text:
The Legion of Super-Villains
+ Saturn Queen
Created by Jerry Siegel	and Curt Swan
From Final Crisis: Legion of Three Worlds #1 (2008); art by George Pérez and Scott Koblish.
» FIRST APPEARANCE:
Adult: Superman vol. 1 #147 (Aug. 1961)
Teen/canonical: Adventure Comics #372 (Sept. 1968)
The Adult Legion of Super-Villains
The Canonical/Teen LSV (and Its Kin)
Legion vol. 4: The Post-Crisis Saturn Queen
Light Speed Vanguard
Appearances / References
See things from Superman's perspective: from his time in the 20th century, he encounters the Legion of Super-Heroes — exciting teen visitors from the distant future. (Adventure Comics #247) The next time he meets them, they're dressed differently. (Adventure #267) Later in life he meets adult versions of those teens — who are evil. (Superman #147) 'I guess they came from different points in time.'
Then after fighting a cosmic Crisis, Superman forgets all of these people, and meets strangers called "Brainiac 5" and "Blok." Again later he meets the original three Legionnaires. (Adventures of Superman #476) After another crisis in time (the "Zero Hour"), these founders were suddenly much younger, and one was now called "Live Wire" instead of "Lightning Lad." (Legion vol. 4 #85)
What is Superman left to conclude? Either that the future is constantly changing in response to his current events, or there are multiple realities. In the Legion's case, both are true.
This applies to the Legions of Super-Villains, too. There are two distinct versions of the LSV: the original group was comprised of adult villains who are now considered to have come from an "alternate reality," or parallel Earth. The membership of the second group was mostly teenagers (like the Legion itself) and existed within the "official," or canonical Legion timeline.
The appearances of the adult LSV were all made within a series of Superman stories dubbed the "Adult Legion." Their first appearance was Superman #147 (1961), and those stories continued sporadically through 1967. These adventures all took place in Superman's time, not Superboy's. In retrospect, it's safe to say that the adult Legion of Super-Villains came from a parallel universe, and were thus eliminated by the Crisis on Infinite Earths (though the adult Saturn Queen apparently survived).
The reason this adult LSV is considered noncanonical to the regular Legion is that the stories depicted events in the Legion's future that never came to pass, or seemed unlikely to occur as Legion history unfolded. For example, the Adult stories predicted that Light Lass and Timber Wolf would marry; that Quantum Queen would become a Legionnaire; and that the Legionnaires would all change their names to "Man" or "Woman." Then again some Adult Legion predictions did come true: the marriages of Cosmic "Man" to Night Girl and the death of Chemical King.
A large group of villains also opposed the 20th century L.E.G.I.O.N., but were never named "Super-Villains," per se. » SEE: L.E.G.I.O.N.s of Super-Villains
The Super-Villains reveal their motivations. From Superman #147 (1961); art by Curt Swan and Sheldon Moldoff.
The adult Legion of Super-Villains was seemingly formed a decade after the Legion of Super-Heroes began. By this time, the teen heroes had grown to adulthood. The LSV's founders mirrored the Legion's own:
Cosmic King was an alchemist and inventor who discovered a ray that would transmute elements. He accidentally stumbled into the ray's path and was imbued with this ability himself! Instead of lauding him, his planets governors cast him out because transmutation was considered evil.
Lightning Lord, the brother of Lightning Lad, shared the same super-origin as his twin siblings. On the planet Korbal, they were attacked by lightning beasts and left them with the power to wield lightning themselves.
And Saturn Queen grew tired of the idyllic crime-free life on Saturn. She ventured out in search of adventure and quickly settled into a life of crime with the LSV. Like all on Titan, she had formidable telepathic abilities.
They were defeated with the timely intervention of the adult Legion of Super-Heroes: Cosmic Man, Lightning Man and Saturn Woman. Note: The Super-Villains' first appearance in print, Superman #147 (Aug. 1961), was also the Adult Legion's first appearance.
The Super-Villains with Echo and Beauty Blaze! From Adventure Comics #355 (1967); art by Curt Swan and George Klein.
Chameleon Chief and Sun Emperor replace Saturn Queen. From Superman's Pal Jimmy Olsen #63 (1962); art by Curt Swan and George Klein.
The Super-Villains come to gawk at history. From Action Comics #583 (1986); art by Curt Swan and Kurt Schaffenberger.
Profile illustration from Who's Who #13 (Mar. 1986); art by Steve Lightle.
Their first 30th century caper involved the manipulation of Douglas Nolan, the twin brother of the deceased Legionnaire, Ferro Lad. Nolan, who was under the mental domination of the Villains, attacked on their behalf. When the adult Legion and Superman freed Nolan of this control, the villains retreated to plot again. (Adventure #354) They returned with reinforcements — the mysterious Echo and fiery Beauty Blaze. This team of five was quite formidable, but the Legionnaires were rescued by benevolent 30th-century descendants of Lex Luthor and Mr. Mxyzptlk. (Adventure #355) Note: Legion vol. 2 #300 (1983) paid homage to these stories by returning to the teen Douglas Nolan, showing his glimpses of possible futures.
The Villains ultimately concluded that they were doomed to failure against the adult Legion of Super-Heroes, and took their crimes into other points in time. Their subsequent campaigns mostly included attempts to conquer the 20th century, during the time of Superman.
In Superboy's time, the young Lex Luthor once speculated that "if a Legion of Super-Heroes will exist centuries from now... then a Legion of Super-Villains probably exists in the future, too!" He vowed to one day contact them to help him destroy Superboy. (Superboy vol. 1 #86) A decade later he eventually succeeded. After Superboy had become Superman, and Luthor had grown into his biggest enemy, Luthor succeeded in contacting the three original Villains, who agreed to help Luthor with a plan to defeat Superman.
But Superman called in his own reinforcements: the adult Legion of Super-Heroes. Superman won this battle by cueing into something that Saturn Woman said: that her life of crime had begun only after she left Titan. Superman took a guess and moved some of the debris from Saturn's rings to Earth. Eve was miraculously rid of her evil urges, and turned on her teammates. She vowed to return to Saturn and reform. (Superman vol. 1 #147) NOTES: The Super-Villains in this case told Luthor that they were from the 21st century, but then Lightning Lord went on to say he was Lightning Lad's brother. The LSV made their next appearance in one of Superman's dreams. (Action #286) There were even a few early Legion of Super-Heroes appearances in which they claimed they were from the 21st century. (Action #287)
Without Saturn Queen, the remaining Villains filled out their ranks with two new members: Chameleon Chief and Sun Emperor. This time they were defeated by Supergirl, with help from Jimmy Olsen. (Jimmy Olsen #63) The next time they returned, Lex Luthor and Brainiac worked with the LSV to attempt to brainwash Jimmy into killing Superman. (Jimmy Olsen #87) Note: Curiously, Saturn Queen appears again with the group in this tale.
Many years later, as Superman's greatest foes converged on him, the adult Legion of Super-Villains appeared on the scene to watch; they fled when they realized they may be injured. (Action #583) Note: This was Superman's final pre-Crisis storyline, but was billed as an "Imaginary Story." The Superman from this imaginary timeline also appeared in the story arc from Superman/Batman #14-18 (below).
Why were the Legion of Super-Villains so persistent in their attempts to invade the 20th century, despite their repeated failure? Perhaps it was because during the so-called "Crisis on Infinite Earths," they got a glimpse of their own fate. When the Anti-Monitor's antimatter wave swept across the multiverse, all time periods collapsed on top of one another and the LSV found themselves battling alongside the 20th century's Lex Luthor and Brainiac in an army of super-villains. (Crisis #9) Surely this must have been an eye-opening experience for Lightning Lord and Cosmic King. They witnessed the fragility and near destruction of their own timeline from ground zero itself. As a result, did they redouble their efforts?
The Post-Crisis Adult LSV
The Legion of Super-Parents. From Superman/Batman #14 (2005); art by Carlos Pacheco and Jesús Merino.
The LSV doubles down with the help of Ra's al Ghul. From Superman/Batman #17 (2005); art by Carlos Pacheco and Jesús Merino.
Saturn Queen explains how she and Ultraman of Earth-Three survived the Crises. From Supergirl vol. 5 #8 (2006); art by Ron Adrian and Norm Rapmund.
Whatever the motivation, the Villains' final plan was specifically designed to ensure the survival of their own timeline. To this end, the adult Legion of Super-Villains made one last trip to the distant past. Somehow they had concluded that in order for their own timeline to survive, they would need to dramatically alter past events. This time Lightning Lord and Cosmic King brought Saturn Queen back into the fold and traveled to the 20th century — to murder the parents of Batman and Superman and raise the heroes as their own. (Superman/Batman #14)
The LSV raised Superman and Batman as their children. They also sabotaged and murdered those who were destined to become Earth's other major heroes — Hal Jordan, Aquaman, Barry Allen, Martian Manhunter. While the parents remained mostly in their moon-based home, their mighty sons subjugated the Earth completely. Their reign was eventually challenged by Princess Diana and a band of Freedom Fighters. Diana slayed Batman, and was slain in turn by Superman. Batman's death set the LSV's carefully executed time-tinkering into a spin. (#15)
Following this new change to the timestream, Superman and Batman began reliving their deaths throughout numerous realities. They learned that the only way to stop the time fluctuations was to restore the damage at the source: at the Super-Villains' initial departure from the 31st century.
When the Villains returned to their future to devise a new ploy, they found that the absence of a Batman in the timeline had resulted in Ra's al Ghul's global domination. The LSV allied with Ra's and also summoned their past members Echo and Beauty Blaze to help defend against the return of their "sons." (#17) Superman and Batman were victorious and returned the villains to their proper time—at a moment before their initial departure. There they also met the adult Legion of Super-Heroes, who were waiting to take the LSV into custody.
After the heroes had left, the villains' worst fears became reality. Superman and Batman had succeeded in cementing their own timeline — one which did not lead to the LSV's. As a result of Superman and Batman's efforts, the adult Legion of Super-Villains' timeline gave way to the antimatter wave of the "Crisis on Infinite Earths" and ceased to exist. (#18)
Ever resourceful, Saturn Queen somehow managed to also survive the elimination of her timeline by the Crisis and Infinite Crisis. Like very few others, she slipped through the cracks; she wound up in the shrunken city of Kandor. There she teamed with another cosmic survivor, Ultraman of Earth-3, to take over the city. They found themselves in conflict with Supergirl, to whom Saturn Queen gave information about the fate of her Kryptonian home, Argo City. (Supergirl vol. 5 #6-8)
Adult Legion of Super-Villains: List of Members
Within the scope of the original adult LSV stories (1961–67), none of its members' real names were ever revealed.
Name (Aliases)
1st app.
Final Status
Cosmic King Superman vol. 1 #147 (Aug. 1961) From an indefinite timeline; status unknown
Saturn Queen Pre-New 52: Active in the 21st Century "Kandor"
Chameleon Chief Superman's Pal Jimmy Olsen #63 (Sept. 1962) From an indefinite timeline; status unknown
Sun Emperor
Beauty Blaze Adventure #355 (Apr. 1967)
The duplicitous Dynamo Boy sends the Legion of Super-Villains off on a nefarious mission. From Adventure Comics #331 (1965); art by Jim Mooney.
Colossal Boy goes undercover and joins Tarik the Mute's new Legion of Super-Villains. From Adventure Comics #372 (1968); art by Jim Mooney.
The first appearance of the canonical Legion of Super-Villains was Adventure Comics #331 (1965). But it featured the same familiar trio of villains … so why was this appearance different? It was the first time they appeared in a story involving the mainstream teen Legion of Super-Heroes — all previous LSV appearances had taken place in the Legion's future (Superman's time).
Lightning Lord, Cosmic King and Saturn Queen came from "a few years in the Legion's future" pretending to have reformed. They hoped to take advantage of the fact that Vorm, the Dynamo-Boy — a boy space-pirate — had conned his way into the Legion. Dynamo Boy had engineered the expulsion of all other members and named himself leader. (Adventure #330) To fill out the ranks of his new Legion, he was screening potential new members when the Super-Villains arrived. Their partnership with Dynamo-Boy was fruitful — too fruitful. They grew jealous of Vorm's success and accolades among his pirate-people, and plotted to betray him. They tricked him into a space ship and stranded him a trillion years into the future. But with Vorm gone, the Legionnaires returned in full-force and sent the Villains packing back to their own era. (Adventure #331) This was the canonical Legion's only meeting with Super-Villains from the future.
Not long after this, the Legion first met the teen version of the Legion of Super-Villains. These villains were organized by a man called Tarik the Mute. Benno Tarik had been the victim of a stray bullet fired by the Science Police. The bullet left him without a larynx, and embittered towards law enforcement. He formed a school for super-villains and recruited young applicants in much the same fashion as the Legion. When he and his LSV kidnapped Colossal Boy's parents, the Legion sent several members, including its brand new recruits Chemical King and Timber Wolf, undercover to take Tarik down. Among Tarik's recruits were several Legion rejects and other terrible enemies: Lightning Lord, Nemesis Kid (a rogue Legionnaire himself), Radiation Roy, Ron-Karr, and Spider Girl. (Adventure #372)
After this, Lightning Lord took command of the Super-Villains and recruited the teen Chameleon Chief and Sun Emperor (in addition to Nemesis Kid, Spider Girl and Radiation Roy). (Superboy #208) They were handily defeated. Lightning Lord continued to resurface over the years, (Superboy #172, Legion vol. 2 #241, 302) but the LSV laid dormant for quite some time.
The Super Rejects
The Rejects: Calorie Queen, Chameleon Kid, Esper Lass, Magno Lad and Phantom Lad. From Superboy vol. 1 #212 (1975); art by Mike Grell.
Among the League of Super-Assassins, an unlikely Legionnaire. From the left: Blok, Lazon, Neutrax, Mist Master, Titania and Silver Slasher. From Superboy & the Legion of Super-Heroes #254 (1979); art by Mike Grell.
Profile picture from Who's Who #13 (1986); art by Joe Staton and Dick Giordano.
That same year, a similar group called the Legion of Super-Rejects rose to challenge the Legion. The Rejects were also made up of heroes who'd been turned down for Legion membership. This group of six in particular were rejected because they all duplicated the powers of existing Legionnaires. (Superboy #212) Some of these applicants (Esper Lass, Magno Lad and Micro Lad) were so incensed that they later turned to crime and joined the Legion of Super-Villains. (Legion vol. 3 #1) Calorie Queen (Taryn Loy of Bismoll) went on to a more productive career, (Legion vol. 4 #10) and Chameleon Kid (Toog Lintens of Durla) and Phantom Lad (Solon Darga of Bgtzl) were never heard from again.
The League of Super-Assassins
The League of Super-Assassins was formed by disgruntled teens from the planet Dryad. These teens were manipulated by the Dark Man into believing that the Legion was responsible for the destruction of their homeworld several years earlier. The League—Blok, Lazon, Mist Master, Neutrax, Silver Slasher and Titania—swiftly took out some of the most powerful Legionnaires. (Superboy & the Legion #253) They were defeated by Brainiac 5 and the Legion of Substitute-Heroes. (#254) Blok went on to repent and even join the Legion! But even though the League learned that Legion was innocent, the others chose the path of evil and later joined the Legion of Super-Villains under Nemesis Kid.
A Super-Sized LSV
Calling the roll of the new LSV. From Legion of Super-Heroes vol. 3 #2 (1984); art by Keith Giffen and Larry Mahlstedt.
All-out war near Orando! From Legion of Super-Heroes vol. 3 #3 (1984); art by Steve Lightle and Larry Mahlstedt.
Profile picture from Who's Who #13 (1986); art by Steve Lightle.
When at last they returned, the Legion of Super-Villains boasted a membership that finally rivaled the Legion in strength and number. Nemesis Kid led them; his second-in-command was Lightning Lord, with many other past members rejoining them. New to the LSV were foes from the Super-Rejects and the League of Super-Assassins.
The LSV freed the rogue Daxamite, Ol-Vir, from Takron-Galtos; they allied with terrorists Zymyr and Terrus; they recruited the Legion's old foes, the Hunter II and Tyr (Legion vol. 3 #1); and Nemesis Kid found a promising new ally in Cosmic King of Venus (his first appearance as a young man). (#2)
Dream Girl foresaw this epic clash — and the death of a Legionnaire. (#1) But before she could really warn her fellow Legionnaires, the LSV struck hard and fast. They chose the planet Orando for their staging grounds and captured its resident Legionnaires, Queen Projectra and Karate Kid. From there, they began stealing key items from around the galaxy. Earth's polymer shield was erected around Orando instead, and various power sources helped them to move the planet into another dimension. (#2-3)
Meanwhile, the Legion had its hands full with a full-scale attack by LSV members freed from the prison planet, Takron-Galtos. But Lightning Lord created his own Achilles heel when he captured and electrocuted his sister, Light Lass. Instead of killing her, the attack restored her original lightning powers and she freed herself and the other captive Legionnaires on Orando. Karate Kid was mortally wounded when he battled Nemesis Kid; he sacrificed his life to destroy the Super-Villains' dimension-warping equipment. (#4) In revenge, Projectra killed Nemesis Kid as well. Without their leader and their technology, the LSV was soon defeated. (#5)
LSV Affiliation
Lightning Lord (Mekt Ranzz of Winath) From future: Adventure #331 (Apr. 1965); in-continuity: Superboy #147 (May 1968) Adventure #331, 372, Superboy #208, Legion vol. 3 #1–5, Legion of Three Worlds #1–6 Active
Nemesis Kid (Hart Druiter of Myar) Adventure #346 (July 1966) Adventure #372, Superboy #208, Legion vol. 3 #1-5 Killed by Projectra, Legion vol. 3 #5 (Dec. 1984)
Radiation Roy (Roy Travich of Earth) Adventure #320 (May 1964) Adventure #372, Superboy #208, Legion vol. 3 #1-5 Active
Ron-Karr of Neptune Adventure #314 (Nov. 1963) Adventure #372, Legion vol. 3 #1-5 Active
Spider Girl (Sussa Paka of Earth) Adventure #323 (Aug. 1964) Adventure #372, Superboy #208, Legion vol. 3 #1-5 Active
Tarik the Mute (Benno Tarik of Earth) Adventure #372 (Sept. 1968) Adventure #372 Deceased Best of DC Digest #24 (Aug. 1982)
Chameleon Chief (Jall Tennuz) Superboy #208 (Apr. 1975) Superboy #208, Legion vol. 3 #1-5 Active
Sun Emperor (Nigal Douglous of Earth) Superboy #208, Legion vol. 3 #1-5 Active
LEGION OF SUPER-REJECTS
Esper Lass (Meta Ulnoor of Titan) Superboy #212 (Oct. 1975) Legion vol. 3 #1-5 Status unknown
Magno Lad (Kort Grezz of Braal) Status unknown
Micro Lad (Lalo Muldroon of Imsk) Killed by Akka, Legion of Super-Villains #1 (May 2011)
LEAGUE OF SUPER-ASSASSINS
Lazon (Chey-Nu of Dryad) Superboy & the Legion #253 (Sept. 1979) Legion vol. 3 #1–5 Active
Mist Master (Yer Sti-Tuan of Dryad) Active
Neutrax (Wi Kan Muur of Dryad) Active
Silver Slasher (Ki-Lan of Dryad) Active
Titania (Titania of Dryad) Active
LSV under Nemesis Kid
Cosmic King (Laevar Bolto of Venus) From future: Adventure #331 (Apr. 1965); in-continuity: Legion vol. 3 #2 (Sept. 1984) Adventure #331, Legion vol. 3 #1–5, Legion of Three Worlds #1-6 Killed by Variable Lad, Adventure #529 (Oct. 2011)
Hunter II (Adam Orion of Simballi) Superboy #199 (Nov. 1973) Legion vol. 3 #1–5 Active
Ol-Vir of Daxam Legion vol. 2 #294 (Dec. 1982) Killed by Darkseid, Legion vol. 3 Annual #2 (1986)
Terrus Legion vol. 3 #1 (Aug. 1984) Active
Tyr of Tyrraz Superboy #197 (Sept. 1973) Active
Zymyr of Gil'dishpan Legion vol. 3 #1 (Aug. 1984) Active
future member
Saturn Queen (Eve Ares of Titan) From future: Adventure #331 (Apr. 1965); in-continuity: Final Crisis: Legion of Three Worlds #1 (Oct. 2008) Adventure #331, Legion of Three Worlds #1-6, Legion vol. 6 #12-16, Legion of Super-Villains #1 Active
Legion vol. 4: The Glorith Reality Saturn Queen
In the timeline that began with Legion of Super-Heroes vol. 4 and leading up to the Legion's total reboot in Zero Hour, there were no LSV appearances but key members were cast in unexpected roles. This period of Legion history is no longer a part of any extant continuity.
Tenzil Kem helps defeat Evillo and free Saturn Queen. From Legion of Super-Heroes vol. 4 #49 (1993); art by Darryl Banks and Pam Eklund.
Sussa Paka (Spider Girl) escapes from and vexes the Dominators on Earth. From Legion of Super-Heroes vol. 4 #27 (1992); art by Jason Pearson and Tony Harris.
The adult Legion of Super-Villains had always claimed they were from the Legion's future. And as it turns out, their existence as adults, visiting the younger Legion might have been very possible. At this point in Legion history, the now-older Lightning Lord and Cosmic King were purported to have coerced Eve Aries of Titan to join them as Saturn Queen and travel into the Legion's past.
Eve Aries was a member of the royal family of Titan. Her personal mission was to recover her family's Hypno-Stone (which was stolen by Universo). Eve hoped that an alliance with the Legion of Super-Villains would help her find the Stone. In her only documented adventure with the LSV, the three traveled back in time and joined with Dynamo-Boy to try to destroy the Legion. (Adventure #331, 2995: The Legion of Super-Heroes Sourcebook)
After their return, the Legion of Super-Villains came to an end. Its members found Earth increasingly under control of the alien Dominators. Mekt Ranzz was imprisoned again and began a successful regimen for criminal rehabilitation on Labyrinth. (Legion vol. 4 #10) Two years later, he was released and returned to live with his brother on Winath. He never returned to a life of crime. (#3)
Saturn Queen continued on her quest to restore the Hypno-Stone of Ouranos to her family on Titan. She learned that the stone had been stolen by the thief Magpie for Prince Evillo of Tartarus. But because of a disagreement, Magpie hid the stone from Evillo. (Legion vol. 4 #49) Eve traveled to Tartarus and became Prince Evillo's twelfth wife. As Queen, she had the resources needed to track down the Stone. While there, she also met the former Legionnaire, Tenzil Kem (Matter-Eater Lad). Tenzil was imprisoned by Evillo but escaped and ultimately destroyed his realm of darkness. (#14)
After this, Evillo's power was greatly curtailed and Eve finally found the Magpie's hiding place for the Hypno-Stone. But Evillo was watching. He snatched it from her hands and used it to enslave his world once again. Eve reached out mentally for Tenzil's aid again, and he helped her to overthrow Evillo for good. (Legion vol. 4 #49) Back on Titan, Eve's family was on the brink of losing the throne. For her to resume her position, she would have to marry immediately. For better or worse, she chose Tenzil as her husband. (#50)
During this period of Legion publishing history, only a few former LSV members were seen:
The Hunter was hired by the Dominion to find R.J. Brande. (Legion vol. 4 #33)
Lightning Lord was reformed and reunited with his family on Winath. (Legion vol. 4 #10)
Radiation Roy was captured by the Dominators. (Legion vol. 4 #27)
Ron-Karr joined up with the underground Subs, to fight the Dominators on Earth. (Legion vol. 4 #26)
Spider-Girl reformed and became romantically involved with Ultra Boy, and eventually joined the Legion itself. (Legion vol. 4 #27)
In 1994's Zero Hour, the Legion's history and timelines were completely wiped and "rebooted." (Legion of Super-Heroes vol. 4 #0) The post-Zero Hour period (1994–2004, here referred to as the Reboot Legion) produced no new Super-Villains team. Some classic characters such as Lightning Lord, Spider Girl and Tyr were reintroduced, however.
Likewise, the Legion universe that was created after this one (here referred to as "Threeboot") had no Legion of Super-Villains. This Legion's primary adversaries were led by Mekt Ranzz, and were called the Wanderers. (Supergirl and the Legion #25)
Retroboot: Infinite Crisis
In the Infinite Crisis (2005–2006), the DC multiverse was reestablished. The original Legion timeline was more-or-less restored as the Legion of the mainstream DC universe. The events from the "Legion vol. 4" era above are not a part of Earth-0 continuity; this put Saturn Queen back into play as a villain.
In the new multiverse there were now three Legions, each from different universes. The "Reboot" team were left homeless — their universe was destroyed in the Infinite Crisis. A "Threeboot" team were said to reside on Earth-Prime.
All three teams were brought together when the Time Trapper sent Superboy Prime to the 31st century, where he freed all of the inmates from the prison planet Takron-Galtos, then destroyed it. (Final Crisis: Legion of Three Worlds #1)
Superboy led this new LSV to recruit Legion foes from across their history, including: Beauty Blaze, Black Mace, Dr. Regulus, Echo, Grimbor, Universo, and the members of the "Justice League" (Earth-Man, Golden Boy, Storm Boy, Tusker), and the Fatal Five (the Emerald Empress, Mano, the Persuader, Tharok, Validus). Superboy made his LSV wear rings bearing an 'S.' They killed the Legion ally Rond Vidar. NOTES: Before this appearance, Echo and Beauty Blaze had not appeared in mainstream DCU continuity. (#2)
More Legionnaires fell as Radiation Roy killed Myg (Karate Kid II), and Superboy Prime killed the Sun Boy of Earth-Prime, (#3) then Kinetix and Element Lad of Earth-Prime. Eventually the Time Trapper himself was revealed — as Superboy Prime! (#4)
The Legionnaires rallied at the End of Time, where they convinced the Time Trapper to confront his younger self. When they made contact, Prime was returned to his 21st century home, Earth-Prime. remainder of the Super-Villains were defeated by the timely intervention of Legion allies Connor Kent (Superboy) and Bart Allen (Impulse). The White Witch eliminated Mordru. (#5)
The Queen's Coronation: Legion vol. 5
Left to right: Immortus, Akka, Zymyr, Saturn Queen, Questor, Lightning Lord, Hunter, Sun Killer. From Legion of Super-Villains #1 (2011); art by Francis Portela.
The ancient Adversary imparts his will to Saturn Queen. From Legion of Super-Heroes vol. 6 #13 (2011); art by Yildiray Cinar and Jonathan Glapion.
The next iteration of the LSV came under the direction of Saturn Queen. She had emerged during the meeting of the three Legions, but without any details of her origins. When she resurfaced, it was returning to her homeworld, Titan, which was destroyed during a scientific inquiry at the Time Institute. (Legion vol. 6 #1) While there, she recalled her motivations; she had always found her people too "soft," and she distant from their peaceful society. She craved a destiny of dominion instead, using her great mental powers. At Titan, she surprised and overpowered several Legionnaires. (#2, Adventure Comics #527) NOTE: Outside Legion of Three Worlds, this is the first contemporary in-continuity appearance of Saturn Queen (not the "adult" version). Her given name, Eve Aries, was created in Legion vol. 4 #14 (Jan. 1991), but not referenced here. That version of Saturn Queen was never truly evil.
Legion reinforcements got the jump on Saturn Queen and she was captured. (#3) In her captivity, Saturn Queen was able to make contact with an ancient cosmic entity, which manifested as a blue-colored baby. Dyogene, the keeper of the Green Lantern Corpse legacy, sensed it and referred to it as the Adversary. The Legionnaires had encountered it just before this, when it struck and gravely injured Dawnstar. (Adventure Comics #521)
Sun Killer (Kodama of Earth) answered a telepathic summons and attempted to free Saturn Queen. Though the escape was unsuccessful, the Adversary imparted some power to Saturn Queen. (#522)
On the rebuilt Takron-Galtos, Saturn Queen managed to overcome her bonds and influence other inmates. She curated a group to plot a massive jailbreak. Her allies included Lightning Lord, the Hunter, Micro Lad and Zymyr, new recruits:
Akka, a martial arts expert from Sklar who killed Micro Lad for interrupting Saturn Queen
Immortus, and unknown person whose brain was preserved inside in Inertron armor that was constructed on Zuun. He reputedly came from Superman's time, and plotted his survival, eventually becoming indestructible, strong and fast with his armor
Questor of Colu was known to Brainiac 5, an outcast from his family
Earth's Kodama, the Sun Killer, who was a solar vampire and claimed to have "stole the heart of a star"
and Stegus of Saurius aided them on Rimbor
The Adversary's mission became that of the new LSV: to remake the universe by eliminating three "immortal worlds." They destroyed the Rock of Eternity (a center of faith) by having Zymyr teleport two planets to crush it from either side. (Legion of Super-Villains #1)
The Legion caught up to them on Rimbor, (#12) but the Adversary's power helped them persevere. (#13) He enabled the Hunter to find their next target, the secret world of wisdom, Utopia. It was protected by a mystical barrier and guarded by Master Kong (the ancient Confucius). (#14) There the Adversary grew to his fullest, and the Super-Villains tested the entire Legion's might by summoning more troops from Takron-Galtos. (#15)
Dyogene confronted the Adversary himself, and the Legion doubled down on their powers by granting them to their new member, Earth-Man. He forced the Adversary into retreat — at the expense of his own life. (#16)
The list below includes only characters who had not previously been members of the LSV.
Legion OF Three Worlds
Superboy Prime (Clark Kent of Earth Prime) DC Comics Presents #87 (Nov. 1985) Final Crisis: Legion of Three Worlds #1 Active
Storm Boy (Myke Chypurz) Adventure #301 (Oct. 1962) Final Crisis: Legion of Three Worlds #2 Active
Golden Boy (Klint Stewirt) Adventure #331 (Apr. 1965) Active
Tusker (Horace Lafeaugh) Active
Dr. Zaxton Regulus Adventure #348 (Sept. 1966) Active
Universo Adventure #349 (Oct. 1966) Active
The Fatal Five: Emerald Empress, Mano, Persuader, Tharok, Validus Adventure #352 (Jan. 1967) Active
Beauty Blaze Adventure #355 (Apr. 1967) Active
Echo Active
Mordru Adventure #369 (June 1968) Active
Black Mace Adventure #374 (Nov. 1968) Active
Earth-Man (Kirt Niedrigh, Absorbency Boy) Superboy vol. 1 #218 (June 1976) Legion vol. 6 #16 (Oct. 2011)
Grimbor the Chainsman Superboy vol. 1 #221 (Nov. 1976) Active
Saturn Queen's LSV
The Adversary Adventure Comics #522 (Mar. 2011) Legion of Super-Villains #1 (May 2011), Legion vol. 6 #11–16 Active
Sun Killer (Kodama of Earth) Active
Akka of Sklar Legion of Super-Villains #1 (May 2011) Killed by Saturn Queen, Legion vol. 6 #14
Questor (Xart Prax of Colu) Active
Stegus of Saurius Legion vol. 6 #12 (June 2011) Legion vol. 6 #12 Active
Mekt Ranzz leads the Light Speed Vanguard. From Legion season 1, episode 9 (2007).
In 2006, on the Legion of Super-Heroes animated series, the villains who opposed our heroes were cleverly named the Light Speed Vanguard. The name's initials are an obvious homage to the LSV.
It was led by Lightning Lad's brother, Mekt Ranzz. When both siblings competed in the Intergalactic Games, Mekt tried unsuccessfully to sabotage his brother. (Legion of Super-Heroes S01.E05)
Mekt returned as the leader of a rival group of young heroes called the "Light Speed Vanguard." Now as Lightning Lord, he recruited Wave (Spider Girl), Ronn Karr, Esper, Tyr, and Hunter. Lightning Lad was actually impressed with the LSV's performance and joined them on a mission. Naturally, the brothers came to odds when Mekt demanded payment for the group's services. (S01.E08)
They returned once, without Lightning Lord, but alongside members of the Fatal Five. At this time Brainiac 5 referred to them as the "Legion of Super-Villains." (S02.E02)
Adult LSV:
Adventure Comics #331, 354–355
Crisis on Infinite Earths #9
Justice Society of America vol. 3 #4 (cameo)
Superman/Batman #14–18
Superman's Pal Jimmy Olsen #63, 87
Adult Saturn Queen:
Legion vol. 4 #14, 49, 50
Canonical LSV:
Adventure #372
Final Crisis: Legion of Three Worlds #1–5
Legion vol. 3 #1–5
Legion vol. 6 #11–16
Superboy #208
Legion of Super-Villains #1, one-shot (May 2011)
Adult Legion of Super-Heroes
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This project contains a simple example shows how to add routes to Jaguar using Controller classes.
The repository can be found at: https://github.com/jaguar-examples/boilerplate. One would typically want to clone it and edit it in IntelliJ IDE.
Contains code to access the example server.


text:
\section{Introduction}\label{sec-intro}

If a group admits a quasi-isometric embedding into the product of
finitely many trees then one can say a lot about the large scale
geometry of the group.  For example, this bounds the (linearly
controlled) asymptotic dimension of the group.  It has implications for
the Hilbert compression of the group, and also the topological dimension
of any asymptotic cone.  Such embeddings have recently been constructed
for hyperbolic groups \cite{Bu-05-asdim-capdim, BuDrSc-07-hyp-grp-prod-trees}.

In this paper we show that if every peripheral group of a relatively
hyperbolic group embeds into the product of finitely many trees, then
the entire group does.  Moreover, we give explicit bounds for the number
of trees required.

We begin by recalling what is known for hyperbolic groups.
In this case, the situation has been completely understood by work of Buyalo and Lebedeva
\cite[Theorems 1.5, 1.6]{BuLe-selfsim}.  (For further refinements, see \cite{BuDrSc-07-hyp-grp-prod-trees}.)
\begin{theorem}\label{thm-buyalo-schroeder}
	Let $G$ be a Gromov hyperbolic group.  Then $G$ admits a quasi-isometric embedding into the product
	of $n+1$ metric trees, where $n = \dim \partial_\infty G$ is the topological dimension of the boundary.
	Moreover, $G$ does not embed into any product of $n$ metric trees.
\end{theorem}

Our main result is the following.
(For the definition of a relatively hyperbolic group, see Section~\ref{sec-rel-hyp-defs}.)
\begin{theorem}\label{thm-main1}
	Suppose the group $G$ is hyperbolic relative to subgroups $H_1, \ldots, H_n$.
	If each $H_i$ quasi-isometrically embeds into a product of $m$ metric trees, then
	$\asdim(G) < \infty$ and
	$G$ quasi-isometrically embeds into a product of $M$ metric trees, where
	\[
		M = \max\{\asdim(G), m+1\}+m+1 < \infty.
	\]
	Conversely, if $G$ quasi-isometrically embeds into a product of $N$ metric trees, then each peripheral group does also.
\end{theorem}
The assertion $\asdim(G) < \infty$ follows from \cite{Osin-asdimrelhyp}.

Both these theorems use results that study the relationship between the asymptotic dimension of a hyperbolic space
and the linearly controlled metric dimension of its boundary.  We now proceed to define these and related concepts.
(See \cite{BuSc-07-asymp-geom-book} for more discussion.)

Suppose $\mathcal{U}$ is a family of subsets of a metric space $X$.
We say $\mathcal{U}$ is \emph{$D$-bounded} if the diameter of every $U \in \mathcal{U}$ is at most $D$.
The \emph{$r$-multiplicity} of $\mathcal{U}$ is the infimal integer $n$ so that every subset of $X$ with diameter
less than or equal to $r$ meets at most $n$ subsets of $\mathcal{U}$.

The \emph{asymptotic dimension} of $X$, denoted by $\asdim(X)$, is the smallest $n \in \mathbb{N} \cup \{\infty\}$ so that
for all $r > 0$, there exists $D(r) < \infty$ and a $D(r)$-bounded cover $\mathcal{U}$ of $X$ with $r$-multiplicity at most $n+1$
\cite[1.E]{Gro-91-asymp-inv}.
The \emph{linearly controlled asymptotic dimension} of $X$, denoted by ${\ell\text{-}\!\asdim}(X)$, is the smallest $n$
so that there exists $L < \infty$ with the property that for all sufficiently large $r<\infty$, there exists an $Lr$-bounded
cover $\mathcal{U}$ of $X$ with $r$-multiplicity at most $n+1$ \cite{LaSc-05-nagata,BDHM-09-asdiman-via-lip}.
(This is sometimes called ``asymptotic Assouad-Nagata dimension'' in the literature.)

The following definition simplifies our discussion of how metric spaces embed in trees.
(For the related notion of ``$t$-rank'', see \cite[page 176]{BuSc-07-asymp-geom-book}.)
\begin{definition}\label{def-ecodim}
Given a metric space $X$, let $\ecodim(X)$ be the smallest $n \in \mathbb{N}$ so that
$X$ quasi-isometrically embeds in the product of $n$ metric trees, and
set $\ecodim(X)= \infty$ if no such embedding exists.
\end{definition}
The inequalities $\asdim(X) \leq {\ell\text{-}\!\asdim}(X) \leq \ecodim(X)$ hold for any metric space $X$.
These three quantities are equal for hyperbolic groups, as shown in \cite{BuLe-selfsim}.
Both equalities can fail in general: there are groups $G$ with finite asymptotic dimension but infinite
linearly controlled asymptotic dimension~\cite{Nowak-07-asdim-an}.
The discrete Heisenberg group $H$ has $\asdim(H)={\ell\text{-}\!\asdim}(H)=3$,
but $\ecodim(H)=\infty$ (see the discussion in Section~\ref{sec-three-mfld}).

As an aside, Lang and Schlichenmaier show that if ${\ell\text{-}\!\asdim}(X, d) < \infty$, then for sufficiently
small $\epsilon>0$, the snowflaked space $(X, d^\epsilon)$ has
$\ecodim(X, d^\epsilon) \leq {\ell\text{-}\!\asdim}(X, d)+1$ \cite[Theorem 1.3]{LaSc-05-nagata}.

For hyperbolic groups, these asymptotic invariants are related to the local properties of the boundary.
The \emph{linearly controlled metric dimension} of $X$,
denoted by ${\ell\text{-}\!\dim}(X)$, is the smallest $n$ so that there exists $L < \infty$ with the property that for
all sufficiently small $r>0$, there exists an $Lr$-bounded
cover $\mathcal{U}$ of $X$ with $r$-multiplicity at most $n+1$ \cite[Prop.\ 3.2]{Bu-05-asdim-capdim}.  (This is also
referred to as ``capacity dimension''.)
Buyalo shows the following embedding theorem.
\begin{theorem}[{\cite[Theorem 1.1]{Bu-05-cdim-prod-trees}}]\label{thm-buyalo-cdim-trees}
	Suppose $X$ is a visual, Gromov hyperbolic metric space, and ${\ell\text{-}\!\dim}(\partial_\infty X) < \infty$.
	Then $\ecodim(X) \leq {\ell\text{-}\!\dim}(\partial_\infty X) +1$.
\end{theorem}

The following proposition gives a simple bound for the linearly controlled metric dimension of the boundary of $X$.
\begin{varthm}[Proposition \ref{prop-cdim-asdim}]
	Suppose $X$ is a Gromov hyperbolic geodesic metric space.  Then ${\ell\text{-}\!\dim}(\partial_\infty X) \leq \asdim(X)$.
\end{varthm}
This proposition does not seem to be recorded in the literature, possibly because in the case when the space
admits a cocompact isometric action the stronger equality $\asdim(X) = {\ell\text{-}\!\dim}(\partial_\infty X) + 1$ holds \cite{BuLe-selfsim}.
Even without such an action, Buyalo shows the inequality $\asdim(X) \leq {\ell\text{-}\!\dim}(\partial_\infty X) +1$ \cite{Bu-05-asdim-capdim}.

In the case of a relatively hyperbolic group $(G,\{H_i\})$, a natural choice for $X$ is $X(G)$, the Bowditch space associated to
$(G,\{H_i\})$, see Definition~\ref{def-rel-hyp}.  This is a visual, Gromov hyperbolic, geodesic metric space.
We bound the asymptotic dimension of $X(G)$ in terms of the asymptotic dimension
of $G$ and the linearly controlled asymptotic dimension of the peripheral groups.
\begin{varthm}[Proposition \ref{prop-asdim-bowditch}]
	Let $G$ be hyperbolic relative to $H_1,\dots, H_n$, and
	let $m = \max_{i=1, \ldots, n} {\ell\text{-}\!\asdim}(H_i)$.
	Then
	\[
		\max\{\asdim(G),m\} \leq \asdim(X(G)) \leq \max \{\asdim(G), m+1\}.
	\]
\end{varthm}
Osin had earlier shown that $\asdim(G)$ is finite if $\asdim(H_i) < \infty$ for each $i$ \cite[Theorem 1.2]{Osin-asdimrelhyp}.

The product of $n$ (unbounded) metric trees has linearly controlled asymptotic dimension $n$ (see, e.g., \cite{LaSc-05-nagata}).
Therefore, Theorem~\ref{thm-buyalo-cdim-trees} and Propositions~\ref{prop-cdim-asdim} and \ref{prop-asdim-bowditch}
combine to show the following.
\begin{corollary}\label{cor-bowditch-trees}
	Suppose the group $G$ is hyperbolic relative to $H_1, \ldots, H_n$.
	Let $m = \max \{ \ecodim(H_i) \}$.
	Then $\ecodim(X(G)) \leq \max\{\asdim(G), m+1\}+1$.
\end{corollary}

In order to prove Theorem~\ref{thm-main1}, we use work of Bestvina, Bromberg and Fujiwara to combine the
embeddings of the peripheral groups and of $X(G)$ into a single embedding of $G$ into a product of trees.
\begin{varthm}[Theorem \ref{prodqtrees}]
	 Let $G$ be hyperbolic relative to $H_1,\dots, H_n$, and suppose that each $H_i$ admits a quasi-isometric
	 embedding into the product of $m$ trees. Then $G$ admits a quasi-isometric embedding into the product of
	 $m$ trees and either $X(G)$ or the coned-off graph $\hat{G}$.
\end{varthm}
This theorem completes the proof of Theorem~\ref{thm-main1}.  Note that the converse statement is automatic,
as peripheral groups of a relatively hyperbolic group are undistorted in the ambient group
\cite[Lemma 4.15]{DSp-05-asymp-cones}.

Rather than using the Bowditch space $X(G)$, to improve the bounds in Theorem~\ref{thm-main1}
one might hope to use instead the (hyperbolic) coned-off graph $\hat{G}$.
However, the non-locally finite nature of $\hat{G}$ leads to a non-compact boundary, and
much of the machinery used above no longer applies.

The embeddings we consider are not and cannot required to be equivariant,
because there exist hyperbolic groups with property (T) (for example, any cocompact lattice in $Sp(n,1)$),
which in particular cannot act interestingly on trees.
We remark that the inverse problem of equivariantly embedding trees into hyperbolic spaces is
treated in~\cite{BIM-05-embed-trees-in-hyp}.

Theorem~\ref{thm-main1} has the following immediate corollary.
\begin{corollary}
	If the peripheral groups of a relatively hyperbolic group $G$ each quasi-isometrically embed
	into the product of finitely many trees, then ${\ell\text{-}\!\asdim}(G) < \infty$.
\end{corollary}
If the group $G$ satisfies ${\ell\text{-}\!\asdim}(G) < \infty$ then it has Hilbert compression $1$ \cite[Theorem 1.1.1]{Gal-08-asdim-hilbert-compression}.
For more general estimates on the Hilbert compression of a relatively hyperbolic group, see \cite{Hume}.

In Section~\ref{sec-three-mfld} we apply our results to $3$-manifold groups, and show, amongst other results,
the following.
\begin{varthm}[Theorem \ref{thm-three-mfld}]
	Let $G = \pi_1(M)$, where $M$ is a compact, orientable $3$-manifold whose (possibly empty) boundary is a union of tori.
	Then $\ecodim(G) < \infty$ if and only if no manifold in the prime decomposition of $M$ has $\mathrm{Nil}$ geometry;
	in this case, $\ecodim(G) \leq 8$.

	In any case, ${\ell\text{-}\!\asdim}(G) \leq 8$.
\end{varthm}

Finally, in Appendix \ref{sec-hnn} we prove a bound on the asymptotic dimension of HNN extensions,
following work of Dranishnikov.

\subsection{Recent developments}
Since the appearance of the first version of this paper, Hume \cite{Hu-bbf-qtrgr} has shown that the construction of Bestvina, Bromberg and Fujiwara that we use in Section \ref{sec-embed-prod-trees-and-bow} gives rise to spaces that are quasi-isometric to a tree-graded space. Tree-graded spaces are defined in \cite{DSp-05-asymp-cones}; roughly speaking, they are tree-like arrangements of certain specified subspaces called pieces (for example, consider a Cayley graph of a free product). It is known \cite{BrHi-09-AN-tree-graded} that if $X$ is tree-graded and $\mathcal{P}$ is its collection of pieces then
$${\ell\text{-}\!\asdim}(X)\leq\max_{P\in\mathcal{P}}\{1,{\ell\text{-}\!\asdim}(P)\}$$
when, say, there are finitely many $(L,C)$-quasi-isometry types of pieces, for some given $L,C$.

As noticed in \cite{Hu-bbf-qtrgr}, the arguments in Section \ref{sec-embed-prod-trees-and-bow} (just considering $\pi_Y$ instead of $f_{i,Y}\circ \pi_Y$ in Subsection \ref{ssec-prodqtrees}), combined with \cite{Hu-bbf-qtrgr}, give the following.

\begin{theorem}[cf.\ Theorem~\ref{prodqtrees}]
 Let $G$ be hyperbolic relative to $H_1,\dots,H_n$. Then $G$ quasi-isometrically embeds in the product of a space $\mathcal{T}$ and either the coned-off graph $\hat{G}$ or the Bowditch space $X(G)$, where $\mathcal{T}$ is tree-graded and each piece is uniformly quasi-isometric to some $H_i$.
\end{theorem}

In particular, once again as noticed in \cite{Hu-bbf-qtrgr}, using the aforementioned result from \cite{BrHi-09-AN-tree-graded} one can get the following.

\begin{corollary}
 A relatively hyperbolic group has finite ${\ell\text{-}\!\asdim}$ if and only if its peripheral subgroups do.
\end{corollary}

\subsection*{Acknowledgements}
We thank Alexander Dranishnikov, David Hume, Urs Lang, Enrico Le Donne and the referee for helpful comments,
and Marc Lackenby for providing a fix for the proof of Proposition \ref{nonclosed}.

\section{Relatively hyperbolic groups}\label{sec-rel-hyp-defs}

In this section we define relatively hyperbolic groups and their (Bow\-ditch) boundaries.

\subsection{Definitions}
 There are many definitions of relatively hyperbolic groups.
 We will give one in terms of actions on a cusped space.
 First we define our model of a horoball.

 \begin{definition}
 	Suppose $\Gamma$ is a connected graph with vertex set $V$ and edge set $E$,
 	where every edge has length one.
 	The \emph{horoball} $\mathcal{H}(\Gamma)$ is defined to be the graph with
 	vertex set $V \times \mathbb{N}$ and
 	edges $((v, n), (v, n+1))$ of length $1$, for all $v \in V$, $n \in \mathbb{N}$,
 	and edges $((v, n), (v', n))$ of length $e^{-n}$, for all $(v, v') \in E$.
 \end{definition}

 Note that $\mathcal{H}(\Gamma)$ is quasi-isometric to the metric space
 constructed from $\Gamma$ by gluing to each edge in $E$ a copy of the strip
 $[0,1] \times [1, \infty)$ in the upper half-plane model of $\mathbb{H}^2$,
 where the strips are attached to each other along $v \times [1,\infty)$.

 As is well known, these horoballs are hyperbolic with boundary a single point.
 Moreover, it is easy to estimate distances in horoballs. We will write $A\approx B$ if the quantities $A$ and $B$ differ by some constant.
 \begin{lemma}\label{distestim}
 	Suppose $\Gamma$ and $\mathcal{H}(\Gamma)$ are defined as above.
 	Let $d_\Gamma$ and $d_\mathcal{H}$ denote the corresponding path metrics.
 	Then for each $(x,m), (y,n) \in \mathcal{H}(\Gamma)$, we have
 	\[
 		d_\mathcal{H}((x,m),(y,n)) \approx 2 \ln(d_\Gamma(x,y)e^{-\min\{m,n\}}+1)+|m-n|.
 	\]
 \end{lemma}
 \begin{proof}
	We may assume that $m \leq n$.
 	We can suppose $d_\Gamma(x,y)\geq e^m$, as if not $|m-n| \leq d_\mathcal{H}((x,m),(y,n))\leq 1+|m-n|$. In particular
 	$\ln(d_\Gamma(x,y)/e^m+1)\approx \ln d_\Gamma(x,y)-m$. By construction of $\mathcal{H}(\Gamma)$, any geodesic
 	$\gamma$ between $(x,m)$ and $(y,n)$ in $\mathcal{H}(\Gamma)$ must go from $(x,m)$ to $(x,t)$ changing
 	only the second coordinate,
 	then follow a geodesic $\gamma' \subset \Gamma \times \{t\} \subset \mathcal{H}(\Gamma)$
 	to $(y,t)$, then back to $(y,n)$.
 	Thus
 	\begin{equation}\label{eq-horoball-distance-estimate}
 		d_\mathcal{H}((x,m),(y,n)) \leq 2(t-m)+|n-m|+ e^{-t} d_\Gamma(x,y),
 	\end{equation}
 	with equality for the best choice of $t$. It is readily seen that this value is attained for the least $t$ so that $l_t=e^{-t} d_\Gamma(x,y)$ satisfies $l_t/e+2\geq l_t$, that is, $l_t\leq 2e/(e-1)=\rho$.
 	So the best choice of $t$ is $t = \lceil \ln (d_\Gamma(x,y)/\rho)\rceil$,
 	and the right hand side of \eqref{eq-horoball-distance-estimate}
 	is $2(\ln d_\Gamma(x,y)-m)-2\ln\rho + |n-m| +\epsilon$, where $|\epsilon|\leq 2+\rho$.
 \end{proof}

\begin{definition}\label{def-rel-hyp}
	Suppose $G$ is a finitely generated group, and $\{H_i\}_{i=1}^n$
	a collection of finitely generated subgroups of $G$.
	Let $S$ be a finite generating set for $G$, so that $S \cap H_i$ generates
	$H_i$ for each $i=1, \ldots, n$.

	Let $\Gamma(G,S)$ be the Cayley graph of $G$ with respect to $S$.
	Let $X(G) =  X(G, \{H_i\}, S)$ be the space resulting from gluing to $\Gamma(G,S)$
	a copy of $\mathcal{H}(\Gamma(H_i, S \cap H_i))$ to each coset $g H_i$ of
	$H_i$, for each $i= 1, \ldots, n$.  We call $X(G)$ the \emph{Bowditch space} associated to $(G, \{H_i\}, S)$.

	We say that $(G, \{H_i\})$ is \emph{relatively hyperbolic} if $X(G)$ is Gromov hyperbolic,
	and call the members of $\{H_i\}$ \emph{peripheral subgroups}.
\end{definition}

This is equivalent to the other usual definitions of (strong) relative hyperbolicity;
see \cite{Bow-99-rel-hyp}, \cite[Theorem 3.25]{Gro-Man-08-dehn-rel-hyp}.

\subsection{Visual metric}
\label{visual-metric}
Let $X$ be a geodesic, Gromov hyperbolic space (not necessarily proper), with fixed base point $0 \in X$.
Suppose all geodesic triangles are $\delta$-slim.
One equivalent definition of the boundary $\partial_\infty X$ is as the set of equivalence classes of $(1,20\delta)$-quasigeodesic
rays $\gamma: [0, \infty) \rightarrow X$, with $\gamma(0)=0$, where two rays are equivalent if they are at finite Hausdorff distance
from each other.
Let $(x | y) = (x | y)_0$ denote the Gromov product on $\partial_\infty X$ with respect to $0$.
Up to an additive error, $(x|y)$ equals the distance from $0$ to
some (any) $(1,20\delta)$-quasigeodesic line from $x$ to $y$ \cite[Remark 2.16]{KB-02-boundaries-survey}.

A metric $\rho$ on $\partial_\infty X$ is a visual metric if
there exists $C_0, \epsilon >0$ so that $\frac{1}{C_0} e^{-\epsilon (x|y)} \leq \rho(x,y) \leq C_0 e^{-\epsilon (x|y)}$
for all $x, y \in \partial_\infty X$. Boundaries of Bowditch spaces will always be endowed with a visual metric.
The linearly controlled metric dimension of such boundary is independent of the choice of visual metric \cite{Bu-05-cdim-prod-trees}.

For more results on the geometric properties of $\partial_\infty X(G)$, see \cite{MS-12-relhypplane}.

\subsection{Distance formula}
Let $G$ be a relatively hyperbolic group and let ${\bf Y}$ be the collection of all left cosets of peripheral subgroups.
Fix a Cayley graph of $G$ with path metric $d$.
For $Y\in{\bf Y}$, let $\pi_Y:G \rightarrow Y$ be a closest point projection map onto $Y$ with respect to $d$.
Denote by $\hat{G}$ the \emph{coned-off graph} of $G$, that is to say the metric graph obtained from the Cayley graph of $G$ by adding an edge
of length one connecting each pair of (distinct) vertices contained in the same left coset of peripheral subgroup.
The path metric on $\hat{G}$ is denoted by $d_{\hat{G}}$.
Let $\big\{\big\{x\big\}\big\}_L$ denote $x$ if $x>L$, and $0$ otherwise.
We write $A\approx_{\lambda,\mu} B$ if $A/\lambda-\mu\leq B\leq \lambda A+\mu$. The following is proven in \cite{projrelhyp}.
\begin{theorem}[Distance formula for relatively hyperbolic groups]\label{distform}
There exists $L_0$ so that for each $L\geq L_0$ there exist $\lambda,\mu$ so that the following holds. If $x,y\in G$ then
\begin{equation}\label{eqn-distform}
	d(x,y)\approx_{\lambda,\mu} \sum_{Y\in\bf{Y}} \big\{\big\{d(\pi_Y(x),\pi_Y(y))\big\}\big\}_L+d_{\hat{G}}(x,y).
\end{equation}
\end{theorem}

\section{Large scale and boundary dimension estimates}\label{sec-dim-estimates}

In this section we bound the asymptotic dimension of the Bowditch space of a relatively hyperbolic group.
We also bound the linearly controlled metric dimension of the boundary of a Gromov hyperbolic space by its asymptotic dimension.

Observe that at the cost of a slight relaxation in the value of $r$,
a collection of subsets $\mathcal{U}$ has $r$-multiplicity at most $m$ if and only if we can write
$\mathcal{U} = \bigcup_{i=1}^m \mathcal{U}_i$, where each $\mathcal{U}_i$ is \emph{$r$-disjoint}: it is a collection of subsets pairwise separated by
a distance of at least $r$.
This follows from an application of Zorn's lemma.
We will use this alternative characterization throughout this section.

\subsection{Asymptotic dimension of Bowditch spaces}

First we bound the asymptotic dimension of a horoball.
\begin{proposition}\label{asdimhor}
 $\asdim(\mathcal{H}(\Gamma))\leq {\ell\text{-}\!\asdim}(\Gamma) +1$.
\end{proposition}

\begin{proof}
 We will use the Hurewicz theorem for asymptotic dimension.
\begin{theorem}[{\cite[Theorem 1]{BeDr-Hurewicz}}]
 Let $f:X\to Y$ be a Lipschitz map between geodesic metric spaces, and suppose that for each $R$ the family $\{f^{-1}(B_R(y))\}_{y\in Y}$ has uniform asymptotic dimension $\leq m$. Then $\asdim(X)\leq \asdim Y+m$.
\end{theorem}

Choose some vertex $x \in \Gamma$, and let $\gamma$ be the geodesic ray in $\mathcal{H}(\Gamma)$ obtained by concatenating
the edges between $(x,i)$ and $(x,i+1)$ for all $i\in \mathbb{N}$.
Let $f:\mathcal{H}(\Gamma)\to \gamma$ be the natural $1$-Lipschitz map given by $f(z,i)=(x,i)$.
Fix some $R>0$. The preimages under $f$ of all balls of radius $R$ in $\gamma$ are quasi-isometric
(with constants depending on $R$ only) to $(\Gamma,d_n)$ for some $n$,
where $d_n(x,y)=2\ln (d_\Gamma(x,y)e^{-n}+1)$, see Lemma \ref{distestim}.
So, we only need to show that the uniform asymptotic dimension of $\{(\Gamma,d_n)\}$ is at most $m$.
Fix any $R$ large enough. There exists $C=C(\Gamma)$ so that the following holds. Let $m={\ell\text{-}\!\asdim}(\Gamma)$.
For each $n$ there exists a covering $\mathcal{U}(n)=\mathcal{U}_1(n)\cup\dots\cup \mathcal{U}_{m+1}(n)$ of $(\Gamma,d_\Gamma)$ such that
each $\mathcal{U}_i(n)$ is $e^{n+R}$-disjoint and $Ce^{n+R}$-bounded.
So, $\mathcal{U}_i(n)$ is $(2R-M)$-disjoint and $(2R+M)$-bounded with respect to $d_n$, where $M$ depends on $\Gamma$ only.
\end{proof}

Proposition \ref{asdimhor} has a weak converse.
\begin{proposition}\label{prop-asdimhor-converse}
 ${\ell\text{-}\!\asdim}(\Gamma)\leq \asdim(\mathcal{H}(\Gamma))$.
\end{proposition}

\begin{proof}
 Set $m=\asdim(\mathcal{H}(\Gamma))$. There exists a covering $\mathcal{U} = \mathcal{U}_1\cup\dots\cup \mathcal{U}_{m+1}$ of $\mathcal{H}(\Gamma)$
 such that each $\mathcal{U}_i$ is $r$-disjoint and $R$-bounded, for some $r,R$ large enough.
 Up to increasing $R$ and decreasing $r$ we have that for each $n$ there exists a covering
 $\mathcal{U}_1(n)\cup\dots\cup \mathcal{U}_{m+1}(n)$ of $(\Gamma,d_n)$ with the same properties,
 where $d_n(x,y)=2\ln (d_\Gamma(x,y)e^{-n}+1)$ as in the previous proof.
 So, each $\mathcal{U}_i(n)$ is $e^n(e^{r/2}-1)$-disjoint and $e^n(e^{R/2}-1)$-bounded in $(\Gamma, d_\Gamma)$.
 We are done as $$ \frac{e^n(e^{R/2}-1)}{e^n(e^{r/2}-1)}$$ is bounded independently of $n$.
\end{proof}

Now we can bound the asymptotic dimension of the Bowditch space.
\begin{proposition}\label{prop-asdim-bowditch}
	Let $G$ be hyperbolic relative to $H_1,\dots, H_n$, and
	let $m = \max_{i=1, \ldots, n} {\ell\text{-}\!\asdim}(H_i)$.
	Then
	\[
		\max \{ \asdim(G), m \} \leq \asdim(X(G)) \leq \max \{\asdim(G), m+1\}.
	\]
\end{proposition}
\begin{proof}
 The lower bound by $m$ follows from Proposition~\ref{prop-asdimhor-converse}.
 The lower bound by $\asdim(G)$ follows as $X(G)$ contains $G$ with a proper metric.

 For the upper bound, we will use the union theorem for asymptotic dimension.
\begin{theorem}[{\cite[Theorem 1]{BeDr-unionthm}}]
 Let $Y$ be a geodesic metric space and suppose that $Y=\bigcup_{i\in\mathbb{N}} A_i$. Also, suppose that $\{A_i\}$ has uniform asymptotic dimension $\leq n$ and for each $R$ there exists $Y_R\subseteq Y$ so that $\asdim Y_R\leq n$ and $\{A_i\backslash Y_R\}$ is $R$-disjoint. Then $\asdim Y\leq n$.
\end{theorem}

We apply the theorem with $\{A_i\}$ the family of horoballs of $X(G)$ and
for each $R$, take $Y_R$ to be a suitable neighborhood of an orbit of $G$.
By Proposition \ref{asdimhor}, $\asdim(A_i) \leq m+1$, and the constants are uniform as there
are only finitely many different isometry types.
Finally, $\asdim(Y_R) = \asdim(G)$ because the action of $G$ on $X(G)$ is proper.
\end{proof}

\subsection{Linearly controlled metric dimension estimate}
\begin{proposition}\label{prop-cdim-asdim}
	Suppose $X$ is a Gromov hyperbolic geodesic metric space.  Then ${\ell\text{-}\!\dim}(\partial_\infty X) \leq \asdim(X)$.
\end{proposition}
\begin{proof}
	We fix the notation of Subsection \ref{visual-metric} regarding visual metrics,
	and write $d$ for the metric on $X$.

	Each $x \in \partial_\infty X$ is the limit of some $(1,20\delta)$-quasigeodesic $\gamma$ \cite[Remark 2.16]{KB-02-boundaries-survey}.
	Given $R > 0$, define the projection $\pi_R : \partial_\infty X \rightarrow X$ by $\pi_R(x) = \gamma(R)$.
	This is well defined up to an error of $C_1=C_1(\delta)$.
	By considering a quasi-geodesic triangle between $a, b$ and $0$, observe that
	there exists $C_2=C_2(C_0, C_1, \delta)$ so that
	if $d(\pi_R(a), \pi_R(b)) \geq s > 2C_1$, then
	$\rho(a,b) \geq \frac1{C_2} e^{-\epsilon(R-s/2)}$.
	Similarly, if $d(\pi_R(a), \pi_R(b)) \leq t$ then $\rho(a,b) \leq C_2 e^{-\epsilon(R-t/2)}$.

	If $\asdim(X) \leq n$, then, given $s = 3C_1$, there exists $t < \infty$ and a cover
	$\mathcal{U} = \bigcup_0^n \mathcal{U}_i$ so that each $U \in \mathcal{U}$ has diameter at most $t$, and $\mathcal{U}_i$ is $s$-disjoint.

	Suppose some small $r>0$ is given.  Let $R = -\frac{1}{\epsilon} \ln r$.
	For $U \in \mathcal{U}_i \subset \mathcal{U}$, let $\hat{U} \subset \partial_\infty X$ be the set of points $z$ so that there exists a
	$(1,20\delta)$-quasigeodesic $\gamma$ from $0$ to $z$ with $\gamma(R) \in U$.
	Let $\hat{\mathcal{U}_i} = \bigcup_{U \in \mathcal{U}_i} \hat{U}$,
	and $\hat{\mathcal{U}} = \bigcup_{i=1}^n \hat{\mathcal{U}_i}$.

	By the above estimates, $\hat{\mathcal{U}_i}$ is $(e^{\epsilon s/2} r/C_2)$-disjoint, and
	$\hat{\mathcal{U}}$ is $C_2 r e^{\epsilon t/2}$-bounded.
	Since the ratio of these distances is bounded by $C_2^2 e^{\epsilon(t-s)/2}$, we have ${\ell\text{-}\!\dim}(\partial_\infty X) \leq n$.
\end{proof}

\section{Embedding in a product of quasi-trees and the Bowditch space}\label{sec-embed-prod-trees-and-bow}
The aim of this section is to find an embedding of a given relatively hyperbolic group into
a product of trees ``stabilized'' by the Bowditch space or the coned-off graph.
\begin{theorem}\label{prodqtrees}
  Let $G$ be hyperbolic relative to $H_1,\dots, H_n$.
  Suppose there exists $k$ so that
  each $H_i$ admits quasi-isometric embedding into the product of $k$ trees.
  Then $G$ admits a quasi-isometric embedding into the product of $k$ trees and either $X(G)$ or the coned-off graph $\hat{G}$.
\end{theorem}

We prove this theorem in subsection~\ref{ssec-prodqtrees}.

\subsection{Quasi-trees of spaces}\label{ssec-dist}

To prove Theorem~\ref{prodqtrees} we use a result by Bestvina, Bromberg and Fujiwara described below.

Let $\bf{Y}$ be a set and for each $Y\in\bf{Y}$ let $\mathcal{C}(Y)$ be a geodesic metric space.
For each $Y$ let $\pi_Y:{\bf Y}\backslash\{Y\}\to\mathcal{P}(\mathcal{C}(Y))$ be a function (where $\mathcal{P}(\mathcal{C}(Y))$ is the collection of all subsets of $\mathcal{C}(Y)$). Define
$$d^{\pi}_Y(X,Z)=\diam\{\pi_Y(X)\cup\pi_Y(Z)\}.$$
Using the enumeration in \cite[Sections 3.1, 2.1]{BBF}, consider the following Axioms. There exists $\xi < \infty$ so that:
\begin{itemize}
	\item[(0)] $\diam(\pi_Y(X))<\xi$ for all distinct $X, Y \in {\bf Y}$,
	\item[(3)] for all distinct $X, Y, Z \in {\bf Y}$ we have $\min\{d^{\pi}_Y(X,Z),d^{\pi}_Z(X,Y)\}\leq \xi$,
	\item[(4)] $\{Y:d^\pi_Y(X,Z)\geq \xi\}$ is a finite set for each $X,Z\in\bf{Y}$.
\end{itemize}

Using the functions $d_Y^\pi$, the authors of \cite{BBF} define a certain complex
$\mathcal{P}_K(\{(\mathcal{C}(Y),\pi_Y)\}_{Y\in\bf{Y}})$ with vertex set ${\bf Y}$, which they denote as $\mathcal{P}_K({\bf Y})$,
and turns out to be a quasi-tree \cite[Section 2.3]{BBF}.

Let $\mathcal{C}(\{(\mathcal{C}(Y),\pi_Y)\}_{Y\in\bf{Y}})$ be the path metric space
consisting of the union of all $\mathcal{C}(Y)$'s and edges of length 1 connecting all points in $\pi_X(Z)$ to
all points in $\pi_Z(X)$ whenever $X,Z$ are connected by an edge in the complex
$\mathcal{P}_K(\{(\mathcal{C}(Y),\pi_Y)\}_{Y\in\bf{Y}})$.
More details of this construction are found in \cite[Section 3.1]{BBF}.

The only result from \cite{BBF} that we need is the following.

\begin{theorem}[{\cite[Theorem 3.10]{BBF}}]\label{bbf}
 If $\{(\mathcal{C}(Y),\pi_Y)\}_{Y\in\bf{Y}}$ satisfies Axioms $(0), (3)$ and $(4)$ and each $\mathcal{C}(Y)$ is a tree then $\mathcal{C}(\{(\mathcal{C}(Y),\pi_Y)\}_{Y\in\bf{Y}})$ is a quasi-tree (i.e. it is quasi-isometric to a tree).
\end{theorem}

To apply this theorem in the context of Theorem~\ref{prodqtrees}, we fix a Cayley graph $\Gamma(G,S)$
with word metric $d$ of the relatively hyperbolic group $G$ as in Definition~\ref{def-rel-hyp}.
Now let $\bf{Y}$ be the collection of all left cosets of peripheral subgroups in $G$.
For each such left coset $Y = g H_i$ we denote by $\mathcal{C}(Y)$ the subgraph $g \Gamma(H_i, S \cap H_i)$ in $\Gamma(G,S)$,
which contains $Y$ and is a copy of the Cayley graph of $H_i$.
Let $\pi_Y$ be a closest point projection on $\mathcal{C}(Y)$ for each $Y\in\bf{Y}$.
Namely, for each $x \in X$, let $\pi_Y(x)$ be the subset of $\mathcal{C}(Y)$ minimizing the function $d(x, \cdot)$,
and let $\pi_Y(X) = \bigcup_{x \in X} \pi_Y(x)$.

As peripheral subgroups are undistorted in $G$ \cite[Lemma 4.15]{DSp-05-asymp-cones} we can, for our purposes,
identify $\mathcal{C}(Y)$ with the corresponding subset of (the Cayley graph of) $G$.
\begin{lemma}
	The collection $\{(\mathcal{C}(Y),\pi_Y)\}_{Y\in\bf{Y}}$ satisfies Axioms $(0), (3)$ and $(4)$.
\end{lemma}
\begin{proof}
	For some constant $C$, the following holds \cite{projrelhyp}:
\begin{itemize}
 \item $\diam(\pi_Y(Y'))\leq C$ whenever $Y\neq Y'$, and
 \item if for some $x,y \in G$ we have $d(\pi_Y(x),\pi_Y(y))\geq C$ then any geodesic from $x$ to $y$ in the Cayley graph of $G$ intersects
 both of the balls of radius $C$ around $\pi_Y(x)$ and $\pi_Y(y)$.
\end{itemize}

The first property clearly implies Axiom $(0)$. The fact that the second property implies Axiom $(3)$ can be considered folklore, see \cite[Lemma 2.5]{contr}. Axiom $(4)$ follows from the fact that the right hand side of the distance formula \eqref{eqn-distform} is finite.
\end{proof}

\subsection{The proof of Theorem~\ref{prodqtrees}}\label{ssec-prodqtrees}
Let us go back to the general setting of Theorem \ref{bbf} for a moment, with
$\{ \mathcal{C}(Y), \pi_Y \}_{Y \in {\bf Y}}$ as in the statement of that theorem.
Suppose that we are also given another collection of geodesic metric spaces $\{ \mathcal{C}'(Y) \}_{Y \in {\bf Y}}$,
indexed by the same set ${\bf Y}$, and coarsely Lipschitz maps
$f_Y:\mathcal{C}(Y)\to \mathcal{C}'(Y)$, for $Y \in {\bf Y}$, with uniform constants.
Observe that if axioms $(0), (3)$ and $(4)$ hold for $\{(\mathcal{C}(Y),\pi_Y)\}_{Y\in\bf{Y}}$, then
they also hold for $\{(\mathcal{C}'(Y),f_Y\circ\pi_Y)\}_{Y\in\bf{Y}}$.

Now suppose that each $H_i$ admits a quasi-isometric embedding into the product of $k$ trees.
We then have coarsely Lipschitz maps $f_{i,Y}$, $i=1,\dots,k$ from $\mathcal{C}(Y)$ to some tree $T_{i,Y}$.
According to the observation we just made and Theorem \ref{bbf}, for each $i$ we have a map
$p_i: G\to {\bf T}_i=\mathcal{C}(\{(T_{i,Y},f_{i,Y}\circ \pi_Y)\})_{Y\in {\bf Y}}$, and ${\bf T}_i$ is a quasi-tree.
The last step in the proof of Theorem \ref{prodqtrees} is the following proposition.
\begin{proposition}
 The map $f=\prod p_i\times c: G\to \prod {\bf T}_i\times \hat{G}$ is a quasi-isometric embedding,
 where $c:G\to \hat{G}$ is the inclusion.
\end{proposition}
\begin{proof}
There are distance formulas for the ${\bf T}_i$'s \cite[Lemma 3.3, Corollary 3.12]{BBF} which summed up give,
for $L,\lambda,\mu$ large enough,
$$d\left(\prod p_i(x),\prod p_i(y)\right)\approx_{\lambda,\mu} \sum_{Y\in\bf{Y}} \big\{\big\{d(\pi_Y(x),\pi_Y(y))\big\}\big\}_L.$$
Comparing this with the distance formula for relatively hyperbolic groups (Theorem~\ref{distform})
immediately gives the required estimates.
\end{proof}

The map $c$ factors through Lipschitz maps $G\to X(G)\to \hat{G}$, so the proposition implies both versions of Theorem \ref{prodqtrees}.

\section{Three dimensional manifold groups}\label{sec-three-mfld}

In this section we prove the following theorem.
\begin{theorem}\label{thm-three-mfld}
	Let $G = \pi_1(M)$, where $M$ is a compact, orientable $3$-manifold whose (possibly empty) boundary is a union of tori.
	Then $\ecodim(G) < \infty$ if and only if no manifold in the prime decomposition of $M$ has $\mathrm{Nil}$ geometry;
	in this case, $\ecodim(G) \leq 8$.

	In any case, ${\ell\text{-}\!\asdim}(G) \leq 8$.
\end{theorem}

The second part of the following proposition is not needed for our purposes, but we think it is of independent interest.

\begin{proposition}\label{nonclosed}
 \begin{enumerate}
  \item If $M$ is a graph manifold with non-empty boundary then ${\ell\text{-}\!\asdim}(\pi_1(M))=2$.
  \item If $M$ is a Haken orientable $3$-manifold with non-empty boundary then $\asdim(\pi_1(M))\leq 2$.
\end{enumerate}
\end{proposition}

Recall that a compact orientable $3$-manifold $M$ is \emph{Haken} if it is irreducible and it contains a $\pi_1$-injective embedded surface $S$. Cutting $M$ along $S$ gives another Haken manifold, and moreover the new $\pi_1$-injective surface can be required to have non-empty boundary. Repeating the cutting procedure finitely many times yields a disjoint union of balls. Such sequence of cuts is called \emph{Haken hierarchy}.

\begin{proof}
 $1)$ The lower bound follows from the existence of undistorted copies of $\mathbb{Z}^2$ in $\pi_1(M)$. Up to passing to a finite-sheeted cover of $M$, we can assume that $M$ fibers over $S^1$ by \cite{WaYu-graphfiber}, so that we have a short exact sequence
$$1\to F\to \pi_1(M)\to \mathbb{Z}\to 1,$$
where $F$ is a free group.
 What is more, the content of \cite[Theorem 0.7]{WaYu-graphfiber} is that a fiber intersects all Seifert components and it splits them open into products of a surface with an interval. In particular, the monodromy of the fiber bundle is a product of Dehn twists along disjoint simple closed curves (the connected components of the intersection of the fiber with the boundary of the Seifert components), and so $F$ is undistorted in $\pi_1(M)$. The result then follows from a direct application of \cite[Theorem 0.2]{Bretal-Hurewicz-AN}.
\par
$2)$ We proceed by induction on the length of a Haken hierarchy for $M$ so that all surfaces involved have non-empty boundary. If $M$ is a ball, then the result trivially holds. Otherwise, $\pi_1(M)$ can be written as an amalgamated product $\pi_1(N_1)*_{\pi_1(S)}\pi_1(N_2)$ or an HNN extension $\pi_1(N_1)*_{\pi_1(S)}$, where $N_i$ is a Haken manifold admitting a strictly shorter Haken hierarchy (of the type described above) than $M$ and $S$ is a compact surface with non-empty boundary. We can then use the induction step and the fact that, when $A,B,C$ are finitely generated groups, $\asdim(A*_C B)\leq \max\{\asdim A,\asdim B,\asdim C+1\}$ and $\asdim(A*_C)\leq \max\{\asdim A,\asdim C+1\}$ (see \cite{Dr-amalg-form} and Theorem \ref{asdimhnn}).
\end{proof}

\begin{remark}
 The second part of the proposition (in the case of toric boundary) also follows from recent results about virtual specialness of fundamental groups of $3$-manifolds \cite{Wi-qconv-hier,Liu-virtcubgraph,PW-mixedvirtspec} in view of a criterion for virtual fibering discovered by Agol \cite{Ag-critvirtfibr}.
\end{remark}

We now show the following easy lemma and then prove the theorem.

\begin{lemma}
 $\ecodim G_1 *G_2=\max\{\ecodim G_1,\ecodim G_2\}$.
\end{lemma}

\begin{proof}
 The inequality $\geq$ follows from the fact that $G_1,G_2$ are undistorted in $G=G_1*G_2$. Suppose that, for $i=1,2$, $f_i:G_i\to \prod_{j=1}^n T^i_j$ are quasi-isometric embeddings. We have to show that $G$ embeds in the product of $n$ trees as well. Denote by $T$ the Bass-Serre tree of $G$. For each vertex $v$ of $T$ denote by $T_k^v$ a copy of $T^{i(v)}_k$, where $i(v)$ equals $i$ if the stabilizer of $v$ is conjugate to $G_i$.
When $e$ is an edge of $T$ with endpoints $v_1$, $v_2$, we let $p_e$ be the only element of $G$ in the intersection of the left cosets of $G_1, G_2$ corresponding to $v_1,v_2$. Finally, we let $T_k$ be the tree obtained from $\bigcup T_k^v$ by adding an edge of length, say, $1$ connecting $f_{1,k}(p_e)$ to $f_{2,k}(p_e)$ for each edge $e$ of $T$, where $f_{i,k}$ is the $k$-th component of $f_i$ and we identify $G_i$ with its left coset corresponding to an endpoint of $e$.
\par
There is a natural map $f:G\to\prod_{i=1}^n T_k$, which (up to an error bounded by $1$) restricts to $f_i$ on every left coset of $G_i$. It is readily checked that this map is a quasi-isometric embedding (using more sophisticated technology than needed, one can use the distance formula and observe that, as we added edges of length $1$ connecting $T_{v_1}$ to $T_{v_2}$ when $v_1$ is adjacent to $v_2$, we have $d(f(x),f(y))\geq d_T(x,y)$ and $d_T(x,y)$ is approximately $d_{\hat{G}}(x,y)$).
\end{proof}

\begin{proof}[Proof of Theorem \ref{thm-three-mfld}]
	Let $G_i$, for $i=1, \ldots, n$, be the fundamental groups of the prime factors $M_i$ of $M$.
	Then $G$ is the free product of the $G_i$, so that $\ecodim(G)=\max\{\ecodim(G_i)\}$ in view of the previous lemma, and also ${\ell\text{-}\!\asdim}(G) = \max \{ {\ell\text{-}\!\asdim}(G_i) \}$ by \cite{BrHi-09-AN-tree-graded}. In particular, we can just study the case when $M$ is prime.
\par
We report below a list of previously known cases. Except for the last two cases, the first column indicates the geometry (of the universal cover) of $M$.
The values in the table are justified below.
	\begin{center}
	\begin{tabular}{c|c|c}
		$M$ & ${\ell\text{-}\!\asdim}(\pi_1(M))$ & $\ecodim(\pi_1(M))$ \\
		\hline
		$S^3$			& $0$	& $0$ \\
		$\mathbb{R}^3$			& $3$	& $3$ \\
		$\mathbb{H}^3$, closed			& $3$	& $3$ \\
		$S^2 \times \mathbb{R}$	& $1$	& $1$ \\
		$\mathbb{H}^2 \times\mathbb{R}$, closed & $3$	& $3$ \\
		$\mathbb{H}^2 \times\mathbb{R}$, non-closed & $2$	& $2$ \\
		$\widetilde{SL_2\mathbb{R}}$ & $3$	& $3$ \\
		$\mathrm{Nil}$	& $3$	& $\infty$ \\
		$\mathrm{Sol}$	& $3$	& $3$ or $4$\\
graph manifold, closed & $3$ & $3$\\
graph manifold, non-closed & $2$ & $2$ or $3$
	\end{tabular}
	\end{center}
	The bounds on $S^3$, $S^2 \times \mathbb{R}$ and $\mathbb{R}^3$ are trivial.
	The calculation of ${\ell\text{-}\!\asdim}(\mathbb{H}^n) = \ecodim(\mathbb{H}^n) = n$ is found in
	\cite{BuSh-05-hyp-tree-bound-above,BuSh-07-hyp-tree-bound-below}.
	This also gives $\ecodim(\mathbb{H}^2 \times \mathbb{R}) \leq 3$, and ${\ell\text{-}\!\asdim}(\mathbb{H}^2 \times \mathbb{R}) = 3$
	\cite[Theorem 4.3]{DrSm-07-an-asdim-prod-r}.
	The spaces $\mathbb{H}^2 \times\mathbb{R}$ and $\widetilde{SL_2\mathbb{R}}$ are quasi-isometric. If $M$ is not closed and has geometry $\mathbb{H}^2 \times\mathbb{R}$ then $\pi_1(M)$ is virtually the product of a free group and $\mathbb{Z}$.

	The discrete Heisenberg group $H$ is quasi-isometric to $\mathrm{Nil}$.
	The result $\asdim(H)={\ell\text{-}\!\asdim}(H)=3$ has been obtained by several people, for example see \cite{DyHi-08-ascone-an-dim}.
	On the other hand, $\ecodim(H)=\infty$ as $H$ does not admit a quasi-isometric embedding into the product
	of finitely many metric trees, or indeed any CAT(0) space \cite{Pauls-01-qi-heis}.
	Thus the proof in the ``only if'' direction is complete.

	For ${\ell\text{-}\!\asdim}(\mathrm{Sol})=3$ see \cite{HiPe-09-andim-lie}.
	As $\mathrm{Sol}$ quasi-isometrically embeds in $\mathbb{H}^2 \times \mathbb{H}^2$ (see, for example, \cite[Section 9]{deCor-08-dim-as-cone-lie}),
	$\ecodim(\mathrm{Sol}) \leq 4$.

	If $M$ is a graph manifold then ${\ell\text{-}\!\asdim}(\pi_1(M))\leq\ecodim(\pi_1(M)) \leq 3$ by \cite{HuSi-11-andim-graph-manifold}, and it is observed in the same paper that the equalities hold if $M$ is closed. We handled the non-closed case for ${\ell\text{-}\!\asdim}$ in Proposition \ref{nonclosed}.

There are only two cases left. First, when $M$ is a finite volume, non-closed hyperbolic manifold
it is well-known that $\pi_1(M)$ is hyperbolic relative to virtually $\mathbb{Z}^2$
subgroups \cite{Farb-98-rel-hyp}.  (Notice also that $\asdim(\pi_1(M))\leq 3$ as $\pi_1(M)$ admits a coarse embedding in $\mathbb{H}^3$.)
As $X(\pi_1(M))$ is quasi-isometric to $\mathbb{H}^3$,
by Theorem~\ref{prodqtrees},
\[
	2 \leq {\ell\text{-}\!\asdim}(\pi_1(M)) \leq \ecodim(\pi_1(M)) \leq 5.
\]

The second case is when $M$
is non-geometric and its geometric decomposition contains at least one hyperbolic component.
In this case $\pi_1(M)$ is hyperbolic relative to virtually $\mathbb{Z}^2$ and graph manifold groups,
as a consequence of the combination theorem \cite[Theorem 0.1]{Dah-03-combination}
and aforementioned fact that the hyperbolic component groups are hyperbolic relative to virtually $\mathbb{Z}^2$ groups
(a full statement is given in \cite[Theorem 9.2]{AFW-12-3manifoldgroups}).
Therefore, by the graph manifold groups bound of \cite{HuSi-11-andim-graph-manifold}, every peripheral group
has $\ecodim$ at most $m=3$.
Note too that $\asdim(\pi_1(M))\leq 4$ holds by \cite{BeDr-04-asdim-trees}.
Thus by Theorem~\ref{thm-main1},
\begin{align*}
	\ecodim(\pi_1(M) & \leq \max\{\asdim(\pi_1(M)),m+1\}+m+1 \\ & = \max\{ 4, 3+1\} + 3+ 1 = 8.
\end{align*}
Therefore, we have
\[
	2 \leq {\ell\text{-}\!\asdim}(\pi_1(M)) \leq \ecodim(\pi_1(M)) \leq 8,
\]
when $M$ is non-closed, and
\[
	3 \leq {\ell\text{-}\!\asdim}(\pi_1(M)) \leq \ecodim(\pi_1(M)) \leq 8,
\]
when $M$ is closed.  The lower bound in the last case follows from the fact that the asymptotic dimension is
greater or equal to the virtual co-homological dimension for groups of type FP \cite{Dr-cohom-asym}.
\end{proof}

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My name is Carol Chike. I am a Registered Nurse and work as a certified Nurse Assessment Coordinator at ManorCare Health Services -Silver Spring. I am barely into my first year at HCR ManorCare, and I feel so blessed to be a part of this great company. I write this letter to you, even though words alone cannot express my gratitude toward the company, especially toward our grant program, the Hug Fund.
The Hug Fund came to my rescue when my apartment home was gutted by fire last June. The fire started after a lightning strike just minutes after I got home from work. My family and I made it out safely, but nothing besides my work laptop, my handbag and the pouch for my official documents were saved.
I had nobody and nowhere to turn to – or so I thought. But then I remembered hearing about the Hug Fund in my orientation. It was the Hug Fund that came to my rescue in my family's time of need.
I want to use this opportunity to thank each person who has contributed any amount to the Hug Fund. Your generosity was used to provide me and my children the necessary household supplies we needed at that crucial moment in our life. We thank you immensely.


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African Journal of Food, Agriculture, Nutrition and Development
Rural Outreach Program
Vol. 20, No. 1, 2020, pp. 15252-15266 Bioline Code: nd20007
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African Journal of Food, Agriculture, Nutrition and Development, Vol. 20, No. 1, 2020, pp. 15252-15266
en DETERMINANTS OF COMPETENCE IN SOYBEAN PROCESSING AMONG SECONDARY SCHOOL STUDENTS IN IBADAN, OYO STATE, NIGERIA
Fadairo, AO; Oyegbami, A & Oyedokun, MO
Competence in the processing of soybean is constrained by many factors which directly influence the consumption pattern in Nigeria. Youth involvement in soybean processing is a strategy to successfully incorporate soybean utilization in households across South- west Nigeria. The study identified factors that influence competence in soybean processing among secondary school students in Ibadan metropolis of Oyo state. The study is an offshoot of a training activity embarked by Institute of Agricultural Research and Training, (IAR&T) Ibadan to improve the knowledge of soybean processing among secondary students. Thus, study purposively selected three secondary schools adopted by the Institute and all the senior secondary students formed the study population. However, 30% of the senior students were randomly sampled which gave a total of 203 respondents. Independent variables such as personal characteristics of students, knowledge of benefit of soybean and competence in the processing of soybean were collected using validated questionnaire. Competence was measured in a four-point scale of high competence, average competence, low competence and no competence. Data were described using percentage and mean, and were analyzed using Analysis of Variance (ANOVA), Chi square, Pearson Product Moment Correlation (PPMC), and multiple regression analysis. Findings revealed that the mean age of respondents was 15.9 years and majority (57.6%) was female. Highest education attained by mothers (57.1%) was secondary education and majority (62.6%) of mothers was involved in trading. The knowledge of benefit of soybean was high among majority (64.1%) but majority (62.0%) also had low competence in soybean processing. Data analysis revealed a significant relationship (p<0.05) between gender (χ 2=17.7), class of students (χ 2=20.3) and competence in soybean processing. Level of mothers' education (65%) and knowledge of benefits of soybean (42%), respectively had positive effect on students' competence. The study concludes that competence in soybean processing among secondary students is limited to some few products such as soymilk and it is associated with many factors like knowledge of its benefit and mothers' education
Competence; Knowledge; Soybean processing; nutrition; secondary schools students; adopted schools
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A new form of cancer treatment that directs the body's own immune system to fight tumours has been named 'Breakthrough of the Year' by one of the world's leading scientific journals.
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Cancer immunotherapy is causing a "paradigm shift" in the way researchers and clinicians think about treating people with cancer, the journal editors say. The treatment targets the immune system, not the tumour itself. It is the body's own immune defences – antibodies and T-cells – that are directed by the treatment to seek out and destroy tumour cells wherever they appear.
"At the Malaghan Institute, we are currently in the early stages of a Phase 1 clinical trial of a therapeutic vaccine for melanoma. This cancer immunotherapy treatment is part of the worldwide movement to harness our own immune system to fight cancer more effectively," he says.


The 2019 Toyota Tacoma is prepared to elevate your commute, but which trim has the exact features you need? Orland Toyota in Tinley Park, IL, has a new inventory with every available model you could want. Simply browse the following guide to discover the must-have specs of the Tacoma you'll love. Don't hesitate to contact our team online or call (708) 336-7288 and we will gladly assist you.
For a closer look at what the 2019 Toyota Tacoma has to offer, browse through the following interior and exterior specs.
Towing and payload capacity can vary depending on the Toyota Tacoma that you choose. Listed below are the payload and towing capabilities of all available models in the pickup truck lineup.
SR- Towing: 3,500 lbs./Payload: 1,620 lbs.
SR5- Towing: 3,500 lbs./Payload: 1,620 lbs.
TRD Sport- Towing: 6,800 lbs./Payload: 1,540 lbs.
TRD Off-Road- Towing: 6,700 lbs./Payload: 1,420 lbs.
Limited- Towing: 6,600 lbs./Payload: 1,370 lbs.
TRD Pro- Towing: 6,400 lbs./Payload: 1,155 lbs.
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SWBAT factor quadratic equations when appropriate. SWBAT connect the solutions of quadratic equations to x-intercepts. SWBAT distinguish between quadratic equations that are not factorable and still have x-intercepts with those that do not have x-intercepts.
Students continue to develop their understanding of factoring and connect this work to the x-intercepts of quadratic functions.
I begin class with a Factoring Activity warm up that will review what students have learned about factoring so far. This will be the first time students see quadratics that have positive and negative signs in terms of factoring. During the warm up, I circulate and check for understanding, working with any students that might need one-on-one help or small group instruction. Ib ring the class together when most students have finished all five questions and ask for volunteers to share out their work.
During this discussion, I try to elicit from students if they have any tricks when working with different signs. I try to get students to talk about why if the quadratic has all positive signs the two binomials must also have two positive signs. Then we talk about what happens when the "b" term is negative but the "c" term is positive. I find it's more powerful to have students come up with these ideas rather than me presenting them as the teacher.
X-intercepts: Not just the "opposites"
Next, I want to show students why factored form is useful and how it can be a quick way to see the x-intercepts for a quadratic. I show students a quadratic that is written in factored form. For example: y = (x – 2) (x + 5). I remind them that the x-intercepts are the values of x that make y = 0. Ask them what values of x will make the product (x – 2)( x + 5) equal to 0? I review the Zero Product Property here by telling students that if ab = 0, if and only if a =0 or b = 0. Students should be able to see that x could equal 2 or -5 and therefore, the x-intercepts of the quadratic are (2, 0) and (-5, 0). I then use an area model to rewrite the quadratic in standard form which will look more familiar to students.
I will make sure to go over the two Questions that are not factorable. I will try to elicit from my students how the two equations are different. One, Question 2 is not factorable, but has two x-intercepts. Question 5 is not factorable and does not cross the x-axis. Be sure to elicit from students how to determine this information. They may need to review Vertex Form in order to get a sense of the how the parabola looks.
For Question 6, I will make sure bring out that although the original equation and the equation divided by 2 have different graphs, they have the same x-intercepts. I will probably use graphing calculators to give students the opportunity to explore and confirm this relationship.
How would you decide what method to use for solving a quadratic equation? Explain your thinking.


Axis Of Oversteer: Pure racing: Overtaking class on Mount Panorama.
Pure racing: Overtaking class on Mount Panorama.
Some of the best racing you never hear anything about.
Fast, new cars are glamorous and exciting but even Saint Ayrton said the best, purest racing he ever did was as an amateur.
So you probably seen Sabine "Queen of the Ring" Schmitz passing a spectacular number of cars in a VLN race before. Amazing stuff but the (hopefully) future Top Gear host is a full time professional, driving the latest model from Porsche, prepared by a very experienced team, on a track she literally grew up driving on.
With that in mind, I give you Joseph Lethal, killing it at a support race for the 2016 Bathurst 12 hours. Lenthall built and prepared his first gen RX7 and this was his first time racing on the mountain course!
Love the dog leg box and how the engine seemingly revs to 4 million rpm.
That car is hooked up.
Damn, the track is narrow, makes passing at the Nurburgring seem simple.
So why did Joseph start 49th anyway ?
"Last weekend, I travelled to Bathurst to race at Mount Panorama for the first time. After walking the track and being told where not to stick it into the fence, I came back to a prepared car thanks to my crew.
"Practice was great and we finished P14 out of a 61- car field. Qualifying was also good, with another P14 and some time left in the car.
"I got a woeful start in race 1 and dropped back to about 18th. After making my way past a few cars, the alternator belt flicked off and the car stopped coming out of turn 3. Unfortunately, between myself and the marshals, we then badly bent the front of the car while trying to load it on the truck.
"Enter my crew again, who through a drill, hammer, cable ties and race tape straightened the front and got the car ready for Race 2. I started 49th and around the last corner; eventually I finished 15th with a straight and running car.


Skyhawk had the most liked content!
I have a question, Will there be another batch produced for ships that have Restock expected, or is it just the one batch? Edit: Disregard my question, I looked at the full description of the ships. They are going to be restocked. My bad.
Perseus cruisers and Hauberk heavy cruisers on the coming soon list. Perfect opportunity to make some of my Hercules heavy cruisers!
MWAHAHAHAHAHAHAHHAHAHAHAHAHHAHAHAHAHAHA!!!!!! FINIALLY SOMETHING TO GET ME EXCITED ABOUT FIRESTORM AGAIN!!!!!!
Thanks, I'll be sure to give it a look. In the mean time I shall look forward to more of your drawings.
Those are some really cool drawings @Bad Idea Comics! I wish I could draw that well! I'll just have to practice more.
Warcradle would have a riot from all the Sorylian players not getting to be space Lizards anymore. We can get a rough Idea of Warcradle is likely to change lore wise by looking at the Dystopian Age. I think its safe to say that there won't be any race changes, it would be change for the sake of change which nether goes over well. @Charistoph that is an interesting idea. I think a timeline jump would work well for Warcradle. It would help smooth ruffled feathers.
Hang on I think we got some spare sledge hammers lying around. Being that it is a different Company I do understand them wanting to put their own twist on it, but I also would like to be able to recognize it as Firestorm universe ( Fluff wise and aesthetically at least).
I try to name the different Squadrons but only my Tier 1s have names painted on them. also did different markings on every battlegroup. The Tier 2 ships have Roman numerals and Regular numbers for the tier 3s. Also the different Battlegroups have their own markings . So players can tell which ship is which and to which squadron it belongs. Saves a lot of hassle trying to name all my ships.
About ten years after. Hehehehehehe. On a more serious note I don't completely understand your faction pairings @Small Mek. For example, Terquai + Works Raptor? I think that the Terquai should go with the Aquans and Works Raptor with the Directorate. Basically keep the original natural Allies. My personal opinion.
Yes it is, I even have a Christmas list! Sentinel heavy cruisers. A Praetorian battleship. Or maybe the ships necessary to build some of my Hercules heavy cruisers....... Alas I'll just have to wait and see!
If they have the render then maybe we'll get to see her in the future!
Very nice fleet @ChibiPotatoes. I look forward to seeing your Final scheme on your Razorthorn. I will also have to have a look at your lore.
We have the DVD version, And I agree it is a good movie.
But which ending is it?
The true weapon used to kill Mr Body has been discovered by Authorities.


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