text:
The Near-infrared Tip of the Red Giant Branch. II. An Absolute Calibration in the Large Magellanic Cloud
Taylor J. Hoyt, Wendy L. Freedman, Barry F. Madore, Mark Seibert, Rachael L. Beaton, Dylan Hatt, In Sung Jang, Myung Gyoon Lee, Andrew J. Monson, Jeffrey A. Rich
We present a new empirical JHK absolute calibration of the tip of the red giant branch (TRGB) in the Large Magellanic Cloud (LMC). We use published data from the extensive Near-Infrared Synoptic Survey containing 3.5 million stars, 65,000 of which are red giants that fall within one magnitude of the TRGB. Adopting the TRGB slopes from a companion study of the isolated dwarf galaxy IC 1613, as well as an LMC distance modulus of μ0 = 18.49 mag from (geometric) detached eclipsing binaries, we derive absolute JHK zero points for the near-infrared TRGB. For a comparison with measurements in the bar alone, we apply the calibrated JHK TRGB to a 500 deg2 area of the 2MASS survey. The TRGB reveals the 3D structure of the LMC with a tilt in the direction perpendicular to the major axis of the bar, which is in agreement with previous studies.
Astrophysical Journal
https://doi.org/10.3847/1538-4357/aab7ed
Magellanic clouds
absolute zero
dwarf galaxies
Space and Planetary Science
Hoyt, T. J., Freedman, W. L., Madore, B. F., Seibert, M., Beaton, R. L., Hatt, D., ... Rich, J. A. (2018). The Near-infrared Tip of the Red Giant Branch. II. An Absolute Calibration in the Large Magellanic Cloud. Astrophysical Journal, 858(1), [12]. https://doi.org/10.3847/1538-4357/aab7ed
Hoyt, Taylor J. ; Freedman, Wendy L. ; Madore, Barry F. ; Seibert, Mark ; Beaton, Rachael L. ; Hatt, Dylan ; Jang, In Sung ; Lee, Myung Gyoon ; Monson, Andrew J. ; Rich, Jeffrey A. / The Near-infrared Tip of the Red Giant Branch. II. An Absolute Calibration in the Large Magellanic Cloud. In: Astrophysical Journal. 2018 ; Vol. 858, No. 1.
@article{89b53b8dafca41e1bd277d4cf318b25f,
title = "The Near-infrared Tip of the Red Giant Branch. II. An Absolute Calibration in the Large Magellanic Cloud",
abstract = "We present a new empirical JHK absolute calibration of the tip of the red giant branch (TRGB) in the Large Magellanic Cloud (LMC). We use published data from the extensive Near-Infrared Synoptic Survey containing 3.5 million stars, 65,000 of which are red giants that fall within one magnitude of the TRGB. Adopting the TRGB slopes from a companion study of the isolated dwarf galaxy IC 1613, as well as an LMC distance modulus of μ0 = 18.49 mag from (geometric) detached eclipsing binaries, we derive absolute JHK zero points for the near-infrared TRGB. For a comparison with measurements in the bar alone, we apply the calibrated JHK TRGB to a 500 deg2 area of the 2MASS survey. The TRGB reveals the 3D structure of the LMC with a tilt in the direction perpendicular to the major axis of the bar, which is in agreement with previous studies.",
author = "Hoyt, {Taylor J.} and Freedman, {Wendy L.} and Madore, {Barry F.} and Mark Seibert and Beaton, {Rachael L.} and Dylan Hatt and Jang, {In Sung} and Lee, {Myung Gyoon} and Monson, {Andrew J.} and Rich, {Jeffrey A.}",
doi = "10.3847/1538-4357/aab7ed",
journal = "Astrophysical Journal",
Hoyt, TJ, Freedman, WL, Madore, BF, Seibert, M, Beaton, RL, Hatt, D, Jang, IS, Lee, MG, Monson, AJ & Rich, JA 2018, 'The Near-infrared Tip of the Red Giant Branch. II. An Absolute Calibration in the Large Magellanic Cloud', Astrophysical Journal, vol. 858, no. 1, 12. https://doi.org/10.3847/1538-4357/aab7ed
The Near-infrared Tip of the Red Giant Branch. II. An Absolute Calibration in the Large Magellanic Cloud. / Hoyt, Taylor J.; Freedman, Wendy L.; Madore, Barry F.; Seibert, Mark; Beaton, Rachael L.; Hatt, Dylan; Jang, In Sung; Lee, Myung Gyoon; Monson, Andrew J.; Rich, Jeffrey A.
In: Astrophysical Journal, Vol. 858, No. 1, 12, 01.05.2018.
T1 - The Near-infrared Tip of the Red Giant Branch. II. An Absolute Calibration in the Large Magellanic Cloud
AU - Hoyt, Taylor J.
AU - Freedman, Wendy L.
AU - Madore, Barry F.
AU - Seibert, Mark
AU - Beaton, Rachael L.
AU - Hatt, Dylan
AU - Jang, In Sung
AU - Lee, Myung Gyoon
AU - Monson, Andrew J.
AU - Rich, Jeffrey A.
N2 - We present a new empirical JHK absolute calibration of the tip of the red giant branch (TRGB) in the Large Magellanic Cloud (LMC). We use published data from the extensive Near-Infrared Synoptic Survey containing 3.5 million stars, 65,000 of which are red giants that fall within one magnitude of the TRGB. Adopting the TRGB slopes from a companion study of the isolated dwarf galaxy IC 1613, as well as an LMC distance modulus of μ0 = 18.49 mag from (geometric) detached eclipsing binaries, we derive absolute JHK zero points for the near-infrared TRGB. For a comparison with measurements in the bar alone, we apply the calibrated JHK TRGB to a 500 deg2 area of the 2MASS survey. The TRGB reveals the 3D structure of the LMC with a tilt in the direction perpendicular to the major axis of the bar, which is in agreement with previous studies.
AB - We present a new empirical JHK absolute calibration of the tip of the red giant branch (TRGB) in the Large Magellanic Cloud (LMC). We use published data from the extensive Near-Infrared Synoptic Survey containing 3.5 million stars, 65,000 of which are red giants that fall within one magnitude of the TRGB. Adopting the TRGB slopes from a companion study of the isolated dwarf galaxy IC 1613, as well as an LMC distance modulus of μ0 = 18.49 mag from (geometric) detached eclipsing binaries, we derive absolute JHK zero points for the near-infrared TRGB. For a comparison with measurements in the bar alone, we apply the calibrated JHK TRGB to a 500 deg2 area of the 2MASS survey. The TRGB reveals the 3D structure of the LMC with a tilt in the direction perpendicular to the major axis of the bar, which is in agreement with previous studies.
U2 - 10.3847/1538-4357/aab7ed
DO - 10.3847/1538-4357/aab7ed
JO - Astrophysical Journal
JF - Astrophysical Journal
M1 - 12
Hoyt TJ, Freedman WL, Madore BF, Seibert M, Beaton RL, Hatt D et al. The Near-infrared Tip of the Red Giant Branch. II. An Absolute Calibration in the Large Magellanic Cloud. Astrophysical Journal. 2018 May 1;858(1). 12. https://doi.org/10.3847/1538-4357/aab7ed
10.3847/1538-4357/aab7ed


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text:
Sophie Wahnich est une historienne française. Elle est directrice de recherche au CNRS et spécialiste de la Révolution française.

Biographie

Parcours professionnel
Agrégée (1988) et docteur en histoire (1994), habilitée à diriger des recherches (2007), elle est directrice de recherche au CNRS directrice de l'Institut interdisciplinaire d'anthropologie du contemporain (IIAC) . Son travail porte sur la Révolution française et le temps présent. Sa thèse avait pour sujet la notion d'étranger dans le discours de la Révolution française, son HDR s'intitulait, Histoire des émotions et présents de l'histoire, une approche politique et anthropologique du sensible en politique.

Engagement
Sophie Wahnich est aussi une intellectuelle engagée, notamment dans le milieu éditorial. Elle est membre du comité de rédaction de la revue politique et culturelle Vacarme et elle dirige une collection dédiée à l'histoire (L'histoire rejouée) au sein de la maison d'édition indépendante Les Prairies ordinaires.

Elle est aussi engagée dans divers collectifs. Elle a par exemple appartenu au bureau du Comité de vigilance face aux usages publics de l'histoire, et elle a participé en 2015 à la création des conseils d'urgence citoyenne.

Elle dénonce aussi une « privatisation du savoir » qui l'amène à être la candidate du Parti pirate dans la  des Hauts-de-Seine à l'occasion des élections législatives 2012.

Elle cosigne des tribunes dans différents médias, dont une tribune publiée en 2017 dans Mediapart intitulée « Faire gagner la gauche passe par le vote Mélenchon ».

En 2019, Sophie Wahnich participe à l'université d'été de Révolution permanente (courant communiste révolutionnaire du Nouveau Parti anticapitaliste) où elle donne une conférence intitulée « 1789 : Révolution bourgeoise ou révolution populaire ? ».

Publications

Sur la Révolution française

Ouvrages
L'Impossible Citoyen, l'étranger dans le discours de la Révolution française, Éditions Albin Michel (1997). Réédité en 2010 dans la collection « Bibliothèque de l'Évolution de l'Humanité », avec une postface de l'auteur.
 Lyon en Révolution, des objets qui racontent l'histoire, Lyon, EMCC, .
 La Liberté ou la mort — Essai sur la Terreur et le terrorisme, Éditions La Fabrique (2003). Réédité en anglais, In Defence of Terror, Liberty or Death in the French Révolution, London, New York, Verso, 2012.
 La Longue Patience du peuple, 1792, naissance de la République, Paris, Payot, 2008.
 Les Émotions, la Révolution française et le présent : Exercices pratiques de conscience historique, Paris, CNRS Éditions, 2009.
 La Révolution française, un événement de la raison sensible, Paris, Hachette, 2012.
 L'intelligence politique de la Révolution française, Paris, Textuel, 2013.
 La Révolution française n'est pas un mythe, Paris, Klincksieck, collection « Critique de la politique », 2017.
 Le Radeau démocratique. Chroniques des temps incertains, Paris, Éditions Lignes, 2017.
 Le Ruban tricolore. Un lien politique, Paris,  Bayard, 2022.

Direction d'ouvrages
 Gilles Sauron, Andrzej Turowski, Sophie Wahnich, ed, L'art et le discours face à la Révolution, Dijon, Éditions universitaires de Dijon, 1998
 Histoire d'un trésor perdu, transmettre la Révolution française, Paris, Les Prairies ordinaires, 2013.

Publication de textes
 Les Voix de la Révolution, projets pour la démocratie, avec Yannick Bosc, Notes et études documentaires, Paris, la Documentation française, 1990.
 Pour le bonheur et pour la liberté, choix et présentation de discours de Robespierre, avec Florence Gauthier et Yannick Bosc, Paris, Éditions La Fabrique, 2000.
 Préface pour Octobre 17, la Révolution trahie. Un retour critique sur la Révolution russe de Daniel Bensaïd, Paris, Éditions Lignes, 2017.

Sur des enjeux politiques contemporains
 Martine Kaluszynski et Sophie Wahnich (éd.), L'Etat contre la politique ? Les expressions historiques de l'Etatisation, Actes des journées d'études sur l'Étatisation, Paris, L'Harmattan, collection « Logiques de la politique », 1998.
 Les Musées d'histoire des guerres du , des lieux du politique ?, Paris, Kimé, 2001.
 Fictions d'Europe, la guerre au musée en France, en Grande Bretagne et en Allemagne, Paris, Éditions des archives contemporaines, 2003
 Lumières l'actualité d'un esprit, Contretemps, en collaboration avec Philippe Corcuff, Paris, Textuel, 2006.
 Une histoire politique de l'amnistie, Paris, PUF, 2007
 Sophie Wahnich, Barbara Lasticova, Andrej Findor (eds), Politics of collective Memory, Cultural Patterns in Post War  Europe, Wien, LIT, 2008
 Marie Cuillerai et Sophie Wahnich eds, Université, ramener la foi en l'impossible, l'Homme et la Société, Paris, Éditions L'Harmattan, 2011,
 Sophie Wahnich (dir.), Culture et musée 20, Réfléchir l'histoire des guerres au musée, Arles, Éditions Actes Sud, 2013,
 Sophie Wahnich et Ratiba Hadj-Moussa dir, Mondes méditerranéens, l'émeute au cœur du politique, L'homme et la société, 187-188, Paris, Éditions L'Harmattan, 2013,
Stridences en conjoncture de trouble. Notre situation historique, photographie de Serge d'Ignazio, Romainville, Éditions Excès, coll. « Voix publiques », 2021,

Notes et références

Liens externes

 Présentation sur le site du laboratoire IIAC/TRAM du CNRS

Historien français du XXe siècle
Historien français du XXIe siècle
Revenu de base
Historien de la Révolution française
Date de naissance non renseignée (XXe siècle)
Candidat aux élections législatives françaises de 2012
Historienne française
Directrice de recherche au CNRS

redpajama_set_name: RedPajamaWikipedia


text:
On Air 4:25AM
A day in the life of Kendra Scott, the CEO of a billion-dollar jewelry empire
Between being a mom to three boys and running her company, designing jewelry isn't the only thing she does.
Author: Rebecca Flores
Published: 12:31 PM CDT September 17, 2019
Updated: 12:31 PM CDT September 17, 2019
AUSTIN, Texas — From sun up to sun down, Austin's own Kendra Scott is a busy woman balancing being an entrepreneur, designer and, not to mention, the CEO of a jewelry empire.
After launching her first collection in 2002, the mom of three quickly scaled her business to several stores across the U.S. According to Forbes, Scott's company is valued at a whopping $1 billion.
If you see a woman walking around Downtown Austin, chances are she's wearing a piece of Scott's jewelry.
KVUE sat down with her to see what a day in her life is like, and to no surprise, the jewelry queen of Texas has jam-packed days.
Kendra Scott pictured in front of a design sketch wall at her Austin office. (Photo courtesy of Kendra Scott)
Courtesy of Kendra Scott
Kendra Scott's Morning Routine
The driven entrepreneur's mornings are usually off to a busy start getting her kids ready for school.
"In the morning I have three sons. I have two in high school and I have one in kindergarten, so it's a crazy morning getting everyone up and out to school. That's my No. 1 priority – getting them fed and ready to go."
Then it's time for her to get ready. Scott said multi-tasking is key to getting herself out the door.
"Usually I'm doing at least one or two conference calls while I put my makeup if you can believe it with my puppies at my feet," she said.
After that, it's time to continue getting dressed while still doing conference calls.
Clocking in at Her Austin Office
Once she makes it to her Austin office, she clocks in and goes straight to work.
"First thing in the morning is I go through my morning brief, which is kind of what the day looks like, what meetings we got, what's happening and I meet with my executive assistant and we go through what the day is going to look like," Scott said.
Next on her list is a stop at the design room – her favorite part of her day.
"I love going in and seeing if any prototypes or samples have arrived or if there are any new stones to play with," she said.
The jeweler explained her company is always designing because it drops a new collection every month.
Although the designing room is her favorite place to be, Kendra said being CEO means she has other things to operate during the day.
PHOTOS: Inside Kendra Scott's Austin office
Kendra Scott's entry way near the front desk at her Austin office. (Courtesy of Kendra Scott)
Kendra Scott design room © Copyrighted image 2016 Thomas McConnell, all rights reserved
Kendra Scott's entryway and front desk at her Austin office. (Courtesy of Kendra Scott)
Kendra Scott's Austin Office. (Courtesy of Kendra Scott)
Dining room at Kendra Scott's Austin office. (Courtesy of Kendra Scott)
One of the meeting rooms at Kendra Scott's Austin office. (Courtesy of Kendra Scott)
Main meeting room behind the entryway at Kendra Scott's Austin office. (Courtesy of Kendra Scott)
Kendra Scott's office at her Austin headquarters. (Courtesy of Kendra Scott)
"Everything from logistics to fulfillment to product development and accounting, you name it, I'm stopping into those different departments every day to see how they're doing," Scott said.
Designing jewelry isn't the only thing Scott does in her role as CEO. Philanthropy has been a big part of her company's mission.
"Philanthropy for us is our purpose," the jeweler said. "We love to give back to our community. Since I first started this business 18 years ago, we have never turned anybody away."
Some of the sweet events Scott has participated in the past include making a stop during her day to read to children and even spending time at a local hospital to decorate a treatment room that inspires young patients.
The passionate designer told KVUE the company has given away $30 million to charities in the United States since 2010.
According to the philanthropy page on her website, the company has participated in more than 8,000 fundraising and awareness events, volunteering more than 2,100 hours.
Kendra Scott's Night Routine
Trying to get home to her boys as fast as she can is always the goal for Scott.
"I never go out to eat lunch. I always eat it in a meeting so I can save the last bit of time at the end of the day to get home to my three boys," Scott explained.
As Scott trades in her CEO job for the night to her role as a mom, it's straight to helping her kids with homework.
"Usually they're working on homework and I'm doing the best to help them as we prepare dinner," Scott added. "A lot of times they're doing homework in the kitchen and I'm making dinner."
Once her kids are washed up and ready for bed, it's reading time. Scott said she usually tucks in her 5-year-old and reads bedtime stories to him before grabbing one of her own favorite novels.
"I think for me that's the way I kind of shut down at night is I get a good book," Scott explained. "I'm one of those people who needs a real book. I like to feel the paper and pages in my hand. I have an old-school reading light so I don't wake up my husband and that's what really puts me to sleep. I'll read a couple of chapters of a really good book and I'll go to bed."
What kind of novels is the jeweler into? Mysteries are her favorite, the jeweler said.
By 9:30 p.m., the jeweler is usually fast asleep.
"I have to go to bed pretty early since I have to get up pretty early. By 9:30 p.m. or 10 p.m., Kendra Scott is asleep," she said laughing.
WATCH: Kendra Scott discusses her success
VERIFY: No gold star on your license? Here's what that means for Texans
Phantom of The Paramount? Artist performing at Austin theater shares chilling photos
Austin's River Place Trail shouldn't charge $10 fee, Texas Parks & Wildlife says

redpajama_set_name: RedPajamaCommonCrawl


text:
Gingerbread Lady
Breakdown: 2W
Maureen Stapleton won Broadway's Tony Award as Best Actress for the role of Evy Meara, a popular singer who has hit the skids. After a short stay in rehab, her best friend age-defying Toby, her daughter Polly, and Jimmy Perry, a gay actor try to help her adjust to sobriety with a jolly birthday party. Enter Lou Tanner, a former lover, who ends up giving her a black eye. The party is a wash out, the "gingerbread lady" falls off the wagon and hits the ropes once again. Later rewritten by Neil Simon as the film Only When I Laugh starring Marsha Mason.
Evy Meara
Lou Tanner
Toby Landau
Polly Meara
Jimmy Perry


text:
Ziad Fahed
Ziad Fahed has a Doctorate in Theology. He is Director of University Mission Office & Full Professor at Notre Dame University-Louaize, Lebanon. His main research interests are religions and politics, Interreligious dialogue and pluralism, the role of the Church during the war in Lebanon, the theology of the mission, and the contribution of the Church in building the Lebanese political system.
As a Fulbright scholar, he focused on 'Religious Pluralism in the US in 2006 in the University of Santa Barbara California. In 2008 he has been awarded a second Fulbright and he focused on Interreligious Dialogue in Temple University, Philadelphia and Auburn Theological Seminary, New York. At the present, he is studying the impact of monotheistic religions on Lebanese society, as well as the challenges they pose to the coexistence among different Lebanese religious communities.
Religious Pluralism in the US
Fulbright Award
Interfaith Community Action Program
Fulbright Award Fall 2008
For the Human Library project
Association of Student Affairs for Catholic College and University – San Diego University
Mission Integration Award
Scholarship Board for a Fulbright Community Action Program
Alumni Engagement Innovation Fund Award
Networking Moderate Religious Leaders in the North of Lebanon
Management System International -MSI-, USAID
Award for the courage and commitment in building respect among members of the human family
"Outstanding Leader in Interreligious Dialogue"
From The Dialogue Institute at Temple University - Philadelphia
Recognition of his
Excellency Ambassador David Hale,
US Ambassador in Lebanon, August 25, 2014 in the State Alumni gathering
Mission Integration Award for Diversity and Inclusion
received form the American Student Affairs for Catholic Colleges and Universities, ASACCU
July 2018 – At San Diego University.
"Land of the Innovators" Award
under the patronage of the Lebanese Ministry of Culture
QUICK ID
ziadfahed@gmail.com
www.ziadfahed.com
In April 2022 I will be receiving KAAD Foundation Peter Hünermann KAAD award for 2022 - Germany.
DOWNLOAD MY FULL RESUME
DOWNLOAD MY FULL FRENCH RESUME
Follow me on social netwroks


text:
\section{Introduction}
 It is well-known that computation of Gauss's
arithmetic-geometric mean involves iteration of a simple
step, whose algebro-geometric interpretation is the
construction of an elliptic curve isogenous to a given
one, specifically one whose period is double the original
period (for a modern survey see \cite{cox}). A higher
genus analogue should involve the explicit
construction of a curve whose jacobian is isogenous to
the jacobian of a given curve. The doubling of the
period matrix means that the kernel of the isogeny should
be a lagrangian subgroup of the group of points of order
$2$ in the jacobian. In  genus $2$ such a construction was
given classically by Humbert \cite{hum} and was studied
more recently by Bost and Mestre \cite{bome}. In this article
we give such a construction for general curves of
genus $3$. We also give a similar but simpler
construction for hyperelliptic curves of genus $3$.
We show that the hyperelliptic construction is
a degeneration of the general one, and we prove that the kernel
of the induced isogeny on jacobians is a lagrangian subgroup
of the points of order $2$. We show that for $g \geq 4$ no similar
construction exists, and we also reinterpret the
genus $2$ case in our setup.

To construct these correspondences we use the bigonal
and the trigonal constructions, familiar in the
theory of Prym varieties (\cite{don}). In genus $2$
Bost and Mestre note that Humbert's construction induces
on jacobians an isogeny whose kernel is of type
$(\ZZ/2\ZZ)^2$. We show that Humbert's construction
is an instance of the bigonal construction, and prove
that the above kernel is a lagrangian subgroup of the
points of order $2$. In fact Bost and Mestre use Humbert's
construction to give a variant of Richelot's genus
$2$ arithmetic-geometric mean.
In light of the clear analogy, in particular the
fact that a generic principally polarized abelian
threefold is a jacobian, one might hope that our
construction could be used in a similar way.

We work throughout over an algebraically closed field of
characteristic $0$. However, our methods clearly extend more
generally. For example, the results of Section~\ref{genustwo}
hold if the characteristic is not $2$, and those of
Sections~\ref{hyperelliptic} and \ref{genusthree} if it is $>3$.

The first author thanks the Hebrew University of Jerusalem
and the Institute for Advanced Studies in Princeton for their
hospitality during the time this work was done.
The second author thanks the University of Pennsylvania for
its hospitality while this article was being written.

\section{Preliminaries} \label{prel}
\noindent {\bf Polarizations.} For an abelian variety
$A$ denote by $A[n]$ the kernel of multiplication by $n$.
In the sequel we will need the following standard facts and
notation.
\begin{enumerate}
\item A polarization $\Theta$ on an abelian variety $A$
induces by restriction a polarization $\Theta_B$ on any
abelian subvariety $B$ of $A$.
\item Recall that the type of a polarization $\Theta$ on a
$g$\/-dimensional abelian variety is a g\/-tuple of
positive integers $d_g|\dots|d_2|d_1$. We say that
$\Theta$ is a principal polarization if it is of type
$1^g = (1,\dots,1,1)$ ($g$\/ times). In that case, suppose
that $p$ is a prime, and that $K$ is a subgroup of $A[p]$
isomorphic to $({\bbB Z}/p{\bbB Z})^r$ and isotropic for the Weil
pairing ${\rm w}_p$. Then $\Theta$ induces a polarization
on $A/K$, characterized by the property that its pull back
to $A$ is $p\Theta$. Its type is then $p^{g-r}\cdot 1^r$.
In this situation we will say that $K$ is a lagrangian
subgroup of $A[p]$ if $r=g$.
\item The type of a polarization is preserved under continuous
deformations.
\end{enumerate}

\vspace*{0.1cm}

\noindent {\bf Double covers.} Given a double cover, i.e. a
finite morphism $\pi: \tilde{C} \rightarrow C$ of degree $2$ between smooth
projective curves, the Prym variety ${\rm Prym}\,(\tilde{C}/C)$ is defined
to be the connected component of the kernel of the norm map
\[ \pi_*: {\rm Jac}\,(\tilde{C}) \rightarrow {\rm Jac}\,(C). \]
It is an abelian variety, and it has a natural principal
polarization when $\pi$ is unramified, namely one half of
the polarization induced on it as an abelian subvariety of
${\rm Jac}\,(\tilde{C})$ (\cite{mum2}). This definition extends to
singular curves $C$, $\tilde{C}$, if we interpret ${\rm Jac}\,$ as
the (not necessarily compact) generalized jacobian. This was
studied by Beauville \cite{bea}.
Particularly important for us will be the cases when
1. $C$, $\tilde{C}$ have only ordinary double points,
2. $\pi^{-1}(C_{\rm sing}) = \tilde{C}_{\rm sing}$, and
3. for each $x\in C_{\rm sing}$ the inverse
image $\pi^{-1}(x)$ consists of a single point,
and each branch of $\pi^{-1}(x)$ maps to a different branch
of $x$ and is ramified over it.
(We shall then say that $\pi$ is of Beauville type at $x$.)
In such cases ${\rm Prym}\,(\tilde{C}/C)$ is compact, and the following
three conditions are equivalent:
\begin{enumerate}
\item $\pi$ is unramified away from $C_{\rm sing}$.
\item The arithmetic genera satisfy $g(\tilde{C}) = 2g(C)-1$.
\item The cover $\tilde{C}/C$ is a flat limit of smooth unramified
double covers.
\end{enumerate}
We shall call a cover satisfying these conditions allowable;
from the third condition we see that the ${\rm Prym}\,$ is prinicipally
polarized in such a case.

Let $C$ be a curve having only ordinary double points as
singularities, and let $\nu_x: N_x \rightarrow C$ be the normalization
map of exactly one such singular point $x$. We denote by $L(x)$
the line bundle of order $2$ in ${\rm Ker}\,\nu_x^\ast$. (It is
obtained from the trivial line bundle on $N_x$ by gluing the
fibers over the two inverse images of $x$ with a twist of $-1$
relative to the natural identification.)

\begin{lemma} \label{polar}
Let $\pi: \tilde{C} \rightarrow C$ be an allowable double cover, $\nu \pi:
\nu \tilde{C} \rightarrow \nu C$ its (partial) normalization at $r \geq 1$
ordinary double points $x_1,\dots,x_r$. Let $g$ be the (arithmetic)
genus of the partial normalization $\nu C$, so the
arithmetic genus of $C$ is $g+r$. Then ${\rm Prym}\, (\tilde{C}/C)$
has a principal polarization,
${\rm Prym}\,(\nu \tilde{C}/ \nu C)$ has a polarization of type
$2^{g}1^{r-1}$, and the pullback map
$\nu^*:{\rm Prym}\,(\tilde{C}/C) \rightarrow {\rm Prym}\,(\nu\tilde{C}/\nu C)$ is an isogeny
of degree $2^{r-1}$. The kernel of $\nu^*$ is the subgroup
of ${\rm Prym}\,(\tilde{C}/C)[2]$ generated by the pairwise differences
of the line bundles $L(x_i)$ defined above. This subgroup is
isotropic for the mod $2$ Weil pairing $w_2$.
\end{lemma}
\begin{pf} The generalized jacobians fit in short exact
sequences
\[\begin{array}{ccccccccc}
0 & \rightarrow & \GG_m^r & \rightarrow & {\rm Jac}\,(\tilde{C}) & \rightarrow & {\rm Jac}\,(\nu \tilde{C}) &\rightarrow & 0 \\
  &     &  \downarrow &     & \downarrow         &     & \downarrow             &    &    \\
0 & \rightarrow & \GG_m^r & \rightarrow & {\rm Jac}\,(C)     & \rightarrow & {\rm Jac}\,(\nu C)     &\rightarrow & 0
\end{array} \]
where the vertical maps are the norm maps induced by $\pi$
and by $\nu\pi$. We
compare the kernels: to begin with, the kernel of the norm
map is connected for ramified double covers (in particular
for $\nu \tilde{C}/\nu C$), and has two components for unramified
covers. This is shown in \cite{mum2} in the nonsingular case,
and so by continuity this holds also for allowable singular
covers (in particular for $\tilde{C}/C$). The multiplicative
groups parametrize extension data
and the norm is the squaring map. So the short exact sequence
of kernels gives
\[0 \rightarrow (\ZZ/2\ZZ)^r \rightarrow \Prym(\Ctil/C) \times \ZZ/2\ZZ \stackrel{\nu^*}{\rightarrow}
\Prym(\nu \Ctil/\nu C) \rightarrow 0,\]
and the first part of the lemma follows.

To prove that the subgroup $(\ZZ/2\ZZ)^{r-1}$ of
$\Prym(\Ctil/C)[2]$ is isotropic for ${\rm w}_2$, notice that
its generators are reductions modulo $2$ of the
vanishing cycles for $\tilde{C}$, and vanishing cycles for distinct
ordinary double points are disjoint. Therefore these vanishing
cycles have $0$ intersection number in ${\bbB Z}$\/- (or ${\bbB Q}$\/-)
homology. By the definition of the polarization of $\Prym(\Ctil/C)$
in Section~\ref{prel}, and the well-known expression for the
Weil pairing in terms of the intersection (or cup product)
pairing (see e.g. \cite[theorem 1, Ch. 23]{mum}), the
rest of the lemma follows.
\end{pf}

\section{The bigonal and the trigonal constructions}
\label{bigonal}
There are several elementary constructions which associate a
double cover of some special kind with another cover (or curve)
with related Prym (of Jacobian).
We now review the bigonal and the trigonal constructions,
following (\cite{don}). Assume we
are given smooth projective curves $\tilde{C}$, $C$ and $K$ and
surjective maps $f:C\rightarrow K$ and $\pi:\tilde{C}\rightarrow C$, so that
$\deg\pi = \deg f=2$ over any component. The bigonal
construction associates new curves and maps of the same
type $\tilde{{C'}}\sra{\pi'} C'\sra{f'}K$  as follows. Let
$U\subset K$ be the maximal open subset over which $f\pi$
is unramified. Then $\tilde{{C'}}$ represents over $U$ the sheaf of
sections, in the complex or the \'etale topology, of
$\pi:(f\pi)^{-1}U\rightarrow f^{-1}U$. It is a $4$\/-sheeted
cover of $U$. We then view $\tilde{{C'}}_{|U}$ as a locally
closed subvariety of $\tilde{C} \times \tilde{C}$ and define
$\tilde{{C'}}$ as the closure.
The projection to $U$ extends to a morphism $\tilde{{C'}}\rightarrow K$,
and the involution $\iota$ of $\tilde{{C'}}_{|U}$
which sends a section to the complementary section extends
to $\tilde{{C'}}$. We define $C'=\tilde{{C'}}/\iota$ and $f'$ and
$\pi'$ as the quotient maps.

We will need to extend this construction  to allowable covers
of curves with ordinary double points; however in a family
acquiring a singularity of Beauville type the arithmetic genus
of the resulting $\tilde{{C'}}$ is not locally constant.
More technically, the naive construction as the
closure of $\tilde{{C'}}_{|U}$ in $\tilde{C}\times\tilde{C}$ is not
flat in families, which is not adequate for our purposes: for
example, we want the bigonal construction to be symmetric.

To achieve this, we define the bigonal construction for
singular allowable covers by {\em choosing} a flat family
of smooth covers whose limit is our allowable cover, and
{\em defining} the construction to be the limit of the
construction for the nonsingular fibers. Beauville's results
imply that this is well defined, and does give a symmetric
construction: this is more or less clear except at a
singularity of Beauville type. There the problem reduces to
a local calculation whose answer, which we record in
\ref{type}. below, is visibly symmetric.

We will need a few properties of this construction (see
\cite[Section 2.3]{don})
\begin{enumerate}
\item As we said, the construction over $U$ is symmetric:
starting with $\tilde{{C'}},\dots,f'$ gives back $\tilde{C},\dots,f$.
\item
\label{type}
Denote the type of $\tilde{C}/C$ at a point $k\in K$ by
\begin{itemize}
\item $\ssub \seq \sslash \smin\smin $ if $C$ is unramified over $k$ and $\tilde{C}$ is
ramified over exactly one point in $f^{-1}(k)$;
\item  $\stackrel{\subset}{\ssub} \sslash \ssub $  if $C$ is ramified
over $k$ but $\tilde{C}$ is unramified over the point
$f^{-1}(k)$;
\item $\ssub \ssub \sslash \smin\smin $ if $C$ is
unramified over $k$ and $\tilde{C}$ is ramified over both
branches of $C$ over $k$;
\item $\btf$ if both $C$,
$\tilde{C}$ have ordinary double points above it, and $\tilde{C}/C$
is of Beauville type there.
\end{itemize}
If $\tilde{C}/C$ is of type $\ssub \seq \sslash \smin\smin $, $\stackrel{\subset}{\ssub} \sslash \ssub $, $\ssub \ssub \sslash \smin\smin $, $\btf$ at
$k$ then $\tilde{{C'}}/C'$ is respectively of type
$\stackrel{\subset}{\ssub} \sslash \ssub $, $\ssub \seq \sslash \smin\smin $, $\btf$,$\ssub \ssub \sslash \smin\smin $ there.

Notice that normalization takes type $\btf$ to type $\ssub \ssub \sslash \smin\smin $.
\item The natural $2$-$2$ correspondence between $\tilde{C}$
and $\tilde{{C'}}$ induces an isogeny
$\Prym(\Ctil/C)\rightarrow\Prym(\tilde{{C'}}/C')$, whose kernel is the same as the kernel
of the natural isogeny
$\Prym(\Ctil/C)\rightarrow\Prym(\Ctil/C)^\vee$ induced by the polarization from
$\Prym(\Ctil/C)$ to its dual abelian variety $\Prym(\Ctil/C)^{\vee}$.
In other words we get an isomorphism
$\Prym(\Ctil/C)^\vee\sra{\sim}\Prym(\tilde{{C'}}/C')$ (cf. Pantazis \cite{pan},
at least
when $K={\bbB P}^1$ which is all we need). As a check, let
$a$, $b$, $c$, and $d$
be the numbers of points where $\tilde{C}/C$ is of type
$\ssub \seq \sslash \smin\smin $, $\stackrel{\subset}{\ssub} \sslash \ssub $, $\ssub \ssub \sslash \smin\smin $, and $\btf$ respectively.
Then by Lemma~\ref{polar} the polarization type
for $\Prym(\Ctil/C)$ is $1^{\frac{a+2c}{2}-1}2^{\frac{b+2d}{2}-1}$.
Similarly the polarization type for
$\Prym(\tilde{{C'}}/C')$ is obtained by interchanging $a$ with $b$ and
$c$ with $d$, and this gives exactly the type dual to the
one of $\Prym(\Ctil/C)$.
\end{enumerate}

For Recillas's trigonal construction start with
$K$, $C$, $\tilde{C}$, $\pi$, and $f$ as before except that $f$
now has degree $3$. We get a cover $g:X\rightarrow K$
of degree $4$ by making over the smooth unramified part
$U$, defined as before, a
construction analogous to what we previously did to get $C'$.
Namely, let $\tilde{X}/U$ represent the sheaf of sections of
$\pi:  (f\pi)^{-1}U\rightarrow f^{-1}U$, and define $X/U$ as the
quotient of $\tilde{X}$ divided by $\iota$ (which is
defined as before). In the nonsingular case we define $X$
as the closure of $X/U$ in $X\times X\times X$, and in the
general allowable case by taking a flat limit of the construction
for smooth, unramified covers. Here we have
(\cite[Section 2.4]{don})
\begin{enumerate}
\item Over $U$ the construction is reversible: $C_{|U}$
represents the sheaf of partitions of $X_{|U}$ to two pairs of
sections, and $\tilde{{C'}}$ represents the choice of one of these
pairs.
\item Denote the type of $\tilde{C}/C$ at a point $k\in K$
by
\begin{itemize}
\item $\tto$ for $C$, $\tilde{C}$ if $C$ has exactly one simple
branch point over $K$ and $\pi$
is unramified over $f^{-1}(k)$;
\item $\ssub\ssub\seq \sslash \smin\smin\smin $ if $f$ is unramified at $k$ and $\pi$ is branched
over two of the branches of $f$ and unramified over the third;
\item $\tttp$ if two branches of
$C$ over $K$ cross normally, the third is unramified, and
moreover, if $\tilde{C}/C$ is of Beauville type over the double
point and unramified over the unramified branch.
\end{itemize}
Then $X$ has exactly one simple branch point at a point
$k\in K$ of type $\tto$ for $\tilde{C}/C$, and we denote
by $\ssub \smin\smin $ the type of $X$ over $k$. Conversely, if $X$
is of type $\ssub \smin\smin $ at $k$
then $\tilde{C}/C$ is of type $\tto$ there. If $\tilde{C}/C$ is
of type $\ssub\ssub\seq \sslash \smin\smin\smin $ at $k$ then $X$ has two simple branch
points over $k$, which we denote by type $\ssub \ssub $.
Here the situation is not reversible:
if $X$ is of type $\ssub \ssub $ at $k$ then $\tilde{C}/C$ is of
type $\tttp$ there. Notice that normalization takes type
$\tttp$ to type $\ssub\ssub\seq \sslash \smin\smin\smin $.
\item If $K\simeq{\bbB P}^1$ and $\tilde{C}/C$ is allowable, then
$X$ is smooth and ${\rm Jac}\,(X)\simeq{\rm Prym}\,(\tilde{C}/C)$. This is
due to Recillas when $\tilde{C}/C$ is smooth unramified, and
again limiting arguments imply this in general.
\end{enumerate}

\section{The genus $2$ case}
\label{genustwo}
Humbert's correspondence of curves of genus $2$ was studied
 by Bost and Mestre (see \cite{hum},
\cite{bome}). We shall show how to make this correspondence
via the bigonal construction, and use this to determine the
type of the isogeny.

Humbert's construction starts with a conic $C$ in ${\bbB P}^2$
with $6$ general points on it (see Remark~\ref{gen} below),
which are given as $3$
unordered pairs $\{P'_i,P_i''\}$, $i=1,2,3$. It associates
to these $3$ new unordered
pairs of points, all distinct, on $C$ as follows. Let
$\overline{P'_iP_i''}$ be the $3$ lines joining paired points,
and let  $l_k$ be the intersection of $\overline{P'_iP_i''}$ and
$\overline{P'_jP_j''}$ if $\{i,j,k\}=\{1,2,3\}$. The new
$3$ unordered pairs of points on $C$ are then the pairs of
points of tangency to $C$ from the $l_k$\/'s.

For our purposes it is more convenient to view the new points
as lying on the conic $C^*$ dual to $C$ in the dual plane
${\bbB P}^{2*}$. A point of ${\bbB P}^{2*}$ is a line in ${\bbB P}^2$; it
is in $C^*$ if and only if this line is tangent to $C$. Let
$\phi:{\bbB P}^2\rightarrow{\bbB P}^{2*}$ be the isomorphism defined by $C$;
namely, for $P\not\in C$ there are two tangents to $C$
through $P$, and
$\phi(P)$ is the line joining their points of tangency. For
$P\in C$, $\phi(P)$ is the tangent to $C$ at $P$. Under the
isomorphism $\phi_{|C}:C\sra{\sim} C^*$, Humbert's new pairs
go to the pairs $L'_k,L''_k$ of tangents to $C$ through $l_k$.
\begin{theorem}
Let $\pi:H\rightarrow C$ and $\pi^*:H^*\rightarrow C^*$ be double covers
branched over the old and new sets of points respectively. Then
there is an isogeny ${\rm Jac}\,(H)\rightarrow{\rm Jac}\,(H^*)$ whose kernel
is a lagrangian subgroup of ${\rm Jac}\,(H)[2]$.
\end{theorem}
\begin{pf} Choose some $k\in\{1,2,3\}$. The set $L^* = L_k^*$ of
lines through $l_k$ is the line in ${\bbB P}^{2*}$ dual to $l_k$.
Let $f:C\rightarrow L^*$ be the ``projection'' sending each point of
$C$ to the line joining it to $l_k$.
Dually, let $f^*:C^*\rightarrow L = \overline{P_k'P_k''}$ be
the ``projection'' sending each tangent line of $C$ to its
intersection with $L$. Let $\psi:L\rightarrow L^*$ be the
isomorphism sending a line through $l_k$ to its intersection
with $L$. The maps $f$ and $f^*$ have degree $2$, and hence
also $g=\psi f^*$ has degree $2$. Both coverings
$H\sra{\pi}C\sra{f}L^*$ and $H^*\sra{\pi^*}C^*\sra{g}L^*$
are unramified over the complement in $L^*$ of the six points
\begin{itemize}
\item The tangents $L'_k,L''_k$ to $C$ through $l_k$; there
$H/C$ is of type $\stackrel{\subset}{\ssub} \sslash \ssub $ and $H^*/C^*$ is of type $\ssub \seq \sslash \smin\smin $.
\item The lines $\overline{P_i'P''_i}$ and $\overline{P_j'P''_j}$
whose intersection defines $l_k$; there both $H/C$ and
$H^*/C^*$ are of type $\ssub \ssub \sslash \smin\smin $.
\item The lines $\overline{P'_kl_k}$ and $\overline{P''_kl_k}$;
there $H/C$ is of type $\ssub \seq \sslash \smin\smin $ and $H^*/C^*$ is of type
$\stackrel{\subset}{\ssub} \sslash \ssub $.
\end{itemize}
It follows that if we perform the bigonal construction
on $H\rightarrow C\rightarrow L^*$, the two points $\overline{P_i'P''_i}$ and
$\overline{P_j'P''_j}$ of $L^*$ are of Beauville type for the
resulting cover $H'\rightarrow C' \rightarrow L^*$ and there are no
other singularities (see Section~\ref{bigonal}). The
preceding analysis of the ramification of $H^*/C^*$,
combined with the one for the bigonal construction
$H'/C'$ in Section~\ref{bigonal}
shows that the normalization of $H'/C'$ is isomorphic to
$H^*/C^*$. It remains to determine the kernel of the induced
isogeny on jacobians; by Pantazis's result recalled above,
it factors as
\[ \begin{array}{rcl}
{\rm Jac}\, H & \simeq &  {\rm Prym}\,(H/C) \sra{\sim} {\rm Prym}\,(H/C)^{\vee}
\sra{\sim} {\rm Prym}\,(H'/C') \\
& \sra{\nu^{\ast}} & {\rm Prym}\,(H^*/C^*) \simeq {\rm Jac}\, H^* \,.
\end{array}\]
To compute the kernel of $\nu^{\ast}$ we cannot use
Lemma~\ref{polar} directly, since $H'/C'$ is
not allowable, being ramified over two points
$x'\in g^{-1}(L'_k)$, $x''\in g^{-1}(L''_k)$. Instead
glue $x'$ to $x''$ to obtain a curve with one more double
point $C''$ and glue their inverse images in $H'$ to get
a curve $H''$, which is now an allowable cover of $C''$
($H''$ is obtained from $H$ by gluing the Weierstrass
points in pairs). We have maps of covers $H^*/C^*\rightarrow
H'/C'\rightarrow H''/C''$ inducing maps of Prym varieties. Applying
Lemma~\ref{polar} twice now gives that the kernel of
${\rm Prym}\,(H''/C'') \rightarrow {\rm Prym}\,(H/C)$ is an isotropic subgroup
isomorphic to $(\ZZ/2\ZZ)^2$ and that ${\rm Prym}\,(H''/C'')$ is isomorphic
to ${\rm Prym}\,(H^*/C^*)$. This implies that ${\rm Ker}\,\nu^{\ast}$ is as
asserted, completing the proof of the Theorem.
\end{pf}
\begin{remark}
\label{gen} {\rm
The points $\{P'_i,P''_i\}$ are assumed general only to
guarantee that they are distinct and that the resulting
new $6$ points are also distinct (for which it suffices
that the tangents to $C$ from $l_k$ in the proof do not
touch $C$ at $P'_k$ nor at $P''_k$). In the case considered
in \cite{bome} this holds, because they assume that
$C$ and the points are real and satisfy some ordering
relations.      }
\end{remark}

\section{The hyperelliptic genus $3$ case}
\label{hyperelliptic}
In this section we will solve our problem in the hyperelliptic
case: we will construct a correspondence between
the generic hyperelliptic curve of genus $3$ and a certain
non-generic curve of genus $3$ (which is not hyperelliptic).
Let $H$ be a hyperelliptic curve of genus $3$ and
let $\pi_1:H\rightarrow{\bbB P}^1$ be the hyperelliptic double cover.
Choose a grouping in pairs of the $8$ branch points
$w_1,\dots,w_8\in{\bbB P}^1$ of $\pi_1$. We claim that there
exists a map $g_1:{\bbB P}^1\rightarrow{\bbB P}^1$, of degree $3$, which
identifies paired points. This can be seen in several
ways. Firstly, let $T$ be the curve obtained from ${\bbB P}^1$
by identifying paired points to ordinary double points.
We think of $T$ as a curve of genus $4$ and take its
canonical embedding to ${\bbB P}^3$. As in the nonsingular
case, the canonical map is well behaved, and in particular the
canonical image of $T$ lies on a unique, generically
nonsingular quadric by the Riemann-Roch theorem. Projecting
via either of the two ruling of this quadric will give the
desired map $g_1$. Notice that by its construction $g_1$
factors as ${\bbB P}^1\sra{\nu} T \sra{g} {\bbB P}^1$, where $\nu$ is
a normalization map.

Another way to get $g_1$ is to embed ${\bbB P}^1$ in ${\bbB P}^3$
as a rational normal curve. We look for a projection
from ${\bbB P}^3$ to ${\bbB P}^1$ which identifies paired points.
The center of this projection is a line $L$ which
must meet the $4$ lines joining the pairs. The
grassmanian $G(1,{\bbB P}^3)$ of lines in ${\bbB P}^3$ is
naturally a quadric in ${\bbB P}^5$ and the condition to meet
a line is a linear condition. We see again that there is
always at least one such $L$, and generically two.

We now perform the trigonal construction. This gives a map
of degree $4$ $f:C\rightarrow{\bbB P}^1$ sitting in a diagram
\begin{equation}
\label{hyp1}
\begin{array}{rcl}
&&H \\
&&\downarrow \! \mbox{$\scriptstyle{\pi_1}$} \\
C&&{\bbB P}^1\\
&\sf\!\!\searrow\;\;\swarrow\!\!\mbox{$\scriptstyle{g_1}$}&\\
&{\bbB P}^1&   \end{array}
\end{equation}
Let $w_{12},\dots,w_{78}$ be the $4$ images of the $w_i$\/'s
under $g_1$, with the indices indicating the grouping.  By
the Riemann-Hurwitz formula there are generically $4$ points
$a_1,\dots,a_4$ in ${\bbB P}^1$ over which $g_1$ is branched, with
a simple branch point over each. Hence $H/{\bbB P}^1$ is of type
$\tto$ at each $a_i$ and of type $\ssub\ssub\seq \sslash \smin\smin\smin $ at each $w_{2i-1,2i}$.
>From the properties of the trigonal construction we get
$2-2g(C)=8-8-4$, so that $C$ has genus $3$.
The trigonal construction gives a birational correspondence between
\begin{itemize}
\item The moduli of the data
$(H \sra{\pi_1}{\bbB P}^1 \sra{g_1} {\bbB P}^1)$ with
$4$ points of type $\tto$ and $4$ points of type $\ssub\ssub\seq \sslash \smin\smin\smin $.
\item A component of the Hurwitz scheme parametrizing
$4$-sheeted covers $f:C\rightarrow {\bbB P}^1$ with $4$ simple branch
points and $4$ double branch points.
\end{itemize}

Each of these moduli spaces is $5$ dimensional. (Another
component of this Hurwitz scheme parametrizes bielliptics,
namely maps $f:C\rightarrow {\bbB P}^1$ which factor through a double cover
$E \rightarrow {\bbB P}^1$ where $E$ is elliptic. Curves in this latter
component are taken by the trigonal construction to towers
$H\rightarrow \overline{H} \rightarrow {\bbB P}^1$ where $\overline{H}=A \cup B$
is reducible, with $A$, $B$ of degrees $1$, $2$ respectively
over ${\bbB P}^1$. We shall not need this component in what follows.)

The key point for us is that the trigonal construction induces
an isogeny ${\rm Jac}\,(C)\rightarrow{\rm Jac}\,(H)$ whose kernel is lagrangian
in ${\rm Jac}\,(C)[2]$. More precisely we have the following
\begin{proposition}
Let ${\bbB P}^1\sra{\nu}T\sra{g}{\bbB P}^1$ be as before, and let $\tilde{T}$
be the curve obtained by identifying the Weierstrass points in
$H$ to ordinary double points with the same grouping as the one
we chose to get $T$. Then
\begin{enumerate}
\item Diagram (\ref{hyp1}) extends to
\[ \begin{array}{rcccl}
&&\tilde{T}&\sla{\tilde{\nu}}&H\\
&&\mbox{$\scriptstyle{\pi}$}\!\downarrow\;&&\;\downarrow\!\mbox{$\scriptstyle{\nu}$}\mbox{$\scriptstyle{\pi}$}\\
C&&T&\sla{\nu}&{\bbB P}^1\\
&\sf\!\!\searrow\;\;\swarrow\!\!\mbox{$\scriptstyle{g}$}&\\
&{\bbB P}^1&   \end{array}\,. \]
Here $\tilde{\nu}:H\rightarrow\tilde{T}$ is the normalization map, and we view
$\nu\pi:=\pi_1:H\rightarrow {\bbB P}^1$ as the normalization of
$\pi:\tilde{T}\rightarrow T$.
\item $\tilde{\nu}$ induces an isogeny of polarized abelian varieties
$\tilde{\nu}^*:{\rm Prym}\,(\tilde{T}/T)\rightarrow{\rm Jac}\,(H)$ whose kernel is
lagrangian in ${\rm Prym}\,(\tilde{T}/T)[2]$.
\item Let $\phi:{\rm Jac}\,(C)\rightarrow{\rm Jac}\,(H)$ be the isogeny obtained
by composing $\tilde{\nu}^*$ with the isomorphism
${\rm Jac}\,(C)\simeq{\rm Prym}\,(\tilde{T}/T)$.
Then the kernel of $\phi$ is lagrangian in ${\rm Jac}\,(C)[2]$, and the
kernel of the dual isogeny $\phi^*:{\rm Jac}\,(H)\rightarrow{\rm Jac}\,(C)$ is the
lagrangian subgroup of ${\rm Jac}\,(H)[2]$ generated by the differences of
identified Weierstrass points.
\end{enumerate}
\end{proposition}
\begin{pf}
Part 1. holds because $\tilde{T}/T$ is allowable. The pairs
of points of $H$ identified by $\nu$ lie over points of
type $\tttp$ for $T$, $\tilde{T}$. Hence they are branch points
for $\pi$, namely Weierstrass points. The rest follows from
Lemma~\ref{polar}.
\end{pf}

\section{The generic genus $3$ case}
\label{genusthree}
Let $C$ be a generic curve of genus $3$. In this section we shall
give a construction of a curve $C'$ of genus $3$ and an isomorphism
${\rm Jac}\, (C)/L \simeq {\rm Jac}\, (C')$ where $L$ is a lagrangian subgroup of
${\rm Jac}\, (C)[2]$. Let $f:C\rightarrow{\bbB P}^1$
be a map of degree $4$, and let $b_1$, $b_2$ be points in
${\bbB P}^1$ such that $f$ has two simple branch points over each
$b_i$. It is easy to show such $f$, $b_1,b_2$ exist, and in fact we will
parametrize the space of such $f$\/'s in the end of this section.

We perform the trigonal construction on $f$. This gives curves
$T$, $\tilde{T}$ and maps $g:T\rightarrow{\bbB P}^1$ and $\pi:\tilde{T}\rightarrow T$, with
$\deg g = 3$ and $\deg \pi=2$. Let $\tilde{\nu}:\nu \tilde{T}\rightarrow\tilde{T}$
and $\nu:\nu T\rightarrow T$ be normalization maps and let
$\nu\pi:\nu\tilde{T}\rightarrow\nu T$ be the map induced by $\pi$. The
properties of the trigonal construction show the following.
Firstly, $T$ and $\tilde{T}$ have each two ordinary double points,
one over each $b_i$, and no other singularities. Next, the
map $g\nu:\nu T\rightarrow{\bbB P}^1$ has exactly $8$ branch points, all
simple, one over each $a_i$. It follows that the genus
$g(\nu T)$ is $2$ and therefore the arithmetic genus $g(T)$
is $4$. The map $\nu\pi$ has exactly $4$ ramification points
$P_i$, $Q_i$, two over each $b_i$ for $i = 1,2$, and hence
$g(\nu\tilde{T})=5$ and $g(\tilde{T})=7$.

Since $\nu T$ has genus $2$, it is hyperelliptic. Let
$h:\nu T\rightarrow {\bbB P}^1$ be the hyperelliptic double cover, and
let $w_1,\dots,w_6\in{\bbB P}^1$ be the branch points of $h$.
The bigonal
construction gives curves and maps of degree $2$
$\nu\tilde{T}'\sra{\nu\pi'}\nu T'\sra{h'}{\bbB P}^1$. The points
in ${\bbB P}^1$ over which $h\nu\pi$ is not \'etale are the $6$
$w_i$\/'s, which are of type $\stackrel{\subset}{\ssub} \sslash \ssub $ for $\nu T$ and $\nu\tilde{T}$,
and the $4$ points $h(P_i)$\/, $h(Q_i)$\/,  $i=1,2$, which
are of type $\ssub \seq \sslash \smin\smin $. The types get reversed for $\nu T'$ and
$\nu\tilde{T}'$, and in particular $\nu T'$ is ramified exactly
over the $h(P_i)$\/'s and the $h(Q_i)$\/'s. It follows that
$g(\nu T')=1$. We also see that $\nu\pi'$ has $6$ branch points,
say $w'_1,\dots,w'_6$, one over each of the $w_i$\/'s, and
hence $g(\nu\tilde{T}')=4$. The curves $\nu T'$ and $\nu\tilde{T}'$
are nonsingular.

Choose a grouping of the $w_i$\/'s in $3$ pairs. Identify
the corresponding $w'_i$\/'s in $\nu T'$ to get a curve
$T'$ with $3$ ordinary double points, say $w'_{12}$, $w'_{34}$,
$w'_{56}$, the indices indicating the groupings. $T'$ has
arithmetic genus $4$. Likewise identify the corresponding
points above the $w'_i$\/'s on $\nu\tilde{T}'$ to obtain a curve
$\tilde{T}'$ with $3$ ordinary double points and arithmetic
genus $7$.

As in the nonsingular case, the canonical embedding sends $T'$
to ${\bbB P}^3$ and the image sits on a unique, generically smooth
quadric. Choosing one of the two rulings of this quadric gives
a map $g':T'\rightarrow{\bbB P}^1$. This map is of degree $3$, because the
canonical curve is a curve of type $(3,3)$ on the quadric. The
map $g'\nu':\nu T'\rightarrow{\bbB P}^1$ is ramified over $n=6$ points, since
$2-2g(\nu T')=0=3(2-2g({\bbB P}^1))-n$. Over these the pair $\nu T'$,
$\nu\tilde{T}'$ is of type  $\tto$. There are also $3$ points of
type $\ssub\ssub\seq \sslash \smin\smin\smin $, the images under $g'$ of the identified pairs $w'_{12}$,
$w'_{34}$, $w'_{56}$.  The
trigonal construction performed on
$\nu\tilde{T}'\slra{\nu\pi'}\nu T'\slra{g'\nu'}{\bbB P}^1$ gives a
curve $C'$ and a map $f':C'\rightarrow{\bbB P}^1$ of degree $4$. We
readily see it has genus $3$. The following diagram summarizes
the procedure:

\[  \begin{array}{rcccccccccl}
&&{}_7\tilde{T}&\sla{\mbox{$\scriptstyle{\tilde{\nu}}$}}&{}_5\nu\tilde{T}&&{}_4\nu\tilde{T}'&
\sra{\mbox{$\scriptstyle{\tilde{\nu}}$}'}& {}_7\tilde{T}'&&\\
&&\mbox{$\scriptstyle{\pi}$}\!\downarrow&&\downarrow\mbox{$\scriptstyle{\nu}$}\mbox{$\scriptstyle{\pi}$}&&\mbox{$\scriptstyle{\nu}$}\mbox{$\scriptstyle{\pi}$}'\!\downarrow&&\downarrow\mbox{$\scriptstyle{\pi}$}'&&\\
{}_3 C&&{}_4 T&\sla{\nu}&{}_2\nu T&&{}_1\nu T'&\sra{\nu'}&
{}_4 T'&&{}_3 C'\\
&\sf\!\searrow\;\swarrow\!\mbox{$\scriptstyle{g}$}&&&& \mbox{$\scriptstyle{h}$}\!\searrow\;\swarrow\!\mbox{$\scriptstyle{h}$}'&&&&
\mbox{$\scriptstyle{g}$}'\!\searrow\;\swarrow\!\sf'&\\
&{\bbB P}^1&&&&{\bbB P}^1&&&&{\bbB P}^1&
\end{array}\,.    \]

Before stating our main result we need to discuss the choices
made in the construction. Writing $f^{-1}(b_i) = 2(P_i+Q_i)$,
we obtain a point of order $2$
\[ \alpha = \alpha(f) = P_1 + Q_1 - P_2 - Q_2.  \]
in ${\rm Jac}\,(C)$. The trigonal isomorphism
${\rm Jac}\,(C) \simeq {\rm Prym}\,(\tilde{T}/T)$ maps $\alpha$ to the
difference $L(b_1) -L(b_2)$ (defined in the discussion
preceeding Lemma \ref{polar}), which is the nontrivial
element in ${\rm Ker}\,\nu^\ast$. Moreover the only choice
made other than $f$ is the grouping of $w_1,\dots,w_6$
under $\nu'$. The differences of the corresponding
paired points in $\nu T$ are the nonzero elements of a
lagrangian subgroup $L_0$ of ${\rm Jac}\,(\nu T)[2]$.

Now observe that the pullback to $\nu\tilde{T}$ by $\nu\pi$ of a
line bundle of order $2$ on $\nu T$ is in the kernel of the
norm map to $\nu T$. This gives a symplectic embedding
\[ \iota:{\rm Jac}\,(\nu T)[2] \hookrightarrow {\rm Prym}\,(\nu\tilde{T}/\nu T)[2]\,. \]
The image ${\rm Im}\,(\iota)$ of $\iota$ can be described in two ways.
On the one hand, it is the kernel of the polarization map
${\rm Prym}\,(\nu\tilde{T}/\nu T) \rightarrow  {\rm Prym}\,(\nu\tilde{T}/\nu T)^\vee$.
(Observe that ${\rm Prym}\,(\nu\tilde{T}/\nu T)$ has a polarization
of type $221$ by Lemma~\ref{polar}, whose kernel is then
isomorphic to $({\bbB Z}/2{\bbB Z})^4$.) On the other hand, let
$\alpha^\perp$ denote the orthogonal complement to (the image
of) $\alpha$  in ${\rm Prym}\,(\tilde{T}/T)$ for the Weil
pairing $w_2$. Then ${\rm Im}\,(\iota)$ is
also the pullback of $\alpha^\perp$ by the normalization
map. Indeed, for $u\in {\rm Jac}\,(\nu T)[2]$ we have
$\iota(u)\in\alpha^\perp$ because
\[ w_2(\iota(u),\alpha) = w_2(u,{\rm Nm}\,_{\tilde{T}/T}(\alpha)) = 0\,, \]
and as both groups have cardinality $16$ they coincide. Hence
this image is isomorphic to
$\alpha^\perp / \langle\alpha\rangle$. In particular, the
inverse image $L$ of $L_0$ in ${\rm Jac}\,(C)$ is a lagrangian
subgroup of ${\rm Jac}\,(C)[2]$ containing $\alpha$.

Conversely, let $L\subset {\rm Jac}\,(C)[2]$ be a lagrangian
subgroup. We will say that the choices $f$, $\nu'$ made
in the course of the construction are compatible with $L$
if $\alpha = \alpha(f)$ is in $L$ and $\nu'$ corresponds
to $L/\langle \alpha \rangle$ as above.
We can now formulate our main theorem, to which we shall
give two proofs:
\begin{theorem}
\label{main}
Let $C'$ be the result of the construction applied to a curve
$C$ of genus $3$ compatibly with a lagrangian subgroup
$L\subset {\rm Jac}\,(C)[2]$. Then there is an induced isomorphism
$Jac(C)/L \sra{\sim} {\rm Jac}\,(C')$. In particular $C'$ is
independent of the (compatible) choices made in the construction.
\end{theorem}
\begin{pf}

The construction induces isogenies whose degrees are marked
below:
\[
\begin{array}{rcl}
 {\rm Jac}\,(C) & \simeq & {\rm Prym}\,(\tilde{T}/T) \sbth{\nu^\ast}{\longrightarrow}{2}
{\rm Prym}\,(\nu\tilde{T}/\nu T) \sbth{\delta}{\longleftarrow}{4}
{{\rm Prym}\,(\nu \tilde{T}'/\nu T')}\\
&\sbth{{\nu'}^{\ast}}{\longleftarrow}{4}& {\rm Prym}\,(\tilde{T}'/T')
\simeq  {\rm Jac}\,(C')\,.
\end{array}
\]
Here the middle step $\delta$ is identified with the
polarization map from an abelian variety of polarization
type $211$ to its dual. As before we identify $\nu^\ast$
with the quotient by $\alpha$, so to construct our isomorphism
${\rm Jac}\,(C)/L \simeq {\rm Jac}\,(C')$ it would suffice to produce
a natural map
$\epsilon: {\rm Prym}\,(\nu\tilde{T}/\nu T) \rightarrow {\rm Prym}\,(\tilde{T}{}'/ T')$
whose kernel is the subgroup $L/\langle \alpha \rangle$ of
${\rm Prym}\,(\nu\tilde{T}/\nu T)$.

One way to do this is to define $\epsilon$ as the dual map
of $\nu'{}^\ast$, using $\delta$ to identify the dual of
${\rm Prym}\,(\nu\tilde{T}'/\nu T')$ with ${\rm Prym}\,(\nu\tilde{T}/\nu T)$, and
using the principal polarization on ${\rm Prym}\,(\tilde{T}'/T')$ to
view it as its own dual. Tracing through the definitions
one verifies that ${\rm Ker}\,(\epsilon)$ is indeed
$L/\langle \alpha \rangle$ as asserted.

An alternative, and more geometric approach, is to show that
the hyperelliptic case treated in Section~\ref{hyperelliptic}
is a specialization of our present general construction. In
fact the hyperelliptic case is obtained when the $4$-sheeted
cover $f: C\rightarrow {\bbB P}^1$ happens to have $4$, rather than the
generic $2$, double branch points. We shall see that this
determines a preferred gluing $\nu'$. In going to this
special case we have to note that the limits of the curves
$\nu T$, $\nu \tilde{T}$ (which we continue to denote with the
same symbols) are no longer non-singular: they are now
only partial normalizations of $T$, $\tilde{T}$, and the map
$\nu \tilde{T} \rightarrow \nu T$ now has $2$ points of Beauville type,
at the singularities which were not normalized. The full
normalizations, say $\nu \nu T$ and $\nu \nu \tilde{T}$, now have
genera $0$ and $3$ respectively, and the resulting diagram
\[\begin{array}{rcl}
&& \nu\nu\tilde{T} \\
&& \mbox{$\scriptstyle{\pi}$}\!\downarrow \\
C && \nu\nu T \\
&\sf\!\!\searrow\;\!\swarrow\!\!\mbox{$\scriptstyle{g}$}' &\\
& {\bbB P}^1 &
\end{array} \]
clearly coincides with diagram~(\ref{hyp1}).

Reagardless of the singularities of the intermediate curves,
we will see that each of the abelian varieties in the
diagram specializes to an abelian variety. In particular,
the limit of ${\rm Prym}\,(\nu\tilde{T}/\nu T)$ is, by Lemma~\ref{polar},
a $4$\/-sheeted cover of
${\rm Prym}\,(\nu\nu\tilde{T}/\nu\nu T) \simeq {\rm Jac}\,(H)$, whose kernel is
$L/\langle \alpha \rangle$. Below we will also
identify the limit of ${\rm Prym}\,(\tilde{T}{}'/T')$ with ${\rm Jac}\,(H)$.
This will produce the desired map $\epsilon$ in this special
case, and hence in general.

For this, we note that the bigonal data
$\nu \tilde{T} \rightarrow \nu T \rightarrow {\bbB P}^1$ has $4$, $2$, $0$ and $2$
points of types $\ssub \seq \sslash \smin\smin $, $\stackrel{\subset}{\ssub} \sslash \ssub $, $\ssub \ssub \sslash \smin\smin $, and $\btf$
respectively, which turn into points of types $\stackrel{\subset}{\ssub} \sslash \ssub $, $\ssub \seq \sslash \smin\smin $,
$\btf$ and $\ssub \ssub \sslash \smin\smin $, respectively, for
$\nu \tilde{{T'}} \rightarrow \nu T' \rightarrow {\bbB P}^1$.
To obtain $\tilde{{T'}} \rightarrow T'$ we need to pair the $6$ ramification
points of $\nu \tilde{{T'}} \rightarrow \nu T'$. There are $15$ ways to do
this, of which one is distinguished: each pair of Beauville
branches gets paired, as do the remaining two ramification
points. Let $T'_{h}$ be the intermediate object, obtained by
gluing only the Beauville branches but not the remaining
pair. It is a singular hyperelliptic curve of genus $3$, and
is a partial normalization of $T'$ at the double point $p$.
Let $\tilde{\Tph}$ be the corresponding $1$-point partial
normalization of $\tilde{{T'}}$, of arithmetic genus $6$.

To continue our construction, we need to identify the two
$g_3^1$'s on $T'$: these two turn out to coincide, and the
unique $g_3^1$ is in fact given by the $g_2^1$ on $T'_{h}$
plus a base point at the double point $p$. To see this we
examine what happens to our general construction of the
$g_3^1$ in this case. The unique quadric surface through
the canonical model of $T'$ is now a quadric cone, with
vertex at (the image of) $p$, because projection from
$p$ gives the canonical image of the hyperelliptic $T'_{h}$,
which is the double cover of a conic. Therefore the two
rulings, hence the two $g_3^1$'s, coincide and have a
base point at $p$, as asserted.

At this point we need to turn the $g_3^1$ into a morphism,
which requires us to blow up the point $p$. This
results in a reducible trigonal curve $T'_{t} := T'_{h} \cup P$,
where $P$ is a copy of ${\bbB P}^1$ intersecting $T'_{h}$ in the two
inverse images $p_1$, $p_2$ of $p$ in $T'_{h}$. The trigonal
map has degrees $2$ and $1$ respectively on the two
components $T'_{h}$ and $P$. This curve is indeed a flat limit,
in the family of triple covers of ${\bbB P}^1$, of the trigonal
curves encountered in the non-hyperelliptic situation. The
corresponding double cover $\tilde{\Tpt} \rightarrow T'_{t}$ is of Beauville
type at all $4$ of the singular points (the two singularities
of $T'_{h}$ plus $p_1$, $p_2$). Here
$\tilde{\Tpt} = \tilde{\Tph} \cup \tilde{P}$, where $\tilde{P}$ is another copy
of ${\bbB P}^1$, double cover of $P$ branched at the points glued
to $p_1$ and $p_2$.

Now that we have identified the trigonal data, we can complete
the construction. By example~2.10(iii) of \cite{don}, or by
inspection, we see that the result $C'$ of applying the
trigonal construction to the reducible trigonal data
$(\tilde{\Tph} \cup \tilde{P}) \rightarrow (T'_{h} \cup P) \rightarrow {\bbB P}^1$ is the
$4$-sheeted cover of ${\bbB P}^1$ obtained by applying the bigonal
construction to $\tilde{\Tph} \rightarrow T'_{h} \rightarrow {\bbB P}^1$. But since the
bigonal construction is reversible this is nothing but the
hyperelliptic curve $H= \nu\nu \tilde{T}$ which resulted from
the construction of Section~\ref{hyperelliptic}, as claimed.

The degeneration just described involves a flat family of
abelian varieties, so the polarization type and the type of
the kernel of the isogeny  on jacobians remain constant. From
the hyperelliptic case we now see that $L$ is the kernel of
our isogeny in the general case. By Torelli's theorem, $C'$
is determined by its polarized jacobian, which is ${\rm Jac}\,(C)/L$.
Hence $C'$ is indeed independent of the choices (compatible
with $L$) made during the construction. This concludes the
proof of Theorem~\ref{main}.
\end{pf}

We now make some further comments on the
choices we made in the course of the construction.
Starting on the left, we fix the curve C, the Lagrangian
subgroup L and an element $\alpha \in L$. Our
$g^1_4$\/'s $f: C \rightarrow {\bbB P}^1$ with $\alpha =\alpha(f)$
are determined by a divisor class in the intersection
\[Z = Z_\alpha =
\Theta_C \cap(\alpha + \Theta_C) \subset {\rm Pic}\,^2(C)\,.\]
More accurately $Z$ parametrizes the family of $g^1_4$'s
(with the specified $\alpha$), together with a marking of
the two singular points $b_1$, $b_2$. Interchanging these
two points gives an involution $i$ of $Z$ induced by the
involution $x \mapsto x+ \alpha$ of ${\rm Pic}\,^2(C)$, and it
is the quotient of $Z$ by $i$ which parametrizes the
$g^1_4$'s alone. $Z$ is also invariant under the involution
$j: x \mapsto K_C- x$ (where $K_C$ is the canonical class)
and $i$ and $j$ commute. In addition, since $\Theta_C$ is
an ample divisor, $Z$ is connected.
Counting fixed points shows that the respective quotients of
$Z$ by $i$, $j$, $k=ij$ have genera $4$, $1$, $4$, and that the
common quotient $\overline{\overline{Z}}:=Z/\langle i,j\rangle$ has genus $1$.

These quotients clearly have the following interpretations as
parameter spaces:

\begin{enumerate}
\item $Z$ parametrizes the $g^1_4$\/'s $f:C\rightarrow {\bbB P}^1$ (equivalently,
via the trigonal construction,
towers of double covers $\tilde{T} \rightarrow T \rightarrow {\bbB P}^1$ of the indicated type),
{\rm with} a choice of a double branch point $b_1$.
\item $Z/i$ parametrizes the $g^1_4$\/'s $f:C\rightarrow {\bbB P}^1$ (equivalently,
towers of double covers $\tilde{T} \rightarrow T\rightarrow{\bbB P}^1$ of the indicated type).
\item $Z/j$ parametrizes the double covers $\tilde{T}\rightarrow T$ of the indicated
type together with a singular point of $T$.
\item $\overline{\overline{Z}}$ parametrizes the double covers $\tilde{T}\rightarrow T$ of the
indicated type, hence it also parametrizes their normalizations, as well
as the maps $\nu\pi': \nu\tilde{T}' \rightarrow  \nu T'$.
\end{enumerate}

We will now discuss what choices we make when we perform the
construction in reverse order, and how the choices from the
two directions are related.

Starting with the genus $3$ curve $C'$, we now assume given a lagrangian
subgroup $L'$\/($={\rm Jac}\,(C)[2]/L)$ of ${\rm Jac}\,(C')[2]$, and a subgroup
$G \subset L'$ of order 4 (which corresponds to
$\alpha^\perp/L$). A marking
of the three double branch points $b'_i$ of $f'$  is equivalent to a
choice of a basis  $\beta' = P'_2 + Q'_2 - P'_1 -Q'_1$ and
$\gamma' = P'_3 + Q'_3 - P'_1 -Q'_1$ of $G$. Let
$\Theta' \subset {\rm Pic}\,^2(C')$ be the theta divisor of $C'$, and for
a class  $u\in{\rm Jac}\,(C')$ let $\Theta'_u$ denote the translation of
$\Theta'$ by $u$. Consider a line bundle $L$ in the intersection
$S = \Theta \cap \Theta_{\beta'} \cap \Theta_{\gamma'}$.
Since the canonical bundle is the only degree 4 bundle on $C'$
with $h^0>2$, there are only two possibilities: either $L^{\otimes 2}$ gives a
$g^1_4$\/ $f':C'\rightarrow {\bbB P}^1$ with three marked
double branch points $b'_i\in {\bbB P}^1$, $i=1,\dots,3$, or else $L$
must be a theta characteristic on $C'$. We claim
the following:\\

\noindent (1) $S$ consists of six points \\

\noindent (2) $S$ is closed under $v\rightarrow K_{C'} - v$.\\

\noindent (3) Four of the points of $S$ are theta characteristics, and two are not.\\

\begin{pf}
(1) holds because $6=g!$.  For (2), suppose that $f',f'':C' \rightarrow {\bbB P}^1$
correspond to $2v$, $2K_{C'} - 2v$ respectively. Then for each double
ramification point $P'_i$, $Q'_i$ of $f'$ we get a unique double
ramification point $P''_i$, $Q''_i$ for $f''$ by imposing the condition
$P''_i + Q''_i + P'_i + Q'_i = K_{C'}$.

For (3) , one checks that there is a unique coset $G'$ of
$G$ in the set of odd theta characteristics
on $C'$: indeed, in coordinates we may take the set of theta characteristics
to be $V = ({\bbB Z}/2{\bbB Z})^6$ with coordinates $x_1,\dots,x_6$, and we may suppose that  $h^0(C',O_{C'}(x))$ mod 2 for $x = (x_1,\dots,x_6)$ is given by
$q(x) = x_1x_2 + x_3x_4 + x_5x_6$. Also we may simultaneously identify
${\rm Jac}\,(C')[2]$ with $V$, with the Weil pairing given by
$w_2(x,y) = q(x+y)-q(x) -q(y)$. Without loss of generality we can also
take $\beta' = e_1$ and $\gamma' = e_3$, with $e_i$ the standard $i$\/th
unit vector. Then  $G' = \{(a,0,b,0,1,1)\}$.

Part (3) is now clear:   $G'$ is contained in $S$, and no other theta characteristics
appear in $S$. This establishes our claim.
\end{pf}

We can now describe all the choices made when we start from the right side.
Our data $C',L',G$ determines a complementary pair of maps $f',f"$. These
determine the data
$\tilde{T}'\sra{\pi'} T'$ uniquely (the two resulting maps $g',g"$ are the
usual two $g^1_3$'s on the genus 4 curve $T'$). The normalization
$\nu\pi': \nu\tilde{T}' \rightarrow  \nu T'$ is therefore also uniquely
determined. So the {\em only} choice made is that of $h'$, given by
an arbitrary point of
$ Pic^2(\nu T') \approx  \nu T'$. Comparing with what we found starting
from the left, we discover that $\nu T'$ is precisely identified with the double
quotient $\overline{\overline{Z}}$.

\section{The case of genus$\geq 4$.}
One might try to generalize our
construction to higher genus by finding, for a generic curve
$C$ of genus $g$, a correspondence with another generic curve
$C'$ of genus $g$ such that ${\rm Jac}\,(C')\simeq{\rm Jac}\,(C)/K$, with
$K$ a lagrangian subgroup of ${\rm Jac}\,(C)[2]$. We shall show
that this is not possible.
\begin{theorem}
Let $K$ be a lagrangian subgroup in ${\rm Jac}\,(C)[p]$, where $C$ is a
generic curve of genus $g\geq 4$ and $p$ is a prime. Then
${\rm Jac}\,(C)/K$ with its induced principal polarization is not a jacobian.
\end{theorem}
\begin{pf}
Let ${\cal T}$, ${\cal S}$, ${\cal M}$ and ${\cal A}$ denote respectively the
Teichm\"uller space, the Siegel space, the moduli space of
curves and the moduli space of principally polarized
abelian varieties, all of genus $g$. The mapping class group
$M=M(g)$ acts on ${\cal T}$ with quotient ${\cal M}$ and the modular
group $\Gamma=\Sp(2g,{\bbB Z})$ acts on ${\cal S}$ with quotient ${\cal A}$.
Moreover $\Gamma$ is naturally a quotient of $M$, because
$M$ acts on symplectic bases for $H_1(C,{\bbB Z})$ through its
action on $\pi_1(C)$, and the period map $\tau:{\cal T}\rightarrow{\cal S}$
is $M$-equivariant for these actions. Passing to the
quotient, we get Torelli's map $\ov{\tau}:{\cal M}\rightarrow{\cal A}$, which is
injective (Torelli's theorem) and exhibits ${\cal M}$ as a locally closed
subvariety of ${\cal A}$. Since ${\cal T}$ is irreducible it follows
that the {\em Torelli space} $\ov{{\cal T}}=\tau({\cal T})$ is a locally closed
irreducible analytic subvariety of ${\cal S}$.

\[ \begin{array}{rcl}
   {\cal T} & \slra{\tau} & {\cal S} \supset \overline{{{\cal T}}}=\tau ({\cal T}) \\
   M \downarrow && \downarrow \Gamma \\
   {\cal M} & \slra{\overline{{\tau}}} & {\cal A}
   \end{array}  \]

Let $W$ be the finite cover
of ${\cal S}$ obtained by taking over each marked abelian
variety $A$ the lagrangian subgroups of $A[p]$. Since $W$ is
unramified over the contractible space ${\cal S}$, it is in fact a
union of copies of ${\cal S}$. Our generic isogeny
${\rm Jac}\,(C)\rightarrow{\rm Jac}\,(C)/K$ translates to the following data. The
curve $C$ lives over an open subset of ${\cal M}$, hence of $\ov{{\cal T}}$.
The subgroup $K$ corresponds to a sheet of $W_{|\ov{{\cal T}}}$.
Therefore our isogeny extends to the quotient map by
the subgroup, still denoted $K$, corresponding to the
``same'' sheet over all of ${\cal S}$. Now recall that ${\cal S}$
is the  space of symmetric
$g\times g$ complex matrices $\Omega$ with positive imaginary
part, and the abelian variety over $\Omega$ is
$A_\Omega = {\bbB C}^g/({\bbB Z}^g+\Omega{\bbB Z}^g)$. Since  monodromy
(i.e. $\Gamma$) acts transitively on the lagrangian subgroups of
$A_\Omega$, we may
take $K = (\frac{1}{p}{\bbB Z}/{\bbB Z})^g$ for convenience. Then
$A_\Omega/K \simeq A_{p\Omega} = A_{s\Omega}$, with
\[ s = \matr{pI_{g\times g}}{0}{0}{I_{g\times g}}
                                 \in\Sp(2g,{\bbB R})\,.\]

If ${\rm Jac}\,(C)/K$, with its principal polarization, were a jacobian,
it would follow that the Torelli locus $\ov{{\cal T}}$ was invariant
under the subgroup $\Delta$ of $\Sp(2g,{\bbB R})$ generated by
$\Gamma$ and by $s$. We claim that $\Delta$ is dense
in $\Sp(2g,{\bbB R})$. Indeed, consider the subgroup $N_+$ of $\Sp(2g, {\bbB R})$
consisting of the matrices
$n(x) = \matr{I_{g\times g}}{x}{0}{I_{g\times g}}$, where
$x$ runs over the real symmetric $g\times g$ matrices. Then
$\Delta$ contains $s^i n(x) s^{-i}=n(p^{-i}x)$ for all integral
symmetric matrices $x$ and integers  $i$. These are dense in $N_+$,
and $\Delta$ likewise contains a
dense subgroup of $N_-={}^t N_+$. It is
well-known (and easy) that $N_+$ and $N_-$ generate $\Sp(2g,{\bbB R})$, so
$\Delta$ is indeed dense in $\Sp(2g, {\bbB R})$.

Therefore, under our assumption, $\ov{{\cal T}}$ would be dense in
${\cal S}$ (in the complex topology),
so that ${\cal M}$ would be dense in ${\cal A}$. This is a contradiction
when $g>3$, because for dimension reasons
${\cal M}$ is not
dense in $A$ even for the Zariski topology then,
and the theorem follows.
\end{pf}

redpajama_set_name: RedPajamaArXiv


When my rep from D'Artagnan texted me to ask if I wanted a suckling pig, how could I say no? Normally they are special order, but he had a cancellation and neither of us wanted it to go to waste. So having never worked with suckling pig, I took a 25 pounder home to experiment.
Confronted with its cute little face and curly tail, D. and I did experience some pangs of guilt, but took some solace in knowing that at least our pig had been humanely raised and slaughtered. We promised to honor the animal by using it to make the most delicious food we could, and that we would not waste an ounce. We then proceeded to carefully break it down into hams, bellies, shoulder, loins, trotters, and yes, head.
I immediately put one of the hams into a brine of salt, sugar and herbs, and salted the bellies to make pancetta. The rest were vacuum sealed and stored in the freezer.
This morning we roasted the ham. It was so small, it only took one hour! During the last few minutes of roasting we brushed the ham with a mixture of dijon mustard, honey, and black pepper and blasted the oven to crisp the skin a bit.
As ham rested, I baked some cheese biscuits and poached a few eggs. Meanwhile, D. was kind enough to mix us a couple Old Fashions that paired perfectly with our lovely boxing day brunch.
Tags: Baking, Comfort Food, Ham, pork, Roasts, Slow Food, Suckling Pig. Bookmark the permalink.
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I read about the vegetable fed chicken at D' Artagnon, but it's only sold to restaurant chefs. Do you purchase them for your restaurant and is there anyway I could buy some form you? Let me know.


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The Weather and Climate in Czech Republic
You are here: Countries / Czech Republic
The temperature in Czech Republic varies. Depending on the different regions of the country, the climate is mixed and is influenced by the topographical locations. At an increasing altitude, the temperature decreases. On the other hand, at the lower elevations, temperatures are mostly uniform. Rainfalls also often occur in some areas. The country's climate is mainly continental, moderate, and transitional climate, perfect for a cozy and an outgoing vacation. Enjoy the uniqueness of the temperature the country offers. Gentle winds, soft rainfalls, fresh summers, and humid winters are naturally enjoyed in Czech Republic.
Bílý Potok in Jizera Mountains is the wettest part of the country and Louny District is the driest area. Furthermore, these certain factors affect the temperature and the climate of Czech Republic: the altitude, mountain distribution, and the location. However, Cheb's annual temperature reaches around 7ºC; Prague reaches about 33ºC at its high temperature and around -1ºC at its lowest temperature in January. The highest point of the country, the Sněžka, has an average temperature of about −0.4°C. In South Moravian Region, its highest temperature is about 10°C.
Do not miss the embrace of the country's relaxing weather; prepare outfits suitable for each and every season depending on your location. Enjoy the humid days and calming nights in May-September; inhale cold air in November to January; and be fond of the sun rays in March and April days. Get your summer shades ready in the warmest months—July, August, and followed by June. Reflect on the falling leaves by September, which is the start of the autumn season.
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Case studies, News
CiviCRM and using open source technology
Thomas Muirhead September 16, 2013
Guest blogger Thomas Muirhead, head of digital and CRM at Leukaemia & Lymphoma Research, shares how open source technology has added real value to his organisation and how you can add exactly the same to yours.
If you're responsible for the digital offering in a charity, you'll be well aware of the ever increasing complexity of the different tools, web services, applications and channels you need to weave together to get your team – and the whole organisation – to function as well as possible.
What this inevitably leads to is inefficiency. The processes involved in pulling data from these different systems are complex and time-consuming. The answer, where possible, is to automate it: integrate systems.
Integrating JustGiving with CiviCRM
One of the biggest third party providers of data and donations for us here at Leukaemia & Lymphoma Research is JustGiving. We analysed our processes and saw that days and days of time every month were being used downloading, tweaking, re-uploading, re-downloading and importing spreadsheets of data to and from JustGiving. As an organisation dedicated to improving things, big and small, it seemed a natural place to focus some attention.
So we've built an integration extension to CiviCRM – our open source CRM (customer relationship management) system – which allows all our JustGiving payments and donors to be pushed directly into our CRM system, thanks to the JustGiving Data APIs. Contributions get correctly assigned to contacts, issues can be resolved and corrected directly in the system and we can push it easily to our accounting software. Also, there's the option to soft credit the fundraiser with the donor's contribution.
It's saved the teams here a lot of time and ensured greater accuracy in our database. Also, the speed with which we understand what our supporters have raised on JustGiving has allowed us to improve our communications with them.
What is CiviCRM and why open source?
Adopting CiviCRM here at Leukaemia & Lymphoma Research was one of the best decisions we've made. Here's why in a nutshell:
As an open source tool we get so much more flexibility out of the system than we would with proprietary providers. We control how the system works.
It's really well integrated with our CMS system – we use Drupal but CiviCRM works out of the box with WordPress and Joomla too. This means all event sign-up, donation and direct debit transactions on the website flow into the CRM directly.
It's a really powerful CRM with everything from donation and form building to case management, grant tracking, automated communications planning and so much more.
We can work with other charities to build functionality together. This is one of the strongest arguments for open source generally – why are we all paying for the same software to be built when we could pay for it once together!
How can other charities do this?
The extension we've built is available to any charity using CiviCRM. It's open source software so anyone can use the code. Visit the CiviCRM site and you can download the extension and start using it immediately. Some charities such as Oasis and STOP THE TRAFFIK are already using a pre-release version of the extension.
CiviCRM is designed specifically for charities, and used by smaller charities like the Merton Music Foundation, medium-sized charities like us at Leukaemia & Lymphoma Research and huge organisations like the New York State Senate. It's free to download and free of monthly subscriptions. It's not straightforward to implement, so you may need to get help depending on the scale of your organisation.
If you want to find out more about CiviCRM or our use of open source technology generally feel free to get in touch with me at tmuirhead@beatingbloodcancers.org.uk
Also, I'm presenting on the growth of open source CRM at the London CiviCon in the first week of October. I'd recommend anyone interested in CRM should attend to find out more about the direction of travel for CRM in the charity sector.
We'd like to continue to see JustGiving integrations built with other open source tools. For example, the front end page building API could be released as a Drupal or WordPress module/plugin.
If you are interested in guest blogging for We make giving social, drop us a line at wemakegivingsocial@justgiving.com
Making the move from everydayhero to JustGiving
By Toni Gregory	| December 22, 2020
Six inspiring examples of charity Christmas campaigns
companiesCRMdata
Thomas is the head of digital and CRM at Leukaemia & Lymphoma Research. He's dedicated to bringing digital tools - particularly open source technology - skills and knowledge in-house, because of the power it puts in the hands of the charity wielding it and the benefit it provides to other charities.


• The Cleveland Browns clearly will be a contending team in 2019, and one of their best players is wide receiver Antonio Callaway. He is one of the most natural receivers in the league, and he also is a dangerous kick returner. If mid-season pickup Breshad Perriman continues to improve, then the trio of Callaway, Perriman and Jarvis Landry will form a terrific receiving group for quarterbacks Baker Mayfield. Add in tight end David Njoku, and the Browns are going to put up points.
• Congratulations to the Philadelphia Eagles for clinching a playoff spot. While Carson Wentz remains their future and franchise quarterback, backup quarterback Nick Foles brings a different and more calming influence to the Eagles offense. One thing he possibly does better than Wentz is playing within himself. Wentz, of course, is a marvelous athlete, and he has a strong belief that he can exert his physical will on every game. While this is admirable, this trait often leads him to play out of control and in a frenzy at times. Foles, on the other hand, takes what the defense gives him and understand precisely what he can or cannot do. As a result, the Eagles have looked more efficient on offense with Foles in recent weeks.
• The Arizona Cardinals played one of their better games of the season but, ironically, it also may have been the last game for head coach Steve Wilks. If that is the case, one issue Cardinals management will have to contend with will be another potential rebuilding of the defensive philosophy. When Wilks came into the job, he started transforming the Cardinals' defensive personnel and that could be an issue that will hinder progress in 2019.
• It has been fashionable lately to criticize Minnesota Vikings quarterback Kirk Cousins for the team's failures. We see it somewhat differently, however. One of the big personnel failures for the Vikings in recent years has been the neglect of their offensive line. This unit will have to be a priority in the offseason. The Indianapolis Colts have shown what a rebuilt offensive line can do, and that is a model for the Vikings to emulate.
• We remain steadfast in our belief that San Francisco 49ers head coach Kyle Shanahan is one of the best young head coaches in the game. However, he desperately needs a strong turn of luck in the injury department. In Sunday's loss to the Rams, the 49ers lost yet another running back when promising undrafted rookie Jeff Wilson was carted off; starting guard Laken Tomlinson also was carted off with what appeared to be a serious leg injury. We see the 49ers as one of the surprise teams for 2019 and a contender for the division… if they can stay healthy.
• The Los Angeles Chargers offense needs to get untracked in a hurry. Once again, they played an uninspiring game on Sunday, letting Denver hang around for far too long. We feel that recent injuries to running backs Melvin Gordon and Austin Ekeler have brought the offense to a halt. As good as quarterback Philip Rivers is, we feel the Chargers need to have a robust running game if they are going to advance in the playoffs.