text:
\section{Introduction}

Let $\mathcal{R}_{g}^{tr}$ denote the moduli space of  non-trivial \'etale double coverings of smooth
trigonal curves of genus $g$ and $\mathcal{M}^{tet}_{g-1,0}$,  the open set of  the moduli space of
tetragonal curves of genus $g-1$ consisting of tetragonal curves whose fibres of the 4:1 map have at least one \'etale
point. The classical trigonal construction due to Recillas \cite{r} gives a canonical
isomorphism
$$
\mathcal{R}_{g}^{tr} \rightarrow \mathcal{M}^{tet}_{g-1,0}
$$
such that for every covering $\widetilde C \rightarrow C$ of $\mathcal{R}_{g}^{tr}$ the Prym variety is
isomorphic to the Jacobian of its image in $\mathcal{M}^{tet}_{g-1,0}$ as principally polarized abelian
varieties.\\

The aim of this paper is to show that a similar statement is valid in the case of double covers  over trigonal curves with
two ramification points. If $f: \widetilde C \rightarrow C$ is a double cover of smooth curves ramified exactly at two points,
the Prym variety of the cover, which we denote by $P(f)$ or $P(\widetilde C/C)$, is a principally polarized
abelian variety (ppav). Apart from the \'etale case,  it is the only way to obtain a ppav from a covering between curves\footnote{
With the exception of non-cyclic triple coverings over a genus 2 curve, whose Prym variety is also a ppav. }.
In the sequel, a {\it ramified double cover} will always  mean a double covering ramified at
exactly two points.
We denote by $\mathcal{R} b^{tr}_g$ the moduli space of ramified double covers
$f: \widetilde C \rightarrow C$ of smooth trigonal covers $h: C \rightarrow \mathbb P^1$ with $g$ the genus of $C$
and the additional property that
the branch locus of $f$ is disjoint from the ramification locus of $h$.

We call an element $\widetilde C \stackrel{f}{\rightarrow} C \stackrel{h}{\rightarrow} \mathbb P^1$ of $\mathcal{R} b^{tr}_g$
{\it special} if the branch locus of $f$ is contained in a fibre of $h$ and {\it general} otherwise.
Let $\mathcal{R} b^{tr}_{g,sp}$ denote the closed subset of $\mathcal{R} b^{tr}_g$ consisting of special coverings
and $\mathcal{R} b^{tr}_{g,gen}$ its complement consisting of general coverings. Moreover, let $\mathcal{M}_{g,1}^{tet}$ denote the subspace of the moduli space of smooth tetragonal curves of genus $g$
as defined in Section 3 and let  $\mathcal{R}^{tet}_{g}$ the moduli space of \'etale double covers of smooth
tetragonal curves of genus $g$.
Then our main theorem is (Theorem \ref{thm4.3} and \ref{thm5.1}),

\begin{thm} \label{mainthm}
\begin{enumerate}
\item[(a)] There is a canonical isomorphism
$$
 \mathcal{R} b^{tr}_{g,sp} \rightarrow  \mathcal{M}_{g,1}^{tet}.
$$
If $\widetilde C \stackrel{f}{\rightarrow} C \stackrel{h}{\rightarrow} \mathbb P^1$ is an element of  $\mathcal{R} b^{tr}_{g,sp}$
and $X'$ the corresponding smooth tetragonal cover, we get an isomorphism of principally polarized
abelian varieties
$$
P(f) \stackrel{\simeq}{\rightarrow} JX'.
$$
\item[(b)]
There is a canonical map
$$
\mathcal{R} b^{tr}_{g,gen} \rightarrow  \mathcal{R}_{g+1}^{tet}.
$$
If $\widetilde C \stackrel{f}{\rightarrow} C \stackrel{h}{\rightarrow} \mathbb P^1$ is an element of  $\mathcal{R} b^{tr}_{g,gen}$
and $\pi: Y \rightarrow X$ the corresponding \'etale double cover, then the principally polarized abelian varieties
$(P(f), \Xi_f)$ and $(P(\pi), \Xi_\pi)$ are canonically isomorphic.
\end{enumerate}
\end{thm}

Furthermore, in the case (b) of the theorem the image of the map is contained
in the subspace  $\mathcal{R}_{g+1,2}^{tet}$, the locus of \'etale double coverings over tetragonal curves $X$, such that the 4:1 map $k: X \rightarrow \mathbb P^1$
has exactly two fibres consisting of two simple ramification points (Proposition \ref{p5.2}). Note that $\mathcal{R} b^{tr}_{g,gen}$ and
$\mathcal{R}_{g+1,2}^{tet}$ are of the same dimension (see Remark
\ref{r5.3}). We do not know the exact image of the map between these moduli spaces nor whether it is generically injective. \\

In Section 2 we define for every $\widetilde C \rightarrow C \rightarrow \mathbb P^1$ the corresponding covering
$Y \rightarrow X$ with $X$ tetragonal and work out its geometric properties. In Section 3 we recall
a special case of Donagi's extension of the trigonal construction which is used in Section 4
for the proof of part (a) of the theorem. Finally in Section 5 we give the proof of part (b).
 \bigskip

{\it Aknowledgements}. We would like to thank  Andrey Soldatenkov who helped us with the translation of \cite{da}.

\section{Double covers of trigonal covers}

Let $C$ be a smooth trigonal curve of genus $g \ge 3$ with trigonal cover $h: C \rightarrow \mathbb P^1$.
According to Hurwitz formula  the ramification divisor $R_h$ of $h$ is of degree $2g+4$. Let
$$
f: \widetilde C \rightarrow C
$$
be a double cover branched over 2 points $p_1,p_2$ of $C$. We assume that $p_1$ and $p_2$
are disjoint from $R_h$. Let $C^{(3)}$ and $\widetilde C^{(3)}$ the third symmetric products of
$C$ and $\widetilde C$ respectively. Let $\mathbb P^1 \simeq g_3^1 \hookrightarrow C^{(3)}$ be the natural embedding
of the trigonal linear system. We define the variety $Y$ by the following left hand cartesian diagram
\begin{equation} \label{d2.1}
\xymatrix{
 Y  \ar@{^{(}->}[r] \ar[d]_{8:1}^{\widetilde k} & \widetilde C^{(3)}  \ar[d]_{8:1}^{f^{(3)}} \\
g^1_3 \ar@{^{(}->}[r] & C^{(3)}
}
\end{equation}
 The fibre of $\widetilde k$ over a point $a \in \mathbb P^1$ consists of the 8 sections $s$
of $f$ over $a$:
$$
s: h^{-1}(a) \rightarrow f^{-1}h^{-1}(a) \quad \mbox{with} \quad f \circ s = id.
$$
(In \cite{d} Donagi denotes $Y$ by $f_*\widetilde C$, since considering $\widetilde C$ as a local system on $C$, this is just the push forward local system
on $\mathbb P^1$). Let denote $\iota$ de involution on $\widetilde{C}$ associated to $f$. There are 2 structures on $Y, $
an involution denoted also by $\iota$:
$$
\iota: Y \rightarrow Y,  \quad q_1 + q_1 + q_3 \mapsto \iota(q_1)+\iota(q_2)+ \iota(q_3),
$$
and an equivalence relation: two sections
$$
s_1,s_2: h^{-1}(a) \rightarrow f^{-1}h^{-1}(a)
$$
are called equivalent if they differ by an even number of changes $q \mapsto \iota(q)$. This defines a branched double cover
$$
O:= O(h \circ f) \rightarrow \mathbb P^1,
$$
called the orientation cover of $h \circ f$. For $n=3$ we have the cartesian diagram (see \cite[Section 2.1
and Lemma 2.1]{d}):
\begin{equation} \label{d2.2}
\xymatrix{
&Y \ar[dl]^{2:1}_\pi \ar[dr]^{\psi}_{4:1}\\
X \ar[dr]^{4:1}_k && O \ar[dl]_{2:1}^\omega\\
& \mathbb P^1 &
}
\end{equation}
with $X := Y/\iota$.

The following lemma is easy to check.

\begin{lem} \label{fixedfree}
The involution $\iota:Y \rightarrow Y$ is fixed-point free.
\end{lem}

Identifying the $\mathbb P^1$ of diagram \eqref{d2.2} with the image of $h$, we have
according to \cite[Lemma 2.3]{d}:

\begin{lem} \label{l2.2}
The cover $O \rightarrow \mathbb P^1$ is branched exactly at $h(p_1)$ and $h(p_2)$.
\end{lem}

Here a branch point
is a point over which the fibre of $O \rightarrow \mathbb P^1$ consists of 1 point, so the fibre might be a singular point
of $O$.  Note that $O \rightarrow \mathbb P^1$ is of degree  2, so Lemma \ref{l2.2} implies that $O$ is connected.

\begin{prop} \label{conn}
Suppose the branch locus of $f$ and the ramification locus of $h$ are disjoint.
Then the curve $Y$ is connected.
\end{prop}

\begin{proof}
It was proven in \cite{w} that $Y$ consists of two connected components when $f$ is \'etale. In that case  all the elements of the
monodromy group of $\widetilde k$ are given by even permutations. The existence of ramification
points for $f$ adds also odd permutations to the monodromy group, and taking into account that $O$ is connected, this
implies that the curve $Y$ is connected.
\end{proof}

\begin{prop} \label{p5.1}
Suppose the branch locus of $f$ and the ramification locus of $h$ are disjoint.
\begin{enumerate}
\item If $h \circ f$ is general, then the curve $Y$ is smooth;
\item If $h \circ f$ is special, the the curve $Y$ is smooth apart from $2$ nodes.
\end{enumerate}
\end{prop}

\begin{proof}
The proof is very similar to the proof of \cite[Proposition on p.107]{w} where Welters proves
that $Y$ is smooth when $f$ is an \'etale double cover of a $d$-gonal curve (with some assumptions on the ramification of $h$).
In particular, $Y$ is smooth for $d=3$.
As we will see, the main difference lies in the fact that,  if a fibre of $h$ contains both branch points $p_1$ and $p_2$ of $f$,
the curve $Y$ acquires two nodes; otherwise, $Y$ is smooth.

Let $D_i$ be the fibre of $h$ passing through $p_i$, we denote by the same letter the divisor
on $C$ as well as the corresponding point of $C^{(3)}$. Of course $D_1 = D_2$ if $h \circ f$ is special.
Let $\widetilde D_i^j$ denote the points of $Y$
defined by $D_i$ for $j = 1, \dots, \nu_i$ ($\nu_i = 2$, respectively $\nu_i=4$, if $h \circ f$ is special,
respectively general).
The same proof as in \cite{w} works for all points of
$Y \setminus (\cup_{i,j} \widetilde D_i^j)$. We shall show that $Y$ is smooth at the points
$\widetilde D_i^j$ if $h \circ f$ is general and nodal if it is special.

So let $D$ be one of the points $D_i \in C^{(3)}$ and $\widetilde D \in \widetilde C^{(3)}$ a point above
it. Suppose
$$
D = p + q + r \quad \mbox{and} \quad \widetilde D = \widetilde p + \widetilde q + \widetilde r.
$$
where $p$ is one of the branch points of $f$, so $f^*(p) = 2\tilde{p}$.
The Zariski  tangent spaces yield the cartesian diagram
\begin{equation} \label{d5.1}
\xymatrix{
T_Y(\widetilde D) \ar@{^{(}->}[r] \ar[d]_{df} & T_{\widetilde C^{(3)}}(\widetilde D) \ar[d]^{df^{(3)}}\\
T_{g_3^1}(D) \ar@{^{(}->}[r]_{dj} & T_{c^{(3)}}(D)
}
\end{equation}
where $j: g_3^1 \hookrightarrow C^{(3)}$ denotes the inclusion map. The curve $Y$ will be smooth at
$\widetilde D$ if and only if $\dim T_Y(\widetilde D) = 1$.
It is easy to see
(\cite[p. 104]{w}) that this is the case if and only if
\begin{equation} \label{e5.2}
T_{C^{(3)}}(D) = dj T_{g_3^1}(D) + df^{(3)} T_{\widetilde C^{(3)}}(\widetilde D)
\end{equation}

According to deformation theory the lower-right triangle of diagram \eqref{d5.1} is given by
\begin{equation} \label{d5.2}
\xymatrix{
&& H^0(\mathcal{O}_{\widetilde D}(\widetilde D) ) \ar[d]^\beta \\
H^0(\mathcal{O}_C(D))/H^0(\mathcal{O}_C) \ar@{^{(}->}[rr]^(.6)\alpha && H^0(\mathcal{O}_D(D))
}
\end{equation}
where $\alpha$ is the canonical map. In order to define the map $\beta$ precisely,
consider $\mathcal{O}_C(D)$ (respectively $\mathcal{O}_{\widetilde C}(\widetilde D)$) as a subsheaf of the rational function field $\mathcal{R}_C$ of the curve $C$ (respectively $\mathcal{R}_{\widetilde C}$ of the curve $\widetilde C$)
by putting for each point $p \in C$ and similarly for each $\widetilde p \in \widetilde C$,
$$
\mathcal{O}_C(D)_p := \mathfrak{m}_{C,p}^{-\nu_p(D)} \quad \mbox{and} \quad
	\mathcal{O}_{\widetilde C}(\widetilde D)_{\widetilde p} := \mathfrak{m}_{\widetilde C,\widetilde p}^{-\nu_{\widetilde p}(\widetilde D)}
$$
where $\mathfrak{m}_{C,p}$ respectively $\mathfrak{m}_{\widetilde C, \widetilde p}$ is the maximal ideal in
$\mathcal{O}_{C,p}$ respectively $\mathcal{O}_{\widetilde C, \widetilde p}$. Translating diagram \eqref{d5.2}
to $\mathcal{O}_C$ and $\mathcal{O}_{\widetilde C}$ gives the diagram
$$
\xymatrix{
&& \mathfrak{m}_{\widetilde C,\widetilde p}^{-1}/\mathcal{O}_{\widetilde C, \widetilde p} \oplus
\mathfrak{m}_{\widetilde C,\widetilde q}^{-1}/\mathcal{O}_{\widetilde C, \widetilde q} \oplus
\mathfrak{m}_{\widetilde C,\widetilde r}^{-1}/\mathcal{O}_{\widetilde C, \widetilde r} \ar[d]^\beta\\
H^0(\mathcal{O}_C(D))/H^0(\mathcal{O}_C) \ar@{^{(}->}[rr]^(.4)\alpha &&
\mathfrak{m}_{ C, p}^{-1}/\mathcal{O}_{ C, p} \oplus \mathfrak{m}_{ C, q}^{-1}/\mathcal{O}_{ C, q} \oplus
\mathfrak{m}_{ C, r}^{-1}/\mathcal{O}_{ C, r}
}
$$

where  $\beta$ is given by the transposition of the natural map
$$
\beta^{*}: \Omega^1_{C,p} / \mathfrak{m}_{C,p}^{1} \oplus  \Omega^1_{C,q} / \mathfrak{m}_{C,q}^{1} \oplus \Omega^1_{C,r} / \mathfrak{m}_{C,r}^{1}
\rightarrow \Omega^1_{\widetilde{C},\tilde p} / \mathfrak{m}_{\widetilde{C},\tilde p}^{1} \oplus  \Omega^1_{\widetilde{C},\tilde q} / \mathfrak{m}_{\widetilde{C},\tilde q}^{1} \oplus \Omega^1_{\widetilde{C},
\tilde r} / \mathfrak{m}_{C,r}^{1}
$$
induced by $f$. Let $t_p, t_q, t_r$ be local parameters around $p,q$ and $r$ respectively.
Then $ t_{\tilde p} := t^{\nu(p)} = t^2$ is a local parameter around  $\tilde p$, since $f$ is has
ramification index 2 at $\tilde p$.  We claim that $\beta^{*}$ restricted to the summand  $\Omega^1_{C,p} / \mathfrak{m}_{C,p}^{1} $ is
not injective, hence $\beta$ restricted to $\Omega^1_{\widetilde{C},\tilde p} / \mathfrak{m}_{\widetilde{C},\tilde p}^{1} $ cannot be surjective.
Indeed, if $dt_p$ is a generator of the one-dimensional space $\Omega^1_{C,p} / \mathfrak{m}_{C,p}^{1} $, then
$$
\beta^{*} (dt_p) = dt_{\tilde p} =  d(t_p^2) = 2t_p dt_p
$$
vanishes modulo $\mathfrak{m}_{C, p}^{1} $, therefore the restriction is zero. In particular, the restriction of  $\beta$
to $\Omega^1_{\widetilde{C},\tilde p} / \mathfrak{m}_{\widetilde{C},\tilde p}^{1} $ is zero, since the target is one-dimensional.

\bigskip

{\it (1) Suppose $h\circ f$ is general}.  In this case $f$ is \'etale at $\tilde q$ and $\tilde r$.
Let $t_{\tilde q}$ and   $t_{\tilde r}$ local parameters around $\tilde q$ and $\tilde r$.
We have that
$$
\beta ( t_{\tilde p}) =0,  \qquad \beta ( t_{\tilde q}) = t_q, \qquad \beta ( t_{\tilde r}) = t_r.
$$
Then $\beta $ is surjective onto the summands $\mathfrak{m}_{ C, q}^{-1}/\mathcal{O}_{ C, q} $ and $\mathfrak{m}_{ C, r}^{-1}/\mathcal{O}_{ C, r}$,
but not onto the first one, where $\beta$ vanishes.

The curve $Y$ is smooth at $\tilde D$ if and only if the composition of $\alpha$ with the cokernel of $\beta$ is surjective,
i.e. if and only if the map
\begin{equation} \label{surj-map1}
H^0(\mathcal{O}_C(D))/H^0(\mathcal{O}_C) \rightarrow \mathfrak{m}_{ C, p}^{-1}/\mathcal{O}_{ C, p}
\end{equation}
is surjective. As a vector space the target is generated by $t_p^{-1} $. Let  $\psi \in H^0(\mathcal{O}_C(D))$ be a section with corresponding
divisor $\bar D= \divi{\psi} +D $. Its image $\bar \psi$ in $\mathfrak{m}_{C,p}^{-1}/\mathcal{O}_{ C, p} $ can be written
as  $\bar \psi = c_k t^{-k}$ with  $c_k \neq 0. $ So the map \eqref{surj-map1} is surjective if such function exists with $k = 1$.
As $\nu_p(\bar D) = \nu_p (\psi) + \nu_p(D) = - k +  1 \geq 0$, it is enough to take a divisor $\bar D$ with $\nu_p(\bar D)=0$, i.e.
which does not contain $p$.  Therefore $Y$ is smooth at $\tilde D$.

\bigskip

{\it (1) Suppose $h\circ f$ is special}. In this case we can assume $p=p_1$ and $q=p_2$  are the branch points of $f$. Then
$$
\beta ( t_{\tilde p}) = 0, \qquad \beta ( t_{\tilde q}) =0, \qquad \beta ( t_{\tilde r}) = t_r
$$
and  $Y$ is smooth at $\tilde D$ if and only if the map
\begin{equation}\label{surj-map}
H^0(\mathcal{O}_C(D))/H^0(\mathcal{O}_C) \rightarrow \mathfrak{m}_{ C, p}^{-1}/\mathcal{O}_{ C, p} \oplus  \mathfrak{m}_{ C, q}^{-1}/\mathcal{O}_{ C, q}
\end{equation}
is surjective. Now the target space has as basis the vectors $(t_p^{-1}, 0 ), (0, t_q^{-1})$, so for the surjectivity one requires
the existence of a section $\psi \in H^0(\mathcal{O}_C(D))$ such that its corresponding divisor $\bar D$ satisfies
$$
\nu_p(\bar D) = \nu_p (\psi) + \nu_p(D) = -1 + 1 =0, \qquad  \nu_q(\bar D) = \nu_q (\psi) + \nu_q(D) = 0 + 1 =1,
$$
that is, a divisor $\bar D$ in the fibre of $h$ containing $q$ but not $p$, which gives a contradiction since the linear series $g^1_3$
defining $h: C \rightarrow \mathbb P^1$ is base-point-free\footnote{Assuming that $C$ is non-hyperelliptic}.

In consequence, $Y$ has 2 nodal singularities at the divisors $\tilde D_1^1 $ and $\tilde D_1^2 $ over $D=p_1 +p_2 +r$.
\end{proof}

\begin{prop} \label{genus}
Assume $h\circ f$ is a general, then $g_Y = 2g+1$.
\end{prop}

\begin{proof}
There are 2 types of ramification of $\widetilde k: Y \rightarrow \mathbb P^1$ for a branch point $a$ of
$\widetilde k$. Either $a$ is a
branch point of $h$ of ramification index $i = 1$ or 2. Then $a$ is a branch point for $h$ with
ramification of type $(2,2,1,1,1,1)$ if $i =1$ and $(3,3,1,1)$ if $i =2$.  Or $a = h(p_\nu)$ for
$\nu = 1$ or 2. Then  $a$ is a branch point for $h$ with ramification of type $(2,2,2,2)$.
Since $|R_h| = 2g+4$ and $p_1$ and $p_2$ are disjoint from $R_h$, this gives for the ramification divisor
$R_{\widetilde k}$ of $\widetilde k$:
$$
|R_{\widetilde k}| = 2(2g+4) + 2 \cdot 4 = 4g + 16
$$
and the result follows from the Hurwitz formula.
\end{proof}

We will see in Section 4, that  in the case of special coverings, $Y$ has arithmetic genus $2g+1$, more precisely,
it is the union of two curves of genus $g$ intersecting in two points.

\section{Donagi's extension of the trigonal construction}

The usual trigonal construction for \'etale double covers of smooth trigonal curves is the following theorem, due to Recillas
\cite{r}. We recall it for \'etale double covers $f$ of trigonal curves $C'$ of genus $g+1$
instead of genus $g$, since we need it in this case.

Let $\mathcal{R}_{g+1}^{tr}$ denote the moduli space of \'etale double covers of smooth
trigonal curves of genus $g+1$,  so it consists of triples $(C, \eta, g^1_3)$, where $\eta \in \Pic^0(C)[2] \setminus \{ 0\}$
(which defines the double covering over $C$)  and $g^1_3$ is a linear series on $C$. Let $\mathcal{G}^1_{g,4}$ denote the moduli
space of pairs $(C, g^1_4)$, with $C$ a smooth curve of genus $g$ together with a  $g^1_4$, and  $\mathcal{M}_{g,0}^{tet} \subset \mathcal{G}^1_{g,4} $
denote the open subspace of tetragonal curves $X$ of genus $g$ with the property that above each point of
$\mathbb P^1$ there is at least one \'etale point of $X$. In the sequel trigonal and tetragonal curves are understood as a pair of a curve with its linear series.
In (\cite{r}) Recillas showed

\begin{thm}
The trigonal construction gives an isomorphism
$$
T^0: \mathcal{R}_{g+1}^{tr} \rightarrow \mathcal{M}_{g,0}^{tet}.
$$
Moreover, for each $[h \circ f] \in \mathcal{R}_{g+1}^{tr}$, $T^0$ induces an isomorphism of principally polarized
abelian varieties
$$
\Pr(f) \simeq J(T^0(h \circ f)).
$$
\end{thm}

\begin{rem}
Recall that  curves of genus $g\geq 5$ (respectively  $g\geq 7$) admit at most  a $g^1_3$  (respectively a  $g^1_4$) and they form
a closed subset in the moduli of smooth curves $\mathcal{M}_g$.  So for high genus one can identify the moduli  space  $ \mathcal{M}_{g,0}^{tet}$
with its corresponding closed subspace in $\mathcal{M}_g $ and something similar holds for $\mathcal{R}_{g+1}^{tr}$.
On the other hand, any curve of genus $g\leq  4$ (respectively  $g\leq 6$)  possesses a $g^1_3$ (respectively  a $g^1_4$) and in this case
the isomorphism $T_0$ sends triples $(C, \eta,  g^1_3)$ to pairs $(X, g^1_4)$.
\end{rem}

In \cite{d} Donagi extended the map $T^0$ to the partial compactification
consisting of admissible double covers of trigonal curves of genus $g+1$, whose Prym variety is an element in $\mathcal{A}_g$, the moduli
space of principally polarized abelian varieties of dimension $g$.
Recall that an {\it admissible double cover} of a connected stable curve $C'$  is a double cover $\pi: \widetilde C' \rightarrow C'$  with $\widetilde C'$ stable such that
\begin{itemize}
\item the nodes of $\widetilde C'$ map under $\pi$ exactly to the nodes of $C'$;
\item away from the nodes $\pi$ is an \'etale double cover.
\end{itemize}
Let $\overline \mathcal{R}_{g+1}^{tr}$ denote the moduli space of admissible double covers of curves of
arithmetic genus  $g+1$, whose  Prym variety is in $\mathcal{A}_g$\footnote{In Donagi's work  these coverings are called ``allowable''.
Often the coverings with this property are also referred to as  ``admissible''. We use this notion.}
and let $\mathcal{M}_{g}^{tet}$ denote the moduli space of smooth tetragonal curves  of genus $g$.
Donagi showed in \cite[Theorem 2.9]{d},

\begin{thm}
The trigonal construction gives an isomorphism
$$
T: \overline \mathcal{R}_{g+1}^{tr} \rightarrow \mathcal{M}_{g}^{tet}.
$$
\end{thm}
Moreover, he showed that if $\widetilde C' \stackrel{f'}{\rightarrow} C' \stackrel{h'}{\rightarrow} \mathbb P^1$ corresponds to
the tetragonal curve $k': X' \rightarrow \mathbb P^1$ and $p$ is a point of $\mathbb P^1$, then the following types of fibers over $p$
of $f', k'$ and $h'$ correspond to each other,
\begin{enumerate}
\item $f',k'$ and $h'$ are \'etale;
\item $k'$ and $h'$ have simple ramification points, $f'$ is \'etale;
\item $k'$ and $h'$ each have a ramification point of index 2, $f'$ is \'etale;
\item $k'$ has two simple ramification points, the fibre of $h'$ consists of a simple node and a smooth
point, $f'$ is admissible;
\item $k'$ has a ramification point of index 3, the fibre of $h'$ consists of a node and at exactly one
branch of it $h'$ is ramified, $f'$ is admissible.
\end{enumerate}

In particular, if $h'$ admits nodes of types (4) or (5), the covering $f'$ is admissible. By \cite{b},
its Prym variety, defined again as the component of the norm map containing 0, is a
principally polarized abelian variety of dimension $g$, where $g+1$ is the arithmetic genus of
$C'$.

The {\it trigonal construction} is formally the same as in the smooth case.
Let $\widetilde C' \stackrel{f'}{\rightarrow} C' \stackrel{h'}{\rightarrow} \mathbb P^1$ be an element of
$ \overline \mathcal{R}_{g+1}^{tr}(2)$ and define $Y'$ in the same way as we defined $Y$ in diagram \eqref{d2.1}. The involution of $\widetilde C'$ induces an involution
on $Y'$ of which $X'$ is the quotient. (Donagi defines $X'$ as follows: The restriction
$f'^0: \widetilde C'^0 \rightarrow C'^0$ of $f'$ to the smooth parts can be considered as a local system on $C'^0$.
Its push-forward $h'_*  \widetilde C'^0$ to $\mathbb P^1$ admits an involution $\iota$. Then $X'$ is defined as the completion of the quotient $h'_*  \widetilde C'^0/\langle \iota \rangle$).

Conversely, the inverse $T^{-1}$ of $T$ is given as follows. Given $k': X' \rightarrow \mathbb P^1$ in
$\mathcal{M}_g^{tet}$, we denote by  $k'_0: X'_0 \rightarrow \mathbb P^1_0$ its restriction to  the open part consisting of fibres
of  types (1), (2) and (3). Let $\widetilde C'$ denote the closure of $ S^2_{\mathbb P^1}X'_0$,  the relative second symmetric product
of $X'_0$ over $\mathbb P^1$. Any fibre over a point
$p \in \mathbb P^1_0$ consists of all unordered pairs in $k'^{-1}(p)$ and $\widetilde C'$ admits an obvious
involution of which $C'$ is the quotient. Moreover, the cover $\widetilde C' \rightarrow \mathbb P^1$ factorizes via
maps $f': \widetilde C' \rightarrow C'$ and $h':C' \rightarrow \mathbb P^1$. The tower
$\widetilde C' \stackrel{f'}{\rightarrow} C' \stackrel{h'}{\rightarrow} \mathbb P^1$ is then an element of
$\overline \mathcal{R}_{g+1}^{tr}$.

The same proof as in the smooth case gives an isomorphism
\begin{equation} \label{e3.1}
\Pr(f') \simeq J(T(h' \circ f')).
\end{equation}

In the next section we need the following corollary. Let
$$
\mathcal{S}_{g+1}^{tr} \subset  \overline \mathcal{R}_{g+1}^{tr}
$$
the locally closed subset
of $ \overline \mathcal{R}_{g+1}^{tr}$ consisting of all elements of type (4), i.e. all admissible double
covers $f': \widetilde C' \rightarrow C'$ of
trigonal  curves $C'$ of arithmetic genus $g+1$ such that $C'$ admits exactly one node such that the
fibre of $h'$ containing the node contains also a smooth point.  Let
$\mathcal{M}_{g,1}^{tet}$ be the corresponding
subset of tetragonal covers $g': X' \rightarrow \mathbb P^1$ with exactly one fibre consisting of two simple ramification points and otherwise only fibres of types
(1), (2) and (3). So we get

\begin{cor} \label{c3.3}
Restricting the trigonal construction to $\mathcal{S}_{g+1}^{tr}$ gives an isomorphism
$$
T: \mathcal{S}_{g+1}^{tr} \rightarrow \mathcal{M}_{g,1}^{tet}
$$
inducing for every element of $\mathcal{S}_{g+1}^{tr}$ the isomorphism \eqref{e3.1}.
\end{cor}

\section{Proof of the main theorem for special ramified covers}

Let $C$ be a smooth trigonal curve of genus $g$ with trigonal cover $h:C \rightarrow \mathbb P^1$ and let
$f: \widetilde C \rightarrow C$ be a special ramified double cover, i.e. the two branch points of $f$ are
contained in a fibre of $h$. Suppose $p_1, p_2 \in C$ are the branch points  of $f$ and
$q_1, q_2 \in \widetilde C$ the corresponding ramification points. Let
$\mathcal{R} b^{tr}_{g,sp}$ denote the closed subset of the variety $\mathcal{R} b^{tr}_g$ consisting of special
covers.

Define new curves $\widetilde C'$ and $C'$ by identifying the points:
$$
\widetilde C' := \widetilde C/(q_1 \sim q_2) \quad \mbox{and} \quad C' := C/(p_1 \sim p_2).
$$
The covering $f$ induces a covering $f': \widetilde C' \rightarrow C'$ which clearly is an admissible double
cover. Similarly $h$ induces a trigonal cover $h': C' \rightarrow \mathbb P^1$, such that the tower
$\widetilde C' \stackrel{f'}{\rightarrow} C' \stackrel{h'}{\rightarrow} \mathbb P^1$ is an element of $\mathcal{S}_{g+1}^{tr}$.
So we get a map
$$
n: \mathcal{R} b^{tr}_{g,sp} \rightarrow \mathcal{S}_{g+1}^{tr}.
$$
\begin{lem} \label{l4.1}
The map $n: \mathcal{R} b^{tr}_{g,sp} \rightarrow \mathcal{S}_{g+1}^{tr}$ is an isomorphism.
\end{lem}
\begin{proof}
Normalization gives the inverse map.
\end{proof}

\begin{lem} \label{l4.2}
Let $\widetilde C \stackrel{f}{\rightarrow} C \stackrel{h}{\rightarrow} \mathbb P^1$ be an element of  $\mathcal{R} b^{tr}_{g,sp}$
and $\widetilde C' \stackrel{f'}{\rightarrow} C' \stackrel{h'}{\rightarrow} \mathbb P^1$ the corresponding element of
$\mathcal{S}_{g+1}^{tr}$. Normalization induces the following isomorphism of principally polarized
abelian varieties
$$
N: P(f') \rightarrow P(f).
$$
\end{lem}

\begin{proof}
According to the proof of \cite[Proposition 3.5]{b} the norm map induces the following commutative diagram with exact rows and columns
$$
\xymatrix{
& 0 \ar[d] & 0 \ar[d] & 0 \ar[d] &\\
0 \ar[r] & \mathbb Z/2 \ar[r] \ar[d] & P(f') \times \mathbb Z/2 \ar[r] \ar[d] & P(f) \ar[r] \ar[d] & 0\\
0 \ar[r] & \mathbb C^* \ar[r] \ar[d]_\Nm & J\widetilde C' \ar[r] \ar[d]_\Nm & J\widetilde C \ar[r] \ar[d]_\Nm & 0\\
0 \ar[r] & \mathbb C^* \ar[r] \ar[d] & JC' \ar[r] \ar[d] & JC \ar[r] \ar[d] & 0\\
& 0 & 0 & 0 &
}
$$
This gives the assertion.
\end{proof}

Combining Corollary \ref{c3.3} with Lemma \ref{l4.1} and equation \eqref{e3.1} with Lemma
\ref{l4.2} gives the following theorem

\begin{thm} \label{thm4.3}
The map
$$
T \circ n:  \mathcal{R} b^{tr}_{g,sp} \rightarrow  \mathcal{M}_{g,1}^{tet}
$$
is an isomorphism.

Moreover, if $\widetilde C \stackrel{f}{\rightarrow} C \stackrel{h}{\rightarrow} \mathbb P^1$ is an element of  $\mathcal{R} b^{tr}_{g,sp}$
and $X'$ is the corresponding (smooth) tetragonal cover, then we have an isomorphism of principally polarized abelian
varieties
$$
P(f) \rightarrow JX'.
$$
\end{thm}

\begin{rem} \label{rem4.4}
\begin{enumerate}
\item[(i)]
In \cite{d}, Donagi outlines also explicitely the inverse map $T^{-1}$. Since the inverse of the map
$n$ is clear, this gives an explicit version of the inverse map of Theorem \ref{thm4.3}.
\item[(ii)]
One can also express $X'$ as the Prym variety of an \'etale double cover of $X'$. In fact, the double
cover $Y' \rightarrow X'$ which occurred in Corollary  \ref{c3.3} is trivial. Hence $JX' = P(Y'/X')$.
\end{enumerate}
\end{rem}

\vspace{1cm}

Finally, we shall describe the covering $Y \rightarrow X$ with $X$ and $Y$ defined as in Section 2. As a consequence, it shows that $P(Y/X)$
is an abelian variety of the right  dimension.

Let $D := p_1 + p_2 +r$ be the unique divisor in $g^1_3$ containing the 2 branch points of $f$.
Denote by
$$
\widetilde k:= k \circ \pi:  Y \rightarrow \mathbb P^1
$$
the $8:1$ covering of diagram \eqref{d2.2}.
If $r_1$ and $r_2 = \iota(r_1)$ are the 2 points of $f^{-1}(r)$, then
$$
D_1 := q_1 + q_2 +r_1  \in Y  \quad \mbox{and} \quad D_2 := q_1 + q_2 +r_2 \in Y
$$
are exactly the 2 points of $\widetilde k^{-1}(h(p_1))$.

\begin{lem}
The curve $Y$ consists of $2$  smooth irreducible components
$$
Y = Y_1 \cup Y_2
$$
with
$$
Y_2 = \iota(Y_1), \quad \quad Y_1 \cap Y_2 = \{D_1,D_2\} \quad \mbox{and} \quad \iota(D_1) = D_2.
$$
\end{lem}

\begin{proof}
According to Lemma \ref{l2.2} the cover $\omega: O \rightarrow \mathbb P^1$ is branched exactly at $h(p_1) =
h(p_2)$. Since $\mathbb P^1$ is simply connected, this implies that $O$ consists of 2 components
$O = O_1 \cup O_2$ with $O_i \simeq \mathbb P^1$ intersecting in the point lying over $h(p_i)$.

It follows that $Y = \psi^{-1}(O)$ consists of at least 2 irreducible components. But it cannot consist
of more than 2 components, since the open dense part $\widetilde k^{-1}(\mathbb P^1 \setminus h(p_1))$
is isomorphic to an open dense part of the cover $Y' \rightarrow X'$ of Remark \ref{rem4.4}, which consists of exactly 2 components $Y_1$  and $Y_2$.

Clearly $Y_1$ and $Y_2$ are smooth with $Y_1 \cap Y_2 = \{D_1,D_2\}$ and $\iota(Y_1) = Y_2$.
\end{proof}

\begin{cor}
The double cover
$$
Y \rightarrow X:= Y/\iota
$$
is a Wirtinger cover in the sense of \cite[Example 1.9,(I)]{d}. In particular its Prym variety $P(Y/X)$
is an abelian variety.
\end{cor}

\begin{proof}
This follows from the fact that $\iota:Y \rightarrow Y$ is fixed-point free by Lemma \ref{fixedfree}
and $\iota(D_1)=D_2$.
\end{proof}

\begin{lem} We have
$$
p_a(Y) = 2g+1 , \qquad p_a(X) = g+1.
$$
and hence $\dim P(Y/X) = p_a(Y) - p_a(X) = g$.
\end{lem}

\begin{proof}
For $i = 1$ and 2 the map $\psi_i:= \psi|Y_i: Y_i \rightarrow O_i = \mathbb P^1$ is a 4-fold cover, where we
identify $O_i$ with $ \mathbb P^1$ via the map $\omega$. The  map $\psi_i$ can be considered as the map
$\widetilde k_i := \widetilde k|Y_i: Y_i \rightarrow \mathbb P^1$.

We then have
$$
p_a(Y) = 2g(Y_i) +1
$$
and it suffices to show that $g(Y_i) = g$. Then also $p_a(X) = g+1$, since $X$ admits exactly 1 node and $Y_i$ is the normalization of $X$.

There are 2 types of ramification of $\widetilde k_i: Y_i \rightarrow \mathbb P^1$ for a branch point $a$ of
$\widetilde k_i$. Either $a$ is a
branch point of $h$ of ramification index $i = 1$ or 2. Then $a$ is a branch point for $\widetilde k_i$
with ramification of type $(2,1,1)$ if $i =1$ and $(3,1)$ if $i =2$.

Or $a = h(p_1) = h(p_2).$ Then  $a$ is a branch point for $\widetilde k_i$ with ramification of type $(2,2)$, i.e. $\widetilde k_i^{-1}(a) =
\{2 \widetilde D_1, 2 \widetilde D_2\}$.

Since $|R_h| = 2g+4$ and $p_1$ and $p_2$ are disjoint from $R_h$, this gives for the ramification divisor $R_{\widetilde k_i}$
of $\widetilde k_i$:
$$
|R_{\widetilde k_i}| = 2g+4 + 2  = 2g + 6.
$$
and the Hurwitz formula gives $g(Y_i) = g.$
\end{proof}

\section{The main theorem for general covers}

Let $C$ be a smooth trigonal curve of genus $g$ with trigonal cover $h: C \rightarrow  \mathbb P^1$ and let
$f: \widetilde C \rightarrow C$ be a general ramified double cover branched over 2 points $p_1,p_2 \in C$,
i.e. $p_1$ and $p_2$ do not lie in a fibre of $h$ and are disjoint from the ramification locus of $h$.
Let $Y$ be the curve defined by diagram \eqref{d2.1}. According to Propositions \ref{conn}, \ref{p5.1} and  \ref{genus},
$Y$ is smooth and irreducible of genus $2g+1$.
According to Proposition \ref{fixedfree} it admits a fixed-point free involution $\iota$. As above let
$\pi: Y \rightarrow X$ denote the corresponding \'etale double cover with diagram \eqref{d2.2}.
Let $(P(\pi),\Xi_\pi)$ denote the corresponding principally polarized Prym variety of dimension $g$.
Let $\mathcal{R} b^{tr}_{g,gen}$ and $\mathcal{R}_{g+1}^{tet}$ the moduli spaces as defined in the introduction.

\begin{thm} \label{thm5.1}
 There is a canonical map
$$
\mathcal{R} b^{tr}_{g,gen} \rightarrow  \mathcal{R}_{g+1}^{tet}.
$$
If $\widetilde C \stackrel{f}{\rightarrow} C \stackrel{h}{\rightarrow} \mathbb P^1$ is an element of  $\mathcal{R} b^{tr}_{g,gen}$
and $\pi: Y \rightarrow X$ the corresponding \'etale double cover, then the principally polarized abelian varieties
$(P(f), \Xi_f)$ and $(P(\pi), \Xi_\pi)$ are canonically isomorphic.

\end{thm}

\begin{proof} {\bf Step 1}: {\it The set up.}

Choose a point $y_0 \in Y$ and let $\widetilde \alpha = \widetilde  \alpha_{y_0}: \widetilde C^{(3)}\rightarrow J\widetilde{C}$ and
$\alpha = \alpha_{f(y_0)}: C^{(3)} \rightarrow JC$ be the corresponding Abel-Jacobi maps. Then the following
diagram
is commutative
\begin{equation} \label{d6.1}
\xymatrix{
Y \ar@{^{(}->}[r] & \widetilde C^{(3)} \ar[r]^{\widetilde \alpha} \ar[d]_{f^{(3)}} & J\widetilde C \ar[d]^{\Nm_f}\\
& C^{(3)} \ar[r]^\alpha & JC
}
\end{equation}
This shows that
$$
\widetilde \alpha(Y) \subset \ker \Nm_f = P(f),
$$
since $\alpha$ maps $f^{(3)}(Y) = g_3^1 = \mathbb P^1$ to 0 in $JC$.
Denote by $\varphi$ the restriction $\widetilde \alpha|Y$ as a map into $P(f)$,
$$
\varphi := \widetilde \alpha|Y: Y \rightarrow P(f).
$$
For any
$y = \sum_{i=1}^3 \widetilde c_i \in Y$ we have
\begin{eqnarray*}
(\varphi + \varphi \iota)(y)& = &  \mathcal{O}_Y\left( y
+ \iota (y) -2y_0  \right)\\
& =& \mathcal{O}_Y\left( \sum_{i=1}^3 \widetilde c_i
+ \sum_{i=1}^3 \iota(\widetilde c_i) -2y_0  \right) = \mathcal{O}_Y(f^*g_3^1 -2y_0).\\
\end{eqnarray*}
But this is a constant in $P(f)$. So replacing $\alpha$ by a suitable translate we may assume that we have
$$
\varphi \iota =  - \varphi.
$$
According to the universal property for Prym varieties (valid also for ramified double covers, see
\cite[12.5.1]{bl}) the map $\varphi$ factorizes via the Abel-Prym map $\psi: Y \rightarrow P(\pi)$, i.e. the following diagram is commutative
$$
\xymatrix{
Y \ar[r]^\varphi \ar[d]_\psi & P(f) \ar[d]^{t_{-\varphi(y_0)}}\\
P(\pi) \ar[r]^{\overline \varphi} & P(f)
}
$$
where $\overline \varphi$ denotes the extension of $\varphi$ and the right vertical map the translation by $-\varphi(y_0)$.\\

{\bf Step 2}: {\it We claim that for the proof of Theorem \ref{thm5.1} it suffices to show that}
\begin{equation} \label{e6.1}
\varphi_*[Y] = \frac{2}{(g-1)!} \wedge^{g-1}[\Xi_f] \quad \mbox{in} \quad H^{2g-2}(P(f),\mathbb Z).
\end{equation}

\begin{proof}
Note first that $\varphi(Y)$ generates $P(f)$ as an abelian variety. This implies that
$\overline \varphi$ is an isogeny, $P(\pi)$ and $P(f)$ being of the same dimension. According to
\cite[Welters' criterion 12.2.2]{bl} we have
$$
\psi_*[Y] = \frac{2}{(g-1)!} \wedge^{g-1}[\Xi_\pi] \quad  \mbox{in} \quad  H^{2g-2}(P(\pi),\mathbb Z).
$$
Since we assume \eqref{e6.1}, the commutativity of the diagram gives
$$
\wedge^{g-1} [\Xi_\pi] = \wedge^{g-1} [\Xi_f].
$$
But then \cite[Lemma 12.2.3]{bl} implies that $\overline \varphi$ is an isomorphism of polarized
abelian varieties.
\end{proof}

{\bf Step 3}: {\it Let $i_{P(f)}: P(f) \hookrightarrow J \widetilde C$ denote the canonical embedding.
Then}
$$
\widehat i_{P(f)*} \widetilde \alpha_* [Y] = \frac{8}{(g-1)!} \wedge^{g-1} [\Xi_f].
$$

The proof applies the formula of Macdonald for the class of the curve
$g_3^1$ in the variety $C^{(3)}$ in $H^4(C^{(3)},\mathbb Z)$ (see \cite[Lemma 8.3.2]{acgh}):
$$
[g_3^1] = \sum_{k=0}^2 {2-g \choose k} \eta^k \cdot \frac{\wedge^{2-k}[\alpha^*\Theta]}{(2-k)!} \quad \mbox{in}
\quad H^4(C^{(3)},\mathbb Z)
$$
where $\eta^k \in H^{2k}(C^{(3)},\mathbb Z)$ denotes the fundamental class of $C^{(3-k)}$
in $C^{(3)}$ under the
embedding $\sum_{k=1}^{3-k} p_i \mapsto \sum_{k=1}^{3-k} p_i+ kp$ with some fixed point $p \in C$.
Similarly we define the classes $\widetilde \eta^k \in H^{2k}(\widetilde C^{(3)},\mathbb Z)$.
It is well known that $\eta^k = \wedge^k \eta^1$ and $\widetilde \eta^k = \wedge^k \widetilde \eta^1$ and both are related by
$$
{f^{(3)}}^* \eta^k = 2^k \widetilde \eta^k.
$$
Applying this to Macdolnald's formula, we get with the commutativity of diagram \eqref{d6.1}
\begin{equation} \label{e6.3}
[Y] = {f^{(3)}}^*[g_3^1] = \sum_{k=0}^2 {2-g \choose k}2^k \widetilde \eta^k \cdot \frac{\widetilde \alpha^*
\Nm_f^*\wedge^{2-k}[\Theta]}{(2-k)!} \quad \mbox{in} \quad H^4(\widetilde C^{(3)},\mathbb Z).
\end{equation}
Now Poincar\'e's formula \cite[11.2.1]{bl} in this case says
$$
\widetilde \alpha_* \widetilde \eta^k = \frac{\wedge^{2g+k-3} [\widetilde \Theta]}{(2g+k-3)!}.
$$
Applying $\widetilde \alpha_*$ to \eqref{e6.3}, Poincar\'e's formula and the projection formula give
\begin{equation} \label{e6.4}
\widetilde \alpha_*[Y] = \sum_{k=0}^2 {2-g \choose k} 2^k \frac{\wedge^{2g+k-3} [\widetilde \Theta]}{(2g+k-3)!} \cdot
\frac{ \Nm_f^*\wedge^{2-k}[\Theta]}{(2-k)!}
\quad \mbox{in} \quad H^{4g-2}(J\widetilde C,\mathbb Z).
\end{equation}
According to \cite[Proposition 12.3.4]{bl} we have
$$
2 [\widetilde \Theta] = \Nm_f^*[\Theta] + \widehat \iota_{P(f)}^*[\Xi_f].
$$
 Applying this and the binomial formula to \eqref{e6.4}, we get
\begin{equation} \label{e6.5}
\widetilde \alpha_* [Y] = 2^{3-2g} \sum_{k=0}^2 {2-g \choose k} \sum_{j=0}^{2g+k-3}
{2g+k-3 \choose j} \frac{\Nm_f^* \wedge^{j+2-k}[\Theta]}{(2-k)!} \cdot \frac{\widehat \iota_{P(f)}^* \wedge^{2g+k-3-j}[\Xi_f]}{(2g+k-3)!}
\end{equation}
Now we claim that
$$
\widehat \iota_{P(f)_*} \Nm_f^* \bigwedge^n \frac{[\Theta]}{n!} = \left\{ \begin{array}{ccl}
                                                                                2^{2g} & if & n= g,\\
                                                                                0   &if& 0\leq n \leq g-1.
                                                                                \end{array}  \right.
$$
For this consider the composed map
$$
\widehat  \iota_{P(f)_*} \circ \Nm_f^*: H^{2n}(JC,\mathbb Z) \rightarrow H^{2n}(J \widetilde C,\mathbb Z) \rightarrow H^{2n-2g}(P(f),\mathbb Z).
$$
By degree reasons, $\widehat  \iota_{P(f)_*} \Nm_f^* \equiv 0$ for $n \neq g$. For $n=g$ note that
$\widehat \iota_{P(f)}$ is multiplication by 2 on $P(f)$. So we get, denoting by $[0]$ the class of the point 0 in $JC$,
$$
\widehat  \iota_{P(f)*} \Nm_f^* \frac{\wedge^{g}[\Theta]}{g!} =
\widehat  \iota_{P(f)*} \Nm_f^* [0] = \widehat  \iota_{P(f)*} [P(f)] = \deg (2_{P(f)}) = 2^{2g}.
$$
Inserting this into equation \eqref{e6.5} we get
\begin{eqnarray*}
\widetilde \iota_{P(f)*} \widetilde \alpha_* [Y] & =& 2^{3-2g} \sum_{k=0}^2 {2-g \choose k} {2g+k-3 \choose g+k -2}
\frac{2^{2g} g!}{(2-k)!} \cdot \frac{\wedge^{g-1} [\Xi_f]}{(2g+k-3)!}\\
&=& \frac{2^{3}}{(g-1)!} \wedge^{g-1} [\Xi_f] \sum_{k=0}^2   {2-g \choose k}
{g \choose 2-k}\\
&=& \frac{2^{3}}{(g-1)!} \wedge^{g-1} [\Xi_f],
\end{eqnarray*}
since $\sum_{k=0}^2   {2-g \choose k} {g \choose 2-k}= \frac{1}{2}g(g-1) + (2-g)g + \frac{1}{2} (2-g)(1-g) = 1$, which is also the coefficient of
$x^2$ in the product $(1+x)^{2-g}(1+x)^g$.

{\bf Step 4}: Now $\alpha_*(Y) \subset P(f)$ and $\widehat \iota_{P(f)}$ restricted to $P(f)$ is multiplication by 2, so
$$
\widetilde \iota_{P(f)*} \widetilde \alpha_* [Y] = 2_{P(f)*}\widetilde \alpha_*[Y] = 4 \widetilde \alpha_*[Y] = \frac{8}{(g-1)!} \wedge^{g-1} [\Xi_f]
\quad \mbox{in} \quad H^{2g-2}(P(f),\mathbb Z).
$$
Since $H^{2g-2}(P(f),\mathbb Z)$ is torsion free, we obtain equation \eqref{e6.1}. This completes the proof of Theorem \ref{thm5.1}.
\end{proof}
\bigskip

\begin{prop}  \label{p5.2}
Let  $\mathcal{R}_{g+1,2}^{tet}$ be the locus of double coverings $[Y \rightarrow X] \in \mathcal{R}^{tet}_{g+1}$ such that the tetragonal covers $k: X \rightarrow \mathbb P^1$ have
exactly two fibres consisting of two simple ramification points and otherwise  only fibres containing at least one smooth point of $X$.
 Then the image of the canonical map of Theorem \ref{thm5.1} is contained in $\mathcal{R}_{g+1,2}^{tet}$.
\end{prop}

\begin{proof}
According to the proof of Proposition \ref{genus} the fibres over $h(p_\nu)$, the images of the 2 branch points of $f$, are the
only ones of type $(2,2)$ in the corresponding tetragonal cover
$k:X \rightarrow \mathbb P^1$.
\end{proof}

\begin{rem} \label{r5.3}
According to Theorem \ref{thm5.1} and Proposition \ref{p5.2} the trigonal construction in the general ramified case is given by the map
$$
\mathcal{R} b^{tr}_{g,gen} \rightarrow \mathcal{R}_{g+1,2}^{tet}.
$$
Both moduli spaces are of dimension $2g+3$.
It would be interesting to compute the exact image as well as the degree of this map. We hope to come back to this question.
\end{rem}

\begin{rem}
During the preparation of this paper we came across the article \cite{da}, where it is claimed that Prym variety of a double
covering of a
trigonal curve ramified at  two points is isomorphic as ppav to the Jacobian of certain tetragonal  curve $X$. The author constructs
a tower of curves,  which links the curves in the double covering and $X$.
It seems to us that in \cite{da} all curves are supposed to be smooth, even in the case
of a special cover. So we do not know whether this construction agrees
with our construction, meaning that it gives the same map of the corresponding moduli spaces.  Certainly they
give isomorphic Prym varieties, but taking into account Donagi's tetragonal construction it is not clear
that the coverings are the same.
\end{rem}

redpajama_set_name: RedPajamaArXiv


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text:
York City cancels 45th annual Olde York Street Fair
Although the Olde York Street Fair was scheduled for mid-May, York City has canceled the event to stay ahead of growing concerns about the coronavirus.
The annual event was scheduled for May 10 to coincide with Mother's Day. And even though models predict COVID-19 cases will peak in two to four weeks, that doesn't mean the threat will vanish, said Philip Given, the city's acting director of community and economic development.
"At this point, there's a lot of uncertainty with when restrictions will be lifted at the federal, state and local levels," Given said. "So just because of the logistics surrounding an event of this size, it's just in everybody's best interest."
More:York City to enforce Supreme Court-mandated halt on evictions
More:COVID-19 Pa. cases jump by nearly 1,000 in a day, 11 new deaths
May 10 would have marked the 45th year for the event that brings food, games and handmade goods to downtown York.
Pennsylvania had 5,805 confirmed cases of COVID-19 and 11 new virus-related deaths as of noon Wednesday, representing by far the greatest single-day jump since the beginning of the outbreak, the state Department of Health reported.
There have been 79 confirmed cases in York County.

redpajama_set_name: RedPajamaCommonCrawl


Aww, he looks so sorry for himself and deflated. It's part of a dog's "job" to roll in inappropriate things.
Kathryn – Shhhh, don't encourage him!
Some are worse than others! But I think spaniels in general are really good at sneaking off & rolling in something nasty. I put Ricky (now 11 years old) into a icy cold stream once, because I didn't want whatever he rolled in to be in my car. Yucky yuck! Now I have a crate, so at least it would be confined to the crate & I wouldn't have to put him in the stream again….
Boy did he look miserable. Didn't connect the dots either as to why this was happening….


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text:
Giving girls more opportunities to compete and create sporting memories with their teammates
Sisters n Sport Competitions
Sisters n Sport believe through taking part in sporting competitions with our teammates ....... and gain many memories on the way - which helps us fall in love with our sport even more and mean we continue to keep playing long after we leave school.
Queen Ethelburga's senior netball team thoroughly enjoyed competing in the Sister n Sport Competition . It gives school teams the opportunity to play schools they would never usually play offering different skills, styles and flair, while making new friends along the way. The structure of the competition is great because it is 'Knock-out' which is both challenging, contesting and rewarding while building experience. "I wouldn't have it any other way" We made it through to the semi-finals, which was a great achievement and WILL be entering this competition again next year. Melissa Bessell - Queen Ethelburga's & Severn Stars NSL Head Coach
Over 750 teams entered this year and growing
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text:
Dynamic of accumulation and redistribution of various carbohydrate forms and nitrogen in organs of tomatoes under treatment with retardants
V. V. Rogach Vinnytsia Mykhailo Kotsiubynskyi State Pedagogical University http://orcid.org/0000-0002-8916-8349
O. O. Kravets Vinnytsia Mykhailo Kotsiubynskyi State Pedagogical University http://orcid.org/0000-0003-1977-1617
O. I. Buina Vinnytsia Mykhailo Kotsiubynskyi State Pedagogical University http://orcid.org/0000-0003-2506-648X
V. G. Kuryata Vinnytsia Mykhailo Kotsiubynskyi State Pedagogical University http://orcid.org/0000-0002-7801-933X
Keywords: Solanum lycopersicum; antigibberellin compounds; leaf apparatus; sugar; starch; total nitrogen; productivity
The regulation of plant growth and development under application of antigibberellin compounds is accompanied by changes in the donor-acceptor system that in turn affects the accumulation and redistribution of carbohydrates and nitrogen-containing compounds. We studied the influence of tebuconazole 0.025% (EW-250) and 0.05% and 0.07% esphon (2-CEPA) on growth processes, leaf apparatus and content of various carbohydrate forms and total nitrogen content in vegetative organs of tomato hybrids Bobkat and Solerosso. A field-based micro-trial setup was established from 2013 to 2017. The plants were treated via foliar spraying with aqueous solution of retardants at the time of initiation of budding. Phytometric measurements were determined every 10 days, the content of various carbohydrate forms and total nitrogen in organs of plants was determined by using biochemical methods in a fixed dry material (the content of sugars and starch – by iodometric method, and the total nitrogen content – by Kjeldahl). It was found that both retardants reduced the linear growth of tomato plants. The results indicate that 2-CEPA treatment significantly decreased the plant height of both hybrids. Application of drugs increased the number of leaves per plant. The leaf dry matter weight increased, the most significant effect was created by EW-250. The leaf area of the retardant-treated tomatoes hybrid Solerosso increased and decreased in the hybrid Bobcat. Application of antigibberellin compounds intensified the outflow of sugars and remobilization of starch in roots and stems of both tomato hybrids. The content of starch, reducing and non-reducing sugars in leaves of hybrid Solerosso was higher after EW-250 treatment. 2-CEPA increased the content of reducing form of sugars, and EW-250 increased non-reducing form in the hybrid Bobkat. EW-250 intensified the accumulation of carbohydrates in fruits due to reducing sugars in both tomato hybrids, whereas 2-CEPA reduced their content compared with control. Growth inhibitors increased the starch content in fruits at the beginning of vegetation and intensified its reutilization at the end. This process was most intensive was in variant of EW-250-treated plants regardless of hybrids. We proved that there was a clear tendency to decrease the ratio of sugar content to starch in almost all organs of both tomato hybrids during the growing season. The most significant was the reduction of this ratio after EW-250 treatment. At the fruitification phase, the proportion of starch in retardant-treated plants increased compared to the control. The increased accumulation of carbohydrates in tomato fruits treated with EW-250 caused an increase in the average weight of the fruit. We have found that the total nitrogen content in vegetative organs decreased in the control and in the experimental variant of both tomato hybrids during vegetation. EW-250 accelerated the outflow of total nitrogen from roots and stems and increased its content in leaves. This drug retarded the accumulation of this element in fruit. Such changes in the dynamics of total nitrogen under the influence of tebuconazole resulted in enhanced formation of fruit on tomato plants. We concluded that the antigibberellin compounds EW-250 and 2-CEPA due the action on growth processes and leaf apparatus of tomato plants, changed the character of donor-acceptor relation through an increase in outflow of plastic substances from vegetative organs to fruits, which optimized the crop production.
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The hardest thing to do for me, at times, is to consider the price of silk. It is so very expensive!!! I usually can't wrap my mind around purchasing it. There are several sources out there that offer silk and about a year ago I purchased around 2 lbs of natural silk to dye. A brave move for me considering the price! Eeek! I knew I had a pattern and I knew what colors I wanted......now to make the next step and get the project started! I selected four hanks to be dyed and Chris and I got busy dying them in a variegated blue and a variegated green keeping them monochromatic since I was going to be using an echo weave pattern.
Once the ends were counted out, the loom was dressed and I was happily weaving. I wove two with a solid black weft......the weft was not dyed here in the studio, it was commercially dyed.


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Raising Dion Fans Just Got The Worst News
By Mike Bedard/April 26, 2022 5:20 pm EST
Netflix is going through a bit of a slashing spree at the moment.
Amid reports that subscriber numbers are down, Netflix has canceled numerous projects. This includes "Archive 81" and "The Dark Crystal: Age of Resistance," which were both canceled after a single season. However, at least those TV series managed to get some episodes made. Netflix made heavy cuts to its animation sector, with the biggest casualty of the cuts being the long-gestating "Bone" TV series. The comic book creator, Jeff Smith, even took to Twitter to voice his frustrations with the various attempts to get "Bone" made.
While shows like "Stranger Things" and "The Witcher" are probably safe for the time being, fans of Netflix series have quite a lot to be concerned about right now. And it looks like another show has bitten the dust in the wake of this mass exodus. Per Variety, fans of "Raising Dion" have good reason to feel sad today as it's just been announced the superhero drama series was getting the ax after only two seasons.
Star Sammi Haney announced the news on Instagram
Season 2 of "Raising Dion" skyrocketed to the number one spot of the most-viewed TV series when it first debuted in February 2022. This was coming on the heels of a massively popular first season, which came out three years prior in 2019. Even with such an immense gap, it appears the show maintained a sizable fanbase, but that wasn't enough to save the show from Netflix's mighty blow.
News of the cancellation came from Sammi Haney, who plays Esperanza Jimenez in "Raising Dion." She posted the announcement on her Instagram profile, writing, "Sad to say that Raising Dion is CANCELED [sic] Thank you for all of the amazing support we got from all of our wonderful fans! Season 2 was a success, equal to Season 1, even if just looking at how many people watched all of it and wanted a season 3!"
However, Haney tried to look at the positives: "The only 'silver lining' is that this does free me up to pursue other roles. My Netflix contract kept me from accepting other major roles offered to me (1 actually written with me in mind) on other tv shows!" While the talented actors of "Raising Dion" will likely do all right for themselves, it's still disappointing news for everyone who had followed the superhero series from the beginning. And it's a stark reminder that in these unprecedented times, no show is safe on the streaming platform.


Paralympic World Cup. This international multi-sport event for top athletes with disability was held annually in Manchester, England, from 2005 to 2012. No World Cups were held in 2013 and 2014, and there is no indication that the event could be resumed in the near future.
Invictus Games. The world of disabled competitive sports might have lost a lot with the discontinuation of the Paralympic World Cup but in 2014, it gained another major event - the Invictus Games. This international multi-sport event for wounded/injured military personnel was hosted by the Queen Elizabeth Olympic Park in London.
Asian Para Games. Held for the first time in 2010, the Asian Para Games are the "successor" of the FESPIC Games (Far East and South Pacific Games for the Disabled) and as such, the central multi-sport event for disabled athletes in Asia.
Parapan American Games. Held every four years since 1999, the Parapan American Games are basically the Pan American Games for athletes with disabilities. In addition to competing for medals, athletes who participate in the Parapan American Games also compete to qualify for the Paralympic Games.
IBSA World Championship and Games. Organised by the IBSA (International Blind Sports Federation), this multi-sport event has been held every fourth year since 1998 with an aim to provide blind and partially sighted athletes with the opportunity to compete in a variety of sports.
Deaflympics. The longest running international multi-sport event for disabled athletes, the Deaflympics is an elite competition for deaf and hearing impaired athletes. It is held every four years.
Special Olympic World Games. The largest and most prestigious multi-sport event for children and adults with special needs is the flagship event of the Special Olympics. It has been held ever since 1968.
Disability World Cups and Championships. In addition in various multi-sport competitions, disabled athletes also compete in numerous world cups and championship such as the IPC Athletics World Championship, IPC Disabled Alpine Skiing World Cup, UCI Para-Cycling Track World Championship, INAS World Football Championship, IPC Powerlifting World Championship, World Para Table Tennis Championship, and many other events that are held at local, national and international levels.


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To the blind an unrequited love.
The Christian myth tells of a wrongness that has entered into creation. The ultimate cause of this wrongness is unknown. It would seem to have been a collective wrong choice of the human race. We see it manifest in three ways: In the hatred of Eros, in the will to dominate others, and in the will to claim for oneself a disproportionate share of the worlds resources. These are the knots that have entered into creation – the knots that must be unraveled if we are to survive. In the self these knots are created when love goes out into the world and there encounters hostility and rejection. This love then turns back into the self where it creates psychic entanglements that prevent the free flow of love between the self and others. This, for example is what creates what in psychoanalytic terms is called the "sadistic super-ego." Similar knots are found in society on all levels.
Death is a part of the creative process. We encounter it in personal death, in the death of our social forms, and perhaps in the death of the earth itself. The faith of the person who is practicing the yoga of politics is not grounded in the wisdom and power either of the self or of some revered leader or teacher, but in the Absolute. It is the faith that those things which have been truly good will be taken up and will serve as starting points for new creations.


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The Morning's Top 5 Pop Culture Stories
1. In the wake of its recent success at the Golden Globes, Downton Abbey creator Julian Fellows is said to be mulling over the idea of doing a feature film version of the TV series. Yes, please — provided that Maggie Smith is on board with the idea. [via The Mirror]
2. YouTube has announced plans to partner with the Venice Film Festival and Ridley Scott's production company on Your Film Festival, a new online film festival/global competition aimed at shorts that offers a $500,000 prize. [via ArtsBeat]
3. Rihanna and Coldplay will be performing "Princess of China" together live at next month's Grammy Awards; Chris Martin says that her appearance on the album is his favorite bit of Mylo Xyloto. [via MTV News]
4. According to the results of a new Harris Poll, Johnny Depp is America's favorite actor for the second year straight, followed by Denzel Washington and Clint Eastwood. How much do you want to bet that some studio exec is currently trying to figure out how to get them all in one movie together? [via Deadline]
5. Bruce Springsteen's "angry" new album has been appropriately titled Wrecking Ball, and is due out in the US on March 6th. Check out a full tracklist, as well as the cover art, here.
Bonus Buzz: 11 Bacteria Photos Presented As Abstract Art


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Q: FactorSquareFree::lrgexp: Exponent is out of bounds for function FactorSquareFree. Error I'm trying to make a two-dimensional contour plot of the temperature of a plate. This is my initial function:
T[x_, y_] :=
 (20/Pi) Sum[((-1)^(n + 1)/
    n) Exp[(-n Pi y)/10] Sin[(n Pi x)/10], {n, 1, i}]

The function has been acting how I expect it to, so I believe everything is correct with that.
When I tried to plot it, I used:
ContourPlot[T[x, y] /. i -> 1, {x, 0, 10}, {y, 0, 10}]

I keep getting the error FactorSquareFree::lrgexp: Exponent is out of bounds for function FactorSquareFree. This will display three times before the systems suppresses the error and then runs without output until I quit the kernel.
I tried to look at the FactorSquareFree function itself to see if I could figure out where it was going out of bounds. Doing that, I got:
FactorSquareFree[T[x, y] /. i -> 1]
-(1/Pi)
 10 I E^(-(1/5) I Pi (x + I y) -
   1/5 Pi (I x + y)) (E^((2 I Pi x)/5 +
     1/10 I Pi (x + I y)) - E^(
    1/5 I Pi (x + I y) + 1/10 Pi (I x + y)) +
    E^(1/5 I Pi (x + I y) + 1/5 Pi (I x + y))
      Log[1 + E^(-(1/10) I Pi (x - I y))] -
    E^(1/5 I Pi (x + I y) + 1/5 Pi (I x + y))
      Log[E^(-(1/10) I Pi (x - I y)) (1 + E^(
         1/10 I Pi (x - I y)))] -
    E^((2 I Pi x)/5) Log[1 + E^(1/10 I Pi (x + I y))] +
    E^(1/5 I Pi (x + I y) + 1/5 Pi (I x + y))
      Log[1 + E^(1/10 I Pi (x + I y))])

Looking through that, I don't see anywhere where the exponent should get very large running from {x,0,10} and {y,0,10}.
Everything I have looked up related to this problem has been too complicated for me to understand what solution people were giving. Especially since the issue is with an internal function for ContourPlot, I'm at a loss of what to change or try.
Thank you for your help!

redpajama_set_name: RedPajamaStackExchange


Integrated Device Technology (IDT) took its time to reveal one of the more interesting energy-saving technologies at IDF 2011.
During Mooly Eden's keynote address, IDT showcased its Panel Self Refresh (PSR) technology that the creators claim can potentially increase battery life by up to 45 minutes.
The panel works in conjunction with a specialized Timing Controller (TCON), which stores the current static image locally in an integrated frame buffer. During this time, the display will refresh the static image from the local frame buffer instead of from the GPU, so the GPU is allowed to power down to a more idle state, at least until the next onscreen movement occurs. Essentially, the power savings comes from eliminating some of the workload by the GPU altogether. To prove its point, IDT even detached the connection between the mainboard and the specialized TCON during a static image, only to still see the idled image on the showcased display.
As readers are probably predicting, the PSR technology will not be all that effective during gaming sessions or movie playback where images fly by quickly. It should, however, make a noticeable difference during web surfing or PowerPoint presentations when more static images are displayed throughout a time span.
Our newest PSR-ready eDP timing controller helps our customers meet these [longer battery life] requirements, making IDT's solution the ultimate high-performance, low-power interface for use in next generation notebooks, Ultrabooks, and tablets," said vice president Ji Park of IDT Video and Display Operations division in the press release.
Exactly when the technology will be implemented in laptops and its associated costs were not announced, but Intel is likely already a big supporter of the energy-saving technology, especially when it can potentially extend the battery life of future Ultrabooks.
SAN JOSE, Calif.--(BUSINESS WIRE)-- Integrated Device Technology, Inc. (IDT®) (NASDAQ:IDTI), the Analog and Digital Company™ delivering essential mixed-signal semiconductor solutions, today announced that it has developed and demonstrated the world's first Embedded DisplayPort (eDP) 1.3-compliant Timing Controller (TCON) with integrated Panel Self Refresh (PSR) technology for significant system-level power savings and extended battery life in portable applications.
IDT's new eDP timing controllers with PSR technology can save significant power when a static image is displayed on the screen of a system with a PSR-enabled graphics processor. Unlike traditional timing controllers that receive and process the display data continuously, IDT's PSR technology identifies a static image and stores it local to the TCON in an integrated frame buffer. It then seamlessly displays the image from the local frame buffer allowing the eDP main link and a majority of the GPU functions to be powered-down, resulting in significant power savings.
IDT's latest solution incorporates novel design techniques and strategic process selection for the most efficient operation and seamless transition between modes. It is fully compliant with the VESA eDP v1.3 standard. eDP solutions offer the lowest system cost and longest battery life versus the LVDS solutions currently being used in portable embedded display applications. The large reduction in power consumption is key to accelerating the adoption of eDP in portable displays and the new IDT solution extends this power advantage.
This new eDP technology is being demonstrated to qualified customers only. For more information, please contact your local IDT sales representative. To learn more about IDT's broad portfolio of display products, visit www.idt.com/go/display.


ECB day. Do we care? No, not really. Draghi has already pushed hard against the already open door of bank lending, and after March's meeting there is not much to say. However, recent caustic criticism of the ECB from German politicians may be raised.
Sweden also has a policy decision – unlike the Euro area the Swedish economy is a functioning monetary union. Like the Euro area it has negative rates. Negative rates are likely to impact the Swedish economy differently from the way negative rates impact the Euro area.
The UK is giving us retail sales figures. These might be more interesting in the future as the unemployment rate did increase very slightly, but overall we would still regard the UK consumer as being in a good place.
There is little to get really excited about in the US, with the Philly Fed business sentiment survey and initial jobless claims. Of course the US labour market is very strong and that is being reflected in the rise in wage growth, but weekly claims are not going to change perceptions that much.


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Julia Buckingham
Julia Buckingham Wonkhe
Julia Buckingham CBE is Universities UK's President and Vice Chancellor and President of Brunel University London.
Throughout her career Julia has combined research and education with supporting the broader aspects of academic life through work with the research councils, medical charities and learned societies. She has published widely in her field, served on numerous national and international review panels and received a number of prestigious awards and honours for her work; she was appointed to a CBE in 2018 for services to Biology and Education and elected to a Fellowship of the Academy of Medical Sciences in 2019.
She is currently a Director of Imperial College Health Partners, Chair of The Concordat Strategy Group, supporting the career development of researchers, and Chair of the Athena SWAN Review Steering Group.
Latest articles by Julia Buckingham...
The new Stepchange is an opportunity to renew our efforts on mental health
see all articles...


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Geddy Lee of Rush to perform with Yes at Rock Hall induction
UPDATE: The Rock And Roll Hall Of Fame has released a statement denying that Rush bassist Geddy Lee will be performing with Yes at their induction in New York next month following comments by singer Jon Anderson as reported below.
Read the updated news here.
Original Yes singer Jon Anderson says both he and Rush bassist Geddy Lee will perform with the band at their induction into the 2017 Rock And Roll Hall Of Fame in New York next month.
Anderson tells Billboard that Lee will join Yes for a performance of their 1971 classic, "Roundabout", in place of the late Chris Squire, who passed away in 2015 at the age of 67 from Acute Erythroid Leukemia (AEL).
"Chris is going to be there in spirit," says Anderson. "He always believed that we should have been in the Hall of Fame 10, 29 years ago. It wasn't high on my agenda; 'It happens when it happens' was my mantra. But we know Chris will be there in spirit, in energy. I loved the guy very much. Without him I wouldn't be here, that's for sure.
"And I can't believe it's nearly 50 years ago when me and Chris started the band. You think you're gonna be together two, three years maybe, if you're lucky, and here we are. Yes' music is still surviving."
Squire's widow, Scotland, will be on hand to represent him at the event.
Yes alumnus Billy Sherwood stepped in to fill Squire's role on bass since his passing.
The band will be inducted by Lee and Rush guitarist Alex Lifeson before the Rock Hall performance, which Anderson hints may also include 1971's "I've Seen All Good People" and the 1983 US No. 1 hit, "Owner Of A Lonely Heart."
The singer – who split from the progressive rock pioneers due to health issues in 2008 – will accept the honor alongside former bandmates Bill Bruford, Steve Howe, Trevor Rabin, Rick Wakeman, Alan White and Tony Kaye.
"It's a family. There's always animosity," explains Anderson. "People that you love you don't always like, and there's always going to be that. But when you're celebrating who you truly are, you forget about all that and just get on with meeting each other and seeing each other, and it's just one of those things. It's not a problem; we'll just get together and have fun. Music is a healing force on every level."
The April 7 ceremony at New York's Barclays Center will see Yes inducted as part of the Class of 2017 alongside Pearl Jam, Journey, Electric Light Orchestra, Joan Baez and Tupac Shakur.
Yes: Rick Wakeman updates plans for Rock Hall induction
Yes: Rick Wakeman comments on upcoming Rock Hall induction
Yes members comment on 2017 Rock Hall induction
Pearl Jam and Journey among 2017 Rock Hall Of Fame inductees
Search Yes at hennemusic
Labels: Rush, Yes


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At what point does data become valuable?
By Salvador Malo
By Salvador Malo • November 2, 2020
Across virtually every discipline, the world today is swimming in data. Its value, however, doesn't lie in the data itself but in what people do with it―in its interpretation. Surprisingly, this has always been the case, but the fact is so rarely emphasized that many people think of it as a modern challenge.
Consider the skies above us. The sun, moon, stars, and planets have circled overhead since long before humans evolved. And once people appeared on the scene, the relevant orbits were there for everyone to see, in what may be the most democratic data set that ever existed, to which nobody enjoyed privileged access, at least until telescopes were invented. However, for thousands of years humans repeatedly drew the wrong insights from this data. Plainly visible as it was, it provided little value until correctly interpreted by Copernicus.
Even when exploited to the point of providing what some would call "predictive analytics" (pre-Copernican models allowed to forecast solar eclipses quite well), the lack of a correct interpretation prevented the data from adding value beyond the obvious. Being able to predict the motions of heavenly bodies was of little use beyond, well … predicting the motions of the bodies.
The correct interpretation, on the other hand, is so valuable that it is now part of elementary school curricula (the earth turns about its axis, the moon revolves around the earth, the earth and planets around the sun, with the stars as distant background). Its value resides in its spillover effect, for science as a whole would have been held back for centuries in the absence of Copernican insight, which in turn hinges on how the motions of the various celestial objects are linked to one another.
In the context of HR analytics―an admittedly younger discipline than astronomy―we face similar situations. To be sure, patterns are now visible that were previously hidden, but the value to be drawn from those patterns is far from self-evident. Data becomes valuable only when it informs business decisions through correct interpretations of observed patterns.
Let us look at an example.
We know from an April 2020 survey that two of the three most frequently measured variables in HR analytics are Employee Engagement (mentioned by 90% of respondents) and Mobility & Attrition (mentioned by 83%). Companies usually want engagement to be high and want attrition―at least that of high-performing employees―to be low. But the two variables tend to be tracked and analyzed in isolation. Subtle but rich connections between the two are hard to quantify and therefore all too often overlooked. The problem with this approach is that it is like tracking the motions of two planets while neglecting the way in which those motions are linked.
Engagement―no matter how a company chooses to define it―is usually held to be inversely linked to attrition, yet the nature of this link is frequently glossed over. Six years ago, a global company with 120,000 employees in the networks and telecommunications industry decided to explore the nature of this link. Its HR analytics team segmented the company's 2013 "engagement index" data into 6 groups corresponding to teams with scores of 91-100, 81-90, 71-80, 61-70, 51-60, and below 50 (every team with at least six respondents received a score). Each segment contained several hundred data points, thus ensuring statistical significance if any link was found to other variables.
The first results were disappointing: engagement levels at the start of the year showed absolutely no correlation with attrition by year end. In hindsight, this shouldn't have been a surprise, since attrition is driven by many factors, including commuting distance, life stage, job role, compensation level, etc. If the analysis had found a straightforward link with engagement scores, it should have aroused suspicion.
But patience paid off. The following year, survey data corresponding to 2014 was processed. Again, the correlation between attrition and engagement was found to be nil. However, due to a business downturn, company-wide attrition had in the meantime increased by a few percentage points. What was interesting was that it had not increased by the same amount across all segments. For teams with an engagement index either below 50 or in the 51-60 range, it had jumped by 10 percentage points, a huge change for such a large company. But at higher engagement levels, attrition had increased less and less, until for teams with engagement scores of 91-100 it was nearly identical to the previous year's attrition. Moreover, the correlation between engagement level and increase in attrition was -0.92 (as good as it gets in this sort of analysis).
In summary, while the link between static (single-year) attrition and engagement levels may have been completely undetectable, there was a solid, quantifiable link between dynamic (year-over-year) changes in attrition and team engagement levels. Teams with an engagement score above 90 were "protected" against company-wide increases in attrition.
This was the point at which the data became valuable, for it justified a continued investment to keep engagement levels high (or increase them if low). And what was even more interesting, using rough but well-known estimates of the average cost of having to replace an employee, the analysis went further to show how many regrettable resignations would have been prevented by boosting every team's engagement to a score of at least 80. For the first time, a financial value had been placed on what had formerly been a rather "soft" HR metric: employee engagement.
Tweets by HRAnalyticsTT


TCDC seen performing above, led by our resident dance instructor, Trevor Miles.
Central Outreach Wellness Center parked their Outreach Van in our lot and encouraged everyone over the age of 18 to get STI tested for free! COWC is an LGBTQ+ health center with resources such as drug/alcohol rehabilitation, transgender hormone therapy, and STI testing and treatment, including HIV/AIDS testing and referral.
We welcomed the beautiful drag performers Christopher Eagleton, Orchid Bloomfield, Ace Phoenix, and Nikki Danielle to strut across our Baverso stage.
Persad Center: strengthening the region's LGBTQ communities and their allies.


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Rapper RZA says arrested A$AP Rocky is a hostage in Sweden
Sat., July 27, 2019timer1 min. read
BEVERLY HILLS, Calif. - RZA (RIHZ'-uh) of the Wu-Tang Clan says he's concerned over what he calls a hostage situation for fellow rapper A$AP Rocky, who's charged with assault in Sweden.
RZA told The Associated Press at an event Friday promoting the Hulu drama series "Wu-Tang: An American Saga" that he doesn't condone violence, but celebrities have to protect themselves, and Rocky is not dangerous or a criminal.
He says he's "disappointed in Sweden" and "disappointed that a judge cannot discern that this is not a man you hold hostage."
Swedish prosecutors say Rocky and two others attacked a man there. Rocky says he acted in self-defence. He's been in custody since July 3.
President Donald Trump has also spoken out against Rocky's detention.
RZA says if Trump helped, "that'd be a beautiful thing."


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Terry Morgan
Nobody puts Terry in a corner. That line has absolutely nothing to do with anything (besides the not-so-subtle 'Dirty Dancing' reference), but there was no other way to start this bio. An avid science fiction fan, Terry spends most of his time dreaming of escaping in a little blue box known as the TARDIS, creating his own Lightsaber, or giant fighting alien robots from the planet Cybertron. Surprisingly, he's actually married with children.
Two Vehicle Crash On I-84 Outside of Wendell
A two-vehicle crash involving a semi truck on I-84 west of Wendell hospitalized a mother and her infant.
Multiple Vehicle Accident Blocks Traffic on US 93
We have early reports of a multiple vehicle accident on US 93. The Idaho State Police, Twin Falls Police, and emergency crews on currently on the scene.
West Nile Virus Positive Pools Found In Twin Falls County
The mosquito-born West Nile Virus has been positively identified in Twin Falls County, in between Twin Falls and Kimberly.
Winner Announced for the Hot Dogs & Cool Cats Pet Photo Contest
It was truly a joy to see all of your pets chillaxing in the Valley View Veterinary Clinic Hot Dogs & Cool Cats Pet Photo Contest, but only one could win, and we let you decide. Find out who won!
Top 5 Tips For Watching The Annual Perseid Meteor Shower In Twin Falls
The annual Perseid Meteor Shower is going on now, but peaks tonight (Aug. 12)! This is one of those cool events that you can see with the unaided eye, and nearly everyone will be able to see a shooting star or two with enough time and patience.
Confirmed: The Universe Is Slowly Dying
Researchers have found that galaxies are losing energy at a an alarming rate, and confirmed that the Universe is dying.
Single Vehicle Rollover Accident on I-86
An 18 year old girl from Pocatello was involved in a rollover accident on Interstate 86 earlier today.
Vote for the Top 10 in the Hot Dog & Cool Cat Photo Contest! [Contest]
The entries are in and the Top 10 photos have been selected in the Valley View Veterinary Clinic Hot Dogs & Cool Cats Photo Contest. It's time for you to vote!
Get Ready For The Annual Perseid Meteor Shower This August
The annual Perseid Meteor Shower is going on now, but peaks August 12th! This is one of those cool events that you can see with the unaided eye, and nearly everyone will be able to see a shooting star or two with enough time and patience.
Pluto: Dwarf Planet and New Horizon
85 years ago, American astronomer Clyde Tombaugh discovered Pluto: a distant, frozen, tiny rocky planet that quickly became the 9th planet of our solar system. It held this position for 76 years until it was demoted to "Dwarf Planet" (or "Plutoid") in 2006.
Can't We Come Up With Something Better Than Chip Sealing Our Roads?
It's 2015. We just explored Pluto. Can't we come up with something better than Chip Sealing our roads?
#MCCASH2015 Instagram Gallery [Photos]
Check out this amazing collection of photos we and others have shared on Instagram during McCash for Kids. Photos can be added by anyone!
$25 Gift Certificate to Upper Crust Bistro


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Category Archives: "Newsletter"
Changing Documentation in the Medical Record
in Articles Clinical Documentation Excellence Newsletter
by Richard Schamp, MD, Chief Executive Officer
When documenting in the medical record, providers should take care to capture their thoughts and findings as completely as possible. Best practice is to document the patient encounter at the time, or shortly afterwards. Inevitably, situations require some changes in the medical record, of which four types are recognized. (more…)
New ICD-10 Guidelines for the Term "With"
in Articles Newsletter	| tagged: Capstone Performance Systems CMS Encounter Data Reporting System (EDRS)
by Katina Yablinsky, CRN-C
In October 2016, there was an update to the ICD-10 coding guidelines that allows a coder more flexibility in assigning certain codes. Previously, the guidelines stipulated that if two conditions are related to each other, the provider needs to document their relationship for them to be coded as such. (more…)
ON TARGET: Senior TLC
in Articles Newsletter
On Target recognizes a client who is showing process improvement from putting Capstone's processes and advice into action.
ICD-10 was the tipping point. CMS's implementation of the new edition of the coding system was the point of no return for Senior Total Life Care (STLC), a PACE organization serving 175 participants in a rural area less than 20 miles outside of Charlotte, North Carolina. (more…)
Dr. David Wilner Joins Capstone Performance Systems
in Newsletter Press Releases	| tagged: Capstone Performance Systems David Wilner MD
David Wilner, MD, FACP, AGSF, has joined Capstone Performance Systems as senior medical consultant. A clinical professor at University of Massachusetts Medical School, Dr. Wilner previously served as medical director and vice president of Summit ElderCare and Fallon Health and as coordinator of geriatric medicine training at St. Vincent's Hospital in Worcester, Mass. He is a past member of the National Pace Association (NPA) board of directors and former chairman of the NPA's…
Medical Necessity and Internal Auditing Compliance
by Richard Schamp, M.D., C.M.D., C.H.C.Q.M., Chief Executive Officer
Several times in 2016, my comments in this column have emphasized elements related to corporate compliance with regulations and rules governing documentation and coding. As 2016 comes to a close, I wish to share two key messages from a recent compliance and auditing conference.
Listening to presentations from several prominent healthcare lawyers, I was reminded of the priority (which we have mentioned repeatedly here) for all healthcare organizations, especially health plans, to have…
Ulcers: Pressure and Non-Pressure (Chronic)
by Angie Hlad, CRN-C
When documenting ulcers, it is important to clarify if the ulcer is a pressure or non-pressure ulcer. When a provider only documents the stage of the ulcer, such as "stage 2 ulcer, left heel", it is does not provide enough information for the coder to determine the proper code assignment. Once the provider makes the determination of chronic non-pressure or pressure ulcer, it should be documented as such. The documentation for these two types of ulcers…
Coding and Documenting Fractures in ICD-10
by Christy Wills, CPC, CRC
Fracture coding has become increasingly more specific in ICD-10. As a result of this specificity, providers must document, in some form, whether the fracture is traumatic or pathologic (non-traumatic). A traumatic fracture is caused by some type of accident, fall, or other kind of force. For example, a traumatic fracture can occur after a car accident or when a person is struck with a heavy object. A pathologic, or non-traumatic fracture, is the break of a…
Correct ICD-10 Documentation: MDD
by Tiffany Brookshire, RN, BSN, CRN-C
The implementation of ICD-10 has increased the specificity for many, if not all, disease processes. One such condition is Major Depressive Disorder, a commonly diagnosed condition in the PACE participant population. The correct documentation of MDD will result in a more specific and risk adjusted code, while incomplete documentation will result in an unspecified code that does not map to an HCC.
Updates to ICD-10 and Official Coding Guidelines
by Dr. George W. Brett
The multi-year freeze on changes to ICD codes ended with the launch of ICD-10 last October. We are all up to speed with ICD-10, right? Good, because on October 1, 2016, the 2017 edition became effective. Providers, coders and billers need to be knowledgeable with these changes as there is no grace period. Using codes that are deleted or changed will impact claims and risk adjustment.
Compliance is Front and Center for OIG
While attending HealthCon 2016 this month, sponsored by the American Academy of Professional Coders, I was struck by several facts and trends and want to pass these on to our PACE community friends. Most of these have to do with compliance issues related to documentation, coding and payment. As mentioned previously, I've selected compliance our dominant theme in 2016, and much of what I heard at HealthCon reinforces this.
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By Jamie Stengle
Technology brings images of Holocaust survivors to life
DALLAS (AP) — Max Glauben was 17 and had already lost his mother, father and brother at the hands of the Nazis when U.S. troops rescued him while he was on a death march from one German concentration camp to another.
The recollections of the Dallas resident who as a Jew in Poland survived the Warsaw Ghetto and Nazi concentration camps are now being preserved in a way that will allow generations to come to ask his image questions. Glauben, who turns 91 on Monday, is the latest Holocaust survivor recorded in such a way by the University of Southern California Shoah Foundation . The Los Ange...


With Double Trouble fast approaching I've started to get a little anxious about completing my Ravenwing. I think the bikes are achievable but the Dark Talon is looking less likely, but I'll give it a go. With that in mind when I went into the ManCave intent on 2nd highlights on the Marines I immediately found a diversion task and started highlighting the black on my Imperial Knight! But I go with the flow, so however much I'm telling myself this is not a priority I do it anyway as that's where my muse is. So I was working on the shin pads, I've done the 1st and 2nd highlights and with the shin on the right you should just be able to see how far down the highlight goes. Whereas the shin on the left has since been shadow blended, with a mix of thinned black and dry brushing, admittedly the picture isn't perfect but you should be able to see the difference.
As I was about to get back onto the bikes I thought it was a unfair not to do the Pauldrons as well. This time the shadow blended one is on the right and the highlighted one on the left.
Sure, its very subtle but it's definitely better with the shadow blend.
Luckily after this I was able to bring myself to highlight some Ravenwing, I was able to complete three sets of 2nd Dark Reaper highlights and work on the cloak of the second Sarge so I wasn't completely naughty. Now I just have two marines to highlight and I can move on to a new stage, hopefully without any detours, at least not just yet.
Awesome work on this Dark Angels models :) need to see how they turn !
Cheers Benjamin 'progress is progress'.
It's OK I'm back on the bikes, of course that meant starting to make a Ravenwing Company Master but I quickly binned him off and shaded all but two of the bikers. Just them and it;s the pin highlights.
I enjoyed this comment a lot!
You are doing a really neat job with that blending, but I can't take my eyes off the rivets :-) Is there a secret to them or something simple?
Simple - washes. Having done all the washes you just stipple on extra highlights around them, keeping the shadows. They're a bit too big but I'll need room for the Verdigris as well, you don't want to cover all of the shading up.
That's a good point, I hadn't though of that [files away for later]. When I do rivets I just use black wash, but your shadows look a little lighter/more subtle. Do you use black or something else?
Warplock bronze is the base, then it's Army Painter Strong tone, which is Devlan Mud/Agrax Earth shade. Then it's Warplock again, followed by Balthasar Gold, which is quite warm. The next highlight is Valejo Shining Gold which is probably a big shift towards a lighter gold and then some extreme highlights where I've mixed Gunmetal into it.
With all the stippling in places and the texture built up with all the fine metal particles in the paint a subsequent Soft Tone [sepia] wash picks up all that texture. It also helps unify the highlights, warms it up and makes the shadows matt, leaving the highlights just a little shiny [as it would be].


If you are considering a wisdom teeth removal in Franklin, call our office and schedule a consultation. Not everyone needs to have their wisdom teeth removed. If there is room in your mouth for this third set of molars and they grow in properly, you could continue to live with them without experiencing any irritation. However, most people do not have enough room, and they grow in impacted as a result. This can lead to discomfort while also creating space for food to become trapped.
If your teeth appear to be shifting out of place, it could be because your wisdom teeth are pushing them forward. If you have already had braces to straighten and align your teeth, this is a risk you may not want to take. When you visit us for a wisdom teeth removal in Franklin, we can examine your teeth and determine if they are moving due to your wisdom teeth or another issue like a tongue thrust or even oral cancer.