Proof. This is a standard consequence of the Riemann-Hurwitz formula for the quotient morphism $C \to C/G$ :

\begin{equation*}2g-2=\lvert G\rvert(2g_q-2)+\sum_p\! (e_p-1),\end{equation*}

where p varies over the branch points, $e_p$ are the ramification indices, and $g_q$ is the genus of the quotient. Recall that the stabilisers of all points must be cyclic, so we get a contribution of the form $\lvert G\rvert(r-1)/r=e_p-1$ for each G-orbit (free orbits contributing 0). Solving for g, we see that $g_q = 0$ or else $g \ge \frac{1}{4}\lvert G\rvert$ .

Proof. The assertion (i) follows from Table 1.

(ii) Regard $\mathbb{P}^4$ as the projectivisation of a vector space $V=\mathbb{C}^5$ and consider a lifting $\tilde G\subset {\text{SL}}(V)$ , where $\tilde G$ is quasi-simple. Since G is not simple, $z(\tilde G)$ is not a subgroup of scalar matrices, i.e. there exists a non-trivial decomposition $V= V'\oplus V''$ of $\tilde G$ -modules, where $\dim V'>\dim V''$ . Then $\dim V''\le 2$ and so the action of $\tilde G$ on V ^′′ must be trivial and on V ^′ it is faithful with $\dim V'=3$ or 4. Then $\tilde G$ has an invariant of degree $\le d$ on V ^′. From Table 1, the only possibility is $\tilde G\simeq {\text{SL}}_2(\textbf{F}_7)$ and $d=4$ .

(iii) and (iv) follow from Table 1.

Proof. First, consider the case where $P\in X$ is Gorenstein. The group G faithfully acts on the tangent space $T_{P,X}$ . If $P\in X$ is smooth, then $\dim T_{P,X}=3$ and $G\simeq 3.{\mathfrak{A}}_6$ by Theorem 2·4. If $P\in X$ is singular, then there exists an analytic equivariant embedding $(X,P) \subset (T,0)$ , where $T\simeq \mathbb{C}^4$ and the action on T is linear. Let $\phi(x_1,\dots,x_4)=0$ be the (invariant) equation of X in T. Write $\phi=\sum \phi_d$ , where $\phi_d$ is homogeneous of degree d. By the classification of terminal singularities [ Reference ReidRei87 ], we conclude $\phi_2\neq 0$ . If moreover $G \hookrightarrow {\text{GL}}_4(\mathbb{C})={\text{GL}}(T)$ is irreducible, then the group $G/{\text{z}}(G)$ faithfully acts on $\mathbb{P}(T)=\mathbb{P}^3$ . In this case $\phi_2=0$ defines an invariant quadric $Q\subset \mathbb{P}^3$ which must be smooth. Thus $Q\simeq \mathbb{P}^1\times \mathbb{P}^1$ and then the simple group $G/{\text{z}}(G)$ embedds into ${\text{PGL}}_2$ ; impossible for G in the list (1·4·1). Let $G \hookrightarrow {\text{GL}}_4(\mathbb{C})={\text{GL}}(T)$ be reducible. We have a decomposition $T=T'\oplus T''$ , where T ^′ is irreducible faithful with $\dim T'<4$ . If $\dim T'=3$ , then $G\simeq 3.{\mathfrak{A}}_6$ . Again $3.{\mathfrak{A}}_6$ has no invariants of degree $\le 3$ on T ^′, so $\phi_2=x_4^2$ and $\phi_3=\lambda x_4^3$ . This contradicts the classification of terminal singularities. Hence $\dim T'=2$ and $G\simeq 2.{\mathfrak{A}}_5$ . Again we have a contradiction.

Consider the case where (X, P) is a singularity of index $r>1$ . Let $\pi\,:\, (X^\sharp, P^\sharp)\to (X,P)$ be the index one cover and let $G^\sharp\subset \text{Aut}(X^\sharp, P^\sharp)$ be the natural lifting of G. We have an exact sequence

\begin{equation*}1 \longrightarrow{\boldsymbol{\mu}}_r \longrightarrow G^\sharp \overset{\nu}\longrightarrow G \longrightarrow 1.\end{equation*}

Since G is a quasi-simple group and $\text{Aut}({\boldsymbol{\mu}}_r)$ is abelian, this is a central extension. Let $Z^\sharp\subset G^\sharp$ be the preimage of ${\text{z}}(G)$ . Since ${\text{z}}(G)$ is cyclic, $Z^\sharp$ is an abelian group with two generators and one of these generators is of order 2 or 3. In this situation, the automorphism group $\text{Aut}(Z^\sharp)$ is solvable. Hence $Z^\sharp$ coincides with the center of $G^\sharp$ . Thus either $G^\sharp={\boldsymbol{\mu}}_r\times G$ or $Z^\sharp\simeq {\boldsymbol{\mu}}_6$ and $G/{\text{z}}(G)\simeq {\mathfrak{A}}_6$ or ${\mathfrak{A}}_7$ . In both cases there exists a quasi-simple subgroup $G'\subset G^\sharp$ such that $\nu(G')=G$ . By the above $G'\simeq 3.{\mathfrak{A}}_6$ and $P^\sharp\in X^\sharp$ is smooth. Since the representation of G ^′ on $T_{P^\sharp, X^\sharp}$ is irreducible, $r=2$ .

Proof. The points of the total space of the tangent bundle TX are of the form (x, t) where $x \in X$ and $t \in T_xX$ . For $g \in G$ , $g(x,t)\,:\!=\,(g(x),dg(t))$ defines an action on TX which is linear on the fibers. Thus TX is linearizable, and so is the canonical bundle.

Proof. The lattice $\text{Pic}(X)$ is equipped with a pairing

\begin{equation*}(D_1,D_2)=-K_X\cdot D_1\cdot D_1.\end{equation*}

which is G-invariant and non-degenerate (by the Hodge index theorem). Hence the group G acts on the orthogonal complement $W\,:\!=\,K_X^{\perp}$ of $K_X$ in $\text{Pic}(X)$ . According to [ Reference ProkhorovPro13b ] we have $\text{rk}\, \text{Pic}(X)\le 4$ and so $\text{rk} W\le 3$ . Since $\text{rk}\, \text{Pic}(X)^G=1$ , we have $W^G=0$ . But the groups from the list (1·4·1) have no rational representations of dimension $\le 3$ , a contradiction.

Proof. Let $A=-K_X/2$ and ${\text{d}}(X)\,:\!=\,A^3$ . Consider the possibilities for ${\text{d}}(X)$ case by case. We use the classification of del Pezzo threefolds [ Reference FujitaFuj90 ], [ Reference ShinShi89 ].

Case ${\text{d}}(X)=1$ . Here $\text{Bs}|A|=\{P\}$ and G faithfully acts on $T_{P,X}$ . Hence $G\simeq 3.{\mathfrak{A}}_6$ by Theorem 2·4. Let $\tilde G$ be the universal central extension of G (see, e.g., [ Reference AschbacherAsc00 , section 33]). Then the action of G on X lifts to an action of $\tilde G$ on $H^0(X,tA)$ for any t. Hence $\tilde G$ acts on the graded algebra

\begin{equation*}R(X,A)\,:\!=\,\bigoplus_{t\ge 0} H^0(X,tA).\end{equation*}

In our case R(X, A) is generated by its elements $x_1,x_2,x_2,y,z$ with $\deg x_i=1$ , $\deg y=2$ , $\deg z=3$ and a unique relation of degree 6.

There exists a natural isomorphism $H^0(X,A)\simeq T_{P,X}$ . The subspace

\begin{equation*}S^2 (H^0(X, A))\subset H^0(X,-K_X)\end{equation*}

is invariant. Since $\dim H^0(X,-K_X)=7$ and $\dim S^2 (H^0(X, A))=6$ , the element $y\in H^0(X,-K_X)$ can be taken to be a relative invariant of $\tilde G$ . Similarly, the subspace of the 10-dimensional space $H^0(X,3A)$ generated by $x_1,x_2,x_3,y$ is invariant and of codimension 1. Hence the element $z\in H^0(X,3A)$ can be taken to be a relative invariant of $\tilde G$ . Since the action on $x_1,x_2,x_3$ has no invariants of degree $<6$ and has a unique invariant $\phi_6$ of degree 6, $X\subset \mathbb{P}(1^3,2,3)$ is given by the equation

\begin{equation*}z^2+ y^3 +\phi_6\end{equation*}

and so we are in the situation of Example 1·6(iii).

Case ${\text{d}}(X)=2$ . In this case the map given by the linear system $|A|$ is a finite morphism $\Phi_{|A|}\,:\, X\to \mathbb{P}^3$ of degree 2 branched over a quartic $B\subset \mathbb{P}^3$ . The group G acts non-trivially on $\mathbb{P}^3$ . Therefore, $G/{\text{z}}(G)$ is either ${\text{PSL}}_2(\textbf{F}_7)$ or ${\mathfrak{A}}_6$ . In the case $G/{\text{z}}(G)={\mathfrak{A}}_6$ , the group G has no invariant quartic. Hence $G/{\text{z}}(G)={\text{PSL}}_2(\textbf{F}_7)$ and $G\simeq {\text{SL}}_2(\textbf{F}_7)$ .

Similar to the above considered case R(X, A) is generated by $x_1,\dots,x_4,y$ with $\deg x_i=1$ , $\deg y=2$ with a unique relation of degree 4. We may take y to be a relative invariant for $\tilde G$ . Hence $X\subset \mathbb{P}(1^4,2)$ can be given by the equation

\begin{equation*}y^2 +y\phi_2+\phi_4=0\end{equation*}

Since the action on $x_1,\dots,x_4$ has no invariants of degree $\le 2$ , $\phi_2=0$ . Now we see that ${\text{z}}(G)$ acts trivially on X (cf. Lemma 2·12), a contradiction.

Case ${\text{d}}(X)=3$ . This case does not occur by Lemma 2·6(ii).

Case ${\text{d}}(X)=4$ . In this case X is an intersection of two quadrics $Q_1$ and $Q_2$ in $\mathbb{P}^5$ , which is impossible by Lemma 2·6(iv).

Proof. By Proposition 3·1·2 we may assume that $({-}K_X/2)^3\ge 5$ and by Proposition 3·1·1 we have $\text{rk}\, \text{Pic}(X)=1$ . Hence by [ Reference ProkhorovPro13a , theorem 1·7] and [ CS16 , proposition 7·1·10] we have $({-}K_X/2)^3= 5$ , X is smooth, and $\text{Aut}(X)\simeq {\text{PSL}}_2(\mathbb{C})$ . On the other hand, ${\text{PSL}}_2(\mathbb{C})$ does not contain finite non-solvable groups different from ${\mathfrak{A}}_5$ , a contradiction.

Proof. Assume that $\dim \text{Bs}|{-}K_X|=0$ , then by [ Reference ShinShi89 ] $\text{Bs}|{-}K_X|$ is a single point, say P, and X has at P terminal Gorenstein singularity of type $cA_1$ . By Lemma 2·7 this is impossible.

Thus $\dim \text{Bs}|{-}K_X|>0$ , then by [ Reference ShinShi89 ] $\text{Bs}|{-}K_X|$ is a smooth rational curve C contained in the smooth locus of X. The action of G on C must be trivial and we obtain a contradiction as above.

Proof. (cf. [ Reference ProkhorovPro12 , lemma 5·3]). Assume that the linear system $|{-}K_X|$ defines a morphism $\varphi \,:\, X\to \mathbb{P}^{g+1}$ which is not an embedding. Let $Y = \varphi(X)$ and let $\bar G$ be the image of G in $\text{Aut}(X)$ . Then $\varphi$ is a double cover and $Y \subset \mathbb{P}^{g+1}$ is a subvariety of degree $g - 1$ (see [ Reference IskovskikhIsk80a ] and [ Reference Przhiyalkovskij, Chel’tsov and ShramovPCS05 ]). Hence either $\bar G\simeq G$ or $\bar G$ is the quotient of G by a subgroup of order 2. Let H be the class of a hyperplane section of Y and let $B\subset Y$ be the branch divisor. Then

(3·3·1) \begin{equation}-K_X=\varphi^* H,\quad K_X=\varphi^*\!\left(K_Y+\textstyle \frac12 B\right),\quad K_Y+H+\textstyle \frac12 B=0.\end{equation}

Apply Theorem 2·8·1. The case where Y is a quadric (case 2·8·1(ii)) does not occur by Lemma 2·6(ii) and [ Reference ProkhorovPro12 , lemma 3·6].

If $Y\simeq \mathbb{P}^3$ (the case 2·8·1(i)), then the morphism $\varphi \,:\, X \to \mathbb{P}^3$ is a double cover with branch divisor $B \subset \mathbb{P}^3$ of degree 6 by (3·3·1). The groups ${\mathfrak{A}}_7$ and ${\text{PSp}}_4(\textbf{F}_3)$ have no non-trivial invariant hypersurfaces of degree 6. If $\bar G\simeq {\text{PSL}}_2(\textbf{F}_7)$ , then we get Example 1·7(i). Likewise, if $\bar G\simeq {\mathfrak{A}}_6$ , we get the case 1·6(iv).

If Y is a cone over the Veronese surface (case 2·8·1(iv)), then $Y\simeq \mathbb{P}(1,1,1,2)$ and ${\mathscr{O}}_{\mathbb{P}}(B)={\mathscr{O}}_{\mathbb{P}}(6)$ by (3·3·1). Hence $\iota(X)=2$ and X is a del Pezzo threefold of degree 1. This case was considered in Proposition 3·1·2.

Finally consider the case where Y is either a rational scroll, a cone over a rational scroll, or a cone over a rational normal curve. Then Y is the image of $\hat Y\,:\!=\,\mathbb{P}_{\mathbb{P}^1} (\mathscr{E})$ , where $\mathscr{E}$ is a nef rank n vector bundle on $\mathbb{P}^1$ , under the map defined by the linear system $|{\mathscr{O}}(1)|$ . Thus $\nu\,:\,\hat Y\to Y$ is the “minimal” resolution of singularities which is given by the blowup of the maximal ideal of $\text{Sing}(Y)$ . In particular, $\nu$ is G-equivariant. Then we have the following equivariant commutative diagram

where $\hat X$ is the normalization of the fiber product. The morphism $\eta$ is a crepant contraction and $\hat X$ has at worst canonical Gorenstein singularities [Reference Przhiyalkovskij, Chel’tsov and ShramovPCS05, lemma 3·6]. The group $\bar G$ trivially acts on $\mathbb{P}^1$ , so it non-trivially acts on each fiber $F\simeq \mathbb{P}^2$ of $\pi$ . Write $\mathscr{E}={\mathscr{O}}_{\mathbb{P}^1}(d_1)\oplus {\mathscr{O}}_{\mathbb{P}^1}(d_2)\oplus {\mathscr{O}}_{\mathbb{P}^1}(d_3)$ , where $d_3\ge d_2\ge d_1\ge 0$ . If $d_1<d_2$ , then the surjection $\mathscr{E} \to {\mathscr{O}}_{\mathbb{P}^1}(d_2)\oplus {\mathscr{O}}_{\mathbb{P}^1}(d_3)$ defines an invariant subscroll $\hat Y_2\subset \hat Y$ such that $Y_2\cap F$ is a line. According to Theorem 2·4 this is impossible. Thus we may assume that $d_1=d_2$ and, similarly, $d_2=d_3$ . So, $\hat Y\simeq \mathbb{P}^2\times \mathbb{P}^1\simeq Y$ . Note that B intersects F along a quartic curve which must be invariant by (3·3·1). Then the only possibility is that B is a divisor of bidegree (4, 0) or a (reducible) divisor of bidegree (4,2), where $\bar G\simeq {\text{PSL}}_2(\textbf{F}_7)$ . In the former case X is the product of $\mathbb{P}^1$ and del Pezzo surface of degree 2. In the latter case X is described in [ Reference KrylovKry16 , example 1·8] as the threefold $X_1$ . In both cases the group G splits by Lemma 2·12, a contradiction.

Remark 3·3·2. Assume that $-K_X$ is very ample. Then, by Proposition 2·11, our group G acts faithfully on the space $H^0(X,-K_X)^\vee$ so that the induced action on its projectivization $\mathbb{P}(H^0(X,-K_X)^\vee)=\mathbb{P}^{g+1}$ is also faithful. This implies that the representation $H^0(X,-K_X)^\vee$ of G is reducible.

Proof. By our assumption $g\ge 3$ . If $g=3$ , then $X=X_4\subset \mathbb{P}^4$ is a quartic. Inspecting the list (1·4·1) one can see that the only possibility is $G\simeq {\text{SL}}_2(\textbf{F}_{7})$ , which implies that (X, G) is as in Example 1·7(i).

Now assume that $g>3$ . Since $X=X_{2g-2}\subset \mathbb{P}^{g+1}$ is projectively normal [ Reference IskovskikhIsk80a ], the restriction map

\begin{equation*}H^0(\mathbb{P}^{g+1},{\mathscr{O}}_{\mathbb{P}^{g+1}}(2)) \longrightarrow H^0(X,{\mathscr{O}}_X(2))\end{equation*}

is surjective. This allows us to compute that the number of linear independent quadrics passing through X is equal to

\begin{equation*}\textstyle \frac12(g - 2)(g - 3)>0.\end{equation*}

Let $Y\subset \mathbb{P}^{g+1}$ be the intersection of all quadrics containing X. It is known that Y is a reduced irreducible variety of minimal degree (see [ Reference IskovskikhIsk80a ] and [ Reference Przhiyalkovskij, Chel’tsov and ShramovPCS05 ]). Thus Y is described by Theorem 2·8·1.

If $g=4$ , then Y is a (unique) quadric passing through X and X is cut out on Y by a cubic, say Z. We may assume that Z is G-invariant. Thus our group G has invariants of degrees 2 and 3. Hence, ${\text{z}}(G)$ is of order 6 and so $G\supset 6.{\mathfrak{A}}_6$ . But then G has no faithful reducible representations of dimension 6 by Table 1.

Thus it remains to consider the case where Y is either a rational scroll, a cone over a rational scroll, or a cone over a rational normal curve. Arguing as in the proof of Lemma 3·3 and using [PCSO5, lemma 4·7], we conclude that there exists a G-equivariant crepant extraction $\eta\,:\,\hat X\to X$ and a degree 3 del Pezzo fibration $\hat X\to \mathbb{P}^1$ . This gives us a contradiction (see Theorem 2·5).

Proof. Let c be the log canonical threshold of (X, S), that is, the pair (X, cS) is maximally lc. Then $c \le 1$ and $-(K_X + cS)$ is nef.

Consider the case $c<1$ . Then $-(K_X + cS)$ is ample. Let $\Lambda\subset X$ be the locus of lc singularities of (X, cS). By Shokurov’s connectedness principle (see [ Reference ShokurovSho93 ] and [ Reference KollárKol92 , chapter 17]), $\Lambda$ is connected (and clearly G-invariant). If $\dim \Lambda=0$ , then $\Lambda$ must be an invariant point. Suppose $\dim \Lambda=1$ . If there exists a zero-dimensional center of lc singularities, then replacing cS with small invariant perturbation $(c-\epsilon)S+\Delta$ we get a zero-dimensional locus of lc singularities and may argue as above (see [ Reference ProkhorovPro12 , claim 4·7·1]). Otherwise $\Lambda$ must be a minimal center of lc singularities and by Kawamata subadjunction theorem [ Reference KawamataKaw98 , theorem 1] $\Lambda$ is a smooth rational curve, because $-(K_X + cS)$ is ample. Since G cannot act non-trivially on $\mathbb{P}^1$ , the action of G on $\Lambda$ is trivial.

Now consider the case $c=1$ and S is irreducible. Then (X, S) is lc. Let $\nu \,:\, S' \to S$ be the normalisation. Write

\begin{equation*}0 \sim \nu^*\! \left(K_X + S\right)|_ S = K_ {S'} + D',\end{equation*}

where D ^′ is the different, an effective integral Weil divisor on S ^′ such that the pair (S ^′, D ^′) is lc (see [ Reference ShokurovSho93 , section 3], [ Reference KollárKol92 , chapter 16], and [ Reference KawakitaKaw07 ]). The group G acts naturally on S ^′ and $\nu$ is G-equivariant. Now consider the minimal resolution $\mu \,:\, \tilde S \to S'$ and let $\tilde D$ be a uniquely defined) divisor such that

\begin{equation*}K_{\tilde S}+\tilde D=\mu^* \!\left(K_{S'}+D'\right)\sim 0,\qquad \mu_*\tilde D=D'.\end{equation*}

is usually called the log crepant pull-back of D ^′. Here $\tilde D$ is again an effective reduced divisor. Run G-equivariant MMP on $\tilde S$ . Clearly, the whole $\tilde D$ cannot be contracted. We get a model $(S_{\min}, D_{\min})$ such that $(S_{\min}, D_{\min})$ is lc, $(K_{S_{\min}}+D_{\min})\sim 0$ , and $D_{\min}\neq 0$ . Assume that $S_{\min}$ has an equivariant conic bundle structure $\pi\,:\, S_{\min}\to B$ . Then $D_{\min}$ has one or two horizontal components which must be G-invariant. By adjunction any horizontal component of $D_{\min}$ is either rational or elliptic curve. Such a curve does not admit a non-trivial action of G, so the action of G on the corresponding component $\tilde D_1\subset \tilde D$ and $\nu(\mu (\tilde D_1))$ must be trivial. Similarly, if $S_{\min}$ is a del Pezzo surface with $\text{rk}\, \text{Pic}(S_{\min})^G=1$ , then $K_{S_{\min}}^2=2$ or 9 by Theorem 2·5 and so $\tilde D$ has at most 3 components. Arguing as above we get that the action of G on some component $\tilde D_1\subset \tilde D$ is trivial.

Proof. Denote $(Y\ni Q)\,:\!=\, (X\ni P)/A$ . If A acts freely in codimension one, then the assertion is a consequence of [ Reference Kollár and Shepherd–BarronKSB88 , proposition 6·12]. Let $\text{Fix}(A,X)$ contain a divisor, say D. We have an A-equivariant embedding $(X\ni P)\subset (\mathbb{C}^4\ni 0)$ and we may assume that the action on $\mathbb{C}^4$ is diagonalizable. If this action is of type $\frac 12 (1,0,0,0)$ , then $\mathbb{C}^4/A$ is smooth and so Y is Gorenstein. Since the quotient singularities are always rational [ KM98 , proposition 5·13], this implies that $(Y\ni Q)$ is canonical [ KM98 , corollary 5·24]. Thus we may assume that the action is of type $\frac 12 (1,1,0,0)$ and the equation of X is of the form

\begin{equation*}\phi=x_1\phi_1(x_1,\dots,x_4)+ x_2\phi_2(x_1,\dots,x_4)=0.\end{equation*}

But then the fixed point locus is not a $\mathbb{Q}$ -Cartier divisor.

Proof. Let $\pi\,:\, (X^\sharp\ni P^\sharp)\to (X\ni P)$ be the index-one cover, so that $(X\ni P)=(X^\sharp\ni P^\sharp)/{\boldsymbol{\mu}}_2$ where the action of ${\boldsymbol{\mu}}_2$ is free outside $P^\sharp$ [KM98, definition 5·19]. By our assumption $(X^\sharp\ni P^\sharp)\simeq (\mathbb{C}^3,0)$ and the action of A is of type $\frac 12 (1,1,1)$ . The action of $G_P$ lifts to an action of $G_P^\sharp$ on $(X^\sharp\ni P^\sharp)$ , where $G_P^\sharp$ is an extension of $G_P$ by ${\boldsymbol{\mu}}_2$ . Let $A^\sharp\subset G_P^\sharp$ be the preimage of A (a group of order 4).

If $A^\sharp\simeq {\boldsymbol{\mu}}_2\times {\boldsymbol{\mu}}_2$ , then the elements of this group act as follows: $\frac 12 (1,1,1)$ , $\frac 12 (1,1,0)$ , $\frac 12 (0,0,1)$ . It is easy to see that in this case the quotient is canonical Gorenstein.

Assume that the extension $A^\sharp\simeq {\boldsymbol{\mu}}_4$ . Then the action on $\mathbb{C}^3$ is of type $\frac 14 (1,1,1)$ or $\frac 14 (1,1,-1)$ . By our assumption the former case does not occur. In the latter case the quotient is terminal.

Proof. Since $-K_X$ is Cartier, the log pair $(X,2\lambda{\mathscr{M}})$ does not have klt singularities by [ CS14 , lemma 2·2]. Choose $\mu\leqslant 2\lambda$ such that $(X,\mu{\mathscr{M}})$ is strictly lc. Let Z be a minimal center of lc singularities of the log pair $(X,\mu{\mathscr{M}})$ , see [ Reference KawamataKaw97 , Reference KawamataKaw98 ] for a precise definition. Then $\theta(Z)$ is also a minimal center of lc singularities of this log pair for every $\theta\in G$ . Moreover, we have

\begin{equation*}Z\cap \theta(Z)\ne \varnothing\iff Z=\theta(Z)\end{equation*}

by [ Reference KawamataKaw97 , proposition 1·5].

Since ${\mathscr{M}}$ does not have fixed components, the center Z is either a curve or a point. Observe that

\begin{equation*}\mu{\mathscr{M}}\mathbin{\sim_{\scriptscriptstyle{\mathbb{Q}}}} \frac{\mu}{\lambda} nH,\end{equation*}

where ${\mu}/{\lambda}\leqslant 2$ . It would be easier to work with $(X,\mu{\mathscr{M}})$ if it did not have centers of lc singularities that are different from $\theta(Z)$ for $\theta\in G$ . This is possible to achieve if we replace the boundary $\mu{\mathscr{M}}$ by (a slightly more complicated) effective boundary $B_X$ such that

\begin{equation*}B_X\mathbin{\sim_{\scriptscriptstyle{\mathbb{Q}}}} \left(\frac{\mu}{\lambda}n+\epsilon\right)H\end{equation*}

for some positive rational number $\epsilon$ that can be chosen arbitrary small. This is known as the Kawamata–Shokurov trick or the perturbation trick (see [ CS16 , lemma 2·4·10] and the proofs of [ Reference KawamataKaw97 , theorem 1·10] and [ Reference KawamataKaw98 , theorem 1]). By construction, we may assume that

(4·4·1) \begin{equation}\frac{\mu}{\lambda}n+\epsilon\leqslant 2n+\epsilon<2n+1.\end{equation}

Note that the coefficients of $B_X$ depend on $\epsilon$ . But we can chose $\epsilon$ as small as we wish, so that the number ${\mu}n/{\lambda}+\epsilon$ can be as close to 2n as we need.

Let $\Sigma$ be the union of all log canonical centers $\theta(Z)$ for $\theta\in G$ . Then $\Sigma$ is either a G-orbit or a disjoint union of irreducible isomorphic curves, which are transitively permuted by G. In both cases, we have an exact sequence of vector spaces

(4·4·2) \begin{align}& 0\longrightarrow H^0\!\left({\mathscr{O}}_X\big((n+1)H\big)\otimes{\mathscr{I}}_{\Sigma}\right)\longrightarrow H^0\!\left({\mathscr{O}}_X\big((n+1)H\big)\right)\\ &\qquad \qquad \qquad \qquad \qquad \longrightarrow H^0\!\left({\mathscr{O}}_{\Sigma}\otimes{\mathscr{O}}_X\big((n+1)H\big)\right)\longrightarrow H^1\!\left({\mathscr{O}}_X\big((n+1)H\big)\otimes{\mathscr{I}}_{\Sigma}\right),\end{align}

where ${\mathscr{I}}_{\Sigma}$ is an ideal sheaf of the locus $\Sigma$ , and ${\mathscr{O}}_{\Sigma}$ is its structure sheaf. Note that ${\mathscr{I}}_{\Sigma}$ is the multiplier ideal sheaf of the log pair $(X,B_X)$ . Since

\begin{equation*}K_X+B_X\mathbin{\sim_{\scriptscriptstyle{\mathbb{Q}}}} \left(\left(\frac{\mu}{\lambda}-1\right)n+\epsilon\right)H,\end{equation*}

we can apply Nadel’s vanishing (see [ Reference LazarsfeldLaz04 , theorem 9·4·17]) to deduce that

\begin{equation*}h^1\!\left({\mathscr{O}}_X((n+1)H)\otimes{\mathscr{I}}_{\Sigma}\right)=0.\end{equation*}

In particular, if Z is a point, it follows from (4·4·2) that

\begin{equation*}|\Sigma|=h^0\!\left({\mathscr{O}}_X\big((n+1)H\big)\right)-h^0\!\left({\mathscr{O}}_X\big((n+1)H\big)\otimes{\mathscr{I}}_{\Sigma}\right).\end{equation*}

To complete the proof of the proposition, we may assume that $\Sigma$ is disjoint union of irreducible isomorphic curves $C_1=Z,C_2,\ldots,C_r$ , which are transitively permuted by G. In particular, if $r=1$ , then $\Sigma=C_1=Z$ is a G-invariant irreducible curve in X.

Let $d=H\cdot C_i$ . Then $rd\leqslant H^3n^2$ . This immediately follows from Corti’s [ Reference CortiCor00 , theorem 3·1]. Namely, observe that $(X,\mu{\mathscr{M}})$ is not klt at general points of every curve $C_i$ . Let M and $M^{\prime}$ be general surfaces in ${\mathscr{M}}$ . Then, applying [ Reference CortiCor00 , theorem 3·1] to the log pair $(X,\mu{\mathscr{M}})$ at general point of the curve $C_i$ , we obtain

\begin{equation*}\text{mult}_{C_i}\!\left(M\cdot M^{\prime}\right)\geqslant\frac{4}{\mu^2}.\end{equation*}

Then

\begin{equation*}\frac{n^2}{\lambda^2}H^3=H\cdot M\cdot M^{\prime}\geqslant\sum_{i=1}^rH\cdot C_i\text{mult}_{C_i}\!\left(M\cdot M^{\prime}\right)\geqslant rd\frac{4}{\mu^2},\end{equation*}

so that $rd\leqslant H^3n^2 {\mu^2}/{4\lambda^2}\leqslant H^3n^2$ as claimed.

By Kawamata’s subadjunction [ Reference KawamataKaw98 , theorem 1], each curve $C_i$ is smooth. Let g be its genus. Moreover, for every ample $\mathbb{Q}$ -Cartier $\mathbb{Q}$ -divisor A on X, it follows from [ Reference KawamataKaw98 , theorem 1] that

\begin{equation*}\left(K_{X}+B_{X}+A\right)\Big\vert_{C_i}\mathbin{\sim_{\scriptscriptstyle{\mathbb{Q}}}} K_{C_i}+B_{C_i}\end{equation*}

for some effective $\mathbb{Q}$ -divisor $B_{C_i}$ on the curve $C_i$ . Computing the degrees of the left hand side and the right-hand side in this $\mathbb{Q}$ -linear equivalence, we see that $2g-2\leqslant nd$ . In particular, the divisor $(n+1)H\vert_{C_i}$ is non-special on $C_i$ , so that

\begin{equation*}h^0\!\left({\mathscr{O}}_{C_i}\big((n+1)H\vert_{C_i}\big)\right)=(n+1)d-g+1\end{equation*}

by the Riemann–Roch formula. Now using (4·4·2), we get

\begin{equation*}r\!\left((n+1)d-g+1\right)=h^0\!\left({\mathscr{O}}_X\big((n+1)H\big)\right)-h^0\!\left({\mathscr{O}}_X\big((n+1)H\big)\otimes{\mathscr{I}}_{\Sigma}\right),\end{equation*}

which complete the proof of the proposition.

Remark 4·5. In the notations and assumptions of Proposition 4·4, there exists a central extension $\widetilde{G}$ of the group G such that the line bundle H is $\widetilde{G}$ -linearisable. Recall the Riemann–Roch theorem for a divisor D on a smooth threefold X:

\begin{equation*} \chi({\mathscr{O}}_X(D)) = \frac 16 D^3-\frac 14 D^2\cdot K_X +\frac1 {12} D\cdot K_X^2 +\frac1 {12} D\cdot c_2(X)+ \chi({\mathscr{O}}_X).\end{equation*}

Thus, the vector space $H^0({\mathscr{O}}_X((n+1)H))$ is a representation of the group $\widetilde{G}$ of dimension

\begin{equation*}h^0({\mathscr{O}}_X((n+1)H))=\frac{(n+1)(2n+1)(3n+2)}{12}H^3+\frac{2n+2}{n}+1.\end{equation*}

Then the exact sequence (4·4·2) in the proof of Proposition 4·4 is an exact sequence of $\widetilde{G}$ -representations.

Proof of Theorem 4·3. Suppose that there exists a non-biregular G-birational map

such that V is a Fano variety with at most canonical singularities. Applying [ CS16 , theorem 3·2·1], we see that there exists a G-invariant linear system ${\mathscr{M}}$ on the threefold X such that ${\mathscr{M}}$ does not have fixed components, and the singularities of the log pair $(X,\lambda{\mathscr{M}})$ are not terminal, where $\lambda$ is a positive rational number such that $\lambda{\mathscr{M}}\mathbin{\sim_{\scriptscriptstyle{\mathbb{Q}}}} -K_{X}$ . We will obtain a contradiction using Proposition 4·4 and Remark 4·5.

Let $m=\textstyle{(n+1)(n+2)}/{12}H^3+{2}/{n}$ . Note that the divisor H is very ample in each of our cases, and $h^0({\mathscr{O}}_X(H))=m+1$ . Thus, we may identify X with its image in $\mathbb{P}^m$ . Then G is a subgroup in ${\text{PGL}}_{m+1}(\mathbb{C})$ . Let $\widetilde{G}$ be a finite subgroup in ${\text{GL}}_{m+1}(\mathbb{C})$ that maps surjectively to G by the natural projection. We may assume that $\widetilde{G}$ is the smallest group with this property.

The vector space $H^0({\mathscr{O}}_X(H))$ is an irreducible representation of the group $\widetilde{G}$ . From Table 1, this is immediate in all cases except 4·3(vi). In this remaining case, $G\simeq{\text{PSL}}_2(\textbf{F}_{11})$ and X the unique smooth Fano threefold of Picard rank 1 and genus 8 that admits a faithful action of the group ${\text{PSL}}_2(\textbf{F}_{11})$ (see [ Reference ProkhorovPro12 , example 2·9]). The space $H^0({\mathscr{O}}_X(H))$ is isomorphic to the representation $\bigwedge^2 V$ , where V is a 5-dimensional faithful representation of G. Recall that

\begin{equation*}\chi_{\bigwedge^2V}(g) = \frac{1}{2}\left( \chi_V(g)^2 - \chi_V\!\left(g^2\right) \right),\end{equation*}

where $g \in G$ and $\chi_{\bigwedge^2V}$ (resp. $\chi_V$ ) is the character of $\bigwedge^2V$ (resp. V). Evaluating g at any element of order 2 in ${\text{PSL}}_2(\textbf{F}_{11})$ , we conclude that $\bigwedge^2V$ is irreducible from the character table [ Reference Conway, Curtis, Norton, Parker and WilsonCCN+85 ].

Since $H^0({\mathscr{O}}_X(H))$ is an irreducible representation of the group $\widetilde{G}$ , the threefold X does not contain G-invariant subvarieties contained in a proper linear subspace of $\mathbb{P}^m$ .

Applying Proposition 4·4, we see that either X contains a G-orbit $\Sigma$ such that

\begin{equation*}m<|\Sigma|=h^0\!\left({\mathscr{O}}_X\big((n+1)H\big)\right)-h^0\!\left({\mathscr{O}}_X\big((n+1)H\big)\otimes{\mathscr{I}}_{\Sigma}\right),\end{equation*}

where ${\mathscr{I}}_{\Sigma}$ is the ideal sheaf of $\Sigma$ , or there exists a G-irreducible reduced curve C that is a disjoint union of smooth irreducible curves $C_1,\ldots,C_r$ of genus g and degree $d=H\cdot C_i$ such that $rd\leqslant H^3n^2$ , $2g-2\leqslant nd$ and

\begin{equation*}r\!\left((n+1)d-g+1\right)=h^0\!\left({\mathscr{O}}_X\big((n+1)H\big)\right)-h^0\!\left({\mathscr{O}}_X\big((n+1)H\big)\otimes{\mathscr{I}}_{C}\right),\end{equation*}

where ${\mathscr{I}}_{C}$ is the ideal sheaf of the curve C. In the former case, by Remark 4·5, the number $|\Sigma|$ is the dimension of some $\widetilde{G}$ -subrepresentation in $H^0({\mathscr{O}}_X((n+1)H))$ . Likewise, in the latter case, the number $r((n+1)d-g+1)$ is also the dimension of some $\widetilde{G}$ -subrepresentation in $H^0({\mathscr{O}}_X((n+1)H))$ . Moreover, if $r=1$ , then the natural homomorphism $G\to\text{Aut}(C)$ is injective, because C is not contained in a hyperplane in this case. Thus, if $r=1$ , then

(4·5·1) \begin{equation}84(g-1)\geqslant \lvert G\rvert\end{equation}

by Hurwitz’s automorphisms theorem.

Case 4·3(i). Here $G\simeq{\mathfrak{A}}_7$ and X is the unique smooth intersection of a quadric and a cubic in $\mathbb{P}^5$ that admits a faithful action of the group ${\mathfrak{A}}_7$ . We have $n=1$ , $H^3=6$ , $m=5$ , $\widetilde{G}\simeq{\mathfrak{A}}_7$ and $h^0({\mathscr{O}}_X(2H))=20$ . Note that $H^0({\mathscr{O}}_X(H))$ is the irreducible ${\mathfrak{A}}_7$ -representation obtained as the quotient of the standard permutation representation by the trivial representation.

Suppose that there exists a G-orbit $\Sigma$ in X such that $5<|\Sigma|\leqslant 20$ . Using Table 2, we see that $|\Sigma|$ is either 7 or 15. In the case $|\Sigma|=15$ , a stabilizer of a point in $\Sigma$ is isomorphic to ${\text{PSL}}_2(\textbf{F}_7)$ . The restriction of the representation $H^0({\mathscr{O}}_X(H))$ to ${\text{PSL}}_2(\textbf{F}_7)$ is the quotient of a transitive permutation representation so it has no trivial subrepresentations. Since ${\text{PSL}}_2(\textbf{F}_7)$ is simple, it therefore has no one-dimensional subrepresentations. The action cannot fix a point so this case is impossible. In the case $|\Sigma|=7$ , the stabiliser is isomorphic to ${\mathfrak{A}}_6$ . In this case, there is a fixed point in the ambient space $\mathbb{P}^5$ corresponding to the fixed point of the natural permutation action. One checks that this point does not lie on X by explicitly checking the defining equations, which are just elementary symmetric functions.

Table 2. Conjugacy classes of maximal subgroups with indices.

Thus, the threefold X contains a G-irreducible reduced curve C that is a disjoint union of smooth irreducible curves $C_1,\ldots,C_r$ of genus g and degree d such that $rd\leqslant 6$ and $2g-2\leqslant d$ . As above, this shows that $r=1$ , so that $d\leqslant 6$ and $g\leqslant 4$ , which contradicts (4·5·1).

Case 4·3(ii). Here $G\simeq{\mathfrak{A}}_7$ and $X=\mathbb{P}^3$ . We have $n=4$ , $H^3=1$ , $m=3$ , $\widetilde{G}\simeq 2.{\mathfrak{A}}_7$ and $h^0({\mathscr{O}}_X(5H))=56$ . Note that $H^0({\mathscr{O}}_X(H))$ is an irreducible four-dimensional representation of the group $\widetilde{G}$ .

Suppose that $\mathbb{P}^3$ contains a G-orbit $\Sigma$ such that $3<|\Sigma|\leqslant 56$ . Going through the list of subgroups in G of index $\leqslant 56$ , we see that

\begin{equation*}|\Sigma|\in\big\{7,15,21,35,42\big\}.\end{equation*}

Let $G_P$ be the stabiliser of a point $P\in\Sigma$ . Using Table 2, we conclude $G_P$ is isomorphic to one of the following groups: ${\mathfrak{A}}_6$ , ${\text{PSL}}_2(\textbf{F}_7)$ , $\mathfrak{S}_5$ , $({\mathfrak{A}}_4\times{\boldsymbol{\mu}}_3)\rtimes{\boldsymbol{\mu}}_2$ , or ${\mathfrak{A}}_5$ . Let $\widetilde{G}_P$ be a subgroup in $\widetilde{G}$ that is mapped to $G_P$ . We claim that the restriction of the representation $V=H^0({\mathscr{O}}_X(H))$ to $\widetilde{G}_P$ does not contain one-dimensional subrepresentations, which contradicts the fact that $G_P$ fixes the point $P\in\mathbb{P}^3$ . Let $\chi$ be the 4-dimensional representation of $\widetilde{G}$ ; we will consider restricted characters of $\chi$ (see [ Reference Conway, Curtis, Norton, Parker and WilsonCCN+85 ]). We have $\chi(g)=-(1\pm\sqrt{-7})/2$ when $g \in \widetilde{G}$ has order 7; if $G_P \simeq {\text{PSL}}_2(\textbf{F}_7)$ then the only possibility is that $\widetilde{G}_P \simeq {\text{PSL}}_2(\textbf{F}_7)$ and $V|_{\widetilde{G}_P}$ is irreducible. We have $\chi(g)=-1$ when g has order 5, which means that $V|_{\widetilde{G}_P}$ is irreducible if $G_P \cong {\mathfrak{A}}_5$ (and a fortiori $\mathfrak{S}_5$ and ${\mathfrak{A}}_6$ ). It remains to consider $G_P \cong ({\mathfrak{A}}_4\times{\boldsymbol{\mu}}_3)\rtimes{\boldsymbol{\mu}}_2$ , which contains a 3-Sylow subgroup $H \subseteq G$ . For some elements g of order 3, we have $\chi(g)=-2$ meaning that $\chi|_H$ does not have any trivial subrepresentations. Since the group $G_P$ has no non-trivial maps to ${\boldsymbol{\mu}}_3$ , we conclude that there are no one-dimensional subrepresentations. With the claim proved, we see this case is impossible.

Thus, there is a G-irreducible reduced curve C in $\mathbb{P}^3$ with the following properties: C is union of smooth irreducible curves $C_1,\ldots,C_r$ of genus g and degree d, $rd\leqslant 16$ , $2g-2\leqslant 4d$ and

\begin{equation*}r\big(5d-g+1\big)\leqslant 56.\end{equation*}

As above, we see that $r\in\{1,7,15\}$ . If $r=15$ , then $d=1$ , so that $g=0$ and

\begin{equation*}90=r\big(5d-g+1\big)\leqslant 56,\end{equation*}

which is absurd. Likewise, if $r=7$ , then $d\leqslant 2$ , so that $g=0$ and

\begin{equation*}35d+7=7\big(5d+1\big)=r\big(5d-g+1\big)\leqslant 56,\end{equation*}

so that $d=1$ . In this case, the stabilizer of the line $C_1$ is isomorphic to ${\mathfrak{A}}_6$ , which is impossible, since the restriction of the representation $H^0({\mathscr{O}}_X(H))$ to the subgroup $2.{\mathfrak{A}}_6$ is irreducible. Thus, we see that $r=1$ , so that C is irreducible. Using (4·5·1), we see that $g\in\{31,32,33\}$ . By Lemma 2·2, one of the expressions

\begin{equation*}\frac{169}{2^2\cdot 3 \cdot 7}, \quad\frac{5071}{2^3\cdot 3^2 \cdot 5 \cdot 7}, \quad\frac{634}{3^2 \cdot 5 \cdot 7}\end{equation*}

is a non-negative integer combination of expressions of the form $1-{1}/{r}$ for

\begin{equation*}r=1,2,3,4,5,6,7,8,10,12,14.\end{equation*}

Since ${1}/{2}+ {2}/{3}+ {6}/{7}$ is larger than these expressions, this is impossible.

Case 4·3(iii). Here $G\simeq{\text{PSp}}_4(\textbf{F}_3)$ and $X=\mathbb{P}^3$ . We have $n=4$ , $H^3=1$ , $m=3$ , $\widetilde{G}\simeq {\text{Sp}}_4(\textbf{F}_3)$ and $h^0({\mathscr{O}}_X(5H))=56$ . We will see that $H^0({\mathscr{O}}_X(5H))$ is a direct sum of irreducible representations of $\widetilde{G}$ of dimensions 20 and 36. Indeed, $h^0({\mathscr{O}}_X(H))$ is an irreducible 4-dimensional representation V of $\widetilde{G}$ with character $\chi$ . Note that every summand of $S^5V$ must be a faithful representation of $\widetilde{G}$ since 5 is coprime to 2. Via the Newton identities, we have the standard formula for the 5th symmetric power

\begin{align*}S^5\chi(g) & =\frac{1}{120}\Big[\chi(g)^5 + 10\chi\!\left(g^2\right)\chi(g)^3 +15\chi\!\left(g^2\right)^2\chi(g)\\& \quad + 20\chi\!\left(g^3\right)\chi(g)^2 + 20\chi\!\left(g^3\right)\chi\!\left(g^2\right) + 30\chi(g^4)\chi(g)+ 24\chi\left(g^5\right)\Big],\end{align*}

where g is an element of $\widetilde{G}$ . From the character table [ Reference Conway, Curtis, Norton, Parker and WilsonCCN+85 ], we see that $\chi(g)=-1$ for any element g of order 5. We compute that $S^5\chi(g)=1$ and conclude from the character table that the only possibility is a sum of characters of degree 20 and 36 as desired.

Suppose that $\mathbb{P}^3$ contains a G-orbit $\Sigma$ such that $|\Sigma|$ is the dimension of some $\widetilde{G}$ -subrepresentation in $H^0({\mathscr{O}}_X(5H))$ . Then

\begin{equation*}|\Sigma|\in\{20,36,56\}.\end{equation*}

Using Table 3, we see that $|\Sigma|=36$ . Let $G_P$ be the stabiliser of a point $P\in\Sigma$ . Then $G_P\simeq\mathfrak{S}_6$ . The group G contains one such subgroup up to conjugation. Let $\widetilde{G}_P$ be a subgroup in $\widetilde{G}$ that is mapped to $G_P$ . Then $\widetilde{G}_P\simeq2.\mathfrak{S}_6$ , and the restriction of the representation $H^0({\mathscr{O}}_X(H))$ to $\widetilde{G}_P$ does not contain one-dimensional subrepresentations. This contradicts the fact that $G_P$ fixes the point $P\in\mathbb{P}^3$ .

Thus, there is a G-irreducible reduced curve C in $\mathbb{P}^3$ that is a union of smooth irreducible curves $C_1,\ldots,C_r$ of genus g and degree d such that $rd\leqslant 16$ , $2g-2\leqslant 4d$ and

\begin{equation*}r\big(5d-g+1\big)\leqslant 56.\end{equation*}

Arguing as above, we see that $r=1$ , so that $d\leqslant 16$ and $g\leqslant 33$ , which is impossible by (4·5·1).

Case 4·3(iv). Here $G\simeq{\text{PSp}}_4(\textbf{F}_3)$ and X is the Burkhardt quartic in $\mathbb{P}^4$ . We have $n=1$ , $H^3=4$ , $m=4$ , $\widetilde{G}\simeq G$ and $h^0({\mathscr{O}}_X(2H))=15$ . Note that $\mathbb{P}^3$ does not contain a G-orbit $\Sigma$ such that $4<|\Sigma|\leqslant 15$ , because G does not contain subgroups of such index. Thus, there is a G-irreducible reduced curve C in $\mathbb{P}^3$ that is union of smooth irreducible curves $C_1,\ldots,C_r$ of genus g and degree d such that $rd\leqslant 4$ and $2g-2\leqslant d$ . Since there are no subgroups of index 2, 3, or 4, we have $r=1$ . Thus $d\leqslant 4$ and $g\leqslant 3$ , which contradicts (4·5·1).

Case 4·3(v). Here $G\simeq{\text{PSL}}_2(\textbf{F}_{11})$ and X is the Klein cubic threefold in $\mathbb{P}^4$ . We have $n=2$ , $H^3=3$ , $m=4$ , $\widetilde{G}\simeq G$ and $h^0({\mathscr{O}}_X(3H))=34$ . If $\chi$ is the character of the representation $V=h^0({\mathscr{O}}_X(3H))$ of G, then the character of the third symmetric power is given by

\begin{equation*}S^3\chi(g) =\frac{1}{6}\Big[\chi(g)^3 + 3\chi\!\left(g^2\right)\chi(g) + 2\chi\!\left(g^3\right) \Big]\end{equation*}

for $g \in G$ . The trivial character occurs exactly once in $S^3V$ by Table 1. From the character table [ Reference Conway, Curtis, Norton, Parker and WilsonCCN+85 ], for any element $g \in G$ of order 3 we have $\chi(g)=-1$ and thus $S^3\chi(g)=2$ . Note that all irreducible characters $\rho$ satisfy $\rho(g) \in \mathbb{R}$ , but only the trivial and 10-dimensional ones have $\rho(g) > 0$ . Since there is only one trivial subrepresentation, this forces the existence of at least one 10-dimensional irreducible subrepresentation. The possible irreducible characters have degrees 1,5,10,11,12, thus $1+10+12+12$ is the only possibility for $S^3V$ . For any $h \in G$ of order 5, $\chi(h)=0$ and $S^3\chi(h)=0$ while $\rho(h) \ne 0$ for the 12-dimensional irreducible representations. We conclude that the vector space $H^0({\mathscr{O}}_X(3H))$ splits as a sum of two non-isomorphic twelve-dimensional representations, and one ten-dimensional representation.

Suppose that X contains a G-orbit $\Sigma$ such that $|\Sigma|$ is the dimension of some G-subrepresentation in $H^0({\mathscr{O}}_X(3H))$ . Going through the list of subgroups in G of index $\leqslant 34$ , we see that $|\Sigma|=12$ . Let $G_P$ be the stabilizer of a point $P\in\Sigma$ . Then $G_P\simeq{\boldsymbol{\mu}}_{11}\rtimes{\boldsymbol{\mu}}_5$ , and the restriction of the representation $H^0({\mathscr{O}}_X(H))$ to $G_P$ is irreducible. This contradicts to the fact that $G_P$ fixes a point in $\mathbb{P}^4$ .

Thus, there is a G-irreducible reduced curve C in $\mathbb{P}^4$ that is union of smooth irreducible curves $C_1,\ldots,C_r$ of genus g and degree d such that $rd\leqslant 12$ , $2g-2\leqslant 2d$ and $r\big(3d-g+1\big)$ $\leqslant 34$ . Going through the list of subgroups in G of index $\leqslant 12$ , we see that $r\in\{1,11,12\}$ . If $r=12$ or $r=11$ , then $d=1$ , so that $g=0$ , which gives

\begin{equation*}11\times 4\leqslant 4r=r\big(3d+1\big)=r\big(3d-g+1\big)\leqslant 34,\end{equation*}

which is absurd. Thus, we have $r=1$ , so that C is irreducible. Then $d\leqslant 12$ and $g\leqslant 13$ . Using (4·5·1), we see that $g\geqslant 9$ , so that $g\in\{9,10,11,12,13\}$ . Since $\lvert G\rvert=25920=2^6\cdot 3^4 \cdot 5$ and $2g-2$ is never divisible by $2^5$ or $3^3$ , the expression on the left hand side from Lemma 2·2 has $2^2 \cdot 3^2$ in the denominator when written in lowest terms. However, it is a non-negative integer combination of expressions of the form $1-\frac{1}{r}$ for $r=1,2,3,4,5,6,9,12$ . Thus, this case is impossible.

Case 4·3(vi). Here $G\simeq{\text{PSL}}_2(\textbf{F}_{11})$ and X is the unique smooth Fano threefold of Picard rank 1 and genus 8 that admits a faithful action of the group ${\text{PSL}}_2(\textbf{F}_{11})$ . We have $n=1$ , $H^3=14$ , $m=9$ , $\widetilde{G}\simeq G$ and $h^0({\mathscr{O}}_X(3H))=40$ . Going through the list of subgroups in G of index $\leqslant 40$ , we see that either $G_P\simeq{\mathfrak{A}}_5$ or $G_P\simeq{\boldsymbol{\mu}}_{11}\rtimes{\boldsymbol{\mu}}_5$ . Since X is smooth, the embedded tangent space at P is 3-dimensional and has a faithful $G_P$ action. This means that the restriction of the representation $H^0({\mathscr{O}}_X(H))$ to $G_P$ has a trivial representation and a 3-dimensional faithful subrepresentation. This is impossible if $G_P\simeq{\boldsymbol{\mu}}_{11}\rtimes{\boldsymbol{\mu}}_5$ , so $G_P \simeq {\mathfrak{A}}_5$ . From the character table [ Reference Conway, Curtis, Norton, Parker and WilsonCCN+85 ] of ${\text{PSL}}_2(\textbf{F}_{11}$ , we see that any faithful irreducible character $\chi$ of degree $\le 10$ must have $\chi(g)=0$ for all $g \in G$ of order 5. However, we have $\rho(g) = (1\pm \sqrt{5})/2$ for an irreducible character $\rho$ of ${\mathfrak{A}}_5$ of degree 3. Thus there must be two 3-dimensional faithful ${\mathfrak{A}}_5$ -subrepresentations of $H^0({\mathscr{O}}_X(H))$ along with a trivial subrepresentation. The remaining character must be a character $\sigma$ of ${\mathfrak{A}}_5$ of degree 2 such that $\sigma(g)=-2$ . No such characters exist, so this case is impossible.

Thus, there is a G-irreducible reduced curve C in X that is a union of smooth irreducible curves $C_1,\ldots,C_r$ of genus g and degree d such that $rd\leqslant 14$ , $2g-2\leqslant d$ and $r(2d-g+1)$ $\leqslant 40$ . Arguing as above, we see that $r\in\{1,11,12\}$ . As above, denote by $G_1$ the stabiliser of the curve $C_1$ . If $r=12$ , then $d=1$ , so that $C_1$ is a line, which implies that the restriction of the representation $H^0({\mathscr{O}}_X(H))$ to $G_1$ contains a two-dimensional subrepresentation. But we already checked that this is not the case, so that $r\ne 12$ . Similarly, we see that $r\ne 11$ , because $G_1\simeq{\mathfrak{A}}_5$ in this case, and the restriction of the representation $H^0({\mathscr{O}}_X(H))$ to $G_1$ does not have two-dimensional subrepresentations either. Hence, we see that $r=1$ , so that C is irreducible. Then $d\leqslant 14$ and $g\leqslant 8$ , which is impossible by (4·5·1). This completes the proof of Theorem 4·3.

Proof. Let X be a rationally connected threefold. Let G be a subgroup in $\text{Aut}(X)$ . Suppose that G is one of the following groups ${\text{SL}}_2(\textbf{F}_{11})$ , ${\text{Sp}}_4(\textbf{F}_{3})$ , $2.{\mathfrak{A}}_7$ or $6.{\mathfrak{A}}_7$ . We seek a contradiction. We may assume that $G\not\simeq 6.{\mathfrak{A}}_7$ because in this case taking the quotient by a subgroup of order 3 in the center reduces the problem to $2.{\mathfrak{A}}_7$ . Thus ${\text{z}}(G)\simeq{\boldsymbol{\mu}}_2$ . We may assume that X has a structure of G-Mori fiber space $\pi\colon X\to S$ . By Theorem 2·5 the base S is a point, i.e. X is a $G\mathbb{Q}$ -Fano threefold.

Let $Y=X/{\text{z}}(G)$ , let $\pi\,:\, X\to Y$ be the quotient map and let $\bar G\,:\!=\,G/{\text{z}}(G)$ . Then Y has canonical singularities by Lemmas 4·1·1, 4·1·2 and Claim 4·7·2 below. Using Theorem 1·3, we see that Y is $\bar G$ -birational to one of the Fano threefolds listed in Theorem 4·3 and by this theorem Y, in fact, is $\bar G$ -isomorphic to one of the varieties in the list. In all our cases Y is a $\bar G$ -Fano with at worst isolated ordinary double points (in fact, Y is smooth except for the case 4.3(iv)). Moreover, $\text{Cl}(Y)^{\bar G}=\text{Pic}(Y)$ by [ Reference Cheltsov, Przyjalkowski and ShramovCPS16 , lemma 2·2]. The Hurwitz formula gives

(4·7·1) \begin{equation} K_X=\pi^*\!\left(K_Y+\textstyle\frac12 B\right),\end{equation}

where B is the branch divisor. Thus B is non-zero $\bar G$ -invariant and there exists a Cartier divisor D such that $2D\sim B$ . In particular, the Fano index of Y is even. Therefore, we are left with the cases 4·3(ii), 4·3(iii), 4·3(v). But in the case 4·3(ii) we have $Y\simeq \mathbb{P}^3$ , $\bar G\simeq {\mathfrak{A}}_7$ and $\deg B\le 6$ by (5·7·1) which is impossible because the minimal degree of invariants in this case is at least 8 (see Table 1). Likewise we obtain a contradiction in the cases 4·3(iii) and 4·3(v).

Proof. Note that the fixed point locus of ${\text{z}}(G)$ on X is G-invariant and so its divisorial part must be a $\mathbb{Q}$ -Cartier divisor. Hence any Gorenstein point of X satisfies conditions of Lemma 4·1·1. Assume that X is a non-Gorenstein Fano threefold. Let $P\in X$ be a non-Gorenstein point and let $G_P\subset G$ be its stabiliser.

Proof. Assume that there exists a G-invariant divisor $S\in |{-}K_X|$ . By Theorem 2·9·1 the pair (X, S) is not plt. By Lemma 3·5 the surface S is reducible and reduced. Since $\text{Pic}(X)=\mathbb{Z}\cdot [S]$ , the linear system $|{-}K_X|=|S|$ has no fixed components. By Bertini’s theorem this G-invariant surface $S\in |{-}K_X|$ is unique. Write $S=\sum _{i=1}^m S_i$ . Then

\begin{equation*}\sum ({-}K_X)^2\cdot S_i = 2g-2\le 18.\end{equation*}

So the cardinality of the orbit of $S_1$ equals 7 or 14 by Table 2. In both cases $g=8$ so that $\dim H^0(X,{\mathscr{O}}_X({-}K_X))=10$ . We get a contradiction with Table 1.

Proof. Observe from Table 1, that all 6-dimensional irreducible representations U have a unique invariant cubic in $\mathbb{P}(U)$ defined by a polynomial f. Then the quadratic polynomials $\partial f/\partial x_i$ generate a 6-dimensional irreducible subrepresentation $Q \subseteq S^2U^\vee$ which is isomorphic to U. The complement U ^′ of Q in $S^2U^\vee$ is 15-dimensional. If U is faithful, then so must be U ^′ and we conclude immediately from the character table [ Reference Conway, Curtis, Norton, Parker and WilsonCCN+85 ] that $S^2U^\vee\simeq U\oplus U_{15}$ , where $U_{15}$ is a faithful irreducible representation of dimension 15. If U is not faithful, then from Table 1 we see that U ^′ contains a unique trivial subrepresentation. In this case, $S^2U^\vee\simeq \mathbb{C}\oplus U\oplus W_{14}$ , where $W_{14}$ is a non-faithful irreducible representation of dimension 14. From the character table, the only possibility is that U ^′ is a direct sum of an irreducible 14-dimensional representation and a trivial representation. In either case, Q is the the unique 6-dimensional irreducible subrepresentation. One can check that the hypersurface $f=0$ in $\mathbb{P}(U)$ is smooth. Hence quadrics from Q have no common zeros in $\mathbb{P}(U)$ .

Remark 5·2·3. Let $X_3$ a cubic fourfold in $\mathbb{P}^5$ that admits a faithful action of the group ${\mathfrak{A}}_7$ . Then $X_3$ is one of two hypersurfaces that (implicitly) appear in the proof of Lemma 5·2·2. For one of them, the action of ${\mathfrak{A}}_7$ is given by the standard irreducible six-dimensional representation of the group ${\mathfrak{A}}_7$ . For the other, the action is given by an irreducible faithful six-dimensional representation of the group $3.{\mathfrak{A}}_7$ . In the former case, one has $\alpha_{{\mathfrak{A}}_7}(X_3)\leqslant {2}/{3}$ , because $\mathbb{P}^4$ contains an ${\mathfrak{A}}_7$ -invariant quadric hypersurface. Here, $\alpha_{{\mathfrak{A}}_7}(X_3)$ is the ${\mathfrak{A}}_7$ -invariant $\alpha$ -invariant of Tian defined in [ Reference TianTia87 ]. However, in the latter case, one has $\alpha_{{\mathfrak{A}}_7}(X_3)\geqslant 1$ . Indeed, suppose that $\alpha_{{\mathfrak{A}}_7}(X_3)<1$ . Then there exists an effective ${\mathfrak{A}}_7$ -invariant divisor D on $X_3$ such that $D\mathbin{\sim_{\scriptscriptstyle{\mathbb{Q}}}} -K_{X_3}$ and $(X_3,D)$ is not lc. This follows from the algebraic formula for $\alpha_{{\mathfrak{A}}_7}(X_3)$ given in [ Reference Cheltsov and ShramovCS08 , appendix A]. Choose positive rational number $\mu<1$ such that $(X_3,\mu D)$ is strictly lc. Let Z be a minimal center of lc singularities of the log pair $(X_3,\mu D)$ . Then $\dim\!(Z)\leqslant 2$ , because $\mathbb{P}^6$ does not contain ${\mathfrak{A}}_7$ -invariant hyperplanes and quadric hypersurfaces. Using the perturbation trick (see the proofs of [ Reference KawamataKaw97 , theorem 1·10] and [ Reference KawamataKaw98 , theorem 1]), we may assume that all log canonical centers of the log pair $(X_3,\mu D)$ are of the form $\theta(Z)$ for some $\theta\in G$ . Moreover, by [ Reference KawamataKaw97 , proposition 1.5], either $Z\cap\theta(Z)=\varnothing$ or $Z=\theta(Z)$ for every $\theta\in G$ . On the other hand, it follows from [ Reference LazarsfeldLaz04 , theorem 9·4·17] that the union of all log canonical centers of the log pair $(X_3,\mu D)$ is connected, which implies that Z is ${\mathfrak{A}}_7$ -invariant. In particular, since $\mathbb{P}^6$ does not have ${\mathfrak{A}}_7$ -fixed points, the center Z is not a point, and ${\mathfrak{A}}_7$ acts faithfully on Z. However, Kawamata’s subadjunction [ Reference KawamataKaw98 , theorem 1] implies that Z is a normal Fano type subvariety, so that either it is a smooth rational curve or a rational surface with at most quotient singularities. This is impossible, because ${\mathfrak{A}}_7$ is not contained in $\text{Cr}_2(\mathbb{C})$ by [ Reference Dolgachev and IskovskikhDI09 ]. Thus, we see that $\alpha_{{\mathfrak{A}}_7}(X_3)\geqslant 1$ in the case when the action of ${\mathfrak{A}}_7$ on $\mathbb{P}^6$ is given by an irreducible faithful six-dimensional representation of the group $3.{\mathfrak{A}}_7$ . In particular, this hypersurface admits a Kähler–Einstein metric by [ Reference TianTia87 ]. Now, we know from [ Reference LiuLiu22 ] that all smooth cubic fourfolds are Kähler–Einstein. However, when we originally wrote this paper, the only known examples of Kähler–Einstein smooth cubic fourfolds were described in [ Reference Arezzo, Ghigi and PirolaAGP06 ], and our ${\mathfrak{A}}_7$ -invariant cubic fourfold was not one of them: in appropriate homogeneous coordinates on $\mathbb{P}^6$ it is given by

\begin{align*} & x_{1}^3+x_{2}^3+x_{3}^3+x_{4}^3+x_{5}^3+x_{6}^3+x_{1}x_{2}x_{3}+x_{1}x_{2}x_{4}\\& \quad +x_{1}x_{2}x_{5}+x_{1}x_{4}x_{5}+x_{1}x_{5}x_{6}+x_{2}x_{4}x_{5}+x_{2}x_{4}x_{6}+x_{3}x_{4}x_{5}\\& \quad +\omega^2x_{1}x_{3}x_{4}+\omega^2x_{1}x_{4}x_{6}+\omega^2x_{2}x_{3}x_{5}+\omega^2x_{2}x_{5}x_{6}+\omega x_{1}x_{2}x_{6}+\omega x_{1}x_{3}x_{5}\\& \quad +\omega x_{1}x_{3}x_{6}+\omega x_{2}x_{3}x_{4}+\omega x_{2}x_{3}x_{6}+\omega x_{3}x_{4}x_{6}+\omega x_{3}x_{5}x_{6}+\omega x_{4}x_{5}x_{6}=0,\end{align*}

where $\omega$ is a primitive cubic root of unity. This implies that it does not contain planes, while Kähler–Einstein smooth cubic fourfolds found in [ Reference Arezzo, Ghigi and PirolaAGP06 ] always contain many planes.

Proof. Take a point $P\in \text{Sing}'(X)$ . Let r be the index of P, let $\Omega$ be its orbit, and let and $n\,:\!=\,|\Omega|$ . Bogomolov–Miyaoka inequality [ Reference KawamataKaw92 , Reference J. Kollár, Mori and TakagiKMMT00 ] gives us

\begin{equation*}0< -K_X\cdot c_2=24 -\sum (r_i-1/r_i)=24 -3N/2\le 24 -3n/2.\end{equation*}

Then using the list of maximal subgroups from Table 2 one can obtain the following possibilities:

1. (i) $G_P\simeq 3.{\mathfrak{A}}_6$ , $n=7$ ;

2. (ii) $G_P\simeq {\text{PSL}}_2(\textbf{F}_7)\times {\boldsymbol{\mu}}_3$ , $n=15$ .

Proof. As in [ Reference ProkhorovPro12 , claim 6·8·1] one can show that the Weil divisor class group $\text{Cl}(X)$ is torsion free. Thus $\text{Cl}(X)^G\simeq \mathbb{Z}$ . Let A be the ample generator of this group. Write $-K_X=aA$ . Since $K_X$ is not Cartier, a is odd. On the other hand, $a^3A^3=({-}K_X)^3=1$ or $3/2$ . Since $aA^3\in \mathbb{Z}/2$ , this implies $a=1$ , i.e. $\text{Cl}(X)^G\simeq \mathbb{Z}\cdot K_X$ . Since $|{-}K_X|=\varnothing$ the assertion follows.

Proof. Suppose that $(X, {3}{\mathscr{M}}/{2})$ is not lc. We seek a contradiction. Choose $\mu< {3}/{2}$ such that $(X,\mu{\mathscr{M}})$ is strictly lc. Let Z be a minimal center of lc singularities of the log pair $(X,\mu{\mathscr{M}})$ . Then Z is either a point or a curve, because the base locus of $|{-}2K_X|$ does not have surfaces by Lemma 5·7·4.

Observe that $\theta (Z)$ is also a minimal center of lc singularities of this log pair for every $\theta \in G$ . Moreover, for every $\theta \in G$ , either $Z\cap \theta (Z)=\varnothing$ or $Z=\theta (Z)$ by [ Reference KawamataKaw97 , proposition 1·5]. Using the perturbation trick (see [ CS16 , lemma 2·4·10] or the proofs of [ Reference KawamataKaw97 , theorem 1·10] and [ Reference KawamataKaw98 , theorem 1]), for every sufficiently small $\epsilon>0$ , we can replace the boundary $\mu{\mathscr{M}}$ by an effective boundary $B_X$ such that

\begin{equation*}B_X\mathbin{\sim_{\scriptscriptstyle{\mathbb{Q}}}} -2(\mu+\epsilon) K_X,\end{equation*}

the log pair $(X,B_X)$ is strictly lc, and all its (not necessarily minimal) centers of lc singularities are the subvarieties $\theta (Z)$ for $\theta \in G$ . In particular, there are minimal log canonical centers of the log pair $(X,B_X)$ . We may assume that $\mu+\epsilon< {3}/{2}$ , since $\mu< {3}/{2}$ .

Let $\Sigma$ be the union of all log canonical centers $\theta (Z)$ for $\theta \in G$ . Then $\Sigma$ is either a G-orbit or a disjoint union of irreducible isomorphic curves, which are transitively permuted by G. In the latter case, each such curve is smooth by Kawamata’s [ Reference KawamataKaw98 , theorem 1]. Let ${\mathscr{I}}_{\Sigma}$ be the ideal sheaf of the locus $\Sigma$ . Then

\begin{equation*}h^1\!\left({\mathscr{O}}_X\big({-}2K_X\big)\otimes{\mathscr{I}}_{\Sigma}\right)=0,\end{equation*}

by [ Reference LazarsfeldLaz04 , theorem 9·4·17] or [ Reference KollárKol97 , theorem 2·16], because $-2K_X-(K_X+B_X)$ is ample. In particular, if Z is a point, we see that

\begin{equation*}|\Sigma|=h^0\!\left({\mathscr{O}}_{\Sigma}\otimes{\mathscr{O}}_X\big({-}2K_X\big)\right)= h^0\!\left({\mathscr{O}}_X\big({-}2K_X\big)\right)-h^0\!\left({\mathscr{O}}_X\big({-}2K_X\big)\otimes{\mathscr{I}}_{\Sigma}\right)\leqslant 5,\end{equation*}

which must be a point since $3.{\mathfrak{A}}_7$ does not have nontrivial subgroups of index $\le 5$ . This is impossible by Lemma 2·7.

Thus, we see that $\Sigma$ is a disjoint union of irreducible isomorphic smooth curves. Denote them by $C_1=Z,C_2,\ldots,C_r$ . Let $d=-2K_X\cdot C_i$ and let g be the genus of the curve $C_1$ . By Kawamata’s subadjunction [ Reference KawamataKaw98 , theorem 1], for every ample $\mathbb{Q}$ -divisor A on X, we have

\begin{equation*}\left(K_{X}+B_{X}+A\right)\Big\vert_{C_i}\mathbin{\sim_{\scriptscriptstyle{\mathbb{Q}}}} K_{C_i}+B_{C_i}\end{equation*}

for some effective $\mathbb{Q}$ -divisor $B_{C_i}$ on the curve $C_i$ . This gives $d>2g-2$ , so that $d\geqslant 2g-1$ . Thus, we have

\begin{align*}r(d-g+1)=h^0\!\left({\mathscr{O}}_{\Sigma}\otimes{\mathscr{O}}_X\big({-}2K_X\big)\right)=h^0\!\left({\mathscr{O}}_X\big({-}2K_X\big)\right)-h^0\!\left({\mathscr{O}}_X\big({-}2K_X\big)\otimes{\mathscr{I}}_{\Sigma}\right)\leqslant 5\end{align*}

by the Riemann–Roch formula applied to each curve $C_i$ . Now, using $r(d-g+1)\leqslant 5$ and $d\geqslant 2g-1$ , we deduce that $r=1$ and $g\leqslant 5$ . This implies that $Z=C_1$ is pointwise fixed by G, since G cannot act non-trivially on a smooth curve of genus $\leqslant 5$ . This is a contradiction, since G does not fix a point in X by Lemma 2·7.