Pencils of symmetric surfaces in $\pitr$
A.~Sarti
Let $G\subset SO(4)$ denote a finite subgroup containing the Heisenberg group. In these notes we classify all these groups, we find the dimension of the spaces of $G$-invariant polynomials and we give equations for the generators whenever the space has dimension two. Then we complete the study of the corresponding $G$-invariant pencils of surfaces in $\mathbb{P}_3$ which we started in \cite{sa}. It turns out that we have five more pencils, two of them containing surfaces with nodes. %other then the three described there. We give generators for them and we find the singular surfaces. In particular we have two more pencils which contain surfaces with nodes.
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%\title{Pencils of symmetric surfaces in $\pitr$}
%\title{Rotation~groups~and~symmetric~surfaces}

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artjul01_00.tex
\begin{center}
{\Large Polynomial Invariants of the Heisenberg Group\\ with extra Symmetries}\\[0.3cm]
{\rm \today}\\[0.3cm]
 A.Sarti\\[0.3cm]
%July 19, 2001\\[0.3cm]
FBR 17 Mathematik, Universit\"at Mainz\\
Staudingerweg 9, 55099 Mainz\\
Germany\\
e-mail:sarti@mathematik.uni-mainz.de\\
\end{center}



\begin{abstract}
Let $G\subset SO(4)$ denote a finite subgroup containing the Heisenberg group. In these notes we classify all these groups, we find the dimension of the spaces of $G$-invariant polynomials and we give equations for the generators whenever the space has dimension two.  Then we complete the study of the corresponding $G$-invariant pencils of surfaces in $\mathbb{P}_3$ which we started in \cite{sa}. It turns out that we have five more pencils, two of them containing surfaces with nodes.
%other then the three described there. We give generators for them and we find the singular surfaces. In particular we have two more pencils which contain surfaces with nodes.
\end{abstract}

\tableofcontents

\section{Introduction}
Consider the Klein four group $V\subseteq SO(3)$. Let $\tilde{V}$ denotes its inverse image in $SU(2)$ under the universal covering $SU(2)$$\rightarrow$$SO(3)$. The image of the direct product $\tilde{V}\times\tilde{V}$ in $SO(4)$ under the double covering $SU(2)$$\times$$SU(2)$$\rightarrow$$SO(4)$ is the Heisenberg group $H$. In this note we classify all the subgroups $G$ of $SO(4)$ which contain $H$. First we classify all the subgroups of $SU(2)\times SU(2)$ which contain  $\tilde{V}\times\tilde{V}$, then their images in $SO(4)$ are the groups $G$ (cf.\ proposition \ref{list} and section 1.4). They operate in a natural way on $\cix$, the ring of polynomials in four variables with complex coefficients. In section 3 we give generators for the spaces  $\cix^G_j$ of homogeneous $G$-invariant polynomials of degree $j$ whenever this dimension is two. Since the groups $G$ contain $H$, we have invariant polynomials only in even degree. When the dimension is two the generators are the multiple quadric \mbox{$q^{j/2}=(x_0^2+x_1^2+x_2^2+x_3^2)^{j/2}$} (trivial invariant) and another polynomial of degree $j$ which we denote by $f$. We describe  then the pencils
\begin{eqnarray*}
f+\lambda q^{j/2}=0,~~~~~\lambda\in\piu,
\end{eqnarray*}
 of symmetric surfaces in the three dimensional complex projective space $\pitr$ (section 4). In particular we find the singular surfaces contained in it.
%Consider the subgroups of $SO(4)$ which contain the Heisenberg group. With the help of the double cover $SU(2)$$\times$$SU(2)$$\rightarrow$$SO(4)$, described in \cite{tits}, p. 77-78, it is possible to classify them completely (cf. {\bf (1.2.1)}). These groups operate in a natural way on $\cix$, the ring of polynomials in four variables with complex coefficients. Here we compute the dimension of their spaces of invariant homogeneous polynomials and describe the corresponding families of symmetric surfaces in $\mathbb{P}^3(\mathbb{C})$:=$\pitr$ whenever these are two-dimensional. In this last case generators are the quadric \mbox{$q^{j/2}=(x_0^2+x_1^2+x_2^2+x_3^2)^{j/2}$} (trivial invariant) and another  invariant polynomial $f_G$ of degree $j$ where $G$ denotes the group which we consider and $j$ the degree of the two-dimensional invariant space.  The corresponding pencil in $\pitr$ has then equation
%\begin{eqnarray*}
%f_G+\lambda q^{j/2}=0~~~~~\lambda\in\piu.
%\end{eqnarray*}
In \cite{sa} we considered the case of $G=TT$, $OO$, $II$ which are the images in $SO(4)$ of the direct products $\tilde{T}\times\tilde{T}$, $\tilde{O}\times\tilde{O}$, $\tilde{I}\times\tilde{I}$ where $\tilde{T}$ denotes the binary tetrahedral group, $\tilde{O}$ the binary octahedral group, $\tilde{I}$ the binary icosahedral group in $SU(2)$. We denoted the groups there by $G_6$, $G_8$ and $G_{12}$  and we called them bi-polyhedral groups. We found pencils containing surfaces with many nodes (=ordinary double points). In particular the degree twelve $II$-invariant pencil contains a surface with $600$ nodes which improves the previous lower bound for the maximal number of nodes of a surface of degree twelve in $\pitr$ (cf. \cite{chmutov}). Here we describe the other $G$-invariant pencils and show that we have two more pencils which contain surfaces with nodes (the others do not contain surfaces with isolated singularities at all). We list the groups $G$ and the degrees $j$ below, as well as the number of nodes on the singular surfaces. In each pencil we have four of these singular surfaces and the nodes there form just one $G$-orbit. For the convenience of the reader we recall the results about the $TT$-, $OO$-, and $II$-invariant pencils too.
 
\renewcommand{\arraystretch}{1.3}

\begin{center}

\begin{tabular}{l|*{2}{l|}*{4}{l}}
$G$& {\rm order}& $j$ &\multicolumn{4}{c}{nodes}\\
\hline 
$(OO)'$&192&4&4&12&16&8\\
$TT$&288&6&12&48&48&12\\
$OO$&1152&8&24&72&144&96\\
$IO$&2880&12&240&360&240&120\\
$II$&7200&12&300&600&360&60
\end{tabular}

\end{center}
The group $IO$ is the image in $SO(4)$ of the direct product of the binary icosahedral group with the binary octahedral group in $SU(2)$$\times$$SU(2)$ and $(OO)'$ is a subgroup of $OO$ which we describe in section 1.2 and  1.4. This is an index two subgroup of the reflection group [3,3,4] (cf.\ \cite{cox}, p.\ 226) and has the same invariant polynomials. In all the cases but one ($G=IO$), the singular surfaces contain just isolated singularities (i.e.\ the nodes). The surfaces in the $IO$-invariant pencil contain two double lines in the base locus. \\
The [3,3,4]-invariant polynomials of degree two and four were already known by Coxeter in \cite{cox0}. Here we show that in the pencil of degree four we have a surface with the maximal number possible of nodes (=16) which is a so called Kummer surface. Finally in section 6 we give a computer picture of the $IO$-invariant surface of degree 12 with 360 nodes.\\
I thank Prof.\ Wolf Barth at the University of Erlangen for many helpful comments and discussions. 
%In these notes we obtain also the degree four invariant polynomials of the {\it bi-polyhedral dihedral group} $D_3D_3$$\subseteq$$SO(4)$ (cf. remark {\bf 3.1.3}, where $D_3$ denotes the dihedral group with six elements). By Mukai \cite{mukai} the quotient $\pitr$$/D_3D_3$ is isomorphic with the Satake compactification of the moduli space of abelian surfaces with $(1,2)$ polarization, hence this invariant polynomials should be related to modular forms. 
%In \cite{sa} we remarked this last fact for the invariants of the bi-polyhedral groups $TT$, $OO$ and $II$. 

%Observe that in degree four we find a surface with the maximal number 16 of nodes. This surface is a so called Kummer surface.


%We show in this way that the best examples of surfaces with many nodes with Heisenberg symmetry and some extra symmetries are the $TT$-, $OO$-, $II$-invariant ones.



 %We are mostly interested in the singular loci, which we compute in section five. We answer then completely the question if there are subgroups of $SO(4)$ containing the Heisenberg group, which leave invariant surfaces with many nodes. We show that in fact the best examples of surfaces with many nodes are obtained considering the $TT$-, $OO$-, and $II$-invariant pencils described in \cite{sa}. 

\section{Symmetry groups}
Denote by $H\subseteq SO(4)$ the Heisenberg group (with 32 elements). We want to collect in a systematic way all the finite subgroups of $SO(4)$ containing $H$, and their polynomial invariants of low degree. These are invariants of the Heisenberg group with extra symmetries.
\subsection{Ternary groups}
We specify the following matrices in $SO(3)$
\begin{eqnarray*}
A_1:=\left(\begin{array} {ccc}
1&0&0\\
0&-1&0\\
0&0&-1
\end{array}\right ), A_2:=\left(\begin{array} {ccc}
-1&0&0\\
0&1&0\\
0&0&-1
\end{array}\right ),
R_n:=\left(\begin{array} {ccc}
1&0&0\\
0&a&-b\\
0&b&a
\end{array}\right ),\\
S:=\left(\begin{array} {ccc}
0&-1&0\\
0&0&-1\\
1&0&0
\end{array}\right ),U:=\frac{1}{2}\left(\begin{array} {ccc}
\tau-1&-\tau&1\\
\tau&1&\tau-1\\
-1&\tau-1&\tau
\end{array}\right ), 
\end{eqnarray*}
where $\tau:=\frac{1}{2}(1+\sqrt{5})=2\cdot$cos$(\frac{\pi}{5})$, $a:= $cos$\frac{2\cdot\pi}{n}$, $b:= $sin$\frac{2\cdot\pi}{n}$. These matrices generate the following subgroups of $SO(3)$
\begin{center}

\begin{eqnarray*}
\begin{array}{lllll}
&{\rm generators}&{\rm order}&{\rm group}& {\rm name}\\
V&A_1,A_2&4&\mathbb{Z}_2\times\mathbb{Z}_2&{\rm Klein~four}\\
D_n&A_2,R_n&2n&D_n&{\rm dihedral}\\
T&A_1,S&12&Alt(4)&{\rm tetrahedral}\\
O&A_1,R_4,S&24&Sym(4)&{\rm octahedral}\\
I&A_1,S,U&60&Alt(5)&{\rm icosahedral}
\end{array}
\end{eqnarray*}
\end{center}
Where $Alt(4)$ and $Alt(5)$ denote the group of even permutations of four  and five elements and $Sym(4)$ denotes the permutation group of four elements.
Whenever $n\in \mathbb{N}$, $n\not=0$, is {\it even}, $V$ is contained in each of the above groups. By the classification of the finite subgroups of $SO(3)$, these are all such subgroups. \\
The groups $T,O$ and $I$ are the rotation groups of tetrahedron, octahedron and icosahedron. An identification with the permutation groups is given in \cite{cox}, p.\ 46-50 as well as an identification of $D_n$ with the symmetry group of a regular polygon with $n$ vertices, p.\ 46. Sometimes it is useful for the computation to identify the matrices above with the cycles of the permutation groups. Indeed the identification of $T$ and $O$ with subgroups of the permutation  group $Sym(4)$ is obtained by letting them act on the four space diagonals of the unit cube. Let these lines and vectors generating them be
\begin{eqnarray*}
d_1:(1,1,1),~d_2:(-1,1,1),~d_3:(1,-1,1),~d_4:(1,1,-1)\\
\end{eqnarray*}
The matrices in $O$ permute these lines by
\begin{eqnarray*}
A_1:(12)(34),~A_2:(13)(24),~R_4:(1423),~S:(123).
\end{eqnarray*}
Using this correspondence with permutation groups, it is easy to write down their conjugacy classes. We write the conjugacy classes of the dihedral group $D_{n}$, $n=2l$, $l\geq 2$ too. In the next table we characterize a conjugacy class by one of its elements. Under each representative we write the number of elements in the conjugacy class.\\
%\begin{eqnarray*}
\begin{center}
\begin{tabular}{l|lllll}
group &\multicolumn{5}{c}{repr. of a conj. class}\\
&\multicolumn{5}{c}{and number of elements}\\
\hline
$V$&1&$A_1$&$A_2$&$A_3$&\\
&1&1&1&1\\
\hline
$D_{n}$&1&$R_n^k$&$R_n^l$&$A_2$&$A_2R_n$\\
&1&2&1&l&l\\
\hline
T&1&$A_1$&$S$&$S^2$&\\
&1&3&4&4\\
\hline
$O$&1&$A_1$&$R_4A_2$&$R_4$&$S$\\
&1&3&6&6&8
\end{tabular}
\end{center}
where $k=1,\ldots,l-1$.\\
The symmetries of $T$ obviously leave invariant the icosahedron (cf.\ \cite{cox}, p.\ 52) with vertices 
\begin{eqnarray*}
(\pm 1,\pm \tau,0),~(0,\pm 1,\pm \tau),~(\pm \tau, 0,\pm 1)
\end{eqnarray*}
The matrix $U$ permutes these vertices as
\begin{eqnarray*}
\begin{array}{c}
\pm(0,1,\tau)\mapsto\pm(0,1,\tau),\\
\pm(0,1,-\tau)\mapsto\pm(-\tau,0,1)\mapsto\pm(-1,-\tau,0)\mapsto\pm(1,-\tau,0)\mapsto\pm (\tau,0,-1)\mapsto\pm(0,1,-\tau).
\end{array}
\end{eqnarray*}
So together with the group $T$ the symmetry $U$ generates a group of order at least $12\cdot 5=60$ , contained in the symmetry group of the icosahedron specified. Therefore it coincides with the icosahedral group $I\cong A_5$. The action of $I$ on the five cosets of $I/T$ defines the map $I\mapsto A_5$. Using
\begin{eqnarray*}
\begin{array}{c}
A_1\cdot U=U^4\cdot A_1,~A_1\cdot U^2=U^3\cdot A_1,\\
S\cdot U=U^3\cdot A_2,~S\cdot U^2=U\cdot A_2S,~S\cdot U^3=U^2\cdot S^2,~S\cdot U^4=U^4\cdot S^2 A_2,\\
\end{array}
\end{eqnarray*}
one finds that under this map 
\begin{eqnarray*}
A_1\mapsto(14)(23),~S\mapsto(132),~U\mapsto(12345).
\end{eqnarray*}
Using this correspondence, one enumerates the conjugacy classes in $I$\\
%\begin{eqnarray*}
\begin{center}
\begin{tabular}{l|lllll}
group &\multicolumn{5}{c}{repr. of a conj. class}\\
&\multicolumn{5}{c}{and number of elements}\\
\hline
$I$&1&$A_1$&$S$&$U$&$U^2$\\
&1&15&20&12&12
\end{tabular}
\end{center}
%\end{eqnarray*}
\subsection{Subgroups of products of ternary groups}
Here we classify {\it subdirect products} $G\subseteq G_1\times G_2$, where $G_1$ and $G_2$ are finite ternary groups $V,D_n$ ($n$ even), $T,O$ or $I$. We assume that $G$ contains the subgroup $V\times V$$\subseteq G_1\times G_2$. We are interested in these subgroups only up to interchanging the factors $G_1$ and $G_2$. So we assume that we are in one of the following cases
\begin{itemize}
\item $|G_1|\geq|G_2|$ and $G_1,G_2\not=D_n$,
\item $G_1\not=D_n$, $G_2=D_n$, 
\item $G_1=D_n$ and $G_2=D_m$.\\
\end{itemize}
Additionally, passing to smaller subgroups $G_i'$$\subseteq$$G_i$ if necessary, we may assume that both projections 
\begin{eqnarray*}
p_1:G\rightarrow G_1',~p_2:G\rightarrow G_2'
\end{eqnarray*}
are surjective, so that $G\subseteq G_1'\times G_2'$. Finally, we do not distinguish between groups conjugate in $SO(3)\times SO(3)$. %In all the cases but one, $(T\times T)'$, this means conjugate in $G_1\times G_2$. 
In the table below we 
%abbreviate $G_1\times G_2:=G_1G_2$, and 
assume $n\not=m$, $n=2l$, $m=2l'$ and let $s:=${\rm lcm}$(n,m)$.\\
\begin{prop}\label{list}
The following list is a complete list of subgroups $G\subseteq G_1\times G_2$ under the assumptions above:
\pagebreak
\begin{eqnarray*}
\begin{array}{l|l|lll|l|l|lll}
G_1&G_2&G&|G|/16&G/V\times V&G_1&G_2&G&|G|/16&G/V\times V\\
\hline
V&V&V\times V&1&1&V&D_n&V\times D_n&n/2&\zi_l\times 1\\
T&V&T\times V&3&\zi_3\times 1&T&D_n&T\times D_n&3n/2&\zi_3\times\zi_l\\
&T&T\times T&9&\zi_3\times\zi_3&& 3|n &(T\times D_n)'&n/2&\zi_l\\
&&(T\times T)'&3&\zi_3&O&D_n&O\times D_n&6n/2&D_3\times\zi_l\\
O&V&O\times V&6&D_3\times1&&4|n&(O\times D_n)'&3n/2&D_{3l/2}\\
&T&O\times T&18&D_3\times\zi_3&I&D_n& I\times D_n&15n/2&\\ 
&O&O\times O&36&D_3\times D_3&D_n&D_n&D_n\times D_n&n^2/4&\zi_l\times\zi_l\\
&O&O\times O&36&D_3\times D_3&&&(D_n\times D_n)'&n/2&\zi_l\\
&&(O\times O)'&6&D_3&D_n&D_m&D_n\times D_m&nm/4&\zi_l\times\zi_{l'}\\
%O&D_4&(O\times D_4)'&6&D_3\\
&&(O\times O)''&18&\zi_3\times\zi_3\times\zi_2&&&(D_n\times D_m)'&s/2&\zi_s\\
I&V&I\times V&15&&&&&&\\
&T&I\times T&45&&&&&&\\
&O&I\times O&90&&&&&&\\
&I&I\times I&225&&&&&&\\
\end{array}
\end{eqnarray*}
\end{prop}
\bprf
We discuss the cases one-by-one. But before that, we observe that the kernels $K_1\subseteq G_1\times 1$ and $K_2\subseteq 1\times G_2$ of both projections $p_i:G\rightarrow G_i$ are normal subgroups. This follows by conjugating component-wise from the surjectivity of both projections.\\
The groups $G_1\times V$ do not have proper subgroups containing $V\times V$ and mapping surjectively onto $G_1$, (for $G_1=D_n$ too). So we do not need to consider the cases $G_2=V$.\\
First consider the case of $G_1=I$. Since $I$ is simple, the kernel $K_1\subseteq I\times 1$ either coincides with $I\times 1$, or is trivial. The latter cannot be the case, because this kernel contains $V\times 1$. The only possibilities are the product cases $I\times V$, $I\times T$, $I\times O$, $I\times I$ and $I\times D_n$.\\
Let now $Q:=G/V\times V$ and $Q_1:=G_1/V$, $Q_2:=G_2/V$. We consider $Q$ a proper subdirect product of $Q_1\times Q_2$.\\
$T,T$: $Q\subseteq \zi_3\times\zi_3$ mapping surjectively onto both factors is either the diagonal or the anti-diagonal.
% i.e. generated by $(S,S)$ or $(S,S^2)$. 
The two corresponding subgroups are not conjugate in $T\times T$, but in $T\times O$ they are. The inverse image of the diagonal $\zi_3$$\subseteq\zi_3\times\zi_3$ in $T\times T$ is $(T\times T)'$.\\
$O,T$: $Q\subseteq D_3\times \zi_3$ mapping surjectively onto both factors would have order six and be isomorphic with $D_3$ under $p_1$. But there is no epimorphism of $D_3$ onto $\zi_3$. Such a group does not exist.\\
$O,O$: $Q\subseteq D_3\times D_3$ must have order six, twelve or 18. If it has order six, it is a graph of an isomorphism between both factors $D_3$. Then it is conjugate to the diagonal, and this leads to the subgroup $(O\times O)'$$\subseteq$$O\times O$. The case $|Q|$=$12$ cannot occur, because then the kernel $K_1$$\subseteq$$D_3\times 1$ would have order two, and could not be normal. If $Q$ has order $18$ both the kernels $K_1$ and $K_2$ have order three, and coincide with the unique proper normal subgroup of $D_3$.  This implies that $G$ contains $T\times T $$\subseteq$$O\times O$ and is the inverse image of a subgroup $\zi_2$$\subseteq$$\zi_2\times\zi_2$$=O\times O/T\times T$. By surjectivity of projections this can only be the diagonal. Its inverse image is $(O\times O)''$. \\
$T,D_n$: if $Q\subseteq \zi_3\times \zi_l$ maps surjectively onto both the components then three divides $l$. Let its inverse image be $(T\times D_n)'$.\\
%$Q$ is generated by an element $(S,R_n)$(?) where $S$ generates $\zi_3$ and $b$ generates $\zi_l$. Let denote its inverse image by $(T\times D_n)'$
%as in the previous case we have a  proper subgroup of $T\times D_n$$/V\times V$$=\zi_3\times \zi_l$ mapping surjectively onto $\zi_3$ and $\zi_l$ only if $l=3$. In this case $G$ would be the inverse image of the diagonal $\zi_3$, like the group $(T\times T)'$$\subseteq$$T\times T$.\\
$O,D_n$: if $Q\subseteq D_3\times \zi_l$ maps surjectively onto both factors then two divides $l$. We denote the inverse image by $(O\times D_n)'$.\\
$D_n,D_n$: $Q\subseteq \zi_l\times\zi_l$ mapping surjectively onto $\zi_l$ is conjugate to the diagonal. We call its inverse image $(D_n\times D_n)'$.\\
$D_n,D_m$:  $Q\subseteq\zi_l\times\zi_{l'}$ mapping surjectively onto $\zi_l$ and $\zi_{l'}$ is generated by an element of order $s/2$.
%is generated by an element $(a,b)$ where $a$ generates  $\zi_l$ and $b$ generates $zi_{l'}$. 
We call its inverse image $(D_n\times D_m)'$.\\
%and $Q$ is generated by $(S,1)$ and $(R_n^{n/2},R_n)$(?). Its inverse image is $(O\times D_n)$. 
%we have $O\times D_n$$/V\times V$$=D_3\times \zi_l$. Since $K_1$$\subseteq$$1\times \zi_l$, $K_1$ is trivial. Hence $G=D_3$. Now $K_2$ is normal in $D_3\times 1$ hence it has order three, by this fact follows that $l=2$  and we have a proper subgroup $(O\times D_4)'$$\subseteq$$O\times D_4$. 
\eprf 
\subsection{Binary groups}
We consider the double cover of \cite{tits} p.\ 77--78. Let $\tilde{G}$ denotes the pre-image in $SU(2)$ of $G$$\subseteq$$SO(3)$. We specify the following matrices, $\tilde{M}$, which are in the pre-image of $M$$\in$$SO(3)$:
 \begin{eqnarray*}
\begin{array}{lll}
\tilde{A_1}:=\left(\begin{array} {cc}
i&0\\
0&-i
\end{array}\right ), &\tilde{A_2}:=\left(\begin{array} {cc}
0&1\\
-1&0
\end{array}\right ),& \tilde{A_3}:=\left(\begin{array} {cc}
0&i\\
i&0
\end{array}\right ),\\
\tilde{S}:=\frac{1}{2}\left(\begin{array} {cc}
1+i&-1+i\\
1+i&1-i
\end{array}\right ),& \tilde{U}:=\frac{1}{2}\left(\begin{array} {cc}
\tau&\tau-1+i\\
1-\tau+i&\tau
\end{array}\right ),& \tilde{R_n}:=\left(\begin{array} {cc}
e^{\frac{i\pi}{n}}&0\\
0&e^{-\frac{i\pi}{n}}
\end{array}\right ).
\end{array}
\end{eqnarray*}
since $\tilde{M}^{\mbox{ord}(M)}$$=-1$, they have order $2\cdot$ord$(M)$. By an argumentation as in \cite{sa} section 2, we can write the conjugacy classes in the binary groups:
\pagebreak
\begin{center}
\begin{tabular}{c|ccccccccc}
group &\multicolumn{9}{c}{repr. of a conj. class and number of elements}\\
%&\multicolumn{5}{c}{and number of elements}\\
\hline
$\tilde{V}$&1&-1&$\tilde{A_1}$&$\tilde{A_2}$&$\tilde{A_3}$&&&&\\
&1&1&2&2&2&&&&\\
\hline
$\tilde{D_n}$&1&-1&$\tilde{A_2}$&$\tilde{A_2}\tilde{R_n}$&$\tilde{R_n}^l$&$\tilde{R_n}^{k}$&$-\tilde{R_n}^k$&&\\
&1&1&2l&2l&2&2&2&&\\
\hline
$\tilde{T}$&1&-1&$\tilde{A_1}$&$\tilde{S}$&$-\tilde{S}$&$\tilde{S}^2$&$-\tilde{S}^2$&&\\
&1&1&6&4&4&4&4&&\\
\hline
$\tilde{O}$&1&-1&$\tilde{A_1}$&$\tilde{R_4}\tilde{A_2}$&$\tilde{R_4}$&$-\tilde{R_4}$&$\tilde{S}$&$-\tilde{S}$\\
&1&1&6&12&6&6&8&8&\\
\hline
$\tilde{I}$&1&-1&$\tilde{A_1}$&$\tilde{S}$&$-\tilde{S}$&$\tilde{U}$&$-\tilde{U}$&$\tilde{U}^2$&$-\tilde{U}^2$\\
&1&1&30&20&20&12&12&12&12
\end{tabular}
\end{center}
where $k=1,\ldots,l-1$.
\subsection{Quaternary groups}
Here we consider the images of the finite groups $\tilde{G_1}\times\tilde{G_2}\subseteq SU(2)\times SU(2)$ under the double covering map
\begin{eqnarray*}
\begin{array}{cc}
SU(2)\times SU(2)\rightarrow SO(4),&(q,q'):p\mapsto q\cdot p\cdot q^{-1}\\
\end{array}
\end{eqnarray*}
(cf.\ \cite{tits} p.\ 77--78), we abbreviate there $G_1G_2$. Since the corresponding subgroups of $SO(3)\times SO(3)$ contain the group $V\times V$, these contain  the Heisenberg group $VV$$\subseteq$$SO(4)$, and by proposition \ref{list} these are all such subgroups. We specify now the matrices:\\
\begin{eqnarray*}
\begin{array}{ll}
(A_1,1):=\left(\begin{array} {cccc}
0&-1&0&0\\
1&0&0&0\\
0&0&0&-1\\
0&0&1&0
\end{array}\right ),& (1,A_1):=\left(\begin{array} {cccc}
0&1&0&0\\
-1&0&0&0\\
0&0&0&-1\\
0&0&1&0
\end{array}\right ),\\
(A_2,1):=\left(\begin{array} {cccc}
0&0&-1&0\\
0&0&0&1\\
1&0&0&0\\
0&-1&0&0
\end{array}\right ),& (1,A_2):=\left(\begin{array} {cccc}
0&0&1&0\\
0&0&0&1\\
-1&0&0&0\\
0&-1&0&0
\end{array}\right ),\\
\end{array}
\end{eqnarray*}
\begin{eqnarray*}
\begin{array}{ll}
(A_3,1):=\left(\begin{array} {cccc}
0&0&0&-1\\
0&0&-1&0\\
0&1&0&0\\
1&0&0&0
\end{array}\right ), &(1,A_3):=\left(\begin{array} {cccc}
0&0&0&1\\
0&0&-1&0\\
0&1&0&0\\
-1&0&0&0
\end{array}\right ),\\
(R_n,1):=\left(\begin{array} {cccc}
\alpha&-\beta&0&0\\
\beta&\alpha&0&0\\
0&0&\alpha&-\beta\\
0&0&\beta&\alpha
\end{array}\right ), &(1,R_n):=\left(\begin{array} {cccc}
\alpha&\beta&0&0\\
-\beta&\alpha&0&0\\
0&0&\alpha&-\beta\\
0&0&\beta&\alpha
\end{array}\right ),\\
(S,1):=\frac{1}{2}\left(\begin{array} {cccc}
1&-1&1&-1\\
1&1&-1&-1\\
-1&1&1&-1\\
1&1&1&1
\end{array}\right ), &(1,S):=\frac{1}{2}\left(\begin{array} {cccc}
1&1&-1&1\\
-1&1&-1&-1\\
1&1&1&-1\\
-1&1&1&1
\end{array}\right ),\\
(U,1):=\frac{1}{2}\left(\begin{array} {cccc}
\tau&0&1-\tau&-1\\
0&\tau&-1&\tau-1\\
\tau-1&1&\tau&0\\
1&1-\tau&0&\tau
\end{array}\right ), &(1,U):=\frac{1}{2}\left(\begin{array} {cccc}
\tau&0&\tau-1&1\\
0&\tau&-1&\tau-1\\
1-\tau&1&\tau&0\\
-1&1-\tau&0&\tau
\end{array}\right ),
\end{array}
\end{eqnarray*}
where $\alpha:=$ cos$\frac{\pi}{n}$, $\beta:=$ sin$\frac{\pi}{n}$.\\
We can write the conjugacy classes of the groups $G_1G_2$$\subseteq$$SO(4)$ in $SO(4)$ and their number of elements. These are the images of the conjugacy classes of $\tilde{G_1}\times\tilde{G_2}$ in $G_1G_2$$\subseteq$$SO(4)$ under the double covering map. Since the fact (1.1) of \cite{sa} still holds, the matrices $(g_1,g_2)$$\in SO(4)$ with the same eigenvalues are conjugate. This simplifies the computations considerably. In this section and in the next one we omit the groups $D_nD_m$ and $(G_1G_2)'$ with $G_2=D_n$. We return to those groups later.


%We recall some facts of \cite{sa}, which simplify the computations considerably:
%\begin{itemize}
%\item The map $\tilde{G_1}\times\tilde{G_2}\rightarrow$$G_1G_2$ is the quotient map by $(-1,-1)$. So the class of $(g,1)$ has the same image as $(g,-1)$.
%\item All six matrices $\pm A_i$, $i=1,2,3$ are conjugate in $SU(2)$. So the six matrices $(\pm A_i,1)$ are conjugate in $SO(4)$, as well as the matrices $(1,\pm A_i)$.
%\item The matrices $(g_1,g_2)$$\in SO(4)$ with the same eigenvalues are conjugate in $SO(4)$.
%\item The matrices $(g_1,g_2)$ and $(g_2,g_1)$ are conjugate in $SO(4)$ via the matrix:
%\begin{eqnarray*}
%C=\left(\begin{array}{cccc}
%1&0&0&0\\
%0&-1&0&0\\
%0&0&-1&0\\
%0&0&0&-1
%\end{array}\right).
%\end{eqnarray*}
%\end{itemize}
In the tables we use the following conventions:
\begin{itemize}
\item we omit the conjugacy classes of $+1, -1$ (these contain one element each)
\item whenever the conjugacy classes $(g_1,g_2)$ and its $s$th power $(g_1^s,g_2^s)$ are distinct we write them just one time since they have the same number of elements.
\end{itemize}
%Hence with $n=2l$, $n\not=m=2l'$ we have
 
\begin{eqnarray*}
%\begin{array}{c}
\begin{array}{c|c|ccccccccc}
G_1G_2&{\rm order}&A_2,1&A_2,A_2&R_4,1&R_4,A_2&R_4,R_4&S,1&S,A_2&S,R_4&S,S\\
\hline
VV&32&12&18&0&0&0&0&0&0&0\\
TV&96&12&18&0&0&0&8&48&0&0\\
TT&288&12&18&0&0&0&16&96&0&64\\
OV&192&24&54&6&36&0&8&48&0&0\\
OT&576&24&54&6&36&0&16&192&48&64\\
OO&1152&36&162&12&216&36&16&288&96&64\\
IV&480&36&90&0&0&0&20&120&0&0\\
IT&1440&36&90&0&0&0&28&360&0&160\\
IO&2880&48&270&6&180&0&28&600&120&160\\
II&7200&60&450&0&0&0&40&1200&0&400\\
\end{array}
\end{eqnarray*}
\begin{eqnarray*}
\begin{array}{c|c|cccccc}
G_1G_2&{\rm order}&A_2,1&A_2,A_2&R_4,1&R_4,A_2&S,1&S,A_2\\
\hline
VD_n&16n&8+4l&6(2l+1)&0&0&0&0\\
%D_nD_n&8n^2&4(2l+1)&2(2l+1)^2&0&0&0&0\\
%D_nD_m&8nm&2(2l+1)&2(2l+1)(2l'+1)&0&0&0&0\\
%&&+2(2l'+1)&&&&&\\
TD_n&48n&8+4l&6(2l+1)&0&0&8&16(2l+1)\\
OD_n&96n&20+4l&18(2l+1)&6&12(2l+1)&8&16(2l+1)\\
ID_n&240n&32+4l&30(2l+1)&0&0&20&40(2l+1)\\
\end{array}
\end{eqnarray*}
\begin{eqnarray*}
\begin{array}{c|ccccccc}
G_1G_2&U,1&U,A_2&U,R_4&U,S&U,U&U,U^2&U,R_n^k\\
\hline
IV&12&72&0&0&0&0&0\\
IT&12&72&0&96&0&0&0\\
IO&12&216&72&96&0&0&0\\
II&24&720&0&480&144&144&0\\
ID_n&12&24(2l+1)&0&0&0&0&24\\
\end{array}
\end{eqnarray*}

\begin{eqnarray*}
\begin{array}{c|cccc}
G_1G_2&A_2,R_n^k& 1,R_n^k&S,R_n^k&R_4,R_n^k\\
\hline
VD_n&12&2&0&0\\
TD_n&12&2&16&0\\
OD_n&36&2&16&12\\
ID_n&60&2&40&0\\
\end{array}
\end{eqnarray*}
where $k=1,\ldots,l-1$. 
%%%%%%%%%%%%%%%%%
%\begin{eqnarray*}
%\begin{array}{c|cccccccc}
%G_1G_2&A_2,R_n^k&A_2,R_m^{k'}& 1,R_n^k&1,R_m^{k'}&R_n^k,R_n^{k_1}&R_n^k,R_m^{k'}&S,R_n^k&R_4,R_n^k\\
%\hline
%VD_n&12&0&2&0&0&0&0&0\\
%D_nD_n&4(4l+2)&0&4&0&4&0&0&0\\
%D_nD_m&2(4l+2)&2(4l'+2)&2&2&0&4&0&0\\
%TD_n&12&0&2&0&0&0&16&0\\
%OD_n&36&0&2&0&0&0&16&12\\
%ID_n&60&0&2&0&0&0&40&0\\
%\end{array}
%\end{array}
%\end{eqnarray*}
%where $k,k_1=1,\ldots,l-1$, and $k'=1,\ldots,l'-1$. Observe that some of the conjugacy classes of the groups containing $(1,D_n)$ can became equal for some values of $n$.
%%%%%%%%%%%%%%%%%%%%5555
For the groups which are not products it is a little more complicated to write down the sizes of their conjugacy classes. But using the description from proposition \ref{list} one finds\\
\begin{eqnarray*}
\begin{array}{c|c|ccccccc}
{\rm group}&{\rm order}&A_2,1&A_2,A_2&R_4,A_2&R_4,R_4&S,1&S,A_2&S,S\\
\hline
(TT)'&96&12&18&0&0&0&0&32\\
(OO)'&192&12&42&48&12&0&0&32\\
(OO)''&576&12&90&144&36&16&96&64\\
%(OD_4)'&192&12&42&48&12&8&48&0&0&0\\
%(D_nD_n)'&16n&12&18&0&0&0&0&0&16&4\\
%&&&+8(l-1)&&&&&&&\\
\end{array}
\end{eqnarray*}


%%%%%%%%%%%%%%%%%%
%\begin{eqnarray*}
%\begin{array}{c|c|ccccccccc}
%&&A_2,1&A_2,A_2&R_4,A_2&R_4,R_4&S,1&S,A_2&S,S&A_3R_n^k,R_n^k&R_n^k,R_n^k\\
%\hline
%(TT)'&96&12&18&0&0&0&0&32&0&0\\
%(OO)'&192&12&42&48&12&0&0&32&0&0\\
%(OO)''&576&12&90&144&36&16&96&64&0&0\\
%(OD_4)'&192&12&42&48&12&8&48&0&0&0\\
%(D_nD_n)'&16n&12&18&0&0&0&0&0&16&4\\
%&&&+8(l-1)&&&&&&&\\
%\end{array}
%\end{eqnarray*}
%%%%%%%%%%

\section{Poincar\'e series}
In this section we want to find the dimension of the spaces of homogeneous invariant polynomials of a given degree. We consider the Poincar\'e series 
\begin{eqnarray*}
p(t):=\sum_{j=0}^{\infty}{\rm dim}\cix^{G}_j \cdot t^j
\end{eqnarray*}
 where $G$ is a group as in section  1.4. By a theorem of Molien (\cite{ben} p.\ 21) and an easy computation as in \cite{sa}  (2.1), it  can be written as
\begin{eqnarray*}
p(t)=\frac{1}{|G|}\sum\frac{n_g}{{\rm det}(g-1\cdot t)}
\end{eqnarray*}
where the sum runs over all the conjugacy classes of $G$ and $n_g$ denote their number of elements. At the denominator we have the characteristic polynomials. Using the numbers of conjugacy classes (under $SO(4)$) given in section 1.4 and computing their characteristic polynomials, the power series package of MAPLE produces the following table of dimensions $m_d$ of invariant polynomials in degree $d$ of the groups $G$$\subseteq$$G_1G_2$. Observe that since $G$ contains the Heisenberg group we do not have invariant polynomials of odd degree. First we consider the case of $G_i$$\not=D_n$:
\begin{eqnarray*}
\begin{array}{c|cccccc}
{\rm group}&m_2&m_4&m_6&m_8&m_{10}&m_{12}\\
\hline
VV&1&5&6&15&19&35\\
TV&1&1&{\bf 2}&5&5&13\\
TT&1&1&{\bf 2}&3&3&7\\
(TT)'&1&3&4&7&9&15\\
OV&1&1&1&4&4&8\\
OT&1&1&1&{\bf 2}&{\bf 2}&4\\
OO&1&1&1&{\bf 2}&{\bf 2}&3\\
(OO)'&1&{\bf 2}&3&5&6&9\\
(OO)''&1&1&{\bf 2}&3&3&5\\
%(OD_4)'&1&1&2&4&4&8\\
IV&1&1&1&1&1&5\\
IT&1&1&1&1&1&3\\
IO&1&1&1&1&1&{\bf 2}\\
II&1&1&1&1&1&{\bf 2}\\
\end{array}
\end{eqnarray*}
Whenever $G_1\not=D_n$ and  $G_2=D_n$, we can write the finite sums $p(t)$ for each $n$. Here we compute the first coefficients of the Poincar\'e series for  $n=4,6,8$. 
%We omit however the case of $D_nD_m$ for $n\not=m$. We will see in the next section that in this case for each $n$ and $m$ we have a two-dimensional space of degree four invariants. The latter is in fact the same as the space of degree four $D_nD_n$-invariants.
\begin{eqnarray*}
%\begin{array}{ll}
\begin{array}{c|cccccc}
{\rm group}&m_2&m_4&m_6&m_8&m_{10}&m_{12}\\
\hline
VD_4&1&3&3&9&11&19\\
VD_6&1&3&3&6&6&14\\
VD_8&1&3&3&6&6&10\\
\hline
TD_4&1&1&1&3&3&7\\
TD_6&1&1&1&{\bf 2}&{\bf 2}&6\\
TD_8&1&1&1&{\bf 2}&{\bf 2}&4\\
\hline
OD_4&1&1&1&3&3&5\\
OD_6&1&1&1&{\bf 2}&{\bf 2}&4\\
OD_8&1&1&1&{\bf 2}&{\bf 2}&3\\
\hline
ID_4&1&1&1&1&1&3\\
ID_6&1&1&1&1&1&3\\
ID_8&1&1&1&1&1&{\bf 2}
%\end{array}&
%\begin{array}{c|cccccc}
%&m_2&m_4&m_6&m_8&m_{10}&m_{12}\\
%\hline
%D_4D_4&1&2&2&6&7&11\\
%D_6D_6&1&2&2&3&3&7\\
%D_8D_8&1&2&2&3&&\\
%\hline
%(D_4D_4)'&1&3&4&9&11&19\\
%(D_6D_6)'&1&2&&&&\\
%(D_8D_8)'&1&2&&&&\\
%\begin{array}{c|cc}
%&m_2&m_4\\
%\hline
%D_4D_4&1&2\\
%D_6D_6&1&2\\
%D_8D_8&1&2\\
%\hline
%(D_4D_4)'&1&3\\
%(D_6D_6)'&1&2\\
%(D_8D_8)'&1&2
%\end{array}
\end{array}
\end{eqnarray*}
Of course, whenever $m_d=1$, the space of invariant polynomials is spanned by the $d/2$-th power of the invariant quadric ({\it trivial invariant})
\begin{eqnarray*}
q:=x_0^2+x_1^2+x_2^2+x_3^2.
\end{eqnarray*}

\input{artjul01_1}





artjul01_1.tex
%\documentclass[12pt,a4paper,twoside]{article}
%\usepackage[latin1]{inputenc}
%\usepackage{amssymb}
%\usepackage{pictex}
%\usepackage{epsfig}
%\newtheorem{rem}{Remark}[section]
%\newtheorem{prop}{Proposition}[section]
%\newtheorem{lemma}{Lemma}[section]
%\newtheorem{cor}{Corollary}[section]
%\newtheorem{defi}{Definition}[section]
%\newtheorem{theorem}{Theorem}[section]
%\newcommand{\disp}{\displaystyle}
%\newcommand{\bprf}{{\it Proof.~}}
%\newcommand{\eprf}{\hfill $\square$ \bigskip\par}
%\newcommand{\ef}{F_n({\lambda})}
%\newcommand{\bn}{\mathcal{B}_n}
%\newcommand{\pitr}{ \mathbb{P}^3}
%\newcommand{\pir}{ \mathbb{P}^3_{\er}}
%\newcommand{\pit}{ \mathbb{P}^3_{\ci}}
%\newcommand{\piu}{\mathbb{P}_1}
%\newcommand{\id}{ \mathbb{I}}
%\newcommand{\aq}{\tilde{A_4}}
%\newcommand{\sq}{\tilde{S_4}}
%\newcommand{\ac}{\tilde{A_5}}
%\newcommand{\ci}{ \mathbb{C}}
%\newcommand{\er}{ \mathbb{R}}
%\newcommand{\cix}{\ci[x_0,x_1,x_2,x_3]}
%\hoffset=2.0cm 
%\voffset=1.0cm
%\begin{document}
%\begin{center}
%{\Large Pencils of symmetric surfaces in $\pitr$}\\[0.3cm]
%A.Sarti\\[0.3cm]
%June 7, 2001\\[0.3cm]
%Mathematisches Institut der Universit\"at\\
%Erlangen, Bismarckstra\ss e 1$\frac{1}{2}$\\
%Germany\\
%e-mail:sarti@mi.uni-erlangen.de\\
%\end{center}
%{\sf Version of \today}
\section{Invariants} \label{invariants}
In this section we want to compute a system of generators for the spaces $\cix_j^G$ whenever this space has dimension two and $G$$\subseteq$SO$(4)$ is a finite subgroup containing the Heisenberg group. We distinguish two cases.\\ 
{\bf (3.1)}{\it First case}. Assume that $G\subseteq G_1G_2$ with $G_i$$\not=D_n$, $i=1,2$. We do some remarks on the groups $G$ which simplify the computations of the invariant polynomials.
%or $G=(OD_4)'$. By the table in section {\bf 2.} we have eleven such spaces. Observe that:
\begin{itemize}
\item the group $TV$$\subseteq$$TT$.
%\item The generators of  $\cix_6^{WW}$ are obtained from the generators of $\cix_4^{WW}$ by multiplication with the quadric $q:=(x_0^2+x_1^2+x_2^2+x_3^2)$.
\item The group $TT$ is contained in $(OO)''$. In fact the generators modulo $VV$ are 
\begin{eqnarray*}
\begin{array}{l|ll}
{\rm group} &TT&(OO)''\\
\hline
{\rm generators}&(1,S)&(1,S),(S,1)\\
&(S,1)&(R_4A_2,R_4A_2)
\end{array}
\end{eqnarray*}
%(OD_4)'(S,1)(R_4A_2,R_4A_2)
with 
\begin{eqnarray*}
(R_4A_2,R_4A_2)=\left(\begin{array} {cccc}
1&0&0&0\\
0&-1&0&0\\
0&0&0&1\\
0&0&1&0
\end{array}\right )
\end{eqnarray*} 
Similarly to \cite{sa} section  3., $(OO)''$ is a subgroup of index two in the reflection group of the $\{3,4,3\}$-cell (cf.\ \cite{cox} p.\ 149 for the definition of this polytope), if we add the generator 
\begin{eqnarray*}
C=\left( \begin{array} {cccc}
1 & 0& 0 & 0 \\
0 & -1& 0 & 0 \\
0 & 0& -1& 0 \\
0 & 0& 0 & -1 
\end{array} \right)
\end{eqnarray*}
 we get the whole reflection group. 
\item Finally since $OT$$\subseteq$$OO$, they have the same two invariant polynomials in degree eight. Those of degree ten are obtained just by multiplication with the quadric $q$.  
\end{itemize}

By these remarks and considering $G$ as in the assumption, it follows that we have to compute the generators of six two-dimensional spaces $\cix_j^G$ for the following pairs $(G,j)$:
\begin{eqnarray*}
\begin{array}{lllll}
((OO)',4)&(TT,6)&(OO,8)&(IO,12)&(II,12).\\
\end{array}
\end{eqnarray*}
The generators of the $TT$-, $OO$- and $II$-invariant spaces are given in \cite{sa} as well as a description of the corresponding pencils of surfaces in $\pitr$ (base locus and singular surfaces). Here we examine the remaining  cases and in the next section we describe the corresponding pencils of surfaces in $\pitr$. The basic idea to find generators of the invariant spaces is the same as in \cite{sa}. \\
{\it $(OO)'$-invariants}.
We start with the space of Heisenberg invariant quartics. It has dimension five, being spanned by 
\begin{eqnarray*}
\begin{array}{lll}
f_0:=x_0^4+x_1^4+x_2^4+x_3^4,&&\\
f_1:=2(x_0^2x_1^2+x_2^2x_3^2),&f_2:=2(x_0^2x_2^2+x_1^2x_3^2),&f_3:=2(x_0^2x_3^2+x_1^2x_2^2),\\
f_4:=4x_0x_1x_2x_3.&&\\
\end{array}
\end{eqnarray*}
In terms of these invariants
\begin{eqnarray*}
q^2=f_0+f_1+f_2+f_3.
\end{eqnarray*}
%On these invariants the symmetries (S,1) and (R,1) act by 
%\begin{eqnarray*}
%\begin{array}{l|lll}
%(S,1)&f_0&\mapsto&\frac{1}{4}(f_0+3f_1+3f_2+3f_3-6f_4)\\
%&f_1&\mapsto&\frac{1}{4}(f_0-f_1+3f_2-f_3+2f_4)\\
%&f_2&\mapsto&\frac{1}{4}(f_0-f_1-f_2+3f_3+2f_4)\\
%&f_3&\mapsto&\frac{1}{4}(f_0+3f_1-f_2-f_3+2f_4)\\
%&f_4&\mapsto&\frac{1}{4}(f_0-f_1-f_2-f_3-2f_4)\\
%\hline
%(R,1)&f_0&\mapsto&\frac{1}{2}(f_0+3f_1)\\
%&f_1&\mapsto&\frac{1}{2}(f_0-f_1)\\
%&f_2&\mapsto&\frac{1}{2}(f_2+f_3+2f_4)\\
%&f_3&\mapsto&\frac{1}{2}(f_2+f_3-2f_4)\\
%&f_4&\mapsto&\frac{1}{2}(f_2-f_3)\\
%\end{array}
%\end{eqnarray*}
%The action of $(1,S)$ resp. $(1,R)$ is obtained by conjugating this table with the symmetry $C$ from page... Explicitly: change the sign of $f_4$ in the transformed polynomials and then change the sign of the transform of $f_4$.
Modulo $VV$, the group $(OO)'$ is generated by $(R_4A_2,R_4A_2)$ and $(S,S)$.
%\begin{eqnarray*}
%\begin{array}{l|ll}
%{\rm group}&WW&OO'\\
%\hline
%{\rm generators}&(R,1)&(R,R)\\
%&(1,R)&(S,S)\\
%\end{array}
%\end{eqnarray*}
Tracing the action of these generators on $f_0,\ldots, f_4$ one finds the invariants $q^2$ and $f_0$.\\
%\begin{eqnarray*}
%\begin{array}{ll}
%WW&f_0+f_1, f_2+f_3\\
%\hline
%OO'&f_0, f_1+f_2+f_3
%\end{array}
%\end{eqnarray*}

{\it $IO$-invariants}. 
The generators $(U,1)$, $(1, S)$ and $(1, R_4)$ of $IO$ operate on the space $\cix^{VV}_{12}$ which is 35-dimensional. A computation with MAPLE shows that it  contains the non trivial $IO$-invariant polynomial
\begin{eqnarray*}
\begin{array}{lll}
S_{IO}&:=&-\sum_{i} f_i^3+11\sum_{i,j} f_i^2f_j+(f_0^2-14f_4^2)\sum_{i}f_i+30f_0f_4^2-2f_0\sum_{i,j}f_if_j\\
&&-30f_1f_2f_3+3\sqrt{5}f_4(2f_0\sum_{i}f_i-\sum_{i,j}f_if_j-f_0^2-\sum_{i}f_i^2+4f_4^2)\\
&&+6\sqrt{5}F_a
\end{array}
\end{eqnarray*}
where the sums run over all the indices $i,j=1,2,3$, $i\not=j$, and 
\begin{eqnarray*}
F_a:=f_1^2f_2+f_3^2f_1+f_2^2f_3-f_1f_2^2-f_3f_1^2-f_2f_3^2
\end{eqnarray*}
is the anti-symmetric part of $S_{IO}$.\\
%(controllare!)
{\bf(3.2)}{\it Second case}. Assume that $G$$\subseteq$$G_1G_2$ with $G_1=D_n$ or $G_2=D_m$, $n,m\geq 4$, even. With the help of the table in section 1.4 we discuss the following cases:
\begin{itemize}
\item $D_nD_m$: observe that $VD_m$ is contained in $D_nD_m$. A direct computation shows that the generator $(1,R_n)$ let invariant a three-dimensional family of degree four $VV$-invariant polynomials. Generators in this case are $f_2+f_3$, $q^2$, $f_2-f_4$. Now the matrix $(R_n,1)$ of $D_nD_m$ operates on these three-dimensional space leaving invariant $q^2$ and $f_2+f_3$.
%\item $D_nD_n$: We have a two-dimensional family of invariants of degree four for each $n$. Generators are $q^2$ and $f_2+f_3$.
%\item $D_nD_m$: since $VD_m$$\subseteq$$D_nD_m$ we have to consider the action of $(R_n,1)$ on the space of degree four $VD_n$-invariant polynomials generated by $q^2$, $f_2+f_3$ and $f_2-f_4$. As in the previous case the only invariant are $q^2$ and $f_2+f_3$.
\item since $(D_nD_m)'$$\subseteq$$D_nD_m$ these groups have already an at least two-dimensional family of invariant polynomials of degree four, which is generated by $q^2$ and $f_2+f_3$.
%$(D_nD_n)'$, $n\geq 6$: the matrix $(R_n,R_n)$ corresponds to the coordinates transformation: $x_0\mapsto x_0$, $x_1\mapsto x_1$, $x_2\mapsto \alpha_1 x_2-\beta_1 x_3$, $x_3\mapsto \alpha_1 x_3+\beta_1 x_2$, $\alpha_1=\alpha^2-\beta^2$, $\beta_1=2\alpha\beta$. Let now $F:=A_0f_0+A_1f_1+A_2f_2+A_3f_3+A_4f_4$ applying the previous transformation and comparing the coefficients of $F$ and $(R_n,R_n)$$\cdot F$ we find the generators $q^2$ and $f_2+f_3$.
\item $TD_n$, $n\geq 6$: we have a five-dimensional family of $TV$-invariant in degree eight. By the action of the generator $(1,R_n)$, $n\geq 6$ we have a two-dimensional family of invariants. Put 
\begin{eqnarray*}
\begin{array}{l}
K_i:=x_0^2+x_1^2-x_2^2-x_3^2+(-1)^i2x_0(x_2+x_3)+2x_1(x_2-x_3),~i=0,3\\
K_i:=x_0^2+x_1^2-x_2^2-x_3^2+(-1)^i2x_0(x_2+x_3)-2x_1(x_2-x_3),~i=1,2
\end{array}
\end{eqnarray*}
then generators are 
\begin{eqnarray*}
q^4 ~\mbox{and}~ P_8:=\prod_{i=0}^{3}K_i.
\end{eqnarray*}
\item Suppose three divides $n$. Since the groups  $(TD_n)'$ are contained in $TD_n$, they have already at least a two-dimensional family of invariant polynomials in degree eight. We consider now the action of the extra generators $(S,R_n)$ of $(TD_n)'$ on the space of degree four, resp. six $VV$-invariant polynomials. A direct computation with MAPLE shows that we have no invariants other then the quadric $q^2$, resp. $q^3$. 
\item $OD_n$, $n\geq 6$: we have $TD_n$$\subseteq$$OD_n$ and the extra generator $(R_4,1)$ leaves the previous polynomials invariant.
\item Suppose now four divides $n$. Since the groups $(OD_n)'$ are contained in $OD_n$, they have at least a two-dimensional family of invariant polynomials in degree eight. Modulo $VV$ the groups $(OD_n)'$ have generators $(S,1)$ and $(R_4,R_n)$, hence they contains the group $TV$.  This has no non-trivial invariant polynomials of degree four and has a two-dimensional family of invariant polynomials in degree six, generated by $q^3$ and $S_6(x)$ (cf. \cite{sa} p.\ 437). For $n\geq 6$, $S_6(x)$ is not $(OD_n)'$-invariant, for $n=4$ it is. In any case we do not get new invariants.  
\item $ID_n$, $n\geq 8$: we have a five-dimensional family of $IV$-invariant polynomials in degree twelve. The action of $(1,R_n)$, $n\geq 8$ on this space produces a two-dimensional family of invariant polynomials. Put 
\begin{eqnarray*}
\begin{array}{l}
K_0':=x_0^2+x_1^2-x_2^2-x_3^2+2(1-\tau)(x_0x_2-x_1x_3)\\
K_1':=x_0^2+x_1^2-x_2^2-x_3^2+2\tau(x_0x_3+x_1x_2)\\
K_2':=x_0^2+x_1^2-x_2^2-x_3^2-2(1-\tau)(x_0x_2-x_1x_3)\\
K_3':=x_0^2+x_1^2-x_2^2-x_3^2-2\tau(x_0x_3+x_1x_2)\\
K_4':=2(1-\tau)(x_0x_3+x_1x_2)+2(x_0x_2-x_1x_3)\\
K_5':=-2\tau(x_0x_3+x_1x_2)+2(x_0x_2-x_1x_3)
\end{array}
\end{eqnarray*}
then generators are 
\begin{eqnarray*}
q^6~\mbox{and}~P_{12}:=\prod_{i=0}^{5}K_i'.
\end{eqnarray*}
\end{itemize}
In conclusion, the dimension of $\cix^G_j$ is two, for $(G,j)$ equal to
 \begin{eqnarray*}
\begin{array}{lllll}
(D_nD_m,4)~ n,m\geq 4,&(OD_n,8)~n\geq 6,&(ID_n,12)~n\geq 8.\\
\end{array}
\end{eqnarray*}
{\bf Remark 3.3} Observe that the polynomials $q^2$ and $f_2+f_3$ are $D_nD_n$-invariant even if $n$ is an odd integer. As in the even case, generators of these groups are $(A_2,1)$, $(1,A_2)$ and $(R_n,1)$, $(1,R_n)$.

\section{Invariant pencils}
%In this section we describe the three pencils of symmetric surfaces of $\pitr$, with generators the polynomials of section 4. \\
%The basis locus is the union of these four lines with multiplicity two. We investigate the singular surfaces in the pencil by a direct calculation. The map
%\begin{eqnarray*}
%\begin{array}{lclc}
%f:&\pitr&\longrightarrow&\pitr\\
%&(x_0:x_1:x_2:x_3)&\mapsto&(x_0^2:x_1^2:x_2^2:x_3^2)\\
%\end{array}
%\end{eqnarray*}
%of (Cat-Ceresa) 
%is an $8:1$ cover with branch locus the coordinate tetrahedron $x_0x_1x_2x_3=0$. The pencil $(f_2+f_3)+\lambda q^2$ is sent to $(u_0+u_1)(u_2+u_3)+\lambda(u_0+u_1+u_2+u_3)^2$ in $\pitr$ with coordinates $u_i$.
%A singular point with coordinates $u_i\not=0$, $i=0,1,2,3$ corresponds to eight singular points upstairs. First assume that none of the $u_i$ is zero and consider the system given by the partial derivatives and the equation of the pencil downstairs:
%\begin{eqnarray*}
%\begin{array}{l}
%2\lambda(u_0+u_1+u_2+u_3)+(u_0+u_1)=0\\
%2\lambda(u_0+u_1+u_2+u_3)+(u_2+u_3)=0\\
%\lambda(u_0+u_1+u_2+u_3)^2+(u_0+u_1)(u_2+u_3)=0\\
%\end{array}
%\end{eqnarray*}
%We can assume $u_0+u_1+u_2+u_3\not=0$, otherwise we find again the four double lines in the base locus. Hence
%\begin{eqnarray*}
%-\frac{u_2+u_3}{2(u_0+u_1+u_2+u_3)}=\lambda=-\frac{u_0+u_1}{2(u_0+u_1+u_2+u_3)}
%\end{eqnarray*}
%and we get $u_2+u_3=u_0+u_1$, which gives $\lambda=-\frac{1}{4}$. For this value of $\lambda$ we find the multiple quadric $q'^2:(x_0^2+x_1^2-x_2^2-x_3^2)^2=0$. This is projective equivalent to $q$ (take the transformation $x_j\mapsto$$x_j$ for $j=0,1$ and $x_j\mapsto ix_j$ for $j=2,3$). Now we are done. In fact taking the generators $q'^2$ the equation of the pencil becomes
%For $\lambda=\frac{1}{4}$ we find the four planes of equation $f_2+f_3$.

%Whenever some $x_j=0$ we consider the system
%\begin{eqnarray*}
%\begin{array}{l}
%2\lambda x_i(x_0^2+x_1^2+x_2^2+x_3^2)+x_i(x_2^2+x_3^2)=0,~~~~i=0,1 \\
%2\lambda x_i(x_0^2+x_1^2+x_2^2+x_3^2)+x_i(x_0^2+x_1^2)=0,~~~~i=2,3\\
%\lambda(x_0^2+x_1^2+x_2^2+x_3^2)^2+(x_0^2+x_1^2)(x_2^2+x_3^2)=0\\
%\end{array}
%\end{eqnarray*}
%in any case we find again $\lambda=-\frac{1}{4}$ or $\lambda=0$. 

%We collect the results about the $WW$-invariant pencil in the following table


%\renewcommand{\arraystretch}{1.3}

%\begin{eqnarray*}
%\begin{array}{r|c|l}
%\lambda& {\rm equation}& {\rm description}\\ 
%\hline
%-\frac{1}{4}&(x_0^2+x_1^2-x_2^2-x_3^2)^2=0&{\rm double~quadric}\\
%\hline
%0 &(x_0^2+x_1^2)(x_2^2+x_3^2)^2=0&{\rm four~planes}\\
%\hline
%\infty& (x_0^2+x_1^2+x_2^2+x_3^2)^2=0&{\rm double~quadric}
%\end{array} 
%\end{eqnarray*}

%\renewcommand{\arraystretch}{1.0}
\subsection{The pencil of $(OO)'$-invariant quartics}
We take the generators $q^2$ and $f_0:=x_0^4+x_1^4+x_2^4+x_3^4$. The base locus is a double curve of degree eight. Let $C_8$ denote the curve $\{q=0,~f_0=0\}$. The pencil is invariant under the matrix $C$ too, hence by symmetry reasons the curve $C_8$ has bi-degree $(4,4)$ on $q$ (cf.\ \cite{sa}, section 5). Moreover we have the following
\begin{lemma} The curve $C_8$ is smooth and irreducible. \end{lemma}
{\it Proof.} The Jacobian matrix of $C_8$ has rank two in each point, hence the curve is smooth. If $C_8=C'\cup C''$, the curves $C'$, $C''$ would meet in some point on $q$ (observe that they cannot be lines of the same ruling), but this is impossible because $C_8$ is smooth. \eprf
In particular (6.1) of \cite{sa} still holds and by Bertini's theorem the general surface in the pencil is smooth. All the other surfaces but $q^2$ are irreducible and reduced and the singular ones have only isolated singularities. \\
%To find the singular surfaces it is now possible to do a direct computation as in {\bf 5.1}, but we prefer to proceed as in \cite{sa}, paragraph {\bf 9.} and consider some special points on a four-dimensional regular polytopes.\\ 
%As in {\bf 5.1} we do a direct computation to find the singular surfaces in the pencil. We consider again the map $f$, the pencil $f_0+\lambda q^2$ is now sent to $(u_0^2+u_1^2+u_2^2+u_3^2)+\lambda(u_0+u_1+u_2+u_3)^2$$=0$. Consider the system
%\begin{eqnarray*}
%\begin{array}{l}
%u_i+\lambda (u_0+u_1+u_2+u_3)=0,~~~i=0,1,2,3\\
%(u_0^2+u_1^2+u_2^2+u_3^2)+\lambda(u_0+u_1+u_2+u_3)^2=0\\
%\end{array}
%\end{eqnarray*}
%and assume $u_i\not=0$, $i=0,1,2,3$ (which implies $u_0+u_1+u_2+u_3\not=0$), then $\lambda=-\frac{u_i}{u_0+u_1+u_2+u_3}$ $i=0,1,2,3$. This shows $u_0=u_1=u_2=u_3$ and $\lambda=-\frac{1}{4}$. Hence the surface upstairs has eight singularities, lies on the point $(1:1:1:1)$. We consider now the points with some $x_i=0$. The system is now
%\begin{eqnarray*}
%\begin{array}{l}
%x_i^3+\lambda x_i (x_0^2+x_1^2+x_2^2+x_3^2)=0,~~~i=0,1,2,3\\
%(x_0^4+x_1^4+x_2^4+x_3^4)+\lambda(x_0^2+x_1^2+x_2^2+x_3^2)^2=0\\
%\end{array}
%\end{eqnarray*} 
%I. W.l.o.g $x_0=0$. Substituting in the system we find $x_1^2=x_2^2=x_3^2$ and $\lambda=-\frac{1}{3}$. So we find 16 singular points. \\ 
%II. W.l.o.g $x_0=x_1=0$. In this case we get $x_2^2=x_3^2$ and $\lambda=-\frac{1}{2}$. Hence we have a surface with 12 singular points.\\
%III. W.l.o.g $x_0=x_1=x_2=0$. In this case $\lambda=-1$ and we have a surface with four singular points.\\
{\it The symmetry group of $\{3,3,4\}$}. Consider the ``cross polytope'' $\beta_4$$=\{3,3,4\}$ in $\mathbb{R}^4$ with vertices the permutations of $(\pm 1, 0, 0,0)$ as in \cite{cox} p.\ 156 and edge $\sqrt{2}$. The generators of $(OO)'$ permutes these points, hence $(OO)'$ is contained in the symmetry group $[3,3,4]$ of $\{3,3,4\}$.  More precisely it is an index two subgroup, in fact the symmetry group of $\beta_4$ has order $2^4\cdot 4!=384$$=2\cdot 192$ and by adding the generator $C$ to $(OO)'$ we get the whole symmetry group $[3,3,4]$ (cf. \cite{cox} p.\ 226). In particular observe that they have the same invariant polynomials. The polytope $\beta_4$ has $N_0=8$, $N_1=24$, $N_2=32$, $N_3=16$ the reciprocal ``measure polytope'', $\gamma_4$$=\{4,3,3\}$ has $N_0=16$, $N_1=32$, $N_2=24$, $N_3=8$. Hence we get four $[3,4,3]$-orbits of points: the vertices and the middle points of the edges of the $\beta_4$ and the vertices and the middle points of the edges of the  reciprocal $\gamma_4$. These have coordinates the permutation of $(\pm 1,0,0,0)$, $(\pm 1,\pm 1, 0,0)$,resp. $(\pm 1,\pm 1,\pm 1, \pm 1)$, $(\pm 1, \pm 1, \pm 1, 0)$. As points of $\pitr$ these are singular on the surfaces $f_0+\lambda q^2$ for $\lambda=-1$, $-\frac{1}{2}$ resp. $-\frac{1}{4}$, $-\frac{1}{3}$ and a direct computation shows that they are all ordinary double points. We do it for $\lambda=-1$ and $(1:0:0:0)$. In the affine chart $\{x_0\not=0\}$ the equation becomes 
\begin{eqnarray*}
\begin{array}{lll}
0&=&1+x^4+y^4+z^4-(1+x^2+y^2+z^2)^2\\
&=&1-2x^2-2y^2-2z^2+{\rm terms~of~degree} \geq 4,
\end{array}
\end{eqnarray*}
hence the rank of the Hessian matrix at $(0,0,0)$ is three. This shows that $(1:0:0:0)$ is an ordinary double point and so are all the points in its orbit.\\  We collect the results on the singular surfaces in the following table. In the middle column we write just one point, but we mean all its permutations. 


%\renewcommand{\arraystretch}{1.3}
\begin{eqnarray*}
\begin{array}{r|c|l}
\lambda& {\rm nodes}& {\rm number}\backslash {\rm description}\\ 
\hline
-1&(1:0:0:0)&4\\
\hline
-\frac{1}{2}&(\pm 1:\pm 1:0:0)&12\\
\hline
-\frac{1}{3}&(\pm 1:\pm 1:\pm 1:0)&16, {\rm Kummer~surface}\\
\hline
-\frac{1}{4}&(\pm 1:\pm 1:\pm 1:\pm 1)&8\\
\hline
\hline
\infty&-&{\rm double~quadric}
\end{array}
\end{eqnarray*}

%\renewcommand{\arraystretch}{1.0}

In the case of $\lambda=-\frac{1}{3}$, we get a surface with 16 nodes which is the maximal number possible for a surface of degree four. This is a Kummer surface. Observe that in this case too as in \cite{sa} the nodes are fix points under the action of some matrices in $[3,3,4]$, resp. in $(OO)'$ and they are contained on lines of fix points, see  (5.2) and  (6.3) of \cite{sa}. Moreover an estimation as in section 8 of \cite{sa} shows that whenever the lines of fix points do not meet the base locus, they contain exactly four nodes. It is natural to expect that these are all the singular surfaces in the $(OO)'$-invariant pencil (as in the case of the $TT$-, $OO$-, and $II$-invariant pencils). This is a direct consequence of the following\\
\begin{prop} 
1. The conjugacy classes in $(OO)'$ (under $(OO)'$) with eigenvalues $\pm 1$ are the following
\begin{eqnarray*}
\begin{array}{l|c|c|c|c|c}
{\it conj.~class}&(A_2,A_2)&(A_2,A_1)&(R_4A_2,R_4A_2)&(S,S)&(R_4,R_4)\\
\hline
{\it number~of~elements}&6&12&24&32&12\\
{\it number~of~fix~lines}&6&12&24&16&6\\
\end{array}
\end{eqnarray*}
2. The fix lines of the matrices in these conjugacy classes contain the maximal number possible of node.\\
% lines of the elements in the conjugacy class of $(S,S^2)$ are contained in $f_0-\frac{1}{2}q^2$. The other lines contain the maximal number possible of nodes.\\}
\end{prop}
{\it Proof.} Choosing a fix line for each of the representative above and intersecting with the singular surfaces we find%In SO$(4)$ the conjugacy classes of $(A_i,A_i)$, $(A_i,A_j)$, $i\not= j$ and $(R,R)$ resp. of $(S,S)$ and $(S,S^2)$ are the same. Choosing now a fix line for each of the matrices and intersecting with the singular surfaces we find
\vspace*{0.3cm}

\begin{eqnarray*}
\begin{array}{c|c|c|c|c}
{\rm matrix~and~fix~line}&\multicolumn{2}{c|}{{\rm value~of} ~\lambda}&\multicolumn{2}{c}{{\rm int.~points}}\\
\hline
(A_2,A_2): &-1&-\frac{1}{2}&(1:0:0:0)&(\pm 1:0:1:0)\\
x_1=x_3=0&&&(0:0:1:0)&\\
(A_2,A_1):&-\frac{1}{2}&-\frac{1}{4}&(0:1:1:0)&(1:1:1:1)\\
x_0=x_3, x_1=x_2&&&(1:0:0:1)&(-1:1:1:-1)\\
(R_4A_2,R_4A_2):&-1&-\frac{1}{2}&(1:0:0:0)&(0:0:1:1)\\
x_1=0, x_2=x_3&-\frac{1}{3}&&(\pm 1:0:1:1)&\\
(S,S):&-1&-\frac{1}{3}&(1:0:0:0)&(0:1:1:1)\\
x_1=x_2=x_3&-\frac{1}{4}&&(\pm 1:1:1:1)&\\
\end{array}
\end{eqnarray*}
%(S,S^2):&-\frac{1}{2}&&(0:-1:0:1)&{\rm +2~points ~of}~C_8\\
%x_0=0,x_1=-x_2&&&(0:-1:1:0)&\\
%&&&(0:0:-1:1)&\\
%\end{array}
%\end{eqnarray*}

where we do not write the matrix $(R_4,R_4)$, since the fix lines of the matrices in its conjugacy class are the same as those of the matrices in the conjugacy class of $(A_2,A_2)$. From this table follows that:\\
\begin{itemize}
\item the fix lines above meet different surfaces, hence the conjugacy classes of these matrices are in fact, all distinct (cf. also \cite{sa} (7.3)).\\
\item The fix lines contain the maximal number possible (four), of nodes (this shows {\it 2.})
\end{itemize}






%\begin{eqnarray*}
%\begin{array}{l|c|c|c|c|c|c|c|c|c|c|c}
%{\rm matrix}&\multicolumn{2}{c|}{(A_i,A_i)}&\multicolumn{2}{c|}{(A_i,A_j)}&\multicolumn{3}{c|}{(R,R)}&\multicolumn{3}{c|}{(S,S)}&(S,S^2)\\
%\hline
%\lambda &-1&-\frac{1}{2}&-\frac{1}{2}&-\frac{1}{4}&-1&-\frac{1}{2}&-\frac{1}{3}&-1&-\frac{1}{3}&-\frac{1}{4}&-\frac{1}{2}\\
%\hline
%{\rm number~of}&2&2&2&2&1&1&2&1&1&2&3 \\
% {\rm int.~points}&&&&&&&&&&&+2~{\rm of}~C_8\\
%\end{array}
%\end{eqnarray*}

%\item The fix lines of the elements in the conjugacy class of $(S,S^2)$ are contained in $f_0-\frac{1}{2}q^2$. In the other cases the fix lines contain the maximal number possible of nodes, in particular they do not meet $C_8$ (this shows 2).\\

About the number of fix lines: observe that the conjugacy classes of $(S,S)$ has order $32$ and contain the elements $(S^2,S^2)$ hence we have $\frac{32}{2}=16$ distinct fix lines. Finally the conjugacy class of $(A_2,A_2)$ contains six elements, $(A_2,A_1)$ contains twelve elements and $(R_4A_2,R_4A_2)$ contains $24$ elements. Since each element in these conjugacy classes has two fix lines and the elements with minus sign are in the same conjugacy class, the number of fix lines is the same as the number of matrices.
\eprf
 We know that the singular points form $[3,4,3]$-orbits, but we can now show something more. \\
\begin{lemma} the nodes on each singular surface form one $(OO)'$-orbit:

%\renewcommand{\arraystretch}{1.3}
\begin{eqnarray*}
\begin{array}{l|l|l|l|l}
\lambda&-1&-\frac{1}{2}&-\frac{1}{3}&-\frac{1}{4}\\
\hline
{\it orbit}&4&12&16&8\\
\hline
{\it fix~group~mod.}\pm 1&S_4&D_4&D_3&A_4\\      
\hline
{\it order} &24&8&6&12\\
\end{array}
\end{eqnarray*}
\end{lemma}
%\renewcommand{\arraystretch}{1.0}

{\it Proof.} The situation is easy for $\lambda=-1$. In fact the matrix $C$$\in$$[3,4,3]$ let each singular point fix, hence the group $(OO)'$ must permute them. Consider $\lambda=-\frac{1}{2}$, $-\frac{1}{3}$ resp. -$\frac{1}{4}$ and assume that the orbit's length of  singular points is less or equal then $12$, $16$, resp. $8$. Then the fix group in $(OO)'$ mod. $\pm 1$ has order bigger or equal then $8$, $6$ resp. $12$. Checking in the table given in the previous page we see that in fact the only possibility is to have equality.
\eprf 



%I. \underline{ The four nodes on $f_0-q^2$}. Observe that the matrix $C$  let each point fix, hence the group $(OO)'$ must permute them.\\
%II. \underline{The eight nodes on $f_0-\frac{1}{4}q^2$}. The two points on the fix line of $(A_2,A_1)$ form one orbit. So we have a configuration of lines and points (cf. \cite{sa}, paragraph {\bf 11.}). By the formula
%\begin{eqnarray*}
%(*)~~N_0\cdot n_1=N_1\cdot n_0
%\end{eqnarray*}
%with $N_0\leq 8$ we find $n_1\geq 3$. If $n_1\geq 4$ the point is contained in at least one fix line of a matrix in the conjugacy class of $(A_i,A_i)$, but these lines do not contain singularities of $f_0-\frac{1}{4}q^2$. Hence $n_1=3$ and $N_0=8$. This shows that the eight singular points form one $(OO)'$-orbit.\\
%The fix group in $(OO)'$$\subseteq$$\mathbb{P}$GL(3) has order twelve, hence it is isomorphic with $A_4$. Since none of the point is contained on a fix line for some element in the conjugacy class of $(A_i,A_i)$,w.l.o.g. it must be generated by $(A_i,A_j)$  $i\not= j$ and $(S,S)$. Hence each point of the eight is contained in a fix line of some element in  $(A_i,A_j)$ and $(S,S)$. To show that they belong to just one orbit, observe that the matrix $(A_i,1)$ commutes with $(A_i,A_j)$ hence the two points on this line form one orbit. By this fact follows that we have a configuration of lines and points (cf...). Now use the formula  $(*)$$N_0\cdot n_1=N_1\cdot n_0$ to find the length of the orbit of points. We have $N_1=12$, $n_0=2$ and $n_1=3$ since the fix group is isomorphic with $A_4$, hence $N_0=8$, which shows that the $8$ points form one orbit.\\  
%In the case of the conjugacy class of $(S,S)$, take the fix line $<...>$ of $\pi_3\pi_3'$ (notation of ....), this contain the points $(\pm 1:1:-1:1)$ of $f_0-\frac{1}{4}q^2$ and the matrix 
%\begin{eqnarray*}
%(R,R)=\left(\begin{array} {cccc}
%1&0&0&0\\
%0&-1&0&0\\
%0&0&0&1\\
%0&0&1&0
%\end{array}\right )
%\end{eqnarray*}
%interchange these two points (and fix the points of intersection with $f_0+\lambda q^2$, $\lambda=-1,-\frac{1}{3}$). Hence these eight nodes form one orbit too.\\
%III. \underline{The twelve nodes on $f_0-\frac{1}{2}q^2$}. Consider the fix lines of the matrices in the conjugacy class of $(S,S^2)$. These contain three singular points, so by the formula $(*)$ and $N_0$$\leq 12$ we get $n_1\geq 4$. If $n_1>4$ more then four lines meet at the singular point, but this cannot happen (consider the matrices in the fix group of the point). Therefore $N_0=12$.\\ 
%The table.. shows that the fix group of such a points has order eight, hence it is a dihedral group $D_4$. W.l.o.g. this can be generated by $(R,R)$ and $(A_i,A_i)$ or $(R,R)$ and $(A_i,A_j)$, $i\not=j$. In any case a singular point is contained on a fix line of an element in the conjugacy class of $(R,R)$. We have a configuration of lines and points and by using the formula $(*)$, with $N_1=24$, $n_0=1$, and $n_1=2$, we find $N_0=12$. This shows that the points form one orbit under $(OO)'$ and in particular that the fix group is generated by $(R,R)$ and $(A_i,A_j)$, $i\not=j$.\\
%IV. \underline{The sixteen nodes on $f_0-\frac{1}{3}q^2$}. We consider the fix lines of the elements in the conjugacy class of $(S,S)$. The formula $(*)$ shows $n_1\geq 1$. If $n_1\geq 2$ then a singular point is contained in at least one fix line of an element in the conjugacy class of $(A_2,A_2)$ or $(A_2,A_1)$. By the table above, this is not possible hence $n_1=1$ and $N_0=16$.
%\eprf
%The fix group has order six hence it is the dihedral group $D_3$, generated by some elements in the conjugacy class of $(R,R)$ and $(S,S)$. Using the formula $(*)$ with $N_1=16$(=the number of fix lines of the elements in the conjugacy class of $(S,S)$) $n_0=1$ and $n_1=1$ we find $N_0=16$. So these nodes form one orbit too.\\
Put now $N_0=$number of nodes on a surface in the pencil, $N_1=$number of fix lines of matrices in the same conjugacy class, $n_0$=number of nodes on a line, $n_1$=number of line through a point. Knowing the fix groups of the singular points and using the formula
\begin{eqnarray}\label{conf}
N_0\cdot n_1=N_1\cdot n_0
\end{eqnarray}
for a configuration of lines and points (cf.\ \cite{sa} section 11.), we can write the table:


%\renewcommand{\arraystretch}{1.3}
\begin{eqnarray*}
\begin{array}{c|c|c|c|c|c|c}
{\rm Repr.~of~the~conj.~class}&\multicolumn{3}{c|}{{\rm value~of }~\lambda}&\multicolumn{3}{c}{{\rm Configuration}}\\
%{\rm conj.~class}&\multicolumn{2}{c|}{{\rm }}&\multicolumn{2}{c}{}\\
\hline
(A_2,A_2)&-1&-\frac{1}{2}&&(4_3,6_2 )&(12_1,6_2)&\\
(A_2,A_1)&-\frac{1}{2}&-\frac{1}{4}&&(12_2,12_2)&(8_3,12_2)&\\
(R_4A_2,R_4A_2)&-1&-\frac{1}{2}&-\frac{1}{3}&(4_6,24_1)&(12_2,24_1)&(16_3,24_2)\\
(S,S)&-1&-\frac{1}{3}&-\frac{1}{4}&(4_4,16_1)&(16_1,16_1)&(8_4,16_2)\\
\end{array}
\end{eqnarray*}
where in the second column we mean the nodes of the surfaces with the given $\lambda$.
%\renewcommand{\arraystretch}{1.0}


%\begin{eqnarray*}
%\begin{array}{ll}
%\begin{array}{l|c|c|c|c}
%{\rm conj. class}&\multicolumn{2}{c|}{(A_i,A_i)}&\multicolumn{2}{c}{(A_i,A_j)}\\
%\hline
%\lambda &-1&-\frac{1}{2}&-\frac{1}{2}&-\frac{1}{4}\\
%\hline
%{\rm config.}&(6_2,4_3)&(6_2,12_1)&(12_2,12_2)&(12_2,8_3)\\
%\end{array}
%\end{eqnarray*}

%\vspace{0.5cm}

%\begin{eqnarray*}
%\begin{array}{l|c|c|c|c|c|c}
%{\rm conj. class}&\multicolumn{3}{c|}{(R,R)}&\multicolumn{3}{c}{(S,S)}\\
%\hline
%\lambda &-1&-\frac{1}{2}&-\frac{1}{3}&-1&-\frac{1}{3}&-\frac{1}{4}\\
%\hline
%{\rm config.} &(24_1,4_6)&(24_1,12_2)&(24_2,16_3)&(16_1,4_4)&(16_1,16_1)&(16_2,8_4)\\
%\end{array}
%\end{array}
%\end{eqnarray*}
\subsection{The pencil of $IO$-invariant 12-ics}
We take the generators $q^6$ and $S_{IO}$. First we compute the base locus. We use the terminology of \cite{sa} definitions  (5.1),  (5.2) and the facts (5.3) and (5.4). Moreover we consider the groups $(1,O)$, resp. $(I,1)$, which operate on the two rulings of the quadric $q$. To convenience of the reader we recall the table on the length of the orbits under the action of the octahedral group $O$, and of the icosahedral group $I$:   

\begin{center}

\renewcommand{\arraystretch}{1.3}

\begin{eqnarray*}
\begin{array}{c|c}
{\rm octahedron} &{\rm icosahedron}\\
\hline
24,~12,~8,~6&60,~30,~20,~12\\
\end{array}
\end{eqnarray*}

\end{center}

\renewcommand{\arraystretch}{1.0} 


\noindent
Now we show:\\
\begin{lemma} The variety $q\cap S_{IO}$ consists of an $(I,1)$-orbit of twelve lines and of an $(1,O)$-orbit of six singular lines of $S_{IO}$.\end{lemma}
%the variety $q\cap S_{IO}$ consists of $12$ lines, resp. $6$ lines of $q$, which form one $(-,I)$-orbit, resp. one $(O,-)$-orbit
{\it Proof}. We have deg$(q\cap S_{IO})$=24. By the table above it can only have bi-degree $(12,12)$ or $(12,6)$. An argumentation as in \cite{sa} (5.5),  (b), shows that $q\cap S_{IO}$  splits into the union of lines of the two rulings of $q$. More precisely it contains the $(I,1)$-orbit of twelve lines and the $(1,O)$-orbit of twelve or of six lines. In the last case the lines have multiplicity two in the intersection. 
%Unfortunately we were not able to find an argumentation which use just the group action to check which case in fact occur. We use explicitly the equations of the surfaces. 
So take the fix line $(\lambda:\mu:i\lambda:i\mu)$, $(\lambda:\mu)$$\in$$\piu$, of the matrix $(1,A_2)$ in the orbit of length six. A direct computation (with MAPLE) shows that it is singular on $S_{IO}$, so we are done.
%The lines in the orbit of length six have fix group of order four, and this contains matrices $(1,A_i)$. A direct computation (with MAPLE) shows that the fix line $(\lambda:\mu:i\lambda:i\mu)$, $(\lambda:\mu)$$\in$$\piu$, of $(1,A_2)$ is singular on $S_{IO}$. This completes the proof.
\eprf

%{\it Proof}. First we show that\\
%{\it the variety $q\cap S_{IO}$ is reduced or is the union of a reduced curve and some singular lines of the $(I,-)$-invariant ruling of $q$.}\\
%The proof is similar to the proof of {\bf (5.5)}, (a), in \cite{sa}. Let $V\subseteq$$q\cap S_{IO}$ a component with multiplicity at least two, then $V$ is singular on $S_{IO}$ or $q$ and $S_{IO}$ are tangent at $V$. Assume that there is a line $L$$\subseteq$$q$, $L\not\subseteq$$q\cap S_{IO}$, $(I,-)$-invariant with $L\cap V$ not empty. Let $x\in$$L\cap V$, then mult$_x(L\cdot S_{IO})$$\geq 2$. Now $x$ has orbit of length $\geq 12$ hence $L$$\subseteq$$S_{IO}$, but this contradicts the assumption. Hence such a $V$ must be a line in the ruling of $L$, i.e. an $(I,-)$-invariant line.\\
 %We are now able to conclude the proof. We have deg$(q\cap S_{IO})$=24. By the table above, $q\cap S_{IO}$ can only have bi-degree $(12,12)$ if it is reduced or $(12,6)$ if it is singular at the six points. In any case with an argumentation as in \cite{sa}, {\bf (5.5)},  (b), the $12$ lines which form an $(I,-)$-orbit are contained in $q\cap S_{IO}$. If the latter is reduced the 12 lines which form an $(-,O)$-orbit are contained in $q\cap S_{IO}$ too, if not the six lines which form an $(-,O)$-orbit are double lines in $q\cap S_{IO}$. To check which case in fact can occur we were unfortunately not able to find an argumentation which use just the group action (as in \cite{sa}). For this we use explicitly the equations of the surfaces. The lines in the orbit of length six have fix group of order four, and this contains matrices $(-,A_i)$. A direct computation shows that the points $(0:1:0:\pm i)$ of $q\cap S_{IO}$ are  fix points of $(-,A_2)$. Hence all the lines in the orbit of length six are in the base locus. Moreover the two points are on the fix line $x_0=0,~x_2=0$ of $(A_2,A_2)$ and have multiplicity two in the intersection. If $S_{IO}$ and $q$ are tangent at the six lines then the fix line of $(A_2,A_2)$ is contained in both the tangent planes at $q$ in the points $(0:1:0:\pm i)$, but this is not possible.
%Now take a fix line of an elements $(Ai,Aj)$, e.g. the line $x_0=0,~x_2=0$ of $(A_2,A_2)$. Substituting in the equation of $S_{IO}$ we find
%\begin{eqnarray*}
%x_0^2x_2^2(x_0^8+x_2^8-2x_0^4x_2^4)=x_0^2x_2^2(x_0^4-x_2^4)^2=0
%\end{eqnarray*}
%hence the intersection points are  $(0:\pm 1:0:0)$, $(0:0:0:\pm 1)$ and the two points of $q\cap S_{IO}$: $(0:1:0:\pm i)$. This shows that the lines of the $(-,O)$-orbit of length $6$ are in the base locus (meglio!!). Moreover the points on the quadric have multiplicity two in the intersection, hence 
\begin{lemma}\label{Li} Let $p$ be a singular point on a surface of the pencil $S_{IO}+\lambda q^6$ and assume that $p$ is on the quadric $q$. Then $p\in L_i$, $i=1,\cdots, 6$, where the $L_i$'s denote the singular lines of the base locus.\end{lemma}
{\it Proof.} As in the proof of  (6.1) in \cite{sa}, if $p$ is singular on a surface in the pencil and $p\in q$, then $p$ is a singular point of $q\cap S_{IO}$. Hence $p\in L_i$.
\eprf
By this fact and Bertini's theorem follows that:\\
\begin{lemma}  1. the general surface in the pencil is smooth away from the lines $L_1,\ldots,L_6$.\\
Moreover we have:\\
2. the singular surfaces have only isolated singularities away from the singular lines $L_1,\ldots,L_6$. In particular they are irreducible and reduced.\end{lemma}
{\it Proof of 2.} 
The surface $q^6$ is the only not reduced surface in the pencil. Indeed, assume that there is another not reduced surface $\{f_{\alpha}^m=0\}$, $\alpha\cdot m=12$, and deg$f_{\alpha}=\alpha$. Then the twelve lines of the base locus in the intersection of $f_{\alpha}^m$ and $q$ would be singular too, which is not the case. Assume now that there is a surface $S$ in the pencil which contain a singular curve $C$. The latter meets the quadric in at least one point, which by lemma \ref{Li} is on some $L_i$. This point is a ``pinch point'' of $S$ (for the definition of ``pinch point'' cf.\ \cite{g-h} p.\ 617). We compute explicitly the pinch points along a line $L_i$. Because of symmetry reason it is enough to do it just for one line. We take $L_1$:$\{x_2=ix_0,x_3=ix_1\}$. First we consider the transformation $x_0\mapsto y_0$, $x_1\mapsto y_1$, $x_2\mapsto y_2+i\cdot y_0$, $x_3\mapsto y_3+i\cdot y_1$, which maps $L_1$ to $\{y_2=y_3=0\}$. We call the transform $L_1$ again.\\
Let $F:=F(y_0,y_1,y_2,y_3,\lambda)$ be the equation of the pencil after the transformation. We can write the Taylor expansion of $F$ along $L_1$
\begin{eqnarray*}
F=A(y_0,y_1,\lambda)\cdot y_2^2+B(y_0,y_1,\lambda)\cdot y_2\cdot y_3+C(y_0,y_1,\lambda)\cdot y_3^2\\
+{\rm terms~of~order}\geq 3,
\end{eqnarray*}
the pinch points are solution of 
\begin{eqnarray*}
{\rm det}\left( \begin{array} {cc}
A& \frac{1}{2}B\\
\frac{1}{2}B& C
\end{array} \right)=0
\end{eqnarray*}
which splits into the product
\begin{eqnarray*}
\begin{array}{l}
(x_1^4+2x_1^3x_0+2x_0^2x_1^2-2x_0^3x_1+x_0^4)(x_1^4-2x_1^3x_0+2x_0^2x_1^2+2x_0^3x_1+x_0^4)\\(x_0^2+3x_0x_1+\sqrt{5}x_0x_1-x_1^2)(x_0^2-3x_0x_1-\sqrt{5}x_0x_1-x_1^2)\\
(3x_0^4-2\sqrt{5}x_0^2x_1^2+3x_1^4)(x_0^4+16x_0^2x_1^2-6\sqrt{5}x_0^2x_1^2+x_1^4)=0.\\
\end{array}
\end{eqnarray*}
this has twenty distinct solutions (independent from $\lambda$), therefore we have ``simple'' pinch points on $L_i$, hence no singular curves.


 %which form one orbit under the action of the group $(I,-)$. We analyze just one point, e.g. consider $p=(1:a:0:0)$ where $a:=\frac{1}{3}\sqrt{6i+3\sqrt{5}}$. Now we proceed as follows:\\
%1)Consider the transformation: $y_0\mapsto w_0$, $y_1\mapsto w_1+a\cdot w_0$, $y_2\mapsto w_2$, $y_3\mapsto w_3$. The pinch point goes to $(1:0:0:0)$. Let now $F_1(w_1,w_2,w_3,l)$$=F(1,w_1+a,w_2,w_3,l)$ be the equation of   the pencil in the chart $U_0:\{x_0 \not= 0\}$\\
%2)Blow up the line $L_1$$\{w_2=w_3=0\}$ in $\pitr$. This is the variety defined by $\{((w_1,w_2,w_3),(l_1:l_3))\in \mathbb{A}^3\times \piu|l_3w_2=l_2w_3\}$. In the chart $V_2:\{l_2\not= 0\}$ we get $\{((w_1,w_2,w_3),t)\in \mathbb{A}^3\times V_2|tw_2=w_3\}$. Let $F_3(w_1,w_2,t,l)$$=F_1(w_1,w_2,tw_2,l)$ the blow up of the pencil along the line $L_1$. Then a calculation shows that $F_3=w_2^2\cdot F_4(w_1,w_2,t)$ where $\{F_4(w_1,w_2,t)=0\}$ define the strict transform of the surface. Let $\pi$ denote the blow up, then $\pi^{-1}((0,0,0))=((0,0,0),t_1):=\tilde{p}$ with $t_1=a\cdot \frac{\sqrt{5}-2i}{3}$.\\
%3)If the singular curve contain the point $p$ the the strict transform $\tilde{C}$ contains the point $\tilde{p}$. And this would be singular on the surface $\tilde{S}$. But $(\partial _{w_i}F_4)(\tilde{p})\not=0$, $i=2,3$.
\eprf

%Assume that there is a surface $S$ in the pencil which contain a singular curve $C$ and let $p\in$$C\cap q$. By {\bf (5.3.2)}, $p\in L_i$. We can assume $C$ irreducible and write $C=m C'$ with $C'$ a reduced curve, $m\geq 2$. It holds $C\cdot 2L_i=$$2m (C'\cdot L_i)$, hence $p$ has multiplicity at least four in the intersection of $C$ and $L_i$ on $S$. Now the $(I,-)$-orbit of $p$ on $L_i$ has length at least $12$. Since deg$(\partial_i^2 S$)$=10$, $L_i$ contains at least $10$ points of multiplicity $\geq 3$ of $S$, hence $3L_i$$\subseteq$$S_{IO}$, but this is not possible.
%\eprf
Observe that (6.3) of \cite{sa} holds for this pencil too. In particular, we get the estimate for the number of isolated singularities as follows (we use the same notation as there): deg$(S\cap \partial_i S)=12\cdot 11$ and $S\cdot \partial_i S=C+2L_1+\ldots+2L_6$ hence deg$C=12\cdot 11-12$. Since $C$ cannot be singular there is a $j=0,1,2,3$ s.\ t.\ $C$$\not\subseteq$$\partial_j S$. Hence deg$(C\cap \partial_j S)=(12\cdot 11-12)(12-1)=120\cdot 11$. Since the singular points are computed two times in the intersection their number is $\leq \frac{120\cdot 11}{2}$$=660$.\\
Now  (7.1) of \cite{sa} holds with mult$_{z_{ij}}(L\cdot S)=2$. In fact a line like $L$ there can be only a fix line of a matrix in the conjugacy class of $(A_2,A_2)$ and it meets the base locus at some line $L_i$. Similar assertion as (7.2),  (7.3),  (7.5) and the first part of section 8. hold here too. About the last part of section  8.: in the case of the map $\bar f$, the points $z_{13}$ and $z_{24}$ have order $\frac{12}{2}-2$, hence the degree of the ramification locus is $2\cdot 12 -10$ and without the two points on the quadric, it is $12-4=8$. In particular we have again an estimation for the number of singular points on the fix lines, i. e. this is $\leq$$12$ if they do not meet the base locus, $\leq 8$ otherwise.\\
{\it The singular surfaces.} %In this case we were unfortunately not able to recognize any kind of action of $IO$ on some four-dimensional regular polytope. Hence 
We proceed by a direct computation using the lines of fix points. 
%Observe that $IO$ is not contained in any reflection group which operates in $\mathbb{R}^4$, since these are only $[3,3,3]$, $[3,3,4]$, $[3,4,3]$, $[3,3,5]$ (cf. cox 2 p. 780), and we were not able to recognize some action of $IO$ on some four-dimensional regular potopes[D[D[D[D[Dlytopes (cf....), hence we use the directly fix lines to find the singular surfaces in the pencil. 
The matrices of $IO$ with fix lines containing singular points are in the conjugacy class of $(A_2,A_2)$, $(S,S)$ or $(A_2,R_4A_2)$. The total number of distinct fix lines is 90, 80,  resp. 180. In the table below we give the singular surfaces and the singular points on the fix lines. We choose a representative in each conjugacy class and a fix line of it, moreover we put $a':=\sqrt{2}-1$, $a:=\sqrt{2}+1$,$\gamma_1=2\sqrt{2}-\sqrt{5}$, $\gamma_1'=2\sqrt{2}+\sqrt{5}$, $\gamma_2:=2+\sqrt{2}+\sqrt{5}+\sqrt{10}$, $\gamma_2':=-2+\sqrt{2}-\sqrt{5}+\sqrt{10}$,  $\gamma_3:=-2+\sqrt{2}+\sqrt{5}-\sqrt{10}$, $\gamma_3':=2+\sqrt{2}-\sqrt{5}-\sqrt{10}$, $\alpha_1:=\frac{1}{2}(-1+\sqrt{2}-\sqrt{10})$, $\alpha_1':=\frac{1}{2}(1+\sqrt{2}-\sqrt{10})$, $\beta_1:=\frac{1}{2}(4-3\sqrt{2}-2\sqrt{5}+\sqrt{10})$, $\alpha_2:=\frac{1}{2}(1+\sqrt{2}+\sqrt{10})$, $\alpha_2':=\frac{1}{2}(-1+\sqrt{2}+\sqrt{10})$, $\alpha_3:=-2+\frac{3}{2}\sqrt{2}-\sqrt{5}+\frac{1}{2}\sqrt{10}$, $\alpha_3':=2+\frac{3}{2}\sqrt{2}+\sqrt{5}+\frac{1}{2}\sqrt{10}$, $\beta_2:=\frac{1}{2}(-10+7\sqrt{2}-4\sqrt{5}+3\sqrt{10})$, $\beta_3:=\frac{1}{2}(\sqrt{2}-2\sqrt{5}+\sqrt{10})$, $\beta_4:=\frac{1}{2}(2+\sqrt{2}+\sqrt{10})$, $c_1:=-\frac{74}{972}+\frac{4}{243}\sqrt{10}$, $c_2:=-\frac{74}{972}-\frac{4}{243}\sqrt{10}$:

%\newpage

\vspace*{0.3cm}

\begin{eqnarray*}
\begin{array}{c|c|c|c|c}
{\rm matrix~and~fix~line}&\multicolumn{2}{c|}{{\rm value~of} ~\lambda}&\multicolumn{2}{c}{{\rm int.~points}}\\
\hline
(A_2,A_2): &0&-\frac{1}{8}&(1:0:0:0)&(\pm a':0:1:0)\\
x_1=x_3=0&&&(0:0:1:0)&(1:0:\pm a':0)\\
&&&(\pm 1:0:1:0)&\\
(A_2,R_4A_2):&-\frac{1}{8}&&(-a:1:0:0)&\\
x_0=-ax_1, x_3=a'x_2&&&(0:0:1:a')&\\
&&&(1:\pm a':1:a')&\\
&c_1&c_2&(\alpha_1:-\beta_1:1:a')&(\alpha_1':-\beta_3:1:a')\\
&&&(-\alpha_2:\beta_1:1:a')&(-\alpha_2':\beta_3:1:a')\\
&&&(\alpha_3:\beta_2:1:a')&(\alpha_3':\beta_4:1:a')\\
&&&(-\alpha_3:-\beta_2:1:a')&(-\alpha_3':-\beta_4:1:a')\\
(S,S):&0&&(1:0:0:0)&\\
x_1=x_2=x_3&&&(0:1:1:1)&\\
&&&(-1:1:1:1)&\\
&&&(-\sqrt{5}:1:1:1)&\\
&&&(\sqrt{5}\pm 2:1:1:1)&\\
&c_1&c_2&(\gamma_1:1:1:1)&(\gamma_1':1:1:1)\\
&&&(\gamma_2:1:1:1)&(\gamma_2':1:1:1)\\
&&&(\gamma_3:1:1:1)&(\gamma_3':1:1:1)\\
\end{array}
\end{eqnarray*} 
Here the fix line of $(A_2,A_2)$ contain two points of the base locus. Observe that the number of singular points is maximal on the lines, hence the four given $\lambda$ are the only values corresponding to singular surfaces.\\
\begin{prop} In the pencil $S_{IO}+\lambda q^6$ we have the following $IO$-orbits of nodes

\begin{center}
\begin{tabular}{l|*{4}{l}}
$\lambda$&$0$&$c_1$&$-\frac{1}{8}$& $c_2$\\
\hline
orbit&$120$&$240$&$360$&$240$\\
\hline
fix group mod. $\pm 1$& $A_4$&$D_3$&$\mathbb{Z}_2\times\mathbb{Z}_2$&$D_3$\\   
\hline
order&$12$&$6$&$4$&$6$\\
\end{tabular}
\end{center}
 \end{prop}
\bprf 
 We analyze case for case the four different surface. A direct computation  as in section  4.2, then shows that the singular points are nodes.\\
I. \underline{$\lambda=0$}. Observe that $(1:0:0:0)\in$$S_{IO}$ is contained in three fix lines of elements in the conjugacy class of $(A_2,A_2)$. Moreover \\

(*) there is a matrix of order four $(1,\gamma)$$\in IO$, with $\gamma^2=A_2$, \\

Hence $(1,\gamma)$ and $(A_2,A_2)$ commute and the four points on the fix line of $(A_2,A_2)$ form one orbit under the action of $(1,\gamma)$. By this fact follows that we have a configuration of lines and points. By the formula (\ref{conf}), we get $N_0\cdot 3=90\cdot 4$, hence $N_0=120$. The point $(1:0:0:0)$ is on the fix line of $(S,S)$ too, more precisely it is contained on four such lines. The formula $120\cdot 4=80\cdot n_0$ shows $n_0=6$ so the six points on the fix line are in the orbit of length 120 too. In conclusion we have just one $IO$-orbit of singular double points.\\
II. \underline{$\lambda=-\frac{1}{16}$}. The four points on the fix line of $(A_2,A_2)$ form one orbit by (*). The formula (\ref{conf}) gives $N_0\cdot n_1=360$. If $n_1\geq 2$ then $N_0\leq 180$ and the fix group in $IO$ mod. $\pm 1$ has order $\geq 8$. Checking in the table on page 24 this is not possible. Hence $N_0=360$ and $n_1=1$. The fix group has order four and it is isomorphic with  $\mathbb{Z}_2\times\mathbb{Z}_2$. Each singular point is contained in two fix lines of matrices in the conjugacy class of $(A_2,R_4A_2)$. Now the formula (\ref{conf})) with $N_0=360$, $n_1=2$ and $N_1=180$ gives $n_0=4$. So we conclude that the singular points on these fix lines are in the orbit of length $360$ too.\\  
III. \underline{$\lambda=c_i,~i=1,2$}. The three points on the fix line of $(S,S)$ form one orbit under the action of $(S,1)$. The formula (\ref{conf}) gives $N_0\cdot n_1=240$. If $n_1\geq 2$ then $N_0\leq 120$ and the fix group mod. $\pm 1$ has order $\geq 12$, which is not possible (check in the table on page 24 again). Hence $N_0=240$ and $n_1=1$. The fix group has order six and it is isomorphic with $D_3$, hence three fix lines of matrices in $(A_2, R_4A_2)$ contain a singular points. The formula (\ref{conf}) in this case gives $n_0=4$, which shows that the singular points on these fix lines are in the previous orbit too.

%The four points on the fix line of $(A_2,R)$ form one orbit by (**) of I.. By $N_0\cdot n_1=180\cdot 4$  and Miyaoka's bound $N_0\leq 645$ (see \cite{mi}) we get $n_1\geq 2$. If $n_1\geq 3$ then $N_0\leq 240$. The fix group of such a point in $IO$ modulo $\{\pm id\}$ has order $\geq 6$. Hence it contains at least one element of order three. By the table above no element of order three have this singular fix point. We conclude that $n_1=2$ and $N_0=360$.
%which in fact gives $N_0\leq 360$. In this situation the fix group of a point is generated by at least by two elements of order two in the conjugacy class of $(A_2,R)$. Since there are not fix lines of elements in the conjugacy class of $(S,S)$ which contain singular points of this surface, it must be exactly two. 
%In particular such a fix point  is contained in a fix line of an element in the conjugacy class of $(A_2,A_2)$ and in fact by (**) it has orbit of length four. This shows that the singular points on these fix lines are in the orbit of length $360$ too. \\
%III. \underline{$\lambda=c_i,~i=1,2$}. Again the four singular points on the fix line of $(A_2,R)$ form one orbit and $n_1\geq 3$. Otherwise, if $n_1=2$, there would be a fix line of an element in the conjugacy class of $(A_2,A_2)$ which contain the singular point (as in II.) and this is not the case. The formula (*) gives  $N_0\leq 240$ and the points are contained in at least one fix line of matrices in the conjugacy class of $(S,S)$ (observe that here they have orbit of length three under the action of $(S,-)$). A direct calculation using the formulas $N_0\cdot n_1=180\cdot 4$ and $N_0\cdot n_1'=80\cdot 3$ shows that the only possibility is $N_0=240$.
\eprf

%{\bf (5.3.7) The singular points are nodes}. It is a direct computation as in .....\\. 
We give now the table of the configurations of lines and points:

\vspace*{0.3cm}

\renewcommand{\arraystretch}{1.3}

\begin{eqnarray*}
\begin{array}{c|c|c|c|c}
{\rm Repr.~of~the~ conj.~class}&\multicolumn{2}{c|}{{\rm value~of }~\lambda}&\multicolumn{2}{c}{{\rm Configuration}}\\
%{\rm conj.~class}&\multicolumn{2}{c|}{{\rm }}&\multicolumn{2}{c}{}\\
\hline
(A_2,A_2)&0&-\frac{1}{8}&(120_3,90_4)&(360_1,90_4)\\
(A_2,R_4A_2)&-\frac{1}{8}&c_i&(360_2,180_4)&(240_3,180_4)\\
(S,S)&0&c_i&(120_4,80_6)&(240_1,80_3)
\end{array}
\end{eqnarray*}

\subsection{The pencil of $D_nD_m$-invariant quartics}
We take the generators $q^2$ and $f_2+f_3=(x_0-ix_1)(x_0+ix_1)(x_2-ix_3)(x_2+ix_3)$. The latter is the union of four complex planes meeting each other at the four complex lines $\{x_0\pm ix_1=0,~x_2\pm ix_3=0\}$ on $q$. 
The pencil contain the multiple quadric $q'^2$$:=q^2-2(f_2+f_3)$$=(x_0^2+x_1^2-x_2^2-x_3^2)^2$ too. Hence with this new generator the equation of the pencil becomes
\begin{eqnarray*}
q'^2+\lambda q^2=(q'+i\sqrt{\lambda}q)(q'-i\sqrt{\lambda}q).
\end{eqnarray*}
A surface in the pencil is the union of two quadrics which meet each other at the four lines given above. An easy computation shows that none of the surface in the pencil contains isolated singularities.



\input{artjul01_22}

\input{i_o}

%\section{On the number of nodes in the invariant pencils}
%[...]\\
%In the case of the pencil $f_0+\lambda q^2$ the base locus (set theoretically) is the curve $C_8$ of bi-degree $(4,4)$ on $q$. This has genus $g=9$ (cf. Hart..p...). Hence $e(C_8)=2-2\cdot g$$=-16$. Moreover
%\begin{eqnarray*} 
%4+e(C_8)=e(\Gamma)=4+e(X_{{\rm gen}})-\nu
%\end{eqnarray*}
%so $\nu=e(X_{{\rm gen}})-e(C_8)$$=24+26=40$. Which is in fact the total number of ordinary double points in the pencil.


 
%\end{document}

\addcontentsline{toc}{section}{  \hspace{0.5ex} References}
%{6  \hspace{0.5ex} References}-all'inizio e' il numero del capitolo-
\begin{thebibliography}{EK}
\bibitem [BS] {basa} Barth, W., Sarti, A., Polyhedral Groups and Pencils of K3-Surfaces with Maximal Picard Number, math. AG/0301329.
\bibitem[B]{ben} Benson, D. J., Polynomial Invariants of Finite Groups, London Math. Society LNS 190, Cambridge University Press (1993).
\bibitem [C] {chmutov}
Chmutov, S. V., Examples of Projective Surfaces with many Singularities, Journal of Alg. Geom. 1, (1992) 191--196.
\bibitem[Co1]{cox0}
Coxeter, H. M. S., The product of the generators of a finite group generated by reflections, Duke Math. J. Vol. 18 (1951) 765--782.
\bibitem[Co2]{cox} Coxeter, H. M. S., Regular polytopes (second edition), The Macmillan company, New York (1963).
\bibitem [GH]{g-h} Griffiths, P., Harris, J., Principles of Algebraic Geometry. New York, Chichester, Brisbane, Toronto, Singapore, John Wiley and Sons, Inc. (1978).
\bibitem[M]{mukai}
Mukai, S., Moduli of Abelian  surfaces and regular polyhedral groups, in: Proceedings of Moduli of algebraic varieties Symposium, Sapporo January 1999.
%\bibitem[M]{mi} Miyaoka, Y.,The Maximal Number of Quotient Singularities on Surfaces and Given Numerical Invariants, Math. Ann 268, 159-171 (1984).
\bibitem[S]{sa} Sarti, A., Pencils of symmetric surfaces in $\mathbb P^3(\mathbb C)$, J. of Alg. 246, 429--452 (2001).
\bibitem[T]{tits} Tits, J., Liesche Gruppen und Algebren, Springer (1983).

\end{thebibliography}

artjul01_22.tex
\subsection{The pencil of $OD_n$-invariant octics ($n\geq 6$) and of $ID_n$-invariant 12-ics ($n\geq 8$).}
We take generators $q^4$, $q^6$ and $P_8=\prod_{i=0}^{3}K_i$, $P_{12}=\prod_{i=0}^{5}K_i'$ (cf. {\bf (3.1.2)}) and we denote the pencils by $\Pi_m(\lambda)$$:=P_m+\lambda q^{\frac{m}{2}}$, $m=8,12$. We have\\
\begin{lemma} The base locus of the pencil is $\frac{m}{2}$-times the intersection $q\cap P_m$, $m=8,12$. Where (as sets)
\begin{center}
\begin{eqnarray*}
\begin{array}{lll}
q\cap P_8&=& {\rm fix~lines~of~}(1,R_n),\\ && {\rm orbit~of~eight~lines~under}~(O,1),\\
%\hline
q\cap P_{12}&=&{\rm fix~lines~of~}(1,R_n), \\ && {\rm orbit~of~twelve~lines~under}~(I,1).\\
\end{array}
\end{eqnarray*}
\end{center}
\end{lemma}
\bprf
The element $(1,R_n)$ has fix lines $L_1:\{x_0=-ix_1,~x_2=ix_3\}$, $L_2:\{x_0=ix_1,~x_2=-ix_3\}$ for each $n$. These are contained in each quadric $K_s$, $s=0,\ldots,3$, resp. $K_t'$, $t=0,\ldots,5$ hence they have at least multiplicity four, resp. six in the intersection $q\cap P_m$, $m=8,12$. By this fact and an argumentation as in \cite{sa}, (5.5), it follows that the lines of the length $m=8,12$ orbit under $(O,1)$, resp. $(I,1)$ are in the base locus too. Moreover the previous multiplicity of intersection are exactly four, resp. six. \eprf

\begin{lemma} The pencil does not contain surfaces with isolated singular points. \end{lemma}
\bprf
The groups have order $96 n$, resp.\ $240n$, since an isolated  singular point has orbit of finite length, it is fixed by $(1,R_n)$. This shows that it is  on the lines of the base locus, hence not isolated. A contradiction to the assumption.\eprf
\begin{prop} In the pencil we have the following singular surfaces, with one orbit of double lines which are fix lines for the elements of some conjugacy class:
\begin{eqnarray*}
\begin{array}{c|cccc}
{\rm value~of}~\lambda& OD_n:& -1&\frac{1}{3}&0\\
&ID_n:&0&&\\
\hline
{\rm orbit}&OD_n:&6&8&12\\
&ID_n:&30&&\\
\hline
{\rm fix~lines~in }&&(A_3,R_n^{\frac{n}{2}})&(S,R_6^2)&(R,R_n^{\frac{n}{2}})\\
{\rm the~conj.~class~of}&&&&\\
\end{array}
\end{eqnarray*}
\end{prop}
\bprf
It is a direct computation, using the equations $x_0=x_2,x_1=-x_3$; $x_0=-(1+\sqrt{3})x_3+x_1, x_2=x_3+(1-\sqrt{3})x_1$ and $x_2=(1-\sqrt{2})x_0,x_3=(\sqrt{2}-1)x_1$ of a fix line of ($A_3,R_n^{\frac{n}{2}}$), ($S,R_6^2$), resp.\ ($R,R_n^{\frac{n}{2}} $). \eprf
%{\bf Remark.} With some more argumentation it should be possible to show that the above given surfaces are all the singular surfaces in the pencils and that they do not contain more singularities.
\section{Final Remarks}
1) By \cite{cox} p.\ 292 there are sixteen regular polytopes $\{p,q,r\}$ in four dimensions. These polytopes correspond to four distinct symmetry groups $[p,q,r]$ listed in the table below. Three of these symmetry groups can be obtained from groups which we described above by adding extra generators $C$ and $C'$, where $C$ denotes the matrix given in section \ref{invariants}, $C'$ is the matrix of \cite{sa} p.\ 433.
\begin{eqnarray*}
\begin{array}{c|c|c|c|c}
{\rm symmetry~groups}&[3,3,3]&[3,3,4]&[3,4,3]&[3,3,5]\\
\hline
{\rm our~groups}&&(OO)'&TT&II\\
\hline
{\rm extra~generators}&&C&C,C'&C\\
\end{array}
\end{eqnarray*}
The group $[3,3,3]\cong Sym(5)$ does not contain the Heisenberg group $H$, in fact it has some invariant polynomials of odd degree (cf.\ \cite{cox0} p.\ 780). In these notes we complete the description of the $G$-invariant pencils of surfaces whenever $G$ is the symmetry group of a regular  four dimensional polytope and $G$ contains $H$.\\
2) In remark 3.3, with $n=3$, we give the degree four invariant polynomials of the {\it bi-polyhedral dihedral group} $D_3D_3$$\subseteq$$SO(4)$. By Mukai \cite{mukai} the quotient $\pitr$$/D_3D_3$ is isomorphic with the Satake compactification of the moduli space of abelian surfaces with $(1,2)$-polarization, hence this invariant polynomials should be related to modular forms.\\
3) In \cite{basa} the quotients $X/G$, $G=TT,OO,II$ are described. It would be  interesting to examine the quotients in the remaining cases. 




i_o.tex
%\documentclass[12pt]{article}
%
%\usepackage{a4}
%\usepackage{epsfig}
\section{Computer Picture}
We exhibit a computer picture of the $I\times O$ symmetric surface of degree 12 with 360 nodes. This has been realized with the program SURF written by S.\ Endra\ss.
%\textheight30cm
%
%\begin{document}
%\pagestyle{empty}
\hoffset=1.5cm
\voffset=1.0cm
%
\begin{center}
%
\hspace*{-4cm}
\epsfig{file=i_o1.eps}
\hspace*{-4cm}\ \\[3ex]
 {\large $I\times O$-symmetric surface of degree 12}\\[0.5cm]
{\large with 360 nodes}
%{\large (Erlangen 1999)}\\[0.5cm]
% {\large %$x_0^8+x_1^8+x_2^8+x_3^8+14\left(x_0^4x_1^4+x_0^4x_2^4+x_0^4x_3^4\right)+$}\\[0%.3cm]
%\right)\left(\tau^2y^2-z^2\right)
    %\left(\tau^2z^2-x^2\right)$}\\[0.3cm]

%{\large %$+14\left(x_1^4x_2^4+x_1^4x_3^4+x_2^4x_3^4\right)+168x_0^2x_1^2x_2^2x_3^2=0$}\\%[0.5cm]
%{\large $\tau=\frac{1+\sqrt{5}}{2}$}
%
\end{center}
%
%\end{document}

math0303364.extracted_bib
\begin{thebibliography}{EK}
\bibitem [BS] {basa} Barth, W., Sarti, A., Polyhedral Groups and Pencils of K3-Surfaces with Maximal Picard Number, math. AG/0301329.
\bibitem[B]{ben} Benson, D. J., Polynomial Invariants of Finite Groups, London Math. Society LNS 190, Cambridge University Press (1993).
\bibitem [C] {chmutov}
Chmutov, S. V., Examples of Projective Surfaces with many Singularities, Journal of Alg. Geom. 1, (1992) 191--196.
\bibitem[Co1]{cox0}
Coxeter, H. M. S., The product of the generators of a finite group generated by reflections, Duke Math. J. Vol. 18 (1951) 765--782.
\bibitem[Co2]{cox} Coxeter, H. M. S., Regular polytopes (second edition), The Macmillan company, New York (1963).
\bibitem [GH]{g-h} Griffiths, P., Harris, J., Principles of Algebraic Geometry. New York, Chichester, Brisbane, Toronto, Singapore, John Wiley and Sons, Inc. (1978).
\bibitem[M]{mukai}
Mukai, S., Moduli of Abelian  surfaces and regular polyhedral groups, in: Proceedings of Moduli of algebraic varieties Symposium, Sapporo January 1999.
%\bibitem[M]{mi} Miyaoka, Y.,The Maximal Number of Quotient Singularities on Surfaces and Given Numerical Invariants, Math. Ann 268, 159-171 (1984).
\bibitem[S]{sa} Sarti, A., Pencils of symmetric surfaces in $\mathbb P^3(\mathbb C)$, J. of Alg. 246, 429--452 (2001).
\bibitem[T]{tits} Tits, J., Liesche Gruppen und Algebren, Springer (1983).

\end{thebibliography}