2$, and the splitting semi-invariant becomes $2x_2(1+z_0)/(x_1-x_2^2)$. Hence $E$ sits over $\infty$ in hyperspace and is mapped to $(-1,1) \in \cal{H}_\infty$.

Since the blow-up is an isomorphism away from $Z$, nothing else in hyperspace moves from its previous position. There is a new minimal $G$-germ $Y = E \inter \tilde{F}$ with coloured data $\cal{V}_Y = \{\nu_E,\nu_{\tilde{F}}\}, \cal{D}^B_Y = \{D_p \mid p\neq \infty,-4\}$. The coloured hyperfan for $X$ is thus:

[ultra thick](-2.5,0) – (0,0);
[semithick][->] (0,0) – (2.5,0);
[semithick][->] (0,0) – (0,3.5);
[ultra thick](0,0) – (-2.5,2.5);
[fill=black] (0,3) circle [color=black,radius=0.05];
[right] at (0,3) $3$;
[fill=black] (-2,0) circle [color=black,radius=0.05];
[below] at (-2,0) $-2$;
[below] at (2.5,0) $\ell$;
[above] at (0,3.5) $h$;
[below] at (0,-0.5) $p_f=\infty$;
[fill=black] (-1,1) circle [color=black,radius=0.15];
[right] at (-1,1) $E$;
(-2,3) circle [radius=0.15];
[right] at (-2,3) $\tilde{D}_{\infty}$;
[on background layer]
[pattern=my north east lines](-2.5,0) to (-2.5,2.5) to (0,0);

[ultra thick](-2.5,0) – (0,0);
[semithick][->] (0,0) – (2.5,0);
[ultra thick][->] (0,0) – (0,3.5);
[below] at (2.5,0) $\ell$;
[above] at (0,3.5) $h$;
[fill=black] (0,2) circle [color=black,radius=0.05];
[right] at (0,2) $2$;
[fill=black] (1,0) circle [color=black,radius=0.05];
[below] at (1,0) $1$;
[below] at (0,-0.5) $p_v=-4$;
(1,2) circle [radius=0.15];
[right] at (1.1,2) $\tilde{D}_{-4}$;
[fill=black] (0,1) circle [color=black,radius=0.15];
[right] at (0,1) $\tilde{F}$;
[on background layer]
[pattern=my north west lines](-2.5,0) to (-2.5,3.5) to (0,3.5) to (0,0);

[ultra thick](-2.5,0) – (0,0);
[semithick][->] (0,0) – (2.5,0);
[semithick][->] (0,0) – (0,3.5);
[ultra thick] (0,0) – (1.75,3.5);
[below] at (2.5,0) $\ell$;
[above] at (0,3.5) $h$;
[fill=black] (0,2) circle [color=black,radius=0.05];
[right] at (0,2) $2$;
[fill=black] (1,0) circle [color=black,radius=0.05];
[below] at (1,0) $1$;
[below] at (0,-0.5) $p_e=0$;
(1,2) circle [radius=0.15];
[right] at (1.1,2) $\tilde{D}_0$;
[on background layer]
[pattern=my north west lines](-2.5,0) to (-2.5,3.5) to (1.75,3.5) to (0,0);

[ultra thick] (-2.5,0) – (0,0);
[thick][->] (0,0) – (2.5,0);
[ultra thick][->] (0,0) – (0,3.5);
[dashed][domain = -2.5:0] plot (, -);
[below] at (2.5,0) $\ell$;
[above] at (0,3.5) $h$;
[right] at (0.1,1) $\tilde{D}_p$;
[fill=black] (-1,0) circle [color=black,radius=0.05];
[below] at (-1,0) $-1$;
[below] at (0,-0.5) $p \neq p_f,p_v,p_e$;
(0,1) circle [radius=0.15];
[on background layer]
[pattern=my north east lines](-2.5,3.5) to (0,3.5) to (0,0) to (-2.5,0);

figureColoured hyperfan of the blow-up of $\proj{3}$ along the twisted cubic

§.§ Blow-up of the Quadric Threefold (2.21)

§.§.§ Homogeneous Space and Structure

Let $G = \sl{2}$ act on $\proj{4} = \bb{P}(S^4\bbk^2)$ as in <Ref>. Then the quadric hypersurface $Q = \cal{Z}(3x_2^2-4x_1x_3+x_0x_4)$ is a smooth, $G$-stable threefold. The stabiliser $G_P$ of the point $P = [0:1:0:0:1] \in Q$ is the binary tetrahedral group $\widetilde{T}$, hence the orbit $G\cdot P$ is 3 dimensional and thus open. The same point has the group of sixth roots of unity as its $B$-stabiliser, so has a 2-dimensional $B$ orbit, so that $Q$ is a quasihomogeneous complexity one $G$-variety containing the homogeneous space $G/H = \sl{2}/\tilde{T}$.

Let $Z$ be the rational normal curve of degree 4 in $\proj{4}$, i.e. $Z = \cal{Z}(x_0x_2-x_1^2,x_0x_3-x_1x_2,x_1x_4-x_2x_3,x_2x_4-x_3^2)$. Then $Z$ is a closed $G$-orbit in $Q$ by rnc. We will see that $Z$ is the unique minimal $G$-germ in $Q$, and is contained in a unique $G$-divisor.

§.§.§ (Semi-) Invariant Functions

We know from <Ref> that there are semi-invariant regular functions $f_e, f_f, f_v$ in $\bbk[G]^{(B \times H)}$ of respective biweights $(6\alpha,1), (4\alpha, \epsilon^{-1})$ and $(4\alpha,\epsilon)$, where $\epsilon$ is a primitive cube root of unity.

On $Q$, $x_4$ and $x_2x_4-x_3^2$ have $B$-weight $4\alpha$ by semi, and checking on the homogeneous space $G/H = G\cdot[0:1:0:0:1]$ we see that acting by $\begin{psmallmatrix}\epsilon & 0 \\ 0 & \epsilon^{-1}\end{psmallmatrix}$ on the right, they have $H$-weights $\epsilon^{-1}$ and $\epsilon$ respectively. Hence $x_4 = f_f$, $x_2x_4-x_3^2 = f_v$. Finally, the function $2x_3^3+x_1x_4^2-3x_2x_3x_4$ is $B$-semi-invariant of weight $6\alpha$ and $H$-invariant, so this is $f_e$.

Now $f_v^3/f_e^2$ is a $B$-invariant rational function on $Q$, so the $B$-quotient map $\pi$ is given by $\pi(P) = [f_v^3(P):f_e^2(P)]$ for $P \in Q$, and defines a family of regular colours $D_p = \pi^*(p)$ for all $p=[\alpha:\beta] \in \proj{1}$ except:
\[p = 0: \pi^*(p) = \cal{Z}(f_e^2) = D_0,\]

\[p = \infty: \pi^*(p) = \cal{Z}(f_v^3) = D_\infty,\]

\[p = -4: \pi^*(p) = \cal{Z}(f_f^3)\union\cal{Z}(4x_2^3+x_1^2x_4+x_0x_3^2-6x_1x_2x_3) = D_{-4} \union F\]
where $F$ is a $G$-invariant divisor. Note that the subregular colours $D_0, D_\infty$ and $D_{-4}$ have multiplicities 2, 3 and 3, respectively, as we should expect from the hyperspace of $G/H$.

Finally, we choose as a splitting semi-invariant $e_{2\alpha} = f_vf_f/f_e \in K^{(B)}_{2\alpha}$.

§.§.§ Coloured Data and Hyperfan of Q

The minimal $G$-germ $Z$ is contained in $F$ and in every colour except $D_{-4}$, so has coloured data $\cal{V}_Z = \{\nu_F\}$, $\cal{D}^B_Z = \{D_p \mid p\neq -4\}$. Hence it defines a supported coloured hypercone of type II in $\cal{H}$.

Our choice of invariant and splitting functions mean that $D_0$ sits at $(\ell,h) = (-1,2)$ in hyperspace, $D_{-4}, D_\infty \mapsto (1,3)$, $F \mapsto (0,1)$ and as always for regular colours we have $D_p \mapsto (0,1)$ for $p\neq 0,\infty,-4$.

Hence the polytope defined by $Z$ is given by $\cal{P} = \cal{P}_0 + \cal{P}_\infty + \cal{P}_{-4} = \{-1/2 + 1/3 + 0\} = \{-1/6\}$. Therefore the full coloured hyperfan defined by $Q$ is as follows:

[ultra thick](-2.5,0) – (0,0);
[semithick][->] (0,0) – (2.5,0);
[ultra thick][->] (0,0) – (0,3.5);
[dashed][domain = -2.5:0] plot (, -);
[below] at (2.5,0) $\ell$;
[above] at (0,3.5) $h$;
[right] at (0.2,1) $D_p$;
[fill=black] (-1,0) circle [color=black,radius=0.05];
[below] at (-1,0) $-1$;
[below] at (0,-0.5) $p \neq p_f,p_v,p_e$;
(0,1) circle [radius=0.15];
[on background layer]
[pattern=my north east lines](-2.5,3.5) to (-2.5,0) to (0,0) to (0,3.5);

[ultra thick](-2.5,0) – (0,0);
[semithick][->] (0,0) – (2.5,0);
[semithick][->] (0,0) – (0,3.5);
[dashed][domain = -2.5:0] plot (, -);
[ultra thick](0,0) – (-1.75,3.5);
[fill=black] (0,2) circle [color=black,radius=0.05];
[right] at (0,2) $2$;
[fill=black] (-1,0) circle [color=black,radius=0.05];
[below] at (-1,0) $-1$;
[fill=black] (-1,0) circle [color=black,radius=0.05];
[below] at (2.5,0) $\ell$;
[above] at (0,3.5) $h$;
[below] at (0,-0.5) $p_e=0$;
(-1,2) circle [radius=0.15];
[right] at (-1,2) $D_0$;
[on background layer]
[pattern=my north east lines](-2.5,0) to (-2.5,3.5) to (-1.75,3.5) to (0,0);

[ultra thick] (-2.5,0) – (0,0);
[semithick][->] (0,0) – (2.5,0);
[ultra thick][->] (0,0) – (0,3.5);
[below] at (2.5,0) $\ell$;
[above] at (0,3.5) $h$;
[fill=black] (1,0) circle [color=black,radius=0.05];
[below] at (1,0) $1$;
[fill=black] (0,3) circle [color=black,radius=0.05];
[right] at (0,3) $3$;
[fill=black] (0,1) circle [color=black,radius=0.15];
[right] at (0.1,1) $F$;
[below] at (0,-0.5) $p_f=-4$;
(1,3) circle [radius=0.15];
[right] at (1.1,3) $D_{-4}$;
[on background layer]
[pattern=my north west lines](-2.5,0) to (-2.5,3.5) to (0,3.5) to (0,0);

[ultra thick] (-2.5,0) – (0,0);
[semithick][->] (0,0) – (2.5,0);
[semithick][->] (0,0) – (0,3.5);
[ultra thick] (0,0) – (3.5/3,3.5);
[below] at (2.5,0) $\ell$;
[above] at (0,3.5) $h$;
[fill=black] (1,0) circle [color=black,radius=0.05];
[below] at (1,0) $1$;
[fill=black] (0,3) circle [color=black,radius=0.05];
[right] at (0,3) $3$;
[below] at (0,-0.5) $p_v=\infty$;
(1,3) circle [radius=0.15];
[right] at (1.1,3) $D_\infty$;
[on background layer]
[pattern=my north west lines](-2.5,0) to (-2.5,3.5) to (3.5/3,3.5) to (0,0);

figureColoured hyperfan of the quadric threefold $Q$

§.§.§ Blow-up

The variety we want is obtained by blowing up $Z$. We calculate the effect on the hyperfan by blowing up in an affine chart. Since $Z$ defines a coloured hypercone of type II, it has a minimal $B$-chart $U_Z$ with the same coloured data. Indeed $Z$ is contained in every colour except $D_{-4} = \cal{Z}(x_4)$, so $U_Z = \proj{4}\setminus{\cal{Z}(x_4)}$. Setting $x_4 = 1$ allows us to eliminate $x_0$ using the equation for $Q$, so $U_Z \inter Q \cong \aff{3} = \Spec{\bbk[x_1,x_2,x_3]}$. Then $U_Z \inter Z = \cal{Z}(x_1-x_2x_3,x_2-x_3^2)$.

Now $X = \bl_{U_Z \inter Z}(\aff{3}) = \cal{Z}(z_0(x_2-x_3^2)-z_1(x_1-x_2x_3)) \sub \aff{3} \times \proj{1}$. The exceptional divisor $E$ is given by $\cal{Z}(x_1-x_2x_3,x_2-x_3^2)$. Now we take another affine chart $V$ by setting $z_1=1$, which allows us to eliminate $x_1$ using the equation for $X$. Thus $V \inter X = \aff{3} = \Spec{\bbk[x_2,x_3,z_0]}$, and $E \inter V = \cal{Z}(x_2-x_3^2)$.

On $V \inter X$, the invariant function $f_v^3/f_e^2$ becomes $(x_2-x_3^2)/(z_0-2x_3)^2$, and the splitting semi-invariant is $1/(z_0-2x_3)$, so we see that $E \mapsto (p,\ell,h) = (\infty,0,1)$ in hyperspace.

Since the blow-up is an isomorphism away from $Z$, all other colours and $G$-divisors lie at the same points in hyperspace as before. The blow-up introduces a new minimal $G$-germ $Y = E \inter \tilde{F}$ which must have coloured data $\cal{V}_Y = \{\nu_E,\nu_{\tilde{F}}\}$, $\cal{D}^B_Y = \{D_p \mid p\neq \infty,-4\}$. Hence the coloured hyperfan for $X$ looks like:

[ultra thick](-2.5,0) – (0,0);
[semithick][->] (0,0) – (2.5,0);
[ultra thick][->] (0,0) – (0,3.5);
[dashed][domain = -2.5:0] plot (, -);
[below] at (2.5,0) $\ell$;
[above] at (0,3.5) $h$;
[right] at (0.2,1) $\tilde{D}_p$;
[fill=black] (-1,0) circle [color=black,radius=0.05];
[below] at (-1,0) $-1$;
[below] at (0,-0.5) $p \neq p_f,p_v,p_e$;
(0,1) circle [radius=0.15];
[on background layer]
[pattern=my north east lines](-2.5,3.5) to (-2.5,0) to (0,0) to (0,3.5);

[ultra thick](-2.5,0) – (0,0);
[semithick][->] (0,0) – (2.5,0);
[semithick][->] (0,0) – (0,3.5);
[dashed][domain = -2.5:0] plot (, -);
[ultra thick](0,0) – (-1.75,3.5);
[fill=black] (0,2) circle [color=black,radius=0.05];
[right] at (0,2) $2$;
[fill=black] (-1,0) circle [color=black,radius=0.05];
[below] at (-1,0) $-1$;
[fill=black] (-1,0) circle [color=black,radius=0.05];
[below] at (2.5,0) $\ell$;
[above] at (0,3.5) $h$;
[below] at (0,-0.5) $p_e=0$;
(-1,2) circle [radius=0.15];
[right] at (-1,2) $\tilde{D}_0$;
[on background layer]
[pattern=my north east lines](-2.5,0) to (-2.5,3.5) to (-1.75,3.5) to (0,0);

[ultra thick] (-2.5,0) – (0,0);
[semithick][->] (0,0) – (2.5,0);
[ultra thick][->] (0,0) – (0,3.5);
[below] at (2.5,0) $\ell$;
[above] at (0,3.5) $h$;
[fill=black] (1,0) circle [color=black,radius=0.05];
[below] at (1,0) $1$;
[fill=black] (0,3) circle [color=black,radius=0.05];
[right] at (0,3) $3$;
[fill=black] (0,1) circle [color=black,radius=0.15];
[right] at (0.1,1) $\tilde{F}$;
[below] at (0,-0.5) $p_f=-4$;
(1,3) circle [radius=0.15];
[right] at (1.1,3) $\tilde{D}_{-4}$;
[on background layer]
[pattern=my north west lines](-2.5,0) to (-2.5,3.5) to (0,3.5) to (0,0);

[ultra thick] (-2.5,0) – (0,0);
[semithick][->] (0,0) – (2.5,0);
[ultra thick][->] (0,0) – (0,3.5);
[below] at (2.5,0) $\ell$;
[above] at (0,3.5) $h$;
[fill=black] (1,0) circle [color=black,radius=0.05];
[below] at (1,0) $1$;
[fill=black] (0,3) circle [color=black,radius=0.05];
[right] at (0,3) $3$;
[fill=black] (0,1) circle [color=black,radius=0.15];
[right] at (0.1,1) $E$;
[below] at (0,-0.5) $p_v=\infty$;
(1,3) circle [radius=0.15];
[right] at (1.1,3) $\tilde{D}_\infty$;
[on background layer]
[pattern=my north west lines](-2.5,0) to (-2.5,3.5) to (0,3.5) to (0,0);

figureColoured hyperfan of the blow-up of $Q$ along $Z$

§.§ $V_5$ (1.15)

§.§.§ Description and Homogeneous Space

Let $G=\sl{2}$ act on $\bbk^2$ with the standard linear action, and hence on $\proj{6} = \bb{P}(S^6\bbk^2)$. By [12], we can realise $V_5$ as the closure in $\proj{6}$ of the $G$-orbit of the point $P = [0:1:0:0:0:-1:0]$. This explicitly shows that $V_5$ is a quasihomogeneous $G$-variety of complexity one. The stabiliser of $P$ in $G$ contains the matrices $y, r$ and $\omega$ which, in <Ref>, were shown to generate the binary cubic group $H = \tilde{C}$. Since no other finite subgroup of $G$ contains $\tilde{C}$, we see that $G\cdot P \cong G/H$. Hence there are three subregular semi-invariants $f_v$, $f_e$ and $f_f$ of respective $(B \times H)$ biweights $(8\alpha,1), (12\alpha,-1)$ and $(6\alpha,-1)$ and multiplicities $3, 2$ and $4$.

The description of $V_5$ in [12] realises it as the subvariety of $\proj{6}$ defined by the equations
\begin{align*} x_0x_4-4x_1x_3+3x_2^2 & = 0, \\ x_0x_5-3x_1x_4+2x_2x_3 & = 0, \\ x_0x_6-9x_2x_4+8x_3^2 & = 0, \\ x_1x_6-3x_2x_5+2x_3x_4 & = 0, \\ x_2x_6-4x_3x_5+3x_4^2 & = 0.\end{align*}
It is clear from the above then that the rational normal curve $Z$ of degree 6 defined as the image of the Veronese embedding $\nu_6 \colon \proj{1} \to \proj{6}$ lies inside $V_5$ as a minimal $G$-germ. We will show later that $Z$ is in fact the unique minimal $G$-germ of $V_5$, contained in a unique $G$-stable divisor.

§.§.§ (Semi)-Invariant Functions

One can find semi-invariants of the correct $B$-weights by using a torus action, under which each co-ordinate function has a given weight. In this case $x_k$ has weight $-6+2k$. It is easy to see immediately that $x_6$ is $B$ semi-invariant of weight 6, so must be the subregular semi-invariant $f_f$. Likewise $x_4x_6-x_5^2$ is $B$ semi-invariant of weight 8, so represents $f_v$. Now from <Ref>, we know that $f_vf_f/f_e$ is a rational function, so has degree 0. Hence $f_e$ must have degree 3 and $B$- (hence $T$-) weight $12\alpha$, and therefore must be a linear combination of the monomials $x_3x_6^2, x_4x_5x_6$ and $x_5^3$. A simple check shows that $f_e = x_3x_6^2-3x_4x_5x_6+2x_5^3$ suffices.

Hence $f_v^3/f_e^2$ is an invariant rational function defining a rational $B$-quotient $\pi \colon V_5 \dashrightarrow \proj{1}$, $P \mapsto [f_v^3(P):f_e^2(P)]$. The pullback of $p = [\alpha:\beta] \in \proj{1}$ defines a regular colour for all $p$ except the following:
\[p = 0: \pi^*(p) = \cal{Z}(f_e^2) = D_0;\]

\[p = \infty: \pi^*(p) = \cal{Z}(f_v^3) = D_\infty;\]

\[p=-4: \pi^*(p) = \cal{Z}(x_6^4) \union \cal{Z}(x_1x_5+3x_3^2-4x_2x_3) = D_{-4} \union F;\]
where $F$ is then a unique $G$-divisor containing the $G$-germ $Z$. Likewise it is straightforward that $Z$ is contained in every colour except $D_{-4}$.

We choose as mentioned the semi-invariant splitting function $e_{2\alpha} = f_vf_f/f_e$ of $B$-weight $2\alpha$, and the colours $D_0, D_\infty$ and $D_{-4}$ are distinguished by it.

§.§.§ Coloured Hyperfan

We already know from previous discussions and our choice of splitting that regular colours $D_p$ for $p \neq 0, \infty -4$ are mapped to points $(p,0,1)$ of hyperspace, and that $D_0 \mapsto (0,-1,2)$, $D_\infty \mapsto (\infty,1,3)$ and $D_{-4} \mapsto (-4,1,4)$. Finally, it is easy to check on the $B$-chart $V_5\setminus{\cal{Z}(x_6)}$ that $F$ is mapped to $(-4,0,1)$.

The coloured data of the minimal $G$-germ $Z$ is $\cal{V}_Z = \{\nu_F\}$, $\cal{D}^B_Z = \{D_p \mid p \neq -4\}$. Hence $Z$ is a $G$-germ of type II and defines a supported coloured hypercone of type II generated by its coloured data and the polytope $\cal{P} = \cal{P}_0 + \cal{P}_\infty + \cal{P}_{-4} = \{-1/2\} + \{1/3\} + \{0\} = \{-1/6\}$. Hence the coloured hyperfan of $V_5$ is as follows:

[ultra thick] (-2.5,0) – (0,0);
[semithick][->] (0,0) – (2.5,0);
[semithick][->] (0,0) – (0,3.5);
[ultra thick] (0,0) – (-1.75,3.5);
[dashed,domain = -2.5:0] plot (, -);
[fill=black] (0,2) circle [color=black,radius=0.05];
[right] at (0,2) $2$;
[fill=black] (-1,0) circle [color=black,radius=0.05];
[below] at (-1,0) $-1$;
[fill=black] (-1,0) circle [color=black,radius=0.05];
[below] at (2.5,0) $\ell$;
[above] at (0,3.5) $h$;
[below] at (0,-0.5) $0 = p_e$;
(-1,2) circle [radius=0.15];
[right] at (-0.9,2) $D_0$;
[on background layer]
[pattern=my north east lines](-2.5,0) to (-2.5,3.5) to (-1.75,3.5) to (0,0);

[ultra thick] (-2.5,0) – (0,0);
[semithick][->] (0,0) – (2.5,0);
[ultra thick][->] (0,0) – (0,3.5);
[below] at (2.5,0) $\ell$;
[above] at (0,3.5) $h$;
[fill=black] (1,0) circle [color=black,radius=0.05];
[below] at (1,0) $1$;
[fill=black] (0,3) circle [color=black,radius=0.05];
[right] at (0,3) $4$;
[below] at (0,-0.5) $-4 = p_f$;
[fill=black] (0,1) circle [color=black,radius=0.15];
[right] at (0.1,1) $F$;
(1,3) circle [radius=0.15];
[right] at (1.1,3) $D_{-4}$;
[on background layer]
[pattern=my north west lines](-2.5,0) to (-2.5,3.5) to (0,3.5) to (0,0);

[ultra thick] (-2.5,0) – (0,0);
[semithick][->] (0,0) – (2.5,0);
[semithick][->] (0,0) – (0,3.5);
[ultra thick] (0,0) – (7/6,3.5);
[below] at (2.5,0) $\ell$;
[above] at (0,3.5) $h$;
[fill=black] (1,0) circle [color=black,radius=0.05];
[below] at (1,0) $1$;
[fill=black] (0,3) circle [color=black,radius=0.05];
[right] at (0,3) $3$;
[below] at (0,-0.5) $\infty = p_v$;
(1,3) circle [radius=0.15];
[right] at (1.1,3) $D_\infty$;
[on background layer]
[pattern=my north west lines](-2.5,0) to (-2.5,3.5) to (7/6,3.5) to (0,0);

[ultra thick] (-2.5,0) – (0,0);
[semithick][->] (0,0) – (2.5,0);
[ultra thick][->] (0,0) – (0,3.5);
[dashed,domain = -2.5:0] plot (, -);
[below] at (2.5,0) $\ell$;
[above] at (0,3.5) $h$;
[fill=black] (0,1) circle [color=black,radius=0.05];
[fill=black] (-1,0) circle [color=black,radius=0.05];
[below] at (-1,0) $-1$;
[below] at (0,-0.5) $p \neq p_f,p_v,p_e$;
(0,1) circle [radius=0.15];
[right] at (0.1,1) $D_p$;
[on background layer]
[pattern=my north east lines](-2.5,0) to (-2.5,3.5) to (0,3.5) to (0,0);

figureColoured hyperfan of $V_5$

§.§ $V_{22}$ (1.10)

§.§.§ Description and Homogeneous Space

Let $G=\sl{2}$ act on $\bbk^2$ with the standard linear action, and hence on $\proj{12} = \bb{P}(S^{12}\bbk^2)$. By [12], we can realise $V_{22}$ as the closure in $\proj{12}$ of the $G$-orbit of the point
\[P = [0:1:0:0:0:0:11:0:0:0:0:1:0].\]
This explicitly shows that $V_{22}$ is a quasihomogeneous $G$-variety of complexity one. The stabiliser of $P$ in $G$ contains $\begin{psmallmatrix}\epsilon & 0 \\ 0 & \epsilon^{-1}\end{psmallmatrix}$ for $\epsilon$ a primitive tenth root of unity. Furthermore, the function $x_{12}$ on $V_{22}$ is a $B$-eigenfunction of weight $12\alpha$. The only finite subgroup $H$ of $\sl{2}$ containing an element of order 10 and such that $\sl{2}/H$ has a semi-invariant of weight $12\alpha$ is $\tilde{I}$, the binary icosahedral group. Hence $V_{22}$ is an embedding of $\sl{2}/\tilde{I}$.

Then by <ref> there are three subregular semi-invariants $f_v$, $f_e$ and $f_f$ of respective $B$-weights $12\alpha, 30\alpha$ and $20\alpha$ and multiplicities $5, 2$ and $3$. Note that $\tilde{I}$ has no weights.

The description of $V_{22}$ in [12] realises it as the subvariety of $\proj{12}$ defined by the equations
\[\sum_{\lambda = 0}^{\rho}{{8 \choose \lambda}{8\choose \rho-\lambda}(x_{\lambda}x_{\rho+4-\lambda} -4x_{\lambda + 1}x_{\rho+3-\lambda} + 3x_{\lambda + 2}x_{\rho+2-\lambda})} = 0\]
for $0 \leq \rho \leq 16$. It is easy to check using these equations that the rational normal curve $Z$ of degree 12, defined as the image of the Veronese embedding $\nu_{12} \colon \proj{1} \to \proj{12}$, lies inside $V_{22}$ as a minimal $G$-germ. We will see that $Z$ is in fact the unique minimal $G$-germ of $V_{22}$, contained in a unique $G$-stable divisor.

§.§.§ (Semi)-Invariant Functions

One can find semi-invariants of the correct $B$-weights by using a torus action, under which each co-ordinate function has a given weight. In this case $x_k$ has weight $(-12+2k)\alpha$. We have mentioned that $x_{12}$ is $B$ semi-invariant of weight $12\alpha$, so must be the subregular semi-invariant $f_v$. Likewise $x_{10}x_{12}-x_{11}^2$ is $B$ semi-invariant of weight $20\alpha$, so represents $f_f$. Now from <Ref>, we know that $f_vf_f/f_e$ is a rational function, so has degree 0. Hence $f_e$ must have degree 3 and $B$- (hence $T$-) weight $30\alpha$, and therefore must be a linear combination of the monomials $x_{10}x_{11}x_{12}, x_{11}^3$ and $x_9x_{12}^2$. A simple check shows that $f_e = 3x_{10}x_{11}x_{12} - 2x_{11}^3 - x_9x_{12}^2$ suffices.

Hence $f_f^3/f_e^2$ is an invariant rational function defining a rational $B$-quotient $\pi \colon V_{22} \dashrightarrow \proj{1}$, $P \mapsto [f_f^3(P):f_e^2(P)]$. The pullback of $p = [\alpha:\beta] \in \proj{1}$ defines a regular colour for all $p$ except the following:
\[p = 0: \pi^*(p) = \cal{Z}(f_e^2) = D_0;\]

\[p = \infty: \pi^*(p) = \cal{Z}(f_v^3) = D_\infty.\]
We also see that the subregular colour $\cal{Z}(f_v)$ lies over $p = -4 = [1:-4]$, so we write $D_{-4} = \cal{Z}(f_v)$, recalling that this colour has multiplicity 5.

We choose as mentioned the semi-invariant splitting function $e_{2\alpha} = f_vf_f/f_e$ of $B$-weight $2\alpha$, and the colours $D_0, D_\infty$ and $D_{-4}$ are distinguished by it.

§.§.§ Coloured Hyperfan

We already know from previous discussions and our choice of splitting that regular colours $D_p$ for $p \neq 0, \infty -4$ are mapped to points $(p,0,1)$ of hyperspace, and that $D_0 \mapsto (0,-1,2)$, $D_\infty \mapsto (\infty,1,3)$ and $D_{-4} \mapsto (-4,1,5)$.

Consider the coloured data of the minimal $G$-germ $Z$: we know that $\cal{D}^B_Z = \{D_p \mid p \neq -4\}$. This is infinite, so $Z$ is a $G$-germ of type II and defines a coloured hypercone generated by its coloured data and the polytope $\cal{P}$. However, the slice $\cal{H}_{-4}$ does not contain a non-central element of $\cal{D}^B_Z$, and hence must contain a non-central element of $\cal{V}_Z$ since $Z$ is of type II. Hence there must be a $G$-divisor $F \sub V_{22}$ containing $Z$ and lying over $-4 \in \proj{1}$. In the slice $\cal{H}_{-4}$, $\cal{C}_Z$ is then generated by  $\cal{P}$, which is central, and $F$, and by completeness it must cover the valuation cone. Hence we must have $F \mapsto (-4,0,h)$ for some $h \geq 0$, and the hyperfan is the same for any such $h$, so take $h=1$. Then $\cal{P} = \cal{P}_0 + \cal{P}_\infty + \cal{P}_{-4} = \{-1/2\} + \{1/3\} + \{0\} = \{-1/6\}$. Hence the coloured hyperfan of $V_{22}$ is as follows:

[ultra thick] (-2.5,0) – (0,0);
[semithick][->] (0,0) – (2.5,0);
[semithick][->] (0,0) – (0,3.5);
[ultra thick] (0,0) – (-1.75,3.5);
[dashed,domain = -2.5:0] plot (, -);
[fill=black] (0,2) circle [color=black,radius=0.05];
[right] at (0,2) $2$;
[fill=black] (-1,0) circle [color=black,radius=0.05];
[below] at (-1,0) $-1$;
[fill=black] (-1,0) circle [color=black,radius=0.05];
[below] at (2.5,0) $\ell$;
[above] at (0,3.5) $h$;
[below] at (0,-0.5) $0 = p_e$;
(-1,2) circle [radius=0.15];
[right] at (-0.9,2) $D_0$;
[on background layer]
[pattern=my north east lines](-2.5,0) to (-2.5,3.5) to (-1.75,3.5) to (0,0);

[ultra thick] (-2.5,0) – (0,0);
[semithick][->] (0,0) – (2.5,0);
[ultra thick][->] (0,0) – (0,3.5);
[below] at (2.5,0) $\ell$;
[above] at (0,3.5) $h$;
[fill=black] (1,0) circle [color=black,radius=0.05];
[below] at (1,0) $1$;
[fill=black] (0,3) circle [color=black,radius=0.05];
[right] at (0,3) $5$;
[below] at (0,-0.5) $-4 = p_v$;
[fill=black] (0,1) circle [color=black,radius=0.15];
[right] at (0.1,1) $F$;
(1,3) circle [radius=0.15];
[right] at (1.1,3) $D_{-4}$;
[on background layer]
[pattern=my north west lines](-2.5,0) to (-2.5,3.5) to (0,3.5) to (0,0);

[ultra thick] (-2.5,0) – (0,0);
[semithick][->] (0,0) – (2.5,0);
[semithick][->] (0,0) – (0,3.5);
[ultra thick] (0,0) – (7/6,3.5);
[below] at (2.5,0) $\ell$;
[above] at (0,3.5) $h$;
[fill=black] (1,0) circle [color=black,radius=0.05];
[below] at (1,0) $1$;
[fill=black] (0,3) circle [color=black,radius=0.05];
[right] at (0,3) $3$;
[below] at (0,-0.5) $\infty = p_f$;
(1,3) circle [radius=0.15];
[right] at (1.1,3) $D_\infty$;
[on background layer]
[pattern=my north west lines](-2.5,0) to (-2.5,3.5) to (7/6,3.5) to (0,0);

[ultra thick] (-2.5,0) – (0,0);
[semithick][->] (0,0) – (2.5,0);
[ultra thick][->] (0,0) – (0,3.5);
[dashed,domain = -2.5:0] plot (, -);
[below] at (2.5,0) $\ell$;
[above] at (0,3.5) $h$;
[fill=black] (0,1) circle [color=black,radius=0.05];
[fill=black] (-1,0) circle [color=black,radius=0.05];
[below] at (-1,0) $-1$;
[below] at (0,-0.5) $p \neq p_f,p_v,p_e$;
(0,1) circle [radius=0.15];
[right] at (0.1,1) $D_p$;
[on background layer]
[pattern=my north east lines](-2.5,0) to (-2.5,3.5) to (0,3.5) to (0,0);

figureColoured hyperfan of $V_{22}$

§ APPLYING KSTABLECENTRAL

In this section we show that Kstablecentral applies to the varieties whose combinatorial data we have just calculated.

§.§ Action with no Fixed Points

The first case of Kstablecentral applies to $\proj{3}$ (1.17), the blow-up of $\proj{3}$ in two lines (3.25) and the blow-up of $\proj{3}$ in three lines (4.6).

Realise $\proj{1}$ as the unit sphere and consider the action of the symmetric group $A = S_3$ on $\proj{1}$ consisting of rotations permuting the vertices of an equilateral triangle inscribed in the equator. This action has no fixed points, one orbit of order 2 consisting of the north and south poles, and various other orbits of orders 3 and 6.

If $X = \proj{3}$, then $S_3 \sub \aut{X} = \pgl{4}$. Recall that the slices of the coloured hyperfan are all identical to each other except for that of the distinguished point, which lies two units to the left on the $\ell$-axis relative to the other slices. Hence we can shift the slice of the distinguished point $p_0$ one unit to the right and shift the slice of some other point $p_1$ one unit to the left. Now $\cal{H}_{p_0}$ and $\cal{H}_{p_1}$ look the same, so we identify these two points with the north and south poles of the sphere. It follows that the $A$-action preserves the coloured hyperfan of $X$, in the sense that for $a \in A$ and $p \in \proj{1}$, we permute the slices of $\cal{H}$ via $\cal{H}_{p} \to \cal{H}_{a\cdot p}$ and the coloured data within each slice are invariant with respect to these permutations. This means that the action of $A$ on $X$ is such that the $B$-quotient is $A$-invariant and Kstablecentral applies.

If $X$ is the blow-up of $\proj{3}$ along two lines, say $Y_q$ and $Y_r$, then let $q$ and $r$ be the north and south poles of the sphere $\proj{1}$. Shift their slices of the hyperspace two units to the left each, then choose two other non-distinguished points $p_1$ and $p_2$ and shift their slices two units to the right each. Then the slices corresponding to $p_1$, $p_2$ and the distinguished point $p_0$ align, and we can choose these to be the vertices of the equilateral triangle acted on by $S_3$. Again, $S_3$ acts on $X$ and preserves the coloured hyperfan up to balanced integral shifts, so Kstablecentral applies.

If $X$ is the blow-up of $\proj{3}$ along three lines, $Y_q, Y_r$ and $Y_s$, the method is the same as when $X = \proj{3}$, only making sure to choose $q, r$ and $s$ to be the vertices of the triangle. Again, Kstablecentral applies.

As an aside, it is worth mentioning that the blow-up of $\proj{3}$ along one line (2.33) has non-reductive automorphism group and is thus not $K$-polystable. It is easy to see that the method above does not work in this case, since the slice of hyperspace corresponding to the blown-up line is fundamentally distinct from all other slices for this variety, so no action on $\proj{1}$ which does not fix the corresponding point could ever preserve the coloured hyperfan.

§.§ Action Interchanging Two Points

The second case of Kstablecentral applies to the blow-up of $\proj{1} \times \proj{2}$ along a curve of bidegree $(1,1)$ (3.17).

Consider the $\bb{Z}_2$-symmetry of $\proj{1}$ given by $[\alpha : \beta] \mapsto [\beta:\alpha]$. This interchanges $0$ and $\infty$, fixes $1$ and $-1$, and puts every other point in an orbit of order 2. Recall that the blow-up of $\proj{1} \times \proj{2}$ has subregular colours lying over 0 and $\infty$, and has a distinguished point $-1$. Since the slices of the coloured hyperfan over $0$ and $\infty$ are identical to each other, the interchange of these points by $\bb{Z}_2$ leaves the hyperfan invariant, hence the $\bb{Z}_2$ action on $X$ respects the $B$-quotient map and Kstablecentral applies.

Oddly, this method seems not to apply to $\proj{1} \times \proj{2}$ itself, even though this variety is known to be $K$-polystable, since in this case, one of the two slices with subregular colours contains a $G$-divisor while the other does not, and the $\bb{Z}_2$-symmetry therefore does not extend. This may require some further exploration.

§.§ Three or More Subregular Colours

We have seen that any smooth Fano $\sl{2}$-threefold whose stabiliser subgroup $H$ is one of $\{\tilde{D}_m, \tilde{T}, \tilde{C}, \tilde{I}\}$ has 3 subregular colours, all lying over distinct points in $\proj{1}$. Then in particular, the third case of Kstablecentral applies to $V_{22}$ (1.10), $V_5$ (1.15), $Q$ (1.16), the blow-up of $Q$ along a twisted quartic (2.21), the blow-up of $\proj{3}$ along a twisted cubic (2.27), $W$ (2.32) and the blow-up of $W$ along a curve of bidegree $(2,2)$ (3.13).

It remains to verify that each of these varieties satisfies the hypotheses of nonint, which we now do one by one.

§.§.§ $V_{22}$ (1.10)

It is known that for $X = V_{22}$, the Mukai-Umemura threefold, we can take $-K_X$ to be a hyperplane section. The $B$-invariant hyperplane section of $X$ for our action is $D_{-4}$. This variety has subregular colours lying over $0, \infty, -4 \in \proj{1}$. With this representative of $-K_X$, we have
\[H_\lambda = -\frac{\lambda}{2}[0] + \frac{\lambda}{3}[\infty] + \min\left\{\frac{\lambda+1}{5},0\right\}[-4].\]
Choosing $\lambda = -5$ gives $\deg{H_\lambda} = \frac{1}{30}$ and non-integral coefficients at all three points.

§.§.§ $V_5$ (1.15)

This time $-K_X = 2D_{-4}$. We have
\[H_\lambda = -\frac{\lambda}{2}[0]+\frac{\lambda}{3}[\infty] + \min\left\{\frac{\lambda+2}{4}, 0\right\}[-4].\]
Again, $\lambda = -5$ works, giving $\deg{H_\lambda} = \frac{1}{12}$ and non-integral coefficients at all three points.

§.§.§ $Q$ (1.16)

$Q$ is a hypersurface of degree 2 in $\proj{4}$ so its anticanonical divisor is given by its intersection with a divisor of degree 3 in $\proj{4}$. The three subregular colours $D_0$, $D_\infty$ and $D_{-4}$ of $Q$ are sections of prime divisors in $\proj{4}$ of degrees 3, 2 and 1, respectively, so we may take $-K_Q = 3D_{-4}$. Then
\[H_\lambda = \frac{-\lambda}{2}[0] + \frac{\lambda}{3}[\infty] + \min\left\{\frac{\lambda+3}{3},0\right\}[-4].\]
Choosing $\lambda = -5$ gives non-integral coefficients at each point and $\deg{H_\lambda} = \frac{1}{6}$.

§.§.§ Blow-up of $Q$ (2.21)

With the same $-K_Q$ as before, blowing up in the twisted quartic, which is contained in $D_0$ and $D_\infty$ but not in $D_{-4}$, gives $-K_X = 3\tilde{D}_{-4}-E$. Adding $\Div(f_0e_{2\alpha}^{-2})$, where $f_0 \in \bb{C}(\proj{1})$ has divisor $[\infty]-[0]$ gives $-K_X = \tilde{D}_{-4} + \tilde{D}_{\infty}$. Then
\[H_\lambda = \frac{-\lambda+2}{2}[0] + \min\left\{\frac{\lambda+1}{3},0\right\}[\infty] + \min\left\{\frac{\lambda+1}{3},0\right\}[-4].\]
Taking $\lambda = -3$ gives non-integral coefficients and $\deg{H_\lambda} = \frac{1}{6}$.

§.§.§ Blow-up of $\proj{3}$ along a twisted cubic (2.27)

$\proj{3}$ with the cubic $\sl{2}$-action has subregular colours $D_0$, $D_\infty$ and $D_{-4}$ of respective degrees 3, 2 and 1. Hence we may choose $-K_{\proj{3}} = 4D_{-4}$. Then blowing up the twisted cubic gives $-K_X = 4\tilde{D}_{-4} - E$, where $E$ is the exceptional divisor and lies over $\infty$. We add $\Div(f_0e_{2\alpha}^{-2})$ to $-K_X$, where $\Div(f_0) = [0]-[\infty]$, giving $-K_X = 2\tilde{D}_{-4}+\tilde{D}_\infty$. We then have
\[H_\lambda = \frac{\lambda}{2}[0] + \min\left\{\frac{1-2\lambda}{3},-\lambda\right\}[\infty] + \min\left\{\frac{\lambda+2}{2},0\right\}[-4].\]
Choosing $\lambda = -3$ gives non-integral coefficients at all points and $\deg{H_\lambda} = \frac{1}{3}$.

§.§.§ $W$ (2.32)

The divisor $W$ on $\proj{2} \times \proj{2}$ of bidegree $(1,1)$ has anticanonical divisor class given by the intersection with $W$ of a class of bidegree $(2,2)$ on $\proj{2} \times \proj{2}$. The subregular colours $D_0$ and $D_\infty$ have bidegrees $(1,0)$ and $(0,1)$, so we take $-K_W = 2D_0 + 2D_\infty$. This gives
\[H_\lambda = \min\left\{\frac{\lambda+2}{2},0\right\}[0] + \min\left\{\frac{\lambda+2}{2},0\right\}[\infty] - \frac{\lambda}{2}[-1].\]
Taking $\lambda = -3$ gives non-integral coefficients at all points and $\deg{H_\lambda} = \frac{1}{2}$.

§.§.§ $\proj{1} \times \proj{2}$ (2.34)

The anticanonical divisor of $\proj{1} \times \proj{2}$ has bidegree $(2,3)$ which we can obtain as $2D_\infty + 3D_{-1}$. This gives
\[H_\lambda = -\frac{\lambda}{2}[0] + \min\left\{\frac{2-\lambda}{2},-\lambda\right\}[\infty] + \min\left\{2\lambda+3,\lambda\right\}[-1].\]
Taking $\lambda = -3$ gives non-integral coefficients at the points $0, \infty$ corresponding to subregular colours and $\deg{H_\lambda} = 1$.

§.§.§ Blow-up of $W$ (3.13)

We blow up $W$ in the curve of bidegree $(2,2)$ obtaining $-K_X = 2\tilde{D}_0 + 2\tilde{D}_\infty - E$. Adding $\Div(e_{2\alpha}^{-1})$ gives $-K_X = \tilde{D}_0 + \tilde{D}_\infty + \tilde{D}_{-1}$. Then
\[H_\lambda = \min\left\{\frac{\lambda+1}{2},0\right\}[0] + \min\left\{\frac{\lambda+1}{2},0\right\}[\infty] + \min\left\{\frac{1-\lambda}{2},-\lambda\right\}[-1].\]
Taking $\lambda = -2$ gives non-integral coefficients at all points and $\deg{H_\lambda} = \frac{1}{2}$.

§.§.§ Blow-up of $\proj{1} \times \proj{2}$ (3.17)

Blowing up $\proj{1} \times \proj{2}$ along the curve of bidegree $(1,1)$ gives $-K_X = 2\tilde{D}_\infty + 3\tilde{D}_{-1} - E$. We can add $\Div(f_0e_{2\alpha}^{-1})$, where $\Div(f_0) = [-1]-[\infty]$ to get $-K_X = \tilde{D}_\infty + 2\tilde{D}_{-1} + \tilde{D}_0$. Then
\[H_\lambda = \min\left\{\frac{1-\lambda}{2},-\lambda\right\}[0] +  \min\left\{\frac{1-\lambda}{2},-\lambda\right\}[\infty] + \min\left\{2\lambda+2,\lambda\right\}[-1].\]
Choosing $\lambda = -2$ gives non-integral coefficients at the points $0, \infty$ corresponding to subregular colours and $\deg{H_\lambda} = 1$.

§ CENTRAL DIVISORS OF $\SL{2}$-THREEFOLDS

Here we prove threefoldsKstable by calculating the $\beta$-invariant for the $G$-stable central prime divisors over each of our list of $\sl{2}$-threefolds. By Kstablecentral, we need to show that (1): $\beta_X(F) \geq 0$ for central divisors $F$ over each variety and (2): if $\beta_X(F) = 0$ then $F$ corresponds to a product configuration. It will turn out that (2) is never the case. Throughout we will mostly use the notation and results of <Ref> for curves, divisors etc. lying in each variety, with any changes to this notation clearly signalled.

§.§ Existence and Uniqueness of Central Divisors

In this subsection we will prove the following:

Let $X$ be a smooth Fano $\sl{2}$-threefold. There exists a unique central $G$-divisor over $X$. If $X$ is of type I, then it contains this central divisor. If $X$ is of type II, the central $G$-divisor over $X$ lies on the type I variety over $X$ whose existence is proved in typeIbirational.

Let $\cal{H}$ be the hyperspace of a $G$-model $X$ of complexity one. The dimension of $\cal{H}$ is the common dimension of each half-space $\cal{H}_p$ for $p \in \proj{1}$. If $\cal{C}_p$ is a coloured cone in $\cal{H}_p$, its dimension is the dimension of a minimal affine subspace of $\cal{H}_p$ containing $\cal{C}_p$. If $\cal{C}$ is a coloured hypercone of type II in $\cal{H}$, set $\dim{\cal{C}} = \max_{p \in \proj{1}}{\dim{(\cal{C} \inter \cal{H}_p)}}$.

Let $X$ be a $G$-model of complexity one and rank $r$, and let $\cal{H}$ be the hyperspace of $X$. We have $\dim{\cal{H}} = 1 + r$.

For each $p \in \proj{1}$, the slice $\cal{H}_p$ of hyperspace corresponding to $p$ is isomorphic to $\Lambda_\bb{Q}^* \times \bb{Q}_{\geq 0}$, where $\Lambda$ is the weight lattice of $X$, which has dimension $1 + \dim{\Lambda_\bb{Q}^*} = 1 + \rk{\Lambda} = 1+r$, by the definition of rank.

Let $X$ be a $G$-model and let $Y \sub X$ be a $G$-germ. The dimension of $\cal{C}_Y$ in $\cal{H}$ is equal to the codimension of $Y$ in $X$.

This follows from the fact that the coloured hypercone corresponding to $X$ itself is $\{0\}$ and that inclusion of coloured hypercones as faces in one another corresponds to the reverse inclusion of the associated $G$-germs.

Let $X$ be a complete $G$-model of dimension $d$, rank $r$ and complexity 1. Then minimal $G$-germs in $X$ must have dimension $d-r-1$.

Since $\cal{V}$ is full dimensional in $\cal{H}$, it follows from completeness that the coloured (hyper)cone corresponding to a minimal $G$-germ $Y$ in $X$ must have the same dimension as $\cal{H}$, i.e. $1 + r$. Then $\codim_X{Y} = \dim{\cal{C}_Y} = 1 + r$, so $\dim{Y} = d - r - 1$.

Now let $G = \sl{2}(\bbk)$ and let $X$ be a complete three dimensional $G$-model of complexity one, i.e. a normal projective threefold with a $G$-action having finite stabilisers. Then $X$ contains an open orbit isomorphic to $G/H$ for $H$ a finite subgroup of $G$.

Let $B$ be the Borel subgroup of $G$ given by the upper triangular matrices. Then $\frak{X}(B)$ is of rank 1, generated by the character $\alpha$ which picks out the upper-left entry. Hence in particular $X$ is a rank 1 variety. The hyperspace $\cal{H}$ of $X$ then has dimension 2, so minimal $G$-germs of $X$ must have codimension 2, i.e. they are curves, and in particular $X$ can contain no $G$-fixed points. The centre $\cal{Z}$ of $\cal{H}$ is a line. Most of the following results arise from this fact. In particular, note that for any finite $H \sub G$, the valuation cone $\cal{V}(G/H)$ intersects $\cal{Z}$ in a ray.

$X$ contains at most one central $G$-divisor.

Any central $G$-divisor must be mapped to the intersection $\cal{V} \inter \cal{Z}$ of the valuation cone and the central hyperplane. As noted, this intersection is a ray. If there were two distinct central $G$-divisors, their coloured (hyper)cones would then both be the same ray, a contradiction.

$X$ contains a central $G$-divisor if and only if $X$ is a model of type I.

Suppose $X$ is of type I, i.e. every $G$-germ of $X$ corresponds to a coloured cone in some slice of the hyperspace. By completeness, these coloured cones must cover $\cal{V}$. Since $\cal{V}$ intersects the central line $\cal{Z}$ in a ray $\rho$, there must be a $G$-germ of $X$ whose coloured cone is $\rho$, i.e. a central $G$-divisor.

Now suppose $X$ has a central $G$-divisor $D$ and that $Y \sub X$ is a $G$-germ of type II with coloured hypercone $\cal{C}_Y$. We know that $\cal{C}_Y$ must intersect $\cal{Z}$ in $\cal{V}$, i.e. it contains the ray $\rho$ corresponding to $D$. Then $\nu_D$ lies in the relative interior of $\cal{C}_Y$, hence in the support $\cal{S}_Y$ of $Y$. But $\nu_D \in \cal{S}_D$, so the supports of $Y$ and $D$ are not disjoint, contradicting separation of $X$. Hence all $G$-germs of $X$ are of type I, as required.

If $X$ is of type II, there exists a central prime divisor over $X$.

We know from typeIbirational that there exists a projective birational morphism $\nu \colon \check{X} \to X$ where $\check{X}$ is of type I. Then by the above proposition there is a central prime divisor $D \sub \check{X}$.

Let $X$ be of type II. Then $X$ has a unique $G$-germ of type II, a curve, which must be a closed $G$-orbit.

We know that $X$ contains at least one $G$-germ of type II. Suppose $Y, Y^\prime \sub X$ are both $G$-germs of type II. Their corresponding coloured hypercones $\cal{C}_Y$, $\cal{C}_{Y^\prime}$ must both intersect the central ray $\rho \sub \cal{V}$, hence their relative interiors must intersect in $\cal{V}$. It follows that $Y = Y^\prime$, and $X$ has exactly one $G$-germ of type II.

Then all other $G$-germs of $X$ are of type I and so each defines a coloured cone in some $\cal{H}_p$. In particular, these coloured cones cannot contain $\cal{C}_Y$, so $Y$ does not contain any other $G$-germ of $X$. Hence $Y$ is a minimal $G$-germ, so in particular a curve and a closed $G$-orbit.

Let $X$ be of type II and suppose that every $G$-divisor of $X$ maps to the boundary of the valuation cone. Then the unique minimal $G$-germ $Y \sub X$ of type II is contained in every $G$-divisor, and $X$ has no minimal $G$-germs of type I. In particular, $Y$ is the unique closed $G$-orbit of $X$.

Let $F \sub X$ be a $G$-divisor mapping to the non-central boundary of $\cal{V}_p = \cal{V} \inter \cal{H}_p$. The ray $\cal{C}_F$ defined by $F$ must then be a face of some coloured cone in $\cal{H}_p$ or coloured hypercone of type II in $\cal{H}$, i.e. $F$ must contain some minimal $G$-germ.

We know that $X$ contains a unique minimal $G$-germ $Y$ of type II, and in particular the coloured hypercone $\cal{C}_Y$ must have $\cal{C}_F$ as a face, i.e. $F$ contains $Y$. Now suppose $F$ contains another $G$-germ $Z$. This must be of type I since $Y$ is the only $G$-germ of type II, so $\cal{C}_Z$ must be a coloured cone in $\cal{H}_p$ with $\cal{C}_F$ as a face. However, $\cal{C}_Z$ must intersect $\cal{V}$ in its relative interior, and since $\cal{C}_F$ is the boundary of $\cal{V}$, it follows that $\cal{C}_Z$ and $\cal{C}_Y$ intersect in their relative interiors, a contradiction.

Hence for each $\cal{H}_p$ containing a $G$-divisor, the only minimal $G$-germ whose coloured (hyper)cone intersects $\cal{H}_p$ is $Y$. If $\cal{H}_p$ does not contain a $G$-divisor, it can also only support one coloured (hyper)cone since it only contains one $B$-divisor, the colour $D_p$.

Therefore $Y$ is indeed the unique minimal $G$-germ of $X$ and hence also the unique closed orbit.

In all cases which we will consider here, we are given a smooth $\sl{2}$-threefold $X$, and either $X$ is of type I and has a central divisor, which is unique, or $X$ is of type II and we can obtain the unique central divisor over $X$ by blowing up a finite sequence of $G$-invariant curves.

Combining uniquecentral with Kstablecentral, we see that to prove threefoldsKstable, we need only find the central divisor over each variety and show that its $\beta$-invariant is positive. In the following subsections we will perform this calculation for each of the examples, thus proving their $K$-polystability.

§.§ $\proj{3}$ and Blow-Ups Along Two or Three Lines

§.§.§ $\proj{3}$ (1.17)

The anticanonical divisor of $\proj{3} = \bb{P}(M_2(\bb{C}))$ is the class of a divisor of degree 4, which we may take to be $2\Delta$, where $\Delta$ is the $G$-invariant divisor of singular matrices. This is also the unique central divisor on $\proj{3}$.

To calculate $\beta_X(\Delta)$ when $X = \proj{3}$, note that $(-K_X)^3 = 64$ and $A_X(\Delta) = 1$ since $\Delta$ lies on $\proj{3}$. It remains to calculate $\vol(\delta)$, where $\delta = -K_X - x\Delta = (2-x)\Delta$. Note that since $\delta$ is $G$-invariant we have $\lambda_\delta = 0$.

We have $\cal{P}(\delta) = \{\lambda \in \Lambda \mid \langle \lambda,\ell_{\Delta}\rangle \geq -(2-x)\}$. Since $\Lambda = \bb{Z}\alpha \cong \bb{Z}$ and $\ell_\Delta = -1$, we get $\cal{P}(\delta) = \{\lambda \in \bb{Z} \mid \lambda \leq 2-x\}$.

\[A(\delta,\lambda) = \sum_{p \in \proj{1}}\left({\min_{p_D = p}{\frac{\langle \lambda,\ell_D\rangle + m_D}{h_D}}}\right).\]
We have $\ell_D = m_D = 0$ for all colours $D$ other than the distinguished colour, which has $m_D = 0$, $\ell_D = 2$ and hence contributes a value of $2\lambda$ to $A(\delta,\lambda)$.

We therefore have $\cal{P}_+(\delta) = \{\lambda \leq 2-x \mid \lambda \geq 0\} = [0,2-x]$. Hence
\[\vol(\delta) = 6\int_{0}^{2-x}{2\lambda\cdot 2\lambda \ \mathrm{d} \lambda} = 8(2-x)^3.\]
We therefore have
\[\beta(\Delta) = 64 - \int_{0}^{2}{8(2-x)^3 \ \mathrm{d}x} = 64-32 = 32 > 0.\]
Hence by Kstablecentral, $\proj{3}$ is $K$-polystable.

§.§.§ Blow-up of $\proj{3}$ Along Two Lines (3.25)

Starting with $-K_{\proj{3}} = 2\Delta$ as before, the anticanonical divisor after blowing up two lines $Y_q$ and $Y_r$ is $2\tilde{\Delta}+E_q+E_r$.

This time we have $(-K_X)^3 = 44$ and again $A_X(\Delta) = 1$. We must now compute $\beta(\tilde{\Delta})$ by calculating $\vol(\delta)$ where $\delta = (2-x)\tilde{\Delta}+E_q+E_r$. We still have $\lambda_\delta = 0$.

Likewise, $\cal{P}(\delta) = (-\infty,2-x]$ as before. This time $A(\delta,\lambda)$ receives the same contribution of $2\lambda$ at the distinguished point, and there is no contribution other than from here and from $q$ and $r$. At $q$, we have the exceptional divisor $E_q$ with $\ell = -1$, $m = 1$ and $h = 1$, and the colour with $\ell = m = 0$. Thus there is a contribution to $A(\delta,\lambda)$ of $-\lambda + 1$ when this is less than or equal to 0, and a contribution of 0 otherwise. The same holds for $r$. Hence we have
\[A(\delta,\lambda) = \begin{cases} 2 & 1 \leq \lambda \leq 2-x \\ 2\lambda & \lambda < 1.\end{cases}\]
Therefore $\cal{P}_+(\delta) = [0,2-x]$, and
\[\vol(\delta) = \begin{cases} 6\int_{0}^{1}{4\lambda^2 \ \mathrm{d}\lambda} + 6\int_{1}^{2-x}{4\lambda \ \mathrm{d}\lambda} & 0 \leq x \leq 1 \\ 6\int_{0}^{2-x}{4\lambda^2 \ \mathrm{d}\lambda} & 1 < x \leq 2.\end{cases}\]
We thus get
\[\beta_X(\tilde{\Delta}) = 44 - \int_{0}^{2}{\vol(\delta) \ \mathrm{d} x} = 44 - 26 = 18 > 0.\]
Hence the blow-up of $\proj{3}$ at two lines is $K$-polystable.

§.§.§ Blow up of $\proj{3}$ Along Three Lines (4.6)

The anticanonical divisor of the blow-up of $\proj{3}$ along three lines $Y_q, Y_r$ and $Y_s$ is $2\tilde{\Delta} + E_q + E_r + E_s$. We have $(-K_X)^3 = 34$, $A_X(\tilde{\Delta}) = 1$, $\delta = (2-x)\tilde{\Delta}+E_q+E_r+E_s$, $\lambda_\delta = 0$ and $\cal{P}(\delta) = (-\infty,2-x]$.

The calculation of $A(\delta,\lambda)$ goes much the same as in the previous case, only $A(\delta,\lambda)$ gains an extra contribution of $-\lambda + 1$ from $E_s$ when $\lambda \geq 1$, giving
\[A(\delta,\lambda) = \begin{cases} 3-\lambda & 1 \leq \lambda \leq 2-x \\ 2\lambda & \lambda < 1.\end{cases}\]
Therefore $\cal{P}_+(\delta) = [0,2-x]$, and
\[\vol(\delta) = \begin{cases} 6\int_{0}^{1}{4\lambda^2 \ \mathrm{d}\lambda} + 6\int_{1}^{2-x}{6\lambda-2\lambda^2 \ \mathrm{d}\lambda} & 0 \leq x \leq 1 \\ 6\int_{0}^{2-x}{4\lambda^2 \ \mathrm{d}\lambda} & 1 < x \leq 2.\end{cases}\]
We thus get
\[\beta_X(\tilde{\Delta}) = 34 - \int_{0}^{2}{\vol(\delta) \ \mathrm{d} x} = 34 - 23 = 11 > 0.\]
Hence the blow-up of $\proj{3}$ at three lines is $K$-polystable.

§.§ Blow-up of $\proj{1} \times \proj{2}$

§.§.§ Central Divisor

Let $X = \cal{Z}(x_0y_0z_2+x_1y_1z_0-x_0y_1z_1-x_1y_0z_1) \sub \proj{1} \times \proj{1} \times \proj{2}$. This variety is the blow up of $\proj{1} \times \proj{2}$ along the $G=\sl{2}$-stable curve $C = \cal{Z}(x_1z_0-x_0z_1,x_0y_1-x_1y_0)$. In $X$ there are $G$-invariant divisors $\Delta = \cal{Z}(x_0y_1-x_1y_0,x_0^2z_2+x_1^2z_0-2x_0x_1z_1)$, $E = \cal{Z}(x_1z_0-x_0z_1,x_0y_1-x_1y_0)$ and $F = \cal{Z}(y_1z_0-y_0z_1,y_0z_2-y_1z_1)$. The curve $Z = F \inter \Delta \inter E$ is $G$-invariant and defined by $\cal{Z}(x_1z_0-x_0z_1,x_1z_1-x_0z_2,x_0y_1-x_1y_0)$.

Taking $x_1 = y_1 = z_2 = 1$ gives a maximal $B$-chart $U$ of $Z$, given by $U = \cal{Z}(x_0y_0+z_0-x_0z_1-y_0z_1) \sub \aff{4}$, so eliminating $z_0$ gives $U = \Spec{\bbk[x_0,y_0,z_1]} \cong \aff{3}$. We have $\Delta \inter U = \cal{Z}(x_0-y_0)$, $F \inter U = \cal{Z}(y_0-z_1)$, $E \inter U = \cal{Z}(z_1-x_0)$ and $Z \inter U = \cal{Z}(z_1-x_0,x_0-y_0)$. We blow up $U$ in this curve to obtain the variety $\tilde{X} = \cal{Z}(u_1(z_1-x_0)-u_0(x_0-y_0)) \sub \aff{3} \times \proj{1}$.

The $B$-invariant rational function
\[f = \frac{x_1^2(z_0z_2-z_1^2)}{(x_0z_2-x_1z_1)^2}\]
on $X$ becomes, on $\tilde{X}$, $f = \frac{z_1-y_0}{x_0-z_1}$. From this one can see that the exceptional divisor $D = \cal{Z}(z_1-y_0,x_0-z_1)$ of the blow-up $\sigma \colon \tilde{X} \to X$ is central.

§.§.§ $\beta_X(F)$ (3.17)

We now want to calculate
\[\beta(D) = A_X(D)(-K_X)^3 - \int_{0}^{\infty}{\vol_X{(-K_X-xD)} \ \mathrm{d}x}.\]
We have $A_X(D) = 2$ since $D$ is the exceptional divisor on a blow-up of $X$, and $(-K_X)^3 = 36$, so
\[\beta(F) = 72 - \int_{0}^{\infty}{\vol_X{(-K_X-xD)} \ \mathrm{d}x} = 72 - \int_{0}^{\infty}{\vol_{\tilde{X}}{(\sigma^*(-K_X)-xD)} \ \mathrm{d}x}.\]

To calculate $\sigma^*(-K_X)$, first let $\Delta = \cal{Z}(x_0^2z_2+x_1^2z_0-2x_0x_1z_1)$ and $F = \cal{Z}(z_0z_2-z_1^2)$ in $\proj{1} \times \proj{2}$. The divisors $\Delta$, $F$ above are the strict transforms of these under the blow-up $\mu \colon X \to \proj{1} \times \proj{2}$. Since $-K_{\proj{1} \times \proj{2}}$ is the class of a divisor of bidegree $(2,3)$, we can represent it by $\Delta + F = (2,1) + (0,2)$.

Then $-K_X = \mu^*(\Delta + F) - E = (\Delta + E) + (F + E) - E = \Delta + F + E$. Hence $\sigma^*(-K_X) = \tilde{\Delta} + \tilde{F} + \tilde{E} + 3D$, and so we need to calculate the volume of $\delta = \tilde{\Delta} + \tilde{F} + \tilde{E} + (3-x)D$. This is given by
\[\vol(\delta) = 6\int\limits_{\lambda_\delta + \cal{P}_+(\delta)}{2\lambda A(\delta,\lambda-\lambda_\delta)\, \mathrm{d}\lambda}.\]

Since $\delta$ is $G$-invariant, we have $\lambda_\delta = 0$. We also have
\[\cal{P}(\delta) = \{\lambda \in \Lambda \otimes \bb{R} \mid \langle \lambda, \ell_D\rangle \geq x-3\} = \{\lambda \mid \lambda \leq 3-x\}\]
\[\cal{P}_+(\delta) = \{\lambda \in \cal{P}(\delta) \mid A(\delta,\lambda) \geq 0\} = \{\lambda \leq 3-x \mid A(\delta,\lambda) \geq 0\}\]
\[A(\delta,\lambda) = \sum_{p \in \proj{1}}{\min_{p_D=p}{\frac{\langle \lambda,\ell_D\rangle + m_D}{h_D}}}.\]

To calculate $A(\delta,\lambda)$, first note that for $p \neq 0,-1,\infty$, the only $B$-divisors with $p_D = p$ are the colours $D_p$ with $\ell_{D_p} = m_{D_p} = 0$, $h_{D_p} = 1$, so there is no contribution in these cases.

For $p = 0$, the two divisors with $p_D = p$ are $E$, with $\ell = -1$, $m = 1$ and $h = 1$, and $D_0$ with $\ell = -1$, $m = 0$ and $h=2$. Hence there is a contribution to $A(\delta,\lambda)$ of
\[\min{\left\{1-\lambda,-\frac{\lambda}{2}\right\}} = \begin{cases} 1-\lambda & \lambda \geq 2 \\ -\frac{\lambda}{2} & \lambda < 2 \end{cases}.\]
For $p = \infty$, the contribution is the same. The two divisors with $p_D = -1$ are $\Delta$ with $\ell = 1$, $m = 1$ and $h = 1$, and $D_{-1}$ with $\ell = -2$, $m = 0$ and $h = 1$, so the contribution to $A(\delta,\lambda)$ is
\[\min{\left\{1+\lambda, 2\lambda\right\}} = \begin{cases} 1+\lambda & \lambda \geq 1 \\ 2\lambda & \lambda < 1 \end{cases}.\]
Hence we have
\[A(\delta,\lambda) = \begin{cases} 3-\lambda & \lambda \geq 2 \\ 1 & 1 \leq \lambda \leq 2 \\ \lambda & \lambda < 1\end{cases}.\]
It follows that $\cal{P}_+(\delta) = \{\lambda \leq 3-x\mid 0 \leq \lambda \leq 3\}$, and since $x \geq 0$, $3-x \leq 3$ and $\cal{P}_+(\delta)$ is empty if $x > 3$. Hence $\cal{P}_+(\delta) = [0:3-x]$ where $0 \leq x \leq 3$.

\[\begin{split} \vol{\delta} & = 6\begin{cases}\int_{0}^{3-x}{2\lambda^2 \ \mathrm{d}\lambda} & 2 \leq x \leq 3 \\ \int_{0}^{1}{2\lambda^2 \ \mathrm{d}\lambda} + \int_{1}^{3-x}{2\lambda \ \mathrm{d}\lambda} & 1 \leq x \leq 2 \\ \int_{0}^{1}{2\lambda^2 \ \mathrm{d}\lambda} + \int_{1}^{2}{2\lambda \ \mathrm{d}\lambda} + \int_{2}^{3-x}{2\lambda(3-\lambda) \ \mathrm{d}\lambda} & 0 \leq x \leq 1 \end{cases} \\ & = \begin{cases} 4(3-x)^3 & 2 \leq x \leq 3 \\ 6x^2-36x+52 & 1 \leq x \leq 2 \\ 4x^3 - 18x^2 + 36 & 0 \leq x \leq 1\end{cases}\end{split}\]
\[\begin{split} \beta(F) & = 72 - \int_{0}^{3}{ \vol{\delta} \ \mathrm{d}x} \\ & = 72 - \int_{0}^{1}{4x^3-18x^2+36 \ \mathrm{d}x} - \int_{1}^{2}{6x^2-36x+52 \ \mathrm{d}x} - \int_{2}^{3}{4(3-x)^3 \ \mathrm{d}x} \\ & = 72 - 31 - 12 - 1 = 28.\end{split}\]
Hence $X$ is $K$-polystable.

§.§ The Divisor $W$ on $\proj{2} \times \proj{2}$ and Its Blow-Up

§.§.§ Central Divisor

Let $W = \cal{Z}(x_0y_2-2x_1y_1+x_2y_0) \sub \proj{2} \times \proj{2}$ . We know that $W$ has $G$-invariant divisors $E_\infty = \cal{Z}(x_0x_2-x_1^2) \inter W$ and $E_0 = \cal{Z}(y_0y_2-y_1^2) \inter W$ whose intersection is a $G$-stable curve $Z$. We obtain the smooth Fano (3.13) by blowing up $W$ along $Z$. The curve $Z$ has a minimal $B$-chart $U = W\setminus{(\cal{Z}(x_2)\union \cal{Z}(y_2))} = \cal{Z}(x_0-2x_1y_1+y_0) \sub \aff{4}$. We eliminate $x_0$ to obtain $U = \Spec{\bbk[x_1,y_0,y_1]} \cong \aff{3}$. Introducing new co-ordinates $x = x_1-y_1$, $y=y_0$, $z=y_1$, the curve $Z \inter U$ is defined by $x = y-z^2 = 0$. The divisors $E_\infty \inter U$ and $E_0 \inter U$ are defined by $z^2 - y - x^2 = 0$ and $y-z^2 = 0$, respectively.

Hence blowing up $U$ along $Z \inter U$ we obtain $X = \cal{Z}(vx-u(y-z^2)) \sub \aff{3} \times \proj{1}$ with exceptional divisor $E = \cal{Z}(x,y-z^2)$ and strict transforms (abusing notation) $\tilde{E}_\infty = \cal{Z}(ux-v,z^2-x^2-y)$ and $\tilde{E}_0 = \cal{Z}(y-z^2,v)$. The intersection of these three $G$-invariant divisors (in fact any two of them) gives a $G$-invariant curve $Y = \cal{Z}(y-z^2,x,v)$, the unique minimal $G$-germ of $X$.

Take a chart $u = 1$ to obtain $X = \cal{Z}(vx-y+z^2)$, then eliminate $y$ so that $X = \Spec{\bbk[x,v,z]} \cong \aff{3}$. Then we have  $\tilde{E}_\infty = \cal{Z}(x-v)$, $\tilde{E}_0 = \cal{Z}(v)$, $E = \cal{Z}(x)$ and $Y = \cal{Z}(x,v)$.

Blowing up this chart along $Y$, we obtain $\tilde{X} = \cal{Z}(wx-sv) \sub \aff{3} \times \proj{1}$. We now have $\tilde{E}_0 = \cal{Z}(v,w)$, $\tilde{E}_\infty = \cal{Z}(x-v,w-s)$, $\tilde{E} = \cal{Z}(x,s)$ and an exceptional divisor $F = \cal{Z}(x,v)$. The $B$-quotient map $W \dashrightarrow \proj{1}$ was originally given by $P \mapsto [y_2^2(x_0x_2-x_1^2):x_2^2(y_0y_2-y_1^2)]$, which on $\tilde{X}$ reduces to $P \mapsto [v-x:v]$. From this one can see that the exceptional divisor $F$ is central.

Since $F$ is $G$-invariant we must have $\ell_F < 0$ (for it to lie in the valuation cone), and since it is a hyperplane we must then have $\ell_F = -1$. Since the coloured hyperfan of any model of type I must consist of strictly convex coloured cones, and the $G$-invariant valuations map injectively into the hyperspace, $F$ is the unique central $G$-invariant prime divisor over $W$.

§.§.§ $\beta_W(F)$ (2.32)

To calculate $\beta_W(F)$, we first must calculate $-K_W$ and its pullback to the model containing $F$. Since $W$ is a hypersurface in $\proj{2} \times \proj{2}$, the adjunction formula gives $-K_W = (-K_{\proj{2} \times \proj{2}}-W)\vert_W$. The anticanonical class of $\proj{2} \times \proj{2}$ is $(3,3)$ where we identify the divisor class group with $\bb{Z} \oplus \bb{Z}$, and since $W$ has bidegree $(1,1)$ we get $-K_W = (2,2)\vert_W$. Represent the divisor class $(2,2)$ on $\proj{2} \times \proj{2}$ by $\cal{Z}(x_0x_2-x_1^2) + \cal{Z}(y_0y_2-y_1^2)$, so that the restriction to $W$ of this class is represented by $E_\infty + E_0 = -K_W$.

After the two blow-ups, this class pulls back to $E_0 + E_\infty + 2E + 4F$, and we must calculate $\beta_W(F) = A_W(F)(-K_W)^3 - \int_{0}^{\infty}{\vol(\delta) \ \mathrm{d}x}$ where $\delta = E_0 + E_\infty + 2E + (4-x)F$. We have $(-K_W)^3 = 48$ and $A_W(F) = 3$ since $F$ is the exceptional divisor of the second of two nested blow-ups of $W$.

We have $\lambda_\delta = 0$ and $\cal{P}(\delta) = (-\infty,4-x]$. To calculate $A(\delta,\lambda)$, first note that there is no contribution at points other than $0$, $\infty$ and $-1$. At $p = 0, \infty$, we have divisors $E_p$ with $\ell = 0$, $h=1$ and $m = 1$, and $D_p$ with $\ell = 1$, $h = 2$ and $m = 0$, so the contribution in each case is $\min\{1,\frac{\lambda}{2}\}$. At $p = -1$ we have $E$ with $\ell = -1$, $h = 1$ and $m = 2$, and $D_{-1}$ with $\ell = -1$, $h = 2$ and $m = 0$, so the contribution is $\min\{2-\lambda,-\frac{\lambda}{2}\}$. Overall, we have
\[A(\delta,\lambda) = \begin{cases} 4-\lambda & \lambda \geq 0 \\ \frac{\lambda}{2} & \lambda < 2.\end{cases}\]
Therefore we have $\cal{P}_+(\delta) = [0,4-x]$, so $0 \leq x \leq 4$, and:
\begin{align*}\vol(\delta) & = 6\int_{0}^{2-x}{2\lambda A(\delta,\lambda) \ \mathrm{d}\lambda} \\ & = \begin{cases}6\int_{0}^{2-x}{\lambda^2 \ \mathrm{d}\lambda} & 2 \leq x \leq 4 \\ 6\int_{0}^{2}{\lambda^2 \ \mathrm{d}\lambda} + 6\int_{2}^{2-x}{8\lambda - 2\lambda^2 \ \mathrm{d}\lambda} & 0 \leq x \leq 2\end{cases} \\ & = \begin{cases} 2(4-x)^3 & 2 \leq x \leq 4 \\ 4x^3-24x^2+80 & 0 \leq x \leq 2.\end{cases}\end{align*}
\[\beta_W(F) = 3\cdot 48 - \int_{0}^{4}{\vol(\delta)\ \mathrm{d}x} = 144-120 = 24 > 0\]
so $W$ is $K$-polystable.

§.§.§ $\beta_X(F)$ (3.13)

We now want to calculate
\[\beta_X(F) = A_X(F)(-K_X)^3 - \int_{0}^{\infty}{\vol_X{(-K_X-xF)} \ \mathrm{d}x}.\]
We have $A_X(F) = 2$ since $F$ is a prime divisor on a blow-up of $X$, and $(-K_X)^3 = 30$. We have $-K_X = \mu^*(-K_W) - E$ where $\mu$ is the blow-up of $W$ in $E_0 \inter E_\infty$, which gives $-K_X = E_0 + E_\infty + E$. Under the next blow-up to the model containing $F$, this pulls back to $E_0 + E_\infty + E + 3F$, so we set $\delta = E_0 + E_\infty + E + (3-x)F$.

We have $\lambda_\delta = 0$ and $\cal{P}(\delta) = (-\infty,3-x]$. To calculate $A(\delta,\lambda)$, first note that for $p \neq 0,-1,\infty$, there is no contribution. For $p = 0, \infty$, the two divisors with $p_D = p$ are $E_p$, with $\ell = 0$, $m = 1$ and $h = 1$, and $D_p$ with $\ell = 1$, $m = 0$ and $h=2$. Hence in each case there is a contribution to $A(\delta,\lambda)$ of
\[\min{\left\{1,\frac{\lambda}{2}\right\}} = \begin{cases} 1 & \lambda \geq 2 \\ \frac{\lambda}{2} & \lambda < 2 \end{cases}.\]
For $p = -1$, the two divisors with $p_D = p$ are $E$ with $\ell = -1$, $m = 1$ and $h = 1$, and $D_p$ with $\ell = -1$, $m = 0$ and $h = 2$, so the contribution to $A(\delta,\lambda)$ is
\[\min{\left\{1-\lambda, -\frac{\lambda}{2}\right\}} = \begin{cases} 1-\lambda & \lambda \geq 2 \\ -\frac{\lambda}{2} & \lambda < 2 \end{cases}.\]
Hence we have
\[A(\delta,\lambda) = \begin{cases} 3-\lambda & \lambda \geq 2 \\ \frac{\lambda}{2} & \lambda < 2\end{cases}.\]
It follows that $\cal{P}_+(\delta) = [0,3-x]$, so $0 \leq x \leq 3$

\begin{align*}\vol{\delta} & = \begin{cases} 6\int_{0}^{3-x}{\lambda^2 \ \mathrm{d}\lambda} & 1 \leq x \leq 3 \\ 6\int_{0}^{2}{\lambda^2 \ \mathrm{d}\lambda} + 6\int_{2}^{3-x}{6\lambda-2\lambda^2 \ \mathrm{d}\lambda} & 0 \leq x \leq 1\end{cases} \\ & = \begin{cases} 2(3-x)^3 & 1 \leq x \leq 3 \\ 4x^3-18x^2+30 & 0 \leq x \leq 1.\end{cases}\end{align*}
\[\beta(F) = 60 - \int_{0}^{3}{\vol(\delta) \ \mathrm{d}x} = 60 - 33 = 27.\]
So $X$ is $K$-polystable.

§.§ Blow up of $\proj{3}$ along the Twisted Cubic

§.§.§ Central Divisor

Let $G=\sl{2}$ act on $\proj{3} = \bb{P}(S^3\bbk^2)$. The twisted cubic curve
\[C = \cal{Z}(x_0x_2-x_1^2,x_0x_3-x_1x_2,x_1x_3-x_2^2)\]
is $G$-invariant. The prime divisor
\[F = \cal{Z}(3x_1^2x_2^2-4x_1^3x_3-x_0^2x_3^2-4x_0x_2^3+6x_0x_1x_2x_3).\]
is $G$-invariant and contains $C$ - indeed $F$ is the secant variety to $C$ and $C$ is the singular locus of $F$, contained with multiplicity 2.

The smooth Fano (2.27) is obtained by blowing up $\proj{3}$ along $C$. We first take the open chart given by $x_3 = 1$, which is the minimal $B$-chart of $C$. In this chart,
\[C = \cal{Z}(x_0x_2-x_1^2,x_0-x_1x_2,x_1-x_2^2) = \cal{Z}(x_0-x_1x_2,x_1-x_2^2),\]
\[F = \cal{Z}(3x_1^2x_2^2-4x_1^3-x_0^2-4x_0x_2^3+6x_0x_1x_2)\]
Now consider the change of co-ordinates
\[(x_0, x_1, x_2) \mapsto (x_0+3x_1x_2+x_2^3, x_1+x_2^2, x_2).\]
It is easily checked to be an isomorphism, and it sends $F$ to $\cal{Z}(x_0^2-4x_1^3)$ and $C$ to $\cal{Z}(x_0,x_1)$. Hence we see that $F$ is isomorphic to the product of a line ($C$) and a cuspidal cubic plane curve. Performing another transformation $x_1 \mapsto x_1/\sqrt[3]{4}$ gives $F = \cal{Z}(x_0^2-x_1^3)$ and leaves $C$ invariant.

Now we blow up $C$, giving $X = \cal{Z}(y_1x_0-y_0x_1) \sub \aff{3} \times \proj{1}$ with exceptional divisor $E = \cal{Z}(x_0,x_1)$. We have
\[\tilde{F} = \cal{Z}(x_0^2-x_1^3,y_1x_0-y_0x_1)\setminus{\cal{Z}(x_0,x_1)}.\]
It is easy to check that this gives
\[\tilde{F} = \cal{Z}(x_0^2-x_1^3,y_1x_0-y_0x_1,y_0x_0-x_1^2y_1,y_0^2-y_1^2x_1).\]
The intersection $\tilde{F}\inter E$ is then given by $\cal{Z}(x_0,x_1,y_0)$. Since we don't yet have a central divisor, we will blow up this curve.

First, take the chart $y_1 = 1$. Then $X$ becomes $\cal{Z}(x_0-x_1y_0) \cong \Spec{\bbk[x_1,x_2,y_0]}$, $F$ becomes $\cal{Z}(y_0^2-x_1)$, and $E$ becomes $\cal{Z}(x_1)$. Hence we obtain $\tilde{X} = \cal{Z}(z_0x_1-z_1y_0) \sub \aff{3} \times \proj{1}$, with exceptional divisor $D = \cal{Z}(x_1,y_0)$. The strict transforms of $\tilde{F}$ and $E$ are $\cal{Z}(y_0^2-x_1,z_0y_0-z_1)$ and $\cal{Z}(x_1,z_1)$, respectively. It is straightforward to check that $\tilde{E}, \tilde{F}$ and $D$ mutually intersect in the curve $\cal{Z}(x_1,y_0,z_1)$. In particular this shows that $D$ is not central, so we must blow up again.

Take the chart $z_0 = 1$, giving $\tilde{X} = \cal{Z}(x_1-z_1y_0) \cong \Spec{\bbk[x_2,y_0,z_1]}$, $\tilde{F} = \cal{Z}(y_0-z_1)$, $\tilde{E} = \cal{Z}(z_1)$ and $D = \cal{Z}(y_0)$. Blowing up the intersection $\cal{Z}(y_0,z_1)$ of these divisors gives $\tilde{X}^\prime = \cal{Z}(u_1y_0-u_0z_1)$ with exceptional divisor $H = \cal{Z}(y_0,z_1)$. Now the strict transforms $\tilde{E}$, $\tilde{F}$ and $\tilde{D}$ all intersect $H$ in different curves and are disjoint from each other: hence $H$ is a central divisor.

§.§.§ $\beta(H)$ (2.27)

We now want to calculate $\beta(H)$. We have $(-K_X)^3 = 38$, and $A_X(H) = 3$ since $H$ is a prime divisor on a variety obtained by two blow-ups of $X$. Hence
\[\beta(H) = 114 - \int_{0}^{\infty}{\vol_X{(-K_X-xH)}\ \mathrm{d}x} = 114 - \int_{0}^{\infty}{\vol_{\tilde{X}^\prime}{(\sigma^*(-K_X)-xH)}\ \mathrm{d}x}\]
where $\sigma \colon \tilde{X}^\prime \to X$ is the birational morphism given by composing the two blow-ups described above.

To calculate $\sigma^*(-K_X)$, first note that the anticanonical class of $\proj{3}$ is the class of a prime divisor of degree $4$, so we can set $-K_{\proj{3}} = F$. Then, blowing up $C$, which is contained in $F$ with multiplicity 2, gives $-K_X = (\tilde{F}+2E) - E = \tilde{F} + E$. The pullback of this class under the blowing up of $\tilde{F} \inter E$ is then $(\tilde{F} + D) + (\tilde{E} + D) = \tilde{F} + \tilde{E} + 2D$. Finally, the second blow-up gives
\[\sigma^*(-K_X) = (\tilde{F}+H) + (\tilde{E} + H) + 2(\tilde{D} + H) = \tilde{F} + \tilde{E} + 2\tilde{D} + 4H.\]

Our next step is the calculate the volume of the divisor $\delta = \tilde{F} + \tilde{E} + 2\tilde{D} + (4-x)H.$ We have $\lambda_\delta = 0$ and $\cal{P}(\delta) = (-\infty,4-x]$  since we must have $\ell_H = -1$. Now we calculate $A(\delta,\lambda)$. The points $p \neq -4, 0, \infty$ contribute nothing as $p_D = p$ in this case only for colours $D_p$ with $m_D = \ell_D = 0$.

At $p = -4$, we have two divisors: $\tilde{F}$, with $m = 1, \ell = 0$ and $h = 1$, and $D_{-4}$, with $m = 0$, $\ell = 1$ and $h = 2$. Hence there is a contribution of
\[\min\left\{1,\frac{\lambda}{2}\right\} = \begin{cases} 1 & \lambda \geq 2 \\ \frac{\lambda}{2} & \lambda < 2\end{cases}.\]

At $p = 0$, we have two divisors: $\tilde{D}$, with $m = 2$, $\ell = 0$ and $h = 1$, and $D_0$, with $m = 0$, $\ell = 1$ and $h = 2$. Hence there is a contribution of
\[\min\left\{2,\frac{\lambda}{2}\right\} = \begin{cases} 2 & \lambda \geq 4 \\ \frac{\lambda}{2} & \lambda < 4\end{cases}.\]

At $p = \infty$, we have two divisors: $\tilde{E}$, with $m = 1$, $\ell = -1$ and $h = 1$, and $D_\infty$, with $m = 0$, $\ell = -2$ and $h = 3$. Hence there is a contribution of
\[\min\left\{1-\lambda, -\frac{2\lambda}{3}\right\} = \begin{cases} 1 - \lambda & \lambda \geq 3 \\ -\frac{2\lambda}{3} & \lambda < 3\end{cases}.\]

All in all, we have
\[A(\delta,\lambda) = \begin{cases} 4-\lambda & \lambda \geq 4 \\ 2-\frac{\lambda}{2} & 3 \leq \lambda < 4 \\ 1-\frac{\lambda}{6} & 2 \leq \lambda < 3 \\ \frac{\lambda}{3} & \lambda < 2\end{cases}.\]
We can then read off $\cal{P}_+(\delta) = [0,4-x]$, so in particular $x \leq 4$. Hence
\[\vol(\delta) = 6\int_{0}^{4-x}{2\lambda A(\delta,\lambda) \ \mathrm{d}\lambda}.\]
That is,
\[\begin{split}\vol(\delta) & = \begin{cases} \int_{0}^{4-x}{4\lambda^2 \ \mathrm{d}\lambda} & 2 \leq x \leq 4 \\ \int_{0}^{2}{4\lambda^2 \ \mathrm{d}\lambda} + \int_{2}^{4-x}{12\lambda - 2\lambda^2 \ \mathrm{d}\lambda} & 1 \leq x < 2 \\ \int_{0}^{2}{4\lambda^2 \ \mathrm{d}\lambda} + \int_{2}^{3}{12\lambda - 2\lambda^2 \ \mathrm{d}\lambda} + \int_{3}^{4-x}{24\lambda - 6\lambda^2 \ \mathrm{d}\lambda} & 0 \leq x < 1\end{cases} \\ & = \begin{cases} -\frac{4}{3}(x-4)^3 & 2 \leq x \leq 4 \\ \frac{32}{3} + \frac{2}{3}(x^3-3x^2-24x+52) & 1 \leq x < 2 \\ 28 + 2(x^3-6x^2+5) & 0 \leq x < 1.\end{cases}\end{split}\]
Hence we have
\[ \beta(H) = 114 - \int_{0}^{4}{\vol(\delta)} = 114 - 59 = 55.\]
So $X$ is $K$-polystable.

§.§ The Quadric Threefold and Its Blow-Up

§.§.§ Central Divisor

Let $Q = \cal{Z}(3x_2^2-4x_1x_3+x_0x_4) \sub \proj{4}$. The twisted quartic curve
\[C = \cal{Z}(x_0x_2-x_1^2, x_0x_3-x_1x_2, x_1x_4-x_2x_3, x_2x_4-x_3^2)\]
is $G$-invariant. The prime divisor
\[F = \cal{Z}(4x_2^3+x_1^2x_4+x_0x_3^2-6x_1x_2x_3) \inter Q\]
is $G$-invariant and contains $C$.

The smooth Fano (2.21) is obtained by blowing up $Q$ along $C$. The minimal $B$-chart of $C$ is given by taking $x_4 = 1$. In this chart, the equation of $Q$ allows us to eliminate $x_0$, so that $Q \cong \Spec{\bbk[x_1,x_2,x_3]}$. Then $C$ is given by
\[C = \cal{Z}(x_1-x_2x_3,x_2-x_3^2)\]
and $F$ by
\[F = \cal{Z}(4x_2^3 + x_1^2 + 4x_1x_3^3 - 3x_2^2x_3^2 - 6x_1x_2x_3)\]
One can read off immediately that there is an isomorphism from this $B$-chart in $Q$ to the $B$-chart in $\proj{3}$ we took in the previous example and that this isomorphism preserves the $G$-invariant subvarieties $F$ and $C$ (again, $F$ is the secant variety of $C$). Hence finding the central divisor is identical in this case to the previous one. Hence keeping the same notation as above, we must blow $Q$ up three times, obtaining exceptional divisors $E$, $D$ and $H$, with the latter being central.

§.§.§ $\beta_Q(H)$ (1.16)

To calculate $\beta_Q(H)$, we must first calculate $-K_Q$ and its pullback to the model containing $H$. Since $Q$ is a hypersurface in $\proj{4}$, the adjunction formula gives $-K_Q = (-K_{\proj{4}}-Q)\vert_Q$. The anticanonical class of $\proj{4}$ is the class of a divisor of degree 5. If we represent this by $Q + \cal{Z}(4x_2^3+x_1^2x_4+x_0x_3^2-6x_1x_2x_3)$, we see that $F$ is an anticanonical divisor of $Q$.

After the three blow-ups described above, this pulls back to $F + 2E + 3D + 6H$, so set $\delta = F + 2E + 3D + (6-x)H$. We have $\lambda_\delta = 0$ and $\cal{P}(\delta) = (\infty,6-x]$. To calculate $A(\delta,\lambda)$, first note that there is no contribution at points $p \neq 0, \infty, -4$. At $p=0$ we have two divisors: $D$ with $\ell = -1$, $m = 3$ and $h = 1$, and $D_0$ with $\ell = -1$, $m = 0$ and $h = 2$. Hence at this point there is a contribution of
\[\min\left\{3-\lambda,-\frac{\lambda}{2}\right\} = \begin{cases} 3-\lambda & \lambda \geq 6 \\ -\frac{\lambda}{2} & \lambda < 6.\end{cases}\]
At $p = -4$ we have $F$ with $\ell = 0$, $m = 1$, $h = 1$, and $D_{-4}$ with $\ell = 1$, $m = 0$, $h = 3$. Hence the contribution is
\[\min\left\{1,\frac{\lambda}{3}\right\} = \begin{cases}1 & \lambda \geq 3 \\ \frac{\lambda}{3} & \lambda < 3.\end{cases}\]
Finally, at $p = \infty$ we have $E$ with $\ell = 0$, $m = 2$, $h = 1$, and $D_\infty$ with $\ell = 1$, $m = 0$, $h = 3$. Hence the contribution is
\[\min\left\{2,\frac{\lambda}{3}\right\} = \begin{cases} 2 & \lambda \geq 6 \\ \frac{\lambda}{3} & \lambda < 6.\end{cases}\]
All in all, we have
\[A(\delta,\lambda) = \begin{cases} 6-\lambda & \lambda \geq 6 \\ 1-\frac{\lambda}{6} & 3 \leq \lambda \leq 6 \\ \frac{\lambda}{6} & \lambda \leq 3.\end{cases}\]
Hence $A(\delta,\lambda) \geq 0$ for $0 \leq \lambda \leq 6$, so we have $\cal{P}_+(\delta) = [0,6-x]$ and $0 \leq x \leq 6$.

Now we have
\begin{align*}\vol(\delta) & = 6\int_{0}^{6-x}{2\lambda A(\delta,\lambda) \ \mathrm{d}\lambda} \\ & = \begin{cases} 6\int_{0}^{6-x}{\frac{\lambda^2}{3} \ \mathrm{d}\lambda} & 3 \leq x \leq 6 \\ 6\int_{0}^{3}{\frac{\lambda^2}{3} \ \mathrm{d}\lambda} + 6\int_{3}^{6-x}{2\lambda-\frac{\lambda^2}{3} \ \mathrm{d}\lambda} & 0 \leq x \leq 3\end{cases}\\ & = \begin{cases} \frac{2}{3}(6-x)^3 & 3 \leq x \leq 6 \\ \frac{2x^3}{3}-6x^2+54 & 0 \leq x \leq 3.\end{cases}\end{align*}
We have $(-K_Q)^3 = 54$ and $A_Q(H) = 4$, since we reached $H$ as the final exceptional divisor after 3 nested blow-ups of $Q$. Therefore
\[\beta_Q(H) = 216-\int_{0}^{6}{\vol(\delta) \ \mathrm{d}x} = 216-135 = 81 > 0\]
and $Q$ is $K$-polystable.

§.§.§ $\beta_X(H)$ (2.21)

We now calculate $\beta_X(H)$, where $X$ is the blow-up of $Q$ in the twisted quartic $C$, i.e. the smooth Fano (2.21). We have $(-K_X)^3 = 28$, and $A_X(H) = 3$.

Since $-K_Q = F$ and the curve $C$ has multiplicity 2 in $F$, we have $-K_X = (F+2E) - E = F + E$. Under thw two subsequent blow-ups to the model containing $H$, this pulls back to $F + E + 2D + 4H$, so we set $\delta = F + E + 2D + (4-x)H$. We have $\lambda_\delta = 0$ and $\cal{P}(\delta) = (\infty,4-x]$. Moving on to calculating $A(\delta,\lambda)$: as before, points $p \neq -4, 0, \infty$ do not contribute.

At $p = -4$, we have two divisors: $F$, with $m = 1, \ell = 0$, $h = 1$, and $D_{-4}$, with $m = 0$, $\ell = 1$, $h = 3$. Hence there is a contribution of
\[\min\left\{1,\frac{\lambda}{3}\right\} = \begin{cases} 1 & \lambda \geq 3 \\ \frac{\lambda}{3} & \lambda < 3\end{cases}.\]

At $p = \infty$, the situation is identical to that at $p = -4$, so we get the same contribution again.

At $p = 0$, we have two divisors: $D$, with $m = 2$, $\ell = -1$ and $h = 1$, and $D_0$, with $m = 0$, $\ell = -1$ and $h = 2$. Hence there is a contribution of
\[\min\left\{2-\lambda,-\frac{\lambda}{2}\right\} = \begin{cases} 2-\lambda & \lambda \geq 4 \\ -\frac{\lambda}{2} & \lambda < 4\end{cases}.\]

Hence all things considered, we have
\[A(\delta,\lambda) = \begin{cases} 4-\lambda & \lambda \geq 4 \\ 2-\frac{\lambda}{2} & 3 \leq \lambda < 4 \\ \frac{\lambda}{6} & \lambda < 3 \end{cases}.\]
Therefore $A(\delta,\lambda) \geq 0$ for $0 \leq \lambda \leq 4$. Hence $\cal{P}_+(\delta) = [0,4-x]$ with $0 \leq x \leq 4$. We thus have
\begin{align*}\vol(\delta) & = 6\int_{0}^{4-x}{2\lambda A(\delta,\lambda) \ \mathrm{d}\lambda} \\ & = \begin{cases} \int_{0}^{4-x}{2\lambda^2 \ \mathrm{d}\lambda} & 1 \leq x \leq 4 \\ \int_{0}^{3}{2\lambda^2 \ \mathrm{d}\lambda} + \int_{3}^{4-x}{24\lambda - 6\lambda^2 \ \mathrm{d}\lambda} & 0 \leq x \leq 1\end{cases} \\ & = \begin{cases} \frac{2}{3}(4-x)^3 & 1 \leq x \leq 4 \\ 18 + 2(x^3-6x^2+5) & 0 \leq x \leq 1.\end{cases}\end{align*}
\[\begin{split} \beta(H) & = 84 - \int_{0}^{4}{\vol(\delta) \ \mathrm{d}x} \\ & = 84 - \int_{0}^{1}{18 + 2(x^3-6x^2+5) \ \mathrm{d}x} - \int_{1}^{4}{\frac{2}{3}(4-x)^3 \ \mathrm{d}x} \\ & = 84 -\frac{49}{2} - \frac{27}{2} = 46 > 0,\end{split}\]
so $X$ is $K$-polystable.

[1]
C. Araujo, A.-M. Castravet, I. Cheltsov, K. Fujita, A.-S. Kaloghiros,
J. Martinez-Garcia, C. Shramov, H. Süss, and N. Viswanathan.
The Calabi problem for Fano threefolds.
MPIM preprint, 2021.

[2]
H. Abban and Z. Zhuang.
$k$-stability of Fano varieties via admissible flags.
arXiv preprint arXiv:2003.13788, 2020.

[3]
H. Abban and Z. Zhuang.
Seshadri constants and $K$-stability of Fano manifolds.
arXiv preprint arXiv:2101.09246, 2021.

[4]
X. Chen, S. Donaldson, and S. Sun.
Kähler-Einstein metrics on Fano manifolds, I-III.
Journal of the American Mathematical Society, 28(1):183–278,

[5]
I. Cheltsov, V. Przyjalkowski, and C. Shramov.
Fano threefolds with infinite automorphism groups.
Izvestiya: Mathematics, 83, 2019.

[6]
I. Cheltsov and K. Shramov.
Extremal metrics on del Pezzo threefolds.
In Proceedings of the Steklov Institute of Mathematics, volume
264, pages 30–44, 2009.

[7]
T. Delcroix.
K-stability of Fano spherical varieties.
arXiv preprint arXiv:1608.01852, 2016.

[8]
S Donaldson.
Kähler geometry on toric manifolds, and some other manifolds with
large symmetry.
In Handbook of geometric analysis. No. 1, pages 29–75.
Somerville, MA: International Press; Beijing: Higher Education Press, 2008.

[9]
S. Donaldson.
Stability of algebraic varieties and Kähler geometry.
In Proceedings of Symposia in Pure Mathematics, volume 97,

[10]
V. Datar and G. Székelyhidi.
Kähler-Einstein metrics along the smooth continuity method.
Geometric and Functional Analysis, 26:975–1010, 2015.

[11]
K. Fujita.
A valuative criterion for uniform $K$-stability of $\bb{Q}$-Fano
arXiv preprint arXiv:1602.00901v2, 2016.

[12]
M. Furushima.
Mukai-Umemura's example of the Fano threefold with genus $12$ as
a compactification of $\mathbb{C}^3$.
Nagoya Math. J., 127:145–165, 1992.

[13]
J. Humphreys.
Introduction to Lie Algebras and Representation Theory.
Graduate Texts in Mathematics. Springer New York, 1972.

[14]
J. Humphreys.
Linear Algebraic Groups.
Springer-Verlag, 1975.

[15]
N. Ilten and H. Süß.
K-stability for Fano manifolds with torus action of complexity $1$.
Duke Mathematical Journal, 166:177–204, 2017.

[16]
F. Knop, H. Kraft, D. Luna, and Th. Vust.
Local properties of algebraic group actions.
In H. Kraft, P. Slodowy, and T.A. Springer, editors, Algebraic
Transformation Groups and Invariant Theory, pages 63–75. Birkhäuser
Basel, 1989.

[17]
J. Kollár and S. Mori.
Birational Geometry of Algebraic Varieties.
Cambridge Tracts in Mathematics. Cambridge University Press, 1998.

[18]
Friedrich Knop.
Über Bewertungen, welche unter einer reduktiven Gruppe
invariant sind.
Mathematische Annalen, 295:333–363, 1993.

[19]
Friedrich Knop.
On the set of orbits of a borel subgroup.
Comentarii Mathematici Helvetici, 70:285–309, 1995.

[20]
R.K. Lazarsfeld.
Positivity in Algebraic Geometry II: Positivity for Vector
Bundles, and Multiplier Ideals.
Ergebnisse der Mathematik und ihrer Grenzgebiete. 3. Folge / A Series
of Modern Surveys in Mathematics. Springer Berlin Heidelberg, 2017.

[21]
C. Li.
$K$-semistability is equivariant volume minimisation.
arXiv preprint arXiv:1512.07205v5, 2017.

[22]
D. Luna and Th. Vust.
Plongements d'espaces homogènes.
Commentarii mathematici Helvetici, 58:186–245, 1983.

[23]
C. Li and C. Xu.
Special test configuration and k-stability of fano varieties.
Annals of Mathematics, 180:197–232, 2014.

[24]
Y. Liu and Z. Zhu.
Equivariant $K$-stability under finite group action.
arXiv preprint arXiv:2001.10557, 2020.

[25]
Y. Matsushima.
Sur la structure du groupe d'homéomorphismes analytiques d'une
certaine variété Kählérienne.
Nagoya Math. J., 11:145–150, 1957.

[26]
H. Süss and I. Cheltsov.
K-polystability of two smooth Fano threefolds.
arXiv preprint arXiv:2107.04797v1, 2021.

[27]
H. Süß.
Kähler-Einstein metrics on symmetric Fano T-varieties.
Advances in Mathematics, 246:100–113, 2013.

[28]
G. Tian.
On Kähler-Einstein metrics on certain Kähler manifolds.
Inventiones Mathematicae, 89:225–246, 1987.

[29]
D. Timashev.
Classification of $G$-varieties of complexity 1.
Izvestiya: Mathematics, 61:363, 10 1997.

[30]
D. Timashev.
Cartier divisors and geometry of normal $G$-varieties.
Transformation Groups, 5:181–204, 01 2000.

[31]
D. Timashev.
Homogeneous Spaces and Equivariant Embeddings.
Springer-Verlag, 2011.

[32]
X.-J. Wang and X. Zhu.
Kähler-Ricci solitons on toric manifolds with positive first
Chern class.
Advances in Mathematics, 188:87–103, 2004.

[33]
C. Xu.
K-stability of Fano varieties: an algebro-geometric approach.
arXiv preprint arXiv:2011.10477v1, 2020.

[34]
Z. Zhuang.
Optimal destabilizing centers and equivariant $K$-stability.
Inventiones Mathematicae, 226:195–223, 2021.