text:
\begin{document}
\begin{abstract}
A singularity is said to be weakly--exceptional if it has a unique purely log
terminal blow up. In dimension $2$, V.~Shokurov proved that
weakly--exceptional quotient singularities are exactly those of types $D_{n}$,
$E_{6}$, $E_{7}$, $E_{8}$. This paper classifies the weakly--exceptional
quotient
singularities in dimensions $3$ and $4$.
\end{abstract}
\maketitle
\setcounter{tocdepth}{1}
\tableofcontents

\section{Introduction}

Let $G\in\slgroup{N}{\complex}$. Then $G$ has a natural linear
action on $\complex^{N}$, and by linearity it induces an action of a group
$\bar{G}$
on \proj{N-1}, where $\bar{G}$ is the image of $G$ in the natural
projection
of \slgroup{N}{\complex} onto \pglgroup{N}{\complex}. Since the two groups
are closely related, I will move from considering $\bar{G}$ to considering
$G$ and back at will, immediately assuming any results proven for $\bar{G}$ as
proven for $G$ (up to scalar multiplication) and vice versa. In this article, I
will assume that the action of $\bar{G}$ is faithful, since any non--faithful
actions will have been looked at when considering a faithful action of a smaller
group.

\subsection{Notation} Here I will define some standard notation that I will be
using throughout this paper:
\begin{itemize}
 \item \cyclgroup{n} is the cyclic group of order $n$.
 \item \dihgroup{2n} is the dihedral group of order $2n$.
 \item \altgroup{n} is the alternating group of degree $n$.
 \item \symgroup{n} is the permutation group of degree $n$.
 \item \bcyclgroup{n}, \bdihgroup{2n}, \baltgroup{n}, \bsymgroup{n} are the
binary versions of the relevant groups, i.e.\ their central extensions by
scalars (see for example~\cite{Springer}).
 \item $\mathbb{V}_{4}$ is the Klein group of size $4$.
 \item $A\times B$ is the direct product of groups $A$ and $B$.
 \item $A\rtimes B$ is the semidirect product of groups $A$ and $B$.
 \item $\zeta_{n}$ is a primitive $n$--th root of unity. Whenever two such are
used in defining generators of the same group, the choice is assumed to be
consistent, i.e.\ $\zeta_{ab}^{a}=\zeta_{b}$ for $a,b\in\integers$.
\end{itemize}
Note that the groups $\mathbb{V}_{4}$, \dihgroup{4} and
$\cyclgroup{2}\times\cyclgroup{2}$
are isomorphic. The notation used for this group will denote the context in
which I am considering it: $\mathbb{V}_{4}$ will be used whenever I am
considering it as a subgroup of \altgroup{4} or \symgroup{4}, and one of the
other two will be used whenever I am considering it on its own.

\subsection{Type of group actions}
Throughout this section, let $H$ be a group acting on $\complex^{N}$. From the
more algebraic point of view, one can can distinguish between several possible
types of action of $H$:
\begin{defin}
The action of $H$ is transitive, if for any non--zero vector $x\in\complex^{N}$,
the $H$--orbit of $x$ spans $\complex^{N}$. Otherwise the action is
intransitive.
\end{defin}
\begin{eg}\label{eg:intrans}
Put the coordinates of $\complex^{4}$ into the form of a $2\times2$ matrix via
\[
\defmap{\left(x,y,u,v\right)}{\left(\begin{array}{cc}x&y\\ u&v
\end{array}\right)}
\]
Now, given $A,B\in\slgroup{2}{\complex}$, can define a
linear action on this space via
\[
\defmapname{[A,B]}{\left(\begin{array}{cc}x&y\\ u&v
\end{array}\right)}{A\left(\begin{array}{cc}x&y\\ u&v
\end{array}\right)B^{T}}
\]
Note that this action preserves a quadric in \proj{3} defined by the vanishing
of the determinant of the matrix form.

For $m,n$ odd positive integers, consider the action of a central extension of the group
$\left(\cyclgroup{n}\times\cyclgroup{m}\right) \rtimes\dihgroup{4}$
generated by:
\[
\left[
\left(\begin{array}{cc}\zeta_{4n}^{2}&0\\0&\zeta_{4n}^{-2}\end{array}\right),
\mbox{id}
\right],\

\left[
\mbox{id},
\left(\begin{array}{cc}\zeta_{4m}^{2}&0\\0&\zeta_{4m}^{-2}\end{array}\right)
\right],
\]\[

\left[
\left(\begin{array}{cc}\zeta_{4n}^{n}&0\\0&\zeta_{4n}^{-n}\end{array}\right),
\left(\begin{array}{cc}\zeta_{4m}^{m}&0\\0&\zeta_{4m}^{-m}\end{array}\right)
\right],\

\left[
\left(\begin{array}{cc}0&1\\-1&0\end{array}\right),
\left(\begin{array}{cc}0&1\\-1&0\end{array}\right)
\right]
\]
(where $\zeta_{k}$ is a primitive $k$--th root of unity). This action is
intransitive, since it preserves the linear subspaces defined by $x=v=0$ and by
$y=u=0$.
\end{eg}
\begin{eg}\label{eg:trans}
In the notation of Example~\ref{eg:intrans}, consider the action of a central
extension of the group $\left(\cyclgroup{n}\times\cyclgroup{m}\right)
\rtimes\dihgroup{4}$
generated by:
\[
\left[
\left(\begin{array}{cc}\zeta_{4n}^{2}&0\\0&\zeta_{4n}^{-2}\end{array}\right),
\mbox{id}
\right],\

\left[
\mbox{id},
\left(\begin{array}{cc}\zeta_{4m}^{2}&0\\0&\zeta_{4m}^{-2}\end{array}\right)
\right],\

\]\[

\left[
\left(\begin{array}{cc}\zeta_{4n}^{n}&0\\0&\zeta_{4n}^{-n}\end{array}\right),
\left(\begin{array}{cc}0&1\\-1&0\end{array}\right)
\right],\

\left[
\left(\begin{array}{cc}0&1\\-1&0\end{array}\right),
\left(\begin{array}{cc}\zeta_{4m}^{m}&0\\0&\zeta_{4m}^{-m}\end{array}\right)
\right]
\]
(cf.\ Lemma~\ref{lemma:sl4:quadric}, \ref{usedin:trans}). It can easily be
checked that this action is transitive.
\end{eg}
Although the groups in Examples~\ref{eg:intrans} and~\ref{eg:trans} look very similar, both being central extensions of quotients of $\dihgroup{4n}\times\dihgroup{4m}$ by normal subgroups isomorphic to \dihgroup{4}, only one of them is transitive and hence interesting for the purpoces of this paper.
\begin{defin}
The action of $H$ is primitive if there does not exist a decomposition of
$\complex^{N}$ into proper linear subspaces, such that the action of $H$
permutes the subspaces. If such a decomposition does exist, then the action is
imprimitive.
\end{defin}
\begin{eg}\label{eg:imprim}
In the notation of Example~\ref{eg:intrans}, consider the action of a central extension of $\left(\mathbb{V}_{4}\times\mathbb{V}_{4}\right)\rtimes\symgroup{3}$ generated by:
\[
\left[
\left(\begin{array}{cc} \zeta_{8}^{2}&0\\ 0&-\zeta_{8}^{2}\end{array}\right),
\mbox{id}
\right],\

\left[
\left(\begin{array}{cc} 0&1\\ -1&0\end{array}\right),
\mbox{id}
\right],\

\]\[

\left[
\mbox{id},
\left(\begin{array}{cc} \zeta_{8}^{2}&0\\ 0&-\zeta_{8}^{2}\end{array}\right)
\right],\

\left[
\mbox{id},
\left(\begin{array}{cc} 0&1\\ -1&0\end{array}\right)
\right],

\]\[

\left[
\frac{1}{\sqrt{2}}\left(\begin{array}{cc} \zeta_{8}^{7}&\zeta_{8}^{7}\\
\zeta_{8}^{5}&\zeta_{8} \end{array}\right),
\frac{1}{\sqrt{2}}\left(\begin{array}{cc} \zeta_{8}^{7}&\zeta_{8}^{7}\\
\zeta_{8}^{5}&\zeta_{8} \end{array}\right)
\right],\

\left[
\left(\begin{array}{cc} 0&\zeta_{8}\\ -\zeta_{8}^{7}&0 \end{array}\right),
\left(\begin{array}{cc} 0&\zeta_{8}\\ -\zeta_{8}^{7}&0 \end{array}\right)
\right]

\]
(cf.\ Lemma~\ref{lemma:sl4:quadric}, \ref{usedin:imprim}). This
group is imprimitive, since its action permutes the four $1$--dimensional
linear subspaces spanned by $\left(1,0,0,1\right)$, $\left(1,0,0,-1\right)$,
$\left(0,1,1,0\right)$ and $\left(0,1,-1,0\right)$ (in the basis used above).
\end{eg}
\begin{eg}\label{eg:prim}
Take the group generated by the group in Example~\ref{eg:imprim} and an
additional element
\[
\left[
\frac{1}{\sqrt{2}}\left(\begin{array}{cc} \zeta_{8}^{7}&\zeta_{8}^{7}\\
\zeta_{8}^{5}&\zeta_{8} \end{array}\right),
\mbox{id}
\right],
\]
turning the group in the previous example into a central extension of
$\left(\altgroup{4}\times\altgroup{4}\right)\rtimes\cyclgroup{2}$
(cf.\ Lemma~\ref{lemma:sl4:quadric}, \ref{usedin:prim}). Then it
can easily be checked that the subspace decomposition given in
\ref{eg:imprim} is no longer valid, and in fact this group action is primitive.
\end{eg}
\begin{defin}
A transitive imprimitive group $H$ is monomial, if any irreducible
representation of $H$ is induced from a linear representation of some subgroup
$H'\leq H$. Otherwise it is non--monomial.
\end{defin}
\begin{eg}\label{eg:imprim:mono}
In the notation of Example~\ref{eg:intrans}, consider the action a central extension of the group
$\left(\left(\mathbb{V}_{4}\times\mathbb{V}_{4}\right) \rtimes\cyclgroup{3}\right) \rtimes\cyclgroup{2}$
generated by transposing the matrix form of $\complex^{4}$ and
\[
\left[
\left(\begin{array}{cc} \zeta_{8}^{2}&0\\ 0&-\zeta_{8}^{2}\end{array}\right),
\mbox{id}
\right],

\left[
\left(\begin{array}{cc} 0&1\\ -1&0\end{array}\right),
\mbox{id}
\right],\

\]\[

\left[
\mbox{id},
\left(\begin{array}{cc} \zeta_{8}^{2}&0\\ 0&-\zeta_{8}^{2}\end{array}\right)
\right],\

\left[
\mbox{id},
\left(\begin{array}{cc} 0&1\\ -1&0\end{array}\right)
\right],

\]\[

\left[
\frac{1}{\sqrt{2}}\left(\begin{array}{cc} \zeta_{8}^{7}&\zeta_{8}^{7}\\
\zeta_{8}^{5}&\zeta_{8} \end{array}\right),
\frac{1}{\sqrt{2}}\left(\begin{array}{cc} \zeta_{8}^{7}&\zeta_{8}^{7}\\
\zeta_{8}^{5}&\zeta_{8} \end{array}\right)
\right]

\]
Take $4$ distinct lines in $\complex^{4}$, spanned by
vectors $\left(1,0,0,1\right)$, $\left(1,0,0,-1\right)$, $\left(0,1,1,0\right)$
and $\left(0,1,-1,0\right)$ (in the basis used above). One can check directly
that the group's action is induced from the action of its subgroup that fixes
these four subspaces. Therefore it is monomial.

The isomorphism class of this group can be given either as
$\left(\cyclgroup{2}\right)^{2}\rtimes\symgroup{4}$ (see
Example~\ref{quadric:example}) or as
$\left(\left(\mathbb{V}_{4}\times
\mathbb{V}_{4}\right)\rtimes\cyclgroup{3}\right)\rtimes\cyclgroup{2}$
(see Lemma~\ref{lemma:sl4:quadric}, \ref{usedin:imprim:mono}).
\end{eg}
\begin{eg}\label{eg:imprim:nonmono}
In the notation of Example~\ref{eg:intrans}, consider the action of a central extension of the group
$\left(\cyclgroup{n}\times\altgroup{4}\right)\rtimes\cyclgroup{2}$
generated by:
\[

\left[
\mbox{id},
\left(\begin{array}{cc} \zeta_{8}^{2}&0\\ 0&-\zeta_{8}^{2}\end{array}\right)
\right],\

\left[
\mbox{id},
\left(\begin{array}{cc} 0&1\\ -1&0\end{array}\right)
\right],

\]
\[
\left[
\mbox{id},
\frac{1}{\sqrt{2}}\left(\begin{array}{cc} \zeta_{8}^{7}&\zeta_{8}^{7}\\
\zeta_{8}^{5}&\zeta_{8} \end{array}\right)
\right],\

\left[
\left(\begin{array}{cc}\zeta_{2n}&0\\0&\zeta_{2n}^{-1}\end{array}\right),
\mbox{id}
\right],\

\]\[

\left[
\left(\begin{array}{cc}0&1\\-1&0\end{array}\right),
\left(\begin{array}{cc} 0&\zeta_{8}\\ -\zeta_{8}^{7}&0 \end{array}\right)
\right]

\]
(cf.\ Lemma~\ref{lemma:sl4:quadric}, \ref{usedin:imprim:nonmono}). The action
of this group is not primitive, since it permutes the $2$--dimensional subspaces
$x=y=0$ and $u=v=0$, but neither is it monomial, since one cannot subdivide
$\complex^{4}$ into four $1$--dimensional linear subspaces that the group
permutes.
\end{eg}
To summarise, the action of $H$ must belong to exactly one of the following
types:
\[
\begin{array}{ll}
\mbox{Intransitive}&\\
\mbox{Transitive}&\left\{
\begin{array}{ll}
\mbox{Primitive}&\\
\mbox{Imprimitive}&\left\{
\begin{array}{l}
\mbox{Monomial}\\
\mbox{Non--monomial}
\end{array}\right.
\end{array}\right.
\end{array}
\]

\subsection{Types of singularities}

From the more geometric point of view, it is more interesting to look at the
variety $\complex^{n}/G$. This variety has a singularity at $G(0)$, which
can be categorised as follows:

\begin{defin}[see~{\cite[Definition~1.14]{Cheltsov-Shramov09}}]
\label{definition:exceptional} Let $(V\ni O)$ be a~germ of a Kawamata log
terminal singularity. The singularity is~said to be exceptional if for every
effective \rationals--divisor $D_{V}$ on the~variety $V$ one of the~following
two conditions~is satisfied:
\begin{itemize}
\item  either the~log pair $(V,D_{V})$ is not log canonical,

\item or for every resolution of singularities \deffunname{\pi}{U}{V}
there exists at most one $\pi$-exceptional divisor $E_{1}\subset
U$~such~that
\[
K_{U}+D_{U}\sim_{\rationals}\pi^{*}\left(K_{V}+D_{V}\right)
+a\left(V,D_{V},E_{1}\right)E_{1}
+\sum_{i=2}^{r}a\left(V,D_{V},E_{i}\right)E_{i},
\]
where $a(V,D_{V},E_{1})=-1$, while $E_{2},\ldots,E_{r}$ are
$\pi$--exceptional divisors different from $E_{1}$, and $D_{U}$ is the~proper
transform of the~divisor $D_{V}$ on the~variety $U$.
\end{itemize}
\end{defin}

\begin{prop}[see~{\cite[Proposition~2.1]{Prokhorov00}}]\label{prop:Exc-prim}
Let $G<\slgroup{N}{\complex}$ be a finite subgroup that induces an exceptional
singularity. Then $G$ is primitive.
\end{prop}
\begin{cor}
For any given $N$, only finitely many finite subgroups of \slgroup{N}{\complex}
induce exceptional singularities.
\end{cor}

One can define a larger class of singularities by:
\begin{defin}
Let $(V\ni O)$ be a~germ of a~Kawamata log terminal singularity,
and let \deffunname{\pi}{W}{V} be~a~birational morphism, such that the~
following hypotheses are satisfied:
\begin{itemize}
\item the~exceptional locus of $\pi$ consists of one irreducible divisor
$E$ such that $O\in\pi(E)$,

\item the~log pair $(W,E)$ has purely log terminal singularities.

\item the~divisor $-E$ is a~$\pi$-ample \rationals--Cartier divisor.
\end{itemize}
We then say that \deffunname{\pi}{W}{V} is a~plt blow up of
the~singularity $(V\ni O)$.
\end{defin}
\begin{thm}[see~{\cite[Theorem~3.7]{Cheltsov-Shramov09}}]
The birational morphism \[\deffunname{\pi}{W}{V}\] does exist.
\end{thm}
\begin{defin}[see~{\cite[Definition~3.8]{Cheltsov-Shramov09}}]
We say that $(V\ni O)$ is weakly--exceptional if it has a unique plt blow up.
\end{defin}
\begin{cor}[see~{\cite[Corollary~3.13]{Cheltsov-Shramov09}}]
If $(V\ni O)$ is exceptional, then $(V\ni O)$ is weakly--exceptional.
\end{cor}
\begin{thm}[see~{\cite[Theorem~1.30]{Cheltsov-Shramov09}}]\label{thm:WExc-trans}
Let $G<\slgroup{N}{\complex}$ be a finite subgroup that induces a
weakly--exceptional singularity. Then $G$ is transitive.
\end{thm}
This result will be used frequently throughout this paper, since for any finite
group $G$, it allows to severely limit the number of actions worth checking.

\begin{eg}[\proj{1}]\label{example:proj1}
As an example, one can consider the case $N=2$. It is a well--known fact that
the only finite subgroups of \Aut{\proj{1}} are:
\begin{itemize}
 \item Cyclic: \cyclgroup{n}, $n\geq1$.
 \item Dihedral: $\dihgroup{2n}=\defsetspanlong{a,b}{a^{n}=b^{2}=\mbox{id},\
bab=a^{-1}}$ ($n\geq2$).
 \item Polyhedral groups \altgroup{4}, \symgroup{4}, \altgroup{5}.
\end{itemize}
Lifting their action to \slgroup{2}{\complex}, get the actions of the following
groups ($\zeta_{n}$ is taken to be a primitive $n$--th root of unity):
\begin{itemize}
 \item Binary cyclic group
\[\bcyclgroup{m}=\defsetspanlong{a}{a^{2n}=1}\]
All its faithful representations are $1$--dimensional, and are of the form
$a\leadsto\zeta_{2n}^{l}$, some $l\in\integers$. Thus a $2$--dimensional
representation has to be a direct sum of two such.
 \item Binary dihedral group
\[\bdihgroup{2n}=\defsetspanlong{a,b}{a^{n}=b^{2}, b^{4}=1, aba^{-1}=a^{-1}}\]
The suitable $2$--dimensional representations of this group are indexed by
different choices of $\zeta_{2n}$. They are:
\[a\leadsto\left(\begin{array}{cc}\zeta_{2n}&0\\0&\zeta_{2n}^{-1}\end{array}
\right), \ b\leadsto\left(\begin{array}{cc}0&1\\-1&0\end{array}\right)\]
 \item Binary tetrahedral group \[\baltgroup{4}=\defsetspan{\zeta_{4}(12)(34),
\zeta_{4}(14)(23), \zeta_{4}(123)}\] (using standard notation for elements of
the symmetric group). Similarly to above, the suitable $2$--dimensional
representations of this group
are determined by the choice of $\zeta_{8}$. They are:
\[
\zeta_{4}(12)(34)\leadsto\left(\begin{array}{cc} \zeta_{8}^{2}&0\\
0&-\zeta_{8}^{2}
\end{array}\right),\ \zeta_{4}(14)(23)\leadsto\left(\begin{array}{cc} 0&1\\ -1&0
\end{array}\right),
\]
\[
\zeta_{4}(234)\leadsto\frac{1}{\sqrt{2}}\left(\begin{array}{cc}
\zeta_{8}^{7}&\zeta_{8}^{7}\\ \zeta_{8}^{5}&\zeta_{8} \end{array}\right)
\]
\item Binary octahedral group \[\bsymgroup{4}=\defsetspan{\zeta_{4}(12)(34),
\zeta_{4}(14)(23), \zeta_{4}(123), \zeta_{4}(34)}\] This group only has $2$
suitable representations, each having a subrepresentation isomorphic to the
representation of $\overline{\altgroup{4}}$ that uses the same value of
$\zeta_{8}$. The extra generator acts as
\[
\zeta_{4}(34)\leadsto\left(\begin{array}{cc} 0&\zeta_{8}\\ -\zeta_{8}^{7}&0
\end{array}\right)
\]
 \item Binary icosahedral group \[\baltgroup{5}=\defsetspan{\zeta_{4}(12345),
\zeta_{4}(12)(34)}\]
\[
\zeta_{4}(12345)\leadsto\left(\begin{array}{cc} \zeta_{5}^{3}&0\\
0&-\zeta_{5}^{2}
\end{array}\right),
\]\[
\zeta_{4}(12)(34)\leadsto\frac{1}{\sqrt{5}}\left(\begin{array}{cc}
-\zeta_{5}+\zeta_{5}^{4}&\zeta_{5}^{2}-\zeta_{5}^{3}\\
\zeta_{5}^{2}-\zeta_{5}^{3}&\zeta_{5}-\zeta_{5}^{4} \end{array}\right)
\]
\end{itemize}

One can see that these group actions are of the following types:
\begin{itemize}
 \item The cyclic groups act intransitively.
 \item \altgroup{4}, \symgroup{4}, \altgroup{5} have primitive actions
 \item Dihedral groups have imprimitive monomial actions.
\end{itemize}

On the geometric side, \altgroup{4}, \symgroup{4} and \altgroup{5} give rise to
exceptional singularities, the dihedral groups give rise to
weakly--exceptional, but not exceptional singularities. The cyclic groups give
rise to singularities that are not weakly--exceptional. For the proofs of these
results, see~\cite{Shokurov92}.
\end{eg}
\begin{cor}
In the $2$--dimensional case, the exceptional singularities are those of types
$E_{6}$, $E_{7}$, $E_{8}$. The weakly--exceptional singularities are those of
types $E_{6}$, $E_{7}$, $E_{8}$, $D_{n}$.
\end{cor}

Since this example fully settles the $N=2$ case, assume for the rest of this paper
that $N\geq3$. When the dimension $N\leq6$, the exceptional singularities have
been fully classified in \cite{Cheltsov-Shramov09}, \cite{Cheltsov-Shramov10}
and \cite{Markushevich-Prokhorov}. The aim of this paper is to obtain a list of
group actions that induce weakly--exceptional singularities in dimension $3$
and $4$. This will be done using the exceptionality criteria obtained by
Y.~Prokhorov, I.~Cheltsov and C.~Shramov in~\cite{Cheltsov-Shramov09}
and~\cite{Prokhorov00}, using the language of Tian's alpha--invariant introduced
in~\cite{Tian87} and~\cite{Tian-Yau87} by G.~Tian and S-T.~Yau.

For dimension $N=3$, the classification of finite subgroups of
\slgroup{3}{\complex} is a classical result by G.~Miller, H.~Blichfeldt and
L.~Dickson (see~\cite{MDB}). A modern exposition of this result has been made
by S.~Yau and Y.~Yu (see~\cite{Yau-Yu}).
The first main result of this paper uses the classification to
obtain an explicit list of groups inducing weakly--exceptional singularities:
\begin{thm}\label{sl3:thm:main}
Let $G<\slgroup{3}{\complex}$ be a transitive finite subgroup. Then $G$ induces
a weakly--exceptional singularity if and only if one of the following holds:
\begin{itemize}
 \item $G$ is conjugate to a monomial group, that is not isomorphic to a central
extension of
$\left(\cyclgroup{2}\right)^{2}\rtimes\cyclgroup{3}$ or
$\left(\cyclgroup{2}\right)^{2}\rtimes\symgroup{3}$.
 \item $G$ is conjugate to an irreducible faithful action of a central extension
of one of:
\begin{itemize}
\item \altgroup{6},
\item Klein's simple group $\mathbb{K}_{168}$ of order $168$,
\item the Hessian group $\mathbb{H}_{648}$ of order $648$,
\item the normal subgroup $F_{216}\lhd\mathbb{H}_{648}$ of order $216$,
\item the normal subgroup $E_{108}\lhd F_{216}$ of order $108$.
\end{itemize}
\end{itemize}
\end{thm}
\begin{rmk}
This is a full classification, since a group that induces a weakly--exceptional
singularity must act transitively (see Theorem~\ref{thm:WExc-trans}).
\end{rmk}
Section~\ref{sect:3D} of this paper is devoted to proving this theorem. The
result settles the $3$--dimensional case, complimenting the earlier result by
D.~Markushevich and Y.~Prokhorov:
\begin{thm}[see~\cite{Markushevich-Prokhorov}]
In the notation of Theorem~\ref{sl3:thm:main}, $G$ induces an exceptional
singularity if and only if it is conjugate to an irreducible
faithful action of a central extension of \altgroup{6}, $\mathbb{K}_{168}$,
$\mathbb{H}_{648}$ or $F_{216}$.
\end{thm}

Since $4$ is not a prime number, \slgroup{4}{\complex} contains significantly
more finite subgroups than \slgroup{3}{\complex} does. The list of finite
primitive subgroups is a classical result, that can be found in H.~Blichfeldt's
book (see~\cite{Blichfeldt}). The list of transitive imprimitive finite
subgroups can be obtained from papers by D.~Flannery (see~\cite{Flannery}) and
B.~H\"{o}fling (see~\cite{Hofling}). In order to avoid reproducing
the tables of groups from the works above, I will instead produce a list of
transitive group actions that give rise to singularities that are not weakly
exceptional. Therefore, the second main result of this paper is:
\begin{thm}\label{sl4:thm:main}
If $G<\slgroup{4}{\complex}$ is a finite transitive subgroup whose action
induces a singularity that is
not weakly--exceptional, then $G$ must be conjugate to a central extension of
one of
(using the notation of Section~\ref{sect:sl4:quadric} where appropriate):
\begin{itemize}

\item A primitive group that is one of:
\begin{itemize}
 \item \symgroup{5}, whose action preserves the diagonal cubic surface.
 \item $\cyclgroup{2}\rtimes\altgroup{5}$, acting as the third symmetric power
of its irreducible faithful $2$--dimensional representation.
 \item Twisted diagonal group
$\left(\altgroup{5},1,\altgroup{5},1\right)_{\alpha}\cong\altgroup{5}$.

\item 9 groups $\left(H_{1},H_{1},H_{2},H_{2}\right)\cong H_{1}\times
H_{2}$, for different choices of
$H_{1},H_{2}\in\defsetshort{\altgroup{4},\symgroup{4},\altgroup{5}}$

 \item 3 families of groups isomorphic to one of:
$\left(\altgroup{4}\times\altgroup{4}\right)\rtimes\cyclgroup{2}$,
$\left(\symgroup{4}\times\symgroup{4}\right)\rtimes\cyclgroup{2}$,
$\left(\altgroup{5}\times\altgroup{5}\right)\rtimes\cyclgroup{2}$.

 \item Groups isomorphic to $\altgroup{5}\rtimes\cyclgroup{2}$.

 \item
$\frac{1}{2}\left[\symgroup{4}\times\symgroup{4}\right]
\cong\left(\altgroup{4}\times\altgroup{4}\right)\rtimes\cyclgroup{2}
\cong\left(\symgroup{4},\altgroup{4},\symgroup{4},\altgroup{4}\right)$.

 \item
$\left(\left(\altgroup{4}\times\altgroup{4}
\right)\rtimes\cyclgroup{2}\right)\rtimes\cyclgroup{2}
\cong\left(\symgroup{4},\altgroup{4},\symgroup{4},\altgroup{4}
\right)\rtimes\cyclgroup{2}$.

\end{itemize}

\item An imprimitive non--monomial group that is one of:
\begin{itemize}
\item 3 families of groups $\dihgroup{2m}\times H_{2}$, where
$H_{2}\in\defsetshort{\altgroup{4},\symgroup{4},\altgroup{5}}$

 \item
$\frac{1}{2}\left[\dihgroup{2m}\times\symgroup{4}\right]
\cong\left(\cyclgroup{m}\times\altgroup{4}\right)\rtimes\cyclgroup{2}
\cong\left(\dihgroup{2m},\cyclgroup{m},\symgroup{4},\altgroup{4}\right)$.

 \item
$\frac{1}{2}\left[\dihgroup{4m}\times\symgroup{4}\right]
\cong\left(\dihgroup{2m}\times\altgroup{4}\right)\rtimes\cyclgroup{2}
\cong\left(\dihgroup{4m},\dihgroup{2m},\symgroup{4},\altgroup{4}\right)$
for $m\geq2$.

 \item
$\frac{1}{6}\left[\dihgroup{6m}\times\symgroup{4}\right]
\cong\left(\cyclgroup{m}\times \mathbb{V}_{4}\right)\rtimes\symgroup{3}
\cong\left(\dihgroup{6m},\cyclgroup{m},\symgroup{4},\mathbb{V}_{4}\right)$.

\end{itemize}

\item A monomial group that is one of:
\begin{itemize}
 \item
$\left(\left(\cyclgroup{3}\right)^{3}\rtimes\cyclgroup{2}
\right)\rtimes\cyclgroup{2}$, acting as shown in
Lemma~\ref{lemma:sl4:cubic:smooth}.
 \item $\left(\cyclgroup{3}\right)^{3}\rtimes\symgroup{4}$, acting as shown in
Lemma~\ref{lemma:sl4:cubic:smooth}.

 \item Twisted diagonal groups $\left(H_{1},1,H_{1},1\right)_{\alpha}\cong
H_{1}$ isomorphic to \altgroup{4} or \symgroup{4}.

 \item 1 family of groups $\dihgroup{2m}\times\dihgroup{2n}$

 \item
$\left(\dihgroup{2n}\times\dihgroup{2n}\right)\rtimes\cyclgroup{2}$

 \item Groups isomorphic to $\dihgroup{4n}\rtimes\cyclgroup{2}$.
 \item Groups isomorphic to $\altgroup{4}\rtimes\cyclgroup{2}$ or
$\symgroup{4}\rtimes\cyclgroup{2}$.

 \item
$\frac{1}{6}\left[\symgroup{4}\times\symgroup{4}\right]
\cong \left(\mathbb{V}_{4}\times \mathbb{V}_{4}\right)\rtimes\symgroup{3}
\cong\left(\symgroup{4},\mathbb{V}_{4},\symgroup{4},\mathbb{V}_{4}\right)$.

 \item
$\frac{1}{3}\left[\altgroup{4}\times\altgroup{4}\right]
\cong\left(\mathbb{V}_{4}\times \mathbb{V}_{4}\right)\rtimes\cyclgroup{3}
\cong\left(\altgroup{4},\mathbb{V}_{4},\altgroup{4},\mathbb{V}_{4}\right)$.

 \item
$\frac{1}{2}\left[\dihgroup{2m}\times\dihgroup{4n}\right]
\cong\left(\cyclgroup{m}\times\dihgroup{2n}\right)\rtimes\cyclgroup{2}
\cong\left(\dihgroup{2m},\cyclgroup{m},\dihgroup{4n},\dihgroup{2n}\right)$
($m,n\geq2$).

 \item
$\frac{1}{4}\left[\dihgroup{4m}\times\dihgroup{4n}\right]_{\alpha}
\cong\left(\cyclgroup{m}\times \cyclgroup{n}\right)\rtimes\dihgroup{4}
\cong\left(\dihgroup{4m},\cyclgroup{m},\dihgroup{4n},\cyclgroup{n}\right)_{
\alpha }$ (where
$\alpha(b)=a_{2n}^{n}$, $\alpha(a_{2m}^{m})=b$).

 \item
$\frac{1}{2}\left[\dihgroup{4m}\times\dihgroup{4n}\right]
\cong\left(\dihgroup{2m}\times\dihgroup{2n}\right)\rtimes\cyclgroup{2}
\cong\left(\dihgroup{4m},\dihgroup{2m},\dihgroup{4n},\dihgroup{2n}\right)$
($m,n\geq2$).

 \item
$\left(\left(\mathbb{V}_{4}\times
\mathbb{V}_{4}\right)\rtimes\symgroup{3}\right)\rtimes\cyclgroup{2}
\cong\left(\symgroup{4},\mathbb{V}_{4},\symgroup{4},\mathbb{V}_{4}
\right)\rtimes\cyclgroup{2}$.

 \item
$\left(\left(\mathbb{V}_{4}\times
\mathbb{V}_{4}\right)\rtimes\cyclgroup{3}\right)\rtimes\cyclgroup{2}
\cong\left(\altgroup{4},\mathbb{V}_{4},\altgroup{4},\mathbb{V}_{4}
\right)\rtimes\cyclgroup{2}$.

 \item
$\left(\left(\cyclgroup{m}\times\cyclgroup{m}\right)\rtimes\dihgroup{4}
\right)\rtimes\cyclgroup{2}
\cong\left(\dihgroup{4m},\cyclgroup{m},\dihgroup{4m},
\cyclgroup{m}\right)_{\alpha}\rtimes\cyclgroup{2}$
(where
$\alpha(b)=a_{2m}^{m}$, $\alpha(a_{2m}^{m})=b$).

 \item
$\left(\left(\dihgroup{2m}\times\dihgroup{2m}\right)\rtimes\cyclgroup{2}
\right)\rtimes\cyclgroup{2}
\cong\left(\dihgroup{4m},\dihgroup{2m},\dihgroup{4m},\dihgroup{2m}
\right)\rtimes\cyclgroup{2}$ for $m\geq2$.

\end{itemize}
\end{itemize}
\end{thm}

\begin{rmk}
As in the $3$--dimensional case, this directly implies a full classification,
since any group inducing a weakly--exceptional singularity must act
transitively
(by Theorem~\ref{thm:WExc-trans}).
\end{rmk}
Section~\ref{sect:4D} is devoted to proving this result. This fully settles the
$4$--dimensional case, since the list of finite subgroups of
\slgroup{4}{\complex} inducing exceptional singularities is known and can be
found in a paper by I.~Cheltsov and C.~Shramov (see~\cite{Cheltsov-Shramov09}).

\section{Preliminaries}

\subsection{The short exact sequence}\label{prelim:SES}
Let $\bar{G}<\pglgroup{N}{\complex}$ be a finite subgroup, and $G$ be its lift
to \slgroup{N}{\complex}. Assume that $S\subset\proj{N-1}$ is a proper
$\bar{G}$--invariant subvariety. Then there exists a natural homomorphism
\deffunname{\pi_{S}}{G}{\Aut{S}}, so get a short exact sequence
\[0\longrightarrow G_{0}\longrightarrow G\stackrel{\pi_{S}}{\longrightarrow}
G_{S}\longrightarrow0\]
where $G_{S}=\pi_{S}\left(G\right)$, $G_{0}=\ker{\pi_{S}}$.

\begin{prop}\label{prop:SES}
Let $G_{0}<\slgroup{N}{\complex}$ be a finite subgroup, and
$\bar{G}_{0}$ the image of its natural embedding into \pglgroup{N}{\complex}.
Let $S\subset\proj{N-1}$ be a subvariety that is not contained in
the union of any two proper linear subspaces of \proj{N-1},
and assume $\bar{G}_{0}$ fixes $S$ pointwise. Then $\bar{G}_{0}$ is trivial.
\end{prop}

\pf  Pick $g\in G_{0}$. Then $\defsetspan{g}$ is a finite
abelian group, and so (in some basis for $\complex^{4}$) consists of diagonal
matrices. Let $\bar{g}$ be the image of $g$ under the natural projection
into \pglgroup{4}{\complex}.

Let $\bar{e}_{1},\ldots,\bar{e}_{N}\in S$ be distinct points, such that
their lifts $e_{1},\ldots,e_{N}$ to $\complex^{N}$ span all of $\complex^{N}$
(these exist, since $S$ is not contained in a proper linear subspace of
\proj{N-1}). Then $e_{i}$ are eigenvectors of $g$, and let $\lambda_{i}$ be the
corresponding eigenvalues. Reordering $e_{i}$--s if necessary, let
$1\leq m\in\integers$ be such that $\lambda_{1}=\ldots=\lambda_{m}$ and
$\lambda_{n}\neq\lambda_{m}$ $\forall m<n\leq N$.

Assume $m<N$. Then take $A\subset\complex^{N}$ to be the linear subspace
spanned by $e_{1},\ldots,e_{m}$ and $B\subset\complex{N}$ to be the
linear subspace spanned by $e_{m+1},\ldots,e_{N}$. Let $\bar{A},\bar{B}$ be
their natural projections into \proj{N-1}. These are proper linear subspaces, so
$\exists \bar{p}\in S\setminus\left(\bar{A}\cup\bar{B}\right)$.

This means there are at least $1+N-m$ distinct linear eigenspaces for $g$ not
contained in $A$, at least one of which is not contained in $B$ either.
Therefore must have $\lambda_{n}=\lambda_{m}$ for some $m<n\leq N$,
contradicting the choice of $m$.

This means $m=N$, and so $g$ is a scalar matrix.\qed

\begin{cor}
Let $G<\slgroup{N}{\complex}$ and let $\bar{G}$ be its natural projection into
\pglgroup{N}{\complex}. Let $S\subset\proj{N-1}$ be a $\bar{G}$--invariant
subvariety, that is not contained in the union of any two proper linear
subspaces of \proj{N-1}. Let \deffunname{\pi_{s}}{\bar{G}}{\Aut{S}} be the
natural homomorphism. Then $\ker{\pi_{S}}=0$.
\end{cor}

\begin{rmk}
If $S\subset\proj{N-1}$ is an irreducible surface, then either it is contained
in an $(N-2)$--dimensional linear subspace of \proj{N-1} or it is not contained
in the union of any two proper linear subspaces of \proj{N-1}.
\end{rmk}

\subsection{Exceptionality criteria}

\begin{defin}[see~{\cite{Cheltsov-Shramov09}}]
Let $X$ be a smooth Fano variety (see~\cite{Iskovskikh-Prokhorov99}) of
dimension $n$, and let $g=g_{ij}$ be a K\"{a}hler metric, such that
\[\omega=\frac{\sqrt{-1}}{2\pi}\sum{g_{ij}\mbox{d}z_{i}\wedge\mbox{d}\bar{z}_{j}
}\in c_{1}\left(X\right) \]
Let $\bar{G}\leq\Aut{X}$ be a compact subgroup, such that $g$ is
$\bar{G}$--invariant.
Let $\mbox{P}_{\bar{G}}\left(X,g\right)$ be the set of $C^{2}$ smooth
$\bar{G}$--invariant functions, such that $\forall
\phi\in\mbox{P}_{\bar{G}}\left(X,g\right)$,
\[\omega+\frac{\sqrt{-1}}{2\pi}\partial\bar{\partial}\phi>0\]
and $\sup_{X}{\phi}=0$. Then the $\bar{G}$--invariant $\alpha$--invariant of
$X$ is
\[
\alpha_{\bar{G}}\left(X\right)=\sup{\defset{\lambda\in\rationals
}{\begin{array}{c}\exists C\in\reals,\ \mbox{such that}\
\forall\phi\in\mbox{P}_{\bar{G}}\left(X,g\right),\\
\int_{X}{e^{-\lambda\phi}\omega^{n}}\leq C \end{array}}}
\]
where $n$ is the dimension of $X$.
\end{defin}
The number $\alpha_{\bar{G}}\left(X\right)$ was introduced in~\cite{Tian87}
and~\cite{Tian-Yau87}.

\begin{defin}[see~{\cite[Definition~2.1]{Cheltsov-Shramov09}}]
Let $X$ be a variety with at most Kawamato log terminal singularities
(see~\cite[Definition~3.5]{Kollar97}) and $D$ an effective $\rationals$--divisor
on $X$. Let $Z\subseteq X$ be a closed non--empty subvariety. Then the log
canonical threshold of $D$ along $Z$ is
\[c_{Z}\left(X,D\right)=\sup{\defset{\lambda\in\rationals}{\mbox{the pair}\
\left(X,\lambda D\right)\ \mbox{is log canonical along}\ Z}}\]
To simplify notation, write $c_{X}\left(X,D\right)=c\left(X,D\right)$.
\end{defin}
There exists an equivalent complex
analytic definition of the log canonical threshold:
\begin{prop}[see~{\cite[Proposition~8.2]{Kollar97}}]
Let $X$ be a smooth complex variety, $Z$ a closed non--empty subscheme of $X$,
and $f$ a non--zero regular function on $X$. Then
\[
c_{Z}\left(X,f\right)=\sup{\defset{\lambda\in\rationals}{ \abs{f}^{-\lambda}\
\mbox{is locally}\ L^{2}\ \mbox{near}\ Z }}
\]
\end{prop}
\begin{rmk}
There exists a very similar definition for real varieties. For more details,
see~\cite[Section~7.2.2]{AGV2} and~\cite{Saito07}.
\end{rmk}

\begin{defin}[see~{\cite[Definition~3.2]{Cheltsov-Shramov09}}]
Let $G$ be a~finite subgroup in \glgroup{N}{\complex},
where $N\geq2$, and let $\bar{G}$ be its image under the natural projection
into \pglgroup{N}{\complex}. Then the global $\bar{G}$--invariant log canonical
threshold of $\proj{N-1}$ is:
\[
\mbox{lct}\left(\proj{N-1},\bar{G}\right)=\inf{\defset{c\left(\proj{N-1},
D\right)}{\begin{array}{l}
D\ \mbox{is a $\bar{G}$--invariant \rationals--Cartier}\\
\mbox{effective \rationals--divisor on \proj{N-1},}\\
\mbox{such that}\ D\sim_{\rationals}-\canclass{\proj{N-1}}
\end{array}}}
\]
\end{defin}
\begin{rmk}[see~{\cite[Theorem~A.3]{Cheltsov-Shramov08}}]
$\mbox{lct}\left(\proj{N-1},\bar{G}\right)=\alpha_{\bar{G}}\left(\proj{N-1}
\right)$
\end{rmk}

\begin{thm}[see {\cite[Theorem~3.15]{Cheltsov-Shramov09}}]
The~singularity $(V\ni O)$ is weakly--exceptional $\iff$
$\mathrm{lct}(\proj{N-1},\bar{G})\geq1$, where $N=\dim{V}$.
\end{thm}
\begin{rmk}
A similar condition is often necessary in order to compute groups of birational
automorphisms of direct products of varieties. For example,
see~\cite[Theorem~6.5]{Cheltsov08}.
\end{rmk}

\begin{thm}[see {\cite[Theorem~3.18]{Cheltsov-Shramov09}}]
\label{thm:criterion:3D} Suppose that
$G<\slgroup{3}{\complex}$ is a finite group and $\bar{G}$ is its image under
the natural projection into \pglgroup{3}{\complex}. Then the~following are
equivalent:
\begin{itemize}
\item the~inequality $\mathrm{lct}(\proj{2},\bar{G})\geq1$ holds,
\item the~group $G$ does not have semi-invariants of degree at most $2$.
\end{itemize}
\end{thm}

\begin{thm}[see {\cite[Theorem~4.1]{Cheltsov-Shramov09}}]
Suppose that $G<\slgroup{4}{\complex}$ is a finite group and $\bar{G}$ is its
image under the natural projection into \pglgroup{4}{\complex}. Then
the~inequality $\mathrm{lct}(\proj{3},\bar{G})\geq1$ holds if and only if
the~following conditions are satisfied:
\begin{itemize}
\item the~group $G$ is transitive,
\item the~group $G$ does not have semi-invariants of degree at most $3$,
\item there is no $\bar{G}$-invariant smooth rational cubic curve in
$\proj{3}$.
\end{itemize}
\end{thm}
Equivalently, the same criterion can be stated as:
\begin{thm}\label{thm:criterion:4D}
Suppose that $G<\slgroup{4}{\complex}$ is a finite group and $\bar{G}$ is its
image under the natural projection into \pglgroup{4}{\complex}. Then
the~inequality $\mathrm{lct}(\proj{3},\bar{G})\geq1$ holds if and only if
the~following conditions are satisfied:
\begin{itemize}
\item the~group $G$ is transitive,
\item the~group $G$ does not have semi-invariants of degree at most $3$,
\item $G$ is not the central extensions of \altgroup{5} acting as the third symmetric power of its irreducible $2$--dimensional representation.
\end{itemize}
\end{thm}
\pf
Assume $\bar{G}<\pglgroup{4}{\complex}$ is a finite transitive subgroup that
does not fix any quadric or cubic surface, and
$C\subset\proj{3}$ is a smooth rational $\bar{G}$--invariant cubic curve.

It is a well--known fact that any smooth rational cubic curve is projectively
isomorphic to the rational normal curve, which is the image of
\[\defmapname{\mathcal{C}}{\left(x:y\right)\in\proj{1}
}{\left(x^{3}:x^{2}y:xy^{2}:y^{3}\right)\in\proj{3}}\]
This means that $\bar{G}$ must be isomorphic to one of the finite automorphism
groups of \proj{1}, with its action induced by the action of \proj{1} via
$\mathcal{C}$. This means $\bar{G}$ must be one of the following:
\begin{itemize}
 \item Cyclic or dihedral group: neither these groups nor their central
extensions by scalar matrices in \slgroup{4}{\complex} have an irreducible
$4$--dimensional representation. Therefore such a $G$ would be intransitive.
 \item \altgroup{4} or \symgroup{4}: The induced action of these two groups
preserves \defset{\left(x:y:u:v\right)\in\proj{3}}{xy-uv=0}, which is a smooth
quadric surface.
 \item \altgroup{5}: This action is primitive. From the embedding of $C$ into
\proj{3}, it is easy to see that the action on $\complex^{4}$ must correspond to
the third symmetric power of the irreducible $2$--dimensional representation of
\baltgroup{5}. or its central extension.
\end{itemize}
\qed
\begin{rmk}
The last condition is necessary, since this action is transitive and does not
preserve any
projective surfaces of degree $2$ or $3$ (can be checked directly or seen in
\cite[Proof of Lemma~4.9]{Cheltsov-Shramov09}).
\end{rmk}

\begin{eg}[cf. Example~\ref{eg:simple:after}]\label{eg:simple:before}
Let
\[S=\defset{\left(x:y:u:v\right)\in\proj{3}}{x^{m}+y^{m}+u^{m}+w^{m}=0},\]
where $m\geq5$. Let $\bar{G}=\Aut{S}$, and $G$ be the lift of $\bar{G}$ to
\slgroup{4}{\complex}. Then $G$ induces a weakly--exceptional singularity.

By Theorem~\ref{thm:criterion:4D}, it is enough to prove that $G$ is
transitive, has no semiinvariants of degree at most $3$, and $\bar{G}$ does not
leave a smooth rational cubic curve invariant. Let $d$ be the minimal degree of
a semiinvariant of $G$.

It can be observed that $G$ is a central extension of
$\left(\cyclgroup{m}\right)^{3}\rtimes\bsymgroup{4}$, acting
by
permuting the basis and multiplying the coordinates by $m$--th roots
of unity. This means $G$ is transitive, i.e.\ $d>1$.

Consider
$\defmapname{\alpha}{\left(x,y,u,v\right)}{\left(\zeta_{m}x,\zeta_{m}^{-1}y,u,
v\right)}$. By applying $\alpha$ and the element of $G$ switching
coordinates~$x$ and~$y$ to the general quadric or cubic polynomial in $4$
variables, it is easy to see that any semiinvariant of degree $2$ or $3$ must be
of the form $xyp_{0}(u,v)+p_{2}(u,v)$ or $xyp_{1}(u,v)+p_{3}(u,v)$ respectively,
where $p_{n}$ are polynomials of degree $n$ in two variables. Applying other
elements of $G$ that permute the basis, one can easily deduce that neither of
these can be semiinvariant of $G$. (By similar considerations, one can see that
any semiinvariant of $G$ of degree $4$ must be a scalar multiple of $xyuv$
which in fact is a semiinvariant of $G$).

It remains to prove that $\bar{G}$ does not leave any smooth rational cubic
curve invariant. Assume it does, and call this curve $C$. Since $C$ is
rational, its automorphism group is isomorphic to the automorphism groups of
\proj{1}. Then let \deffunname{\pi}{\bar{G}}{\Aut{C}} be the natural
homomorphism with image $\bar{G}_{C}$ and kernel $H\unlhd\bar{G}$. Then
$\bar{G}_{C}\cong\bar{G}/H$. Checking all possible normal subgroups of $\bar{G}$
against the list of finite automorphism groups of \proj{1}, it is easy to see
that $\bar{G}_{C}$ must be one of \symgroup{4}, \symgroup{3} and
$\cyclgroup{2}$. But that would mean that $\alpha$ (as defined above)
leaves $C$ pointwise invariant, and hence $C$ must be contained in a proper
linear subspace of \proj{3}, which $G$ would have to preserve, contradicting
its transitivity. Therefore no such $C$ exists, and $G$ induces a
weakly--exceptional singularity.
\end{eg}

\subsection{Effect of transitivity on an action's possible invariants}
Let $\bar{G}<\pglgroup{N-1}{\complex}$, $N>2$, be a finite transitive group and
$G$ its lift to \slgroup{N}{\complex}. Let $S\subset\proj{N-1}$ be a
subvariety left invariant by all of $\bar{G}$.
\begin{rmk}\label{rmk:excluding-1} In the notation above, the following hold:
\begin{itemize}
 \item $\bar{G}$ cannot be cyclic, as in that case, in some basis for
$\complex^{N}$, $G$ must consist of diagonal matrices (as all irreducible
representations of abelian groups are $1$--dimensional), and hence would not be
transitive.
 \item $S$ cannot be contained in a proper linear subspace of \proj{N-1}, since
in that case $G$ would have to fix a non--empty proper linear subspace of
$\complex^{N}$ and so would not be acting transitively. In particular, $\bar{G}$
cannot fix any point, line (if $N\geq3$) or degree $1$ surface (if $N\geq4$) in
\proj{N-1}.
\end{itemize}
\end{rmk}

\begin{rmk}
In the notation above, if $p\in S$ is an isolated singularity, then $\forall
\bar{g}\in \bar{G}$, $\bar{g}(p)\in S$
is an isolated singularity of the same type. For example, if $S$ is a cubic
surface in \proj{3} and $p\in S$ is an $A_{2}$ singularity, then so is
$\bar{g}(p)$ for every $\bar{g}\in\bar{G}$. Similarly, a line on $S$ must be
mapped into another line on $S$.
\end{rmk}

\begin{lemma}\label{lemma:singNo}
In the notation above, let $S$ have only isolated singularities. Then for any
given singularity type, either $S$ has no singularities of this type or is has
at least $N$ of them.
\end{lemma}
\pf
Let $S$ have $k$ ($1\leq k<N$) singularities of some given type, positioned at
$p_{1},\ldots,p_{k}$. Since $S$ is $\bar{G}$--invariant,
\[\bar{G}\left(\defsetshort{p_{1},\ldots,p_{k}}\right)=
\defsetshort{p_{1},\ldots,p_{k}}\]
Therefore, $p_{i}$ span a non--empty invariant linear subspace of
\proj{N}, which is proper since $k<N$. This (by Remark~\ref{rmk:excluding-1})
contradicts the transitivity of $G$.\qed

\section{Three-dimensional case}\label{sect:3D}
Assume $N=3$ throughout this section. This section is devoted to proving
Theorem~\ref{sl3:thm:main}.

The list of all finite subgroups of \slgroup{3}{\complex} have been given
in~\cite{Yau-Yu}. It is:
\begin{prop}[see~{\cite[Theorem~A]{Yau-Yu}}]\label{sl3:prop:classification}
Up to conjugacy, any finite subgroup of \slgroup{3}{\complex} belongs to one of
the following types:
\begin{enumerate}[(A)]
 \item\label{gptype:A} Diagonal abelian group.
 \item\label{gptype:B} Group isomorphic to a transitive finite subgroups of
\glgroup{2}{\complex} and not conjugate to a group of type~(\ref{gptype:A}).
 \item\label{gptype:C} Group generated by the group in (\ref{gptype:A}) and $T$
and not conjugate to a group of types~(\ref{gptype:A}) or (\ref{gptype:B}).
 \item\label{gptype:D} Group generated by the group in (\ref{gptype:C}) and $Q$
and not conjugate to a group of types~(\ref{gptype:A})---(\ref{gptype:C}).
 \item\label{gptype:E} Group of order $108$ generated by $S$, $T$ and $V$.
 \item\label{gptype:F} Group of order $216$ generated by the group in
(\ref{gptype:E}) and an element $P\defby UVU^{-1}$.
 \item\label{gptype:G} Hessian group of order $648$ generated by the group in
(\ref{gptype:E}) and $U$.
 \item\label{gptype:H} Simple group of order $60$ isomorphic to alternating
group \altgroup{5}.
 \item\label{gptype:I} Simple group of order $168$ isomorphic to permutation
group generated by $\left(1234567\right)$, $\left(142\right)\left(356\right)$,
$\left(12\right)\left(35\right)$.
 \item\label{gptype:J} Group of order $180$ generated by the group in
(\ref{gptype:H}) and $W$.
 \item\label{gptype:K} Group of order $504$ generated by the group in
(\ref{gptype:I}) and $W$.
 \item\label{gptype:L} Group $G$ of order $1080$ with its quotient
$G/\defsetspan{W}$ isomorphic to the alternating group \altgroup{6}.
\end{enumerate}
Where the matrices used are given in Figure~\ref{fig:matrices}.
\end{prop}
\begin{figure}[htbp]
\caption{Matrices used in the finite subgroup
classification}\label{fig:matrices}
\[\begin{array}{lll}
S=\left(\begin{array}{ccc}1&0&0\\0&\omega&0\\0&0&\omega^{2}\end{array}\right)&
T=\left(\begin{array}{ccc}0&1&0\\0&0&1\\1&0&0\end{array}\right)&
V=\frac{1}{\sqrt{-3}}\left(\begin{array}{ccc}1&1&1\\1&\omega&\omega^{2}
\\1&\omega^{2}&\omega\end{array}\right)\\
&&\\
U=\left(\begin{array}{ccc}\epsilon&0&0\\0&\epsilon&0\\0&0&\epsilon\omega
\end{array}\right)&
Q=\left(\begin{array}{ccc}a&0&0\\0&0&b\\0&c&0\end{array}\right)&
P=\frac{1}{\sqrt{-3}}\left(\begin{array}{ccc}1&1&\omega^{2}
\\1&\omega&\omega\\\omega&1&\omega\end{array}\right)\\
&&\\
W=\left(\begin{array}{ccc}\omega&0&0\\0&\omega&0\\0&0&\omega\end{array}
\right)&\\
\end{array}\]
\begin{center}
Where $\omega=e^{2\pi i/3}$, $\epsilon^{3}=\omega^{2}$ and $abc=-1$ in such a
way that $Q$ generates a finite group.
\end{center}
\end{figure}
These groups can be put into the form of the algebraic classification of groups
given in the earlier sections as follows:
\begin{rmk} The list of groups in Proposition~\ref{sl3:prop:classification} can
be subdivided as follows:
\begin{itemize}
 \item Groups of types (\ref{gptype:A}) and (\ref{gptype:B}) are intransitive.
 \item Groups of types (\ref{gptype:E})---(\ref{gptype:L}) are primitive.
 \item Groups of types (\ref{gptype:C}) and (\ref{gptype:D}) are monomial.
\end{itemize}
\end{rmk}
\pf Immediate from the lists of generators given in~\cite{Yau-Yu}.\qed

Let $G<\slgroup{3}{\complex}$ be a transitive finite subgroup with image
$\bar{G}$ under the natural projection into \pglgroup{3}{\complex}.
\begin{lemma}\label{lemma:sl3:badGroups}
 Assume the singularity the action of $G$ induces is not weakly--exceptional.
Then:
\begin{itemize}
 \item $\bar{G}$ leaves a smooth conic $C\subset\proj{2}$ invariant.
 \item $G$ is isomorphic to one of \dihgroup{2n},
\altgroup{4}, \symgroup{4}, \altgroup{5} (some $n\geq2$) or to one of their
central extensions by scalar matrices.
\end{itemize}
\end{lemma}
\pf By Theorem~\ref{thm:criterion:3D}, $\bar{G}$ must preserve a conic
$C\subset\proj{2}$. If $C$ is singular, then it must have exactly one isolated
singularity, which is impossible by Lemma~\ref{lemma:singNo}. Therefore, $C$
must be smooth.

Any smooth conic in \proj{2} is rational and is not contained in any proper
linear subspace of \proj{2}, so by Proposition~\ref{prop:SES}, $\bar{G}$ must be
one of the finite transitive subgroups of \Aut{\proj{1}}, and $G$ must be its
central extension.\qed

\pf[ of Theorem~\ref{sl3:thm:main}] Groups of types~(\ref{gptype:A}) and
(\ref{gptype:B}) can be excluded immediately, since they are not transitive.
Assume the singularity $G$ induces is not weakly--exceptional. Then, comparing
the lists in Proposition~\ref{sl3:prop:classification} and
Lemma~\ref{lemma:sl3:badGroups}, $G$ must be conjugate to one of:
\begin{itemize}
 \item A central extension of
$\left(\cyclgroup{2}\right)^{2}\rtimes\cyclgroup{3}\cong\altgroup{4}$, where both
\cyclgroup{2}--s act diagonally and \cyclgroup{3} permutes
the
basis (such a group is of type~(\ref{gptype:C})).
 \item A central extension of
$\left(\cyclgroup{2}\right)^{2}\rtimes\symgroup{3}\cong\symgroup{4}$, where both
\cyclgroup{2}--s act diagonally and \symgroup{3} permutes the basis
(such a group is of type~(\ref{gptype:D})).
 \item A central extension of \altgroup{5} (such a group is of
type~(\ref{gptype:H}) or (\ref{gptype:J})).
\end{itemize}
Let the \proj{2} that $\bar{G}$ is acting on have coordinates
$\left(x:y:z\right)$. In each of the cases, fix $G$ to be the member of its
conjugacy class, whose generators are given in~\cite{Yau-Yu}. Looking at the
generators, it is immediate that the groups of types~(\ref{gptype:C})
and~(\ref{gptype:D}) in this list leave the smooth conic defined by
$x^{2}+y^{2}+z^{2}=0$ invariant. Similarly, the groups of types~(\ref{gptype:H})
or (\ref{gptype:J}) leave the smooth conic defined by $x^{2}-yz=0$ invariant.

The statement of Theorem~\ref{sl3:thm:main} follows by renaming the remaining
groups to fit with the more widely--used notation.
\qed

\section{Four-dimensional case}\label{sect:4D}
The aim of this section is to prove Theorem~\ref{sl4:thm:main}. Throughout the
section, let $G<\slgroup{4}{\complex}$ be a finite transitive subgroup, and
$\bar{G}$ its image under the natural projection
\deffun{\slgroup{4}{\complex}}{\pglgroup{4}{\complex}}.

In view of Theorem~\ref{thm:criterion:4D}, the only transitive $\bar{G}$
that are not weakly--exceptional are the finite subgroups of automorphism
groups
of surfaces of degree $2$ or $3$ and \altgroup{5} with one specific action.
When considering automorphisms of a surface $S$, without loss of generality one
can assume that there is no
$\bar{G}$--invariant surface $S'$ of smaller degree.
\begin{lemma} Let $S\subset\proj{3}$ be a $\bar{G}$--invariant surface of
minimal degree. Then either $\deg{S}\geq4$ or $S$ is smooth.
\end{lemma}
\pf Since $G$ is transitive, $\deg{S}\geq2$. If $\deg{S}=2$ and $S$ is singular,
then $S$ has exactly $1$ isolated singularity. This is impossible (by
Lemma~\ref{lemma:singNo}), so if $\deg{S}=2$ then $S$ must be smooth.

If $\deg{S}=3$ and $S$ has a singular curve $C$, then, since the surface $S$ is
a $\bar{G}$--invariant surface of minimal degree and hence irreducible, by
\cite[Case E]{Bruce-Wall} $C$ is a line. Since $\bar{G}(S)=S$, $\bar{G}(C)=C$,
and so there exists a $\bar{G}$--invariant line, contradicting transitivity of
$G$. Therefore if $\deg{S}=3$ then $S$ must have at worst isolated
singularities.

If $\deg{S}=3$ and $S$ is singular with only isolated singularities,
then by~\cite{Bruce-Wall}, the singularity types form one of the following
collections:
$\left(A_{1}\right)$,
$\left(2A_{1}\right)$,
$\left(A_{1},A_{2}\right)$,
$\left(3A_{1}\right)$,
$\left(A_{1},A_{3}\right)$,
$\left(2A_{1},A_{2}\right)$,
$\left(4A_{1}\right)$,
$\left(A_{1},A_{4}\right)$,
$\left(2A_{1},A_{3}\right)$,
$\left(A_{1},2A_{2}\right)$,
$\left(A_{1},A_{5}\right)$.
Since by Lemma~\ref{lemma:singNo}, $S$ cannot have exactly $1$, $2$ or $3$
singularities of any given type, $S$ has to have $4$~$A_{1}$ singularities.
Since there is only one such surface (see, for example, \cite{Bruce-Wall}), $S$
must be the Cayley cubic, defined by
\[S=\defset{\left(x:y:u:v\right)\in\proj{3}}{xyu+xyv+xuv+yuv=0}\]

But in this case $S$ contains exactly $9$ lines, $6$ of which are going through
pairs of singular points and the other $3$ defined by
\[x+y=0=u+v,\ x+u=0=y+v\ \mbox{and}\ x+v=0=y+u\]
These last three lines are coplanar and must be mapped to each other by all of
$\bar{G}$. Therefore, $\bar{G}$ preserves a plane, contradicting the
transitivity assumption for $G$. Thus if $S$ is a cubic surface, then it must be
smooth.
\qed

Summarising, if $\bar{G}$ is transitive but the singularity it
induces is not weakly--exceptional, any $\bar{G}$--invariant surface $S$ must be
a smooth surface of degree $2$ or $3$. These cases will be considered separately
in the next two sections.

\pf[ of Theorem~\ref{sl4:thm:main}] By Theorem~\ref{thm:criterion:4D} and
the discussion above, the theorem is an immediate consequence of
Lemmas~\ref{lemma:sl4:cubic:smooth} and~\ref{lemma:sl4:quadric}.\qed

\subsection{If $S$ is a smooth $\bar{G}$--invariant cubic surface}
This section is devoted to proving the following lemma:
\begin{lemma}\label{lemma:sl4:cubic:smooth}
If $\bar{G}<\pglgroup{4}{\complex}$ is a finite transitive subgroup that does
not fix any quadric surface, $G$ its lift to \slgroup{4}{\complex} and
$S\subset\proj{3}$ a smooth $\bar{G}$--invariant cubic surface, then $G$ must be
isomorphic to a central extension of one of:
\begin{itemize}
 \item $\left(\left(\cyclgroup{3}\right)^{3}
\rtimes\cyclgroup{2}\right) \rtimes\cyclgroup{2}$. This produces a
monomial action.
 \item $\left(\cyclgroup{3}\right)^{3}\rtimes\symgroup{4}$. This
produces a monomial action.
 \item \symgroup{5}. This produces a primitive action.
\end{itemize}
by scalar matrices, acting as described below.
\end{lemma}
As stated before, $\bar{G}<\Aut{S}$ is a finite subgroup, so by \cite{Hosoh},
$\bar{G}$
must belong to one of the following isomorphism classes:
\begin{enumerate}
 \item\label{groups:cubic:cyclic} \defsetshort{e}, \cyclgroup{2},
\cyclgroup{4}, \cyclgroup{8}.
 \item $\left(\cyclgroup{2}\right)^{2}$
 \item \symgroup{3}, the symmetric group of degree $3$.
 \item $\symgroup{3}\times\cyclgroup{2}$
 \item \symgroup{4}
 \item $\left(\cyclgroup{3}\right)^{2}
\rtimes\cyclgroup{2}$
 \item $\left(\left(\cyclgroup{3}\right)^{3}
\rtimes\cyclgroup{2}\right)
\rtimes\cyclgroup{2}$
 \item \symgroup{5}
 \item $\left(\cyclgroup{3}\right)^{3}\rtimes\symgroup{4}$
\end{enumerate}

\subsubsection{Case $\bar{G}$ cyclic}
The groups in (\ref{groups:cubic:cyclic}) are all cyclic, so have a degree $1$
invariant surface.
\subsubsection{Case $\bar{G}\cong\left(\cyclgroup{2}\right)^{2}$}
This is a dihedral group of order $4$, and neither it nor its central extension
by a group of scalar matrices in \slgroup{4}{\complex} have an irreducible
$4$--dimensional representation. Therefore this group does not act transitively.
\subsubsection{Case $\bar{G}\cong\symgroup{3}$}
This is the dihedral group of order $6$, and neither it nor its central
extension by a group of scalar matrices in \slgroup{4}{\complex} have a
$4$--dimensional irreducible representation. Therefore the action of this group
is intransitive.
\subsubsection{Case
$\bar{G}\cong\symgroup{3}\times\cyclgroup{2}$}
This is the dihedral group of order $12$, so, similarly to the case above, the
corresponding action is intransitive.
\subsubsection{Case $\bar{G}\cong\symgroup{4}$} This group by itself has no
$4$--dimensional irreducible representations, while its central extension has
(up to a choice of a root of unity) only one such, which preserves a quadric
surface (see the twisted diagonal actions).
\subsubsection{Cases $\bar{G}\cong\left(\cyclgroup{3}\right)^{2}
\rtimes\cyclgroup{2}$ and
$\bar{G}'\cong\left(\left(\cyclgroup{3}\right)^{3}
\rtimes\cyclgroup{2}\right)
\rtimes\cyclgroup{2}$}
Write $G'$ for the lift of $\bar{G}'$ to \slgroup{4}{\complex}. As stated in
\cite{Hosoh}, these two cases correspond to groups
$\bar{G}=G_{54}^{9}/C\left(G_{54}^{9}\right)$ and
$\bar{G}'=G_{54}^{9}\rtimes\cyclgroup{2}$. This means there exist
elements
$\bar{\alpha},\bar{\beta},\bar{\gamma},\bar{\delta},\bar{\epsilon} \in\bar{G}'$,
such that $\bar{\alpha},\bar{\beta},\bar{\gamma},\bar{\delta}$ generate
$G_{54}^{9}$, with $\bar{\alpha}$ generating its centre,
$\bar{\alpha},\bar{\beta},\bar{\gamma}$ of order $3$,
$\bar{\delta},\bar{\epsilon}$ of order $2$. Let
$\alpha,\beta,\gamma,\delta,\epsilon$ be lifts of
$\bar{\alpha},\bar{\beta},\bar{\gamma},\bar{\delta}$ (respectively) to
\slgroup{4}{\complex}.

Let $h_{1}\defby\alpha^{3}$, $h_{2}\defby\beta^{3}$. $\bar{\alpha},\bar{\beta}$
commute, so say $\beta\alpha=\alpha\beta h_{3}$. By the structure of the lift,
$h_{i}$ are scalar matrices of order $1$, $2$ or $4$. Then
\[h_{1}^{3}h_{2}=\left(\alpha^{2}\beta\alpha\right)^{3}=
\left(\beta h_{1}h_{3}\right)^{3}=h_{2}h_{1}^{3}h_{3}^{3}\]
and so $h_{3}=\mbox{id}$. Similarly, get $\alpha,\beta,\gamma$ all commuting.
Hence the corresponding matrices can all be taken to be diagonal (by choosing a
suitable basis). It is then easy to see that $\delta$ and $\epsilon$ must act
as elements of a central extension of \symgroup{4} permuting the basis.

Since $\bar{G}'$ has only one normal subgroup of index $2$, and $\bar{G}'$ has
no centre (otherwise $\bar{G}'/C\left(\bar{G}'\right)$ would be on the list of
groups acting on a cubic surface),
$\bar{\delta}\bar{\epsilon}\neq\bar{\epsilon}\bar{\delta}$. Therefore, up to
conjugation, $\delta$ interchanges the first and the second basis vectors, and
$\epsilon$ interchanges the first basis vector with the third one and the second
basis vector with the fourth one.

This means that $G$ is not transitive, while $G'$ is transitive and (up
to conjugation) is generated by
\[
\left(\begin{array}{cccc}
\zeta_{3}&0&0&0\\ 0&1&0&0\\ 0&0&1&0\\ 0&0&0&\zeta_{3}^{-1}
\end{array}\right),\

\left(
\begin{array}{cccc}
1&0&0&0\\ 0&\zeta_{3}&0&0\\ 0&0&1&0\\ 0&0&0&\zeta_{3}^{-1}
\end{array}
\right),\

\left(\begin{array}{cccc}
1&0&0&0\\ 0&1&0&0\\ 0&0&\zeta_{3}&0\\ 0&0&0&\zeta_{3}^{-1}
\end{array}\right),\
\]
\[
\zeta_{8}\left(\begin{array}{cccc}
0&1&0&0\\ 1&0&0&0\\ 0&0&1&0\\ 0&0&0&1
\end{array}\right),\

\left(\begin{array}{cccc}
0&0&1&0\\ 0&0&0&1\\ 1&0&0&0\\ 0&1&0&0
\end{array}\right)
\]

This group clearly leaves the cubic polynomial $x^{3}+y^{3}+z^{3}+w^{3}$ (in
coordinates $\left(x,y,z,w\right)$ for $\complex^{3}$) semi--invariant. This
action of $G'$ is monomial.
\subsubsection{Case $\bar{G}\cong\symgroup{5}$}
According to~\cite{Segre}, this is the automorphism group of the
irreducible diagonal cubic surface
\[
S=\defset{\left(x_{0}:x_{1}:x_{2}:x_{3}:x_{4}\right)\in\proj{4}}{
\begin{array}{c} x_{0}^{3}+x_{1}^{3}+x_{2}^{3}+x_{3}^{3}+x_{4}^{3}=0,\\
x_{0}+x_{1}+x_{2}+x_{3}+x_{4}=0\end{array}}
\]
which is embeddable into \proj{3} via
\[
\defmap{\left(x_{0}:x_{1}:x_{2}:x_{3}:x_{4}\right)}{
\left(x_{0}-x_{1}:x_{1}-x_{2}:x_{2}-x_{3}:x_{3}-x_{4}\right) }
\]
In this case $G$ acts by permuting the basis of the $\complex^{5}$ containing
the \proj{4}. This action descends to the relevant $\complex^{4}$ as
\[
(12)\leadsto\zeta_{8}\left(\begin{array}{cccc} -1&1&0&0\\ 0&1&0&0\\ 0&0&1&0\\
0&0&0&1
\end{array}\right),\
(2345)\leadsto\zeta_{8}^{7}\left(\begin{array}{cccc} 1&0&0&0\\ 1&0&0&-1\\
0&1&0&-1\\ 0&0&1&-1
\end{array}\right)
\]
where $\zeta_{8}$ is a primitive $8$--th root of unity. One should note that
this action is primitive and irreducible. Furthermore, the only other
$4$--dimensional irreducible representation of \symgroup{5} has a
semi--invariant quadric (see the section on quadric surfaces).

\subsubsection{Case
$\bar{G}\cong\left(\cyclgroup{3}\right)^{3}\rtimes\symgroup{4}$}
As stated in \cite{Segre}, this group acts by permuting the basis vectors of
$\complex^{4}$ arbitrarily and multiplying them by arbitrary cube roots of
unity. Hence (up to cinjugation) $G$ is a central extension of such a group by
scalar matrices.

This group clearly leaves the cubic polynomial $x^{3}+y^{3}+z^{3}+w^{3}$ (in
coordinates $\left(x,y,z,w\right)$) semi--invariant. The action of this group is
monomial.

\subsection{If $S$ is a smooth $\bar{G}$--invariant quadric surface}\label{sect:sl4:quadric}
Let the group $\bar{G}<\pglgroup{4}{\complex}$ be a finite irreducible
transitive
subgroup, $G$ its lift to \slgroup{4}{\complex} and $S\subset\proj{3}$ a smooth
$\bar{G}$--invariant quadric surface. This section is devoted to creating a
list of the possible values that $G$ (equivalently, $\bar{G}$) can take in this
situation. The final list is presented in Lemma~\ref{lemma:sl4:quadric}.

In this case there exists a basis for \proj{3}, in which $S$ is the image of
the Veronese embedding
\[\defmapname{\mathcal{V}}{\left(\left(a:b\right),\left(c:d\right)\right)}
{\left(ac:ad:bc:bd\right)}\]
of $\proj{1}\times\proj{1}$ into \proj{3}. The image of this embedding is
\[S=\defset{\left(x:y:u:v\right)\in\proj{3}}{xv-yu=0}\]

It is possible to present \proj{3} as the set of non--zero $2\times2$ matrices.
From now on, consider the ``matrix form'' of \proj{3} to be equivalent to its
standard presentation. Name the coordinates as follows:
\[\left(\begin{array}{cc}x&y\\u&v\end{array}\right)\equiv \left(x:y:u:v\right)\]
Then $S$ takes the form of the set of matrices with zero determinant, and the
automorphism group of $S$ is the group of operations on the $2\times2$ matrix
leaving the determinant semiinvariant. $\bar{G}$ must then be conjugate to the
image of a finite subgroup $\hat{H}$ of this group.

$\proj{1}\times\proj{1}$ has exactly two rulings, which $H$ can either preserve
or interchange. Consider the short exact sequence
\[1\longrightarrow H\longrightarrow \hat{H}\longrightarrow
\symgroup{2}\longrightarrow1\]
where \symgroup{2} permutes the two rulings. Then $H\unlhd\hat{H}$ is the
maximal subgroup that preserves the ruling. Let $\tau\in\hat{H}\setminus H$ if
this set is non--empty. Note that if $\tau$ exists, it can be chosen to be an
involution. Let \deffunname{\pi_{i}}{H}{H_{i}} be the projections of $H$ on the
two components of $\proj{1}\times\proj{1}$. Then have two more short exact
sequences
\[1\longrightarrow K_{2}\longrightarrow H\longrightarrow H_{1}\longrightarrow1\]
\[1\longrightarrow K_{1}\longrightarrow H\longrightarrow H_{2}\longrightarrow1\]
It is clear that $K_{1}\cap K_{2}=\defsetshort{1}$. Therefore, for $i\neq j$
($i,j\in\defsetshort{1,2}$),
\[
K_{i}\cong\hat{K}_{i}\defby\defset{kK_{j}}{k\in K_{i}}\unlhd H/K_{j}=H_{i}
\]
In this notation, $H_{1}/\hat{K}_{1}\cong H/\left(K_{1}K_{2}\right)\cong
H_{2}/\hat{K}_{2}$, so the group can be defined completely by the 5--tuple
$\left(H_{1},K_{1},H_{2},K_{2},\alpha\right)$, where $\alpha$ is an isomorphism
\deffun{H_{1}/\hat{K}_{1}}{H_{2}/\hat{K}_{2}}. In return, if $H$ is known, one
can reconstruct $\alpha$ by making it map
\defmap{\pi_{1}(a)\hat{K}_{1}}{\pi_{2}(a)\hat{K}_{2}}.

In the matrix form described above, $H_{1}$ acts on \proj{3} by left matrix
multiplication, and $H_{2}$ acts by transposed right matrix multiplication.
The involution switching the two rulings of $\proj{1}\times\proj{1}$ corresponds
to transposing the matrix. If $h_{1}\in H_{1}$ acts on the first component of
$\proj{1}\times\proj{1}$ and $h_{2}\in H_{2}$ acts on the second one, then
write $\left[h_{1},h_{2}\right]$ to denote this action. Explicitly, for any
$2\times2$ matrices $A$ and $B$,
\[
\left[A,B\right]\left(
\left(\begin{array}{cc}x&y\\u&v\end{array}\right)\right) =
A\left(\begin{array}{cc}x&y\\u&v\end{array}\right)B^{T}
\]

It is worth noting that if $\tau$ exists, then conjugation by $\tau$ provides
isomorphisms $H_{1}\cong H_{2}$ and $K_{1}\cong K_{2}$.

\subsubsection{Types of group action}
The actions of the finite automorphism groups of the quadric surface on
$\complex^{4}$ will need to be put into one of the following four categories:
\begin{itemize}
 \item Intransitive.
 \item Transitive monomial.
 \item Transitive non--monomial imprimitive.
 \item Primitive (hence transitive).
\end{itemize}
This can be done directly by looking at the representations used to build the
action. To make the explanations more simple, it will be assumed that $H_{1}$
and $H_{2}$ both contain a non--scalar diagonal matrix. This will mean that any
proper invariant subspace must have a basis, which is a subset of the chosen
basis for $\complex^{4}$. The representations used are all listed in
Example~\ref{example:proj1}, and clearly each of them contains such a matrix.
Now fix this basis for the remainder of this section.

\subsubsection{Transitivity:}
Assume first that $\tau$ (an element of $\hat{H}$ interchanging the rulings)
does not exist. Then it can be seen that in order for $G$ to be
transitive,
both $H_{1}$ and $H_{2}$ need to be transitive: If $H_{1}$ is intransitive,
then (in the matrix presentation of \proj{3}) the rows of \proj{3} give
invariant lines, i.e.
\[\defsetshort{x=y=0},\defsetshort{u=v=0}\subset
\defsetshort{\left(x:y:u:v\right)\in\proj{3}}\]
are invariant subspaces. Similarly, if $H_{2}$ is intransitive, then there
exist $H$--invariant lines corresponding to the columns of the matrix, i.e.
\[\defsetshort{x=u=0},\defsetshort{y=v=0}\subset
\defsetshort{\left(x:y:u:v\right)\in\proj{3}}\]
are invariant subspaces. Even when both $H_{i}$ are transitive, $G$ can still
be intransitive, as illustrated in Examples~\ref{eg:intrans}
and~\ref{eg:trans}. However, this is essentially the only configuration where
this happens.

This means that for the action of $G$ to be transitive, need (for any $i\neq
j\in\defsetshort{1,2})$) one of the following to hold:
\begin{itemize}
 \item $H_{i},H_{j}\in\defsetshort{\altgroup{4},\symgroup{4},\altgroup{5}}$
 \item $H_{i}\in\defsetshort{\altgroup{4},\symgroup{4},\altgroup{5}}$,
$H_{j}=\dihgroup{2n}$
 \item $H_{1}=\dihgroup{4m}$, $H_{2}=\dihgroup{4n}$ and the action contains
elements of the
form $[a^{l},b]$ or $[b,a^{k}]$, not just $[a^{l}b,a^{k}b]$ ($k,l\in\integers$).
\end{itemize}

If an involution $\tau$ exists, then it must keep one of the diagonals of the
matrix form of \proj{3} invariant, i.e.
\[\defsetshort{x=v=0},\defsetshort{y=u=0}\subset
\defsetshort{\left(x:y:u:v\right)\in\proj{3}}\]
are left invariant by $\tau$. However, if $\tau$ exists, then $H_{1}\cong
H_{2}$, so the only two intransitive cases where it can potentially make the
action transitive are
when $H_{1}\cong H_{2}\cong\dihgroup{2n}$ (the action remains
intransitive, as $[a^{l}b,a^{k}b]$ leaves the same two proper subspaces
invariant as $\tau$ does) or when $H_{1}\cong H_{2}\cong\cyclgroup{n}$
(the action remains intransitive, as $\tau$ only permutes pairs of coordinates,
so the group has a pair of distinct invariant proper subspaces)

The discussion above has been summarised in
Table~\ref{sl4:quadrics:transitivity}.
\begin{table}[htbp]
\caption{Transitivity of $G$}\label{sl4:quadrics:transitivity}
\begin{center}
\begin{tabular}{cr|ccc}
&&\multicolumn{3}{c}{$H_{1}$}\\
&&\altgroup{4}, \symgroup{4}, \altgroup{5}& \dihgroup{2m}&
\cyclgroup{m}\\
\hline
\multirow{3}{*}{$H_{2}$}&\altgroup{4}, \symgroup{4},
\altgroup{5}&Transitive&Transitive&Intransitive\\
&\dihgroup{2m}&Transitive&Depends on action&Intransitive\\
&\cyclgroup{n}&Intransitive&Intransitive&Intransitive
\end{tabular}
\end{center}
\end{table}

\subsubsection{Primitivity:}
Assume the action of $G$ is transitive. Then the question of the action
being monomial, imprimitive nonmonomial or primitive becomes of interest.

By direct computation, it is easy to see that the place of $G$ in this
classification depends on the matrices in $H_{i}$ that have $3$ or more
non--zero entries, and how these matrices are combined in $G$, i.e.\ on the
isomorphism \deffunname{\alpha}{H_{1}/\hat{K}_{1}}{H_{2}/\hat{K}_{2}} (as
defined above).

Throughout the remainder of this section,
say elements of a given type are ``coupled'', if $\forall[g,g']\in G$ with $g$
an element of this type, then $g'$ must be either be an element of the same type
or a product of such an element and an element of a different type. Otherwise
say that elements of this type are ``separated''. For example, if
$H_{1}=H_{2}=\dihgroup{2n}$, then elements $b$ (see generators for \dihgroup{2n}
given in Example~\ref{example:proj1}) are separated in $G$ if $G$ contains an
element $[a^{k},ba^{l}]$ for some $k,l\in\integers$. Otherwise they are coupled.

With this in mind, direct computation provides the following criteria (putting
$i\neq j\in\defsetshort{1,2}$):
\begin{itemize}
 \item If $H_{1},H_{2}$ dihedral and $G$ transitive, then $G$ acts monomially.
 \item If $H_{i}\in\defsetshort{\altgroup{4}, \symgroup{4}, \altgroup{5}}$ and
$H_{j}=\dihgroup{2n}$, then the action of $G$ is non--monomial imprimitive.
 \item If $H_{1}\cong H_{2}\cong\altgroup{5}$ then the action of $G$ is
primitive.
 \item If $H_{1},H_{2}\in\defsetshort{\altgroup{4},\symgroup{4}}$ and the
$3$--cycles are separated, then the action of $G$ is primitive.
 \item If $H_{1}\cong H_{2}\cong\altgroup{4}$ and the $3$--cycles are coupled
then the action of $G$ is monomial.
 \item If $H_{i}\cong\symgroup{4}$, $H_{j}\cong\altgroup{4}$ and the
$3$--cycles are coupled, then the action is imprimitive non--monomial.
 \item If $H_{i}\cong\symgroup{4}$, $H_{j}\cong\symgroup{4}$, the $3$--cycles
are coupled and the odd permutations are coupled, then the action is monomial.
 \item If $H_{i}\cong\symgroup{4}$, $H_{j}\cong\symgroup{4}$, the $3$--cycles
are coupled, but the odd permutations are separated, then the action is
primitive.
\end{itemize}
This list is clearly not exhaustive, but it is sufficient for determining the
nature of all the groups below. The (non--)existence of $\tau$ does not
influence this classification (can be checked by direct computation).

\subsubsection{Possible isomorphism classes of $\bar{G}$}
Since $\bar{G}$ is a finite group leaving a smooth quadric $S$ invariant, its
action must be equal (as shown in section~\ref{prelim:SES}) to a suitable of one
of the finite automorphism groups of a smooth $2$--dimensional quadric. Thus
$\bar{G}$ must be conjugate to the image of one of the finite groups given
in~\cite{D-I}, Theorem~4.9.

In order to make the structure of each of the groups slightly more explicit,
the group structure will also be given in the notation
\[\left(H_{1},K_{1},H_{2},K_{2}\right)_{\alpha},\] where $H_{i}$, $K_{i}$ are as
before, and $\alpha$ is the gluing isomorphism between $H_{1}/\hat{K}_{1}$ and
$H_{2}/\hat{K}_{2}$. Where only one such isomorphism exists, $\alpha$ will be
omitted. For each isomorphism class, several representations of the group can
be chosen. However, looking at the list of possible choices of representations,
it is clear that the different representations will differ by at most an outer
automorphism, so all the properties that are of interest in this discussion
will be the same for all of them. Therefore, for each isomorphism class, any
faithful representation of $H_{i}$ can be chosen. For any group
$\left(H_{1},K_{1},H_{2},K_{2}\right)_{\alpha}$, there also exists a group
$\left(H_{2},K_{2},H_{1},K_{1}\right)_{\alpha^{-1}}$, which corresponds to the
same group with the components of the ruling of the quadric swapped. These two
groups are conjugate to each other.

\begin{lemma}\label{lemma:sl4:quadric}
If $\bar{G}<\pglgroup{4}{\complex}$ is a finite irreducible transitive subgroup,
$G$ its lift to \slgroup{4}{\complex} and $S\subset\proj{3}$ a smooth
$\bar{G}$--invariant quadric surface, and $\tau\in\bar{G}$ and element
interchanging the ruling of $S$ (if such exists), then $\bar{G}$ must be
conjugate to one of the groups in the list below.
\end{lemma}

First assume $\tau$ does not exist, i.e.\ $\bar{G}$ leaves the ruling of
$\proj{1}\times\proj{1}$ invariant. Bearing the analysis above in mind, for
$G$ to be transitive, it must be conjugate to one of the following:
\begin{enumerate}
 \item Product subgroups $\left(H_{1},H_{1},H_{2},H_{2}\right)\cong H_{1}\times
H_{2}$ for some finite groups $H_{i}\in\Aut{\proj{1}}$. Taking different
choices for $H_{1},H_{2}$,
get the following groups of the form $H_{1}\times H_{2}$:
\begin{enumerate}
\item 9 primitive groups when
$H_{1},H_{2}\in\defsetshort{\altgroup{4},\symgroup{4},\altgroup{5}}$
\item 3 families of nonmonomial imprimitive groups $\dihgroup{2m}\times H_{2}$,
where $H_{2}\in\defsetshort{\altgroup{4},\symgroup{4},\altgroup{5}}$
\item 1 family of monomial groups $\dihgroup{2m}\times\dihgroup{2n}$
\end{enumerate}
 \item Twisted diagonal subgroups $\left(H_{1},1,H_{1},1\right)_{\alpha}\cong
H_{1}$ for some finite group $H_{1}\in\Aut{\proj{1}}$. This gives $3$ families
of groups,
indexed by the choice of isomorphism $\alpha$. They are:
\begin{enumerate}
 \item Monomial groups isomorphic to \altgroup{4} or \symgroup{4}.
 \item Primitive groups isomorphic to \altgroup{5}.
\end{enumerate}
The twisted diagonal groups isomorphic to the dihedral groups do not act
transitively, as the relevan central extensions do not have any
$4$--dimensional irreducible representations.

 \item\label{usedin:prim}
$\frac{1}{2}\left[\symgroup{4}\times\symgroup{4}\right]
\cong\left(\altgroup{4}\times\altgroup{4}\right)\rtimes\cyclgroup{2}
\cong\left(\symgroup{4},\altgroup{4},\symgroup{4},\altgroup{4}\right)$,
a primitive group generated by elements corresponding to
$[(12)(34),\mbox{id}]$,
$[\mbox{id},(12)(34)]$,
$[(123),\mbox{id}]$,
$[\mbox{id},(123)]$ and
$[(12),(12)]$.

 \item\label{usedin:imprim:nonmono}
$\frac{1}{2}\left[\dihgroup{2m}\times\symgroup{4}\right]
\cong\left(\cyclgroup{m}\times\altgroup{4}\right)\rtimes\cyclgroup{2}
\cong\left(\dihgroup{2m},\cyclgroup{m},\symgroup{4},\altgroup{4}\right)$,
a family of imprimitive nonmonomial groups generated by
$[a_{m},\mbox{id}]$,
$[\mbox{id},(12)(34)]$,
$[\mbox{id},(123)]$ and
$[b,(12)]$.

 \item
$\frac{1}{2}\left[\dihgroup{4m}\times\symgroup{4}\right]
\cong\left(\dihgroup{2m}\times\altgroup{4}\right)\rtimes\cyclgroup{2}
\cong\left(\dihgroup{4m},\dihgroup{2m},\symgroup{4},\altgroup{4}\right)$
$\left(m\geq2\right)$,
a family of imprimitive nonmonomial groups generated by the action of
$[a_{2m}^{2},\mbox{id}]$,
$[b,\mbox{id}]$,
$[\mbox{id},(12)(34)]$,
$[\mbox{id},(123)]$ and
$[a_{2m}^{m},(12)]$.

 \item
$\frac{1}{6}\left[\dihgroup{6m}\times\symgroup{4}\right]
\cong\left(\cyclgroup{m}\times \mathbb{V}_{4}\right)\rtimes\symgroup{3}
\cong\left(\dihgroup{6m},\cyclgroup{m},\symgroup{4},\mathbb{V}_{4}\right)$,
a family of imprimitive non--monomial groups generated by
$[a_{3m}^{3},\mbox{id}]$,
$[\mbox{id},(12)(34)]$,
$[\mbox{id},(13)(24)]$,
$[a_{3m}^{m},(123)]$ and
$[b,(12)]$.

 \item\label{usedin:imprim}
$\frac{1}{6}\left[\symgroup{4}\times\symgroup{4}\right]
\cong \left(\mathbb{V}_{4}\times \mathbb{V}_{4}\right)\rtimes\symgroup{3}
\cong\left(\symgroup{4},\mathbb{V}_{4},\symgroup{4},\mathbb{V}_{4}\right)$,
a monomial group generated by
$[(12)(34),\mbox{id}]$,
$[(13)(24),\mbox{id}]$,
$[\mbox{id},(12)(34)]$,
$[\mbox{id},(13)(24)]$,
$[(123),(123)]$ and
$[(12),(12)]$.

 \item
$\frac{1}{3}\left[\altgroup{4}\times\altgroup{4}\right]
\cong\left(\mathbb{V}_{4}\times \mathbb{V}_{4}\right)\rtimes\cyclgroup{3}
\cong\left(\altgroup{4},\mathbb{V}_{4},\altgroup{4},\mathbb{V}_{4}\right)$,
a monomial group generated by
$[(12)(34),\mbox{id}]$,
$[(13)(24),\mbox{id}]$,
$[\mbox{id},(12)(34)]$,
$[\mbox{id},(13)(24)]$ and
$[(123),(123)]$.

 \item
$\frac{1}{2}\left[\dihgroup{2m}\times\dihgroup{4n}\right]
\cong\left(\cyclgroup{m}\times\dihgroup{2n}\right)\rtimes\cyclgroup{2}
\cong\left(\dihgroup{2m},\cyclgroup{m},\dihgroup{4n},\dihgroup{2n}\right)$
(for $m,n\geq2$),
a family of monomial groups generated by
$[a_{m},\mbox{id}]$,
$[\mbox{id},a_{2n}^{2}]$,
$[\mbox{id},b]$ and
$[b,a_{2n}^{n}]$.

 \item\label{usedin:trans}
$\frac{1}{4}\left[\dihgroup{4m}\times\dihgroup{4n}\right]_{\alpha}
\cong\left(\cyclgroup{m}\times \cyclgroup{n}\right)\rtimes\dihgroup{4}
\cong\left(\dihgroup{4m},\cyclgroup{m},\dihgroup{4n},\cyclgroup{n}\right)_{
\alpha }$ (where
$\alpha(b)=a_{2n}^{n}$, $\alpha(a_{2m}^{m})=b$),
a family of monomial groups generated by
$[a_{2m}^{2},\mbox{id}]$,
$[\mbox{id},a_{2n}^{2}]$,
$[a_{2m}^{m},b]$ and
$[b,a_{2n}^{n}]$.

 \item
$\frac{1}{2}\left[\dihgroup{4m}\times\dihgroup{4n}\right]
\cong\left(\dihgroup{2m}\times\dihgroup{2n}\right)\rtimes\cyclgroup{2}
\cong\left(\dihgroup{4m},\dihgroup{2m},\dihgroup{4n},\dihgroup{2n}\right)$
(for $m,n\geq2$),
a family of monomial groups generated by
$[a_{2m}^{2},\mbox{id}]$,
$[b,\mbox{id}]$,
$[\mbox{id},a_{2n}^{2}]$,
$[\mbox{id},b]$ and
$[a_{2m}^{m},a_{2n}^{n}]$.

\suspend{enumerate}

Now assume that $\tau\in\bar{G}$ exists. Then the normal subgroup of $\bar{G}$
fixing the ruling of the quadric must also be a group of automorphisms of $S$.
Furthermore, it must have $H_{1}\cong H_{2}$ and $K_{1}\cong K_{2}$, since
conjugation by $\tau$ provides the two isomorphisms. That means that $\bar{G}$
can be isomorphic to one of the following groups:
\resume{enumerate}

 \item $\left(H_{1}\times H_{1}\right)\rtimes2\cong
\left(H_{1},H_{1},H_{1},H_{1}\right)\rtimes\cyclgroup{2}$
($H_{1}\in\Aut{\proj{1}}$).
Taking different choices for $H_{1}$ and bearing in mind that choosing
$H_{1}$ to be \cyclgroup{n} produces an intransitive group (see discussion
above), get $1$ family of monomial groups
\begin{enumerate}
 \item
$\left(\dihgroup{2n}\times\dihgroup{2n}\right)\rtimes\cyclgroup{2}$
\suspend{enumerate}
and $3$ families of primitive groups, all of them indexed by the possible
involutions acting on $H_{1}$:
\resume{enumerate}
 \item $\left(\altgroup{4}\times\altgroup{4}\right)\rtimes\cyclgroup{2}$
 \item $\left(\symgroup{4}\times\symgroup{4}\right)\rtimes\cyclgroup{2}$
 \item $\left(\altgroup{5}\times\altgroup{5}\right)\rtimes\cyclgroup{2}$
\end{enumerate}

 \item $H_{1}\rtimes\cyclgroup{2}\cong \left(H_{1},1,H_{1},1\right)_{\alpha}$
($H_{1}\in\Aut{\proj{1}}$). This gives $3$ families of groups,
indexed by the choice of isomorphism $\alpha$. They are:
\begin{enumerate}
 \item Monomial groups isomorphic to $\dihgroup{4n}\rtimes\cyclgroup{2}$.
 \item Monomial groups isomorphic to $\altgroup{4}\rtimes\cyclgroup{2}$ or
$\symgroup{4}\rtimes\cyclgroup{2}$.
 \item Primitive groups isomorphic to $\altgroup{5}\rtimes\cyclgroup{2}$.
\end{enumerate}

 \item
$\left(\left(\altgroup{4}\times\altgroup{4}
\right)\rtimes\cyclgroup{2}\right)\rtimes\cyclgroup{2}
\cong\left(\symgroup{4},\altgroup{4},\symgroup{4},\altgroup{4}
\right)\rtimes\cyclgroup{2}$,
a family of primitive groups.

 \item
$\left(\left(\mathbb{V}_{4}\times
\mathbb{V}_{4}\right)\rtimes\symgroup{3}\right)\rtimes\cyclgroup{2}
\cong\left(\symgroup{4},\mathbb{V}_{4},\symgroup{4},\mathbb{V}_{4}
\right)\rtimes\cyclgroup{2}$,
a family of monomial groups.

 \item\label{usedin:imprim:mono}
$\left(\left(\mathbb{V}_{4}\times
\mathbb{V}_{4}\right)\rtimes\cyclgroup{3}\right)\rtimes\cyclgroup{2}
\cong\left(\altgroup{4},\mathbb{V}_{4},\altgroup{4},\mathbb{V}_{4}
\right)\rtimes\cyclgroup{2}$,
a family of monomial groups.

 \item
$\left(\left(\cyclgroup{m}\times\cyclgroup{m}\right)\rtimes\dihgroup{4}
\right)\rtimes\cyclgroup{2}
\cong\left(\dihgroup{4m},\cyclgroup{m},\dihgroup{4m},
\cyclgroup{m}\right)_{\alpha}\rtimes\cyclgroup{2}$
(where
$\alpha(b)=a_{2m}^{m}$, $\alpha(a_{2m}^{m})=b$),
a family of monomial groups.

 \item
$\left(\left(\dihgroup{2m}\times\dihgroup{2m}\right)\rtimes\cyclgroup{2}
\right)\rtimes\cyclgroup{2}
\cong\left(\dihgroup{4m},\dihgroup{2m},\dihgroup{4m},\dihgroup{2m}
\right)\rtimes\cyclgroup{2}$
($m\geq2$),
a family of monomial groups.
\end{enumerate}

\begin{eg}\label{quadric:example}
When one thinks about finite groups in \slgroup{4}{\complex} that leave a smooth
quadric invariant, one of the first examples that come to mind is the monomial
group $G=\left(\cyclgroup{2}\right)^{3}\rtimes\bsymgroup{4}$, where the normal
subgroup $H=\left(\cyclgroup{2}\right)^{3}$ acts
diagonally, and the binary symmetric group permutes the basis. Here, one of the
size two subgroups of $H$ acts by scalar matrices, so the action on the
projective quadric surface is isomorphic to
$\left(\cyclgroup{2}\right)^{2}\rtimes\symgroup{4}\cong\left(\left(\mathbb{V}_{4
}\times
\mathbb{V}_{4}\right)\rtimes\cyclgroup{3}\right)\rtimes\cyclgroup{2}$, which is
in
position~(\ref{usedin:imprim:mono}) in the list above.
\end{eg}

\begin{eg}[cf.\ Example~\ref{eg:simple:before}]\label{eg:simple:after}
Let
\[S=\defset{\left(x:y:u:v\right)\in\proj{3}}{x^{m}+y^{m}+u^{m}+w^{m}=0},\]
where $m\geq5$. Let $\bar{G}=\Aut{S}$, and $G$ be the lift of $\bar{G}$
to \slgroup{4}{\complex}. Then $G$ induces a weakly--exceptional singularity.
\end{eg}
\pf $G$ is a central extension of
$\left(\cyclgroup{m}\right)^{3}\rtimes\bsymgroup{4}$, acting by
permuting the basis and multiplying the coordinates by arbitrary $m$--th roots
of unity. Thus $G$ is transitive. $G$ is not isomorphic to any of the groups in
Lemmas~\ref{lemma:sl4:cubic:smooth} or~\ref{lemma:sl4:quadric}, so it does not
have an invariant quadric or cubic surface. Furthermore,
$\bar{G}\not\cong\altgroup{5}$. Therefore, $G$ induces a weakly--exceptional
singularity.\qed

\def\cprime{$'$}