Proof. From [Reference Arezzo, Pacard and SingerAPS11, Reference SzékelyhidiSzé15], there is ${\varepsilon }_0$ such that $({\rm Bl}_Z(X), {\Omega }_{\varepsilon })$ admits an extremal Kähler metric for ${\varepsilon }\in (0,{\varepsilon }_0)$ . We only need to show the vanishing of the Futaki invariant ${\rm Fut}_{{\Omega }_{\varepsilon }}$ . This will follow from its equivariance, as already used in [Reference Sektnan and TiplerST24, Proposition 2.1]. Define $\tilde X={\rm Bl}_Z(X)$ . By $(i)$ and $(ii)$ , $G$ lifts to a subgroup of ${{\rm Aut}}(\tilde X)$ such that ${\Omega }_{\varepsilon }$ is $G$ -invariant. Then, by equivariance of ${\rm Fut}_{{\Omega }_{\varepsilon }} \in ({\rm Lie} ( {{\rm Aut}}(\tilde X)))^*$ (see e.g. [Reference FutakiFut88, Chapter 3] or [Reference LeBrun and SimancaLS94, § 3.1]), we have, for all $g\in G$ and $v\in ({\rm Lie} ( {{\rm Aut}}(\tilde X)))$ ,

\begin{align*} {\rm Fut}_{{\Omega }_{\varepsilon }}({\rm Ad}_g(v))={\rm Fut}_{{\Omega }_{\varepsilon }}(v). \end{align*}

Let $\tilde w\in {\rm Lie} ( {{\rm Aut}}(\tilde X))$ . Then $\tilde w$ is the lift of an element $w\in {\rm Lie} ( {{\rm Aut}}( X))$ that vanishes on $Z$ . Moreover,

\begin{align*} \tilde w_0 := \sum _{g\in G} {\rm Ad}_g(\tilde w) \end{align*}

is a fixed point under the adjoint action of $G$ on ${\rm Lie} ( {{\rm Aut}}(\tilde X))$ . As $Z$ is $G$ -invariant, we deduce that it is the lift of a fixed point $w_0\in {\rm Lie} ( {{\rm Aut}}( X))$ under the adjoint action of $G$ . From $(iii)$ , we deduce that $\tilde w_0=0$ . Hence, by equivariance of ${\rm Fut}_{{\Omega }_{\varepsilon }}$ ,

\begin{align*} 0={\rm Fut}_{{\Omega }_{\varepsilon }}(\tilde w_0)=\sum _{g\in G} {\rm Fut}_{{\Omega }_{\varepsilon }}({\rm Ad}_g(\tilde w))=\vert G\vert \; {\rm Fut}_{{\Omega }_{\varepsilon }}(\tilde w) \end{align*}

where $\vert G \vert$ is the cardinal of $G$ . This concludes the proof.

Proof of Proposition2.3 . The tangent space at $X$ , denoted $\tilde H^1(TX)$ , of the base of a semi-universal family of complex deformations of $X$ compatible with the polarisation $\Omega$ is constructed in [Reference Chen and SunCS14, Lemma 6.1] (see also [Reference SzékelyhidiSzé10]). In general, the construction of $\tilde H^1(TX)$ goes as follows. First, the usual Kuranishi complex for deformations of complex structures on $X$ is

\begin{align*} \cdots \to {\Omega }^{0,k}(TX) \stackrel {\overline \partial }{\to } {\Omega }^{0,k+1}(TX) \to \cdots , \end{align*}

where the extension of $\overline \partial$ to $(0,p)$ -forms is given in local coordinates, for $\beta = \sum _j \beta _j \otimes \frac {\partial }{\partial z^j}$ , by

\begin{align*} \overline \partial \beta = \sum _j \overline \partial \beta _j \otimes \frac {\partial }{\partial z^j}. \end{align*}

We are interested in deformations that are compatible with some cscK metric ${\omega }\in {\Omega }$ . Following [Reference Fujiki and SchumacherFS90], we define maps

\begin{align*} \begin{array}{cccc} \iota ^k_\bullet : & {\Omega }^{0,k}(TX) & \to & {\Omega }^{0,k+1}\\ & \beta & \mapsto & \iota _\beta {\omega } \end{array} \end{align*}

where $\iota _\beta {\omega }$ is obtained by the composition of first contraction and then alternation operators. For $k\geqslant 1$ , we then set

\begin{align*} {\Omega }_{{\omega }}^{0,k}(TX) := \ker \iota ^k_\bullet , \end{align*}

while we define

\begin{align*} {\Omega }_{{\omega }}^{0,0}(TX):= {\mathscr{C}}^\infty (X,{\mathbb{C}}). \end{align*}

Together with the restriction of $\overline \partial$ , and the map $\overline \partial^0:=\mathscr{D}$ defined, for $f\in {\mathscr{C}}^\infty (X,{\mathbb{C}})$ , by

\begin{align*} \mathscr{D}(f):=\overline \partial (\nabla _{{\omega }}^{1,0} f), \end{align*}

where $\nabla _{{\omega }}^{1,0} f$ is given by $\overline \partial f = {\omega }( \nabla _{{\omega }}^{1,0}f , \cdot )$ , we obtain another elliptic complex $({\Omega }_{{\omega }}^{0,\bullet }(TX), \overline \partial ^\bullet )$ . We denote by $\tilde H^{\bullet }(TX)$ the associated cohomology groups. Then, from this complex, following Kuranishi’s techniques, one can build a semi-universal family $(\mathscr{X},\mathscr{L})\to Z$ of polarised deformations of $(X,\Omega )$ , where $Z\subset B\subset \tilde H^1(TX)$ is an analytic subspace (corresponding to integrable infinitesimal deformations) of some open ball $B$ centred at zero (again, we refer to [Reference Chen and SunCS14, Lemma 6.1] for the detailed construction). As in [Reference DoanDoa22], setting $K:={\rm Aut}(X,{\omega })$ , the group of holomorphic isometries of $(X,\omega )$ , this construction can actually be made $K$ -equivariantly, and even locally $K^{\mathbb{C}}$ -equivariantly on $Z$ , for the natural $K^{\mathbb{C}}$ -action on $\tilde H^1(TX)$ . From [Reference Dervan and NaumannDN20, § 3], $\mathscr{M}_X$ is given by an open neighbourhood of $0$ in the analytic GIT quotient $W_X//G$ , where $G=K^{\mathbb{C}}$ , and where $W_X\subset \tilde H^1(TX)$ is the smallest $G$ -invariant Stein space containing $Z$ . As noted in [Reference Rollin and TiplerRT14, Lemma 2.10], as $X$ is toric, by Bott, Steenbrink and Danilov’s vanishing of $h^{0,1}(X)$ and $h^{0,2}(X)$ (see [Reference Cox, Little, Schenck and varietiesCLS11, Theorem 9.3.2]), the vector space $\tilde H^1(TX)$ is $G$ -equivariantly isomorphic to $H^1(X,TX)$ . Also, for a toric surface, ${{\rm Aut}}(X)$ is a linear algebraic group, so that $G\simeq {{\rm Aut}}(X)$ , as the Lie algebra of ${{\rm Aut}}(X)$ has no parallel vector field (see [Reference GauduchonGau, Chapter 2]). Finally, $H^2(X,TX)=0$ by [Reference IltenIlt11, Corollary 1.5], so that any small element of $H^1(X,TX)$ is integrable, and $Z=B\subset H^1(X,TX)$ . Hence $W_X=H^1(X,TX)$ , and putting all this together, we obtain the first part of Proposition 2.3. The statement about klt type singularities follows directly from [Reference Braun, Greb, Langlois and JoaquinBGLJ24, Theorem 1], which asserts that the GIT quotient of a klt type singularity is of klt type, and which implies in particular that the GIT quotient of $H^1(X,TX)$ (which is smooth) by ${{\rm Aut}}(X)$ is of klt type.

Proof. By Matsushima and Lichnerowicz’s theorem [Reference MatsushimaMat57, Reference LichnerowiczLic58], ${{\rm Aut}}(X)$ is reductive. The result then follows from Nill’s result [Reference NillNil06, Theorem 1.8] together with Demazure’s structure theorem [Reference DemazureDem70, Proposition 11, p. 581], which we now recall. Demazure’s root system for $(N,\Sigma )$ , which we denote here by $\mathscr{R}$ (it corresponds to ${\rm Rac}(\Sigma )$ in Demazure’s notation [Reference DemazureDem70]), is the set

\begin{align*} {\mathscr{R}}:=\lbrace m\in M\:\vert \: \exists \rho \in \Sigma (1)\:\textrm{with}\: \langle m, u_\rho \rangle=1 \:\textrm{but}\:\forall \rho '\in \Sigma (1),\: \rho '\neq \rho , \langle m , u_{\rho '}\rangle \leqslant 0\rbrace , \end{align*}

where we use the notation $u_\rho \in N$ for the primitive generator of the ray $\rho$ . Demazure’s structure theorem then asserts that the identity component ${{\rm Aut}}(X)^0$ of ${{\rm Aut}}(X)$ is generated by $T$ and some unipotent elements $\lbrace U_m, m\in {\mathscr{R}}\rbrace$ . In particular,

\begin{align*} {{\rm dim}}({{\rm Aut}}(X))={{\rm dim}}(T)+\vert {\mathscr{R}}\vert . \end{align*}

Then, the quotient ${{\rm Aut}}(X)/{{\rm Aut}}(X)^0$ is finite and isomorphic to a quotient of ${{\rm Aut}}(N,\Sigma )$ by a subgroup $W(N,\Sigma )\subset {{\rm Aut}}(N,\Sigma )$ that is the Weyl group of a maximal reductive subgroup of ${{\rm Aut}}(X)$ with root system ${\mathscr{R}}\cap -{\mathscr{R}}$ . Then Nill’s theorem [Reference NillNil06, Theorem 1.8] gives an upper bound for the dimension of the automorphism group of a smooth complete toric variety, when it is reductive. In the case of a smooth toric surface $X$ with reductive automorphism group, the outcome is that either $X$ is a product of projective spaces or ${{\rm dim}}({{\rm Aut}}(X))=2$ . In the latter case, from the previous discussion on Demazure’s result, this implies that the set of Demazure’s roots $\mathscr{R}$ for $(N,\Sigma )$ is empty. Hence, ${{\rm Aut}}(X)^0=T$ and ${{\rm Aut}}(X)/{{\rm Aut}}(X)^0\simeq {{\rm Aut}}(N,\Sigma )$ , which concludes the proof.

Proof. Fix an isomorphism $N\simeq \mathbb{Z}^2$ and identify $g$ with an element of ${\rm GL}_2(\mathbb{Z})$ . The characteristic polynomial of

\begin{align*}g =\left [ \begin{array}{cc} a &\quad b \\ c &\quad d \end{array} \right ]\end{align*}

then reads

\begin{align*} \chi (Y)=Y^2-(a+d)Y+(ad-bc). \end{align*}

On the other hand, Jordan’s normal form for the $\mathbb{C}$ -linear extension of $g$ implies that it is conjugated (in ${\rm M}_2({\mathbb{C}})$ ) to

\begin{align*} \left [ \begin{array}{cc} \alpha &\quad \delta \\ 0 &\quad \beta \end{array} \right ] \end{align*}

for $(\alpha ,\beta )\in {\mathbb{C}}^2$ and $\delta \in \lbrace 0,1 \rbrace$ . As $g$ is both invertible and of finite order, we deduce that $\delta=0$ and $g$ is conjugated to

\begin{align*} \left [ \begin{array}{cc} \alpha &\quad 0 \\ 0 &\quad \beta \end{array} \right ]. \end{align*}

Moreover, $\alpha ^m=\beta ^m=1$ , so that $\alpha$ and $\beta$ are $m$ -th roots of unity in $\mathbb{C}$ . On the other hand, by conjugation invariance, we also have

\begin{align*} \det (g)=\alpha \beta =ad-bc\in \lbrace -1,+1 \rbrace \end{align*}

as $g\in {\rm GL}_2(\mathbb{Z})$ and

\begin{align*} {\rm trace}(g)=\alpha +\beta =a+d \in \mathbb{Z}. \end{align*}

If $\alpha$ and $\beta$ belong to $\lbrace -1, +1 \rbrace$ , that is if $m=2$ , then we are done. If not, at least one of $\alpha$ or $\beta$ is not real, and thus $\alpha$ and $\beta$ must be complex conjugated roots of $\chi (Y)$ . Hence

\begin{align*}{\rm trace}(g)=\alpha +\overline {\alpha }\in \mathbb{Z},\end{align*}

but also

\begin{align*} \vert \alpha +\overline {\alpha } \vert \lt 2\end{align*}

so that

\begin{align*}\alpha +\overline {\alpha } \in \lbrace -1,0, 1 \rbrace . \end{align*}

Hence

\begin{align*}\alpha \in \lbrace \pm i, \pm j, \pm j^2 \rbrace ,\end{align*}

from which the result follows easily.

Proof. As ${{\rm Aut}}(N,\Sigma )$ is finite, we can fix an ${{\rm Aut}}(N,\Sigma )$ -invariant euclidean metric on $N_{\mathbb{R}}$ (again considering the $\mathbb{R}$ -linear extension of ${{\rm Aut}}(N,\Sigma )$ ). Then, by the classification of finite subgroups of the group of orthogonal transformations of the plane, we deduce that ${{\rm Aut}}(N,\Sigma )$ is isomorphic to $C_p$ or $D_p$ , for some $p\in \mathbb{N}^*$ . As those groups admit elements of order $p$ , by Lemma 3.2, the result follows.

Proof. The proof is straightforward, although a bit tedious, so we will only give a sketch of it. Fix an orientation of ${\mathbb{R}}^2$ , and pick a two dimensional cone $\sigma$ of $\Sigma _p$ (the proof is the same for $\Sigma _p'$ ). Let $g\in {{\rm Aut}}(\mathbb{Z}^2,\Sigma _p)$ . Then $g\cdot \sigma \in \Sigma _p(2)$ . Moreover, $g$ maps the two facets of $\sigma$ to the two facets of $g\cdot \sigma$ . As the fan $\Sigma _p$ is explicitly described, for each possible $\tau =g\cdot \sigma \in \Sigma _p(2)$ , we deduce the possible matrix coefficients for $g$ ,

\begin{align*} g=\left [ \begin{array}{cc} a_\tau &\quad b_\tau \\ c_\tau &\quad d_\tau \end{array} \right ]\in {\rm GL}_2(\mathbb{Z}). \end{align*}

For any possible $\tau$ , one can then explicitly compute the images of the rays of $\Sigma _p$ by $g$ . They should all belong to $\Sigma _p(1)$ , and this enables us to select the allowed cones $\tau$ for $g$ to be a lattice automorphism of $\Sigma _p$ . The result then follows easily.

Proof. Through its representation

\begin{align*}C_p \to {\rm GL}_{\mathbb{R}}(N_{\mathbb{R}}), \end{align*}

the subgroup $C_p$ of ${{\rm Aut}}(N,\Sigma )$ acts on $N_{\mathbb{R}}$ by rotations (once an invariant euclidean metric is fixed), and hence has no fixed points. We deduce that the action of $C_p$ on $\Sigma (1)$ is free, and then $p$ divides $\vert \Sigma (1)\vert$ , the number of rays of $\Sigma$ .

On the other hand, from the classification of toric surfaces (see e.g. [Reference Cox, Little, Schenck and varietiesCLS11, Chapter 10]), we must have $ \vert \Sigma (1)\vert \geqslant 3$ . Moreover, if $\vert \Sigma (1)\vert=3$ , then $X\simeq \mathbb{P}^2$ , while if $\vert \Sigma (1)\vert=4$ , then $X\simeq {\mathbb{F}}_n$ for some $n\in \mathbb{N}$ . By Remark 3.10, we see that $X\simeq \mathbb{P}^1\times \mathbb{P}^1$ in that case.

So we may assume that $\vert \Sigma (1)\vert \geqslant 5$ , and that $p$ divides $\vert \Sigma (1)\vert$ . By the classification of toric surfaces, we know that $X$ is obtained by successive blow-ups from ${\mathbb{F}}_n$ , $n\geqslant 0$ , or from $\mathbb{P}^2$ . Then, there is a ray $\rho _i$ in $\Sigma (1)$ generated by the sum of two primitive elements

\begin{align*}u_i=u_{i-1}+u_{i+1}\end{align*}

generating adjacent rays $\rho _{i-1}$ and $\rho _{i+1}$ in $\Sigma (1)$ (that is both $\rho _i+\rho _{i-1}$ and $\rho _i+\rho _{i+1}$ belong to $\sigma (2)$ ). Note that

\begin{align*} \det (u_{i+1},u_{i-1})=\det (u_i-u_{i-1},u_{i-1})=\det (u_i,u_{i-1})=1 \end{align*}

so the contraction to a point of the divisor associated to the ray $\rho _i$ is smooth. By linearity, for any $g\in C_p$ we have

\begin{align*}g\cdot u_i=g\cdot u_{i-1}+g\cdot u_{i+1}, \end{align*}

and hence we can contract to points the $C_p$ -orbit of the torus invariant divisor $D_{\rho _i}$ associated to $\rho _i$ and obtain a smooth toric surface whose fan still admits $C_p$ as a subgroup of its lattice symmetry group. By an inductive argument on $\vert \Sigma (1)\vert$ , we obtain the result.

Proof. This is simply an induction on the number of $C_p$ -equivariant blow-ups in Proposition 3.13, using Theorem2.2 at each step, and the fact that $\mathbb{P}^1\times \mathbb{P}^1$ , $\mathbb{P}^2$ and ${\rm Bl}_{p_1,p_2,p_3}(\mathbb{P}^2)$ admit cscK metrics in $C_p$ -equivariant classes (for ${\rm Bl}_{p_1,p_2,p_3}(\mathbb{P}^2)$ , this follows again from Theorem2.2 applied to $\mathbb{P}^2$ ).

Proof. Note that, by Example 4.1, $\mathbb{P}^2$ , $\mathbb{P}^1\times \mathbb{P}^1$ and ${\rm Bl}_{p_1,p_2,p_3}(\mathbb{P}^2)$ are rigid. If $X$ is not one of those three foldable surfaces, then, from Proposition 3.13, it either belongs to $\lbrace Y_2, Y_3, Y_4 \rbrace$ or is obtained from $Y_j$ by successive $C_p$ -equivariant blow-ups, where $p=2$ for $j\in \lbrace 2,4 \rbrace$ and $p=3$ for $j=3$ . From [Reference IltenIlt11, Corollary 1.6], $H^1(Y_j,TY_j)$ injects in $H^1(X,TX)$ . Hence, from the computations of Example 4.1, we see that $X$ is not rigid, and the result follows.

Proof. If $V=0$ , then there is nothing to prove. We then assume that $X$ is not rigid. From Lemma 4.3, $X$ is obtained from $Y_j$ by successive $C_p$ -equivariant blow-ups, where $j\in \lbrace 2, 3, 4\rbrace$ , for suitable $p\in \lbrace 2,3 \rbrace$ .

Denote by $d={{\rm dim}}(V)$ and by $\tilde N\simeq \mathbb{Z}^d$ the lattice of one-parameter subgroups of $\tilde T \simeq ({\mathbb{C}}^*)^d$ acting on $V\simeq {\mathbb{C}}^d$ by multiplication on each coordinate:

\begin{align*} \forall (t,x)\in \tilde T \times V,\: t\cdot x =(t_i x_i)_{1\leqslant i \leqslant d}. \end{align*}

The $T$ -module structure of $V$ is then equivalent to the injection $T \to \tilde T$ , or the lattice monomorphism

\begin{align*} B : N \to \tilde N \end{align*}

defined by

\begin{align*}B(u)=(\langle m_i , u \rangle )_{1\leqslant i \leqslant d}\end{align*}

where the weights $m_i$ run through all the weights in the weight space decomposition of $V$ (with multiplicities when ${{\rm dim}}(V_m)\geqslant 2$ ). The fact that $B$ is injective comes again from the injection $V_j \subset V$ (cf. [Reference IltenIlt11, Corollary 1.6]) and the explicit computation of $V_j$ in Example 4.1. We then consider the quotient

\begin{align*} N':=\tilde N / N, \end{align*}

where we identify $N$ with $B(N)$ by abuse of notation. We claim that $N'$ is again a lattice, that is, that $N$ is saturated in $\tilde N$ . If $X=Y_j$ , for $j\in \lbrace 2,3,4\rbrace$ , then this can be checked directly using the description of $V_j$ in Example 4.1. If $X$ is a blow-up of $Y_j$ , then $V_j \subset V$ , so that $B$ can be written $B=(B_j, B_+)$ , where

\begin{align*}B_j : N \to \tilde N_j\end{align*}

is the injection corresponding to the weight space decomposition of $V_j$ , with

\begin{align*}\tilde N_j\simeq \mathbb{Z}^{d_j},\end{align*}

for $d_j={{\rm dim}}(V_j)$ . Then, the fact that $B_j(N)$ is saturated in $\tilde N_j\subset \tilde N$ implies that $B(N)$ is saturated in $\tilde N$ . Thus, we have a short exact sequence of lattices

(4) \begin{align} 0\longrightarrow N \stackrel {B}{\longrightarrow }\tilde N \stackrel {A}{\longrightarrow } N' \longrightarrow 0 \end{align}

where

\begin{align*}A : \tilde N \to \tilde N /N\end{align*}

denotes the quotient map.

Introduce $(\tilde e_i)_{1\leqslant i\leqslant d}$ , the basis of $V$ dual to the coordinates $\chi ^{m_i}$ of the weight spaces $V_{m_i}$ . This corresponds to a $\mathbb{Z}$ -basis of $\tilde N$ , still denoted $(\tilde e_i)_{1\leqslant i \leqslant d}$ . The cone

\begin{align*}\tilde \sigma =\sum _{i=1}^d {\mathbb{R}}_+\cdot \tilde e_i\subset \tilde N_{\mathbb{R}}\end{align*}

satisfies

\begin{align*} {\rm Spec}({\mathbb{C}}[\tilde \sigma ^\vee \cap \tilde M])=V, \end{align*}

where

\begin{align*}\tilde M={{\rm Hom}}_{\mathbb{Z}}(\tilde N, \mathbb{Z}). \end{align*}

Consider the cone of $N'_{\mathbb{R}}$

\begin{align*} \sigma = A(\tilde \sigma )= \sum _{i=1}^d {\mathbb{R}}_+\cdot A(\tilde e_i) \end{align*}

(we will still denote by $A$ and $B$ their $\mathbb{R}$ -linear extensions). Our goal is then to show that $W$ is isomorphic to the affine toric variety defined by $\sigma$ . We first need to prove that $\sigma$ has the required properties to define such a variety. The cone $\sigma$ is clearly rational, convex and polyhedral. We now prove that $\sigma$ is strictly convex, that is

\begin{align*}-\sigma \cap \sigma =\lbrace 0 \rbrace .\end{align*}

So let $v\in -\sigma \cap \sigma$ . Then there is $(u_+, u_-)\in (\tilde \sigma )^2$ such that $A (u_\pm )=\pm v$ . We deduce that

\begin{align*}u_+ + u_-\in \ker (A)\cap \tilde \sigma = {{\rm Im}} (B)\cap \tilde \sigma .\end{align*}

We will then show that

(5) \begin{align} {{\rm Im}} (B) \cap \tilde \sigma = \lbrace 0 \rbrace , \end{align}

which implies $u_+=u_-=0$ by strict convexity of $\tilde \sigma$ . Notice that it is enough to show (5) for $X=Y_j$ . Indeed, using the decomposition $B=(B_j,B_+)$ as before, if

\begin{align*}u=B(x)\in \tilde \sigma ,\end{align*}

then

\begin{align*}u_j=B_j(x)\in \tilde \sigma _j,\end{align*}

where $\tilde \sigma _j \subset (\tilde N_j)_{\mathbb{R}}$ is the cone corresponding to $V_j$ . If

(6) \begin{align} {{\rm Im}}(B_j)\cap \tilde \sigma _j =\lbrace 0 \rbrace , \end{align}

then by injectivity of $B_j$ we deduce that $x=0$ and then $u=0$ . It remains to prove (6), which follows again from the explicit description of the weights of the $T$ action on $V_j$ . For example, if $j=2$ , and if

\begin{align*}B_2(x_1,x_2)=(x_2,-x_2,-x_1+x_2,x_1-x_2) \in \tilde \sigma _2,\end{align*}

then by definition of $\tilde \sigma _2$ we deduce

\begin{align*} \left \{ \begin{array}{ccc} x_2 & \geqslant & 0, \\ - x_2 & \geqslant & 0, \\ -x_1 +x_2 & \geqslant & 0, \\ x_1 - x_2 & \geqslant & 0, \end{array} \right . \end{align*}

and thus $x_1=x_2=0$ . The cases $j\in \lbrace 3, 4 \rbrace$ are similar. We have just proved that $\sigma$ is a strongly convex rational polyhedral cone, and it thus defines an affine toric variety.

We now claim that

\begin{align*} W\simeq {\rm Spec}({\mathbb{C}}[\sigma ^\vee \cap M']) \end{align*}

with

\begin{align*}M'={{\rm Hom}}_{\mathbb{Z}}(N', \mathbb{Z}). \end{align*}

It is equivalent to show that

\begin{align*} {\mathbb{C}}[\sigma ^\vee \cap M']\simeq {\mathbb{C}}[\chi ^{m_1},\ldots ,\chi ^{m_d}]^T. \end{align*}

The latter equality follows easily from the definitions. Indeed, considering the dual sequence to (4),

(7) \begin{align} 0\longrightarrow M' \stackrel {A^*}{\longrightarrow }\tilde M \stackrel {B^*}{\longrightarrow } M \longrightarrow 0 \end{align}

for $m'\in M'$ , if $\tilde m=A^* (m')$ , then

\begin{align*} \begin{array}{ccc} \chi ^{m'} \in \sigma ^\vee & \Longleftrightarrow & \forall i\in [\![ 1 , d ]\!],\: \langle m', A(\tilde e_i) \rangle \geqslant 0\\ & & \\ & \Longleftrightarrow & \forall i\in [\![ 1 , d ]\!],\: \langle A^*( m'), \tilde e_i \rangle \geqslant 0\\ & & \\ & \Longleftrightarrow & \left \{ \begin{array}{c} B^*(\tilde m)=0 \\ \forall i\in [\![ 1 , d ]\!],\: \langle \tilde m, \tilde e_i \rangle \geqslant 0\end{array} \right . \\ & & \\ & \Longleftrightarrow & \chi ^{\tilde m}\in {\mathbb{C}}[\chi ^{m_1},\ldots ,\chi ^{m_d}]^T, \end{array} \end{align*}

where in the last equivalence we used the facts that, from the definition of the basis $(\tilde e_i)_{1\leqslant i\leqslant d}$ ,

\begin{align*} \chi ^{\tilde m}\in {\mathbb{C}}[\chi ^{m_1},\ldots ,\chi ^{m_d}] \Longleftrightarrow \forall i\in [\![ 1 , d ]\!],\: \langle \tilde m, \tilde e_i \rangle \geqslant 0, \end{align*}

and the $T$ -module structure on ${\mathbb{C}}[\chi ^{m_1},\ldots ,\chi ^{m_d}]$ reads, for any one-parameter subgroup $\lambda _u:{\mathbb{C}}^*\to T$ generated by some $u\in N$ ,

\begin{align*} \lambda _u(t)\cdot \chi ^{\tilde m}=t^{\langle \tilde m, B(u)\rangle }\;\chi ^{\tilde m}. \end{align*}

This proves the claim, and the result follows.

Proof. We will exclude the rigid cases, and, as in the proof of Proposition 4.4, assume that $X$ is obtained from $Y_j$ by successive $C_p$ -equivariant blow-ups, where $j\in \lbrace 2, 3, 4\rbrace$ , for suitable $p\in \lbrace 2,3 \rbrace$ .

We use the characterisation in Proposition 4.7, and first need to describe the rays of $\sigma$ and their primitive generators. Note that, by construction, the rays of $\sigma$ belong to the set $\lbrace {\mathbb{R}}_+\cdot A(\tilde e_i),\: 1\leqslant i\leqslant d \rbrace$ . Let $\rho _i={\mathbb{R}}_+\cdot A(\tilde e_i)$ be such a ray. We claim that $A(\tilde e_i)$ is primitive in $N'$ . The argument is similar to the one used to prove strict convexity in the proof of Proposition 4.4. Suppose, for a contradiction, that there is $a\in \mathbb{N}$ , $a\geqslant 2$ , and $\tilde e \in \tilde \sigma \cap \tilde N$ such that $A(\tilde e_i)=a A(\tilde e)$ . Note that there is $x\in N$ such that

(8) \begin{align} B(x)=(B_j(x),B_+(x))=\tilde e_i - a \tilde e. \end{align}

By injectivity of $B_j$ , if $B_j(x)=0$ , then $x=0$ hence $\tilde e_i = a\tilde e$ . This is absurd as $\tilde e_i$ is primitive. So we may assume $B_j(x)\neq 0$ . Similarly, using Equation (6), we may assume as well that $B_j(x)\notin - \tilde \sigma _j$ and $B_j(x)\notin \tilde \sigma _j$ . Hence $B_j(x)$ must have at least one positive and one negative coordinate in the basis $(\tilde e_k)_{1\leqslant k \leqslant d_j}$ . As $\tilde e \in \tilde \sigma \cap \tilde N$ , its coordinates in the basis $(\tilde e_k)_{1\leqslant k\leqslant d}$ are non-negative integers, and, as $a\geqslant 2$ , we can write

\begin{align*} a\tilde e = \sum _{k=1}^d a_k \tilde e_k \end{align*}

with $a_k=0$ or $a_k\geqslant 2$ . Then, from Equation (8), $B_j(x)$ has exactly one coordinate equal to $1$ , while its other non-zero coordinates are all less than or equal to $-2$ . A case-by-case analysis using the description of $V_j$ in Equation (3) (cf. Example 4.1) shows that this is impossible. Indeed, for $V_2$ , the map $B_2$ is

\begin{align*} B_2(x_1,x_2)=(x_2,-x_2, -x_1+x_2, x_1-x_2), \end{align*}

with $x=(x_1,x_2)\in {\mathbb{R}}^2=N_{\mathbb{R}}$ , so if one coordinate of $B_2(x)$ equals $1$ , there is another coordinate that equals $-1$ . A similar argument leads to the same conclusion for $B_4$ . For $B_3$ , we can simply use the fact that the weight spaces are two dimensional, and we have

\begin{align*} B_3(x)=(-x_2,-x_2,x_1,x_1,-x_1+x_2,-x_1+x_2), \end{align*}

so the value $1$ would appear at least twice in the coordinates of $B_3(x)$ , if it ever does.

We will then prove that there is $m' \in M'$ such that

\begin{align*} \forall i\in [\![1, d ]\!]\: ,\:\langle m', A(\tilde e_i) \rangle=1 ,\end{align*}

which implies that $W$ is Gorenstein. So let $m'\in M'$ . Set

\begin{align*}\tilde m=A^*(m')\in \ker (B^*).\end{align*}

Then,

\begin{align*} \begin{array}{ccc} \forall i\in [\![1, d ]\!]\:,\: \langle m', A(\tilde e_i) \rangle=1 & \Longleftrightarrow & \forall i\in [\![1, d ]\!]\:,\: \langle \tilde m, \tilde e_i \rangle=1 \\ & & \\ & \Longleftrightarrow & \tilde m = (1, \ldots ,1) \end{array} \end{align*}

where we used the basis $(\tilde e_i)$ to produce the coordinates of $\tilde m$ . Hence, it is equivalent to show that

\begin{align*} (1, \ldots , 1)\in \ker (B^*), \end{align*}

which, by definition of $B$ , is equivalent to

\begin{align*} \sum _{i=1}^d m_i=0\in M \end{align*}

where the weights $m_i$ describe the $T$ -action on $V$ . This is where we use the fact that $X$ is foldable. Recall that $G={{\rm Aut}}(N,\Sigma )$ . The $G$ -action on $N$ naturally induces a $G$ -action on $M$ by duality, in such a way that the duality pairing is $G$ -invariant. Explicitly, for $m\in M={{\rm Hom}}(N,\mathbb{Z})$ , and $g\in G$ , we have

\begin{align*} g\cdot m := m(g^{-1}\,\cdot ) \end{align*}

and thus

\begin{align*}\langle g\cdot m , g\cdot u \rangle =\langle m , u\rangle \end{align*}

for any $u\in N$ . By the characterisation of the weights $(m_i)$ in Equation (2), we see that this $G$ -action preserves the set of weights appearing in the decomposition $V=\bigoplus _{m\in M} V_m$ . As $G$ contains a non-trivial cyclic group, there is an element $g\in {{\rm Aut}}(N)$ of order $p$ with no fixed point but $0$ . This element generates a cyclic group that acts freely on $\lbrace m_i,\: 1\leqslant i \leqslant d \rbrace$ (freeness comes from the fact that the action is free on the whole of $M_{\mathbb{R}}$ ). Hence

\begin{align*} \sum _{i=1}^d m_i=\sum _{i'} \sum _{k=0}^{p-1} g^k\cdot m_i' \end{align*}

where we picked a single element $m_i'$ in each orbit under this action. As $g\neq {{\rm Id}}$ , we have

\begin{align*} {{\rm Id}} + g + \cdots + g^{p-1}=0 \end{align*}

so that for each orbit

\begin{align*} \sum _{k=0}^{p-1} g^k\cdot m_i'=0. \end{align*}

We then deduce the existence of the required $m'$ , and that $W$ is Gorenstein.

We proceed to the proof of the fact that $W$ has terminal singularities. Let

\begin{align*} \Pi _\sigma := \left \{ \sum _{\rho \in \sigma (1)} \lambda _\rho u_\rho \:\vert \: \sum _{\rho \in \sigma (1)} \lambda _\rho \leqslant 1,\: 0\leqslant \lambda _\rho \leqslant 1 \right \}. \end{align*}

As described above, we may fix a subset of

\begin{align*}\lbrace A(\tilde e_i),\: 1\leqslant i\leqslant d \rbrace \end{align*}

as a set of generators for the rays in $\sigma (1)$ . Let $u'\in \Pi _\sigma \cap N'$ be given by

\begin{align*} u'=\sum _{i=1}^d \lambda _i A(\tilde e_i), \end{align*}

with

\begin{align*}0\leqslant \lambda _i \leqslant 1,\end{align*}

and

\begin{align*} \lambda _1 +\cdots +\lambda _d \leqslant 1,\end{align*}

assuming $\lambda _i=0$ when $A(\tilde e_i)$ is not in our fixed chosen set of ray generators. Assume that there is $i_0$ with $\lambda _{i_0}\neq 0$ . We need to show that $\lambda _{i_0}=1$ and $\lambda _i=0$ for $i\neq i_0$ . By the assumptions on the $\lambda _i$ , it is enough to show that $\lambda _i\in \mathbb{Z}$ for all $i$ . As $u'\in N'$ , there exists $\tilde u\in \tilde N$ such that

\begin{align*} A (\tilde u )= A\left(\sum _{i=1}^d \lambda _i \tilde e_i\right), \end{align*}

and then there is $x\in N_{\mathbb{R}}$ such that

\begin{align*} \tilde u - \sum _{i=1}^d \lambda _i \tilde e_i=B(x)=(B_j(x), B_+(x)), \end{align*}

where we recall that we still denote by $A$ and $B$ their $\mathbb{R}$ -linear extensions to $\tilde N_{\mathbb{R}}$ and $N_{\mathbb{R}}$ . By injectivity of $B_j$ again, we only need to consider the case when $B_j(x)\neq 0$ and $1\leqslant i_0\leqslant d_j$ . Indeed, if $B_j(x)=0$ then by injectivity $x=0$ , and $\tilde u$ being in $\tilde N$ forces the $\lambda _i$ to be integral, whereas if $\lambda _i=0$ for all $i\leqslant d_j$ , then $B_j(x)\in \tilde N$ (recall that $B_j$ maps to the first $d_j$ -coordinates) and so by saturation of $B_j(N)$ we have $x\in N$ and thus $B(x)\in \tilde N$ which, together with $\tilde u\in \tilde N$ , implies that the $\lambda _i$ are integral. Hence, we have reduced the problem to the case when $X=Y_j$ . We will again use the explicit descriptions of the spaces $V_j$ from Example 4.1. For $V_3$ , we find the system

\begin{align*} \left \{ \begin{array}{l} \tilde u_1 - \lambda _1 = -x_2,\\ \tilde u_2 - \lambda _2 = -x_2,\\ \tilde u_3 - \lambda _3 = x_1,\\ \tilde u_4 - \lambda _4 = x_1,\\ \tilde u_5 - \lambda _5 = -x_1+x_2,\\ \tilde u_6 - \lambda _6 = -x_1+x_2, \end{array} \right . \end{align*}

where $x=(x_1,x_2)\in {\mathbb{R}}^2=N_{\mathbb{R}}$ and $\tilde u=(\tilde u_1,\ldots ,\tilde u_6)\in \mathbb{Z}^6\simeq \tilde N$ (using the basis $(\tilde e_i)_{1\leqslant i \leqslant 6}$ ), from which we obtain

\begin{align*} ( \lambda _1 + \lambda _3+\lambda _5, \lambda _1 + \lambda _3+\lambda _6,\lambda _1 + \lambda _4+\lambda _6,\lambda _1 + \lambda _4+\lambda _5)\in \mathbb{N}^4 \end{align*}

and

\begin{align*} (\lambda _2 + \lambda _3+\lambda _5, \lambda _2 + \lambda _3+\lambda _6,\lambda _2 + \lambda _4+\lambda _6,\lambda _2 + \lambda _4+\lambda _5)\in \mathbb{N}^4, \end{align*}

where we used the fact that the $\tilde u_i$ are integers and $\lambda _i\geqslant 0$ for all $i$ . Together with

(9) \begin{align} 0 \leqslant \lambda _1 + \cdots + \lambda _6 \leqslant 1, \end{align}

we deduce that actually

\begin{align*} ( \lambda _1 + \lambda _3+\lambda _5, \lambda _1 + \lambda _3+\lambda _6,\lambda _1 + \lambda _4+\lambda _6,\lambda _1 + \lambda _4+\lambda _5)\in \lbrace 0,1\rbrace^4 \end{align*}

and

\begin{align*} (\lambda _2 + \lambda _3+\lambda _5, \lambda _2 + \lambda _3+\lambda _6,\lambda _2 + \lambda _4+\lambda _6,\lambda _2 + \lambda _4+\lambda _5)\in \lbrace 0,1\rbrace^4. \end{align*}

Let us then assume that $i_0=1$ , to fix our ideas, so $\lambda _1\gt 0$ , the other cases being similar. We must then have $\lambda _1 + \lambda _3+\lambda _5=1$ , and summing with $\lambda _2 + \lambda _4+\lambda _6$ , we see with (9) that $\lambda _2+\lambda _4+$ $\lambda _6=0$ . Hence $\lambda _2=\lambda _4=\lambda _6=0$ . Then, $\lambda _1+\lambda _4+\lambda _6=\lambda _1\gt 0$ so $\lambda _1=1$ . Considering $\lambda _1 + \lambda _3+\lambda _6=1$ and $\lambda _1 + \lambda _4+\lambda _5=1$ yields $\lambda _3=\lambda _5=0$ , and the result follows in that case. The cases of $V_2$ and $V_4$ are similar, so we will only treat $V_4$ . In that case, in the coordinates $\tilde x=(\tilde x_1,\ldots ,\tilde x_4)\in {\mathbb{R}}^4\simeq \tilde N_{\mathbb{R}}$ given by $(\tilde e_1,\ldots ,\tilde e_4)$ , the map $A$ is given by

\begin{align*} A(\tilde x_1,\tilde x_2,\tilde x_3,\tilde x_4)=(\tilde x_1+\tilde x_2,\tilde x_3+\tilde x_4). \end{align*}

Hence

\begin{align*} \left \{ \begin{array}{ccc} A(\tilde e_1) & = & A(\tilde e_2), \\ A(\tilde e_3) & = & A(\tilde e_4), \end{array} \right . \end{align*}

so we can pick the ray generators $(A(\tilde e_1), A(\tilde e_3))$ for the two dimensional cone $\sigma$ . The description of $V_4$ then implies

\begin{align*} \left \{ \begin{array}{ccc} \tilde u_1 - \lambda _1 & = & x_1,\\ \tilde u_2 & = & -x_1,\\ \tilde u_3 - \lambda _3 & = & x_2,\\ \tilde u_4 & = & -x_2, \end{array} \right . \end{align*}

hence $x\in N$ and the result follows.