text:
\section{Introduction}
If $C$ is a curve of genus two, equipped with an \emph{optimal} covering $\mapu\colon C\to E$ of degree~$n$ of an elliptic curve then there exists an elliptic curve~$E'$ such that the Jacobian~$\jac(C)$ is isogenous to $E\times E'$ via an isogeny of degree $n^2$. Such a Jacobian is said to be $(n,n)$-split. The classical treatment of cases $n\leq 4$, in terms of elliptic integrals, can be found in the works of Legendre, Jacobi, Goursat, and others; e.g. see \cite{jacobi}, \cite{goursat}, \cite{brioschi}, \cite{bolza}. A modern treatment can be found in e.g.~\cite{kuhn},
\cite{frey},
\cite{freykani},
\cite{shaska2},
\cite{shaska},
\cite{howe},
\cite{bruin},
\cite{shaska3},
\cite{kumar}, dealing with various cases with $n\leq 11$.
The problem of finding the curve $E'$, given the map $\mapu$, was considered in~\cite{kuhn}. Explicit examples and infinite families appear in~\cite{howe}, \cite{kuhn}, \cite{shaska} for $n=3$, which is the case that is the topic of this paper. In the first three sections, we review known results and offer minor corrections and clarifications. In the remaining sections, we consider a similar problem -- given two elliptic curves $E$ and~$E'$, find the isomorphism class of a genus-$2$ curve $C$ such that~$\jac(C)\sim E\times E'$, if it exists. When dealing with both of these problems, one is not only interested in concrete examples, but also in parametrizations of infinite families. In Section~5 we give a parametrization of the Igusa--Clebsch invariants of~$C$ in terms of two parameters that define a pair of elliptic curves from the Hesse pencil. We believe that this approach can be generalized for various $n\geq 4$.

\subsection*{Notations and definitions}
Throughout the paper,~$K$ denotes a field of characteristic $\ch(K)\neq 2$ and~${\newoverline[.75]{K}}$ denotes an algebraic closure of $K$. Unless otherwise specified, all varieties are projective and defined over~$K$. If~$C$ is a genus-$2$ curve over~$K$, we say that its Jacobian $\jac(C)$ is \emph{split} if it is isogenous over~$K$ to a product of two elliptic curves. By a \emph{covering} we mean a finite, surjective, separable morphism. Given a divisor~$D$ on a $K$-variety, we denote by~$L(D)$ the $K$-vector space of global sections of the invertible sheaf~$\mathscr{L}(D)$ associated to~$D$. Given a commutative ring~$R$ and polynomials $F,G\in R[x]$, the resultant of~$F$ and~$G$ is denoted by~$\res_x(F,G)$  and the discriminant of $F$ is denoted by~$\disc_x(F)$.

\section{A curve of genus two covering a curve of genus one}
\label{section:coverings}
Let $C$ be a curve of genus two, defined over~$K$ and equipped with a covering $\mapu\colon C\to E$ of degree~$\degree$, where~$E$ is a curve of genus one and~$\degree$ is coprime to~$\ch(K)$. Recall that~$C$ is hyperelliptic since the linear system defined by its canonical divisor~$K_C$ defines a $2$-to-$1$ map to~$\PP^1$, by Riemann--Roch. Let~$\invol$ denote the hyperelliptic involution on~$C$, so that $C/\invol\cong\PP^1$. For a Weier\-stra{\ss}{ } point \mbox{$W\in C({\newoverline[.75]{K}})$}, one can define the Abel--Jacobi map $\varepsilon:C\inj\jac(C)$, given by~\mbox{$P\mapsto[P-W]$,} and embed $C$ into its Jacobian. One also has a corresponding isomorphism $E\cong\jac(E)$, given by~\mbox{$P\mapsto[P-\mapu(W)]$.} The morphism \mbox{$\mapu : C\to E$} induces a group morphism $\mapu_* : \jac(C) \to E$ so that~$\mapu_* \raisebox{.36pt}{\hspace{.2em}\scalebox{.8}{$\circ$}\hspace{.2em}} \varepsilon = \mapu$. The group morphism~$\mapu_*$ commutes with $[- 1]$ and therefore induces a map $\jac(C)/[- 1] \to E/[- 1]$. Thus we obtain the following commutative diagram
\begin{equation}
\label{diag:mapd2}
\begin{tikzcd}
C \arrow[r,"\mapu"] \arrow{d}{\projj[C]} & E \arrow{d}{\projj[E]}\\
C/\bigindex{\invol} \arrow[r,"\mapdd"] & E/[- 1]
\end{tikzcd}
\end{equation}
Here $\pi_C$ and $\pi_E$ denote the canonical maps.

\begin{remark}
The Abel--Jacobi map $\varepsilon$ is in general only defined over a quadratic extension of $K$, but this issue is easily resolved when $\degree$ is odd, as we shall explain below.
\end{remark}

\begin{assumption}
From now on, we will assume that the covering map $\mapu\colon C\to E$ is always \emph{optimal}, which is to say that it does not factor through a non-trivial isogeny.
\end{assumption}

\subsection{Ramification analysis}
Kuhn~\cite{kuhn} analysed the ramification of the map~$\mapdd$. We recall the main results. The map $\pi_C$ has six geometric ramification points, whereas~$\pi_E$ has four, by Riemann--Hurwitz. These are, of course, the points fixed by the corresponding involutions. Let $W_1,\dots ,W_6$ denote the ramification points of~$\pi_C$ and let $T_1,\dots,T_4$ denote the ramification points of~$\pi_E$. Moreover, let~$w_1,\dots, w_6$ and~\mbox{$t_1,\dots,t_4$} denote their respective images under the corresponding canonical maps~$\pi_C$ and~$\pi_E$. It is clear from the above that $\{\mapu(W_i)\}\subseteq\{T_j\}$ and~$\{\mapdd(w_i)\}\subseteq\{t_j\}$.
\par

By Riemann--Hurwitz, we have that $\mapu$ has ramification degree $2$, meaning that $\mapu$ either doubly ramifies at two distinct points or it has one triple ramification point. We distinguish two cases -- either the ramification of~$\mapu$ occurs above some~$T_j$ or it does not. These are referred to as the ``special'' case and the ``general'' case, respectively. As~$\invol$ acts on the ramification divisor $R_\mapu$ of $\mapu$, if there are two distinct ramification points, they cannot lie above two distinct~$T_j$. In the generic case, the map \mbox{$\pi_E\raisebox{.36pt}{\hspace{.2em}\scalebox{.8}{$\circ$}\hspace{.2em}}\mapu=\mapdd\raisebox{.36pt}{\hspace{.2em}\scalebox{.8}{$\circ$}\hspace{.2em}}\pi_C$} ramifies at~$4n$ double points that lie above the~$T_j$. Since~$\pi_C$ ramifies at six double points, we have that~$\mapdd$ ramifies at $\tfrac{1}{2}(4n-6)=2n-3$ double points above the $t_j$, none of which is any of the $w_i$. By Riemann--Hurwitz, $\mapdd$ has ramification degree~\mbox{$2n-2$}, which means that there is one more doubly ramified point that does not lie above the~$t_j$. In the special case, all of the ramification lies above the~$t_j$. Since~$\mapdd$ is a finite map between smooth varieties, it is flat and every fibre has $\degree=\deg\mapdd$ geometric points, counting with multiplicities.
\begin{lemma}
\label{lemma:kuhn}
With the notations as above, for every $i\in\{1,2,\dots,6\}$ the divisor $\mapdd^{*}(\sum_{j=1}^{4}t_j)$ contains $w_i$ with odd multiplicity and any other points with even multiplicity.
\end{lemma}
\begin{proof}
See Lemma in \S1 of \cite{kuhn} or Lemma 2.1 in \cite{freykani}.
\end{proof}

\begin{lemma}
\label{lemma:main_lemma}
Let $C$ be a curve of genus two and let $\phi_1\colon C\to E_1$ be an optimal covering of an elliptic curve $(E_1,O_1)$ with $\deg\phi_1=n$ coprime to $\ch(K)$. Then, possibly after extending the base field, there exists an elliptic curve $(E_2,\zero2)$, an optimal covering $\phi_2\colon C\to E_2$, and an isogeny $\varphi\colon E_1\times E_2\to\jac(C)$ such that:
\begin{enumerate}[label=(\arabic*)]
\item $\deg\phi_2=n$;
\item $\varphi=\phi_1^*+\phi_2^*$;
\item $\deg\varphi=n^2$;\item $\ker(\varphi)\cong E_1[n]\cong E_2[n]$.
\end{enumerate}
\end{lemma}
\begin{proof}
The covering map $\phi_1$ induces an embedding $\phi_1^*\colonE_1\inj\jac(C)$, with respect to an isomorphism $E_1\cong\jac(E_1)$. The elliptic curve $E_2$ is given as $\ker(\phi_{1*})\subset\jac(C)$, which is connected because $\phi_1$ is optimal. Let $\varepsilon\colon C\inj\jac(C)$ be an embedding, not necessarily defined over $K$. Recalling that Jacobians are (canonically) self-dual, let $\eta\colon\jac(C)\toE_2$ denote the map dual to the inclusion $E_2\inj\jac(C)$. The covering map $\phi_2\colon C\toE_2$ is then obtained as the composition $\eta\raisebox{.36pt}{\hspace{.2em}\scalebox{.8}{$\circ$}\hspace{.2em}}\varepsilon$. The isogeny \mbox{$\varphi\colonE_1\timesE_2\to\jac(C)$} is given by $\varphi=\phi_1^*+\phi_2^*$ and its kernel is the image of~$E_i[n]$ under the embedding $\mapu_i^*\colonE_i\inj\jac(C)$ for both $i\in\{1,2\}$. For more details, see Lemma in~\S2 of \cite{kuhn} or Lemma~1.6 in~\cite{thesis}.
\end{proof}
\begin{dfn}
With the assumptions of Lemma~\ref{lemma:main_lemma}, the Jacobian $\jac(C)$ is said to be \emph{$(\degree,\degree)$-split} and the curves~$E_1$ and~$E_2$ are said to be \emph{glued along their $\degree$-torsion}.
\end{dfn}
\begin{rem}
The constructions in Lemma~\ref{lemma:main_lemma} depend on the choice of the embedding $\varepsilon\colon C\inj\jac(C)$, which need not be $K$-rational.
\end{rem}

\subsection{Optimal coverings of odd degree}
Let $\phi_1\colon C\toE_1$ be an optimal covering of odd degree $\degree$. Let~$\pi_i\colonE_i\to\PP^1$ be the canonical maps, let~$T_j$ be the geometric ramification points of~$\pi_1$ and let $t_j=\pi_1(T_j)$. It follows from Lemma~\ref{lemma:kuhn} that there is a unique ramification point of $\pi_1$, say~$T_4$, such that exactly three of the~$W_i$ map to it under $\phi_1$. Moreover, there is exactly one~$W_i$ above each point in $\{T_1,T_2,T_3\}$. We index the points so that $W_1,W_2,W_3$ lie above~$T_4$. It follows that the divisors $W_1+W_2+W_3$ and $W_4+W_5+W_6$ are~$K$-rational and that the point $T_4$ is $K$-rational. Analogous statements hold for the points~$w_i$ and~$t_j$. Thus we conclude that the curve~$C$ admits an affine plane model~\mbox{$y^2=P(x)Q(x)$,} where~\mbox{$P(x),Q(x)\in K[x]$} are cubics whose roots are~\mbox{$\{w_1,w_2,w_3\}$} and~\mbox{$\{w_4,w_5,w_6\}$}, respectively. Since the class of the canonical divisor~\mbox{$K_C\sim 2W_i$} is~\mbox{$K$-rational}, so is the class of the divisor~\mbox{$W_1-W_2+W_3$}. Since the latter is a divisor of degree one and invariant under the action of the hyperelliptic involution, it follows that~$\phi_1$ induces a canonical \mbox{$K$-rational} embedding~\mbox{$C\inj\jac(C)$}, given by
\begin{equation}
\label{eq:embedding}
 P\mapsto[P-W_1+W_2-W_3],
\end{equation}
 which is compatible with the isomorphism~$E_1\cong\jac(E_1)$, given by~\mbox{$P\mapsto[P-T_4]$}, and the involutions. We summarize with the following (see \S4 in \cite{kuhn} for details).

\begin{lemma}
There is a canonical and $K$-rational choice for $E_2$ and the associated morphisms.
\end{lemma}

\begin{dfn}
\label{def:complement}
We will implicitly assume the embedding~\eqref{eq:embedding} and say that $E_2$, $\phi_2$, and~$f_2$ are \emph{complementary} to $E_1$, $\phi_1$, and $f_1$, respectively.
\end{dfn}

\begin{lemma}
\label{lemma:kuhn_odd}
The roles of the divisors $w_1+w_2+w_3$ and $w_4+w_5+w_6$ are exchanged between the complementary maps~$f_1$ and~$f_2$, and we have $f_{1*}(w_1+w_2+w_3)=3\pi_{1*}(O_1)$ and $f_{2*}(w_4+w_5+w_6)=3\pi_{2*}(\zero2)$, and hence also ${f_1}_*(w_4+w_5+w_6)=\pi_{1*}(E_1[2]\setminus\{O_1\})$ and $f_{2*}(w_1+w_2+w_3)=\pi_{2*}(E_2[2]\setminus\{\zero2\})$, where the identity element~$O_1\inE_1(K)$ is the point~$T_4$.
\end{lemma}

\section{Covering maps of degree three}
We will now focus on the case~$\degree=3$. We therefore assume that $\ch(K)\not\in\{2,3\}$. It follows from Lemma~\ref{lemma:kuhn} and its preceding paragraph that the ramification of the map $f_1\colon C\toE_1$ is restricted to one of two possibilities. This ultimately leads to a parametrization of the ${\newoverline[.75]{K}}$-isomorphism invariants of the curves involved. We will deal with the generic case first.

\subsection{The generic case}
\label{sub:generic}
The results in this subsection appear in~\cite{kuhn}, but here we go through the derivation in full detail. Let \mbox{$\phi_1\colon C\toE_1$} be an optimal covering of degree~$3$. In the generic case, the map $f_1$ is doubly ramified at a $K$-rational point that is not in any of the fibres $f_1^*(t_j)$. Let us denote the image of this point under $f_1$ by~$t_0$. Since~$t_0$ and~$t_4$ are $K$-rational and we are interested in isomorphism classes, we may and do assume that $t_0=0$, $t_4=\infty$, and $f_1^{*}(0)=2\cdot0+\infty$. In summary, we assume that the ramification of~$f_1$ is as depicted in Fig.~\ref{fig:fig1}, where the unramified points above $t_1,\dots,t_4$ are the~$w_i$ (depicted are the ramification indices of the points in the fibres of $f_1$ above the $t_j$).
\begin{figure}[H]%
\hspace{-2.35em}
\centering
\parbox{1em}{
\raisebox{8.4em}{
\begin{tabular}{c}
$\infty$ \\
\raisebox{-0.7em}{$0$}
\end{tabular}
}
}
\;\;
\parbox{\picsize}{%
\centering
\includegraphics[width=\picsize]{deg3ram_generic.pdf}\\

\tikzmark{tikzmark1}{}\\
\hspace{0.6cm}$f_1$\\
\tikzmark{tikzmark2}{}\\
\includegraphics[width=\picsize]{line5pts.pdf}\\
\hspace{.05em} $0$ \hspace{1.41em} $t_1$ \hspace{1.5em}  $t_2$ \hspace{1.5em}  $t_3$ \hspace{1.5em}  $\infty$
}\DrawArrow[->]{tikzmark1}{tikzmark2}
\caption{Ramification of the map $f_1$ in the generic case.}
\label{fig:fig1}
\end{figure}
\par
\noindent
In other words, we assume, without loss of generality, that
$$
f_1(x)=\dfrac{x^2}{P(x)},
$$
where $P(x)=x^3+ax^2+bx+c\in K[x]$ has roots $w_1,w_2,w_3\in{\newoverline[.75]{K}}$. The~$w_i$ are pairwise distinct and none of them equals zero. This can be expressed as
\begin{align*}
\res_x(x,P(x))&=c\neq0,\\
\disc_x(P(x))&=a^2 b^2 - 4 b^3 - 4 a^3 c + 18 a b c - 27 c^2\neq0.
\end{align*}
The pullback of $t_1+t_2+t_3$ corresponds to the roots of $D(x)^2Q(x)$, where~$D(x)$ and~$Q(x)$ are cubics in $K[x]$. Moreover, the roots of~$D(x)$ are the ramification points distinct from~$0$, and the roots of~$Q(x)$ are $w_4,w_5,w_6$. Since
$$
\frac{\df_1}{\d x}(x)=-\frac{x(x^3- b x -2 c )}{P(x)^2}
$$
and the roots of the numerator correspond precisely to the doubly ramified points of~$f_1$, we can take $D(x)=x^3- b x -2 c$. The ramification points are again pairwise distinct so we have
\begin{equation}
\disc_x(D(x))=4 (b^3 - 27 c^2)\neq0.
\end{equation}
From this we can calculate the nonic $D(x)^2Q(x)$ whose roots correspond to the divisor $f_1^*\raisebox{.36pt}{\hspace{.2em}\scalebox{.8}{$\circ$}\hspace{.2em}}{f_1}_*(d_1+d_2+d_3)$, where the $d_i$ are the roots of $D(x)$. In particular, we have the following equality, up to multiplication by constants,
$$
\res_y(x^2P(y)-y^2P(x),D(y))=D(x)^2Q(x).
$$
This resultant is easily found to equal
$$c (x^3- b x -2 c )^2 (4 c x^3+ b^2 x^2 +  2 b c x + c^2),
$$
whence we can take $Q(x)=4 c x^3+ b^2 x^2 +  2 b c x + c^2$. It follows that, up to quadratic twists, the curve~$C$ admits an affine plane model given by
\begin{equation}
\label{eq:generic_C}
y^2=P(x)Q(x)=(x^3+ax^2+bx+c)(4 c x^3+ b^2 x^2 + 2 b c x + c^2).
\end{equation}
By Lemma~\ref{lemma:kuhn_odd}, we may assume
$$
f_2(x)=\frac{(x+d)^2(x+e)}{4 c x^3+ b^2 x^2 + 2 b c x + c^2}
$$
for some $d,e\in K$, up to multiplication by a constant. To determine $d$ and $e$, we apply the procedure used to obtain $Q(x)$ from~$f_1$ to the map~$f_2$. In doing so, we must ultimately obtain a cubic polynomial~$R(x)$ that is a multiple of~$P(x)$, by Lemma~\ref{lemma:kuhn_odd}. Working over the field $K(a,b,c,d,e)$, we can compute the polynomial~$R(x)$ and perform Euclidean division on~$P(x)$ and~$R(x)$. By the argument above, the remainder must equal zero and we obtain three polynomial equations over the ring~$K[a,b,c,d,e]$. More details can be found in~\cite{thesis}, where the computations are performed over $K(a,b,c)[d,e]$. The argument here is only slightly different. Let~$I\subset K[z,a,b,c,d,e]$ denote the ideal generated by
$$
1-z\cdot(d-e)\cdot P(0)\cdot Q(-d)\cdot Q(-e)\cdot\disc_x(P(x))\cdot\disc_x(Q(x))
$$
and the coefficients of the remainder obtained by dividing~$R(x)$ and~$P(x)$. Eliminating the variable $z$ and computing the primary decomposition of the corresponding elimination ideal gives, among others, the following two equations:
\begin{align*}
& bd - 3c = 0,\\
& acd - 4ace + b^2e - bc + 3cde = 0.
\end{align*}
We therefore take
\begin{equation}
f_2(x)=\frac{(bx+3c)^2((b^3 - 4abc + 9c^2)x + b^2c - 3ac^2)}{4 c x^3+ b^2 x^2 + 2 b c x + c^2},
\label{eq:d_and_e}
\end{equation}
which also covers the cases when $\infty$ is a zero of $f_2$. Now one can determine the modular invariants of~$E_1$ and~$E_2$. An affine plane model for $E_1$ can be determined, up to quadratic twists, by requiring that the set of branch points of the canonical map $\pi_1$ is $\{t_1,t_2,t_3,\infty\}$, i.e.~$\infty$ and the image under~$f_1$ of the three roots of~$Q(x)$. Likewise, an affine plane model for~$E_2$ can be determined by requiring that~$\pi_2$ ramifies above~$\infty$ and the image under $f_2$ of the three roots of~$P(x)$. The corresponding cubics can be obtained from the resultants \mbox{$\res_y(xP(y)-y^2,Q(y))$} and \mbox{$\res_y(xQ(y)-(y+d)^2(y+e),P(y))$} and are omitted here. The $j$-invariants of the two elliptic curves can then be obtained from the cubics by a direct computation. We obtain the following two expressions, which appear in~\S6 of~\cite{kuhn}:
\begin{equation*}
j(E_1)=\frac{16 (a^2 b^4 + 12 b^5 - 126 a b^3 c + 216 a^2 b c^2 + 405 b^2 c^2 - 972 a c^3)^3}{(b^3 - 27 c^2)^3 (a^2 b^2 - 4 b^3 - 4 a^3 c + 18 a b c - 27 c^2)^2},
\label{jinv33}
\end{equation*}
\begin{equation*}
j(E_2)=\frac{256 (a^2 - 3 b)^3}{a^2 b^2 - 4 b^3 - 4 a^3 c + 18 a b c - 27 c^2}.
\end{equation*}

\subsection{The special cases}
In this subsection we will deal with the cases in which one or both of the maps $\mapdd_i$ are special, i.e. there is a triple ramification point above the branch locus of $\pi_i$.

\subsubsection{First map special}
\label{subsub:first_special}
Suppose that $f_1$ is special and $f_2$ is not. By passing to a quadratic extension of~$K$ if necessary and applying a suitable automorphism of~$\PP^1$, we may and do assume that $f_1$ has a triple zero at 0 and a simple pole at~$\infty$, so that
$$
f_1(x)=\frac{x^3}{x^2+ax+b},\quadf_2(x)=\frac{(x+c)^2(x+d)}{Q(x)},
$$
The argument used in the previous subsection gives $Q(x)=(a^2 - 4 b) x^3 - 2 a b x^2 - 3 b^2 x$ and therefore, up to quadratic twists,~$C$ admits an affine plane model given by
$$
y^2=x((a^2 - 4 b) x^2 - 2 a b x - 3 b^2)(x^2+ax+b).
$$
As before, we impose the generic ramification on~$f_2$ and use Lemma~\ref{lemma:kuhn_odd}. The difference in this case is that~$R(x)$ is a priori a cubic so we must identify its leading coefficient with zero and then find the remainder by dividing by~$P(x)$. This ultimately gives
\begin{align*}
& a c - 3 b  = 0,\\
& 3 a d - 3 b + c^2 - 4 c d = 0.
\end{align*}

It follows that we can take
\begin{equation*}
\label{eqn:f2_first_special}
f_2(x)=\frac{
(a x + 3 b)^2 (a (a^2 - 4 b) x + b (a^2 - 3 b))
}{
x ((a^2 - 4 b) x^2 - 2 a b x - 3 b^2)
}.
\end{equation*}
Simple resultant computations finally give models of~$E_1$ and~$E_2$ from which we obtain
$$
j(E_1)=\frac{16 (16 a^6 - 144 a^4 b + 405 a^2 b^2 - 324 b^3)^3}{729 b^4 (a^2 - 3 b)^3 (a^2 - 4 b)^2},
\quad
j(E_2)=\frac{256 (a^2 - 3 b)^3}{b^2 (a^2 - 4 b)}.
$$
A simpler parametrization can be obtained by reversing the roles of the maps, as below.

\subsubsection{Second map special}
\label{subsub:second_special}
Suppose that $f_2$ is special and $f_1$ is not. This is analogous to the general case and we may and do assume that
$$
f_1(x)=\frac{x^2}{x^3 + ax^2 + bx + c},\quadf_2(x)=\frac{(x+d)^3}{4cx^3 + b^2x^2 + 2bcx + c^2}.
$$
Following the same argument, we obtain:
\begin{align*}
& b d - 3 c = 0,\\
& -2 b + 3 a d - 3 d^2 = 0,
\end{align*}
whence we can take
$$
f_1(x) = \frac{x^2}
{(b x + 3 c)(9 c x^2+ 2 b^2 x + 3 b c)},
\quad
f_2(x) = \frac{(b x+3c)^3}
{4 c x^3 + b^2 x^2 + 2 b c x + c^2}.
$$
Therefore a quadratic twist of $C$ admits an affine plane model given by
$$
y^2 = (b x + 3 c)(9 c x^2+ 2 b^2 x + 3 b c)(4 c x^3 + b^2 x^2 + 2 b c x + c^2).
$$
The $j$-invariants of $E_1$ and $E_2$ are then easily found to be
\begin{equation}
\label{eqn:second_special_j}
j(E_1) = \frac{64 b^3}{c^2},
\quad
j(E_2) = \frac{64(4b^3 - 27c^2)^3}{729 b^3 c^4}.
\end{equation}
Note that in all cases one must assume that the numerator and the denominator of~$\mapdd_i$ have no common factors with one another. These conditions can be expressed as the non-vanishing of the corresponding resultants, omitted here.

\subsubsection{Both maps special}
\label{subsub:both_special}
Finally, suppose that both $f_1$ and $f_2$ are special. Without loss of generality, we assume
$$
f_1(x)=\frac{x^3}{x^2+ax+b},
\quad
f_2(x)=\frac{1}{Q(x)},
$$
where $b\neq 0$ and $a^2-4b\neq 0$. Applying the usual argument to~$f_1$, we get
\begin{equation*}
Q(x)=(a^2-4b)x^3-2abx^2-3b^2x.
\end{equation*}
Applying the same argument to $f_2$, we conclude that $x^2+ax+b$ must divide
$$
3(a^2 - 4 b)^2 x^2 - 4ab(a^2 - 4 b) x-16  b^2(a^2 - 3 b)
$$
Dividing the latter by the former gives the remainder
$$
-a(3a^2-8b)(a^2-4b)x-a^2b(3a^2-8b).
$$
Given that $a^2-4b\neq 0$ and $b\neq 0$, the remainder is identically zero if and only if~$a=0$ or $b=3a^2/8$. For $a=0$, we obtain
$$
f_1(x)=\frac{x^3}{x^2+b},\quad f_2(x)=\frac{1}{4x^3+3bx},\quad j(E_1)=j(E_2)=1728.
$$
For $b=3a^2/8$, we obtain
$$
f_1(x)=\frac{x^3}{8x^2+8ax+3a^2},
\quad
f_2(x)=\frac{1}{32x^3+48ax^2+27a^2x},
$$
$$
j(E_1)=j(E_2)=-\frac{873722816}{59049}=-\frac{2^6\cdot 239^3}{3^{10}}.
$$
\par
This shows that there are exactly two pairs of isomorphism classes of $E_1$ and $E_2$ such that two covering maps $C\toE_i$ of degree three have a triple ramification point and such that the fibre with three Weier\-stra{\ss}{ } points of one covering has no points in common with the fibre with three Weier\-stra{\ss}{ } points of the other covering. However, only the case $j(E_1)=j(E_2)=1728$ corresponds to a pair of complementary curves and coverings in the sense of Definition~\ref{def:complement}. In the other case we have $E_1=E_2$ and one special covering is obtained from the other one by composing with an automorphism of $C$. A complementary pair of curves with~$j(E_1)=-873722816/59049$ and a special covering $C\to E_1$ is obtained by putting $a=3a^2/8$ in~\S\ref{subsub:first_special}, which yields a generic covering $C\to E_2$ with~\mbox{$j(E_2)=64/9$.} This is elaborated upon in the Appendix (Example~\ref{example:bad_conclusion}).\par
Therefore there is a unique pair of isomorphism classes with two complementary special coverings of degree three. There have been conflicting claims in the literature about the number of such pairs (cf.~\cite{kuhn}, \cite{shaska}) and we hope that this detailed exposition settles the matter -- the statement in~\S6 of~\cite{kuhn} is correct.
\begin{remark}
Suppose that a curve of genus two covers an elliptic curve of \mbox{$j$-invariant~$J_1$} ``generically'' $3$-to-$1$ and that it covers the complementary curve, of \mbox{$j$-invariant~$J_2$,} ``specially'' $3$-to-$1$. Then it is readily verified, using the parametrizations above, that $F(J_1,J_2)=0$, where
\begin{equation}
\label{eq:param_curve}
F(X,Y):= X^3 - 1296 X^2 - 729 X Y + 559872 X - 80621568.
\end{equation}
The \mbox{$j$-invariant} pairs that are given in~\cite{shaska} are then obtained as the intersection of the curves~ \mbox{$F(X,Y)=0$} and \mbox{$F(Y,X)=0$.} However, the points of the intersection do not correspond to the \mbox{$j$-invariant} pairs of curves whose $3$-to-$1$ coverings by a genus-$2$ curve are both special since~\eqref{eq:param_curve} is obtained under the assumption that one of the coverings is generic. For example, the point $(1728,1728)$ lies on both curves. To see that it is on the first curve, we can take $(b,c)=(3t^2,- t^3)$ in~\eqref{eqn:second_special_j}, for $t\neq 0$. However, these parameters do not define a hyperelliptic curve because $t$ is a multiple root of both~$P(x)$ and~$Q(x)$.
\end{remark}

\begin{remark}
Our choices of ramification points and parametrizations are the same as those made by Kuhn~\cite{kuhn}; this provides some context for the analysis of the special cases. However, in terms of elegance and simplicity, the parametrization given in~\cite{howe} is superior. This is particularly evident when it comes to explicitly writing down the elliptic curves $E_i$ and the degree-$3$ morphisms $C\to E_i$ (see~\S\ref{app:explicit_covering} in the Appendix).
\end{remark}

\section{Gluing two elliptic curves}
In the previous section we started with an optimal covering $C\to E_1$ of degree~$3$ and constructed the complementary curve~$E_2$. In the remaining sections we adopt a different approach. We start with two elliptic curves~$E_1,E_2$ and construct a curve of genus two whose Jacobian is isogenous to \mbox{$E_1\timesE_2$} via an isogeny whose kernel is prescribed. This approach can be found in~\cite{freykani}. First we recall some definitions and classical results.
\par
Let $A$ be an abelian variety over $K$ and let $\lambda\colon A\to\chk{A}$ be a polarization. Suppose that $m\in\ZZ$ is coprime to $\ch(K)$ and such that $\ker(\lambda)\subset A[m]$. Let $$e_m\colon A[m]({\newoverline[.75]{K}})\times\chk{A}[m]({\newoverline[.75]{K}})\to\mu_m$$ denote the Weil pairing. Then we can associate to $\lambda$ a skew-symmetric pairing
$$
e_\lambda\colon\ker(\lambda)\times\ker(\lambda)\to\mu_m
$$
that is defined for any pair $(P,Q)$ of geometric points as $e_\lambda(P,Q) = e_m( P, \lambda(R))$, where $R$ is such that $[m]R=Q$. This does not depend on $R$ or $m$ (see~\S16 in \cite{milneav}).
\par
\begin{lemma}
\label{lemma:mumford}
Let $\varphi\colon A\to B$ be an isogeny whose degree is coprime to $\ch(K)$ and let $\lambda\colon A\to\chk{A}$ be a polarization induced by a line bundle~$\mathscr{L}$. Then the following are equivalent:
\begin{enumerate}
\item There exists a line bundle $\mathscr{M}$ on $B$ such that $\mathscr{L}=\varphi^*(\mathscr{M})$, inducing a polarization $\lambda'\colon B\to\chk{B}$,
\item $\ker(\varphi)\subset\ker(\lambda)$ and $e_\lambda$ is trivial on~$\ker(\varphi)\times\ker(\varphi)$.
\end{enumerate}
\end{lemma}
\begin{proof}
See Proposition 16.8 in \cite{milneav} or Theorem 2 and its Corollary in~\S23 of \cite{mumford}.
\end{proof}
\begin{cor}
\label{cor:mumford}
Let $\phi_1\colon C\to E_1$ be an optimal covering of an elliptic curve by a curve of genus two, such that $\deg\phi_1=\degree$ is coprime to $\ch(K)$, and let~$E_2$ be the complementary elliptic curve. Let \mbox{$\alpha\colon E_1[\degree]\stackrel{\sim}{\longrightarrow}E_2[\degree]$} be the induced canonical isomorphism (with respect to an embedding of $C$). Then~$\alpha$ inverts the Weil pairing, i.e.
\begin{equation}
e_\degree(P, Q)=e_\degree\left(\alpha(P),\alpha(Q)\right)^{-1}
\label{eq:isotropy_condition}
\end{equation}
for any $P,Q\in E_1[\degree]({\newoverline[.75]{K}})$.
\end{cor}

In other words, Lemma~\ref{lemma:mumford} provides a criterion for deciding when a polarization ``descends'' through an isogeny. In view of Lemma~\ref{lemma:mumford}, our starting data are two elliptic curves $E_1, E_2$ and an isomorphism $\alpha\colon~E_1[\degree]\stackrel{\sim}{\longrightarrow}~E_2[\degree]$ that is \emph{anti-symplectic} with respect to the Weil pairing, i.e. $\alpha$ satisfies~\eqref{eq:isotropy_condition} for any $P,Q\in E_1[\degree]({\newoverline[.75]{K}})$, where~$\degree$ is coprime to~$\ch(K)$.
\par
We assume the usual principal polarization on $ E_1\times E_2$, given by the divisor
$$
\Theta=\{O_1\}\times E_2+E_1\times\{\zero2\}.
$$
Let $\Gamma_\alpha\subset(E_1\times E_2)[\degree]$ denote the graph of~$\alpha$ and let
$$
\varphi\colon E_1\times E_2\to(E_1\times E_2)/\Gamma_\alpha=:J
$$
be the canonical map, which is clearly an isogeny.
\begin{lemma}
\label{lemma:Jpolarized}
The isogeny $\varphi\colon E_1\times E_2\to J$ induces a principal polarization of $J$.
\end{lemma}
\begin{proof}
See p.~156 in~\cite{freykani}.
\end{proof}

\begin{lemma}
\label{lemma:uniqueC}
Suppose that $\degree$ is odd. Then there exists a unique effective divisor $C$ on $J\otimes{\newoverline[.75]{K}}$ such that $D=\varphi^*(C)$ is linearly equivalent to $\degree\Theta$ and fixed by $[- 1]_{E_1\timesE_2}$. The divisor $C$ is fixed by $[- 1]_J$ and principally polarizes $J$.
\end{lemma}
\begin{proof}
See Proposition 1.1 and Corollary 1.2 in \cite{freykani}.
\end{proof}

It is a well known theorem of Weil (Satz 2 in~\cite{weil}) that over ${\newoverline[.75]{K}}$ any principally polarized abelian surface is either a Jacobian or a product of two elliptic curves (with the usual polarizations). Therefore the question of whether or not $J$ is a Jacobian reduces to the question of whether or not the divisor $C$ is irreducible.
\par
\begin{lemma}
The divisor $C$ is irreducible if and only if the divisor~$D$ is irreducible.
\end{lemma}
\begin{proof}
See Proposition 1.3 in \cite{freykani}.
\end{proof}
\par
\begin{remark}
\label{rem:reducibleC}
There exist examples with $C$ reducible. Let $\gamma\colon E_1\to E_2$ be an isogeny of degree~\mbox{$\degree-1$} and let $\alpha\colon E_1[\degree]\stackrel{\sim}{\longrightarrow} E_2[\degree]$ be the anti-symplectic isomorphism that is the restriction of $\gamma$ to the $\degree$-torsion. Then $C$ is reducible. Moreover, whenever~$C$ is reducible, $E_1$, $E_2$, and the irreducible components of $C$ are all isogenous and therefore $J$ is generically a Jacobian.
\end{remark}

\section{Gluing two elliptic curves along the $3$-torsion}
\label{section:gluing33}
In this section we will deal with the case $\degree=3$, given two elliptic curves from the Hesse pencil. We begin by fixing additional assumptions.
\begin{assumption}
The field $K$ is of characteristic $\ch(K)\neq 2,3$ and it contains a primitive third root of unity.
\end{assumption}
From now on, let ${\omega}\in K$ be such that $1+{\omega}+{\omega}^2=0$.
\par
\subsection{Prerequisites}
The one-dimensional family of curves given by
\begin{equation}
\label{eq:hesse_pencil}
E_a\colon x^3+y^3+z^3+3a xyz=0,
\end{equation}
is called the \emph{Hesse pencil}.\footnote{The $3$ does not usually appear in the definition. We added it for simplicity.} With the exception \mbox{of $a^3=-1$,} each $a\in K$ defines an elliptic curve $E_a$. We choose $[-1:1:0]$ to be the identity element. The group morphisms are given in the Appendix. We denote the set of elliptic curves in the Hesse pencil by~$\mathcal{H}$. The $j$-invariant of $E_a$ is
\begin{equation}
j({E_a})=-\frac{27a^3(a^3-8)^3}{(a^3+1)^3}
\label{eq:j_lambda}
\end{equation}
and $j\colon\mathcal{H}\to\Aff^1$ is $12$-to-$1$, except above $j=0$ \mbox{and $j=1728$}. In particular, the elements of
\begin{align*}
\Biggl\{ & a, a{\omega}, a{\omega}^2, \frac{2-a}{1+a}, \frac{2-a}{1+a}{\omega}, \frac{2-a}{1+a}{\omega}^2, \frac{2{\omega}-a}{{\omega}+a},\\
 &\frac{2{\omega}-a}{{\omega}+a}{\omega}, \frac{2{\omega}-a}{{\omega}+a}{\omega}^2, \frac{2{\omega}^2-a}{{\omega}^2+a}, \frac{2{\omega}^2-a}{{\omega}^2+a}{\omega}, \frac{2{\omega}^2-a}{{\omega}^2+a}{\omega}^2 \Biggr\}
\end{align*}
define isomorphic elliptic curves.
\par
The $3$-torsion subgroup of every elliptic curve in~$\mathcal{H}$ is fully $K$-rational and is given by $xyz=0$. Therefore the same nine points in~$\PP^2$ are the $3$-torsion points of every element of~$\mathcal{H}$ and each of the points can be given by homogeneous coordinates that are a permutation of $\{0,1,u\}$, where $u^3=-1$. Moreover, the Hesse pencil is exactly the family of all cubics passing through these nine points. \par
There is a partial converse in the form of Lemma~\ref{lemma:converse} below. A short direct proof is included in the Appendix (see~\S\ref{app:proof}).
\begin{lemma}
\label{lemma:converse}
Every elliptic curve over $K$ with fully $K$-rational $3$-torsion is isomorphic to a quadratic twist of an element of the Hesse pencil.
\end{lemma}
This can be thought of as an explicit realization of the fact that the modular curve~$X(3)$ has genus zero (see pp. 22--23 in~\cite{shimura}, for example).
\par
Now let $S=[-1:0:1]$ and $T=[-w:1:0]$. These two points generate $E[3]$ for every $E\in\mathcal{H}$. From now on, we are going to fix an isomorphism $\eta\colon E[3]\stackrel{\sim}{\longrightarrow}(\ZZ/3\ZZ)^2$ for every elliptic curve $E$ in~$\mathcal{H}$, given by $\eta(S)=(1,0)$ and $\eta(T)=(0,1)$.
\par
Let $E$ be an elliptic curve in the Hesse pencil. The Weil pairing on $E[3]$ is completely determined by the value $e_3(S,T)$ and we can easily calculate $e_3(S,T)={\omega}.$
For example, using Chapter III~\S8 in~\cite{aec}, we find that $e_3(S,T)=g(P+T)/g(P)$, where
$$
g = \frac{x^2z+y^2x+z^2y}{xyz}\in K(E)
$$
and $P\in E({\newoverline[.75]{K}})\setminus\left(E[3]\cup(P+E[3])\right)$. It follows that the Weil pairing on $E[3]$ is given by
$$
e_3(P,Q)={\omega}^{\det(\eta(P),\eta(Q))}
$$
and we can interpret it as the determinant map
$$
\det\colon\ZZ/3\ZZ\times\ZZ/3\ZZ\to\ZZ/3\ZZ.
$$
Since $\Aut(\ZZ/3\ZZ)\cong \textrm{GL}_2(\ZZ/3\ZZ)$ is a group of order~$48$, an anti-symplectic isomorphism $E_1[3]\stackrel{\sim}\longrightarrow E_2[3]$ corresponds to one of the $24$ elements of the coset $\smallmat{0}{1}{1}{0} \textrm{SL}_2(\ZZ/3\ZZ)$. However, since each isomorphism can be composed with $[- 1]$, we are left with $12$ distinct cases at most.
\par
Before we deal with the general case, we consider a concrete example in which gluing two elliptic curves does not give a Jacobian.
\begin{example}
\label{ex:split_glue}
Let $a=-(1+2t^3)/(3t^2)$ for some~$t\in K$, such that $$t(t^3-1)(8t^3+1)\neq 0.$$ Then the elliptic curve
$$E_a\colon x^3+y^3+z^3-\frac{1 + 2 t^3}{t^2} x y z =0$$
has a rational point $[t:t:1]$ of order two. Let $b=(1-4t^3)/(3t)$. Then the map
$$
\gamma\colon E_a\to E_b,\quad [x:y:z]\mapsto[f_1(x,y,z) : f_2(x,y,z) : f_3(x,y,z)],
$$
where
\begin{align*}
&f_1 = x (-2 t^2 y^2 - t^2 x y + t^2 x^2 - y z + 2 t^3 x z + t z^2),\\
&f_2 = y (-2 t^2 x^2 - t^2 x y + t^2 y^2 - x z + 2 t^3 y z + t z^2),\\
&f_3 = t z (x + y + t z) (x + y - 2 t z),
\end{align*}
is an isogeny whose kernel is the cyclic group of order two that is generated by the point $[t:t:1]$. Restricting $\gamma$ to the $3$-torsion, we obtain the isomorphism \mbox{$\alpha\colon E_a[3]\to E_b[3]$} that corresponds to $\smallmat{1}{0}{0}{2}\in\textrm{GL}_2(\ZZ/3\ZZ)$. We have that \mbox{$J:=(E_a\times E_b)/\Gamma_\alpha$} is isomorphic to $E_a\times E_b$ as a principally polarized abelian surface (recall Remark~\ref{rem:reducibleC}). We note that $a$ and $b$ satisfy
\begin{equation}
3a^2b^2 + a^3 + b^3 - 3ab + 2 = 0,
\label{eq:disc_vanishes}
\end{equation}
which is an equation describing a singular affine curve of genus zero.
\end{example}

We now consider the isomorphism from Example~\ref{ex:split_glue} in full generality. From now on, we fix $\alpha=\smallmat{1}{0}{0}{2}$. Let $E_1$ and $E_2$ be two elliptic curves in $\mathcal{H}$, corresponding to parameters $a$ and $b$, respectively. Let $A$ and $G$ respectively denote the images of~$E_1\times E_2$ and $\Gamma_\alpha$ in $\PP^8$ under the Segre embedding $\sigma$, given by
\begin{equation*}
\left([x_1:y_1:z_1],[x_2:y_2:z_2]\right)\mapsto [x_1x_2:x_1y_2:x_1z_2:y_1x_2:y_1y_2:y_1z_2:z_1x_2:z_1y_2:z_1z_2].
\end{equation*}
The identity element of $A$ is $O_A=[1:-1:0:-1:1:0:0:0:0]$ and the inversion morphism $[- 1]_A$ is given by
\begin{equation}
[X_1:X_2:\cdots:X_9]\mapsto[X_5:X_4:X_6:X_2:X_1:X_3:X_8:X_7:X_9].
\label{eq:linear_minus_id}
\end{equation}
Let $\Theta=\sigma(E_1\times\{O_2\})+\sigma(\{O_1\}\times E_2)$ and let $D$ denote the effective divisor on~$A$ that is linearly equivalent to~$3\Theta$, invariant under $[- 1]_A$, and invariant under the translation by the points of~$G$. Let $\varphi\colon A\to J$ denote the isogeny with kernel $G$ and let $C=\varphi(D)$.

\subsection{The computations}
Now we will go over the steps that lead to the Igusa--Clebsch invariants of a genus-$2$ curve whose Jacobian is isomorphic to $(E_1\times E_2)/\Gamma_\alpha$ as a principally polarized abelian surface, if such a curve exists. All computations were performed using the computational algebra system \textsc{Magma}~\cite{magma}. Naturally, some technical details will be omitted.
\par
The \emph{Igusa--Clebsch invariants} are the invariants $A'$, $B'$, $C'$, $D'$ defined on p.~319 of~\cite{mestre}. The \emph{Igusa invariants} are the invariants $J_2$, $J_4$, $J_6$, $J_8$, $J_{10}$ defined on p.~324 ibid.
By the corresponding \emph{absolute invariants} we mean the values
$$
j_1 = \frac{J_2^5}{J_{10}},\quad j_2 = \frac{J_2^3 J_4}{J_{10}},\quad j_3=\frac{J_2^2 J_6}{J_{10}}.
$$
\par
We start by computing the ideal $I=I(A)$ that defines $A$ as a variety in~$\PP^8$. This is a straightforward computation and is omitted here.
\begin{lemma}
\label{lemma:lemma1}
Let $\mathcal{W}_1$ and $\mathcal{W}_2$ denote the set of geometric points of order two on~$\sigma(E_1\times\{O_2\})$ and the set of geometric points of order two on $\sigma(\{O_1\}\times E_2)$, respectively. Then any hyperplane section on $A$ that is invariant under~$[- 1]_A$ contains either $\mathcal{W}_1\cup \mathcal{W}_2$ or its complement in $A[2]({\newoverline[.75]{K}})$.
\end{lemma}
\begin{proof}
The two eigenspaces of~\eqref{eq:linear_minus_id} are respectively generated by the sets
\begin{equation}
\begin{split}
S_1&=\{X_1+X_5, X_2+X_4, X_3+X_6, X_7+X_8, X_9\},\\
S_2&=\{X_1-X_5, X_2-X_4, X_3-X_6, X_7-X_8\}.
\end{split}
\label{eq:invol_invariants}
\end{equation}
By adding the corresponding linear forms from~\eqref{eq:invol_invariants} to~$I$, we find that~$A[2]({\newoverline[.75]{K}})$ consists of six points that are in the zero locus of the ideal generated by~$S_1$ and ten points that are in the zero locus of the ideal generated by~$S_2$. Since every linear form that is an eigenvector for~$[- 1]_A$ is a linear combination of the elements of exactly one of these two sets, the claim follows.
\end{proof}
\begin{corollary}
\label{cor:cor1}
The quotient $J=A/G$ is a Jacobian if and only if the six geometric points of $\mathcal{W}_1\cup \mathcal{W}_2$ are the only $2$-torsion points on the divisor~$D$.
\end{corollary}
\begin{proof}
The divisor $C$ is either a curve of genus two or a sum of two elliptic curves that meet in a rational $2$-torsion point. Since $[- 1]_J$ induces a hyperelliptic involution~$\invol$ on the irreducible components of~$C$, we conclude that $C({\newoverline[.75]{K}})$ contains exactly six points fixed by~$\invol$ if and only if it is irreducible and that it contains exactly seven points fixed by~$\invol$ if and only if it is reducible. Since $\deg\varphi$ is odd, the restriction of $\varphi$ to the $2$-torsion is an isomorphism and there is exactly one geometric point of $(E_1\times E_2)[2]$ above each point of $C({\newoverline[.75]{K}})$ that is fixed by~$\invol$. Therefore $D({\newoverline[.75]{K}})$ contains at most seven $2$-torsion points. By Lemma~\ref{lemma:lemma1},~$D({\newoverline[.75]{K}})$ contains at least the order-$2$ points of $\sigma(E_1\times\{\zero2\})$ and~$\sigma(\{O_1\}\times E_2)$ and the claim follows.
\end{proof}
A fact crucial to our approach is that the translations by the points of~$A[3]$ are linear. In fact, they can be extended to automorphisms of $\PP^8$. This is a consequence of the fact that $A$ is embedded into $\PP^8$ via an embedding corresponding to $L(3\Theta)$. It can also be shown directly, using the addition formulas. In particular, the group of translations by the points of~$G$ is generated by the following two automorphisms:
\begin{align*}
[X_1:X_2:\dots:X_9]&\mapsto[X_5:X_6:X_4:X_8:X_9:X_7:X_2:X_3:X_1]:\\
[X_1:X_2:\dots:X_9]&\mapsto[X_1:{\omega} X_2:{\omega}^2 X_3:{\omega}^2 X_4:X_5:{\omega} X_6:{\omega} X_7:{\omega}^2 X_8:X_9]
.\end{align*}
From this we immediately determine that the nine effective divisors invariant under the action of~$G$ and linearly equivalent to~$3\Theta$ are the hyperplane sections defined by the following linear forms:
\begin{align*}
L_1 &= X_1 + X_5 + X_9,&
L_6 &= {\omega}^2 X_3 + {\omega} X_4 + X_8,\\
L_2 &= {\omega} X_1 + {\omega}^2 X_5 + X_9,&
L_7 &= {\omega} X_2 + {\omega}^2 X_6 + X_7,\\
L_3 &= {\omega}^2 X_1 + {\omega} X_5 + X_9,&
L_8 &= {\omega} X_3 + {\omega}^2 X_4 + X_8,\\
L_4 &= X_3 + X_4 + X_8,&
L_9 &= {\omega}^2 X_2 + {\omega} X_6 + X_7.\\
L_5 &= X_2 + X_6 + X_7,&
\end{align*}
We note that the divisor~$D$, that is invariant under~$[- 1]_A$, is defined by~$L_1=0$ and does not contain~$O_A$. Now we can compute the scheme that is the intersection of~$D$ and the nine points of~$A[2]({\newoverline[.75]{K}})$ that are not $2$-torsion points on $\sigma(E_1\times\{O_2\})$ or~$\sigma(\{O_1\}\times E_2)$ and apply Corollary~\ref{cor:cor1}. Dehomogenizing by setting $X_9=1$, taking the corresponding ideal in the ring~$K[X_1,\dots,X_8,a,b]$, and eliminating the~$X_i$ gives
\begin{equation}
3a^2b^2 + a^3 + b^3 - 3ab + 2 = 0
\label{eq:disc_vanishes2}
\end{equation}
Note that this matches \eqref{eq:disc_vanishes}. Thus we have obtained the following (cf. Proposition~2.2. and Corollary~2.3 in \cite{freykani}).
\begin{proposition}
The principally polarized abelian surface $J=A/G$ is a product of two elliptic curves if and only if~\eqref{eq:disc_vanishes2} holds, i.e. if and only if~$E_1$ and~$E_2$ are $2$-isogenous.
\end{proposition}
\par
Equation~\eqref{eq:disc_vanishes2} can be thought of as the analogue of~$\Phi_2(j(E_1),j(E_2))=0$ that is specific to our choice of~$\alpha$. Here~$\Phi_2$ denotes the classical modular polynomial
\begin{align*}
\Phi_2(X,Y) = &\:X^3 + Y^3 - X^2Y^2 + 1488(X^2Y+XY^2) - 162000(X^2+Y^2)\:+\\
& + 40773375XY + 8748000000(X+Y) - 157464000000000.
\end{align*}
\par
The abelian surface $J$ can be found explicitly as the quotient of the variety~$A$ under the group action of~$G$, where $G$ acts by point translation (see Lecture~10 in~\cite{harris}, for example). Since~$\varphi^*$ is injective and $\dim_K(L(nC))=n^2$ for every $n\in\NN$, we have that the subspace~$L(nD)^G=\varphi^*(L(nC))$ of~$G$-invariants of~$L(nD)$ is of dimension~$n^2$ for every~$n\in\NN$. In particular, since $\mathscr{L}(3C)$ is very ample, by finding nine linearly independent $G$-invariant elements of~$L(3D)$, we can obtain~$\varphi$ as a map to~$\PP^8$. We may take the nine $G$-invariant forms~$L_i^3$ for this purpose. Unsurprisingly, explicitly computing~$\varphi(A)$ is not feasible.
\par
From now on, let us assume that $3a^2b^2 + a^3 + b^3 - 3ab + 2 \neq 0$ so that~$C$ is irreducible. It follows that the global sections of~$\mathscr{L}(2C)$ define the canonical map~$\kappa\colon J\to \mathcal{K}$, where~$\mathcal{K}=J/[- 1]\subset\PP^3$ is a Kummer surface (Proposition~4.23 in~\cite{16-6}). Therefore the four-dimensional $G$-invariant subspace $L(2D)^G\subset L(2D)$ defines the composition~$\psi=\kappa\raisebox{.36pt}{\hspace{.2em}\scalebox{.8}{$\circ$}\hspace{.2em}}\varphi$. We have that~$\psi(D)$ is a conic in~$\PP^3$ and the image under~$\psi$ of the~\mbox{$2$-torsion} points that lie on~$D$ gives six pairwise distinct (geometric) points on~$\psi(D)$ that are the branch locus of the canonical $2$-to-$1$ map $C\to\psi(D)$. By finding a~$K$-rational point on the conic~$\psi(D)$, we obtain an isomorphism~$\psi(D)\stackrel{\sim}{\longrightarrow}\PP^1$ and the image of the six branch points gives us a sextic that defines a plane model of a hyperelliptic curve that is in the isomorphism class of~$C$. We can then directly compute the absolute invariants from this model. We may take the following four~$G$-invariant forms to define~$\psi$:
\begin{align*}
X_2X_4 + X_3X_7 + X_6X_8,
&\quad
X_2X_3 + X_4X_6 + X_7X_8,\\
X_2X_8 + X_3X_6 + X_4X_7,
&\quad
X_1^2 + X_5^2 + X_9^2.
\end{align*}
An alternative approach is to compute the curve~$C=\varphi(D)$ directly, compute the canonical divisor~$K_C$, and then find the image in~$\PP^1$ of the six points of~$J[2]({\newoverline[.75]{K}})$ that lie on~$C$, under the canonical map defined by~$L(K_C)$. However, this is significantly slower in practice than the Kummer surface approach.
\par
We make an important observation. The absolute invariants of~$C$, as functions of the parameters~$a$ and~$b$, will have certain symmetries. For example, the abelian surface $E_1\times E_2$ is isomorphic to~$E_2\times E_1$ and the isomorphism (which is just a permutation of the coordinates) leaves~$G$ intact so that \mbox{$(E_1\times E_2)/\Gamma_\alpha$} and \mbox{$(E_2\times E_1)/\Gamma_\alpha$} will give the same absolute invariants. Similarly, the same invariants are obtained if one starts with a pair $(E_1,E_2)\in\mathcal{H}^2$ defined by parameters~$(a{\omega},b{\omega}^2)$ or~$(a{\omega}^2,b{\omega})$.
\begin{remark}
Recall that for each curve in~$\mathcal{H}$ there are eleven other curves in~$\mathcal{H}$ that are isomorphic to it, with the exception of the usual two isomorphism classes. It is a natural question to ask which of the $144$ possible isomorphic pairs~$E_1\times E_2$ result in the same isomorphism class of~$C$ when taking the quotient by~$G$. It turns out that the pairs are partitioned into twelve sets of twelve pairs and every pair in the same set gives the same isomorphism class of~$C$. Moreover, all of the remaining eleven choices of~$\alpha$ can be reduced to the case we are considering by applying an isomorphism to a suitable product of two elliptic curves in~$\mathcal{H}$. Our choice of~$\alpha$ is particularly suitable for computations because of the simplicity of the equations defining~$D$ and the equation~\eqref{eq:disc_vanishes2}.
\end{remark}
To obtain the absolute invariants of~$C$ as functions of $(a,b)$, the first thing we do is make several degree estimates. For example, we can take~$a$ and~$b$ to be two large integers of comparable height, such as two large consecutive primes. We can also take~$a$ to be a large integer and~$b\in\{0,1\}$. This gives us estimates for the degrees of particular monomials that appear in the invariants. Then we notice that the so-called discriminant~$J_{10}$, that appears in the denominators, is going to be zero for choices of~$(a,b)$ that either do not define a pair of elliptic curves or do not define a quotient~$J$ that is a Jacobian. By factoring the invariants obtained for various choices of~$a,b\in\ZZ$ and combining this information with the degree estimates, we conclude that, up to a constant, $J_{10}$ equals
\begin{equation}
\label{eq:probable_j10}
(a^3+1)(b^3+1)(3a^2b^2 + a^3 + b^3 - 3ab + 2)^{12}.
\end{equation}
\par
To obtain the numerators, we use interpolation. We compute the absolute invariants of~$C$ for many choices of~$(a,b)$ and multiply by~\eqref{eq:probable_j10} in each case. We conclude from the aforementioned symmetries that the numerators are also linear combinations of monomials~$a^mb^n$, where $m\equiv n$ (mod 3). This significantly reduces the number of non-zero coefficients and makes the computation reasonably fast. Using our empirical bounds on the degrees and the coefficients, we interpolate over finite fields~$\FF_p$ for a suitable set of primes~$p$ and lift the results using the Chinese remainder theorem. Finally, we obtain the Igusa--Clebsch invariants from the Igusa invariants using the formulas in~\cite{mestre}. We summarize our results in the following proposition.
\begin{proposition}
\label{prop:prop2}
Let $E_1$ and $E_2$ be two elliptic curves over $K$, respectively given by the Hesse models
\begin{align*}
E_1\colon x^3+y^3+z^3+3axyz&=0,\\
E_2\colon x^3+y^3+z^3+3bxyz&=0,
\end{align*}
with identity point $O=[-1:1:0]$, where $a,b\in K$ and $3a^2b^2 + a^3 + b^3 - 3ab + 2\neq 0.$ Let $\alpha\colon E_1[3]\stackrel{\sim}{\longrightarrow}E_2[3]$ be the isomorphism defined by
$$[-1:0:1] \mapsto [-1:0:1],\quad[-{\omega}:1:0] \mapsto[-{\omega}^2:1:0],$$
and let~$\Gamma_\alpha$ denote its graph. Then the principally polarized abelian surface \mbox{$(E_1\times E_2)/\Gamma_\alpha$} is isomorphic to the Jacobian of a curve of genus two whose Igusa--Clebsch invariants are as follows:
\allowdisplaybreaks
\begin{align*}
I_2  =\;& 72 (9 a^6 b^6 - 30 (a^7 b^4 + a^4 b^7) - 88 a^5 b^5 + a^8 b^2 + a^2 b^8 + 54 (a^6 b^3 + a^3 b^6) + 65 a^4 b^4+\\
& - 32 (a^7 b + a b^7) - 104 (a^5 b^2 + a^2 b^5) + 40 (a^6 + b^6) +  44 a^3 b^3 + 100 (a^4 b + a b^4)+\\
& - 68 a^2 b^2 + 16 (a^3 + b^3) + 112 a b - 20),\\[.5em]
I_4  =\;& 36 (3 a^2 b^2 + a^3 + b^3 - 3 a b + 2)^4 (9 a^4 b^4 + 240 a^3 b^3 + 8 (a^4 b + a b^4) + 240 a^2 b^2+\\
&+ 160 (a^3 + b^3) + 256 a b + 320),\\[.5em]
I_6  =\;& 72 (3 a^2 b^2 + a^3 + b^3 - 3 a b + 2)^4 (729 a^{10} b^{10} - 3402 (a^{11} b^8 + a^8 b^{11})+\\
&+ 30456 a^9 b^9 +  81 (a^{12} b^6 + a^6 b^{12}) - 70794 (a^{10} b^7 + a^7 b^{10}) -
 201555 a^8 b^8 + \\
&- 2160 (a^{11} b^5 + a^5 b^{11}) + 60 (a^{12} b^3 + a^3 b^{12}) + 106560 (a^9 b^6 + a^6 b^9) +\\
&-  148932 a^7 b^7 - 121608 (a^{10} b^4 + a^4 b^{10}) +
 480 (a^{11} b^2 + a^2 b^{11}) +\\
&- 358740 (a^8 b^5 + a^5 b^8)  - 8 (a^{12} + b^{12}) + 156928 (a^9 b^3 + a^3 b^9) +\\
&+ 336444 a^6 b^6 -  50160 (a^{10} b + a b^{10}) + 81072 (a^7 b^4 + a^4 b^7) +\\
&- 462096 a^5 b^5 - 167112 (a^8 b^2 + a^2 b^8) + 84224 (a^9 + b^9) +\\
&+  455568 (a^6 b^3 + a^3 b^6) + 761040 a^4 b^4 + 181152 (a^7 b + a b^7) +\\
&- 93600 (a^5 b^2 + a^2 b^5) +  219552 (a^6 + b^6) + 383424 a^3 b^3 +\\
&+ 564480 (a^4 b + a b^4) + 88512 a^2 b^2 + 74624 (a^3 + b^3) + 314112 a b - 55040),\\[.5em]
I_{10} =\;& 36864 (a^3+1) (b^3+1) (3 a^2 b^2 + a^3 + b^3 - 3 a b + 2)^{12}.
\end{align*}
\end{proposition}
\par

When $K$ is a number field or a finite field of characteristic $\ch(K)>5$, it is possible to construct a genus-$2$ curve over~$K$ with given Igusa--Clebsch invariants (see~\cite{cardona} and~\cite{mestre}). When~$K$ is a number field, the recent work of Bruin, Sijsling, and Zotine~\cite{sijsling} allows one to verify numerically over~$\CC$ that a curve obtained from Igusa--Clebsch invariants of the form above indeed has a $(3,3)$-split Jacobian.
\par

\begin{ack} The author wishes to thank Robin de Jong and Fabien Pazuki for the fruitful discussions that inspired this paper. The author is also grateful to Ronald van Luijk, Christophe Ritzenthaler, and Jeroen Sijsling for their helpful comments and remarks. Portions of the paper have already appeared in the author's Ph.D. thesis, completed at the universities of Leiden and Bordeaux. Most of the work presented was completed during the author's stay at the University of Ulm.
\end{ack}
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redpajama_set_name: RedPajamaArXiv


Royal Crescent is a crescent in West Glasgow with a high concentration of businesses including restaurants, independent financial advisors and accountants.
Royal Crescent also has the following useful amenity: a pub - Tang's Golden Well Limited.
Andrew Merchant Ltd is an insurance company offering a range of different types of insurance.
Benham Conway is an accountancy company dealing with a range of accounts and other financial matters.
G B M Services (Scotland) Ltd is a commercial cleaning company offering a range of cleaning services for offices and other workplaces.
J R G Financial Consultancy is a firm of independent financial advisers offering whole of market financial advice to clients.
JRG Financial Consultancy is a firm of independent financial advisers offering whole of market financial advice to clients.
Lion Television is a television channel which shows a variety of programmes providing viewers with information and entertainment.
R G Cowley Ltd is a property management company which operates a variety of real estate.
Russell & Callander Ltd offers a variety of accountancy services for various customers.
The PRG Partnership is a solicitors' firm handling various legal matters.
Townhouse Hotel offers a range of accommodation and other facilities for guests.
Whitelaw Wells deals with a range of accounts and financial matters for various customers.
Map showing Royal Crescent in Glasgow.


Messing about on Final Cut Pro - More to come!
Today I went down to Gym 66 to analyse the lighting in preparation for the LIFT-IT Worldwide event which is on the 12th September where I will be filming the event. I'm looking forward to getting in the studio with some of the athletes.


text:
You are at:Home»Execs to Know»Bob Smith Transitions to Consultant Role after 40-year GovCon Career
Bob Smith Transitions to Consultant Role after 40-year GovCon Career
By Staff Writer August 9, 2021
Bob Smith is putting his 40 years in GovCon to work as a newly branded consultant in an industry he spent nearly a lifetime learning. His career began in 1980 as an Air Force contracting officer who happened into that position by a happy twist of fortune.
He went on to hold positions in a number of private sector companies, managing the infrastructure and built sale value, including most recently as chief administrative officer of IntelliBridge, a position he left in December 2020 to focus on his consultancy. He now assists firms through a full range of corporate infrastructure efficiencies, complex mergers and acquisitions, and integration-related consulting advice to help clients advance their own businesses through organic growth and/or acquisition.
Better to be Lucky than Good
In 1980, after two years of college, Smith enlisted in the Air Force. On the strength of a high score on the Armed Service Vocational Aptitude Battery test, he elected to have the Air Force choose his specialty based on its needs. He considers himself lucky to have been slotted as a contract specialist and, ultimately, a warranted contracting officer, paving the way for a long GovCon career.
"To give you an idea of technology in 1980, one of the criteria for the contracting specialty code was 'having use of all 10 fingers' so as to be able to dial a rotary phone," Smith said. "This was certainly fortunate for me as I learned a marketable skill during the era of Reaganomics that allowed a then 20-year-old to learn government contracting in an era of high government spending."
During the next 7 years, Smith managed Air Force service and supply contracts at Keesler Air Force Base in Mississippi and later construction and architect engineer contracts at Andrews Air Force Base in Maryland. He was honorably discharged from the Air Force in 1987 as a staff sergeant.
"I share the perspective of many who started their careers in the military — it gives you the necessary discipline to be successful later in life and, if you're lucky, provides you a technical skill that can springboard a private sector career," Smith said.
Combining Government and Industry Experience – Start Big, then Go Small
After the Air Force, Smith went to work at Northrop Worldwide Aircraft Services, Inc. as superintendent of procurement for a government-owned, contractor-operated facility at Vance Air Force base in Enid, Oklahoma, focused on T-37 and T-38 trainer aircraft.
After two years with Northrop, Smith returned to Washington, D.C., as a contract manager for AT&T Federal Systems, where he managed national-level programs with the Defense Information Systems Agency, from 1989 to 1994. Smith values his time at Northrop and AT&T for the experience he gained learning the inner workings of larger companies — values he would soon transfer to helping build and mature smaller ones.
In 1994, amid the Clinton-era focus on small business in government contracting, Smith shifted his career focus and began a 25-year run as a CAO helping build value and cost savings for growing small business government contractors and assisted five different owners to grow and later exit their businesses in ways that met their goals.
Experience in M&A through Relationships
Between 1994 and 2019, Smith was a key member of multiple management teams that grew and sold preeminent, emerging, high-growth, small business government contracting companies.
Smith joined I-NET, Inc. in 1994, and in 1996, assisted owner Ken Bajaj as vice president of administration in the sale of I-NET, Inc. to Wang Laboratories.
"Ken showed me how to successfully delegate to others to grow an organization's size and scale while keeping indirect expenses low," Smith said.
Smith joined SIGNAL Corp. in 1996, and in 2002, assisted owner Roger Mody as CAO in the sale of SIGNAL to Veridian (now GDIT). He said Mody helped him understand the value of "building a high-energy team who worked hard while playing hard, thereby building mutual trust and creating a force multiplier in the executive suite."
This team also included executives Cass Panciocco, Scott Goss and Barry Kane, who all went on to be key players in government contracting M&A with industry-leading firms. Smith would collaborate with each of them in subsequent growth and sale activities of other GovCon firms.
Smith joined RSIS Information Systems in 2004, and in 2008, was involved as CAO in the sale of RSIS to Wyle Labs (now KBR). While at RSIS, Smith worked for co-owner Ron Trowbridge, a consummate GovCon marketeer who now owns BlackWatch International, and with Panciocco, who was COO, and with Goss who worked as a financial consultant.
Smith also met Ron's daughter, Karen Trowbridge, who was a participant in a rotational training program for high-potential employees Smith managed.
In 2010 after completing the integration of RSIS into Wyle, Smith reunited with Kane, who was president of Catapult Technology and recruited Smith as CAO to assist in the continued growth and the ultimate sale of Catapult to DC Capital, which closed in 2012.
Smith joined Trowbridge in 2014 as CAO, and in 2019, he assisted owner Karen Trowbridge, whom he had mentored at RSIS years earlier in the sale of Trowbridge to Enlightenment Capital.
"From the time I met Karen in 2006, she always said that when she started her own company that she wanted me to come work for her," Smith said. "Frankly, I didn't put a lot of stock in it at the time, but 8 years later in 2014, sure enough it came true. Working for Karen was a poetic career milestone and brought me full circle from working for her father to working for her and assisting both of them in meeting their growth and exit goals."
Panciocco and Goss were also part of the Trowbridge team, Panciocco as chief operating officer and Goss as a financial consultant.
The Importance of a Good character and the Golden Rule
Smith believes long-term success requires more than strong technical and managerial skills.
Staff-level employees are conditioned to say "no" to customers and teammates, he said, and "Invariably, execs must be able to escalate to their C-suite peers to get to 'yes.' This can only be done when one accommodates others when they need help."
"This quid pro quo always serves individuals and companies well," he continued. "No matter how experienced you are, you never know it all. Your staff all have unique knowledge, experience and talents that when utilized and fostered are a force multiplier resulting in profitable, fun and rewarding environments that also promote retention of high performers."
And the best long-term deals?
"Those happen when both parties get something they want," he said. "This fosters an environment more likely to favor long-term customer engagements and a sense of customer loyalty among the employee base. Executives who live [by]this principle seem to have long-term success in working with and for others based on their reputational capital."
Smith lives in Maryland and has three children in their 30s. He credits wife Merle with being the life partner who supported his career success while they raised a family. When they are not in the D.C. area, they enjoy time at their "safe haven" home in Newport, Rhode Island.
In addition to support from his wife, Smith credits his mother, now hale and hearty in her 90s, for giving him a foundation that helped him be successful. She continues to influence him today.
"I'm the youngest of six boys," he said. "My father died when I was 16, and my mother had to raise six boys on her own. She's always been the matriarch of the family."
From her, he learned patience, authenticity and the importance of treating everyone the way he wanted to be treated.
"You should be nice to everybody from the CEO down to the janitor," he said.
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text:
Tag: Horror
Photo: Pixabay.com
Creative Writing, Venue
The Attic Head
The ceramics teacher wanted to make mothers of us, 'Make yourself in your image,' and we all got to work, squeezing these wet slabs, pressing, thumbing, stretching lids over eyes, strands over scalps, building a mirror from the dust that layered every hand. Back from the kiln, the ward, its birthing heat, and I…
03/12/2019	Continue to read
Photo: Flickr
Film, TV
James Wan – Master of Modern Horror
Malaysian-born, Australian film director James Wan is most known for reviving the horror genre since debuting his short film Saw in 2003. After the 1990's slasher craze wore off, it was Saw that encouraged horror enthusiasts to give the genre another try. Its simplistic plot, featuring moralistic ethos of righteous living and a madman's intricate…
Photo: Max Pixel
My Favourite Horror: (The Haunted Place) 1963
It is near impossible to declare a favourite in a genre so heterogenous. Does one choose an example of body horror, or celebrate the nostalgia of slasher films? Should we recognise a cult classic, or applaud the slew of modern horror reclaiming some seriousness? I feel inclined to give the vied-for-top-spot to 1963's 'The Haunted…
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Arts, Venue
Horror favourites: Books and paintings
For me the best horror book is Mary Shelley's Frankenstein. A story about abandonment, power, beauty and what it truly means to be human. This just scrapes the tip of the iceberg of why the book still haunts me. When you factor in that Shelley was 18 when she wrote it, grieving from a miscarriage,…
My introduction to the Gothic
I'm a big fan of the Gothic; literature, art, movies, architecture, you name it. The creatives who cultivated this interest from a relatively young age were filmmaker Tim Burton, and by extension, Helena Bonham Carter, and author Marcus Sedgwick. Many people are familiar with the work of the former two, but, in my opinion, the…
Photo: Pixabay.com, Victoria Borodinova
Fashion, Venue
What your Halloween costume says about you (and other stupid assumptions)
As the trees shed their leaves and the days grow shorter, there's a quiet anticipation for the spookiest day of the year. Yes, people, it's the most wonderful time of the year – Halloween. It means we'll be witnessing this year's additions to the pop-culture pantheon. From Princess Leia to your friendly neighbourhood Spiderman, Britney…
Photo: Needpix
'I'm tired of countless sexist comments'
The other day I was having an internal dispute that I often have: how to be a palatable feminist. In other words, how do I engage actively with feminism, always questioning people's internalised misogyny whilst not being portrayed as an irritating 'man-hater'? I used to be one of those girls that didn't want to cause much…
© Norwich Theatre Royal
Review: The House on Cold Hill at the Norwich Theatre Royal
Halloween is long gone, but thankfully a little bit of horror is always in season. The House on Cold Hill, based on Peter James' novel, centres around a family who have moved from Brighton to an abandoned monk's abbey turned decadent home. Little do they know that their new gothic house is – you guessed…
Escape Room (In Cinemas 4 January) One of the reasons why I'm most interested for this film is because of its release date. The film is set to premiere on the first week of the new year, which is a slot that has an infamy for not being well-received, particularly in the US. I am…


text:
\section{\label{sec:getter} Hydrogen purification failure modes}

The two SAES PS4-MT50-R getter modules installed on the XENON1T gas purification system \cite{XENON1T_overview} use heated sintered zirconium to irreversibly bind to oxygen and other electronegative impurities.  Hydrogen however is reversibly absorbed in hot zirconium, with hydrogen being released as the zirconium is heated \cite{Getters, GARG1990275}.  XENON1T heats their getter to 450$^\circ$C, above the specified operating temperature for these getters, in order to more efficiently crack and purify methane \cite{Getters, DOBI2010594}.  The SAES PS4-MT50-R has a separate 'Hydrogen Removal Unit' is placed inline after the hot getters, but I can not yet report any information on the performance or capacity of these units.

The purification obtainable by a hot zirconium or zirconium alloy getter is limited. The absorption of hydrogen, either that from molecular hydrogen or from water, methane, or other hydrogenated compounds, follows Sievert's law.  Using a xenon gas density of 4.54~g/L at 400$^\circ$C and 1.935~bar passing the getter, this law can be expressed as
\begin{gather}
\text{Concentration of H}_2\text{ remaining} \notag \\
= (1.0 \text{ ppb}) exp\left({\frac{5200 K}{723.15 K} - \frac{5200 K}{T}}\right) \notag \\
\times \left(\frac{\text{Concentration of H}_2\text{ absorbed}}{1\text{ ppb}} \frac{m_{Xe}}{3.2 \text{ ton}} \frac{1.14 kg}{m_{Zr}} \right )^2
\end{gather}
for SAES St 171 getter material and in comparing to the nominal xenon mass and getter temperature in XENON1T \cite{Getters}.

The XENON1T getters would be saturated with hydrogen after absorbing an initial concentration of a few ppb of either hydrogen, methane, or water from the detector.  The hydrogen remaining in the xenon may also be preferentially enriched in tritium due to its lower diffusivity into the bulk of the zirconium getter \cite{*[{There's even a patent for using zirconium getters as an isotope separator, but the effect may be due to differences in absorption depth rather than different equilibrium sorbtion characteristics of tritium and hydrogen.}] patent}.

A significant fraction of the hydrogen in the getters may remain if they are insufficiently baked out prior to service.  For St 171 getters, heating to 900$^\circ$C while efficiently pumping out hydrogen to below 1 millitorr is recommended \cite{Getters}.  Even using this procedure, the remaining hydrogen in the getter at 450$^\circ$C produces a partial pressure of 4.0~microtorr, equivalent to 43 ppt of hydrogen in the xenon.

\section{\label{sec:tritium} Cosmogenic tritium production}

As noted by \cite{aprile2020observation} the concentration of tritium in hydrogen or water is normally at the $10^{-18}$ level.  However, this ratio is only a minimum valid for sources of tritium that are well mixed with natural sources of hydrogen.  Tritium that desorbs from materials that have been cosmogenically activated may be significantly more concentrated, thus any hydrogen off-gassed into the detector should be suspected of containing additional tritium.

While tritium and other radioisotopes can also be produced underground, this is unlikely to explain the full excess observed by XENON1T.

It has been shown that cosmogenic activation from muon showers is largely produced by energetic secondary particles produced from the initial muon \cite{PhysRevC.71.055805}.  In light elements such as seen in Borexino \cite{Bellini_2013}, photonuclear processes dominanate, while the mass normalized production of hadrons, particularly neutrons, is enhanced for heavy target nuclei \cite{PhysRevC.96.014605}.

A full accounting of the expected tritium production in XENON1T will require a dedicated simulation of muon and muon shower spallation similar to what has been modelled for Borexino.  However, a rough estimate can be obtained by assuming that tritium is dominantly produced by neutron secondaries from muon showers.  Neutron spallation is the dominant production mechanism of cosmogenic isotopes at the earth's surface.  Following the same method to estimate the neutron spallation production cross section as in \cite{AGNESE20191} using TALYS 1.8 \cite{talys}, INCL-ABLA \cite{INCL, ABLA07}, and spectra from \citet{Gordon}, the cosmogenic production rate is estimated to be 71.5 atoms/kg/day, significantly above the values quoted by \citet{aprile2020observation} from \citet{:Zhang}. \footnote{An attempt, reported in \cite{AGNESE20191}, to reproduce the tritium production cross sections in \cite{:mei}, as cited by \cite{:Zhang}, using older versions of TALYS could not reproduce their result for neutron energies above 80~MeV.  In addition, \cite{:mei} had mistakenly overstated the exposure of the IGEX detector crystals by nearly a factor of nine \cite{:cebrian_priv}, leading to an apparent but false confirmation of their calculated results.  The apparently larger production rates from the Geant4 simulations in \cite{:Zhang} is therefore not an indication of a conservative calculation, as \cite{aprile2020observation} claims, but rather the result of a gross error in one of the compared calculations.}

\begin{figure}
\includegraphics{h3xs_xe.pdf}
\caption{\label{fig:1} Calculated production cross sections for tritium from neutron spallation in xenon.  TALYS 1.8 is used to calculate the cross section below 100 MeV in xenon, and below 35 MeV in carbon, while INCL-ABLA is used at higher energies.}
\end{figure}

The neutrons from muon spallation at Gran Sasso and at surface have similar spectral shapes between 100 MeV and 5 GeV, where the majority of tritium production occurs.  The neutron fluence on surfaces is however greater by a factor of $8.25\times10^6$ compared to that entering the cavern walls at Gran Sasso \cite{:mei_hime, Gordon}.  The neutron fluence in XENON1T is likely enhanced by a further factor of $(131/22.87)^{0.95}$ due to the greater neutron production efficiency in xenon versus in the rock surrounding Gran Sasso.  Combining these factors, a tritium yield of $0.036\times10^{-25}$~mol/mol$\cdot$yr results.  While this value should be considered uncertain by an order of magnitude, is is still insufficient on its own to explain the observed excess of $6.2\pm2.0\times10^{-25}$~mol/mol in XENON1T.

\section{Assay and mitigation of hydrogen and tritium}

The hot zirconium getters used in XENON1T have limited hydrogen absorption capacity and are likely to be saturated.  It should be possible to measure the amount of hydrogen in the getters by using the temperature dependence of their absorbtion capacity.  After pumping on the getters to $10^{-6}$~torr at room temperature, the getter temperature can be raised while monitoring the vacuum pressure.  A hydrogen pressure of $9.45 \times 10^{-5} \times ([\text{H}_2]/1\text{ ppb})^2$ at 450$^\circ$C can be used to determine the impurity level in the xenon.

If the contamination of the xenon is measured to be $\sim$60~ppb, as XENON1T claims would be required to explain the observed tritium signal, it would require approximately 70 kg of activated zirconium to clean their detector to less than 1~ppb of hydrogen, significantly more zirconium than is contained in the SAES PS4-MT50-R that XENON1T uses.  A purification hydrogen purification campaign, with frequent reactivation of the getters, would be required.  As the zirconium getters are easily saturated with hydrogen, any single reactivation of the getter may produce only a partial reduction of the observed tritium content.  Changing the xenon recirculation flow rate, as was done for Science Run 2 and used as a double check by XENON1T \cite{XENON1T_overview}, would have little effect on the purification of the xenon through a saturated getter.

The absorbtion efficiency of zirconium getters for hydrogen is strongly affected by the temperature in the bulk of the getter.  If a hydrogen contamination level below 1~ppb is found in XENON1T, indicating the presence of tritium enriched hydrogen, the tritium and hydrogen concentrations may be reduced by a factor of 2.9 by lowering the getter temperature from 450$^\circ$C to 400$^\circ$C.  Further reduction of the tritium content could be achieved through flushing the detector with added hydrogen.

\section{Conclusion}

The use of hot zirconium getters can remove hydrogen from nobel gases to very low concentrations, but the capacity of these systems to absorb hydrogen is limited to approximately 3~mg of hydrogen per kg of zirconium, where 3~mg is equivalent to 1 ppb of contamination in XENON1T.  Unlike for other contaminants in liquid nobel gas detectors, hydrogen is reversibly absorbed onto hot zirconium.  Therefore, the getters may be used to assay the hydrogen content of the detectors by measuring the amount of hydrogen effused during getter regeneration.  Given zirconium's low capacity for absorbing hydrogen, a campaign to purify hydrogen and tritium from XENON1T, and other liquid nobel gas detectors, is required.

\begin{acknowledgments}
Thank you to Peter Rowson and Marie-C\'ecile Piro for valuble discussions about xenon gas purification and to Susana Cebri\'an for discussions about cosmogenic activation.  This research was undertaken thanks to funding from the Canada First Research Excellence Fund through the Arthur B. McDonald Canadian Astroparticle Physics Research Institute.
\end{acknowledgments}


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Lady Gaga, Jennifer Lopez to perform at diverse Biden inauguration
Lady Gaga will perform at Joe Biden's inauguration.
Lady Gaga and Jennifer Lopez will take the stage at US President-elect Joe Biden's inauguration ceremony in Washington next week alongside others in what his transition team said would showcase a diverse America.
The two will perform at the January 20th event for the incoming 46th US president, which will also feature remarks from a black firefighter from Georgia, a former Youth Poet Laureate, a Catholic priest, and a pastor from Mr Biden's hometown of Wilmington, Delaware.
Lady Gaga will sing the national anthem and Lopez will give a musical performance.
"They represent one clear picture of the grand diversity of our great nation," his team said in a statement on Thursday.
The presenters reflect Mr Biden and Vice President-elect Kamala Harris' "steadfast vision of a new chapter in our American story in which we are an America united in overcoming the deep divisions and challenges facing our people, unifying the country, and restoring the soul of our nation," it added.
Their roles come as the Democrat takes over the White House from Republican Donald Trump in a scaled-down event amid the Covid-19 pandemic and growing security concerns after a mob of Trump supporters attacked the US Capitol seeking to thwart lawmakers' certification of Mr Biden's victory.
Jennifer Lopez is set to perform at Joe Biden's inauguration.
The public will not be on hand to watch the swearing-in at the West Front of the Capitol building, now fortified by fencing, barriers and thousands of National Guard Troops.
In announcing the inauguration speakers, Mr Biden's transition team cited Lady Gaga's advocacy of LGBTQ rights and health issues, and Lopez's work raising awareness about the disproportionate impact of the novel coronavirus on Latinos.
Other speakers include career firefighter Andrea Hall of Fulton County, Georgia, and Los Angeles native 2017 National Youth Poet Laureate Amanda Gorman. Former Georgetown University President Father Leo O'Donovan and Reverend Dr Silvester Beaman of Bethel African Methodist Episcopal Church — both close to the Biden family — will also speak.
Lady Gaga,
Jennifer Lopez,
Biden inauguration
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Netflix admits rival streaming services 'may be affecting our growth'


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Wyoming: Expert Discussion on Cell Phones, Wireless Radiation and Health: Jackson, 31 July 2017
Expert discussion on cell phones, wireless radiation and health, sponsored by the Environmental Health Trust, 6 to 7:30 p.m. Monday, July 31 at Teton County Public Library, free.
"The forum comes only weeks after the National Frequency Agency of France released research showing nine out of 10 cell phones exceed government radiation limits when researchers tested the way people actually use them—directly next to the body, Davis said. Manufacturers, however, don't have to test phones directly on the body, like in pants pockets, she said. It's a major finding in France and internationally and a topic panelists will discuss."
DON'T MISS: Dangerous Communication
by Kelsey Dayton, planetjh.com, 25 July 2017
Scientists discuss seminal research about cell phones and other health hazards.
JACKSON HOLE, WY – Dr. Devra Davis's first grandchild was born in 2004 and by the time he was nine months old, he could grab a cell phone, turn it on and even start a game. At first, Davis was a proud grandparent and marveled at her clever grandson. But then the founder/president of the Environmental Health Trust realized there was a lot about cell phones she didn't know, especially when it came to the impact on young brains.
In her 2007 book The Secret History of the War on Cancer, she titled the final chapter "Presumed Innocent" and raised questions about cell phone safety.
It was the only topic in her book she received pushback about. No one debated tobacco could kill you, or air pollution was bad for your health. "Because you had dead bodies and statistically significant evidence," she said.
At the time Davis was a typical busy professional, carrying and using multiple cell phones.
Today, there is plenty of evidence, Davis said. She's gone onto write several books on the topic, including Disconnect: The Truth About Cell Phone Radiation. Yet people still don't understand the risk of cell phones when they are used directly on the body, instead of through car systems or on speaker phone. Cell phones emit radiation that can damage the brain, when the phone is placed directly next to the head. The body also absorbs radiation from the phone when it's in a pocket, or tucked in a bra, she said.
On Monday a panel of experts from the United States, Canada and France will discuss the latest research on cell phones and what people should know to keep themselves and their children safe.
The goal is to make people aware of the growing body of experts, from organizations like the American Academy of Pediatrics to researchers and governments that recognize the dangers of cell phones. "The solution isn't to say 'no' to technology, but 'yes' to health and so we want our technology as safe as possible," Davis said.
The Environmental Health Trust has hosted forums on the topic, but not for about five years.
Speakers include Anthony Miller, a doctor and professor emeritus at the Dalla Lana School of Public Health at the University of Toronto, who is also an advisor to the World Health Organization; Iris Udasin, a doctor and medical director of employee health at Rutgers University's Environmental Health and Occupational Health Sciences Institute; Cindy Lee Russell, a surgeon at Palo Alto Medical Foundation and vice president of environmental health for the Santa Clara Medical Association; Annie Sasco, a doctor and team leader of epidemiology for cancer prevention at the Victor Segalen Bordeaux University in France; Frank Clegg, founder and CEO of Canadians for Safe Technology and former president of Microsoft Canada; and Marc Arazi, a French physician. Davis will also speak at the event.
The forum comes only weeks after the National Frequency Agency of France released research showing nine out of 10 cell phones exceed government radiation limits when researchers tested the way people actually use them—directly next to the body, Davis said. Manufacturers, however, don't have to test phones directly on the body, like in pants pockets, she said. It's a major finding in France and internationally and a topic panelists will discuss.
The forum is a way to update people on the latest science and research happening around the world. "Bring your most scientific questions," said Theodora Scarato, director of educational resources with the Environmental Health Trust. But it also will be an introduction to those just now learning of cell phone radiation. "I actually find most people know nothing about this," she said.
Scarato, a social worker and therapist, only learned about the dangers of cell phones a few years ago. She had no idea that phones weren't pre-market checked for radiation exposure for kids. The more she learned, the more she realized the significance of the issue. "The science is very much raising a flag," she said.
Cell phone companies are waging a battle similar to those of tobacco companies that didn't want warning labels on packaging, or more research looking at long-term health effects, she said.
Scarato learned how people in other countries were more aware of the issue than people in the U.S. and how other countries already had policies in place.
She hopes the talk teaches people about the radiation issues and is a way for the experts to talk about what is next in terms of research and needed regulations.
"If to me, as a parent, if there is doubt something is safe and research is showing effects—that's all I need to know because my children's health means everything," she said.
Expert discussion on cell phones, wireless radiation and health, sponsored by the Environmental Health Trust, 6 to 7:30 p.m. Monday, July 31 at Teton County Public Library, free. PJH
http://planetjh.com/2017/07/25/dont-miss-dangerous-communication/
Libellés : Electromagnetic, Phonegate


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'I have never felt so loved in my life': Lavoie sisters from Regina in Amazing Race final
Search the Owen Sound Sun Times
Presented by HAL WRIGHT CHEVROLET CADILLAC GMC BUICK
"I do the one thing you should never do on The Amazing Race: I lost a clue."
More from Regina Leader-Post
Published on: September 11, 2019 | Last Updated: September 11, 2019 10:38 AM EDT
Lauren (left) and Joanne Lavoie are representing Saskatchewan in the latest season of The Amazing Race Canada. Submitted: Bell Media / Saskatoon
Regardless of how they finished the Amazing Race Canada, the Lavoie sisters can get some much-needed sleep knowing they did their best.
It's a philosophy instilled by their dad: Did you do your best? If you did, nothing else matters.
Lauren, 27, and Joanne, 21, were one of three pairs last standing in the seventh season of the competitive reality show. As the show's first team from Saskatchewan, they garnered the support of a whole province.
"I have never felt so loved in my life," Joanne said in an interview. "Everyone in Saskatchewan is like our family."
Speaking from Toronto (where Joanne was disappointed to miss a chance at meeting Jennifer Lopez during TIFF) before Tuesday's finale, the sisters from Regina were excited to stop keeping secrets and share their full experience with family, friends and fans.
"It has been so painful. I'm so bad at lying," Joanne said.
"Maybe Santa Claus feels this way too before he gives everybody presents, but it's very difficult."
They learned to stay on their toes and take a breath before answering prying questions at the grocery store, which isn't easy for Joanne. "I love to blurt out everything," she said.
The sisters first achieved fame on Vine, the now-defunct six-second video app, after Lauren tapped her brakes while Joanne was eating a McFlurry in the passenger seat, causing a spoonful of ice cream to jam into her face. The video racked up millions of views and the pair continued to make daily content on the platform.
Despite the six-year age gap between Lauren, a teacher, and Joanne, a yoga instructor, the two are very close and have leaned on each other for support. Lauren said one of the hardest parts of the race was being cut off from family at home.
Lauren said the finale was incredible and very intense for all remaining teams. They were up against married couple Anthony and James from Edmonton (ed. note: the eventual winners) and track teammates Sarah and Sam living in Toronto. The winning team claims two all-new 2019 Chevrolet Blazers, a once-in-a-lifetime trip for two around the world and a $250,000 cash prize.
While she couldn't spill the beans Tuesday afternoon about how the Lavoies fare in the final leg, she said her most frustrating moment of the whole competition took place in that night's episode.
"I do the one thing you should never do on The Amazing Race: I lost a clue," Lauren said, citing a tense moment shown in previews for the finale.
"It was heartbreaking because I was always so organized, I always had things together. I just got overwhelmed with the idea of racing a pair of Olympians and marathon runners."
It felt like she let her partner down. But neither of them stayed down.
"We're totally fine with having a little bit of a setback," Joanne said. "I think that showed through the entire race. We aren't scared if something goes the wrong way."
The saving grace, they agree, was that the error occurred early in the leg. They didn't quit. Like their perspective on winning the competition, they were inspired by their dad, who is three-weeks removed from spinal surgery to treat a chronic illness that prevents fluid draining from the brain.
"Seeing him struggle and fight through all of this, surgeries he was afraid of, it inspired us endlessly on the race," Lauren said. "Whatever cards you're dealt — if it's a bad cab at the beginning of the leg or losing a clue at the beginning of the leg — you always, always, always keep going."
Regina Piranhas Swimming coaches Joanne, left, and Lauren Lavoie in July 2017. Michael Bell / Regina Leader-Post
© 2020 Owen Sound Sun Times. All rights reserved.


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