text:
\section{Introduction}

A $K3$ surface is a simply connected compact complex surface whose
canonical bundle is trivial. Every Enriques surface appears as a
quotient of a $K3$ surface by a fixed-point-free (shortly, free) involution.
Theoretically, to consider an Enriques surface is equivalent to
consider the pair of the covering $K3$ surface and the free involution.
For example, the period
map for Enriques surfaces is constructed under this description.
But the properties of free involutions on a {\em fixed} $K3$ surface are
rather unclear to us. The existence is already a special property,
their geometric realizations and the isomorphism classes of
the quotient Enriques surfaces are other problems.

For a fixed $K3$ surface $X$, two quotient Enriques surfaces are isomorphic if and only if
the two free involutions are conjugate in $\mathop{\mathrm{Aut}}\nolimits (X)$. In \cite{ohashi} it is shown
that the number of the conjugacy classes of
free involutions (and more generally, of finite subgroups) are finite.
There this number, i.e., the number of isomorphism classes of quotient Enriques surfaces,
is computed for $K3$ surfaces with Picard number $\rho=11$ or for
Kummer surfaces associated with the product of two elliptic curves
whose periods are very general.

The aim of this paper is to study fixed-point-free
involutions on surfaces studied in \cite{keum97,kondo98}.
Let $C$ be a smooth projective curve of genus $2$.
Its Jacobian variety $J(C)$ is the abelian surface parametrizing
divisor classes on $C$ of degree $0$.
The quotient surface $J(C)/\{\pm 1_{J(C)}\}$ has $16$ nodes
and can be embedded into $\bb{P}^3$ as a quartic hypersurface.
We call it the {\em{Kummer quartic surface}} associated with $C$ and
denote by $\overline{Km} (J(C))=:\overline{X}$.
The minimal desingularization
$Km(J(C))=:X$ of $\overline{Km} (J(C))$ is called the {\em{Jacobian Kummer surface}}
associated with $C$, which is a $K3$ surface.
$X$ is {\em{Picard-general}} if the Picard number of $X$ equals $17$,
the minimum possible value. In what follows, $X$ will always be a Picard-general
Jacobian Kummer surface except for Sections \ref{method} and \ref{166}.

In \cite{mukai}, Mukai observed that there exist three kinds of free involutions on $X$.
\begin{itemize}
\item {\em{A switch}} associated with {\em{an even theta characteristic $\beta$}}.
\item {\em{A Hutchinson-G\"{o}pel (shortly HG) involution}} associated
with {\em{a G\"{o}pel tetrad $G$}}.
\item {\em{A Hutchinson-Weber (shortly HW) involution}} associated with {\em{a Weber hexad $W$}}.
\end{itemize}

Essentially these automorphisms date back more than a century, but
their freeness are found only recently in comparison.
Mukai studied HG involutions
in connection with the numerically reflective involutions of Enriques surfaces.
Also he conjectured that these are the all free involutions on $X$.
In this paper we prove the following theorem and confirm the conjecture.
\begin{thm}\label{thm1}
On a Picard-general Jacobian Kummer surface $X$,
there are exactly $31=10+15+6$ free involutions up to conjugacy in $\mathop{\mathrm{Aut}}\nolimits (X)$.
$10$ are switches, $15$ are HG involutions and $6$ are HW involutions.
\end{thm}

In \cite{kondo98}, Kondo proved that $\mathop{\mathrm{Aut}}\nolimits (X)$ is generated by
$32$ translations and switches,
$32$ projections and correlations,
$60$ HG involutions,
and $192$ Keum's automorphisms.
One point of the proof was that $192$ Keum's automorphisms did not
correspond in one-to-one way to
the $192$ facets of the polyhedral cone introduced by
Borcherds and Kondo. Moreover they had infinite order while the others had order $2$.
In this respect, it can be expected that there exist $192$ {\em involutions} which
correspond in one-to-one way to the $192$ facets of the polyhedral cone and
together with
the $32+32+60$ involutions they generate $\mathop{\mathrm{Aut}}\nolimits (X)$.
In fact, the HW involutions work well.
\begin{thm}\label{thm2}
$\mathop{\mathrm{Aut}}\nolimits (X)$ is generated by the following {\em involutions}:
translations, switches, projections, correlations, HG involutions and HW involutions.
\end{thm}
This is a biproduct of the proof of Theorem \ref{thm1}.

The proof of Theorem \ref{thm1} is given in the following way.
In Section \ref{method}
we introduce an invariant of a free involution, called {\em a patching subgroup},
which is a subgroup of $A_{NS(X)}=NS(X)^*/NS(X)$.
This subgroup appears naturally in the light of Nikulin's theory of lattices \cite{nikulin-sym}.
Under some condition,
we can show the invariance of the patching subgroup under conjugations.
Section \ref{invariants}, Proposition \ref{fin} shows conversely two free involutions
are conjugate if their patching subgroups are the same,
when $X$ is a Picard-general Jacobian Kummer surface.
Simultaneously we see that $X$ has no more than $31$ non-isomorphic Enriques quotients.
These two Sections
reduce the proof of Theorem \ref{thm1} to concrete computations of patching subgroups
of free involutions itemized above.
The occurence of $31$ distinct patching subgroups shows Theorem \ref{thm1}.
The computations are worked out in
Sections \ref{switch}-\ref{reye}.
The result shows that the generators of patching
subgroups are expressed in terms of the classical notions.
It is summarized as follows.

In the switch case, let $\beta$ be an even theta characteristic
and $\sigma_{\beta}$ be the switch. $\beta$ corresponds to
a pair of Rosenhain subgroups $R_1,R_2$. Then the patching
subgroup $\Gamma_{\sigma_{\beta}}$ is cyclic of order $4$ and generated by
\[H/4 + \sum_{\alpha\in R_1} N_{\alpha}/2. \]
Of course we obtain the same group after replacing $R_1$ by $R_2$ in this case.

In the HG involution case, let $G$ be a G\"{o}pel tetrad and
$\sigma_G$ be the HG involution. Then the patching subgroup $\Gamma_{\sigma_{G}}$
is $2$-elementary abelian of order $4$ and generated by
\[H/2\text{ and }\sum_{\alpha\in G} N_{\alpha}/2. \]
We remark that this result of HG involution case also follows from the computations
of \cite{mukai}.

In the HW involution case, let $W$ be a Weber hexad and
$\sigma_W$ be the HW involution. Then the patching subgroup $\Gamma_{\sigma_{W}}$
is cyclic of order $4$ and generated by
\[H/4 + \sum_{\alpha\in W} N_{\alpha}/2. \]
The divisors $H, N_{\alpha}\in NS(X)$ and
also the classical notions appeared here will be defined in Section \ref{166},
where we recall the basic properties of Jacobian Kummer surfaces.
After fixing the basis of $A_{NS(X)}$, we can easily check that
there appear $31$ distinct patching subgroups.

\medskip
{\noindent{\textbf{Notation}:}} We refer the readers to \cite{nikulin-sym} for the basic
properties of the finite quadratic form $(A_L,b_L,q_L)$ associated with an even
nondegenerate lattice $L$. By definition, $A_L$ is the finite abelian group $L^*/L$,
$b_L: A_L\times A_L \rightarrow \bb{Q}/\bb{Z}$ is the symmetric bilinear form
and $q_L: A_L\rightarrow \bb{Q}/2\bb{Z}$ is the quadratic form, both naturally induced from
that of $L$.
Usually we denote finite forms by $(A_L,q_L)$, omitting $b_L$, or only by $A_L$.

The hyperbolic plane is denoted by $U$, the root lattices $A_l,D_m,E_n$
are considered to be negative definite.
The rank one lattice $\bracket{2n}$ is also used in this paper.
On finite forms, $u(2)$ is the associated form of the lattice $U(2)$,
$\bracket{1/2n}$ is that of $\bracket{2n}$.
The set of generators $\{e,f\}$ of $u(2)$ satisfying
\[q(e)=q(f)=0, b(e,f)=1/2\]
is called the standard generator.

For a lattice $T$ and $k=\bb{Q},\bb{C}$ we denote the scalar extension
by $T_k$.
If $T$ is a lattice and $T_{\bb{C}}$ is equipped with a Hodge
structure, then $\mathop{\mathrm{Aut}_{\mathrm{Hodge}}}\nolimits (T)$ is
a subgroup of $O(T)$ whose elements preserve the Hodge decomposition.

\section{The method of counting}\label{method}

In this section $X$ is any $K3$ surface.
Let $\sigma$ be a free involution on $X$.
The $(-1)$-eigenspace of the action of $\sigma$ on $NS(X)$ is denoted by $K$.
Then it is well-known that $K$ is negative definite, contains no $(-2)$-element
and the primitive hull of $K\oplus T_X$ in $H^2(X,\bb{Z})$ is isometric to
$U\oplus U(2)\oplus E_8(2)=:N$. We choose a marking $\phi: \overline{K\oplus T_X}\rightarrow N$
for this isometry.

The nonzero global holomorphic $2$-form $\omega_X$ on $X$ determines via $\phi$ a point
in $\mca{D}(N)/O(N)$, which is the period of the Enriques surface $Y:=X/\sigma$,
where
\[ \mca{D}(N):=\{\bb{C}\omega \in \bb{P}(N_{\bb{C}})|
   \omega\in N\otimes \bb{C},\omega\cdot \omega=0,\omega\cdot\overline{\omega}>0\}\]
is the (two copies of)
bounded symmetric domain of type IV associated to lattice $N$ of
signature $(2,10)$. Obviously this period is independent of the choice of $\phi$ and
the Torelli theorem of Enriques surfaces says that
this point determines the isomorphism class of $Y$ uniquely.

Conversely given a primitive embedding $\phi: T_X\rightarrow N$ such that
the orthogonal complement $K$ is free from $(-2)$-elements,
by the surjectivity there exists an Enriques surface $Y$
whose period is exactly $[\phi(\bb{C}\omega_X)]$.
If $\rho (X)\ge 12$ then
\cite[Theorem 1]{keum90} shows that $X$ is isomorphic to the universal double
cover of $Y$. Even if $\rho (X)\le 11$ the same holds,
whose proof is in \cite{ohashi07pre}.

Thus we have shown
\begin{prop}\label{correspondence}
There is a one-to-one correspondence between the sets
\[\{{\mathrm{Enriques\ quotients\ of\ }}X\}/{\mathrm{(isomorphisms)}}\]
and
\[\left\{
\begin{array}{l}
{\mathrm{Primitive\ embeddings\ }}\phi :T_X\rightarrow N\\
{\mathrm{such\ that\ }}K=T_X^{\perp}{\mathrm{contains\ no\ }}(-2){\mathrm{-elements}}
\end{array}
\right\}
\big/{\mathrm{(Hodge\ isometries\ of\ }}N{\mathrm{)}},\]
where for each $\phi$ we equip $N$ with a Hodge structure induced from that of $T_X$
by $\phi$.
\end{prop}

In the following, we identify $\overline{K\oplus T_X}$ with $N$ by $\phi$.
By \cite{nikulin-sym}, there are subgroups $\Gamma_K\subset A_K$ and $\Gamma_{T_X}\subset
A_{T_X}$ and a sign-reversing isometry $\varphi: \Gamma_K \stackrel{\sim}{\rightarrow}
\Gamma_{T_X}$ such that $N$ is the sublattice of $K_{\bb{Q}}\oplus T_{X,\bb{Q}}$
generated by $K$, $T_X$ and $\{(x,\varphi(x))|x\in \Gamma_K\}$.
\begin{dfn}\label{df}
The {\em{patching subgroup}} $\Gamma_{\sigma}$ of the free involution $\sigma$
is the inverse image of $\Gamma_{T_X}$ by the natural sign-reversing isometry
$A_{NS(X)}\stackrel{\sim}{\rightarrow} A_{T_X}$.
\end{dfn}

Under a condition, $\Gamma_{\sigma}$ is an invariant of a conjugacy class
which is very computable.
\begin{prop}\label{invariance}
If $\mathop{\mathrm{Aut}_{\mathrm{Hodge}}}\nolimits (T_X)=\{\pm \mathop{\mathrm{id}}\nolimits\}$, then $\Gamma_{\sigma}$ depends only on the
isomorphism class of the quotient Enriques surface.
\end{prop}
\begin{proof}
By Proposition \ref{correspondence}, conjugate free involutions induce on $N$
an isometric Hodge structure.
Any Hodge isometry of $N$ preserves $K=\omega_X^{\perp}$ and
hence $T_X$. Thus it induces $\pm \mathop{\mathrm{id}}\nolimits$ on $T_X$ and preserves the subgroup $\Gamma_{T_X}$.
\end{proof}

\begin{rem}
The condition above is weak. It is true if $\rho (X)$ is odd, see \cite[p597]{kondo98},
or even if $\rho (X)$ is even, it is true if
$X$ is very general in the period domain (\cite[Proposition 3.1]{ohashi}).
\end{rem}

In general there are free involutions not conjugate each other but with
the same $\Gamma_{\sigma}$. However in the Picard-general Jacobian Kummer case,
$\Gamma_{\sigma}$ completely classifies free involutions.
This will be shown in the next section.

The computation of $\Gamma_{\sigma}$ is done by
\begin{lem}\label{gamma}
Let $\sigma ,K$ as above. Then
\[\Gamma_{\sigma}=\{[x]\in NS(X)^*/NS(X)|\exists [y]\in K^*/K,\ x-y\in NS(X)\}.\]
\end{lem}
\begin{proof}
Let $\rho : A_{NS(X)}\rightarrow A_{T_X}$ be the canonical isomorphism.
Then $\rho ([x])=[z]$ is equivalent to $x+z\in H^2(X,\bb{Z})$.
Since
$\Gamma_{T_X}=\{[z]\in T_X^*/T_X|\exists [y]\in K^*/K,\ y+z\in N\}$,
\begin{eqnarray*}
\Gamma_{\sigma} &=& \{ [x]\in NS(X)^*/NS(X)| \rho ([x])\in \Gamma_{T_X}\}\\
\ &=& \{[x]\in NS(X)^*/NS(X)|\exists [y]\in K^*/K,\ x-y\in NS(X)\}.
\end{eqnarray*}
This is what we need.
\end{proof}

\section{Invariants of free involutions}\label{invariants}

Let $C$ be a genus $2$ curve, $J(C)$ its Jacobian and $Km(J(C))=X$
the associated Jacobian Kummer surface as in the Introduction.
As is well-known,
$J(C)$ contains $C$ as a theta divisor:
\[\Theta=\{[p-p_0] | p\in C \}\subset J(C), \quad p_0\in C.\]
Hence $\mathop{\mathrm{rank}}\nolimits NS(J(C))\ge 1$ and $\mathop{\mathrm{rank}}\nolimits NS(X)\ge 17$
holds. When we have the equality, we call $X$ {\em{Picard-general}}.
In this case, since $T_{J(C)}=U^{\oplus 2}\oplus \bracket{-2}$ we have
$T_X=U(2)^{\oplus 2}\oplus \bracket{-4}$
and $NS(X)=U\oplus D_4^{\oplus 2}\oplus D_7$.

For simplicity, we put $T:=T_X$.
Suppose we are given a primitive embedding of $T$ into
$N$ such that the orthogonal complement is free from $(-2)$-elements,
as in Proposition \ref{correspondence}.
First we determine the orthogonal complement.
\begin{prop}\label{e7}
The lattice $K=T_X^{\perp}$ is isometric to $E_7(2)$.
\end{prop}
\begin{proof}
Consider the unique embedding of $N$ into the abstract $K3$ lattice $L$.
The orthogonal complement is denoted by $M$, $M\simeq U(2)\oplus E_8(2)$.
By \cite{nikulin-sym}, we have the following isomorphism of discriminant
quadratic forms:
\begin{equation}
-q_K \simeq (q_M\oplus q_T|_{\Gamma^{\perp}})/\Gamma\label{nikulin}
\end{equation}
where $\Gamma$ is the pushout (i.e. the graph) of a sign-reversing isometry of subgroups
$\Gamma_M\subset A_M$ and $\Gamma_T\subset A_T$.

For a finite quadratic form $(A,q)$, we denote the quadratic form induced on the
$2$-torsion subgroup $A_2=\{x\in A| 2x=0\}$ by $(A_2,q_2)$.
Note that even if $q$ is nondegenerate, $q_2$ may be
degenerate.

In our equality (\ref{nikulin}),
$A_M$ is $2$-elementary, hence $\Gamma$ is $2$-elementary and $\Gamma_T$ is
contained in $(A_T)_2$. Put $\#\Gamma=2^a$. This shows $a \le 5=l_2(A_T)$, where $l_2$ denotes
the number of minimal generators of the $2$-Sylow subgroup of $A_T$.

 Also it follows
\begin{equation}\label{nikulin2}
((A_M\oplus A_T)_2|_{\Gamma^{\perp}})/\Gamma
 \subset (A_M\oplus A_T|_{\Gamma^{\perp}})/\Gamma=A_K,
\end{equation}
since $\Gamma$ is $2$-elementary.
$(A_M\oplus A_T)_2$ has a radical of order $2$ contained in
$(A_T)_2$. Since $\Gamma$ is a graph, this radical is not contained
in $\Gamma$. This shows that
$\# ((A_M\oplus A_T)_2|_{\Gamma^{\perp}})=2^{15-a}$.
Thus the order of the left-hand-side of (\ref{nikulin2}) is $2^{15-2a}$.
Since $K$ is of rank $7$, we have $15-2a\le 7$ and hence $a=4,5$.

We show that if $a=5$ then $K$ contains $(-2)$-elements and contradicts
the assumption.
For this, first note that in this case
$\Gamma_T=(A_T)_2$ is uniquely determined
and the embedding of $\Gamma_T$ in $A_M\simeq u(2)^{\oplus 5}$ is unique up to isomorphism by
Witt's theorem. So we can compute $q_K$ directly and
get $q_K\simeq u(2)^{\oplus 2}\oplus \bracket{1/4}$.
From this we see that there are inclusions
$K\subset K' \subset \bracket{-1}^{\oplus 7}$
such that $K'$ is an even lattice, $[K':K]=4$ and $[\bracket{-1}^{\oplus 7}: K']=2$.
By the definition of $D_7$, $K'\simeq D_7$.
Consider the Dynkin diagram of $D_7$ and take a subgraph isomorphic to $A_6$
with vertices $e_1,\cdots e_6$ in this order. Put $f_0=0, f_j=e_1+\cdots+e_j$, $1\le j \le 6$.
Any difference of two of these seven elements have self-intersection $(-2)$.
If $K$ has no $(-2)$-elements, then $\{f_j\}_{0\le j\le 6}$ cannot be in
the same residue class of $K'/K$.
Then we must have $[K':K]\ge 7$ and contradiction.

Thus we obtain $a=4$. From (\ref{nikulin2}), we see that
$\#(A_K)_2\ge 2^7$. It follows that $K(1/2)$ is an integral
(may be odd) lattice and $\mathop{\mathrm{det}}\nolimits K(1/2)=-2$. By assumption, the minimal norm
of the positive definite lattice $K(-1/2)$ is greater than $1$.
It follows from \cite[p400, Table 15.8]{conway-sloane} that $K(1/2)\simeq
(\bracket{-2}^{\perp}\ \mathrm{in}\ E_8)\simeq E_7$.
\end{proof}

The following nature of the lattice $K=E_7(2)$ will be used.
\begin{lem}\label{surj}
The canonical homomorphism $\sigma: O(K)\rightarrow O(q_K)$ is surjective.
\end{lem}
\begin{proof}
The same property for the lattice $E_8(2)$ is known by \cite{barth-peters}.
We reduce the lemma to this case.
Firstly, we know the orders of the two groups.
By \cite{bourbaki}, $\# O(E_7(2))= \# O(E_7)= 2^{10}\cdot 3^4 \cdot5 \cdot 7$.
On the other hand, we can easily compute the order of $O(q_K)$
as $\# O(q_K)=\# O(u(2)^{\oplus 2}\oplus \bracket{1/4})= 2^{10}\cdot 3^4 \cdot5 \cdot 7$
(c.f. Lemma \ref{orbit}).
Thus it is enough to show that $\sigma$ is injective.

We take a $(-4)$-element $r$ of $E_8(2)$ and identify $K$ with $r^{\perp}$.
Obviously $g\in \ker \sigma$ can be extended to an isometry $\overline{g}$ of $E_8(2)$
by defining $\overline{g}(r)=r$. It is clear that $\overline{g}$ acts on
the discriminant $A_{E_8(2)}$ trivially. It follows from \cite[Proposition 1.7]{barth-peters}
that $\overline{g}=\pm \mathop{\mathrm{id}}\nolimits.$ Since $\overline{g}(r)=r$, $\overline{g}= \mathop{\mathrm{id}}\nolimits$.
\end{proof}

Because $(A_K,q_K)\simeq u(2)^{\oplus 3}\oplus \bracket{1/4}$, the next
lemma is also used.
\begin{lem}\label{orbit}
Let $(A,q)=u(2)^m \oplus \bracket{1/4}$ be a finite quadratic form.
Then the action of $O(q)$ on $A$ decomposes $A$ into $6$ orbits.
If we denote the standard generator of one of $u(2)$ by $\{e,f\}$ and
that of $\bracket{1/4}$ by $\{g\}$,
they are as in the following table.
\begin{center}
\begin{tabular}{lll} \hline
a representative  & length & square\\ \hline
$0$ & $1$ & $0$ \\
$2g$ & $1$ & $1$ \\
$e$ & $2^{2m}-1$ & $0$ \\
$e+f$ & $2^{2m}-1$ & $1$ \\
$g$ & $2^{2m}+2^m$ & $1/4$ \\
$e+f+g$ & $2^{2m}-2^m$ & $-3/4$ \\ \hline
\end{tabular}
\end{center}
In this table, for a representative $x$, the length is $\# (O(q)\cdot x)$ and the square
is $q(x)\in \bb{Q}/2\bb{Z}$.
\end{lem}
The proof is given by induction on $m$ and we omit it. \\

Now we are going to describe the latter set of Proposition \ref{correspondence},
i.e., we classify the Hodge structures on $N$ induced from embeddings
$T\subset N$ as in the proposition.
We recall Definition \ref{df}, there is an isomorphism
\begin{equation}\label{nikulin3}
q_N \simeq (q_K \oplus q_T)|_{\Gamma^{\perp}}/\Gamma,
\end{equation}
where $\Gamma$ is the pushout of a sign-reversing isometry of subgroups
$\Gamma_K\subset A_K$ and $\Gamma_T\subset A_T$.
By Proposition \ref{invariance}, $\Gamma_T$ is an invariant of the Hodge structure.

We will prove the converse. Namely, suppose we have two embeddings $T\subset N_1$
and $T\subset N_2$ whose orthogonal complements are denoted by $K_i$.
For each embedding we have $(\Gamma_i, \Gamma_{K,i} \Gamma_{T,i})$ and the equality
(\ref{nikulin3}). What we want to show is
\[(*):\Gamma_{T,1}=\Gamma_{T,2}\Rightarrow N_1\simeq N_2 {\rm (Hodge\ isometry)}.\]
The argument goes as follows. Assume we could find an isometry $\sigma_K:
A_{K_1}\rightarrow A_{K_2}$ such that the following commutes.
\[\begin{CD}
 A_{K_1} \supset\ @. \Gamma_{K,1} @>\sim>> \Gamma_{T,1} @. \ \subset A_{T}\\
@V\sigma_K VV @V\sigma_K VV @VV\mathop{\mathrm{id}}\nolimits V @VV\mathop{\mathrm{id}}\nolimits V \\
 A_{K_2} \supset\ @. \Gamma_{K,2} @>\sim>> \Gamma_{T,2} @. \ \subset A_{T}
\end{CD}\]
Then by Lemma \ref{surj} we can lift $\sigma_K$ to $\sigma'_K: K_1
\stackrel{\sim}{\rightarrow} K_2$ and the pair $(\sigma'_K,\mathop{\mathrm{id}}\nolimits_T)$
can be lifted to an Hodge isometry $N_1\stackrel{\sim}{\rightarrow} N_2$.
Thus it is enough to find $\sigma_K$.

By Proposition \ref{e7}, $[N:K\oplus T]=\#\Gamma=4.$
Thus there are two possibilities of underlying groups of $\Gamma_K\simeq \Gamma_T$.
We consider each case separately.

\medskip
First we consider the case $\Gamma_{T,i}\simeq \bb{Z}/4\bb{Z}$.
The square of the generator
$g_T\in \Gamma_{T,1}=\Gamma_{T,2}$
is independent of the
choice and there are two possibilities, $q_T(g_T)=-1/4$ or $3/4$.
Let $(g_{K,i},g_T)\in \Gamma_i$. We have $q_{K_i}(g_{K,i})=1/4$ or $-3/4$ respectively.
By Lemma \ref{orbit}, in these cases we can find $\sigma_K$ and
$(*)$ is proved. We find easily that there are $10$ subgroups $\Gamma_T$
satisfying $q_T(g_T)=-1/4$. Also there are $6$ with $q_T(g_T)=3/4$.

\medskip
Second we consider the case $\Gamma_T\simeq \bb{Z}/2\bb{Z}\oplus \bb{Z}/2\bb{Z}$.
The argument becomes slightly complicated, but the conclusion is the same.
To prove $(*)$ in this case, first we show that
$\Gamma_{i}$ always contains a particular element.
Here, for a clear argument, we take generators $g_i$ and $g'$ of $\bracket{1/4}\subset A_{K_i}$ and
$\bracket{-1/4}\subset A_T$ respectively. We denote an element of
$A_{K_i}\oplus A_T$ by
\[(x,y;z,w)\in A_{K_i}\oplus A_T \ ;\ x\in u(2)^{\oplus 3},
y\in \bracket{1/4}, z\in u(2)^{\oplus 2}, w\in \bracket{-1/4}.\]
Then the claim is that
\[(0,2g_i;0,2g')\in \Gamma_i.\]
In fact, since $\Gamma_i$ is contained in $(A_{K_i}\oplus A_T)_2$,
the radical element $(0,2g_i;0,0)$ of $(A_{K_i}\oplus A_T)_2$
is in $\Gamma^{\perp}_i$.
Hence its residue class $(0,2g_i;0,0)+\Gamma_i$ determines an element of $A_{N_i}$ by the
isomorphism (\ref{nikulin3}). It is nonzero because $q_{K_i}(2g_i)=1.$
Since $A_{N_i}$ is nondegenerate, there exists an element $(x,y;z,w)+\Gamma_i \in A_{N_i}$
with $(0,2g_i;0,0)\cdot (x,y;z,w)=1/2.$ It follows $y=\pm g_i$.
Further since $(q_{K_i}\oplus q_T)(x,y;z,w)\in\bb{Z}$, it follows $w=\pm g'$,
i.e., there exists an element in $\Gamma_i^{\perp}$ of the form $(x,\pm g_i;z,\pm g')$.
Since the residue class of this element is of order $2$ in $A_{N_i}$, we have that
$(0,2g_i;0,2g')\in \Gamma_i$.

Let $\Gamma_{T,1}=\Gamma_{T,2}=\bracket{2g',\alpha}$.
Replacing $\alpha$ by $\alpha+2g'$ if necessary,
we can assume $q_T(\alpha)=0$. Let $(\beta_i,\alpha)\in \Gamma_i$, $q_{K_i}(\beta_i)=0$.
By Lemma \ref{orbit}, we can find $\sigma_K: A_{K_1}\stackrel{\sim}{\rightarrow}
A_{K_2}$ which takes $\beta_1$ to $\beta_2$. This $\sigma_K$ must take $2g_1$ to $2g_2$,
so we have now proved $(*)$.
There are $15$ possible $\Gamma_T$
in this case.

In summary, we have obtained the following.
\begin{prop}\label{fin}
Let $X$ be a Picard-general Jacobian Kummer surface. Then
free involutions $\sigma_1,\sigma_2$ are conjugate
if and only if the patching subgroups $\Gamma_{\sigma_1},\Gamma_{\sigma_2}$ coincide.
There exist (at most) $31=10+15+6$ free involutions.
\end{prop}

The existence of $31$ free involutions is assured by concrete constructions
in the following sections.

\section{The $(16)_6$ configuration on a Jacobian Kummer surface}\label{166}

In this section we recall and prepare notations
concerning the divisors on Jacobian Kummer surfaces.
The content of this section is known,
references are \cite{keum97, kondo98, dolgachev-keum}.

{\textbf{The index set.}}
Let $C$ be a smooth projective curve of genus $2$.
It is a double cover of $\bb{P}^1$
which ramifies at $6$ Weierstrass points $\{p_1,\cdots ,p_6\}\subset C$.
Here we should notice the linear equivalence
\[p_i+p_j+p_k-p_l-p_m-p_n\sim 0 \]
for an arbitrary permutation $\{i,j,k,l,m,n\}$ of $\{1,\cdots,6\}$.
The set of theta characteristics of $C$ is by definition
\[S(C)=\{D\in \mathop{\mathrm{Pic}}\nolimits (C)|2D\sim K_C\}.\]
They are divided into odd theta characteristics
$\{[p_i]|i=1,\cdots ,6\}$ and even ones
$\{[p_i+p_j-p_k]|i,j,k \mathrm{\ are\ distinct\ each\ other}\}$. There are $16$
theta characteristics.

The Jacobian variety $J(C)$ consists of divisor classes of degree $0$ on $C$.
We denote by $J(C)_2$ the set of sixteen $2$-torsion points of $J(C)$. Then
\[J(C)_2=\{0\}\cup \{[p_i-p_j]|i\neq j\}.\]

These $16+16=32$ divisor classes naturally correspond to partitions of the set
$\{1,\cdots 6\}$ into two subsets in the following way.
\begin{gather*}
[p_i]\in S(C) \longleftrightarrow\{i\}\cup \{i\}^c.\\
[p_i+p_j-p_k]\in S(C) \longleftrightarrow\{i,j,k\}\cup \{i,j,k\}^c.\\
[p_i-p_j]\in J(C)_2 \longleftrightarrow\{i,j\}\cup \{i,j\}^c.\\
0\in J(C)_2 \longleftrightarrow \emptyset\cup\{1,\cdots,6\},
\end{gather*}
where the complement is taken in the set $\{1,\cdots,6\}$.
We denote these partitions
by exhibiting one of the subsets, surrounded by $[\ ]$.
For example, $p_1-p_2$ corresponds to $[12]=[3456]$, $p_1+p_2-p_3$ corresponds to $[123]=[456]$,
etc. $[\emptyset]$ is denoted by $[0]$.
In this notation, we see that the symmetric difference of subsets
$\alpha,\beta$ of $\{1,\cdots,6\}$ corresponds to addition or difference in $\mathop{\mathrm{Div}}\nolimits (C)$
as follows.
\begin{gather*}
[\alpha\circleddash \beta]=[\alpha]-[\beta]\ {\rm if}\ [\alpha],[\beta] \in S(C),\\
[\alpha\circleddash \beta]=[\alpha]+[\beta]\ {\rm otherwise}.
\end{gather*}
When we use a partition $[\alpha]$ as an index, $[\ ]$ will be omitted.

\medskip
{\textbf{The $(16)_6$ configuration.}}
The sixteen theta divisors on $J(C)$ corresponding to $\beta\in S(C)$ are
\[\Theta_{\beta}=\{[p-\beta]\in J(C)|p\in C\}.\]
The sixteen nodes $\{n_{\alpha}\in \overline{X}|\alpha\in J(C)_2\}$
on $\overline{X}=J(C)/\{\pm 1\}$ are the images of $\alpha\in J(C)_2$.
On the minimal desingularization $X$ of $\overline{X}$,
$n_{\alpha}$ is blown up to give a smooth rational curve $N_{\alpha}$ on $X$.
The tropes $\overline{T_{\beta}}\subset \overline{X}$ and $T_{\beta}\subset X$
are the strict transforms of ${\Theta}_{\beta}$.
Hence we obtain $32$ $(-2)$-curves $\{N_{\alpha},T_{\beta}\}_{\alpha,\beta}$ on $X$.
The incidence relation between these divisors
is called the $(16)_6$ configuration. It is given explicitly by
\begin{gather*}
(N_{\alpha},N_{\alpha '})=-2\delta_{\alpha,\alpha '},\qquad
(T_{\beta},T_{\beta '})=-2\delta_{\beta,\beta '},\\
(N_{\alpha},T_{\beta})=1 \Leftrightarrow \alpha+\beta\in\{[1],[2],[3],[4],[5],[6]\}.
\end{gather*}

A permutation of the set $\{N_{\alpha},T_{\beta}\}_{\alpha,\beta}$
which preserves the incidence relation above is called an automorphism.
Nikulin \cite{nikulin-jac} showed that the automorphism
group is isomorphic to $(\bb{Z}/2\bb{Z})^{5}\rtimes \mf{S}_6$,
where $(\bb{Z}/2\bb{Z})^{5}$ consists of automorphisms
induced from translations by elements of $J(C)_2\cup S(C)$
and $\mf{S}_6$ acts on the index set $\{1,\cdots, 6\}.$
We took our notations as above because this $\mf{S}_6$-action is best seen.

Translations with respect to $\alpha \in J(C)_2$ are geometrically
realized on $J(C)$. They induce automorphisms $t_{\alpha}$ of $X$.
These are the {\em translations} in the classical terms.
In the next section we will see that translations with respect to
$\beta\in S(C)$ are also geometrically realized by $\sigma_{\beta}\in \mathop{\mathrm{Aut}}\nolimits (X)$.
These $\sigma_{\beta}$ are the {\em switches}.
On the other hand, in general the action of $\mf{S}_6$ cannot be
lifted to an automorphism of $X$.

{\noindent{\textbf{Remark}:}} In \cite{keum97} and \cite{kondo98}, the notations are a little different.
To adjust notations of \cite{kondo98} to ours, first we regard
$p_0$ of \cite{kondo98} as our $p_6$.
Then the correspondence is as in below.
\begin{center}
\begin{tabular}{c|cccccc}
\cite{kondo98} & $N_0$ & $N_i$ & $N_{ij}$ & $T_0$ & $T_i$ & $T_{ij}$ \\ \hline
ours & $N_0$ & $N_{i6}$ & $N_{ij}$ & $T_6$ & $T_i$ & $T_{ij6}$
\end{tabular}
\end{center}
\begin{lem}[\cite{kondo98}]\label{nsstr1}
For $\beta\in S(C)$, let $\Lambda(\beta):=\{\alpha\in J(C)_2|(N_{\alpha},T_{\beta})=1\}.$
Then the divisor class of
\[H=2T_{\beta}+\sum_{\alpha\in \Lambda(\beta)}N_{\alpha}\]
is independent of $\beta$ and coincides with the pullback of the hyperplane section
by the morphism $X\rightarrow \overline{X}\subset \bb{P}^3$.
\end{lem}
\begin{lem}[\cite{kondo98}]\label{nsstr2}
Assume that $X$ is Picard-general.
\begin{enumerate}
\item $NS(X)$ is generated over $\bb{Z}$ by $\{ N_{\alpha},T_{\beta}\}_{\alpha,\beta}$.
\item $\{H,N_{\alpha}\}_{\alpha}$ is an orthogonal basis of $NS(X)_{\bb{Q}}$ over $\bb{Q}$.
\item A generator set of the discriminant group $A_{NS(X)}$ is given by
\begin{gather*}
\msf{e}_1=(N_{26}+N_{12}+N_{36}+N_{13})/2,\msf{f}_1=(N_{16}+N_{12}+N_{46}+N_{24})/2,\\
\msf{e}_2=(N_{26}+N_{12}+N_{46}+N_{14})/2,\msf{f}_2=(N_{16}+N_{12}+N_{36}+N_{23})/2,\\
\msf{g}=H/4+(N_{0}+N_{16}+N_{26}+N_{12})/2.
\end{gather*}
\end{enumerate}
\end{lem}

\medskip
{\textbf{Special sets of nodes.}}
Lastly we mention several special sets of nodes of $\overline{X}$.
See also \cite{dolgachev-keum}.
We identify the set of nodes with $J(C)_2$ which is a $4$-dimensional
vector space over $\bb{F}_2$. We have then the symplectic bilinear form
\[([\alpha],[\alpha'])\mapsto \#(\alpha\cap \alpha') \mod 2.\]

A $2$-dimensional subspace $G$ is called {\em{G\"{o}pel}} if it
is totally isotropic. The translations of G\"{o}pel subgroups are
called {\em{G\"{o}pel tetrads}}. There are $60$ G\"{o}pel tetrads.
A $2$-dimensional subspace $R$ which is not G\"{o}pel is called {\em{Rosenhain}}
and its translations {\em{Rosenhain tetrads}}. There are $80$ Rosenhain tetrads.
A {\em{Weber hexad}} is a $6$-set which can be written as the symmetric difference of
a G\"{o}pel tetrad and a Rosenhain tetrad.
It can be shown that any Weber hexad is of one of the following forms
\begin{equation}\label{web}
\{0,ij,jk,kl,lm,mi\}\ {\mathrm{or}}\ \{ij,jk,ki,il,jm,kn\}
\end{equation}
according to whether it contains $0$ or not.
There are $192$ Weber hexads.

In the following sections, we introduce automorphisms using these special sets.

\section{Switches}\label{switch}

Switches are one kind of automorphisms
found by F. Klein \cite{klein}. The freeness in even cases is an
easy consequence of the description of \cite{maria}, although it is implicit there.
Let $\beta\in S(C)$.
For a smooth point $\overline{a}\in \overline{X}$, which means
that the preimage of $\overline{a}$ in $J(C)$ is $\{a,-a\}$,
the divisors $t_a(\Theta_{\beta})$ and $t_{-a}(\Theta_{\beta})$
intersect at two points, which is of the form
\[t_a(\Theta_{\beta})\cap t_{-a}(\Theta_{\beta})=\{b,-b\}.\]
The switch is defined by $\sigma_{\beta}:\overline{a}\mapsto \overline{b}$.

More precisely, these switches are defined as the composite
of the Gauss map
\[G: \bb{P}^3 \supset \overline{X}\dashrightarrow
\overline{X}^* \subset (\bb{P}^3)^*, \]
which maps a smooth point $\overline{a}$ to $T_{\overline{a}}\overline{X}$,
and the projective linear isomorphism
\[F_{\beta}: \overline{X}^*\rightarrow \overline{X},\]
defined for each $\beta$. See \cite{maria}.

$\sigma_{\beta}$ is a birational involution of $\overline{X}$. Hence it induces an
involution of $X$, which we denote by the same $\sigma_{\beta}$.
We can easily check that $\sigma_{\beta}$ interchanges $N_{\alpha}$ with
$T_{\alpha+\beta}$ for $\forall \alpha\in J(C)_2$.
\begin{prop}
For an even theta characteristic $\beta$, $\sigma_{\beta}$ is a free involution on $X$.
\end{prop}
\begin{proof}
Suppose a smooth point $\overline{a}\in \overline{X}$ is
a fixed point of $\sigma_{\beta}$.
This is equivalent to
$t_a(\Theta_{\beta})\cap t_{-a}(\Theta_{\beta})=\{a,-a\}$
and it is necessary that $a\in t_a(\Theta_{\beta})$, $0\in \Theta_{\beta}.$
This is untrue if $\beta$ is even.

On the other hand, the divisor $N_{\alpha}$ is disjoint from $T_{\alpha+\beta}$,
so $\sigma_{\beta}$ has no fixed points.
\end{proof}

{\noindent{\textbf{Remark}:}} (1) The proof above does not use the assumption of being Picard-general.
Thus switches for even theta characteristics are always free involutions.\\
(2) The fixed point set of a switch for an odd theta characteristic is
a curve of genus $5$, named after Humbert.\\

Let $\sigma_{\beta}$ be a free switch.
In the following computation, we take the case $\beta=[123]$ for simplicity.
We can obtain the result for other cases by the action of $\mf{S}_6$.
Let $K$ be the
$(-1)$-eigenspace of the action of $\sigma_{123}$ on $NS(X)$ as in Section \ref{method}.
\begin{prop}
For Picard-general $X$, $K$ is generated over $\bb{Z}$ by
the following elements.
\begin{gather*}
f=N_{15}-T_{146},\ e_2=T_{145}-N_{16},\ e_3=N_{45}-T_{6},\ e_4=T_{123}-N_{0},\\
e_5=N_{12}-T_{3},\ e_6=T_{124}-N_{34},\ e_7=N_{24}-T_{134}, \\
e_1=-(1/2)(f+2e_2+3e_3+4e_4+3e_5+2e_6+e_7).
\end{gather*}
\end{prop}
\begin{proof}
We can check that $\{f,e_2,\cdots, e_7\}$ spans a sublattice of $K$
isomorphic to $A_7(2)$.
We now show $e_1\in NS(X)$. Modulo $NS(X)$,
\begin{eqnarray*}
e_1 &\equiv & (f+e_3+e_5+e_7)/2\\
&\equiv & (N_{15}+N_{45}+N_{12}+N_{24})/2+(T_{146}+T_{6}+T_{3}+T_{134})/2\\
&=& (N_{15}+N_{45}+N_{12}+N_{24})/2+(\frac{H}{2}\cdot 4 -\frac{1}{2}
\sum_{\alpha\in \Lambda([146])\cup\Lambda([6])\cup\Lambda([3])\cup\Lambda([134])} N_{\alpha})/2\\
&=& (N_{15}+N_{45}+N_{12}+N_{24})/2+H-\frac{1}{4}
\sum_{\alpha\in J(C)_2-\{[15],[45],[12],[24]\}} 2N_{\alpha}\\
&\equiv & \sum_{\alpha\in J(C)_2} N_{\alpha}/2.
\end{eqnarray*}
The blow up $\widehat{J(C)}$ of $J(C)$ at points of $J(C)_2$ is
the double cover of $X$ branched exactly over $\cup_{\alpha} N_{\alpha}$.
Thus $e_1\in NS(X)$ follows.

Then it is easy to check that $\{e_1,e_2,\cdots, e_7\}$ spans a sublattice of $K$
isomorphic to $E_7(2)$. By Proposition \ref{e7}, they coincide.
\end{proof}
\begin{prop}\label{result1}
The patching subgroup of $\sigma_{123}$ is the cyclic group generated by the element
$[x=H/4+(N_{0}+N_{12}+N_{23}+N_{31})/2]\in A_{NS(X)}$.
\end{prop}
\begin{proof}
The facts $x\in NS(X)^*$ and $y:=-(e_1+e_5+e_7)/4+e_5/2+e_6/2\in K^*$ are easily checked.
We use Lemma \ref{gamma}. We first check $x-y\in NS(X)$. This is because
\begin{eqnarray*}
y &=& (1/8)(f+2e_2+3e_3+4e_4+5e_5+6e_6-e_7)\\
 &=& H/4 -(N_{14}+N_{24}+N_{34}+N_{56})/2
\end{eqnarray*}
and
\begin{eqnarray*}
x-y &=& (1/2)(N_{0}+N_{12}+N_{23}+N_{31}+N_{14}+N_{24}+N_{34}+N_{56})\\
&\equiv& T_{123}-T_{4} \equiv 0.
\end{eqnarray*}
Thus $[x]\in \Gamma_{\sigma_{123}}$. Then since $[x]$ is of order $4$ in
$A_{NS(X)}$ and $\#\Gamma_{\sigma_{123}}=4$, $\Gamma_{\sigma_{123}}$ is generated
by $[x]$.
\end{proof}
{\noindent{\textbf{Observation}:}} In the expression of $[x]$, $\{n_{0},n_{12},n_{23},n_{31}\}$ is a
Rosenhain subgroup defined in Section \ref{166}. The class of $-x$ can be
written as $[H/4+(N_0+N_{45}+N_{56}+N_{64})/2]$, where $\{n_{0},n_{45},n_{56},n_{64}\}$
is also a Rosenhain subgroup.
In general, for an even theta characteristic $\beta$, the $6$-set $\Lambda (\beta)$
(see Lemma \ref{nsstr1}) can be uniquely written in the form $R_1\circleddash R_2$
where $R_i$ are Rosenhain subgroups. In our case $\beta=[123]$,
$R_1=\{n_{0},n_{12},n_{23},n_{31}\}$ and $R_2=\{n_{0},n_{45},n_{56},n_{64}\}$.
\begin{prop}\label{sec}
The patching subgroup of $\sigma_{\beta}$ for general $\beta$ is generated by
$[H/4+(\sum_{\alpha\in R}N_{\alpha})/2]$ where $R$ is one of the
two Rosenhain subgroups corresponding to $\beta$.
\end{prop}
\begin{proof} When $\beta=[123]$, this is Proposition \ref{result1}.
Since the action of $\mf{S}_6$ is compatible with the observation above,
the general case follows.
\end{proof}

By Proposition \ref{sec}, we can write down the generator of the
patching subgroup of the switch $\sigma_{\beta}$ for all $\beta$.
We use the notations of
Lemma \ref{nsstr2}.

\begin{center}
\begin{tabular}{c||c|c|c|c|c}
$\beta$     & $[123]$ & [124] & [125] & [126] & [134] \\ \hline
& $\msf{e}_1+\msf{f}_2+\msf{g}$ & $\msf{e}_2+\msf{f}_1+\msf{g}$ &
 $\msf{e}_1+\msf{f}_1+\msf{e}_2+\msf{f}_2+\msf{g}$ & $\msf{g}$ &
 $\msf{f}_1+\msf{f}_2+\msf{g}$
\end{tabular}
\smallskip
\begin{tabular}{c||c|c|c|c|c}
$\beta$     & [135] & [136] & [145] & [146] & [156]  \\ \hline
& $\msf{f}_1+\msf{g}$ & $\msf{e}_1+\msf{g}$ & $\msf{f}_2+\msf{g}$ &
 $\msf{e}_2+\msf{g}$ &$\msf{e}_1+\msf{e}_2+\msf{g}$
\end{tabular}
\end{center}
Since all these are distinct each other, we see that
the ten switches are not conjugate each other in $\mathop{\mathrm{Aut}}\nolimits (X)$
if $X$ is Picard-general.

\section{Hutchinson's involutions associated with G\"{o}pel tetrads}

These
automorphisms appear in \cite{hut01}. The generic freeness is found by
J. H. Keum in \cite{keum90}. We briefly recall the construction.
Let $G$ be a G\"{o}pel tetrad. If we choose $G$ as the reference points of the
homogeneous coordinates of $\bb{P}^3$, the equation of $\overline{X}$
becomes
\begin{eqnarray*}
A(x^2 t^2+y^2 z^2)+B(y^2 t^2+z^2 x^2)+C(z^2 t^2+x^2 y^2)+Dxyzt\\
+E(yt+zx)(zt+xy)+G(zt+xy)(xt+yz)+H(xt+yz)(yt+zx)=0,
\end{eqnarray*}
for suitable scalars $A,\cdots,H$.
$\sigma_G$ is the Cremona involution
\[(x,y,z,t) \mapsto (1/x,1/y,1/z,1/t).\]
For a translation $t=t_{\alpha}$, we have $\sigma_{t(G)}=t\sigma_G t$,
so that we can restrict ourselves to the case $G$ is a G\"{o}pel subgroup.
But any G\"{o}pel subgroup is of the form $\{n_{0},n_{ij},n_{kl},n_{mn}\}$
hence up to $\mf{S}_6$ we can assume
$G_0=\{n_{0},n_{12},n_{34},n_{56}\}$.
By \cite{keum97}, the induced action of $\sigma_{G_0}$ on $NS(X)$ is given by
\begin{gather*}
N_{\alpha}\leftrightarrow H-N_{0}-N_{12}-N_{34}-N_{56}+N_{\alpha},\quad {\textrm{for\ }}
\alpha =[0],[12],[34],[56]\\
T_1\leftrightarrow T_2,\ T_3\leftrightarrow T_4,\ T_5\leftrightarrow T_6, \\
T_{134}\leftrightarrow T_{234},\ T_{123}\leftrightarrow T_{124},\ T_{125}\leftrightarrow T_{126}.
\end{gather*}
\begin{prop}
The $(-1)$-eigenspace $K$ of $\sigma_{G_0}$ is generated over $\bb{Z}$ by the following elements.
\begin{gather*}
g=N_{0}+N_{12}+N_{34}+N_{56}-H,
e_5=T_1-T_2,\ e_1=T_3-T_4,\ e_7=T_5-T_6, \\
f=T_{134}-T_{234},\ e_3=T_{123}-T_{124},\ h=T_{125}-T_{126},\\
e_2=(1/2)(f+g+h-e_3),e_4=(1/2)(f-e_1-e_3-e_5),\\
e_6=(1/2)(f+h-e_5-e_7).
\end{gather*}
\end{prop}
\begin{proof}
$e_1,e_3,e_5,e_7,f,g,h\in K$ generate
a sublattice of $K$ isomorphic to $A_1(2)^{\oplus 7}$.
It is easy to see that $e_2,e_4,e_6\in NS(X)$. For example, modulo $NS(X)$,
\begin{eqnarray*}
e_2&\equiv& (1/2)(H+N_{0}+N_{12}+N_{34}+N_{56}\\
& & + T_{123}+T_{124}+T_{125}+T_{126}+T_{134}+T_{234}) \\
&=& 2H+N_0-(1/2)\sum_{\alpha\in J(C)_2} N_{\alpha}.
\end{eqnarray*}
and as in Section \ref{switch} $e_2\in NS(X)$. $e_4,e_6$ are similar.

Then we see that
$e_1,\cdots, e_7$ span
a lattice isomorphic to $E_7(2)$.
\end{proof}
\begin{prop}\label{result2}
The patching subgroup of $\sigma_{G_0}$ is $2$-elementary abelian and generated by
\[x=(N_0+N_{12}+N_{34}+N_{56})/2, \text{ and }y=H/2.\]
\end{prop}
\begin{proof}
This proposition is proved in the same way as
Proposition \ref{result1}. The corresponding element in $K^*/K$
is $x'=(e_1+e_3)/2, y'=(e_1+e_5+e_7)/2$ and we can check
$x-x', y-y'\in NS(X)$. Then we use Lemma \ref{gamma}.
\end{proof}
By the $\mf{S}_6$-symmetry, we obtain the following.
\begin{prop}
For any G\"{o}pel subgroup $G$, we have
$\Gamma_{\sigma_{G}}=\bracket{H/2,(1/2)\sum_{\alpha\in G}N_{\alpha}}.$
\end{prop}
More generally, ussing the translation relation $\sigma_{t(G)}=t\sigma_G t$,
the generator above is valid for any G\"{o}pel tetrad.\\

There are $15$ G\"{o}pel subgroups. Under the notations of
Lemmas \ref{nsstr1} and \ref{nsstr2}, we deduce the following table.
\begin{center}
\begin{tabular}{ll} \hline
The tetrad  & Patching element corresponding to $x$ \\ \hline
$[0]+[12]+[34]+[56]$ &  $\msf{e}_1+\msf{f}_1+\msf{e}_2+\msf{f}_2$ \\
$[0]+[12]+[35]+[46]$ &  $\msf{f}_1+\msf{e}_2$ \\
$[0]+[12]+[36]+[45]$ &  $\msf{e}_1+\msf{f}_2$ \\
$[0]+[13]+[24]+[56]$ &  $\msf{e}_1+\msf{f}_1+2\msf{g}$ \\
$[0]+[13]+[25]+[46]$ &  $\msf{e}_1+\msf{f}_1+\msf{f}_2+2\msf{g}$ \\
$[0]+[13]+[26]+[45]$ &  $\msf{f}_2$ \\
$[0]+[14]+[23]+[56]$ &  $\msf{e}_2+\msf{f}_2+2\msf{g}$ \\
$[0]+[14]+[25]+[36]$ &  $\msf{f}_1+\msf{e}_2+\msf{f}_2+2\msf{g}$ \\
$[0]+[14]+[26]+[35]$ &  $\msf{f}_1$ \\
$[0]+[15]+[23]+[46]$ &  $\msf{e}_1+\msf{e}_2+\msf{f}_2+2\msf{g}$ \\
$[0]+[15]+[24]+[36]$ &  $\msf{e}_1+\msf{f}_1+\msf{e}_2+2\msf{g}$ \\
$[0]+[15]+[26]+[34]$ &  $\msf{f}_1+\msf{f}_2$ \\
$[0]+[16]+[23]+[45]$ &  $\msf{e}_1$ \\
$[0]+[16]+[24]+[35]$ &  $\msf{e}_2$ \\
$[0]+[16]+[25]+[34]$ &  $\msf{e}_1+\msf{e}_2$ \\ \hline
\end{tabular}
\end{center}
Since all these are distinct each other, we see that
the $15$ Hutchinson involutions are not conjugate each other in $\mathop{\mathrm{Aut}}\nolimits (X)$
if $X$ is Picard-general.

{\noindent{\textbf{Remark}:}} In \cite{mukai} it is shown that if $(C,G)$ is bielliptic, then
the involution $\sigma_G$ cannot be defined.

\section{Hutchinson's involutions associated with Weber hexads}\label{reye}

These automorphisms appear in \cite{hut1,hut2}.
The freeness is found in \cite{dolgachev-keum}.
We fix a Weber hexad $W$.
Then the linear system $L=|\mca{O}_{\overline{X}}(2)-W|$ with the assigned base
points at $W$ defines an another quartic model $\overline{X}_W$ of $X$ in $\bb{P}^4$,
\[\overline{X}_W:\ s_1+\cdots+s_5=0,\ \lambda_1/s_1+\cdots+\lambda_5/s_5=0,\]
where $\lambda_i$ are nonzero constants.

$\sigma_W$ is the Cremona involution
\[\sigma_W: (s_1,\cdots, s_5)\mapsto (\lambda_1/s_1,\cdots, \lambda_5/s_5),\]
preserving $\overline{X}_W$. It is free if $X$ is Picard-general \cite{dolgachev-keum}.
For any translation $t=t_{\alpha}$, we have $\sigma_{t(W)}=t \sigma_W t$
as in the Hutchinson case. Hence we can assume that the Weber hexad
doesn't contain $n_{0}$. Then recalling (\ref{web}) in Section \ref{166},
we have only one Weber hexad
$W_0=\{n_{12},n_{23},n_{31},n_{14},n_{25},n_{36}\}$ up to the action of $\mf{S}_6$.
In the following we discuss this case.

\begin{lem}[\cite{dolgachev-keum}]\label{dk}
$\sigma_{W_0}$ interchanges the following $10$ pairs of smooth rational curves.
\begin{eqnarray*}
(N_{0},T_{123}),(N_{56},T_{1}),(N_{46},T_{2}),(N_{45},T_{3}),(N_{15},T_{124}),\\
(N_{16},T_{134}),(N_{24},T_{125}),(N_{26},T_{146}),(N_{34},T_{136}),(N_{35},T_{236}).
\end{eqnarray*}
\end{lem}

\begin{prop}\label{neg}
The $(-1)$-eigenspace $K$ of $\sigma_{W_0}$ is generated over $\bb{Z}$ by the following elements.
\begin{gather*}
e_1=T_2-N_{46},\ e_2=N_{15}-T_{124},\ e_3=T_1-N_{56},\\
e_4=N_0-T_{123},\ e_5=T_3-N_{45},\ e_6=N_{34}-T_{136},\\
e_7=N_{23}-N_{56}-N_{34}-N_{24}-T_{134}-T_{124}.
\end{gather*}
\end{prop}
\begin{proof}
By computing the determinant, we can see that $10$ divisors
\begin{equation}
N_0+T_{123},\cdots,N_{35}+T_{236}\label{inv}
\end{equation}
from Lemma \ref{dk} span over $\bb{Q}$ the invariant sublattice.
$e_1,\cdots,e_6\in K$ is easy. $e_7\in K$ follows from the fact that $e_7$ is orthogonal
to all of the divisors in (\ref{inv}).
Then $e_1,\cdots,e_7$ spans the lattice $E_7(2)\simeq K$.
\end{proof}
{\noindent{\textbf{Remark}:}} The action of $\sigma_{W}$ on $NS(X)$ is very complicated, but essentially
we can write down this action using the proposition above. In fact we find the
following.

Let $W$ be a general Weber hexad. The ``degree $1$ part'' $W_1$ of $W$ is
the set
\[\{\beta\in S(C)| (T_{\beta},\sum_{\alpha\in W}N_{\alpha})=1\}.\]
$W_1$ consists of $6$ elements. We have a natural bijection
$\mu :W\rightarrow W_1$ defined by $(N_{\alpha},T_{\mu (\alpha)})=1$.
On the other hand, for $\alpha\not\in W$, we have the unique decomposition
\[W=G\circleddash R,\,\, G\cap R=\{n_{\alpha}\}.\]
Let $R^{\perp}$ be the $2$-dimensional affine subspace of $J(C)_2$ which is orthogonal
to $R$ and contains $n_{\alpha}$.
Then $R\circleddash R^{\perp}$ is a Rosenhain hexad, i.e., $R\circleddash R^{\perp}$
is of the form $\Lambda (\beta)$ for some $\beta\in S(C)$.
This defines a bijection $\mu': J(C)_2-W\rightarrow S(C)-W_1$, $\alpha\mapsto
\mu'(\alpha)=\beta$. Using these data, the action of $\sigma_W$
is as follows.
\begin{eqnarray*}
\sigma_W (N_{\alpha})= 3H- (\sum_{\alpha\in J(C)_2}N_{\alpha})/2-(\sum_{\alpha\in W}N_{\alpha})
-T_{\mu(\alpha)}, \text{ if } \alpha\in W.\\
\sigma_W (N_{\alpha})=T_{\mu'(\alpha)} \text{ if } \alpha\not\in W.\\
\sigma_W (H)=9H -(\sum_{\alpha\in J(C)_2}N_{\alpha})-4(\sum_{\alpha\in W}N_{\alpha}).
\end{eqnarray*}

\begin{prop}\label{result3}
The patching subgroup of $\sigma_{W_0}$ is cyclic and generated by
\[x=(3/4)H+(1/2)(N_{12}+N_{23}+N_{31}+N_{14}+N_{25}+N_{36}).\]
\end{prop}
\begin{proof}
The corresponding element in $K^*/K$ is
\[y=\frac{1}{4}e_1+\frac{1}{2}e_2+\frac{1}{2}e_4+\frac{3}{4}e_5+\frac{1}{4}e_7,\]
and we check $x-y\in NS(X)$.
\end{proof}
By the $\mf{S}_6$-symmetry and the translation relation, we obtain
\begin{prop}
For general $W$, the patching subgroup of $\sigma_W$ is
\[\Gamma_{\sigma_W}=\bracket{(3/4)H+ (\sum_{\alpha\in W}N_{\alpha})/2}.\]
\end{prop}

There are $12$ Weber hexads modulo translations.
One more relation is hidden in the remark above. For $\alpha\not\in W$,
we have the unique decomposition $W=G\circleddash R,\,\, G\cap R=\{n_{\alpha}\}$.
Let $R^{\perp}$ be the orthogonal complement of $R$ at $n_{\alpha}$ and let
$W_{\alpha}^{\perp}$ be the Weber hexad $G\circleddash R^{\perp}$.
Then $\sigma_W$ and $\sigma_{W_{\alpha}^{\perp}}$ are conjugate,
related by $\sigma_{W_{\alpha}^{\perp}}= \sigma_{\mu'(\alpha)}\sigma_W\sigma_{\mu'(\alpha)}$.
Modulo this relation,
we have $6$ Weber hexads. Under the notations of Lemmas \ref{nsstr1} and \ref{nsstr2},
their patching subgroups are as follows.
\begin{center}
\begin{tabular}{ll} \hline
Weber hexad  & patchings \\ \hline
$[12]+[23]+[31]+[14]+[25]+[36]$ &  $\msf{e}_1+\msf{f}_1+\msf{e}_2+\msf{g}$ \\
$[12]+[13]+[23]+[24]+[15]+[36]$ &  $\msf{f}_1+\msf{e}_2+\msf{f}_2+\msf{g}$ \\
$[23]+[13]+[12]+[34]+[25]+[16]$ &  $\msf{e}_2+\msf{f}_2+\msf{g}$ \\
$[24]+[23]+[34]+[14]+[25]+[36]$ &  $\msf{e}_1+\msf{f}_1+\msf{f}_2+\msf{g}$ \\
$[25]+[23]+[35]+[54]+[21]+[36]$ &  $\msf{e}_1+\msf{e}_2+\msf{f}_2+\msf{g}$ \\
$[26]+[23]+[36]+[64]+[25]+[13]$ &  $\msf{e}_1+\msf{f}_1+\msf{g}$ \\
\end{tabular}
\end{center}
Thus we see that
there are $6$ HW involutions
that are not conjugate each other in $\mathop{\mathrm{Aut}}\nolimits (X)$
if $X$ is Picard-general.

{\noindent{\textbf{Remark}:}} (1) In a forthcoming paper we will be able to determine when $\sigma_W$ is not free.\\
(2) The $6$ conjugacy classes of HW involutions are naturally ``dual''
to the $6$ Weierstrass points, in the sense that the $\mf{S}_6$ action on both
factors through an outer automorphism. Details are as follows.
There are $20$ Weber hexads $W$ which don't contain $n_0$ and
conjugate each other.
Writing $W$ uniquely as $W=G\circleddash R$ with $G\cap R=\{n_0\}$,
we can associate with such $W$ the G\"{o}pel subgroup $G$.
But a G\"{o}pel subgroup $G=\{n_0,n_{ij},n_{kl},n_{mn}\}$
is determined just by
the ``syntheme'' $(ij)(kl)(mn)\in \mf{S}_6$. Thus we obtain $20$ synthemes from the
conjugacy class. The fact is that there appear only $10$ synthemes, and
the synthemes not appearing here form a ``total'', which is
the classical description of the dual of the $6$-set $\{1,\cdots,6\}$.\\
(3) The method of this paper is applicable to the case of
Picard-general quartic Hessian surfaces
of \cite{dolgachev-keum}. In this case we have exactly one Enriques quotient.

\quad \\
{\textbf{Proof of Theorem \ref{thm2}:}
Let $N'$ be the group generated by $16$ translations $t_{\alpha}$, $16$ switches $\sigma_{\beta}$,
$16$ projections $p_{\alpha}$, $16$ correlations $p_{\beta}$, $60$ HG involutions $\sigma_G$,
$192$ HW involutions $\sigma_W$. The theorem follows from the following lemma
as in \cite[Lemma 7.3]{kondo98}.
\begin{lem}
Let $\phi$ be an isometry of $NS(X)$ that preserves the ample cone.
Then there exists a $g\in N'$ such that $g\phi\in \mathop{\mathrm{Aut}}\nolimits (D')$.
\end{lem}
\begin{proof}
Let $w'=2H-\sum N_{\alpha}/2$ be the projection of the Weyl vector $w$.
Let $v=\phi (w')$ and
let $g\in N'$ be an element that attains the minimum $min\{(g(v), w')|g\in N'\}$.
If $r$ is the Leech root corresponding to \cite[Lemma 4.6, Case (0),(1),(2)]{kondo98},
then as in \cite{kondo98} we have $(r', g(v))>0$.

If $r$ is the Leech root corresponding to \cite[Lemma 4.6, Case (3)]{kondo98},
then it corresponds to some Weber hexad $W$ and \cite[Remark 6.3, (1)]{kondo98}
can be rewritten as
\[4r'=3H-2\sum_{\alpha\in W}N_{\alpha}.\]
Using Proposition \ref{neg} and its Remark, we have
\[\sigma_W (w')=w'+8r'.\]
Thus, we have
\begin{eqnarray*}
(g(v), w')\le (g(v), \sigma_W^{-1} (w'))= (g(v),w')+8(g(v),r'), \\
(g(v),r')>0
\end{eqnarray*}
Hence $g(v)\in D'$.
\end{proof}

{\noindent{\textbf{Remark}:}} Unfortunately, $\mathop{\mathrm{Aut}}\nolimits (X)$ cannot be generated only by the subset
\[S=\{t_{\alpha},\sigma_{\beta},\sigma_{G},\sigma_W|\alpha\in J(C)_2,\beta\in S(C),
G:\text{G\"{o}pel tetrad},W:\mathrm{Weber\ hexad}\},\]
introduced in this paper.
It is easy to see that for any element $\varphi$ written as a product of elements
in $S$, we have $(w',\varphi(w'))\in 4\bb{Z}$. But the projection $p_{\alpha}$
have $(w',p_{\alpha}(w'))=26.$

\medskip
{\noindent{\textbf{Acknowledgement}:}} The author expresses his sincere gratitude to Professor Shigeru Mukai.
He suggested using Torelli theorem for Enriques surfaces
in proving Proposition \ref{correspondence},
which was a better method of counting than that of \cite{ohashi} and
an important step for the computation in this paper.
He explained his study of \cite{mukai} in process and led the author to his conjecture.

Financial support has been provided by the Research Fellowships of the
Japan Society for the Promotion of Science for Young Scientists.

redpajama_set_name: RedPajamaArXiv


text:
| Car Profiles
2003 Mach 1 Mustang - Top Of The Rock
Adam Sampson Runs A Different Type Of Race With His '03 Mach 1
Frank H. Cicerale
Steve Lewis was the other competitor in Adam's class, ironically in another Mustang, that being his black '04 Mach 1. Steve's car showcases a host of Steeda and Maximum Motorsports suspension components, along with an MRT catted H pipe and an SLP Loudmouth exhaust. Steve's Mach 1 is his daily driver, and he has also competed in the Silver State Classic with the SN-95.
Whenever family members or friends come into the New York metro area, there are certain things you tell them to go see and/or do. Whether it is eating "real" pizza, taking a look at the Statue of Liberty, or catching a Broadway show, there are certain landmarks, historical and food finds, that are must-sees when it comes to the New York metropolitan area.
While getting to the top of the Empire State Building is a definite for anyone who is visiting the Big Apple, for the first or the umpteenth time, there is a grand old structure with an address of 30 Rockefeller Center that is just as nice. That structure is the 70-story Rockefeller Building with an observation deck at the top called the Top of the Rock.
For Adam Sampson, knowing the sights and sounds of the New York metro area is a way of life. For the Keansburg, New Jersey, native, though, he visits a different rock, each time he ventures to nearby Pennsylvania to compete in hill climb events with his '03 Mach 1 Mustang.
Yes, you read right-we said hillclimb. While an image of a Mustang with Super Swamper tires may come to mind, by hillclimb we mean Adam's participation in the events put on by the Pennsylvania Hillclimb Association (PHA). The PHA was founded in 1963, and holds events where drivers show fast they can climb a paved road on an uphill grade, usually over the course of one mile or so.
The PHA is a part of the Sports Car Club of America, better known by its acronym, the SCCA. To participate, a willing contestant must join both the PHA and the SCCA. To do so, you need a valid state driver's license, the SCCA membership, and a willingness to earn an SCCA time trials Novice license (either before, or at, your first event). After that, you pick up the required safety gear, such as a helmet, firesuit, Nomex socks, fire retardant shoes, and gloves, pick the class in which your car will fit-of which there are 11-and cruise on out to the next event. It's that simple.
When we chatted with Adam at a recent tech install at Crazy Horse Racing, he told us all about the hillclimb racing he does with his Mach 1. We were eager to tag along for an event and see not only how big of a following hillclimb racing has, but also check out what goes into preparing a car for this type of racing. We also wanted to know why someone who lives in an area rich in drag racing history would choose to turn left and right up a hill with trees and other foliage a mere foot or two from either side of his precious Pony. With cameras in hand, we met with Adam at the last PHA event of '08, the Weatherly Hill Climb.
"I have been into Mustangs ever since I was 14 years old," the 31-year-old explains about the beginning of this interesting journey. "It started when me and a couple of my friends were on our bikes hanging out and we checked out a friend's older brother's Mustang. We all swore that when we got our driver's licenses, we would own one. Some of us bought one, some of us didn't. I picked up a black '85 GT with a five-speed at 18 while I was in the service. The thing was, I bought it without knowing how to drive a stick!"
Being from central New Jersey, it was a given that Adam would spend his spare time on Wednesday and Friday nights banging gears with the rest of the Ford faithful and Mustang maniacs at Old Bridge Township Raceway Park in Englishtown, New Jersey. Eventually, though, Adam tired of the quarter-mile strip, and started searching for a different venue in which to fill his need for speed. "Years later, I picked up the Mach, and I realized I needed to stop drag racing when I tried making the turn for the return road at Englishtown faster and faster," Adam says. "Drag racing was getting boring very quickly due to the amount of cars that would go. You'd make a pass and then sit for hours on end."
Chris Winter of Crazy Horse Racing was doing the wrenching on Adam's '03 Mustang Mach 1, and while at the shop one day, the duo had an in-depth conversation regarding another form of racing. "After talking with Chris, my fianc Jenna and I attended a PDA/HPDE event at Pocono Raceway. I went along with him for a five-lap ride, and right then and there I said that was it, no more drag racing for me."
So, with his sights set on doing the left and right thing, Adam decided to find a style and class before he started modifying the Mach 1 for track duty. "I spent a lot of time searching the Internet looking for exactly what I wanted to do when I came across a video on the Mach 1 Registry website of Steve Lewis and his '04 Mach 1 doing the hillclimb," Adam explains. "I saw it and thought it looked like something I would like to do. I often hit the roads in northern New Jersey and upstate New York for some spirited driving, and this looked like a legal way to go even faster. Jenna and I drove to the Fall '07 Weatherly race, met Steve and his wife, Nancy, and that day set a goal of making the '08 season opener."
With that goal in mind, Adam picked his class, and enlisted the help of Winter to get the car ready for the upcoming season. According to the rules, Adam's Mach 1 would fit into the E/Street Prepared category (E/SP). Ironically, this is the same class that Adam's new friend, Steve Lewis, ran his black Mach 1 in. The Street Prepared classes follow SCCA's Solo 2 rules, and allow for most suspension and drivetrain related bolt-on parts. No major modifications are allowed, though, meaning a supercharger, turbocharger, or porting and polishing of the heads and intake are a big no-no. Additionally, the cars have to be street appearing, meaning the head and taillights must work, as well as the interior needing to be retained, minus the addition of the appropriate safety equipment. With the rulebook in his hand, Adam set about performing the upgrades to get his Mach 1 (which is still his daily driver) ready for competition.
Adam didn't want to have his car down for an extended period of time with a complete overhaul of the naturally aspirated Four-Valve, so he went the bolt-on route with the Azure Blue New Edge. The motor retains its stock bore and stroke, as well as the rotating assembly, cams, cylinder heads, and induction system, with the still-cool Shaker hoodscoop. Improving the induction side of things is a C&L intake elbow, a K&N filter, and a 3/8-inch intake spacer. A switch over to NGK plugs brings a bigger jolt to the party in the combustion chamber, and exhausting the burnt fossil fuel is the stock exhaust manifolds and mufflers that now dump into an IRS-style FR500 cat-back system. Throw in a custom tune from Winter uploaded into the ECM via an SCT tuner, and the Mach 1 pumps out 305 hp and 327 ft-lb of torque at the rear tires.
Hiding underneath the car and lying in wait behind the engine is the stock Tremec five-speed transmission. Inside of the bellhousing resides the stock clutch, though an aluminum Spec flywheel was installed to allow the motor to rev quicker. Adam makes each gear change via an MGW shifter.
It's an ironic twist of fate when it comes to the Mach 1's rearend choice. While most of the Terminator Cobra contingent leans toward swapping out the Snake's IRS rear setup for the Mach 1 or Mustang GT's solid-rear configuration, Adam went the opposite route. "The rules allow for the addition of the production IRS setup found on the Cobra, and Winter agreed that going to the IRS would be extremely beneficial when it came to handling in the corners," Adam explains. "I found a guy on one of the forums with a Cobra looking for a solid rear setup, and I got in touch with him. He agreed to swap his IRS for my solid rear."
One of the coolest parts of the Weatherly hillclimb is a spot the spectators and racers affectionately call the "Jump". This part of the course comes after a hard righthander, and is the spot where you can get some major air. Adam routinely pulled the front wheels of his '03 Mach 1 off the ground at this spot in the course.
Winter performed the swap in the South Amboy, New Jersey, shop, gutting the Mach of all of the stock rear components and replacing them with the IRS parts and pieces. The hind end of Adam's Pony now showcases a Cobra IRS. The stock springs were also bolted in, but gone were the stock shocks, as they were replaced with a set of Bilsteins. Crazy Horse full-length sub-frame connectors with a seat support coincide with an x brace off of an '03 Mustang GT convertible to stiffen up the frame, while Maximum Motorsports IRS differential bushings were installed for longevity. As for the front, the stock springs and Tokico struts handle the weight of the Four-Valve, and work in tandem with Maximum Motorsports caster/camber plates, rack bushings, a solid steering shaft, and urethane front control arm bushings, sway bar bushings, and sway bar endlinks to take corners with authority. The stock 17x8 inch Mach 1 wheels still reside on all four corners, though the footprint of the car is enhanced with the rims being shod in Nitto 555 shoes. With handling comes braking, and to that end, Adam had a set of cross-drilled and slotted rotors put on, sized 13 inches up front and 11.7 inches out back, along with Hawk HPS pads and a Russell braided brake line.
With the drivetrain now up to snuff, Adam moved into the interior of the car to make it legal for competition. A Maximum Motorsports four-point rollbar was installed, and the bar, complete with the diagonal option and a 2.5 pound fire extinguisher, was given the stamp of approval by the SCCA tech inspection. The stock seats remain, though both Adam and passenger are fastened securely with G-Force five-point harnesses. A Speed of Sound pillar gauge pod containing Ford Racing water temp and oil pressure gauges was installed next, along with a Raptor LED shift light that illuminates brightly when the time comes to make the gear change. Before Adam fires the Mach 1 up for a high-speed blast up the hill, he dons a G-Force two-piece firesuit, gloves, and shoes, and then straps on the G-Force full-face skid lid.
When all was said and done, Adam's car was ready for the opening race of the season at Weatherly. While he didn't make all of the races during the season, he did make the final event, also at Weatherly, where we caught up with him. According to Adam and the rest of the drivers, the Weatherly course, which is one mile in length, features six turns and an elevation change of 340 feet, making it one of the more technical ones on the circuit. Two of the more interesting turns on the Weatherly course is a hairpin turn called "the Wall", which has a bank to it that is reminiscent of one seen at a NASCAR track. A little further up the course is "the Jump," which follows a hard righthand turn. This is where a lot of spectators like to go, as the cars will literally fly over the crest of the hill, pulling two, or in some cases all four, tires off of the ground.
While rain hampered the first couple of runs Adam made, the sun came out, the road dried, and he was able to put the hammer down. While only he and Steve Lewis' Mach 1 were running in the E/SP class, Adam was able to knock time off on each successive run up the hill, resulting in a best lap of 69.202 seconds, placing him Second in class. According to Adam, the quicker, more specialized cars can make it up the hill in a minute or less, so his 69-second lap in a street-driven Mach 1 Mustang with bolt-on parts is pretty darn impressive.
If you are looking for an interesting form of racing, then check out the PHA hillclimb events. Information for the PHA can be found online at www.pahillclimb.org, and don't forget to check out the Muscle Mustangs website, as we strapped our video camera into Adam's car for a couple of blasts up the hill.
Who knows-maybe one day you and your Mustang can get to the top of the rock. According to Adam Sampson, it sure is fun trying!

redpajama_set_name: RedPajamaCommonCrawl


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  public void read(org.apache.thrift.protocol.TProtocol iprot) throws org.apache.thrift.TException {
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  public void write(org.apache.thrift.protocol.TProtocol oprot) throws org.apache.thrift.TException {
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  private static class RDF_DataTupleTupleScheme extends org.apache.thrift.scheme.TupleScheme<RDF_DataTuple> {

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redpajama_set_name: RedPajamaGithub


Incentive Stock Options vs. Nonqualified Stock Options. but spread is taxable for alternative minimum tax purposes ("AMT"). Exercise NOT subject to employment tax withholding. 7 Responses to "Incentive Stock Options vs. Nonqualified Stock Options" By Bo Sartain May 16, 2013 - 7:03 am.
The exercise of non-qualified stock options is not an AMT preference item or an AMT adjustment so you didn't "have to pay AMT on the NSO(s)". The ONLY way that could have happened is if you entered the exercise in the "ISO Exercise and Hold" interview, in which case you need to go back and amend 2016.
Non qualified stock options. Non qualified stock options (NQ or NQSO) are one of three types of stock options commonly received by tech employees in San Francisco. Incentive stock options (ISO) and restricted stock units (RSU) are the others.
© Are non qualified stock options subject to amt Binary Option | Are non qualified stock options subject to amt Best binary options.


text:
As a publisher we have a critical role to play in ensuring that all work we publish is not only of a high standard, but that it is above reproach in terms of integrity. We therefore hold ourselves, our editors, and our authors to very high standards. This ranges from anti-plagiarism checking and thorough peer review, to ensuring that all data behind the research is also made available. Above all, we believe that open access is the best possible way to ensure that work is fully open to professional and public scrutiny.
Anti-plagiarism Checking
A combination of pre-screening and open access is the best possible defense against plagiarism. All articles submitted to Elementa are automatically screened for plagiarism by the CrossCheck system from CrossRef. This system compares incoming articles to a large database of academic content, and alerts editors to any possible issues.
Rigorous Peer Review
The Elementa Editors ensure that all research output is thoroughly peer reviewed by external reviewers, and the journal offers the option of open peer review if requested. Publications of a commentary or opinion nature may not be sent for external peer review but will include extensive editorial review and revisions. Via the guidelines at the Committee on Publication Ethics (COPE), our editors and editorial board receive ethical guidance and professional networking. Elementa adheres to the COPE guidelines for best practice.
Open Licenses
All Elementa articles are published under open licenses from Creative Commons. We believe that only CC BY meets the requirements for true open access for journals, and strongly prefer CC0 for open research data.
Elementa also endorses the Panton Principles, which state that "for science to effectively function, and for society to reap the full benefits from scientific endeavors, it is crucial that science data be made open."
The Elementa Editors require authors to make the research objects associated with their publications openly available. This includes research data, software, bioresources and methodologies. This means that peer reviewers are able to better assess the foundations of claims made, and the research community and wider public are able to similarly validate authors' work, and are more easily able to extend and build upon it. Please see our Author Guidelines for further details.
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All our items are graded to make it easier for you to understand their condition before you buy. Any difference to this will be noted above.
This item has been classed as: Grade B.
Grade A - Item is in used condition, will show minimal signs of use. A small scratch or minor blemish can be expected at most.
Grade B - Normal used condition, will show signs of use, some scratches, blemishes, signs of use to be expected.
Grade C - Clearance product, item will show significant levels of use or have a fault as stated in the notes, often for spare or repair.
All required power cables are provided to operate the item fully. Any difference to this will be noted above.
- The warranty covers any hardware fault found with your item.
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Apple Macbook Pro A1398 Retina 15.4"


Travel west on 6th Avenue to the Indiana Street / Denver West / Colorado Mills exit. Turn left (south) onto Indiana Street and travel approximately one block. Turn right (west) onto the 6th Avenue Frontage Road and travel approximately two blocks. The main entrance to the Fairgrounds will be on your left. Follow the signage posted at the main entrance to your desired destination.


text:
Q: Is IBM Cloud available in China? I plan to use IBM Cloud to deploy an application from Shanghai.
Is there any limitation for IBM Cloud in China? Can anyone confirm whether the platform is available without a VPN?

A: IBM Cloud will totally work in China and it will work without VPN as well. The only thing probably need to noted is the closest IAAS data center is located in Japan. hope this help

redpajama_set_name: RedPajamaStackExchange


🎉 congrats to all who had their #kubecon #cloudnativecon talks accepted and thank you to everyone who submitted a talk... ~13% acceptance rate!
The Istio project reached 1.0 this summer, and is mature enough to have LTS releases. It's getting a lot of attention, but in a lot of ways it's still a mystery. You've probably read about it, you might have tried it, but do you really understand it? It promises advances routing, security, and resiliency, all for free! In this session I'll present a practical introduction to the operation of Istio - what features it can bring to your environment.
What's unique about this talk is that we'll be exploring the different parts of Istio by following one plucky little packet into the mesh, through it, and out again. As we meet each component we'll learn why it's there, what it does, and see a demo of how to configure it for common tasks. This will leave you not only with slides showing example configs, but a valuable mental model and a unique insight into the service mesh's operation.
What do you get when you start with a microservice, then add libraries for #ServiceMesh functionality like security, observability, and reliability?
Engineers want to understand how something works, not a high level overview of what it can do. That's why @mt165's talk was so well received.
Thanks for the awesome talk. It's explained in a really good manner how everything works in istio and also why the components needed at a certain point in a very clear understandable way.


text:
U.S. National ArchivesPublic Domain ArchiveNot developed or endorsed by NARA.
Aircraft and Airports - California
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Photographs Relating to National Forests, Resource Management Practices, Personnel, and Cultural and Economic History
national forest service
small airport
resource management practices
The U.S. National Archives
https://catalog.archives.gov/
Explore: airports
Aircraft and Airports - Washington
Aircraft and Airports - Montana
Aircraft and Airports - Idaho
Aircraft and Airports - Minnesota
STS062-90-007 - STS-062 - Earth observations from STS-62 flight
Explore: vintage aircraft
Explore: small airport
U.S. National Archives
The objects in this collection are from The U.S. National Archives. The U.S. National Archives and Records Administration (NARA) was established in 1934 by President Franklin Roosevelt. NARA keeps those Federal records that are judged to have continuing value—about 2 to 5 percent of those generated in any given year. There are approximately 10 billion pages of textual records; 12 million maps, charts, and architectural and engineering drawings; 25 million still photographs and graphics; 24 million aerial photographs; 300,000 reels of motion picture film; 400,000 video and sound recordings; and 133 terabytes of electronic data. All of these materials are preserved because they are important to the workings of Government, have long-term research worth, or provide information of value to citizens.
Disclaimer: A work of the U.S. National Archives is "a work prepared by an officer or employee" of the federal government "as part of that person's official duties." In general, under section 105 of the Copyright Act, such works are not entitled to domestic copyright protection under U.S. law and are therefore in the public domain. This website is developed as a part of the world's largest public domain archive, PICRYL.com, and not developed or endorsed by the U.S. National Archives. https://www.picryl.com


text:
AG charges midwife with practicing medicine without a license
Updated: 3:20 PM CDT Sep 13, 2019
MAN THEY'RE LOOKING FOR IF YOU HIM MAKE SURE TO CALL DEPUTIES A WOMAN FACING CHARGES TONIGHT ACCUSED OF PRACTICING MEDICINE WITHOUT A LICENSE HERE IN OKLAHOMA INVESTIGATORS FOUND OUT THE WOMAN A MIDWIFE PERFORMED A MEDICAL PROCEDURE RECENTLY DURING A HOME BIRTH. SHE'S BEEN PERMANENTLY BANNED FROM GETTING A LICENSE IN ARKANSAS NOW IN OKLAHOMA POLICE PEOPLE DO NOT HAVE TO HAVE A LICENSE TO BE A MIDWIFE, BUT YOU DO HAVE TO HA
Oklahoma Attorney General Mike Hunter charged a Roland midwife with practicing medicine without a license after she allegedly performed a medical procedure during a recent home birth, according to a news release from the attorney general's office. The woman was in labor for nearly three days, according to the release.The charges allege that Debra Disch performed an episiotomy and administered the drug Pitocin in May to slow the woman's bleeding. An episiotomy is a surgical procedure that is performed to facilitate childbirth. Pitocin is a drug that induces labor. A search warrant of Disch's residence found five vials of the drug, according to the release. Oklahoma doesn't require midwives to be licensed, and they aren't regulated under state law. However, only physicians may perform an episiotomy, and Disch didn't have a prescription for Pitocin. Hunter said Disch was irresponsible and put the mother and baby in danger. "The details of this case are disturbing," Hunter said in the release. "Our evidence shows that Disch was reckless in the way she performed this procedure and she was entirely outside the scope of her abilities and the law. The mother and her baby are lucky to be alive."According to court documents, the woman's family repeatedly urged Disch to call an ambulance, and Disch refused. The baby had to be resuscitated, according to court documents, and Disch allegedly dropped the baby while attempting to perform the lifesaving procedure. After the birth, the woman began bleeding, which is when Disch administered Pitocin, which can also be used to control bleeding, according to the court filing. In 2016, Arkansas permanently banned Disch from being licensed as a midwife , according to the release.
Oklahoma Attorney General Mike Hunter charged a Roland midwife with practicing medicine without a license after she allegedly performed a medical procedure during a recent home birth, according to a news release from the attorney general's office. The woman was in labor for nearly three days, according to the release.
The charges allege that Debra Disch performed an episiotomy and administered the drug Pitocin in May to slow the woman's bleeding. An episiotomy is a surgical procedure that is performed to facilitate childbirth. Pitocin is a drug that induces labor.
A search warrant of Disch's residence found five vials of the drug, according to the release.
Oklahoma doesn't require midwives to be licensed, and they aren't regulated under state law. However, only physicians may perform an episiotomy, and Disch didn't have a prescription for Pitocin.
Hunter said Disch was irresponsible and put the mother and baby in danger.
"The details of this case are disturbing," Hunter said in the release. "Our evidence shows that Disch was reckless in the way she performed this procedure and she was entirely outside the scope of her abilities and the law. The mother and her baby are lucky to be alive."
According to court documents, the woman's family repeatedly urged Disch to call an ambulance, and Disch refused. The baby had to be resuscitated, according to court documents, and Disch allegedly dropped the baby while attempting to perform the lifesaving procedure. After the birth, the woman began bleeding, which is when Disch administered Pitocin, which can also be used to control bleeding, according to the court filing.
In 2016, Arkansas permanently banned Disch from being licensed as a midwife , according to the release.


text:
Communicating in Italy
Italian is fun to listen to, but even more fun to speak — let go of your inhibitions and play along.
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Download Rick's Headshot — Color | Black and White
Watching the way Italians interact — like this couple on a bridge — provides a great insight into their culture. (photo: Rick Steves)
Understanding hand gestures is key to communicating in Italy. Here an Italian chef uses extra flourishes to proudly describe his menu. (photo: Dominic Bonuccelli)
I am terrible at foreign languages. Despite having traveled to Europe four months a year since I was a kid, I can barely put a sentence together anywhere east or south of England. But with some creative communication, I manage just fine to write guidebooks, produce TV shows, and simply enjoy Europe on vacation. And nowhere do I have more fun communicating than in Italy.
Because Italians are so outgoing and their language is such fun, interactions are a pleasure. Italians have an endearing habit of speaking Italian to foreigners, even when they know we don't speak their language. If a local starts chattering at you in Italian, don't resist. Go with it. You may find you understand more than you'd expect. Italians want to connect and, by my measure, they try harder than any other Europeans. Play along.
I find Italian beautiful, almost melodic. It's fun to listen to and even more fun to speak. It has a lovely rhythm and flow, from buon giorno to buona sera to buona notte (good day / good evening / good night), and especially words like bellissima (very beautiful). Two of my favorite phrases — and sentiments — in any language are Italian's la dolce vita (the sweet life) and il dolce far niente (the sweetness of doing nothing).
Italians are animated and dramatic, communicating as much with their bodies as with their mouths. You may think two people are arguing, when in reality, they're agreeing enthusiastically. When I'm in Italy, I make it a point to be just as melodramatic and exuberant. Don't just say, "Mamma mia." Say, "MAMMA MIA!" with arms open wide and hands up in the air. It feels liberating to be so uninhibited. Self-consciousness kills communication.
In Italy, hand gestures can say as much as words. For instance, the cheek screw (pressing a forefinger into the cheek and rotating it) is used to mean "cute" or "delicious." A chin flick with the fingers means, "I'm not interested; you bore me." The hand purse (fingers and thumb bunched together and pointed upward) is a gesture for a question, such as, "What do you want?" or, "What are you doing?" It can also be used as an insult to say, "You fool."
The Italian version of the rude middle finger is to clench the right fist and jerk the forearm up, slapping the bicep with the left hand. This jumbo version of "flipping the bird" says, "I'm superior." If Italians get frustrated, they might say, "Mi sono cadute le braccia!" (I throw my arms down!) — sometimes even thrusting their arms toward the floor in an "I give up!" gesture.
Italians appreciate sensuality, which can be heard in their language. Rather than differentiating among the five senses to describe what they're hearing, smelling, or tasting, Italians talk about sensing (sentire):
"Did you sense the ambience as you walked by?"
"Wow, sense this wine."
"Oh, sense these flowers."
Instead of asking, "Are you listening?" an Italian will ask, "Do you sense me?"
One of the best ways to observe Italians communicating — and to communicate with them — is to participate in the passeggiata. This ritual promenade takes place in the early evenings, when shoppers, families, and young flirts on the prowl all join the scene to stroll arm in arm, spreading their colorful feathers like peacocks. In a genteel small town, the passeggiata comes with sweet whispers of "bella" (pretty) and "bello" (handsome). In Rome, the passeggiata is a cruder, big-city version called the struscio (grind). Younger participants utter the words "buona" and "buono" — meaning, roughly, "tasty." As my Italian friend explained, "'Bella' is a woman you admire — without touching. 'Buona' is something you want, something…consumable. 'Bella' is too kind for this struscio."
To really immerse yourself in the culture, it's important to take risks in conversation. Italians appreciate your attempts. Miscommunication can happen on both sides, but it's part of the fun.
Once I was eating at a restaurant in Assisi with a guide named Giuseppe and his wife, Anna. Anna greeted each plate with unbridled enthusiasm. Suddenly, Giuseppe looked at me and said in English, "My wife's a good fork."
Shocked, I thought I must have misheard him.
Giuseppe explained, "Una buona forchetta — a good fork. That's what we call someone who loves to eat."
Rick Steves (www.ricksteves.com) writes European guidebooks, hosts travel shows on public TV and radio, and organizes European tours. You can email Rick at rick@ricksteves.com and follow his blog on Facebook.
Heart of Italy Tour
Italy Guidebook