text:
package org.fusesource.process.fabric.commands;

import org.apache.felix.gogo.commands.Argument;
import org.apache.felix.gogo.commands.Command;
import org.apache.felix.gogo.commands.Option;
import io.fabric8.utils.shell.ShellUtils;
import org.fusesource.process.fabric.ContainerInstallOptions;
import org.fusesource.process.manager.InstallTask;
import org.fusesource.process.manager.Installation;

/**
 * Installs a new process
 */
@Command(name = "process-install-jar", scope = "fabric:", description = "Installs a jar as managed process into this container.")
public class InstallJar extends ContainerInstallSupport {

    @Option(name="-c", aliases={"--classifier"}, required = false, description = "The maven jar classifier")
    protected String classifier;
    @Option(name="-e", aliases={"--extension"}, required = false, description = "The maven extension (defaults to jar)")
    protected String extension = "jar";
    @Option(name="-o", aliases={"--offline"}, required = false, description = "Whether to use offline mode when resolving dependencies")
    protected boolean offline;
    @Option(name="-opt", aliases={"--optional"}, required = false, multiValued = true, description = "List of patterns of optional dependencies to include. Of the form group[:artifact] with * allowed as wildard")
    protected String[] optionalDependencyPatterns;
    @Option(name="-exc", aliases={"--exclude"}, required = false, multiValued = true, description = "List of patterns of dependencies to exclude. Of the form group[:artifact] with * allowed as wildard")
    protected String[] excludeDependencyPatterns;
    @Option(name="-m", aliases={"--main"}, required = false, description = "The Java executable main() class")
    protected String mainClass;

    @Argument(index = 0, required = true, name = "groupId", description = "The maven group Id of the jar")
    protected String groupId;
    @Argument(index = 1, required = true, name = "artifactId", description = "The maven artifact Id of the jar")
    protected String artifactId;
    @Argument(index = 2, required = false, name = "version", description = "The maven version Id of the jar")
    protected String version;

    void doWithAuthentication(String jmxUser, String jmxPassword) throws Exception {
        ContainerInstallOptions options = ContainerInstallOptions.builder()
                .container(container)
                .user(jmxUser)
                .password(jmxPassword)
                .controllerUrl(getControllerURL())
                .groupId(groupId)
                .artifactId(artifactId)
                .version(version)
                .classifier(classifier)
                .extension(extension)
                .offline(offline)
                .optionalDependencyPatterns(optionalDependencyPatterns)
                .excludeDependencyFilterPatterns(excludeDependencyPatterns)
                .mainClass(mainClass)
                .controllerUrl(getControllerURL())
                .build();
        // allow a post install step to be specified - e.g. specifying jars/wars?
        InstallTask postInstall = null;

        Installation install = getContainerProcessManager().installJar(options);
        ShellUtils.storeFabricCredentials(session, jmxUser, jmxPassword);
        System.out.println("Installed process " + install.getId() + " to " + install.getInstallDir());
    }
}


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When you work with a distributor the selection of promotional items is determined by your needs and goal expectations. I will take into consideration the audience you want to reach, budget and deadlines. I know the current trends, what works, what doesn't work, how your logo will best imprint and how to get you the results you want.
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text:
Photos & Memories
The Way It Used to Be!!!
Deceased Classmates
Become Part of the Alumni
Gene Harvey Nerren and Jo Ann (Ricks) Nerren
Gene Harvey Nerren and Jo Ann Ricks Nerren of Lufkin, celebrated their 55th Wedding Anniversary by renewing their wedding vows at Believers Bible Church in Lufkin on September 4, 2010, with their children, Michael and Barbara Nerren of Rockwall, Marilyn and Richard Troth of Grove City, Ohio, Matthew and Toni Nerren of Corpus Christi, granddaughter, Michelle Troth of Grove City, Ohio and grandson, Martin Troth of Grove City, Ohio, as their wedding party. Pastor, Dr. Bob Klund officiated the ceremony before family and friends. John and Cynthia Burnett sang a duet, "The Vows Go Unbroken." Joe Nerren provided the music for the ceremony. Kenneth Ricks, Jo Ann's brother, who was best man at their wedding 55 years ago, in Del Rio on September 3, 1955, read the scripture reading from 1 Corinthians 13:4-13. During the commitment, 1 John 4:7-10 was read by Martin Troth and before the renewing of the vows, Ruth 1:16-17 was read by Michelle Troth.
After the ceremony, a reception, with dinner and dancing, was held at the Courtyard Marriott in Lufkin. Joe Nerren provided the music for the evening.
Gene and Jo Ann entertained their family and out of town guests in their home on Friday evening, September 3. 2010.
Gene was Chief Financial Officer for Brookshire Brothers before retiring in 2000 after 38 years. Jo Ann established Nerren Employment Service in 1969, the first employment service in Lufkin. Gene and Jo Ann have lived in Lufkin for 55 years. Mrs. Nerren is the former Jo Ann Ricks, graduate of Del Rio High School in 1954.
Note: Any Alumni who has been married 50+ years deserves credit and respect. If you are in that elite category please make us aware.
bold>Right-Rail Menu</bold>
Alumni Contributors
List of Deceased Classmtes
© 2013 Del Rio Wildcats - High School Alumni Web

redpajama_set_name: RedPajamaCommonCrawl


text:
Six new electric cars, Daimler invests, Tesla Model 3 pricing proposed, U.S. gas mileage: Today's Car News
Stephen Edelstein March 2, 2017 Comment Now!
Today, we look at six new electric cars coming in 2018 and 2019, Daimler invests in charging network ChargePoint, and a new study indicates U.S. fleet average fuel economy did not increase much over the past 25 years. All this and more on Green Car Reports.
We have additional plug-in electric car sales figures for February 2017, showing that the Chevrolet Volt outsold not only the Bolt EV, but the Nissan Leaf and Toyota Prius Prime as well.
From the second-generation Nissan Leaf to the Porsche Mission E, check out these six new electric cars coming in the next two years.
A lobbying group projects that biofuels would be needed to offset the climate impact of the 3 billion cars it estimates will be on world roads by 2050.
See the inside of a Tesla Model S battery pack, courtesy of this YouTube video teardown.
Mercedes-Benz parent Daimler invested in ChargePoint, joining BMW and providing funding for expansion of the charging network's European operations.
Overall U.S. fleet gas mileage rose only slightly over the last 25 years, according to a new study.
Toyota believes plug-in hybrids will catch on faster than battery-electric cars. We asked our Twitter followers what they thought about that.
How much will Tesla's upcoming "affordable car" actually cost? One reader creates a hypothetical Tesla Model 3 pricing grid for the eagerly anticipated electric car.
Fiat Chrysler Automobiles warned investors about the possible negative impact of recent diesel lawsuits and government inquiries.
Finally, the H600 is a hybrid luxury sedan from Chinese startup Hybrid Kinetic that will debut at the 2017 Geneva Motor Show.
Green Cars News Today in Car News
Electric school buses, Taycan range, Nissan EV platform: Today's Car News
Honda and Isuzu, Lucid Air, cleaner EVs from less coal: Today's Car News
Chevy Bolt EV deals, talking Teslas, Nissan rejects more screens...
Nissan electric crossover tech, lithium-sulfur batteries, EV cost...


text:
Study discovers surprising views on Bernie Sanders
February 10, 2016 Fn_admin Bernie Sanders, Jim Boeheim, presidential campaign, Presidential campaign 2016, satire
By: Andrew Koerner – Alumnus, Class of 2015
A recent study conducted by Pew Research Center in Washington, D.C, was made available to the public this past Monday. The aims of the study were to further understand a common political beliefs among the American public today. The think tank's three-month venture was vast and covered at least one major city population in every state. The key point of the published report involved former Vermont senator and current presidential candidate Bernie Sanders. The study stated that one out of every five Americans confuses the 76-year-old with Hall of Fame basketball coach Jim Boeheim.
Janis Boca, who led the Pew research team's efforts, elaborated on the key discovery: "Every now and then we'd discover someone, typically a college basketball fan, who believed that the two were the same person; either someone believed Boeheim was running for president in his spare time or that Sanders was coaching the University of Syracuse men's basketball team."
Boca went on to explain additional findings of the study, "Oh yeah, nothing has changed," Boca said while lighting a cigarette. "People like Donald Trump because they're angry at Muslims or something, and people still think Hillary Clinton chats dirty with Vladimir Putin on federal email accounts. We didn't wish to waste any time pursuing such predictable areas–this was literally the only original belief we found in major public opinion today."
Greg Barchek, a first-year English major and Sanders supporter, provided Flyer News with some insight on Sanders' popularity: "I think it's so inspiring how someone without a political background can run for president. This dude was a stud baller back in the '60s for the Orange. How can there be any concern about his health? Anybody who can run the point can run the oval office in my opinion."
In the wake of Pew's study, many American news sources have apologized for displaying incorrect photographs of Sanders over the course of the presidential primaries, both on television and on the Internet. ABC News displayed a courtside image of Jim Boeheim on its website after the second democratic debate. Out of 257 comments posted to date, not one mentioned the error. This past week, CBS Sports posted an image of Sanders on its website after Syracuse's 68-60 victory over Virginia Tech University. Similarly, the mistake was not noticed until recently.
"It was an old man pointing angrily at someone," exclaimed Douglas Volk, executive producer of the network. "Production rolled with it, and it was our most read game breakdown of the night."
During a timeout in that same game, protestors with the Black Lives Matter movement took center court with megaphones and homemade signs demanding the basketball players do something more meaningful with their lives than chase after a ball for people's amusement. During postgame interviews, Jim Boeheim was also asked questions about healthcare reform and the wealth gap.
Editor's Note: This article is satirical. At this point and time, Pew has not developed enough research to prove American's confusion between Bernie Sanders and Jim Boeheim.
← Letter to the Editor: Still puzzled
Why we need Black History Month →


text:
Rainbow Six Extraction is just short of being the tactical sci-fi homage to the series it needs to be – Review
An intensely tactical team-based shooter that might be too tough for its own good.
Jamie Moorcroft-Sharp
Rainbow Six Extraction Reviews
Having early access to a multiplayer game like Rainbow Six Extraction is a blessing and a curse. On one hand, we've had the opportunity to try out the majority of the game in advance, but the drawback of that is that it's impossible to tell how most people will play the game. Rainbow Six Extraction is even more unique because it's not a traditional shooter, as most people expected — as a result, it's taken a week or so for players to clock onto how to play or check out entirely because it's just not for them. What we've found is that the game is deeply rooted in the Rainbow Six franchise, but it stumbles in a big way from being a truly fantastic entry in it.
Always online, endlessly replayable missions
The core gameplay of Rainbow Six Extraction is made up of Incursions, missions into one of 12 maps spread across four zones. Three players pick from a roster of 18 Operators to head into an area and complete various objectives that will help a new arm of Team Rainbow called React, battle an alien scourge that's resurfaced on Earth known as the Archaeans. This enemy and gameplay loop was even in Rainbow Six Siege in 2018, and Ubisoft has taken that idea and created a full-blown game out of it.
Maps change each time you see them thanks to the Sprawl, a black goo that the Archaeans use to create nests and make themselves feel at home. While the maps remain the same, doorways can be blocked or broken and many items moved thanks to the randomization of the Sprawl. Higher difficulty levels add mutations to the parasite, such as Caustic Sprawl, injuring any Operator that touches it, or Blinding Spores growing on all Archaeans, forcing you to kill them from afar or suffer.
The Operators you take into each mission also change the way the game feels. Each Operator has a unique gadget or skill that's been borrowed from Rainbow Six Siege. Very little has changed in their transition, meaning Siege players will be familiar with these outcasts, equipped with abilities that make them better for PvE than PvP.
Should you fail a mission, your Operator will be captured by the aliens. They'll then be marked as MIA in your roster, and you'll need to go on an Incursion in that same map to rescue them as part of an MIA objective the next time you play. This makes death feel punishing and forces you to abandon missions if they're going south, which is possible at the extraction point in every sub-zone.
A huge variety in enemy types adds to the random feel of each mission. While many look similar, they behave differently enough that you need to tackle them in different ways. Grunts, for example, can be killed quickly with a couple of shots anywhere, whereas Spikers or Sowers have armor to work around, and Smashers are best left alone unless an objective requires their death.
The game resembles renowned co-op shooters like Left 4 Dead but adds in a tactical element that those games lacked. You can't run in and gun your way through the Archaeans. If you do, you'll alert every enemy on the map, cause the nests to spawn new ones, and kill every member of your team while simultaneously failing each objective. Be stealthy, kill enemies from the shadows, and work as a unit, and you'll breeze through even the toughest difficulty modes.
Constant progression and rewards
Every Operator in the game gets experience points from missions. Completing objectives, killing enemies, scanning enemies, and extracting all add to the final score, which pushes their level up. The higher their level, the stronger and more useful they are thanks to upgrades to their gadget. This allows them to take on higher difficulty levels, which add new modifiers and enemies into the mix.
Each zone has a set of studies to complete, adding to your Milestone level. Higher Milestone levels mean more content you unlock from the game, including Operators, new zones, and new game modes. By Milestone 16, you'll have unlocked endgame content, but the early game missions remain enjoyable no matter how many times you play through them.
In addition to this, you'll unlock new cosmetic items for your Operators as you progress in each area. The toughest challenge in the game, Maelstrom Protocol, awards exclusive gear each week based on your performance in it over the past seven days. This is what should keep you pushing yourself to do better long after you've completed everything else the game has to offer.
One misstep from greatness
Rainbow Six Extraction is an online-only game. While you can play each mission with others online or solo, making for an acutely terrifying tactical shooter experience, there's one thing missing from the game that could have made it amazing.
All the pieces for a randomized single player campaign mode are here. Maps renew thanks to the Sprawl, 18 Operators provide a variety in gameplay, and there are enough objective types to fill out every map in the game and then some. If Ubisoft had implemented some form of roguelike single player experience, giving Rainbow Six fans what they've wanted since Siege transformed it into one of the biggest multiplayer games on the planet, this would be a truly fantastic game.
The endgame content, Assignments and Maelstrom Protocol, add a new layer of variety in the form of unique mutations such as endless waves of enemies or unique mutations for every map and objective. However, they pale in comparison to the closeness you feel with strangers on the internet after a mission. For example, we entered a high-level mission with two random players and smashed the first two objectives. We thought the final one, a Hunt for three strong enemies, would be a piece of cake. We watched on as our teammates were slaughtered and encased in foam, leaving us to sneak through the alerting Archaean stronghold and extract them as quietly as possible. It was incredibly tense and felt like we were playing Alien Isolation, waiting for death to drop down from above at any second. We succeeded though, meaning our teammates wouldn't need to extract an MIA Operative in the future and brought home some serious experience thanks to the plethora of enemies that had been awakened.
That's what every mission in Rainbow Six Extraction feels like with players that understand how to play it. Of course, there will always be those new to the game and others who think they can ignore stealth. But we found that after the initial rush of players in the first week, the people that are left make this a game worth playing.
Rainbow Six Extraction players have themselves mutated since the game's launch. It gives us hope that those still playing understand the game much better than everyone did on release day. This isn't a game you can pick up and run into as you would Call of Duty. It demands your respect, and in return, it respects your time. If you put in the effort to sneak through a mission, save your allies, and extract alive, you'll reap the rewards.
+ Deep, tactical experience for those willing to concentrate and invest their time
+ Endlessly replayable missions that don't feel like a grind if you play them right
– A missed opportunity on the single player front
– Endgame content has more variety but lacks the punch it should have for high-level content
Disclosure: Gamepur was provided with a game code for review purposes.
Can you catch a shiny shadow Registeel in Pokémon Go?
All songs in Theatrhythm: Final Bar Line
Get your final pins in: Rumbleverse is shutting down after six months of service
How to get a shadow Registeel in Pokémon Go


text:
Q: Pointerup event isn't working correctly with Windows Phone 8.1 / IE11 The pointerup event is not working correctly with Win Windows Phone 8.1 and Internet Explorer 11. I mean, it is triggered automatically (just after pointerdown), and not when your finger goes out.
I've done an example here.
If you test it with desktop browsers or Android-Chrome or iOS-Chrome/Safai works correctly
Has anyone any workaround about it?

A: Are you getting a pointercancel event?
If you don't set the touch-action correctly, then pointercancel gets sent through immediately after pointerdown. I think this is because the browser thinks a manipulation such as scrolling has been initiated.
To avoid this we use:
html.is-pointer-events .touch-grip {
    -ms-touch-action: none;
    touch-action: none;
}

Note that we activate this rule only if we are using Microsoft pointer events, because otherwise the touch-action property interferes with other browsers (e.g. Chrome).
We add is-pointer-events to document.documentElement.className if navigator.msPointerEnabled is true (i.e. we add the class when we are using onpointerdown instead of ontouchstart). If you are using modernizr.js then it looks like it provides a pointerevents class you can use.
Unless you have a compelling need for them and you can afford to invest a lot into it, I would strongly recommend against supporting Microsoft pointer events. For us they have been an expensive headache for us to support, and the value returned has been minimal for us.

redpajama_set_name: RedPajamaStackExchange


text:
Dragon Drop
Dragon Drop Slot Review
This game, presented by NextGen Gaming, is full of complete fun, featuring a very animated cartoon of a magical fairytale land with dragons, knights, castles and all the rest thrown in for good measure. It's like something you'd find on a children's cartoon channel, which makes it all that bit more enjoyable – helping you tap in to your inner childhood. The reels feature various different symbols, alongside which you'll find the aforementioned castles and knights, as well as treasure chests full of gold, dragon eggs waiting to hatch, and, oddly enough, sheep.
There's a total of five reels and three rows included in the game, while NextGen have also added a total of 25 different pay lines as well. These can be customised to allow you to wager on as many lines as you'd like, while the bet you're able to place per round can also be altered. This begins from its minimum amount of £0.01 and can be increased up to a maximum level of £2 per active line. This allows all players the possibility of placing a maximum overall wager of £50 per spin.
As with most online video slots, you'll find a wild icon has been included in Dragon Drop. This is represented by the dragon icon, and this appears only on reels two, three and four. If it does so, it has the chance of standing in for any other icon on the reels in ordered to form a win, with the exception of scatters.
And with regards to the scatters, these are represented by the aforementioned dragon eggs. These pay out wherever they appear on the reels, and should three or more of them show up, you'll trigger the eight freespins round. All prizes are doubled in this round, and what's more, there's also the 'Sticky Dragon Feature' included as well. This means that should any dragons appearing during the round, they're held in place for the duration of it.
Finally, you'll notice three dragons at the top of the reels, and during any game these may drop in and turn any symbol into a dragon wild.
This is by far a very entertaining game that NextGen have put together and provided, offering up a cartoon-like set of imagery merged with some entertaining game play.
100% up to €200 + €10 free Play Now


text:
16th Annual Raymond James West Coast Super Senior Grand Prix Tennis Tournament
Feb. 24, 2020 — Mar. 1, 2020
The 16th Annual Raymond James Super Senior Grand Prix is one of the top three senior tennis tournaments held in Florida (USTA Category II National Championship and ITF Grade 1). It attracts close to 300 top players from all over the U.S. and several from outside the country. The seven-day event starts on February 24 for this year. There are eight age divisions in both singles and doubles in 5-year increments, beginning at age 55 through 90. Players earn cash prizes and points for state, national and world rankings.
2000 Country Club Way S
St. Pete, FL 33712
Event Details Venue Information
Start Date: February 24, 2020
End Date: March 1, 2020
St. Pete Country Club
St. Pete , FL 33712


text:
2020 Dodge Challenger SRT Super Stock
ChallengerChallenger SRT Super StockChallenger SRT Hellcat
Challenger SRT Super Stock
2020 Dodge Challenger SRT Super Stock For Sale
Best Muscle Cars
Best Sports Coupes
Dodge Coupes
Dodge Sports Cars
2020 Dodge Challenger SRT Super Stock First Look Review: A Road-Legal Dragster
by Jared Rosenholtz
Just when you think the Challenger SRT Hellcat couldn't get any more powerful, Dodge ups the ante. Last year, the carmaker introduced the Hellcat Redeye, which increased the output from 707 to 797 horsepower. This year, Dodge has managed to squeeze out even more power with another new model based on last year's Redeye. The 2020 Dodge Challenger SRT Super Stock arrives as the most powerful model since the Demon, built to take on America's drag strips.
If you did not have an opportunity to nab one of the 3,000 Demons built in 2018, this new Challenger Super Stock is the next best thing. Unlike the Demon, Dodge says the Super Stock will not have a limited production and will continue into the 2021 model year. This is music to our ears because it means we shouldn't see insane dealer markups or speculators trying to sell cars that haven't even been driven.
Is the 2020 Dodge Challenger a good car?
Exterior Design 10 /10
Performance 10 /10
Fuel Economy 5 /10
Interior & Cargo 7 /10
Infotainment & Features 9 /10
Reliability 8 /10
Safety 7 /10
Value For Money 8 /10
What is BuzzScore?
2020 Dodge Challenger SRT Super Stock Models
SRT Super Stock
6.2L Supercharged V8 Gas
Stealthy Styling?
Dodge says it has designed the Challenger Super Stock based on the "Super Stock" class of vehicle in drag racing. This class is filled with highly modified vehicles that look just like ordinary passenger cars. To that effect, the Super Stock does not have any defining styling features aside from what is already equipped to a Challenger Hellcat. There aren't even any badges signifying that it is a Super Stock. To spot one of these from far away, you'll have to have some serious skills.
The only dead giveaways are the 18-inch-by-11-inch wheels with a Low Gloss Granite finish, which are wrapped in 315/40R18 Nitto NT05R drag radial tires. As for the rest of the styling, Dodge may call this a sleeper car but there is nothing sleepy about the Challenger's aggressive styling, especially when finished in a bright hue like TorRed, Frostbite, or Go Mango.
A Hellcat Interior
By this point, you have either learned to accept the Dodge Challenger for its flaws, or you haven't. The interior, despite having all most of the same features found in luxury cars, such as comfy heated and ventilated seats, and automatic climate control, it doesn't have the high-quality materials to match. There are plenty of cheap plastics and some panels will be questionably fastened down but the money isn't spent on the interior, its spent under the hood.
Even More Power
Dodge has now managed to increase the output of its 6.2-liter supercharged HEMI V8 to 807 horsepower (on pump gas). This is a 10 hp increase over the standard Redeye while torque remains the same at 707 pound-feet. Putting all of this power down is a TorqueFlite 8HP90 eight-speed automatic transmission sending drive to the rear wheels, which are wrapped in sticky drag radials with an increased sidewall. All of these changes help the Super Stock hit 60 mph in just 3.25 seconds with a quarter-mile time of 10.5 seconds at 131 mph. The top speed is limited only by the tires at 165 mph.
This car was clearly built with the drag strip in mind but it still features four-piston Brembo brake calipers with 14.2-inch vented rotors, a new suspension with Bilstein shocks, and an asymmetrical limited-slip differential with a 3.09 final drive ratio. Dodge will also include a slew of performance features such as launch control, line lock, torque reserve, race cooldown, SRT performance pages, and SRT power chiller, which uses the A/C to cool the engine.
Pricing & Competition
Dodge does not appear to have any direct competitors to the Challenger SRT Super Stock. Both the Chevrolet Camaro ZL1 1LE and Ford Mustang Shelby GT500 are too focused on race track performance to keep up with the Super Stock on a drag strip. Those cars should leave the Challenger for dead on the first corner but if you prefer your tracks to be a single long straight, the Super Stock is your car. No pricing information has been announced but we expect the Super Stock to cost more than the Challenger SRT Hellcat Redeye Widebody, which starts at $78,295.
Check other Dodge Challenger Styles
Challenger SRT Hellcat
Dodge Challenger SRT Super Stock Popular Comparisons
The most popular competitors of 2020 Dodge Challenger SRT Super Stock:
2019 Chevrolet Corvette Grand Sport Coupe
2020 Chevrolet Camaro ZL1 Coupe
2020 Chevrolet Corvette Stingray Coupe
Dodge Challenger SRT Super Stock News
Dodge's 807-HP Challenger SRT Super Stock Is Finally Here
The potent pony car has just started shipping out to dealers in the US.
This Baywatch Dodge Challenger Is A Muscle Wagon For The Beach
This is one cool surfboard-carrying lifesaver.
Dodge Reveals Pricing For Challenger SRT Super Stock
This is how much you'll pay for the fastest muscle car in the world


text:
\subsubsection{} laisse encore sous LaTeX2e un espace superflu, le corriger par un \hspace{-0.4cm} (le plus simple pour la soumission etant de faire un \renewcommand\subsubsection[1]{\subsubsection{1} \hspace{-0.4cm}}, l'autre solution etant d'inclure la premiere phrase (en retirant le point de fin) en titre, la phrase etant inseree dans un \textnormal{}, mais cela donne plus de travail a la mise en page specifique)

\documentclass[a4paper,10pt]{amsart}

\usepackage[all]{xy}
\usepackage[T1]{fontenc}
\usepackage[leqno]{amsmath}
\usepackage{amssymb}
\usepackage{amsthm}
\usepackage[english]{babel}

\newcounter{eq}[subsection]
\newcounter{seq}[subsubsection]

\newcommand\eq{\stepcounter{eq}}
\newcommand\num{\eq\tag{\thesubsection.\theeq}}
\newcommand\seq{\stepcounter{seq}}
\newcommand\snum{\seq\tag{\thesubsubsection.\theseq}}

\theoremstyle{plain}
\newtheorem*{theo}{Theorem}
\newtheorem{theoreme}[subsection]{Theorem}
\newtheorem{lemma}[subsection]{Lemma}
\newtheorem{prop}[subsection]{Proposition}
\newtheorem{cor}[subsection]{Corollary}

\theoremstyle{definition}
\newtheorem{rem}[subsection]{Remark}

\theoremstyle{plain}
\newtheorem{stheoreme}[subsubsection]{Theorem}
\newtheorem{slemma}[subsubsection]{Lemma}
\newtheorem{sprop}[subsubsection]{Proposition}
\newtheorem{scor}[subsubsection]{Corollary}
\theoremstyle{definition}
\newtheorem{srem}[subsubsection]{Remark}

\renewcommand{\labelitemi}{--}

\newcommand{\transp}[1]{{}^{\mathit t}\!{#1}}

\newcommand\lra{\longrightarrow}
\newcommand\mso{\mathcal{M}_{\mathbf{SO}_r}}
\newcommand\mo{\mathcal{M}_{\mathbf{O}_r}}
\newcommand\msl{\mathcal{M}_{\mathbf{SL}_r}}
\newcommand\sor{{\mathbf{SO}_r}}
\newcommand\slr{{\mathbf{SL}_r}}
\newcommand\mgl{\mathcal{M}_{\mathbf{GL}_r}}
\newcommand\glr{{\mathbf{GL}_r}}
\newcommand\gl{{\mathbf{GL}}}
\newcommand\ssl{{\mathbf{SL}}}
\renewcommand\o{{\mathbf{O}}}
\newcommand\so{{\mathbf{SO}}}
\newcommand\Sp{{\mathbf{Sp}}}
\newcommand{{\boldsymbol{\mathfrak{so}}}}{{\boldsymbol{\mathfrak{so}}}}
\newcommand{{\boldsymbol{\mathfrak{gl}}}}{{\boldsymbol{\mathfrak{gl}}}}
\newcommand{{\boldsymbol{\mathfrak{sp}}}}{{\boldsymbol{\mathfrak{sp}}}}
\renewcommand\O{\mathcal{O}}

\newcommand\git{/\!\!/}

\newcommand\Q{\widetilde Q}
\newcommand\A{\tilde A}
\newcommand\X{\widetilde X}
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\newcommand\G{{\o_N \times \Sp_{N'}}}
\newcommand\M{\mathrm M}
\newcommand\tr{\mathrm{tr}}
\newcommand\n{{N+N'}}

\pagestyle{plain}

\author{Olivier Serman}
\address{Laboratoire J.-A. Dieudonn\'e \\ UMR 6621 du CNRS \\ Universit\'e de Nice \\ Parc Valrose \\ F-06108 Nice Cedex 02}
\email{serman@math.unice.fr}
\title{Moduli spaces of orthogonal bundles over an algebraic curve}

\begin{document}

\begin{abstract}
We prove that, given $C$ a smooth projective curve of genus $g \geqslant 2$, the forgetful map $\mo \to \mgl$ from the moduli space of orthogonal bundles to the moduli space of all vector bundles over $C$ is an embedding. Our proof relies on an explicit description of a set of generators for the polynomial invariants on the representation space of a quiver under the action of a product of classical groups.
\end{abstract}

\maketitle

\section*{Introduction}

Let $C$ be a smooth, irreducible, projective algebraic curve of genus $g \geqslant 2$ over an algebraically closed field $k$ of characteristic zero. If $G$ is a reductive group we denote by $\mathcal{M}_G$ the moduli space of semi-stable principal $G$-bundles on $C$.

We focus here on the case $G=\sor$, which amounts to considering the moduli space of semi-stable orthogonal bundles of rank $r$ with an orientation. It is a normal projective variety, composed of two connected components distinguished by the second Stiefel-Whitney class. This space is related to the moduli space $\msl$ of vector bundles of rank $r$ and trivial determinant on $C$ through the forgetful map $\mso \to \msl$ which sends any $\sor$-bundle to its underlying vector bundle. It is natural to ask whether this map is a closed embedding. In fact, when $r$ is even, it even fails to be injective, and it is therefore more convenient to ask the same question about $\mo \to \mgl$.

In the same way we consider the forgetful morphism $\mathcal M_{{\Sp}_{2r}} \to \mathcal M_{\ssl_{2r}}$ defined on the moduli space of symplectic bundles of rank $2r$ on $C$.

Our main result may be stated as follows:

\begin{theo}
{\rm{(i)}} The forgetful map $\mo \rightarrow \mgl$ is an embedding.

\indent {\rm {(ii)}} When $r$ is odd, $\mso \to \msl$ is again an embedding, while, when $r$ is even, it is a $2$-sheeted cover onto its image.

\indent {\rm{(iii)}} The forgetful map $\mathcal M_{{\Sp}_{2r}} \to \mathcal M_{\ssl_{2r}}$ is also an embedding.
\end{theo}

We give the full proof for the orthogonal case, and sketch the obvious modifications required by the symplectic one.

We consider in the first section the injectivity of $\mso \to \msl$: this comes down to a close comparison of the equivalence relations between $\sor$-bundles and vector bundles which define the closed points of the corresponding moduli spaces. We then check that the tangent maps of $\mo \to \mgl$ are injective. This differential point of view is much more involved: it relies on Luna's slice \'etale theorem, which naturally leads to the consideration of representations of quivers. To carry our discussion to its end we need an auxiliary result relative to the invariant theory of these representations for the action of a product of classical groups: this is the aim of the second section (note that a characteristic-free proof of this result can be found in \cite{lopatin}). In the third one we show first how this computation results in our main theorem. We then give a few complements about the local structure of the moduli space $\mo$, in the same way as Laszlo did with $\mgl$ (see \cite{localstr}).

\section{Injectivity of $\mathcal M _{\sor} \rightarrow \mathcal M _{\slr}$}

In this section we study the injectivity of the forgetful map $\mathcal M _{\sor} \lra \mathcal M _{\slr}$, which is already known to be finite (e.g. by \cite[8.5]{balsesh}). The closed points of $\mathcal M _G$ are in a one-to-one correspondence with the set of equivalence classes of semi-stable $G$-bundles (cf. \cite{ramanathanthesis}). When $G=\slr$ one easily recovers from this notion Seshadri's definition of $S$-equivalence for vector bundles. We proceed here in the same way to link together equivalence between $\sor$-bundles and $S$-equivalence between their underlying vector bundles.

\subsection{} Let $Q$ be a quadratic form on a $r$-dimensional vector space $V$. Any parabolic subgroup of $\so(Q)$ is the stabilizer of an isotropic flag of $V$. If $0=N_0 \subset N_1 \subset \cdots \subset N_l$ is such a flag then its stabilizer $\Gamma$ stabilizes $0=N_0 \subset N_1 \subset \cdots \subset N_l \subset N_l^\bot \subset \cdots \subset N_1^\bot \subset V$ too. In a basis adapted to the first flag, in which $Q$ is represented by the matrix $\begin{pmatrix} 0 & 0 & \mathrm I_{\dim N_l} \\ 0 & Q' & 0 \\ \mathrm I_{\dim N_l} & 0 & 0 \end{pmatrix}$,  $\Gamma$ must be a subgroup of the group of matrices of the form
$${M=\left( \begin{array}{ccc}
\begin{array}{cccc}{A_1} & \ast  &\cdots  &\ast   \\ 0     & { A_2}&\ddots  &\vdots  \\ \vdots&\ddots & \ddots & \ast  \\ 0  & \cdots&  0 &  A_l \end{array} & \begin{array}{c} \ast \\ \vdots \\ \vdots\\ \ast \end{array} &\textrm{\huge $\ast$} \\
\begin{array}{cccc} 0\ &\cdots \ &\cdots \!&\ 0 \end{array}  & B & \begin{array}{cccc} \ast \quad  & \ \cdots\!  &\ \quad \cdots &\quad \ast  \end{array} \\
\textrm{\LARGE  $0$}  & \begin{array}{c} 0\\ \vdots\\ \vdots\\0\end{array} & \begin{array}{cccc} \ \transp{A_1}\! ^{-1} & 0  &\cdots& 0 \\  \ast &\ \transp{A_2}\! ^{-1}&\ddots& \vdots    \\ \vdots & \ddots & \ddots& 0  \\ \ast & \cdots & \ast &\ \transp{A_l}\! ^{-1} \end{array}
\end{array}\right)},$$
\noindent where $A_i \in \mathbf{GL}_{r_i - r_{i-1}}$, $B \in \mathbf{SO}(Q')$ (and $r_i=\dim N_i$).

Hence, if $P$ is a semi-stable $\sor$-bundle and $E=P(\slr)$ its associated vector bundle, a reduction of structure group $\sigma$ of $P$ to a parabolic subgroup $\Gamma$ defined by an isotropic flag  $0=N_0 \subset N_1 \subset \cdots \subset N_l$ induces isotropic subbundles $E_i = \sigma^\ast E (N_i) \subset E$ of rank $r_i$ giving a filtration  $0=E_0 \subset E_1 \subset \cdots \subset E_l \subset E_l^\bot \subset \cdots \subset E_1^\bot \subset E$. But one knows how to construct such a filtration, with the extra conditions that $E_i/E_{i-1}$ is a stable bundle of degree $0$ and  $E_l ^\bot / E_l$ a stable orthogonal bundle. By \cite[4.5]{ramanan} the latter splits as a direct orthogonal sum of mutually non-isomorphic stable bundles. The graded object $\mathrm{gr} E_{\bullet}$ is then precisely the Jordan-H\"older one, which is known to characterize the point of $\msl$ corresponding to $E$. \label{jordan-holder}

\subsection{} A representative of the equivalence class of $P$ as an orthogonal bundle is given by the $\sor$-bundle $\mathrm{gr} P$ obtained from a suitable reduction of structure group of $P$ to a parabolic subgroup of $\sor$ (cf. \cite[3.12]{ramanathanthesis}). Let us check that the reduction $\sigma$ attached to the above filtration satisfies the conditions of (\textit{loc. cit.}). As any character $\chi$ on $\Gamma$ which is trivial on the neutral component $Z^0$ of its center is of the form $M \mapsto \prod \det(A_i)^{\alpha_i}$ with $\sum (\dim N_i / N_{i-1}) \alpha_i=0$, the assertion $\deg(\chi_\ast \sigma^\ast E (k))=0$ is equivalent to $\mu(E_i / E_{i-1})=0$, $i=1,\ldots,l$, and $\sigma$ definitely is an admissible reduction.

On the other hand the unipotent radical  $\mathrm{R}_u(\Gamma)$ of $\Gamma$ is the unipotent part of the (neutral component of the) intersection of all of its Borel subgroups. Since these are the stabilizers of the flags adapted to the one giving $\Gamma$ whose length is maximal, $\mathrm{R}_u(\Gamma)$ is a subgroup consisting of matrices $M$ with $A_i=\mathrm{id}$ for all $i$ and $B=\mathrm{id}$. We deduce from the preceding a Levi decomposition of $\Gamma$, the Levi component $L$ being isomorphic to  the product  $\prod \mathbf{GL}_{r_i-r_{i-1}} \times \mathbf{SO}_{r-2r_l}$. Let $p \colon \Gamma \rightarrow L$ be the projection on this Levi component. The required stability of the $L$-bundle  $p_\ast \sigma^\ast E$ is then a consequence of the stability of the successive quotients $E_i / E_{i-1}$ together with the following easy fact:

\begin{lemma}
Let $(E_i)_{i=1,2}$ be two stable $G_i$-bundles. Then the  $G_1\times G_2$-bundle $E_1 \times_C E_2$ is also stable.
\end{lemma}

According to \cite[11.14\! (1)]{borel} any parabolic subgroup of $G_1 \times G_2$ is a product  $\Gamma=\Gamma_1 \times \Gamma_2$ where $\Gamma_i$ is a parabolic subgroup of $G_i$. It follows that maximal proper parabolic subgroups may be written $\Gamma_1 \times G_2$ or $G_1 \times \Gamma_2$, $\Gamma_i$ being any maximal proper parabolic subgroup of $G_i$. Therefore the associated bundle $E/\Gamma$ is isomorphic to $E_1/\Gamma_1$ or $E_2/\Gamma_2$, and the lemma is a consequence of the very definition of stability.

\subsection{} The class of $P$ is therefore defined by the $\sor$-bundle $\mathrm{gr} P = p_\ast \sigma^\ast P (\sor)$, and $(\mathrm{gr} P)(\slr)$ is again the Jordan-H\"older graded object. Two semi-stable $\sor$-bundles $P$ and $P'$ are sent to the same point of $\msl$ if and only if $\mathrm{gr} P$ and $\mathrm{gr} P'$ are both obtained from reduction of structure group to $\sor$ of the same polystable vector bundle $E$. Such a reduction amounts to a section of $E/\sor \to C$, and two of them give the same bundles if and only if they are conjugated by the action of  $\mathrm{Aut}_{\slr}(E)$ on $\Gamma(C , E / \sor)$. Elements of $\Gamma(C , E / \sor)$ correspond to isomorphisms $\iota \colon E \buildrel\sim\over\longrightarrow E^\ast$ such that $\iota^\ast=\iota$ and $\det \iota$ is the square of the trivialisation of $\det E$ inherited from the $\slr$-torsor structure. The action of $\mathrm{Aut}_{\slr}(E)$ simply is $$(f , \iota) \in \mathrm{Aut}_{\slr}(E)\times\Gamma(C , E / \sor) \mapsto   f^\ast \iota f.$$

\label{pointsfermes} Since $E$ is polystable $\mathrm{Aut}_{\glr}(E)$ acts transitively on the set $\Gamma(C,E/\o_r)$ of all symmetric isomorphisms from $E$ onto $E^\ast$ : indeed the Jordan-H\"older fitration allows us to split $E$ as
$$E=\bigoplus\limits_i \Bigl(F^{(1)}_{i}\otimes V^{(1)}_{i}\Bigr) \oplus \bigoplus\limits_j \Bigl(F^{(2)}_{j} \otimes V^{(2)}_{j}\Bigr) \oplus \bigoplus\limits_k \Bigl((F^{(3)}_{k} \oplus {F^{(3)}_{k}}{}^\ast)\otimes V^{(3)}_{k}\Bigr),$$
\noindent where $V^{(l)}_{i}$ are finite-dimensional vector spaces and the $F^{(1)}_{i}$ (resp. $F^{(2)}_{j}$, resp. $F^{(3)}_{k}$) are orthogonal (resp. symplectic, resp. non isomorphic to their dual) mutually non isomorphic stable vector bundles, in such a way that any symmetric isomorphism $E \to E^\ast$ is equivalent to the data of orthogonal (resp. symplectic, resp. non degenerate) forms on each one of the $V^{(1)}_i$ (resp. $V^{(2)}_j$, resp. $V^{(3)}_k$). The action of
$$\mathrm{Aut}_\glr(E)= \prod \gl(V^{(1)}_{i}) \times \prod \gl(V^{(2)}_{j}) \times \bigl(\prod \gl(V^{(3)}_{k}) \times \gl(V^{(3)}_{k}) \Bigr)$$
\noindent on the set of these collections is obviously transitive.

Any two elements $\sigma$ and $\sigma'$ of $\Gamma(C , E / \sor)$ are then conjugate under the action of  $\mathrm{Aut}_\glr(E)$, by an automorphism whose determinant equals to $\pm 1$. When $r$ is odd $-\mathrm{id}_E$ is an $\o_r$-isomorphism which exchanges orientation, and the action of $\mathrm{Aut}_{\slr}(E)$ on $\Gamma(C , E / \sor)$ is still transitive. On the contrary when $r$ is even this action fails to remain transitive. For example let $F$ be a vector bundle of rank $r/2$, non isomorphic to its dual, and consider the two orthogonal bundles $F \oplus F ^\ast$ and $F ^\ast \oplus F$, equipped with the standard hyperbolic pairing: these bundles cannot be $\sor$-isomorphic (in fact any orthogonal isomorphism must exchange the orientation). We have proven so far:

\begin{prop}
When $r$ is odd the map $\mathcal M _{\sor}(k) \rightarrow \mathcal M _{\slr}(k)$ is injective; when $r$ is even this is a finite map of degree $2$.
\end{prop}

\begin{rem} \label{giraud}
The distinction between the odd or even case relies on the fact that the semi-direct product $1 \to \sor \to \mathbf{O}_r \to \mathbb Z / 2 \mathbb Z \to 0$ may be direct or not. When it is direct the map  $H^1(C,\sor) \to H^1(C,\mathbf{O}_r)$ is injective. But as soon as $r$ is even the section is no longer compatible with the action of $\sor$ on the previous exact sequence by inner automorphisms; we know then how to compute the fibers of  $H^1(C,\sor) \to H^1(C,\mathbf{O}_r)$ by twisting this sequence (cf. \cite[III 3.3.4]{giraud}). If $r$ is even we have just chosen a bundle $E$ which induces a non trivial connecting homomorphism (since $\mathrm{Aut}_{\sor}(E)\to \mathrm{Aut}_{\mathbf O _r}(E)$ is then an isomorphism), whence the lack of injectivity.
\end{rem}

\section{Invariant theory of representations of quivers}

We refer the reader to \cite{LBP} about the notion of representations of a quiver $Q$ of given dimension $\alpha \in \mathbb N^n$ and their isomorphism classes. Let $Q$ stand for a quiver consisting of $n=n_1+n_2+2n_3$ vertices
$$s_1, \ldots, s_{n_1}, t_1,\ldots,t_{n_2}, u_1,u_1^\ast,\ldots,u_{n_3},u_{n_3}^\ast,$$
\noindent and $\alpha \in \mathbb N^n$ be an admissible dimension vector (that is a vector such that  $\alpha_{t_j}$ is even and $\alpha_{u_k}=\alpha_{u_k^\ast}$). We define $\Gamma_\alpha$ to be the group
$$\Gamma_\alpha=\prod \o_{\alpha_{s_i}} \times \prod \Sp_{\alpha_{t_j}} \times \prod \gl_{\alpha_{u_k}},$$
\noindent which is actually thought here as a subgroup of $\gl(\alpha)=\prod_{i=1}^n \gl_{\alpha_i}$ via the inclusions $P \in \gl_{\alpha_{u_k}} \mapsto (P,\transp P^{-1}) \in \gl_{\alpha_{u_k}} \times \gl_{\alpha_{u_k^\ast}}$. The natural action of $\gl(\alpha)$ on the space $R(Q,\alpha)$ of all representations of $Q$ of dimension $\alpha$ restricts to an action of $\Gamma_\alpha$ on $R(Q,\alpha)$.

Le Bruyn and Procesi have shown in (\textit{loc. cit.}) that the algebra of polynomial invariants  $k[R(Q,\alpha)]^{\gl(\alpha)}$ is generated by traces along oriented cycles in the quiver $Q$. Following their proof, we produce here a set of generators for the algebra of invariants under the action of $\Gamma_\alpha$. The local study of the map $\mo \to \mgl$ made later heavily rests on this description.

\subsection{First fundamental theorem for $\G$}
In this section we adapt the argument of \cite[Appendix 1]{abp}. We will denote by $\M_n$ the space of $n \times n$ matrices. Let $M$ be the matrix  $\left({I_N \atop 0}{0 \atop J_{N'}}\right)$, with $J = \left({0 \atop -I}{I \atop 0}\right)$; it represents a bilinear pairing, given as the standard orthogonal sum of a quadratic form of rank $N$ and a symplectic form of rank $N'$. The key lemma becomes (note that $\M_N \times \M_{N'}$ is identified with its image in $\M_\n$):

\begin{slemma}
Any polynomial function $f \colon (\M_N \times \M_{N'})(k) \to k$ such that $f(BA)=f(A)$ for all $B \in \G$ may be written $f \colon A \mapsto F(\transp AMA)$ with $F$ a polynomial map on $(\M_N \times \M_{N'})(k) \subset \M_\n(k)$.
\end{slemma}

In other words $f$ factors through the application
$$\pi \colon A \in (\M_N \times \M_{N'})(k) \mapsto \transp AMA \in (\M_N \times \M_{N'})(k).$$
\noindent Let $\Psi_{N,N'}$ be its image, which is nothing else than the product of the space of symmetric $N \times N$ matrices and the space of antisymmetric $N' \times N'$ matrices. The restriction of $\pi$ to   $\gl_N \times \gl_{N'}$ identifies the open subset $\Psi_{N,N'}^{\circ}$ consisting of non-degenerate forms with the geometric quotient $(\gl_N \times \gl_{N'}) \git (\G)$ (say by \cite[Proposition 0.2]{GIT}). The lemma then follows from the commutative diagram
$$\xymatrix@C=50pt{
(\M_N \times \M_{N'}) \git (\G) \ar[r]_-{\pi} & \Psi_{N,N'} \ {\vphantom{\ ;}}\\
(\gl_N \times \gl_{N'}) \git (\G) \ar@{}[u]|{\bigcup} \ar[r]^-\sim_-\pi & {\vphantom{}}\  \Psi_{N,N'}^{\circ} \,;  \ar@{}[u]|{\bigcup}
}$$
\noindent the restriction to $(\gl_N \times \gl_{N'}) \git (\G)$ of a map $f$ defined on the good quotient $(\M_N \times \M_{N'}) \git (\G)$ must indeed be induced by a function of the form $A \in \gl_N \times \gl_{N'} \mapsto F(\transp AMA)/H(\transp AM A)$ with $F$ and $H$ two coprime polynomials (defined on $(\M_N \times \M_{N'})(k)$). The equality $F(\transp AMA)=f(A)H(\transp AMA)$ finally ensures that $H$ is invertible.

\subsubsection{} \label{firstmaintheorem} We are now in a position to establish the first main theorem of invariant theory for $\G$. Let us start with the case of multilinear invariants, again after \cite{abp}. Let $V$ be a vector space of dimension $\n$ endowed with the non-degenerate bilinear form  $\langle\cdot\,,\cdot\rangle$ given by the matrix $M$. So $V = V_1 \buildrel\bot\over\oplus V_2$, $V_1$ being a quadratic space of dimension $N$ and $V_2$ a symplectic space of dimension $N'$.

\begin{stheoreme} \label{firsttheorem}
Any linear $\G$-invariant morphism $V^{\otimes 2i} \to k$ is a linear combination of functions of the form
$$\varphi_\sigma \colon v_1 \otimes \cdots \otimes v_{2i} \longmapsto \langle v_{\sigma(1)},v_{\sigma(2)} \rangle \cdots \langle v_{\sigma(2i-1)},v_{\sigma(2i)} \rangle ,$$
with $\sigma \in \mathfrak S _{2i}$.
\end{stheoreme}

Let $\varphi \colon V^{\otimes 2i} \to k$ be any linear $\G$-invariant map, and consider the following polynomial function:
$$f \colon (A,\omega) \in (\mathrm{End}\, V_1 \oplus \mathrm{End}\, V_2) \times V^{\otimes 2i} \mapsto \varphi(A\omega) \in k.$$
By the previous lemma there exists a polynomial $F$ on $(\mathbf S^2 V_1^\ast \oplus \boldsymbol \Lambda^2 V_2^\ast) \times V^{\otimes 2i}$, linear in the second variable, such that $f(A,\omega)= F(\transp A M A,\omega)$. This polynomial certainly is invariant for the natural action of $\gl (V_1) \times \gl (V_2)$ on $(\mathbf S^2 V_1^\ast \oplus \boldsymbol \Lambda^2 V_2^\ast) \times V^{\otimes 2i}$: for any $\Gamma \in \gl(V_1) \times \gl(V_2)$, we have $F({}^{\mathit{t}}\Gamma^{-1} \transp A M A \Gamma^{-1}, \Gamma \omega)=F(\transp A M A,\omega)$.

The assertion results, by polarization, from the description of linear forms on ${V_1^\ast}^{\otimes a_1} \otimes V_1^{\otimes b_1} \otimes {V_2^\ast}^{\otimes a_2} \otimes V_2^{\otimes b_2}$ which are invariant for the action of  ${\gl(V_1)\times\gl(V_2)}$: $F$ is an homogeneous function of degree $i$ in its first variable, which therefore arises from linear forms on $(\mathbf S^2V_1^\ast)^{\otimes k} \otimes V_1^{\otimes 2k} \otimes (\boldsymbol\Lambda^2V_2^\ast)^{\otimes i-k}\otimes V_2^{\otimes 2i-2k}$ (via the projections $(\mathbf S^2 V_1^\ast \oplus \boldsymbol \Lambda^2 V_2^\ast)^{\otimes i} \times V^{\otimes 2i} \to (\mathbf S^2V_1^\ast)^{\otimes k} \otimes V_1^{\otimes 2k} \otimes (\boldsymbol\Lambda^2V_2^\ast)^{\otimes i-k}\otimes V_2^{\otimes 2i-2k}$). Since $\varphi(\omega)=F(M,\omega)$, one then just has to evaluate $F$ on $M$.

\subsubsection{} \label{procesitrick} One easily deduces from the foregoing a family of generators for the algebra of polynomial invariants under the diagonal action (by conjugation) of $\G$ on $\M_{\n}(k)^i$: according to \cite[\S7]{PrAdv} it is enough to work out the behaviour of the composition, the trace and the adjunction (denoted by  $A \mapsto A^\ast=M^{-1}\transp A M$) via the identification $\mathrm{End}\,V \simeq V \otimes V$ induced by the bilinear pairing. If $v=v_1+v_2 \in V=V_1 \oplus V_2$ (cf. \ref{firstmaintheorem}), we have the following identities:

\begin{itemize}
\item $(v \otimes w) \circ (u \otimes t)=\langle v,t\rangle u \otimes w$,
\item $\tr(v \otimes w)=\langle v,w \rangle$,
\item $((v_1+v_2) \otimes w)^\ast = w \otimes (v_1-v_2)$.
\end{itemize}

This relations allow us to translate the functions $\varphi_\sigma$ occuring in theorem \ref{firsttheorem} in a way leading to the following statement:

\begin{stheoreme} \label{maintheorem}
Any $\G$-invariant function defined on $\M_{\n}(k)^i$ is a polynomial in the
$$(A_1,\ldots,A_i) \mapsto \tr(U_{j_1} U_{j_2} \cdots U_{j_k}),$$
with $U_j \in \{A_j,{A_j}^\ast\}$.

The ring of $\G$-equivariant morphisms from $\M_{\n}(k)^i$ to $\M_{\n}(k)$ is generated as an algebra over $k[\M_{\n}(k)^i]^{\G}$ by the elements $(A_1,\ldots,A_i) \mapsto A_j$ or $(A_1,\ldots,A_i) \mapsto {A_j}^\ast$.
\end{stheoreme}

(The second assertion is implied by the first exactly as in \cite{PrAdv}.)

\subsection{Generalization of the main result of \cite{PrJofalg}}
The main result of \cite{PrJofalg} asserts that, if $R$ is a $k$-algebra with trace satisfying the $n$-th Cayley-Hamilton identities, then there exists a universal map $R \to \M_{n}(A)$ which induces an isomorphism $R \to \M_n(A)^{\gl_n}$. In this section we state a similar result dealing with algebras with a trace and an antimorphism of order dividing $4$.

\subsubsection{} \label{foncteur} Let $R$ be a $k$-algebra endowed with an antimorphism $\tau$ of order $4$ (from now on we will write ``of order $4$'' instead of ``of order dividing $4$''). When $R$ is the algebra $\M_\n(B)$ of all matrices with coefficients in a commutative ring $B$ we choose $\tau$ to be the adjunction map (for the considered bilinear form)  $\iota \colon A \in \M_\n(k) \mapsto M^{-1}\transp A M$. As soon as $N$ or $N'$ is zero $\iota$ is in fact of order $2$, and we could restrict ourselves to algebras with anti-involution.

Let $j \colon R \to \M_\n(A)$ be the universal map corresponding to the functor
$$X_{R,\n} \colon B \mapsto \mathrm{Hom}_k(R,\M_\n(B))$$
\noindent (cf. \cite[\S3]{LBP}). The existence of this morphism easily leads to the representability of the functor $\X_{R,\n}$ which associates to any commutative algebra $B$ the set of all morphisms of $k$-algebras from $R$ to $\M_\n(B)$ preserving the antimorphisms. This functor is actually represented by a closed subscheme of $X_{R,\n}$, still called $\X_{R,\n}$: the map $r \in R \mapsto \iota{j\tau^3(r)} \in \M_\n(A)$ comes from a morphism $t \colon A \to A$ of order $4$, and the induced map $\tilde j \colon R \to \M_\n(\A)$ (where $\A$ is the quotient of $A$ by the action of $t$) is universal among the morphisms $R \to \M_\n(B)$ commuting with $\tau$ and $\iota$.

The group $\G$ acts by conjugation on $\M_\n(B)$, inducing a right action on $\A$, hence an action of $\G$ on $\M_\n(\A)$. The universal map $\tilde j$ maps $R$ to the algebra $\M_\n(\A)^{\G}$ of $\G$-equivariant morphisms from $\X_{R,\n}$ to $\M_\n(k)$ (cf. \cite[1.2]{PrJofalg}).

\subsubsection{} A $k$-algebra with trace and antimorphism of order $4$ is an algebra with trace (cf. \cite{PrJofalg}) equipped with an antimorphism $\tau \colon R \to R$ of order $4$ commuting with $\tr$. The algebra $\M_\n(B)$ carries its usual trace and the antimorphism $\iota$ described earlier.

Our purpose is to generalize the main theorem of (\textit{loc. cit.}) to any algebra which is a quotient of the algebra $T_{N,N'}$ of $\G$-equivariant morphisms $\M_{\n}(k)^i \to \M_{\n}(k)$. Note that a set of generators for $T_{N,N'}$ has been given in \ref{maintheorem}.

\begin{sprop}
Let $R$ be a $k$-algebra with trace and antimorphism of order $4$. If $R$ is a quotient of $T_{N,N'}$ then the universal map $\tilde j \colon R \to \M_\n(\A)^{\G}$ is an isomorphism.
\end{sprop}

It follows from an immediate adaptation of Procesi's proof.

\subsection{Generators for $k[R(Q,\alpha)]^{\Gamma_\alpha}$}

\subsubsection{} \label{associatequiver} Let us go back to the quiver $Q$, and the action of $\Gamma_\alpha$ on its representation space. Consider the quiver $\Q$ obtained from $Q$ by adding one new arrow $a^\ast \colon \sigma(v') \to \sigma(v)$ for any arrow  $a \colon v \to v'$, where we called $\sigma$ the involution of the set of vertices fixing the $s_i$ and $t_j$, and permuting $u_k$ and $u_k^\ast$. Let $R$ (resp. $\R$) be the path algebra of the opposite quiver $Q^{\mathrm{op}}$ (resp. $\Q^{\mathrm{op}}$). There is quite a natural way to endow $\R$ with an antimorphism $\tau$ of order $4$ whose action on the idempotents (who are associated to the constant paths) comes from the one of $\sigma$: $\tau$ fixes the $e_{s_i}$ and $e_{t_j}$, permutes $e_{u_k}$ and $e_{u_k^\ast}$, while it sends an arrow $a$ to $\varepsilon a^\ast$, where $\varepsilon$ equals $-1$ if $a$ starts from $s_i$, $u_k$ or $u_k^\ast$ and ends at $t_j$, and $1$ otherwise.

We need here to adapt the map $\iota$ previously defined on $\M_\n(k)$: $\iota$ still associates to a map its adjoint, but the bilinear pairing has to be replaced by the one represented by a matrix of the form
$$\Phi=\left(\begin{array}{ccc} I_{N_1} &0 &0\\0& J_{N_2} &0\\0 &0&\begin{array}{cc} 0 & I_{N_3} \\ I_{N_3} &0 \end{array} \end{array}\right).$$
Put $N_1 = \sum_{i=1}^{n_1} \alpha_{s_i}$, $N_2 = \sum_{j=1}^{n_2} \alpha_{t_j}$, $N_3=\sum_{k=1}^{n_3} \alpha_{u_k}$, and $N=N_1+N_2+N_3$, and consider the decomposition of $k^N$ into pairwise orthogonal subspaces $k^{N_1} \oplus k^{N_2} \oplus k^{2N_3}$ given by $\Phi$. Note that representations of $\R$ of dimension $\alpha$ commuting with $\tau$ and $\iota$ adapted to the previous decomposition correspond bijectively to representations of $R$ of the same dimension adapted to this decomposition. This allows us to identify $R(Q,\alpha)$ with the subspace of $R(\Q,\alpha)$ consisting of all representations which preserve the preceding antimorphisms, that is to a subspace of $\X_{\R,N}(k)$, where $\X_{\cdot ,N}$ is the functor introduced in \ref{foncteur} (once the bilinear form has been replaced by $\Phi$).
\subsubsection{} Consider now the algebra $\St_n$ defined as the quotient of $k[e_{s_i},e_{t_j},e_{u_k},e_{u_k^\ast}]$ by the ideal $J$ generated by the relations $e_v^2=e_v$, $e_v e_{v'}=0$ if $v \neq v'$, $\sum_v e_v=1$. This algebra is contained in $\R$. The restriction of $\tau$ to this algebra is exactly the antiinvolution described in \ref{associatequiver}, and we have a fairly nice description of $\X_{\St_n,N}$:
$$\X_{\St_n,N}=\bigcup_{\sigma,\omega} \X_{\sigma,\omega}\, ,$$
\noindent where $\sigma$ and $\omega$ range over pairs of admissible vectors in $\mathbb N^n$ such that $\sum \sigma_j=N_1+2N_3$ and $ \sum \omega_j=N_2$, the component $\X_{\sigma,\omega}$ being isomorphic to
$$\bigl(\o_{N_1+2N_3}\times\Sp_{N_2}\bigr) / \Bigl(\prod \bigl(\o_{\sigma_{s_i}} \times \Sp_{\omega_{s_i}}\bigr) \times \prod \bigl(\o_{\sigma_{t_j}} \times \Sp_{\omega_{t_j}}\bigr) \times \prod \bigl(\gl_{\sigma_{u_k}} \times \gl_{\omega_{u_k}}\bigr)\Bigr).$$

It induces a decomposition of $\X_{\R,N}$ as the union $\bigcup_{\sigma,\omega} \varpi^{-1}\X_{\sigma,\omega}$, where we call $\varpi \colon \X_{\R,N} \to \X_{\St_n,N}$ the map induced by the inclusion $\St_n \subset \R$. By applying the argument of \cite[\S3]{LBP} to the component corresponding to the dimensional vectors $\sigma$ and $\omega$ whose coordinate are $\sigma_{s_i}=\alpha_{s_i}$, $\sigma_{t_j}=0$, $\sigma_{u_k}=\alpha_{u_k}$, $\omega_{t_j}=\alpha_{t_j}$ and $\omega_{s_i}=\omega_{u_k}=0$, one gets the expected assertion:

\begin{stheoreme} \label{quiver}
The algebra of polynomials on $R(Q,\alpha)$ invariant under the action of $\prod \o_{\alpha_{s_i}} \times \prod \Sp_{\alpha_{t_j}} \times \prod \gl_{\alpha_{u_k}}$ is generated by the functions $$(f_a)_{{}_a} \mapsto \tr (f_{{\tilde a}_p} \cdots f_{{\tilde a}_1}),$$
\noindent where ${\tilde a}_i$ is an arrow in the associated quiver $\Q$ equals to either $a_i$ or ${a_i}^\ast$, in such a way that $({\tilde a}_1,\ldots,{\tilde a}_p)$ forms an oriented path in that quiver and $f_{{\tilde a}_i}$ means $f_{a_i}$ or its adjoint according to whether ${\tilde a}_i$ is $a_i$ or ${a_i}^\ast$.
\end{stheoreme}

\begin{srem}  It is now easy to deal with the case where we let the whole linear group act (by conjugation) above some of the unpaired vertices: let us call $r_1,\ldots,r_{n_4}$ this vertices, $Q'$ the quiver obtained by adding $n_4$ new vertices ${r}_l^\ast$ and $\alpha ' \in \mathbb N^{n_1+n_2+2(n_3+n_4)}$ the admissible vector naturally deduced from any given admissible dimension vector $\alpha \in \mathbb N^{n_1+n_2+2n_3+n_4}$. The group
$$\Gamma_\alpha=\prod \o_{\alpha_{s_i}} \times \prod \Sp_{\alpha_{t_j}} \times \prod \gl_{\alpha_{u_k}} \times \prod \gl_{\alpha_{r_l}}$$
\noindent acts on $R(Q,\alpha)$ and $R(Q',\alpha ')$ (the action of $g \in \gl_{\alpha_{r_l}}$ on ${r}_l^\ast$ being $f \mapsto \transp g^{-1} f \transp g$), and the $\Gamma_\alpha$-equivariant projection $k[R(Q',\alpha')] \to k[R(Q,\alpha)]$ allows us to compute the ring of $\Gamma_\alpha$-invariants of $k[R(Q,\alpha)]$.
\end{srem}

\section{Local study of the forgetful map}

In order to simplify the local study of $\mso \to \msl$ it is convenient to investigate separately the injective morphism $\mo \to \mgl$ and the natural map from $\mso$ to the subscheme $\mo^{\O} \subset \mo$ consisting of all orthogonal bundles with trivial determinant. This distinction seems to be quite valuable since the direct differential study of $\mso$ would involve invariant theory for special orthogonal groups, which is far more difficult to deal with (see \ref{gitforso}). We show here that the former is an embedding, while the later is an isomorphism (resp. a $2$-sheeted cover) when $r$ is odd (resp. even).

\subsection{Differential behaviour of $\mo \to \mgl$}

\subsubsection{}\label{luna} Let us now briefly point out the classical way to analyse the local behaviour of $\mo \to \mgl$. Recall first that this application arises as a quotient (by a general linear group $\Gamma=\gl_M$) of an equivariant map between two well-known parameters spaces $R_{\o_r} \to R_{\glr}$ (cf. \cite[7.3]{BLS}). Luna's \'etale slice theorem and deformation theory then allow us to grasp the local structure of these good quotients (cf. \cite[2.5]{kls}): at any polystable vector bundle $E$, $\mgl$ is \'etale locally isomorphic to an \'etale neighbourhood of the origin in the good quotient
$$\mathrm{Ext}^1(E,E) \git \mathrm{Aut}_{\glr}(E),$$
while $\mo$ is, at any polystable orthogonal bundle $P$, \'etale locally isomorphic to an \'etale neighbourhood of the origin in
$$H^1(C,\mathrm{Ad}(P)) \git \mathrm{Aut}_{\o_r}(P),$$
where $\mathrm{Ad}(P)$ stands for the vector bundle $P \times^{\o_r} {\boldsymbol{\mathfrak{so}}}_r$ associated to the adjoint representation of $\o_r$, which is nothing else than the vector bundle of germs of endomorphisms $f$ of $E$ such that  $\sigma f + f^\ast \sigma=0$, where $\sigma \colon E \to E^\ast$ is the symmetric isomorphism given by the quadratic structure on $E$; in other words the adjoint vector bundle $\mathrm{Ad}(P)$ is canonically isomorphic to $\mathbf{\Lambda}^2E^\ast$.

Then, if $P \in \mo$ is a polystable orthogonal bundle with associated vector bundle $E \in \mgl$, the application $\mo \to \mgl$ coincides at $P$, through the preceding local isomorphisms (in the \'etale topology), with the natural map $$H^1(C,\mathrm{Ad}(P)) \git \mathrm{Aut}_{\o_r}(P) \to \mathrm{Ext}^1(E,E) \git \mathrm{Aut}_{\glr}(E)$$ at the origin. In particular the corresponding tangent maps are identified.

\subsubsection{}\label{immersionorthogonal} A more explicit description of the vector spaces $H^1(C,\mathrm{Ad}(P))$ and $\mathrm{Ext}^1(E,E)$ is then strongly needed in order to understand their quotients; we show here that $H^1(C,\mathrm{Ad}(P)) \git \mathrm{Aut}_{\o_r}(P)$ is a closed subscheme of $\mathrm{Ext}^1(E,E) \git \mathrm{Aut}_{\glr}(E)$, which implies our main theorem.

According to \ref{pointsfermes} the orthogonal structure on the polystable vector bundle $E$ associated to any point $q \in \mo$ gives rise to a splitting of $E$ as a direct orthogonal sum of the form
\begin{align*} \label{polystable} \snum E=\bigoplus\limits_{i=1}^{n_1} E^{(1)}_i \oplus \bigoplus\limits_{j=1}^{n_2}E^{(2)}_j \oplus \bigoplus\limits_{k=1}^{n_3}E^{(3)}_k,
\end{align*}
\noindent where each direct summand $E^{(a)}_l$ may be written as
\begin{itemize}
\item $E^{(1)}_i={F^{(1)}_{i}}\otimes V^{(1)}_{i}$, where $(F^{(1)}_{i})^{}_i$ are mutually non isomorphic stable orthogonal bundles and $(V^{(1)}_{i})^{}_i$ some quadratic vector spaces,
\item $E^{(2)}_j={F^{(2)}_{j}}\otimes V^{(2)}_{j}$, where $(F^{(2)}_{j})^{}_j$ are mutually non isomorphic stable symplectic bundles and $(V^{(2)}_{j})^{}_j$ some symplectic vector spaces,
\item $E^{(3)}_k=({F^{(3)}_{k}} \oplus {F^{(3)}_{k}}{}^\ast) \otimes V^{(3)}_{k}$, where $(F^{(3)}_{k})^{}_k$ are mutually non isomorphic stable bundles such that $F^{(3)}_{k} \not\simeq {F^{(3)}_{k'}}{}^\ast$ and $(V^{(3)}_{k})^{}_k$ some vector spaces carrying a non degenerate bilinear form.
\end{itemize}
\noindent Let us denote by $\psi^{(a)}_{l} \colon F^{(a)}_{l} \to {F^{(a)}_l}{}^\ast$ the duality isomorphism (when $a=1,2$), and $\sigma_l^{(a)} \colon E_{l}^{(a)} \to {E_{l}^{(a)}}{}^\ast$ the symmetric isomorphism defined on $E^{(a)}_l$ (note that ${F^{(3)}_{k}} \oplus {F^{(3)}_{k}}{}^\ast$ has been tacitly endowed with the hyperbolic form).

The two isotropy groups are then easily identified: we find that
$$\mathrm{Aut}_{\glr}(E) \simeq \prod\limits_{i=1}^{n_1} \gl({V^{(1)}_{i}}) \times \prod\limits_{j=1}^{n_2} \gl({V^{(2)}_{j}}) \times \prod\limits_{k=1}^{n_3}\Bigl(\gl(V^{(3)}_{k}) \times \gl({V^{(3)}_{k}})\Bigr),$$
\noindent while $\mathrm{Aut}_{\o_r}(P) \subset \mathrm{Aut}_{\glr}(E)$ is isomorphic to the subgroup
$$\prod\limits_{i=1}^{n_1} {\o}({V^{(1)}_{i}}) \times \prod\limits_{j=1}^{n_2} \Sp(V^{(2)}_{j}) \times \prod\limits_{k=1}^{n_3} \gl(V^{(3)}_{k}),$$
\noindent where $\gl({V^{(3)}_{k}})$ stands for its image in $\gl({V^{(3)}_{k}}) \times \gl({V^{(3)}_{k}})$ by the morphism $g \mapsto (g, \transp g^{-1})$.

The space $\mathrm{Ext}^1(E,E)$ splits into a direct sum of the spaces $\mathrm{Ext}^1(E^{(k)}_i,E^{(l)}_j)$, and each of these summands is isomorphic to $\mathrm{Ext}^1(F^{(k)}_{i},F^{(l)}_{j}) \otimes \mathrm{Hom}({V^{(k)}_{i}},{V^{(l)}_{j}})$ when neither $k$ nor $l$ equals $3$, or to a sum of summands of this form otherwise. The isotropy groups act on each of those spaces via the natural actions of $\gl(V) \times \gl(V')$ on $\mathrm{Hom}(V,V')$.

An element $\omega=\sum \omega^{(k,l)}_{i,j} \in \mathrm{Ext}^1(E,E) \simeq \bigoplus \mathrm{Ext}^1(E^{(k)}_i,E^{(l)}_j)$ belongs to the space $H^1(C,\mathrm{Ad}(P))$ if and only if $\omega^{(i,i)}_{k,k} \in H^1(C,\boldsymbol{\Lambda}^2E^{(i)}_k{}^\ast) \subset \mathrm{Ext}^1(E^{(i)}_k,E^{(i)}_k)$ and, for $(i,k) \neq (j,l)$, $\sigma^{(l)}_j \omega^{(k,l)}_{i,j} + {\omega^{(l,k)}_{j,i}}{}^\ast \sigma^{(k)}_i=0$. So, identifying $\mathrm{Ext}^1(E_i^{(k)},E_j^{(l)})$ with its image in $\mathrm{Ext}^1(E_i^{(k)},E_j^{(l)}) \oplus \mathrm{Ext}^1(E_j^{(l)},E_i^{(k)})$ by the application $\omega_{i,j}^{(k,l)} \mapsto \omega_{i,j}^{(k.l)} - {\sigma_i^{(k)}}{}^{-1}{\omega_{i,j}^{(k,l)}}{}^\ast \sigma_j^{(l)}$, it appears that $H^1(C,\mathrm{Ad}(P))$ is the subspace of $\mathrm{Ext}^1(E,E)$ isomorphic to the direct sum \stepcounter{seq}
\begin{multline*}
\snum \label{H1} \bigoplus\limits_{k} \left(\bigoplus\limits_i  H^1(C,\boldsymbol{\Lambda}^2E^{(i)}_k{}^\ast) \oplus \bigoplus\limits_{i<j} \mathrm{Ext}^1(E^{(k)}_i,E^{(k)}_j) \right) \oplus
\\
\bigoplus\limits_{k<l} \bigoplus\limits_{i,j}  \mathrm{Ext}^1(E^{(k)}_i,E^{(l)}_j),
\end{multline*}
\noindent each one of the diagonal summand being more precisely expressed as:
\begin{multline*} \snum   H^1(C,\boldsymbol{\Lambda}^2E^{(1)}_i{}^\ast)= \Bigl( H^1(C,{\mathbf{S}}^2F^{(1)}_i{}^\ast) \otimes {\boldsymbol{\mathfrak{so}}}({V^{(1)}_{i}})\Bigr)  \oplus\\ \Bigl(H^1(C,\boldsymbol{\Lambda}^2F^{(1)}_i{}^\ast) \otimes \mathbf{S}^2 {V^{(1)}_{i}}{}^\ast\Bigr),
\end{multline*}
\begin{multline*}
\snum  H^1(C,\boldsymbol{\Lambda}^2F^{(2)}_j{}^\ast)=\Bigl( H^1(C,\boldsymbol{\Lambda}^2F^{(2)}_j{}^\ast) \otimes {\boldsymbol{\mathfrak{sp}}}({V^{(2)}_{j}})\Bigr) \oplus \\
\Bigl( H^1(C,\mathbf{S}^2F^{(2)}_j{}^\ast) \otimes \boldsymbol{\Lambda}\!^2 {V^{(2)}_{j}}{}^\ast\Bigr),
\end{multline*}
\begin{multline*}
\snum \label{contributionhyperbolique}  H^1(C,\boldsymbol{\Lambda}^2E^{(3)}_k{}^\ast)= \Bigl(\mathrm{Ext}^1(F^{(3)}_{k},F^{(3)}_{k}) \otimes {\boldsymbol{\mathfrak{gl}}}({V^{(3)}_{k}})\Bigr) \oplus \\ \Bigl(H^1(C,\mathbf{S}^2F^{(3)}_k{}^\ast) \otimes \boldsymbol{\Lambda}^2V^{(3)}_{k}{}^\ast\Bigr) \oplus \Bigl(H^1(C,\boldsymbol{\Lambda}^2F^{(3)}_k{}^\ast) \otimes \mathbf{S}^2({V^{(3)}_{k}}{}^\ast)\Bigr) \oplus \\ \Bigl(H^1(C,\mathbf{S}^2F^{(3)}_k) \otimes \boldsymbol{\Lambda}^2V^{(3)}_{k}{}^\ast\Bigr) \oplus \Bigl(H^1(C,\boldsymbol{\Lambda}^2F^{(3)}_k) \otimes \mathbf{S}^2{V^{(3)}_{k}}{}^\ast\Bigr),
\end{multline*}
\noindent where $\mathrm{Ext}^1(F^{(3)}_{k},F^{(3)}_{k})$ has been identified with its image in $\mathrm{Ext}^1(F^{(3)}_{k},F^{(3)}_{k}) \oplus \mathrm{Ext}^1({F^{(3)}_{k}}{}^\ast,{F^{(3)}_{k}}{}^\ast)$ by the map $\omega \mapsto \omega-\omega^\ast$. Note that the dimensions of all the extension spaces under consideration are trivially available.

\subsubsection{} This pretty intricate situation suitably expresses itself in terms of representations of quivers. Indeed let us consider the quiver $\mathcal Q$ whose set of vertices is
$${{\mathcal Q}_0=\lbrace{s_{1}^{(1)},\ldots,s^{(1)}_{n_1},s^{(2)}_{1},\ldots,s^{(2)}_{n_2},s^{(3)}_{1},s^{(3^\ast)}_{1},\ldots,s^{(3)}_{n_3},s^{(3^\ast)}_{n_3}\rbrace}},$$
\noindent these vertices being connected by $\dim \mathrm{Ext}^1(F_{i}^{(k)},F_{j}^{(l)})$ arrows from $s^{(k)}_i$ to $s^{(l)}_j$ (where we have set $F_{i}^{(3^\ast)}={F_{i}^{(3)}}{}^\ast$). Next define $\alpha \in \mathbb{N}^{n_1 + n_2 +2 n_3}$ according to the dimensions of the corresponding vector spaces $V^{(a)}_{l}$. Therefore the $\mathrm{Aut}_{\gl_r}(E)$-module $\mathrm{Ext}^1(E,E)$ is exactly the $\gl(\alpha)$-module $R(\mathcal Q,\alpha)$ composed of all the representations of $\mathcal Q$ of dimension $\alpha$, and the result of \cite{LBP} recalled earlier provides us with a description of the algebra $k[\mathrm{Ext}^1(E,E)]^{\mathrm{Aut}_{\gl_r}(E)}$.

The inclusion $H^1(C,\mathrm{Ad}(P)) \hookrightarrow \mathrm{Ext}^1(E,E)$ is an $\mathrm{Aut}_{\o_r}(P)=\prod\limits_{i=1}^{n_1} {\o}({V^{(1)}_{i}}) \times \prod\limits_{j=1}^{n_2} \Sp(V^{(2)}_{j}) \times \prod\limits_{k=1}^{n_3} \gl(V^{(3)}_{k})$-equivariant application, so that we have an exact sequence $$k[\mathrm{Ext}^1(E,E)]^{\mathrm{Aut}_{\o_r}(P)} \to k[H^1(C,\mathrm{Ad}(P))]^{\mathrm{Aut}_{\o_r}(P)} \to 0.$$
\noindent This sequence and the theorem \ref{quiver} result in a description of a set of generators for the algebra $k[H^1(C,\mathrm{Ad}(P))]^{\mathrm{Aut}_{\o_r}(P)}$, namely the $(f_a)_{{}_a} \mapsto \tr (f_{{\tilde a}_p} \cdots f_{{\tilde a}_1})$, where $f_{{\tilde a}_i}$ stands for either $f_{a_i}$ or its adjoint map. Now, according to (\ref{H1}),  $H^1(C,\mathrm{Ad}(P))$ is a subspace of $R(\mathcal Q,\alpha)$ made up of representations having the following property: if $f_a \colon V_v \to V_{v'}$ denotes the map associated to an arrow $a \colon v \to v'$, then its adjoint map $f_a^\ast \colon V_{v'}^\ast \to V_{v}^\ast$ is, up to the sign, the map associated to one of the arrows from $v'$ to $v$. So the algebra $k[H^1(C,\mathrm{Ad}(P))]^{\mathrm{Aut}_{\o_r}(P)}$ is generated by traces along oriented cycles in the quiver $\mathcal Q$. This exactly means that the application $k[\mathrm{Ext}^1(E,E)]^{\mathrm{Aut}_{\gl_r}(E)} \to k[H^1(C,\mathrm{Ad}(P))]^{\mathrm{Aut}_{\o_r}(P)}$ is onto.

In view of what has been discussed in \ref{luna} this proves the injectivity of the tangent map of $\mo \to \mgl$ at $q=[P]$. But we have shown in \ref{pointsfermes} that the map $\mo(k) \to \mgl(k)$ is injective. This implies the following:

\begin{stheoreme}
The forgetful map $(E,q) \in \mo \mapsto E \in \mgl$ is an embedding.
\end{stheoreme}

 One easily gets in the very same way the corresponding assertion relative to the moduli of symplectic bundles:
\begin{stheoreme}
The forgetful map $\mathcal M_{{\Sp}_{2r}} \to \mathcal M_{\ssl_{2r}}$ is an embedding.
\end{stheoreme}
The point is that any closed point of $\mathcal{M}_{{\Sp}_{2r}}$ represents a polystable vector bundle of the form
$$E=\bigoplus\limits_i \Bigl(F^{(1)}_{i}\otimes V^{(1)}_{i}\Bigr) \oplus \bigoplus\limits_j \Bigl(F^{(2)}_{j} \otimes V^{(2)}_{j}\Bigr) \oplus \bigoplus\limits_k \Bigl((F^{(3)}_{k} \oplus F_{k}^{(3) \, \ast})\otimes V^{(3)}_{k}\Bigr),$$
\noindent where $(F^{(1)}_{i})_i$ (resp. $(F^{(2)}_{j})_j$, resp. $(F^{(3)}_{k})_k$) is a family of mutually non isomorphic symplectic (resp. orthogonal, resp. not self-dual) bundles (which are stable as vector bundles), and $(V^{(1)}_{i})_i$ (resp. $(V^{(2)}_{j})_j$, resp. $(V^{(3)}_{k})_k$) a family of quadratic (resp. symplectic, resp. endowed with a non-degenerate bilinear pairing) vector spaces ($F^{(3)}_{k} \oplus F^{(3) \, \ast}_{k}$ being now equipped with the standard symplectic form). Let us denote by $\sigma \colon E \to E^\ast$ the resulting symplectic form on $E$. The bundle $\mathrm{Ad}(P)$ is now isomorphic to the bundles of germs of symmetric endomorphisms of $E$  (that is endomorphisms verifying $\sigma f + f^\ast \sigma=0$), and both the space $H^1(C,\mathrm{Ad}(P))$ and the considered isotropy groups can be described in a manner analogous to that of \ref{immersionorthogonal} (one only has to switch the factors $\boldsymbol{\Lambda}^2F^{(a)}_l{}^\ast$ and $\mathbf{S}^2F^{(a)}_l{}^\ast$, and of course to redefine in the obvious way every map of the form $\mathrm{Ext}^1(F,F') \to \mathrm{Ext}^1(F,F') \oplus \mathrm{Ext}^1(F'^\ast,F^\ast)$). The theorem \ref{quiver} then allows us to conclude again.

\subsection{About $\mso \to \mo$} \label{gitforso}

\subsubsection{} We have recalled in \ref{giraud} how to compute the fibers of the finite morphism from $\mso$ onto $\mo^{{\O}}=\det^{-1}({\O}_C)$. A point $[P] \in \mo$ in its image has two antecedents if and only if $\mathrm{Aut}_\sor(P) \hookrightarrow \mathrm{Aut}_{\o_r}(P)$ is an isomorphism, that is if and only if every orthogonal bundle $F_{i}^{(1)}$ appearing in the splitting (\ref{polystable}) of $E$ has even rank.

Luna's theorem reduces once again the differential study of this application to an invariant calculus: the tangent map of $\mso \to \mo$ at $[P] \in \mso$ is indeed identified with that of  $H^1(C,\mathrm{Ad}(P)) \git \mathrm{Aut}_\sor(P) \to H^1(C,\mathrm{Ad}(P)) \git \mathrm{Aut}_{{\o}_r}(P)$ (at the origin).

Therefore, if $r$ is odd, $\mso \to \mo^{\O}$ is an isomorphism.

\subsubsection{} Let us consider now the even case. The morphism $\mso \to \mo$ is then a $2$-sheeted cover, which is \'etale above the locus of points having two antecedents. A branched point corresponds to an orthogonal polystable bundle $E$ containing at least one subbundle isomorphic to $F^{(1)}_{i} \otimes V^{(1)}_{i}$ where $F^{(1)}_{i}$ is an orthogonal bundle of odd rank: we then have to understand the inclusion
$$k[H^1(C,\mathrm{Ad}(P))]^{\mathrm{Aut}_{\o_r}(P)} \hookrightarrow k[H^1(C,\mathrm{Ad}(P))]^{\mathrm{Aut}_{\sor}(P)}.$$
\noindent It is easy to produce a primitive element for the generic extension (which is of degree $2$). First note that the vector space $W$ obtained as the direct sum of the $V^{(1)}_i$ corresponding to the orthogonal bundles $F^{(1)}_i$ of odd rank has even dimension, and has an orthogonal structure inherited from the ones of the $V^{(1)}_i$. The space composed of all the antisymmetric endomorphisms of $W$ may be identified with a direct summand of $H^1(C,\mathrm{Ad}(P))$, and mapping any element $\omega \in H^1(C,\mathrm{Ad}(P))$ to the pfaffian of the endomorphism of $W$ induced by $\omega$ then defines a function belonging to $k[H^1(C,\mathrm{Ad}(P))]^{\mathrm{Aut}_\sor(P)}$ which is not $\mathrm{Aut}_{\o_r}(P)$-invariant; this function certainly generates the generic extension.

It is more difficult to give a convenient description of this algebra: in the (simplest) case where $P$ is isomorphic to $\O \otimes V$ with $V$ a quadratic vector space of even dimension, we have to understand the action of $\mathrm{Aut}_{\sor}(P) \simeq \sor$ on $H^1(C,\mathrm{Ad}(P)) \simeq H^1(C,\O) \otimes {\boldsymbol{\mathfrak{so}}}(V)$. This can be solved again thanks to Procesi's trick (cf. \ref{procesitrick}): the computation has been carried out in \cite{aslaksen}, and provides a set of generators for the $k[H^1(C,\mathrm{Ad}(P))]^{\mathrm{Aut}_{\o_r}(P)}$-algebra $k[H^1(C,\mathrm{Ad}(P))]^{\mathrm{Aut}_{\sor}(P)}$ in terms of \textit{polarized pfaffians}.

Let us finally mention that in the general case we can easily infer from the main result of \cite{lopatin} a family of generators for $k[H^1(C,\mathrm{Ad}(P))]^{\mathrm{Aut}_\sor(P)}$ which are also obtained as polarized pfaffians.

\subsection{About the multiplicity at stable points}

\subsubsection{} The discussion held in \ref{luna} contains in fact a more precise statement, related to the completed local rings of $\mo$ and $\mgl$. Indeed, if $q$ is a point of $\mo$ representing a polystable bundle $P$ whose image in $\mgl$ is a point $s=[E]$, we have the following commutative diagram, where the rings of the second row are the completions of the local rings (of the involved algebras of invariants) at the origin,
\begin{align} \snum  \label{lunadiag}
\begin{array}{c}
\xymatrix@C=35pt{
\widehat{\O}_{\mgl ,\, s} \ar[d]^-{\wr} \ar[r]  &\widehat{\O}_{\mo ,\, q} \ar[d]^-\wr\\
 \Bigl(k[\mathrm{Ext}^1(E,E)]^{\mathrm{Aut}_{\gl_r}(E)}\Bigr)^{\widehat{\vphantom{}}}\ar[r] & \Bigl(k[H^1(C,\mathrm{Ad}(P))]^{\mathrm{Aut}_{\o_r}(P)}\Bigr)^{\widehat{\vphantom{}}}.}
\end{array}
\end{align}
\noindent This description of the completed local rings of $\mo$ provides us with additional informations about the local structure of $\mo$, at least at the points where the situation is not too bad (see \cite{localstr} for the case of $\mgl$): the more we know about the second main theorem of invariant theory for the isotropy group of $P$, the easier our calculations will be.

\subsubsection{} Let $P$ be an orthogonal bundle whose underlying vector bundle is of the form $E=E_1 \oplus E_2$, with $E_1$ and $E_2$ two non-isomorphic $\gl$-stable orthogonal bundles. The description of the inclusion $H^1(C,\mathrm{Ad}(P)) \hookrightarrow \mathrm{Ext}^1(E,E)$ given in (\ref{H1}) here reduces to
$$\begin{array}{ccccccc}
H^1(C,\mathrm{Ad}(P)) &=& H^1(C,\boldsymbol{\Lambda}^2E_1{}^\ast) & \oplus & \mathrm{Ext}^1(E_1,E_2) & \oplus & H^1(C,\boldsymbol{\Lambda}^2E_2{}^\ast) \\
\cap                  & &    \cap                 &        &          \cap           &        &   \cap     \\
\mathrm{Ext}^1(E,E)   &=& \mathrm{Ext}^1(E_1,E_1) & \oplus & \mathrm{Ext}^1(E_1,E_2)\  \oplus \ \mathrm{Ext}^1(E_2,E_1) & \oplus & \mathrm{Ext}^1(E_2,E_2); \end{array}$$
\noindent the isotropy subgroup $\mathrm{Aut}_{\o_r}(P)$, isomorphic to $\mathbb Z / 2 \mathbb Z \times \mathbb Z/2\mathbb Z$, acts trivially on $H^1(C,\boldsymbol{\Lambda}^2E_i{}^\ast)$ and by multiplication by $\pm 1$ on $\mathrm{Ext}^1(E_1,E_2)$ (while $(\alpha_1,\alpha_2) \in \mathrm{Aut}_\glr(E) \simeq \mathbb G_m \times \mathbb G_m$ acts on $\mathrm{Ext}^1(E_i,E_j)$ by $\alpha_j \alpha_i^{-1}$).

Let $(X_k^{(i)})^{}_k$ (resp. $(Y_l)_l$) be a basis of $ H^1(C,\boldsymbol{\Lambda}^2E_i{}^\ast)^\ast$ (resp. $\mathrm{Ext}^1(E_1,E_2)^\ast \subset H^1(C,\mathrm{Ad}(P))^\ast$). Then $k[H^1(C,\mathrm{Ad}(P))]^{\mathrm{Aut}_{\o_r}(P)}$ is the subring of $k[X_k^{(i)},Y_l]$ generated by all the $X_k^{(i)}$ and the products $Y_l Y_{l'}$; if $\mathcal V$ denotes the affine cone over the Veronese variety $\mathbb P(\mathrm{Ext}^1(E_1,E_2)) \subset \mathbb P\left(\mathbf{S}^2\mathrm{Ext}^1(E_1,E_2)\right)$ we get the following isomorphism:
$$\mathrm{Spec}(k[H^1(C,\mathrm{Ad}(P))]^{\mathrm{Aut}_{\o_r}(P)}) \buildrel\sim\over\lra \Bigl(H^1(C,\mathrm{Ad}(P_1))\oplus H^1(C,\mathrm{Ad}(P_2))\Bigr) \times \mathcal V.$$

Using the identification $\widehat{\O}_{\mo,\, q} \simeq \Bigl(k[H^1(C,\mathrm{Ad}(P))]^{\mathrm{Aut}_{\o_r}(P)}\Bigr)^{\widehat{\vphantom{}}}$ we have the following result:

\begin{scor}
The tangent space to $\mo$ at a point $[P]$ given as the direct sum $E_1 \oplus E_2$ of two non-isomorphic stable orthogonal bundles is isomorphic to
$$ H^1(C,\boldsymbol{\Lambda}^2E_1{}^\ast)\oplus  H^1(C,\boldsymbol{\Lambda}^2E_2{}^\ast) \oplus (\mathbf{S}^2\mathrm{Ext}^1(E_1,E_2)),$$
and the multiplicity of $\mo$ at this point is equal to $\displaystyle{2 ^{r_{1} r_{2}(g-1)-1}}$, where $r_i$ is the rank of $E_i$.
\end{scor}

\begin{srem} (i) The general case of a stable point $q \in \mo$ is more difficult: such a bundle corresponds to a vector bundle which splits as a direct sum of $n$ mutually non-isomorphic $\gl_{r_i}$-stable orthogonal vector bundles. We can use \ref{quiver} to try to get some additional informations about the local structure at $q$, for instance by computing the multiplicity at this point. One can easily check that, if $n=3$ (resp. $4$) this multiplicity is equal to $\displaystyle{2\prod_{i<j}2^{r_ir_j(g-1)-1}}$ (resp. $\displaystyle{8\prod_{i<j}2^{r_ir_j(g-1)-1}}$).

\noindent(ii) It is not hard to deal with the case of an orthogonal (non stable) bundle of the form $F \oplus F^\ast$ with $F \not\simeq F^\ast$: we see that $\mathcal M_{\o_r}$ is, at such a point, \'etale locally isomorphic to $\mathrm{Ext}^1(F,F) \oplus \mathcal S$, where $\mathcal S$ is the affine cone over the Segre variety
$$\mathbb P(H^1(C, \boldsymbol{\Lambda}^2F^\ast))\times \mathbb P(H^1(C,\boldsymbol{\Lambda}^2F))\subset \mathbb P\left(H^1(C,\boldsymbol{\Lambda}^2F^\ast)\otimes H^1(C,\boldsymbol{\Lambda}^2F)\right).$$
\end{srem}

\nocite{*}
\bibliographystyle{amsplain}

redpajama_set_name: RedPajamaArXiv


Below are some of the most frequently asked questions about dental implants and the services we provide.
Do I need to be referred by my general dentist to come to your office?
Not at all. You can come to us directly and should call us to make an appointment for an examination or consultation. This is true whether you have a current dentist or not. If there are any records that we may need from the general dentist who last treated you, we can easily obtain them.
What happens in my first full examination?
Our full examinations for new patients are very thorough. We take complete digital X-rays, intraoral and extraoral photos and when needed, a 3D CT scan of your jaw, teeth and surrounding structures.
Dr. Rapoport or Dr. Schuler will conduct a full oral examination which also includes an oral cancer screening, a bite and jaw joint check and a review of your medical and dental history. The doctor will discuss your goals with you, thoroughly explain your treatment options and answer all your questions.
Learn more about our comprehensive new patient exam and dental implant consultation here.
We currently have a comprehensive exam and dental implant consultation which contains all the above for $99. This is a $590 value, and your $99 payment will be applied to any future treatment. This makes it very easy to find out if you are a candidate for dental implants and discover all the treatment options available to you. You can easily schedule an appointment by calling our office or requesting one online.
Should I always try to save my teeth?
Not in every instance. Dr. Rapoport and Dr. Schuler are experts at saving teeth and often do so even when teeth are severely compromised. However, there can be times when teeth are so deteriorated that the work needed to try to save them is very costly, very extensive and still runs the risk of failing completely in the near future. Our doctors can examine you and let you know what they recommend in your particular situation. For teeth that are broken-down and should be replaced, Dr. Rapoport or Dr. Schuler can extract your teeth, place a dental implant and give you a functional replacement tooth all on the same day.
Learn more about what can be done with broken-down teeth here.
How will I know whether a denture, a partial denture or an implant is the best option if I have broken-down teeth?
Although implants are almost always the best option, our doctors will only be able to make a final determination after a complete examination. The exam will include full X-rays and 3D imaging plus a thorough physical evaluation of your teeth, bone levels and gum health. They will also obtain a full dental and medical history. After all the information is analyzed they will explain your options in detail, give you the pros and cons of each, and answer all your questions.
Learn more about our new patient exam and implant consultation here.
Due to major advancements in technology, most people are now able to safely receive and enjoy the benefits of dental implants. This even includes patients who have experienced severe bone loss in their jaw, who smoke heavily or who have medical problems such as diabetes or heart disease. You can receive dental implants at any age, the only exception being young people whose jaws have not yet fully developed.
Learn more about who can receive dental implants here.
Who are candidates for implants and replacement teeth on the same day?
People who need their teeth extracted and people who already have missing teeth but don't have severe bone loss can be candidates for implant placement and functional new teeth in the same visit.
Patients can replace individual teeth with single implants mounted by crowns, and they can replace several missing teeth with implant-supported bridges. For patients who need all their teeth replaced, Dr. Rapoport or Dr. Schuler can extract any remaining teeth, place the necessary implants and provide a full-arch set of new replacement teeth on the same day, using the All-on-4® New Teeth in One Day procedure.
Learn more about same-day tooth replacement here.
That all depends on your particular circumstances. Dr. Rapoport or Dr. Schuler will determine if any bone grafting is needed after a complete examination and evaluation. For patients receiving extractions and immediate implants with replacement teeth on the same day, or for patients eligible for the All-on-4 New Teeth in One Day procedure, the need for bone grafting is extremely rare. You can find out what would be needed in your case in a new patient exam and consultation.
The essence of the All-on-4 procedure is to place four specially designed implants in strategic positions and at precise angles in the jaw where the bone is of the highest volume and density, thus providing the greatest stability and support. These implants support a set of full-arch replacement teeth that are firmly fixed in your mouth and can only be removed by the dentist.
Learn more about All-on-4 New Teeth in One Day here.
Would the All-on-4 procedure be good for me if I'm already wearing dentures?
Absolutely. Patients who have already been wearing dentures can benefit just as much from the All-on-4 New Teeth in One Day procedure as patients who still have some remaining teeth that need to be extracted. Come to our office for a full examination and consultation with one of our doctors to see if you are a candidate for All-on-4.
Learn more about discovering if you are a candidate for All-on-4 here.
How long does the All-on-4 procedure take on the day I receive my implants and new teeth?
The entire procedure for both arches will generally take between six to eight hours from the time you are seated in the treatment room. You will walk out the door that day with all implants placed and a complete set of new replacement teeth. We schedule patients early in the morning so they can leave our office by the end of the afternoon.
Learn more about our All-on-4 Procedure here.
How soon after extractions would it be best to receive an implant?
It would almost always be best to receive the implant immediately after your teeth are extracted, as missing teeth can cause deterioration of your bone and gums. Implants need a certain amount of bone to be present in your jaw to allow for stable placement. Our advanced tooth extraction procedures preserve maximum bone and gum during extraction and reduce or eliminate the need for any bone grafting.
Learn more about our Same-Day Tooth Replacement here.
How much downtime and recovery time is involved when receiving an individual implant?
If you are receiving anywhere from one to four implants at the same time, you are not out of action at all– you can be up and about the same day if you don't take sedation. If you undergo the procedure under sedation, you should take the day of the procedure off and have someone drive you to and from the office. You will normally be able to go back to work the next day.
How much downtime and recovery time is involved for the All-on-4 procedure?
With All-on-4 treatment, most people have minimal downtime–they are able to resume normal activities the same day when they get home from the procedure. However, you will need to schedule some downtime for the two days following your procedure as there is some swelling that needs to come down and you will need time to get used to your new teeth. You will be given specific instructions to follow regarding medications, diet, and cleaning your new teeth over the next several weeks, as your implants heal.
Is there discomfort associated with receiving implants?
For individual implants, there is little to no discomfort during the procedure, and post-operative discomfort is minimal. This would be true for receiving one implant and all the way to getting four implants in one sitting.
For the All-on-4® New Teeth in One Day procedure, there is virtually no discomfort during the procedure as you are under sedation or general anesthesia. You will receive specific instructions for analgesic medications on a precise schedule for the next ten days to minimize any potential discomfort that could occur afterwards.
How much do implants or All-on-4 cost?
There is no way to give a flat cost for implants or the All-on-4® procedure, as each case is different and the exact procedural requirements will only be known after a complete oral exam which includes full digital X-rays and 3D scans.
We use only the highest quality brands of implants and components, thus guaranteeing maximum stability and success rate. This proves to be more cost-effective over time, as these implants and materials provide a very stable foundation and usually last a lifetime with proper care.
Learn more about the cost of dental implants here.
Yes. We accept cash, checks and all major credit cards. We also work with two major health care financing services that provide zero-interest financing to qualified applicants. Our treatment coordinator will help you with all financial options to make paying for your treatment as convenient as possible.
Even if we are not in-network with your insurance, we will happily verify your coverage when you see us. Since certain procedures can be covered by medical insurance, we request that you bring us the information for any medical as well as dental insurance policies. We will file your claim on your behalf to help you get the maximum reimbursement from your insurer for covered procedures.
Learn more about how we work with your insurance here.
All insurance policies have different coverage for different procedures. We will happily verify your insurance coverage when you come to see us.
We are conveniently located at the Southcenter Professional Plaza, 411 Strander Blvd, Suite 302, Seattle, WA 98188-2964. We are at the corner of Strander Blvd. and Andover Park E. Our office is just four miles due east of Seattle-Tacoma International Airport, about one mile east of Interstate 5 and about one mile south of Interstate 405. Across the street, to the north on Strander Blvd., is Albert Lee Appliances, and directly west of us across our parking area is a Barnes & Noble.
See the map on our Contact Us page for further detail.
To make an appointment, call 206‑575‑1086 or click here to request an appointment online.


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