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\title[Extensions of maximal symplectic actions on K3 surfaces]{Extensions of maximal symplectic actions on K3 surfaces}
\alttitle{Extensions d'actions symplectiques maximales sur les surfaces K3}
\subjclass{14J28, 14J50}
\keywords{K3 surface, automorphism, Mathieu group}
\author[\initial{S.} \lastname{Brandhorst}]{\firstname{Simon} \lastname{Brandhorst}}
\address{Fakult\"at f\"ur Mathematik und Informatik,\\
Universit\"at des Saarlandes,\\
Campus E2.4, 66123 Saarbr\"ucken, (Germany)}
\email{brandhorst@math.uni-sb.de}
\thanks{S.B. is funded by the Deutsche Forschungsgemeinschaft (DFG, German Research Foundation) – Project-ID 286237555 – TRR 195. K.~H. was partially supported by Grants-in-Aid for Scientific Research (17K14156).}
\author[\initial{K.} \lastname{Hashimoto}]{\firstname{Kenji} \lastname{Hashimoto}}
\address{Graduate School of Mathematical Sciences,\\
The University of Tokyo,\\
3-8-1 Komaba, Maguro-ku, Tokyo,\\
153-8914, (Japan)}
\email{kenji.hashimoto.math@gmail.com}
\begin{abstract}
We classify pairs $(X,G)$ consisting of a complex K3 surface $X$ and a finite group $G~\leq~\Aut(X)$
such that the subgroup $G_s \lneq G$ consisting of symplectic automorphisms is among the $11$ maximal symplectic ones as classified by Mukai.
\end{abstract}
\begin{altabstract}
Nous classifions les paires $(X, G)$ formées d'une surface K3 complexe $X$ et d'un groupe fini $G~\leq~\Aut(X)$ pour lesquelles le sous-groupe $G_s \lneq G$ des automorphismes symplectiques appartient aux $11$ sous-groupes symplectiques maximaux classifiés par Mukai.
\end{altabstract}
\datereceived{2020-05-04}
\daterevised{2020-10-07}
\dateaccepted{2020-11-23}
\editor{S. Cantat}
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\begin{DefTralics}
\DeclareMathOperator{\Aut}{Aut}
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\dateposted{2021-08-26}
\begin{document}
\maketitle
\section{Introduction}
A (complex) \emph{K3 surface} is a compact, complex manifold $X$ of dimension~$2$ which is simply connected and admits a nowhere degenerate holomorphic symplectic form $\sigma_X \in \H^0(X,\Omega^2_X)$ unique up to scaling. An automorphism of a K3 surface is called \emph{symplectic} if it leaves the $2$-form invariant and non-symplectic else.
Finite groups of symplectic automorphisms of K3 surfaces were classified by Mukai up to isomorphism of groups. Namely, a group acts faithfully and symplectically on some complex K3 surface if and only if it admits an embedding into the Mathieu group $M_{24}$ which decomposes the $24$ points into at least $5$ orbits and fixes a point (in particular it is contained in $M_{23}$)~\cite{kondo:symplectic,mukai:symplectic}. This leads to a list of $11$ maximal subgroups (with $5$ orbits) among the subgroups of $M_{24}$ meeting these conditions. A finer classification, namely up to equivariant deformation, was obtained in~\cite{hashimoto:symplectic}. There are $14$ maximal finite symplectic group actions (see Table~\ref{tbl:symplectic}).
However not every automorphism of a K3 surface is symplectic. Let $X$ be a K3 surface and $G \leq \Aut(X)$ a group of automorphisms. We remark that $G$ is finite if and only if there is an ample class on $X$ invariant under $G$. Denote by $\sym{G}$ the normal subgroup consisting of symplectic automorphisms. Let $G$ be finite. Then we have a natural exact sequence
\[
1 \rightarrow \sym{G} \rightarrow \nsym{G} \overset{\rho}{\rightarrow} \mu_n \rightarrow 1
\]
where $n \in \{k \in \bbN \mid \varphi(k) \leq 20 \}$ and $\varphi$ is the Euler totient function. The homomorphism $\rho$ is defined by $g^* \sigma_X=\rho(g)\cdot\sigma_X$. In the present paper, we classify finite groups $\nsym{G}$ of automorphisms of K3 surfaces, under the condition that $\sym{G}$ is among the $11$ maximal groups and $G_s \lneq G$. As it turns out, this forces the underlying K3 surface $X$ to have maximal Picard number $20$, i.e. it is a singular K3 surface. In particular it has infinite automorphism group. Moreover, those K3 surfaces (with $G$) are rigid (i.e. not deformable). Let $(X,\nsym{G})$ and $(X',\nsym{G}')$ be two pairs of K3 surfaces with a group of automorphisms. They are called isomorphic if there is an isomorphism $f \colon X \rightarrow X'$ with $f G f^{-1} = G'$.
\begin{theo}\label{thm:classification}
Let $X$ be a K3 surface and $\nsym{G}\leq \Aut(X)$ a maximal finite group of automorphisms such that the symplectic part $\sym{G}$ is isomorphic to one of the $11$ maximal groups and $\sym{G} \lneq \nsym{G}$. Then the pair $(X,\nsym{G})$ is isomorphic to one of the $42$ pairs listed in Section~\ref{sec:classification}.
\end{theo}
The representations of the groups $G$ on the K3 lattices $\Lambda\cong \H^2(X,\Z)$ are given in the ancillary file to~\cite{brandhorst-hashimoto:arxiv} on arXiv.
The proof goes via a classification, up to conjugacy, of suitable finite subgroups of the orthogonal group of the K3 lattice. Then the strong Torelli type theorem~\cite{shafarevic:torelli,burns-rapoport:1975} and the surjectivity of the period map~\cite{todorov:1980} abstractly provide the existence and uniqueness of the pairs $(X,G)$.
Note that a K3 surface admitting a non-symplectic automorphism of finite order must be projective~\cite[Theorem~3.1]{nikulin:auto}. Thus, at least in principle, it is possible to find projective models of the K3 surfaces and the automorphisms. For $25$ out of the $42$ pairs $(X,G)$ we list explicit equations in Section~\ref{sec:classification}. Using the Torelli-type theorem Kond\=o proved in~\cite{kondo:maximal} that the maximal order of a finite group of automorphisms of a K3 surface equals $3840$. Section~\ref{sect:maximal} is devoted to deriving its equations for the first time.
For the full story of symplectic groups of automorphisms we recommend the excellent survey~\cite{kondo:symplectic_survey}. Non-symplectic automorphisms of prime order are treated in~\cite{artebani_sarti_taki:non-symplectic}. In~\cite{frantzen:automorphisms}, a similar classification with different methods is carried out, albeit under the restrictive condition that $\nsym{G} = \sym{G} \times \mu_2$ and the action by $\mu_2$ has fixed points. Note that the author misses cases \textbf{70d} and \textbf{76a} (see Section~\ref{sec:classification}).
\begin{rema}
Let $(X,\nsym{G})$ be as in Theorem~\ref{thm:classification}. It turns out that the non-symplectic part $\nsym{G}/\sym{G}\cong \mu_n$ is of even order and the pair $(X,G)$ is determined up to isomorphism already by $\sym{G}$ and any involution in $\nsym{G}/\sym{G}$. See Section~\ref{sect:extensions} for details.
\end{rema}
\subsection*{Open problems}
\noindent We close this section with some interesting problems concerning groups of automorphisms of K3 surfaces.
\begin{enumerate}
%\itemsep0pt
\item Find the remaining $17$ missing equations among the $42$ K3 surfaces and their automorphisms.
\item Give generators of the full automorphism group of the corresponding K3 surfaces. Since a Conway chamber in the nef cone of this surface has large symmetry, chances are that one can find a nice generating set for the automorphism group.
\item Find a projective model of the K3 surface with a linear action by $M_{22}$ in characteristic $11$. Its existence is proven by Kond\=o~\cite{kondo:m22} using the crystalline Torelli type theorem.
\item Use the present classification to study finite groups of automorphisms of Enriques surfaces beyond the semi-symplectic case~\cite{mukai-ohashi:m12}.
\end{enumerate}
Finding equations for the surface is often much easier than for the automorphisms. Should you find equations or relevant publications on one of the surfaces treated here, please notify the first author. We will update the arXiv version of this paper with your findings.\\
\subsection*{Acknowledgements}
We would like to thank the organizers of the conference Moonshine and K3 surfaces in Nagoya in 2016 where the idea for this work was born. The first author would like to thank the University of Tokyo and Keiji Oguiso for their hospitality. Thanks to Matthias Sch\"utt for encouragement and discussions. We warmly thank Cédric Bonnafé, Noam Elkies, Hisanori Ohashi and Alessandra Sarti for sharing explicit models of symmetric K3 surfaces with us. We also thank the anonymous referee for carefully reading our manuscript and suggesting many improvements.
\section{Lattices} In this section we recall the basics on integral lattices (equivalently quadratic forms) and fix notation. The results are found in~\cite{conway-sloane:sphere_packings,nikulin:quadratic_forms}.\\
A \emph{lattice} consists of a finitely generated free $\Z$-module $L$ and a non-degenerate integer valued symmetric bilinear form
\[
\langle \cdot \,, \cdot \rangle \colon L \times L \rightarrow \Z.
\]
Given a basis $(b_1,\,\dots,\,b_n)$ of $L$, we obtain the \emph{Gram matrix} $Q= (\langle b_i,b_j \rangle)_{1\,\leq\,i,\,j\,\leq\,n }$. The determinant $\det Q$ is independent of the choice of basis and called the \emph{determinant} of the lattice $L$; it is denoted by $\det L$. We display lattices in terms of their Gram matrices. The \emph{signature} of a lattice is the signature of its Gram matrix. We denote it by $(s_+,s_-)$ where $s_+$ (respectively $s_-$) is the number of positive (respectively negative) eigenvalues. We define the \emph{dual lattice} $L^\vee$ of $L$ by $L^\vee=\{x \in L \otimes \Q | \langle x, L \rangle \subseteq \Z \} \cong Hom(L,\Z)$. The \emph{discriminant group} $L^\vee/L$ is a finite abelian group of cardinality $|\det L|$. We call a lattice \emph{unimodular} if $L=L^\vee$, and we call it \emph{even} if $\langle x, x \rangle$ is even for all $x \in L$. The discriminant group of an even lattice carries the \emph{discriminant form}
\[
q_L \colon L^\vee / L \rightarrow \Q / 2\Z, \quad \bar x \mapsto \langle x,x\rangle + 2\Z.
\]
An \emph{isometry} of lattices is a linear map compatible with the bilinear forms. The \emph{orthogonal group} $O(L)$ is the group of isometries of $L$ and the special orthogonal group $\SO(L)$ consists of the isometries of determinant $1$. Discriminant forms are useful to describe embeddings of lattices and extensions of isometries. A sublattice $L \subseteq M$ is called \emph{primitive}, if $L = (L \otimes \Q) \cap M$. By definition, the orthogonal complement $S^\perp\subseteq M$ of a (not necessarily primitive) sublattice $S$ is a primitive sublattice. For $L_1$ primitive and $L_2 = L_1^\perp$ we call $L_1 \oplus L_2 \subseteq M$ a \emph{primitive extension}. Now, suppose that $M$ is even, unimodular, then
\[
H_M = M/(L_1 \oplus L_2) \subseteq \left(\disc{L_1}\right) \oplus \left(\disc{L_2}\right)
\]
is the graph of a so called \emph{glue map} $\phi_M\colon \disc{L_1} \rightarrow \disc{L_2}$, that is, any element in $H_M$ is of the form $x \oplus \phi_M(x)$ for $x \in \disc{L_1}$. This isomorphism is an anti-isometry, namely, it satisfies $q_{L_2} \circ \phi_M = -q_{L_1}$. Conversely given such an anti-isometry $\phi$, its graph $H_\phi$ defines a primitive extension $L_1\oplus L_2\subseteq M_\phi$ with $M_\phi$ even, unimodular.
Given an isometry $f \in O(L_1)$, it induces an isometry $\bar f \in O(\disc{L_1})$ of the discriminant group. Let $g\in O(L_2)$ be an isometry on the orthogonal complement. Then $f\oplus g \in O(L_1 \oplus L_2)$ extends to $M$ if and only if $(\bar f \oplus \bar g)(H_M)=H_M$, or equivalently, $\phi_M \circ \bar f = \bar g \circ \phi_M$.
\begin{lemm}\label{lem:extend}
Let $L \subseteq M$ be a primitive sublattice of an even unimodular lattice $M$. Set $O(M,L)=\{f \in O(M) | f(L) = L\}$ and $K = L^\perp$. If the natural map $O(K) \rightarrow O(\disc{K})$ is surjective, then the restriction map $O(M,L)\rightarrow O(L)$ is surjective. In other words: any isometry of $L$ can be extended to an isometry of $M$.
\end{lemm}
\begin{proof}
Denote the glue map by $\phi=\phi_M$, and let $g \in O(K)$ be a preimage of $\phi \circ \bar f \circ \phi^{-1}$. Then $\phi \circ \bar f = \bar g \circ \phi$. Hence, $f \oplus g$ extends to $M$.
\end{proof}
Let $L$ be a lattice and $G \leq O(L)$. We define the \emph{invariant} and \emph{coinvariant lattices} respectively by
\[
L^G = \left\{x \in L \middle|\forall g \in G\colon g(x)=x \right\} \quad\text{and}\quad L_G = \left(L^G\right)^\perp.
\]
Then, by definition, $L^G \oplus L_G \subseteq L$ is a primitive extension. Two lattices are said to be in the same genus, if they become isometric after tensoring with the $p$-adics $\Z_p$ for all primes $p$ and the reals $\R$. A genus is denoted in terms of the Conway--Sloane symbols~\cite[Chapter~15]{conway-sloane:sphere_packings}. For instance the genus of even unimodular lattices of signature $(3,19)$ is denoted by $\II_{3,\,19}$. In fact all lattices in this genus are isometric.
\section{K3 surfaces and the Torelli type theorem} In this section we recall standard facts about complex K3 surfaces. All results can be found in the textbooks~\cite{BHPV:compact_complex_surfaces,huybrechts:k3-book}.
Let $X$ be a K3 surface. Its second integral cohomology group $\H^2(X,\Z)$ together with the cup product is an even unimodular lattice of signature $(3,19)$. It comes equipped with an integral weight $2$ Hodge structure. Such a Hodge structure is given by its Hodge decomposition
\[
\H^2(X,\Z) \otimes \C = \H^2(X,\C)= \H^{2,\,0}(X)\oplus \H^{1,\,1}(X) \oplus \H^{0,\,2}(X)
\]
with $\H^{i,\,j}(X) = \overline{\H^{j,\,i}}(X)$ and natural isomorphisms $\H^{i,\,j} \cong \H^j(X,\Omega_X^i)$. The corresponding Hodge numbers are $h^{2,\,0}=h^{0,\,2}=1$ and $h^{1,\,1}=20$. We can recover the entire Hodge structure from $\H^{2,\,0}(X)$ via $\H^{0,\,2}(X)=\overline{\H^{2,\,0}(X)}$ and $\H^{1,1}(X) = \left(\H^{2,\,0}(X) \oplus \H^{0,\,2}(X)\right)^\perp$.
The \emph{transcendental lattice} of a K3 surface is defined as the smallest primitive sublattice $T_X$ of $\H^2(X,\Z)$ such that $T_X\otimes \C$ contains the period $\H^{2,\,0}(X)=\C \sigma_X$. By the Lefschetz theorem on $(1,1)$-classes, the \emph{Néron--Severi lattice} $\NS_X$ of a K3 surface is given by $\H^{1,\,1}(X) \cap \H^2(X,\Z)$. Note that $\NS_X$ and $T_X$ can be degenerate~\cite[(3.5)]{nikulin:auto}. But if $X$ is projective, then they are (non-degenerate) lattices of signatures $(1,\rho-1)$ and $(2,20-\rho)$ respectively, and we have $\NS_X = T_X^\perp$.
As a next step we want to compare Hodge structures of different K3 surfaces. For this we fix a reference frame, namely a lattice $\Lambda \in \II_{3,\,19}$.
\begin{definition}
A \emph{marked} K3 surface is a pair $(X,\eta)$ consisting of a complex K3 surface $X$ and an isometry $\eta \colon \H^2(X,\Z) \rightarrow \Lambda$. We call $\eta$ a \emph{marking}.
\end{definition}
We associate a marked K3 surface $(X,\eta)$ with its \emph{period}
\[
\eta_\C\left(\H^{2,\,0}(X)\right) \in \clP_\Lambda:=\left\{ \C \sigma \in \P(\Lambda \otimes \C) \middle| \langle \sigma,\bar \sigma \rangle > 0, \langle \sigma,\sigma \rangle =0 \right\}.
\]
Here we extend the bilinear form on $\Lambda$ linearly to that on $\Lambda \otimes \C$. We call $\clP_\Lambda$ the \emph{period domain}. As it turns out, the concept of marking works well in families. This allows one to define the moduli space $\clM_\Lambda$ of marked K3 surfaces and a period map
\[
\clM_\Lambda \rightarrow \clP_\Lambda, \quad (X,\eta) \mapsto \eta_\C\left(\H^{2,\,0}(X)\right).
\]
The period map is in fact holomorphic, and it turns out to be surjective as well (the surjectivity of the period map for K3 surfaces~\cite{todorov:1980}). The moduli space $\clM_\Lambda$ is not very well behaved. For example it is not Hausdorff. This can be healed by taking into account the K\"ahler (resp.\ ample) cone.
The \emph{positive cone} $\mu_X$ is the connected component of the set
\[
\left\{x \in \H^{1,\,1}(X,\R) \middle| \langle x, x \rangle > 0\right\}
\]
which contains a K\"ahler class. Set $\Delta_X = \{x \in \NS_X \mid \langle x,x\rangle = -2\}$. An element in $\Delta_X$ is called a root. For $\delta \in \Delta_X$, either $\delta$ or $-\delta$ is an effective class by the Riemann--Roch theorem. In fact the effective cone is generated by the effective classes in $\Delta_X$ and the divisor classes in the closure of the positive cone (i.e. $\NS_X \cap \overline{\mu_X}$). The connected components of $\mu_X \setminus \bigcup_{\delta\,\in \,\Delta_X} \delta^\perp$ are called the \emph{chambers}. The hyperplanes $\delta^\perp$ for $\delta \in \Delta_X$ are called the \emph{walls}. One of the chambers is the K\"ahler cone. For a root $\delta \in \Delta_X$, the reflection with respect to the wall $\delta^\perp$ is given by $r_\delta(x) = x + \langle x, \delta \rangle \delta$. The \emph{Weyl group} is the subgroup of $O(\H^2(X,\Z))$ generated by the reflections $r_\delta$ for $\delta \in \Delta_X$. The action of the Weyl group on the chambers is simply transitive. So by composing the marking with an element of the Weyl group, we can ensure that any given chamber in the positive cone of $\Lambda$ corresponds to the K\"ahler cone.
\begin{defi}
Let $X,X'$ be K3 surfaces. An isometry $\phi\colon \H^2(X,\Z) \rightarrow \H^2(X',\Z)$ is called a \emph{Hodge isometry} if $\phi_\C(\H^{i,\,j}(X))\subseteq \H^{i,\,j}(X')$ for all $i,j$. It is called \emph{effective}, if it maps effective (resp. K\"ahler, resp. ample) classes on $X$ to effective (resp. K\"ahler, resp. ample) classes on $X'$.
\end{defi}
The following Torelli type theorem for K3 surfaces is the key tool for our classification of automorphisms.
\begin{theo}[{\cite{burns-rapoport:1975,shafarevic:torelli}}]
Let $X$ and $X'$ be complex K3 surfaces. Let
\[
\phi \colon \H^2(X,\Z) \rightarrow \H^2(X',\Z)
\]
be an effective Hodge isometry. Then there is a unique isomorphism $f \colon X' \rightarrow X$ with $f^* = \phi$.
\end{theo}
We thus obtain a Hodge theoretic characterization of the automorphism group of a K3 surface.
\begin{coro}
Let $X$ be a complex K3 surface. Then the image of the natural homomorphism
\[
\Aut(X) \rightarrow O\left(\H^2(X,\Z)\right)
\]
consists of the isometries preserving the period and the K\"ahler cone.
\end{coro}
\section{Symplectic automorphisms}
In this section we review known facts on symplectic automorphisms needed later on.
Let $X$ be a complex K3 surface. We obtain an exact sequence
\[
1 \rightarrow \sym{\Aut(X)} \rightarrow \Aut(X) \xrightarrow{\rho} \GL(\C \sigma_X).
\]
(Recall that we have $\C \sigma_X = \H^0(X,\Omega_X^2)$.) The elements of the kernel $\sym{\Aut(X)}$ of $\rho$ are the symplectic automorphisms. An automorphism which is not symplectic is called non-symplectic. If $G\leq \Aut(X)$ is a group of automorphisms, we denote by $\sym{G}$ the kernel of $\rho|_G$ and call it the symplectic part of $\nsym{G}$. In order to keep the notation light, we identify $\nsym{G}$ and its isomorphic image in $O(\H^2(X,\Z))$.
Recall that if $L$ is a lattice and $G\leq O(L)$, then $L^G$ is the invariant and $L_G=(L^G)^\perp$ the coinvariant lattice. For the sake of completeness we give a proof of the following essential lemma.
\begin{lemm}[cf.~\cite{nikulin:auto}]\label{lem:symplectic}
Let $\sym{G}\leq \sym{\Aut(X)}$ be a finite group of symplectic automorphisms of some K3 surface $X$. Then
\begin{enumerate}
\item\label{lemm4.1.1} $T_X \subseteq \H^2(X,\Z)^\sym{G}$ and $\H^2(X,\Z)_\sym{G} \subseteq \NS_X$;
\item\label{lemm4.1.2} $\H^2(X,\Z)^\sym{G}$ is of signature $(3,k)$ for some $k\leq 19$;
\item\label{lemm4.1.3} $\H^2(X,\Z)_\sym{G}$ is negative definite;
\item\label{lemm4.1.4} $\H^2(X,\Z)_\sym{G}$ contains no vectors of square $-2$;
\item\label{lemm4.1.5} if $\sym{G}$ is maximal (that is, $G_s$ is isomorphic to one of the 11 maximal finite groups of symplectic automorphisms), then $\sym{G}\cong\ker \left(O(H) \rightarrow O(\disc{H})\right)$ where $H=\H^2(X,\Z)_\sym{G}$.
\end{enumerate}
\end{lemm}
\begin{proofc}\ \\*[-1em]
\eqref{lemm4.1.1} The elements of $\sym{G}$ are all symplectic, i.e. they fix the $2$-form $\sigma_X$. Thus $\C \sigma_X \subseteq \H^2(X,\Z)^\sym{G} \otimes \C$. By minimality of the transcendental lattice and primitivity of the invariant lattice, we get $T_X \subseteq \H^2(X,\Z)^\sym{G}$. Taking orthogonal complements yields the second inclusion.
\noindent\eqref{lemm4.1.2} Let $\kappa'$ be a K\"ahler class. Since automorphisms preserve the K\"ahler cone, the class $\kappa = \sum_{g\,\in\,G} g^*\kappa'$ is a $\sym{G}$-invariant K\"ahler class. Thus $\kappa, (\sigma_X + \bar \sigma_X)/2$ and $(\sigma_X - \bar \sigma_X)/(2i)$ span a positive definite subspace of dimension~$3$ of $\H^2(X,\R)^\sym{G}$.
\noindent\eqref{lemm4.1.3} Recall that $H^2(X,\Z)_{\sym{G}}=(H^2(X,\Z)^{\sym{G}})^\perp$, and $\H^2(X,\Z)$ has signature $(3,19)$. Now, use~\eqref{lemm4.1.2}.
\noindent\eqref{lemm4.1.4} As before we take a $\sym{G}$-invariant K\"ahler class $\kappa$. If $r \in \NS_X$ is of square $-2$, then either $r$ or $-r$ is effective by the Riemann--Roch theorem. Thus $\langle \kappa,r \rangle\neq 0$. Since $\H^2(X,\Z)_\sym{G}$ is orthogonal to $\kappa$, it cannot contain $r$.
\noindent\eqref{lemm4.1.5} Let $g$ be an element in the kernel. Since $g$ acts trivially on $\disc{H}$, it can be extended to an isometry $\tilde{g}$ on $H^2(X,\Z)$ such that $\tilde{g}|_{H^\perp} = \IDL_{H^\perp}$. As $H^\perp\otimes \C$ contains $\sigma_X$ and a K\"ahler class, $\tilde{g}$ is in fact an effective Hodge isometry. The strong Torelli type theorem implies that it is induced by a symplectic automorphism. Since the coinvariant lattice $H$ is negative definite (by~\eqref{lemm4.1.3}), $O(H)$ is finite. In particular, the group $\tilde{G}$ generated by $\sym{G}$ and $g$ is a finite group. By the maximality of $\sym{G}$, $\sym{G}$ must contain $g$.
\end{proofc}
\begin{theo}[\cite{hashimoto:symplectic}]
Let $\sym{G}$ be a finite group of symplectic automorphisms of a $\Lambda$-marked K3 surface. Identify $\sym{G}$ with its image in $O(\Lambda)$. Then the conjugacy class of $\sym{G}$ is determined by the isometry class of the invariant lattice $\Lambda^\sym{G}$. For maximal $G_s$, the invariant lattices can be found in Table~\ref{tbl:symplectic}, and the coinvariant lattice $\Lambda_{G_s}$ is uniquely determined up to isomorphism by the abstract group structure of $G_s$.
\end{theo}
For maximal $\sym{G}$, we have $\rank \syminv=3$ and $\rank \symco=19$~\cite{mukai:symplectic}. The key observation we take from Lemma~\ref{lem:symplectic}, is that the invariant lattice is definite and so is the coinvariant lattice. Hence, the direct product $O(H^2(X,\Z)^\sym{G}) \times O(H^2(X,\Z)_\sym{G})$ is a finite group. It can be computed explicitly with the Plesken--Souvignier algorithm~\cite{plesken-souvignier:orthogonal} as implemented for instance in PARI~\cite{pari}. As it turns out the groups $G \leq \Aut(X)$ we aim to classify are subgroups of this product.
%%table en array qui fait buguer l'ensemble
\begin{table}
\caption{Maximal finite symplectic groups of automorphisms}\label{tbl:symplectic}
%\bgroup
%\newcommand*\arraystretch{1.8}
%\begin{equation*}
\begin{tabular}{c|c|c|c|c|c|c|c|c}
\hline \Tstrut
No. & $\sym{G}$ & $\#\sym{G}$ & $\det \Lambda^\sym{G}$& genus of $\Lambda^\sym{G}$& $\Lambda^\sym{G}$ & $\SO(\syminv)$ & $\#O(\symco)$ & $\#O\!\left(q_{\symco}\right)$\\
\hline $54$ & $T_{48}$ & $48$ & $384$ & $2^{+1}_1,8^{-2}_\II,3^{+1}$ & $\TQ{2}{16}{16}{8}{0}{0}$ & $D_{6}$ & $9216$ & $192$\\[5pt]
\rowcolor{blue!8} $62$ & $N_{72}$ & $72$ & $324$ & $4^{+1}_7,3^{+2},9^{+1}$ & $\TQ{6}{6}{12}{3}{3}{0}$ & $D_4$ & $20736$ & $288$\\[5pt]
$63$ & $M_9$ & $72$ & $216$ & $2^{-3}_1,3^{+1},9^{+1}$ & $\TQ{2}{12}{12}{6}{0}{0}$ & $D_{6}$ & $5184$& $72$\\[4pt]
\rowcolor{blue!8} &&&&& $\TQ{4}{4}{20}{0}{0}{1}$ & $D_2$ && \\[5pt]
\rowcolor{blue!8} \multirow{-2}{*}{$70$} & \multirow{-2}{*}{$\MF{S}_5$} & \multirow{-2}{*}{$120$} & \multirow{-2}{*}{$300$} & \multirow{-2}{*}{$4^{-1}_5,3^{-1},5^{-2}$} & $\TQ{4}{6}{16}{1}{2}{2}$ & $D_2$ & \multirow{-2}{*}{$5760$} & \multirow{-2}{*}{$48$}\\[5pt]
&&&&& $\TQ{2}{4}{28}{0}{0}{1}$ & $D_2$ & &\\[5pt]
\multirow{-2}{*}{$74$} & \multirow{-2}{*}{$L_2(7)$} & \multirow{-2}{*}{$168$} & \multirow{-2}{*}{$196$} & \multirow{-2}{*}{$4^{+1}_7,7^{+2}$} & $\TQ{4}{8}{8}{1}{2}{2}$ & $D_4$ &\multirow{-2}{*}{$5376$} & \multirow{-2}{*}{$32$}\\[5pt]
\rowcolor{blue!8} $76$ & $H_{192}$ & $192$ & $384$ & $4^{-2}_4,8^{+1}_1,3^{+1}$ & $\TQ{4}{8}{12}{0}{0}{0}$ & $D_4$ & $24576$ & $128$\\[5pt]
$77$ & $T_{192}$ & $192$ & $192$ & $4^{-3}_1,3^{-1}$ & $\TQ{4}{8}{8}{4}{0}{0}$ & $D_{6}$ & $36864$ & $192$\\[4pt]
\rowcolor{blue!8} $78$ & $\MF{A}_{4,4}$ & $288$ & $288$ & $2^{+2}_\II,8^{+1}_7,3^{+2}$ & $\TQ{8}{8}{8}{2}{4}{4}$ & $D_4$ & $36864$ & $128$\\[4pt]
&&&&& $\TQ{2}{8}{12}{0}{0}{1}$ & $D_2$ & & \\[5pt]
\multirow{-2}{*}{$79$} & \multirow{-2}{*}{$\MF{A}_6$} & \multirow{-2}{*}{$360$} & \multirow{-2}{*}{$180$} & \multirow{-2}{*}{$4^{-1}_3,3^{+2},5^{+1}$} & $\TQ{6}{6}{8}{3}{3}{0}$ & $D_4$ & \multirow{-2}{*}{$11520$} & \multirow{-2}{*}{$32$}\\[5pt]
\rowcolor{blue!8} $80$ & $F_{384}$ & $384$ & $256$ & $4^{+1}_1,8^{+2}_2$ & $\TQ{4}{8}{8}{0}{0}{0}$ & $D_4$ & $49152$ & $128$\\[5pt]
$81$ & $M_{20}$ & $960$ & $160$ & $2^{-2}_\II,8^{+1}_7,5^{-1}$ & $\TQ{4}{4}{12}{2}{2}{0}$ & $D_4$ & $92160$ & $96$\\[4pt]
\hline
\end{tabular}\\[3pt]
%\end{equation*}
%\egroup
No.\ denotes the number of the group $\sym{G}\leq O(\Lambda)$ as given in~\cite{hashimoto:symplectic}. It is isomorphic to the corresponding group in the column $\sym{G}$. See~\cite{mukai:symplectic} for the notation. The entry genus is given in Conway and Sloane's~\cite{conway-sloane:sphere_packings} notation. The dihedral group of order $2k$ is denote by $D_k$.
\end{table}
\section{Non-symplectic extensions}\label{sect:extensions}
In this section we prove the classification. The invariant lattices of the symplectic actions play a major role. For a start we observe that the cyclic group $\nsym{G}/\sym{G}$ acts on the invariant lattice. Indeed for $g \in \nsym{G}$ and $x\in \H^2(X,\Z)^{\sym{G}}$, $g \sym{G} (x) = g(x)$ is well defined and lies in the invariant lattice since $\sym{G}$ is normal in $\nsym{G}$ and fixes $x$. This yields a homomorphism
\[
\nsym{G}/\sym{G} \rightarrow O\left(\H^2(X,\Z)^{\sym{G}}\right)
\]
of groups which turns out to be injective.
\begin{lemm}\label{lem:conditions}
Let $X$ be a K3 surface and $G \leq \Aut(X)$ a finite group of automorphisms such that the subgroup $\sym{G} \leq \nsym{G}$ of symplectic automorphisms is among the $11$ maximal ones. Then the homomorphism $\nsym{G}/\sym{G} \rightarrow O(\H^2(X,\Z)^{\sym{G}})$ is injective and its image is a cyclic subgroup of $\SO(\H^2(X,\Z)^{\sym{G}})$. In particular its order is $n \in \{ 1,2,3,4,6\}$.
\end{lemm}
\begin{proof}
By our assumption $\sym{G}$ is maximal. Thus, by Table~\ref{tbl:symplectic}, $\H^2(X,\Z)^\sym{G}$ is of rank $3$. Hence a basis of $\H^2(X,\R)^\sym{G}$ is given by a $G$-invariant K\"ahler class $\kappa$, $(\sigma_X +\bar \sigma_X)/2$ and $(\sigma_X -\bar \sigma_X)/(2i)$ (see the proof of Lemma~\ref{lem:symplectic}). Since $\nsym{G}/\sym{G}$ acts on $\H^{2,0}(X)=\C \sigma_X$ faithfully (by the definition of $\sym{G}$), the injectivity in the statement of the lemma follows. By the same reason, $\nsym{G}/\sym{G}$ is cyclic.
Let $g\sym{G}$ be a generator of $\nsym{G}/\sym{G}$. Then $(x-1)$ divides $\chi(x)=\det(x \IDL - g|_{\H^2(X,\,\Z)^\sym{G}})$. Since $\H^2(X,\R)^\sym{G}$ is actually defined over $\Q$ and $g$ is of finite order, $\chi(x)$ is a product of cyclotomic polynomials. Note that the eigenvectors $\sigma_X$ and $\bar \sigma_X$ have complex conjugate eigenvalues. Hence $\chi(x)\neq (x+1)(x-1)^2$. This leaves us with $\chi(x)/(x-1)$ to be one of $(x \pm 1)^2$ or $\Phi_n$ for $n\in \{3,4,6\}$. We conclude by computing the determinant from the characteristic polynomial.
\end{proof}
Recall that via a marking we may identify $\H^2(X,\Z)$ and $\Lambda$.
\begin{rema}
A choice of basis turns the groups $\SO(\syminv)$ into subgroups of $\SL(\Z^3)$. Finite subgroups of $\SO(3)$ are an essential building block for crystallographic groups. It is known that they are isomorphic to a subgroup of a dihedral group, or one of $(\Z/2\Z)^2 \rtimes C_3 \leq (\Z/2\Z)^2 \rtimes S_3$ (cf.~\cite[Table~I]{auslander-cook:crystallographic}). Up to conjugation there are exactly $3$ subgroups of the second type, i.e. $3$ integral representations of $(\Z/2\Z)^2 \rtimes C_3$. Since the three representations are irreducible, there is, up to homothety, a unique invariant quadratic form for each. Their gram matrices are given by
\[
\begin{pmatrix}
1&0&0\\
0&1&0\\
0&0&1
\end{pmatrix},\;
\begin{pmatrix}
2 &1& 2\\
1 &2 &2\\
2& 2& 4
\end{pmatrix},\;
\begin{pmatrix}
3 &2& 2\\
2 & 4 & 0\\
2 & 0& 4
\end{pmatrix}
\]
and are of determinant $1$, $2^2$ and $2^4$. Obviously none of the invariant lattices in Table~\ref{tbl:symplectic} is homothetic to one of these three. Thus $\SO(\syminv)$ must be a (subgroup of) a dihedral group.
\end{rema}
Part 2 of the next lemma will be used later in the proof of Proposition~\ref{prop:k3fromgroup}.
\begin{lemm}\label{lem:orientation-reversal}
Let $H$ be one of the $14$ symplectic fixed lattices and $g \in \SO(H)$ an involution. Then
\begin{enumerate}
\item \label{lemm5.3.1}$\SO(H)$ is isomorphic to a dihedral group $D_k$ of order $2k$ with $k \in \{2,4,6\}$;
\item \label{lemm5.3.2}there is another involution $f\in \SO(H)$ in the centralizer of $g$ with $\det f|H_g\linebreak=-1$ where $H_g$ denotes the coinvariant lattice of the group generated by $g$.
\end{enumerate}
\end{lemm}
\begin{proof}
The proof of~\eqref{lemm5.3.1} is by a direct computation of $\SO(H)$ for each case using a computer. For the reader's entertainment we calculate No.\ 63 by hand. The fixed lattice is given as the orthogonal direct sum $(2) \oplus A_2(6)$ where $A_2(6)$ is a rescaled hexagonal lattice. The orthogonal group of the hexagonal lattice is that of the hexagon, i.e. it is the dihedral group $D_6$. Since the decomposition of a lattice into irreducible lattices is unique up to ordering, the orthogonal group preserves this decomposition and is isomorphic to $\{\pm 1\} \times D_6$. It contains the special orthogonal group with index $2$. The restriction of the $\SO$ action to the hexagonal plane is by $D_6$. Thus it is in fact isomorphic to $D_6$.
For~\eqref{lemm5.3.2} fix $k\in \{2,4,6\}$ and let $g \in D_k$ be an involution. Then it is not hard to check, that there exists an involution $f$ different from $g$ and commuting with $g$. Now view $f$ and $g$ as elements of $\SO(H)$. The characteristic polynomials of both are equal to $(x-1)(x+1)^2$. If $\det f|H_g = 1$, then $f|H_g = -\IDL$. This implies $f=g$, which we excluded.
\end{proof}
Recall the exact sequence
\[
1 \rightarrow \sym{G} \rightarrow \nsym{G} \rightarrow \mu_n \rightarrow 1.
\]
From the proof of Lemma~\ref{lem:conditions} we obtain that if $\sym{G}$ is maximal, then $n \in \{1,2,3,4,6\}$. We want to reconstruct $\nsym{G} \leq O(\Lambda)$ knowing $\sym{G}$ and $\nsym{G}/\sym{G}\cong \mu_n$. In this situation one speaks of an extension of groups. We are interested not only in the group structure, but also in its action on the K3 lattice. This motivates the next definition.
\begin{defi}
Let $L$ be a lattice, $G \leq O(L)$ a group of isometries and $N \leq G$ a normal subgroup with cyclic quotient $G/N = \langle gN \rangle$. We say that $G$ is an \emph{extension} of $N$ by $g|_{L^N}$ where $L^N$ denotes the invariant lattice of $N\leq O(L)$.
\end{defi}
\begin{rema}
In our setting $L$ is unimodular and $N$ coincides with the kernel of the natural map $O(L, L_N) \rightarrow O(L_N^\vee/L_N$). In this case, $G\leq O(L)$ is uniquely determined by $N$ and $g|_{L^N}$.
\end{rema}
Before extending the group, we first have to extend single elements. We are in the luxurious position that every element extends:
\begin{lemm}[{\cite[Theorem~5.1]{hashimoto:symplectic}\label{lem:surj}
}]
Let $\Lambda_\sym{G}$ be the coinvariant lattice for one of the $11$ maximal finite groups. Then the natural map
\[
\psi \colon O(\Lambda_\sym{G}) \rightarrow O\left(\disc{\Lambda_\sym{G}}\right)
\]
is surjective. In particular any isometry of $O(\Lambda^{\sym{G}})$ can be extended to an element in $O(\Lambda)$ normalizing $\sym{G}$.
\end{lemm}
\begin{rem}
One may double check the theorem as follows: first compute $O(\Lambda_\sym{G})$ with the Plesken--Souvignier backtracking algorithm. Then check by a direct computation that the natural map is surjective. For the reader's convenience we list the orders of the groups involved in Table~\ref{tbl:symplectic}. Note that by Lemma~\ref{lem:symplectic} we have $\# \sym{G} \cdot \#O(q_{\Lambda_{\sym{G}}}) = \#O(\Lambda_{\sym{G}})$, if and only if the natural map $\psi$ is surjective.
\end{rem} In general extensions of a given group of isometries are not unique, not even up to conjugacy. But we are in a particulary nice situation.
\begin{lemm}\label{lem:unique-extension}
Let $\sym{G} \leq O(\Lambda)$ be one of the $11$ maximal symplectic groups. Let $g \in O(\Lambda^\sym{G})$ be an isometry. Then
\begin{enumerate}
\item there is a unique extension of $\sym{G}$ by $g$;
\item if $\tilde{g}\in O(\Lambda^\sym{G})$ is conjugate to $g$, then the corresponding extensions are conjugate in $O(\Lambda)$.
\end{enumerate}
\end{lemm}
\begin{proof}
Recall that $\sym{G}$ is a subgroup of the orthogonal group of the K3 lattice $\Lambda$. In particular we have a primitive extension $\Lambda^\sym{G} \oplus \Lambda_\sym{G} \subseteq \Lambda$. Since the K3 lattice is unimodular, this primitive extension is determined by an anti-isometry
\[
\phi \colon \disc{{\Lambda^\sym{G}}} \longrightarrow \disc{{\Lambda_\sym{G}}}.
\]
The natural map $ \psi \colon O(\Lambda_\sym{G})
\rightarrow O(\disc{{\Lambda_\sym{G}}})$, $f \mapsto \bar f$ is surjective (Lemma~\ref{lem:surj}). Hence, we find an $h \in O(\Lambda_\sym{G})$ such that $\overline{h} = \phi \circ \overline{g}\circ \phi^{-1}$. This means that $\tilde{g}= g \oplus h$ extends to an isometry of $\Lambda$. We set $\nsym{G} = \langle \sym{G},\tilde{g} \rangle$. Any other choice of $h$ is of the form $h \cdot (\IDL_{\Lambda^\sym{G}} \oplus f)$ with $f \in \ker \psi \cong \sym{G}$ (Lemma~\ref{lem:symplectic}$\MK$\eqref{lemm4.1.5}). Then $\nsym{G}$ remains unchanged.
We turn to the second claim. Let $f\in O(\Lambda^\sym{G})$ and let $g^f= f^{-1}g f$ be a conjugate of $g$. Take an extension $\tilde{g}$ of $g$ to an isometry of $\Lambda$. We can extend $f$ to an isometry $\tilde{f}=f \oplus f'$ of $\Lambda$ as well (Lemma~\ref{lem:extend}). Since the restriction $\sym{G}|_{\Lambda_\sym{G}}$ is a normal subgroup of $O(\Lambda_\sym{G})$, conjugation by $f$ preserves $\sym{G}$. Further the restriction of $\tilde{g}^{\tilde{f}}$ to ${\Lambda^\sym{G}}$ is equal to $g^f$. Hence, by part 1, the extensions $G^{\tilde{f}}$ and $\langle \tilde{g^f},\sym{G} \rangle$ are equal.
\end{proof}
If $(X,G) \cong (X',G')$ are isomorphic pairs consisting of a $\Lambda$-marked K3 surface with a group of automorphisms, then $G$ and $G'$ (viewed in $O(\Lambda)$ via the marking) are conjugate. In our case the pairs do not deform, so there is hope for the converse statement to hold.
\begin{prop}\label{prop:k3fromgroup}
Let $(X,\eta)$ and $(X',\eta')$ be marked K3 surfaces and $\nsym{G}\linebreak\leq \Aut(X)$, $\nsym{G'}\leq \Aut(X')$ finite subgroups such that $\sym{G}$ and $\sym{G'}$ are isomorphic to one of the $11$ maximal groups. Suppose that $\eta G \eta^{-1}$ and $\eta' G' \eta'^{-1}$ are conjugate in $O(\Lambda)$, then there is an isomorphism $f\colon X \rightarrow X'$ with $G = f^{-1}G'f$, i.e. the pairs $(X,G)$ and $(X',G')$ are isomorphic.
\end{prop}
\begin{proof}
Changing the marking $\eta$ conjugates $\eta G \eta^{-1}$ in $O(\Lambda)$. To ease notation, we identify $G,G'$ with their image in $O(\Lambda)$. In order to use the strong Torelli type Theorem, we have to produce an effective Hodge isometry conjugating $G$ and $G'$.
Let $n$ be the order of $\nsym{G}/\sym{G}$. We choose a primitive $n$th root of unity $\zeta \in \C$. Then $\nsym{G}/\sym{G}$ comes with a distinguished generator $g\sym{G}$ given by $g\left(\eta_\C(\sigma_X)\right)=\zeta \sigma_X$. And likewise $g'\sym{G'}$. By assumption $\nsym{G}$ and $\nsym{G'}$ are conjugate via some $f \in O(\Lambda)$. If $n=2$, then the generators $g\sym{G}$ and $g'\sym{G'}$ are unique. Otherwise $n=3,4,6$, and then $\SO(\Lambda^{\sym{G}})$ is a dihedral group of order $8$ or $12$ (Lemma~\ref{lem:orientation-reversal}). In any case there is a unique conjugacy class of order $n$. Since we can extend any conjugator of the dihedral group to an element of $O(\Lambda)$ (Lemma~\ref{lem:extend}) preserving $\sym{G}$, we may modify the conjugator $f$ in such a way that it conjugates the distinguished generators $g\sym{G}$ and $g'\sym{G'}$ as well. So after conjugation, we may assume that $\nsym{G'} = \nsym{G}$ and further that $g'\sym{G'} = g \sym{G}$.
Suppose that $n > 2$. Then the periods of $X$ and $X'$ are uniquely determined by the distinguished generators as the (1-dimensional!) eigenspaces with eigenvalue\linebreak$\zeta$ of $g|H(X,\C)^\nsym{G}$, respectively $g'|H(X',\C)^{\nsym{G'}}$. And we are done. (Note that if $\sigma$ is an eigenvector for $\zeta \neq \pm 1$, then $\langle \sigma,\sigma \rangle=\zeta^2 \langle \sigma,\sigma \rangle$ implies $\sigma^2=0$.) If $n=2$ then the eigenspace for $-1$ of $g|H(X,\C)^G$ is of dimension~$2$. However, the period is of square zero. Thus the period is one of the two isotropic lines in the eigenspace. These correspond to the two orientations of the transcendental lattice. By Lemma~\ref{lem:orientation-reversal} one can find an isometry $f$ of $\Lambda^{G}$ centralizing $g$ and reversing the orientation. This\linebreak$f$ extends to an isometry of $\Lambda$ preserving $G$. Thus we have obtained a Hodge isometry conjugating $G$ and $G'$.
Note that $\H^2(X,\Z)^G$ is spanned by an ample class $l$ and likewise for $G'$. Since our Hodge isometry conjugates $G$ and $G'$ it maps $l$ to $l'$ or $-l'$. In the second case our Hodge isometry is not effective. However, we may then replace it by its negative.
\end{proof}
\begin{prop}\label{prop:class}
Let $\sym{G} \leq O(\Lambda)$ be a maximal symplectic group. There is a one to one correspondence between conjugacy classes of non trivial cyclic subgroups of $\SO(\syminv)$ and isomorphism classes of pairs $(X,\nsym{G}')$ consisting of a K3 surface $X$ and $G' \leq \Aut(X)$ a finite subgroup with $\sym{G}\cong \sym{G'}