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The Black Dahlia (novel)

The Black Dahlia (1987) is a crime fiction novel by American author James Ellroy. Its subject is the 1947 murder of Elizabeth Short in Los Angeles, California, which received wide attention because her corpse was horrifically mutilated and discarded in an empty residential lot. The investigation ultimately led to a broad police corruption scandal. While rooted in the facts of the Short murder and featuring many real-life people, places and events, Ellroy's novel blends facts and fiction, most notably in providing a solution to the crime when in reality it has never been solved. James Ellroy dedicated The Black Dahlia, "To Geneva Hilliker Ellroy 1915-1958 Mother: Twenty-nine Years Later, This Valediction in Blood." The epigraph for The Black Dahlia is "Now I fold you down, my drunkard, my navigator, My first lost keeper, to love and look at later. -Anne Sexton."

This book is considered the one that gained Ellroy critical attention as a serious writer of literature, expanding his renown beyond the crime novels of his early career. The Black Dahlia is the first book in Ellroy's L.A. Quartet, a cycle of novels set in 1940s and 1950s Los Angeles. He portrays the city in this period as a hotbed of political corruption and depravity. The Quartet continues with The Big Nowhere, L.A. Confidential, and White Jazz.

Synopsis
In 1946, the LAPD arranges a boxing match between patrolmen Dwight "Bucky" Bleichert and Lee Blanchard as a publicity stunt. The fight is promoted as "Fire versus Ice", reflecting their contrasting boxing styles and personalities. Bucky is cautious and reserved, racked with guilt over his betrayal of two Japanese-American friends (resulting in their wartime internment) and involvement in violently suppressing of the Zoot Suit Riots. By contrast, Lee is outgoing and jovial, already famous for catching the notorious bank robber Bobby De Witt. Lee is openly cohabiting with De Witt's girlfriend, Kay Lake, in violation of LAPD policy.

Bucky and Lee are promoted to plainclothes jobs in the Warrants Division and are subsequently partnered together. Bucky quickly forms a close bond with Lee and Kay, seeing them as a surrogate family. Kay becomes attracted to Bucky and attempts to seduce him, revealing that her relationship with Lee is platonic, but Bucky nevertheless rebuffs her advances in order to avoid upsetting the harmonious friendship between the three of them.

In January 1947, the hideously mutilated body of Elizabeth Short is discovered. The killing, which the press dub "Black Dahlia", reminds Lee of the unsolved disappearance of his younger sister Laurie, prompting him to volunteer himself and Bucky for the investigation. One of the other detectives, the thuggish Fritz Vogel, is in cahoots with Assistant DA Ellis Loew, who hopes to use the case to launch his political career. He and Vogel repeatedly try to frame innocent men in order to close the case quickly. Bucky is initially resentful of his and Lee's involvement in the case, but gradually becomes obsessed with Short as he learns about her chaotic and unhappy life.

One line of enquiry leads Bucky to Madeleine Sprague, a spoiled socialite who closely resembles Short and was also a lesbian lover of hers. In exchange for Bucky suppressing evidence about her involvement, Madeleine has sex with him, causing him to fantasize that she is Short. Bucky meets the rest of Madeleine's family, comprising her tyrannical father Emmett, an amoral property developer; her sickly mother Ramona; and her sister Martha, an aspiring artist. He also learns about George Tilden, an old friend of Emmett's who was badly disfigured in a car crash and subsequently became a reclusive handyman on Emmett's estate.

Lee becomes increasingly agitated; besides being preoccupied with the Short case, he is also worried about De Witt's imminent release from prison. When Bucky finds a pornographic film featuring Short, Lee absconds to Tijuana, ostensibly to track down the man who made the film. Bucky follows and stumbles across the corpse of De Witt, apparently murdered along with Mexican drug trafficker Felix Chasco by corrupt Rurales hired by Lee. Despite De Witt's death, Lee fails to reappear, and Bucky returns to Los Angeles.

Bucky discovers that Vogel has been suppressing evidence in order to hide the fact that Short was hired as a prostitute by his son Johnny shortly before her death. Bucky forces Johnny to confess as much, leading to his being arrested for soliciting, while Vogel commits suicide. Bucky is placed on two weeks' leave and uses the time off to return to Mexico to look for Lee. There, Bucky learns from a San Diego private investigator that Lee was murdered by a Mexican woman, probably Chasco's lover avenging his death. With the PI's help, Bucky finds Lee's body buried on the beach at Ensenada. He returns to L.A. and reveals all to Kay, who in turn reveals that Lee was the real leader of the De Witt gang, and that he turned on his former accomplices in order to stop De Witt's domestic abuse. Bucky is horrified by this revelation, but forgives Lee and marries Kay. He ends his relationship with Madeleine and puts the Dahlia case behind him.

Bucky's superiors punish him for his part in Vogel's downfall by transferring him to the forensics unit, a career dead-end. His initially happy marriage with Kay gradually deteriorates. Two years later, he is assigned to collect evidence surrounding the suicide of Eldridge Chambers, a former friend of Emmett. At Chambers' house, he sees a painting of a grotesque clown with a Glasgow smile, similar to Short's facial mutilations. His interest in the Dahlia case piqued again, Bucky stakes out the Sprague mansion and discovers that Madeleine is frequently making herself up to look like Short and picking up sailors in the same bars which Short once frequented. Bucky confronts Madeleine, who explains that she's been mimicking Short in order to lure Bucky to her, having developed a sexual fixation on him. The two renew their affair, causing Kay to leave Bucky.

During the pageant to celebrate the removal of the last four letters of the "Hollywoodland" sign, a nearby shack is discovered to contain a mattress covered in dried blood. Bucky conducts a forensic examination of the shack and determines that it is the place where Short was tortured and killed. He matches fingerprints at the shack to Tilden and learns that the shack is owned by Emmett.

Bucky confronts Madeleine and Emmett, who are revealed to be lovers, but they explain it is not incestuous as Madeleine is not actually Emmett's daughter but the result of an affair between Ramona and Tilden, for which Emmett got revenge by disfiguring Tilden with a knife. Years later, Madeleine shot the porn film featuring Short in one of Emmett's empty houses, and Tilden witnessed it and became enamoured with Short. He subsequently blackmailed Emmett and Madeleine, threatening to reveal that his disfigurement was the result of Emmett's assault rather than a car crash unless Madeleine arranged for him to go on a date with Short. Moreover, Emmett also reveals that Tilden, the son of an anatomist, has always been obsessed with dead bodies and habitually exhumes corpses in order to collect their organs. Aware that he can't publicly expose Tilden and the Spragues, lest he also be convicted for suppression of evidence, Bucky tracks down Tilden and kills him.

The murder is seemingly solved, but Bucky is bugged by a discrepancy in the case files and goes to speak to Martha Sprague, from whom he discovers that Lee had actually deduced Emmett and Madeleine's involvement in the first few days of the investigation and used this knowledge to blackmail Emmett for $100,000. Bucky is shocked by this revelation, and in particular by the fact that Kay acted as go-between in the deal. He confronts Kay, who is so ashamed that she leaves L.A.

Bucky bumps into Chambers' widow and learns that the clown painting is of a character from the novel The Man Who Laughs. Remembering that a copy of that book was found at the shack where Short was killed, he discovers that Chambers bought the painting from Ramona. Ramona confesses to Bucky that she is the real murderer, having killed Short in Tilden's shack after gatecrashing their date; she persuaded Tilden to go along with it by offering to let him keep Short's organs after the murder. Having modelled her torture of Short upon the novel, she subsequently sold the painting to Chambers "as an act of purging".

While vacillating about whether or not to arrest Ramona, Bucky discovers that Madeleine is again dressing up as Short. Desperate to get his attention, she reveals to Bucky that she, in disguise as a Mexican woman, was the one who murdered Lee, in order to retrieve the $100,000 that he extorted from her father. Bucky arrests her and she is institutionalized; she gets revenge by revealing Bucky's suppression of evidence, causing him to be fired from the LAPD. Bucky doesn't disclose what he knows about Ramona, and the Dahlia case remains officially unsolved. Kay writes to Bucky telling him that she is pregnant, and the two reconcile. The novel ends with Bucky flying out to join Kay, and on the plane, he prays for Short's spirit to watch over him in his new home – Boston, her birthplace.

Reception
The Black Dahlia was one of numerous neo-noir novels published in the late 1970s and 1980s. Ellroy was known as an author of crime fiction but this novel is considered to have gained him critical notice as a serious writer of literature.

Ellroy wrote three other novels in what he termed the L.A. Quartet, a cycle of novels set in 1940s and 1950s Los Angeles. He portrays the city in this period as a hotbed of political corruption and depravity. The Quartet continues with The Big Nowhere, L.A. Confidential, and White Jazz.

Film adaptation
The Black Dahlia was adapted for a 2006 film of the same name by director Brian De Palma. Starring Scarlett Johansson, Josh Hartnett and Aaron Eckhart, it was a critical and commercial failure, with the consensus being that it had been poorly made and acted. It was criticized as sometimes appearing incoherent. The latter fault may have been caused by De Palma's drastic editing of the finished product, which initially ran for three hours and he eventually cut down to two.

Graphic novel
In 2013, Matz and David Fincher adapted James Ellroy's novel into a comic called Le Dahlia noir, with Miles Hyman as the illustrator. Originally published in French, it was published in English in 2016 as The Black Dahlia: A Crime Graphic Novel.

Anachronisms
The Black Dahlia has several references to characters having been committed to Atascadero State Hospital, but this institution did not open until 1954. The character Madeleine is committed there (ch. 3–5), but the hospital's patient population was historically limited to men.

Madeleine tells her father she and Bucky met at an art show at Stanley Rose's bookstore. However, that store closed permanently eight years prior in 1939.

title: The Black Dahlia (novel)


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Bananaman

Bananaman is a fictional character appearing in British comic books. Bananaman is a parody of traditional superheroes, being portrayed as a schoolboy who is transformed into a muscled, caped adult man when he eats a banana. The character originally appeared in Nutty as the back page strip in Issue 1, dated 16 February 1980 drawn by John Geering.

He has since appeared in The Dandy and The Beano.

Original strip
The original strip, by Dave Donaldson and Steve Bright, written and developed by the latter, and mostly drawn by John Geering until his death in 1999, is essentially a parody of Superman and Batman with elements of Captain Marvel and his British twin, Marvelman, and occasionally other Silver Age characters, while also combining comic slapstick with a heavy dose of eccentric British humour similar to Alan Moore's contemporary work on Captain Britain.

After John Geering died in 1999, Barrie Appleby took over and later Tom Paterson. In 2003, the original scriptwriter, Steve Bright, drew it, until 2007. Sporadically from 2007 to 2010 the character appeared in reprinted strips from the John Geering era. For a short time, in the end of 2008, artist Chris McGhie reinvented Bananaman in a series of new strips.

McGhie's other work included The Three Bears for The Beano (in 2002) and the characters on Yoplait's 'Wildlife' product range. Two new strips drawn by Barrie Appleby appeared that year as well.

Following the Dandy revamp of October 2010, Wayne Thompson took over drawing Bananaman in a style reminiscent of French cartoonist Lisa Mandel. Thompson was a popular artist in The Dandy who had previously drawn Jak, Agent Dog 3-Zero and, occasionally, Bully Beef and Chips.

In Issue 3515, Thompson's style changed notably, becoming more cartoonish and detailed. As of spring 2011, Thompson's version of Bananaman appears in full colour over two pages. From 1983 to 1986, Bananaman also had his own annual. This was unusual because, unlike many other comics at the time, Nutty never had an annual.

Unlike Dennis the Menace and Bash Street Kids, which mostly consisted of reprints, all the material in these annuals was new. In Issue 3618, dated 14 January 2012, Bananaman made his debut appearance, as John Geering reprints, in The Beano; however, he continued to appear in The Dandy. Another Beano character, Bananagirl of Super School, was revealed to be his cousin.

The Dandy print comic ended in December 2012, but Bananaman was still seen in the digital version drawn by Andy Janes. New Bananaman strips drawn by Wayne Thompson and written by Nigel Auchterlounie, Kev F Sutherland and lately Cavan Scott continued to run in The Beano throughout 2014.

In 2016, writing duties for the strip were taken over by Tommy Donbavand and Danny Pearson, and since 2018, Bananaman has been written by Ned Hartley.

Character
In the strip, Eric Wimp, an ordinary schoolboy living at 29 Acacia Road, Nuttytown (later changed to Dandytown and then Beanotown when the strip moved to other comics), eats a banana to transform into Bananaman, an adult superhero, sporting a distinctive cowled blue and yellow outfit complete with a yellow two-tailed cape resembling a banana skin.

His superpowers include the ability to fly, superhuman strength (often quoted as "twenty men... twenty big men" but sometimes limitless, with "nerks", "women" and "snowmen" all being used in place of "men"), and seeming invulnerability.

This is offset by the fact that he is just as naive and foolish (if not more so) as his alter ego; as mentioned in the comic once or twice, he has the "muscles of twenty men and the brains of twenty mussels".

If Bananaman needs extra power, bananas can be eaten for strength boosts, provided by his faithful pet crow; if he does not have enough strength to shatter an ice block, for example, after eating another banana, he will have enough. If he eats many bananas in one sitting, he quickly becomes obese in his transformation; if he eats bananas that are not full, he transforms with extra weight in the lower part of his body.

There have also been comics where he has eaten a variant on normal bananas, and transforms differently, reflecting the difference in that banana. The effects of eating the bananas are not consistent from story to story. In one Beano issue with Eric unable to find a banana, he resorted to drinking banana milk, becoming a liquid, totally useless version of Bananaman who later in the story is mopped up by a janitor.

Varying origins
Eric Wimp was rocketed to Earth from the Moon as a baby, and gained his powers because the crescent moon resembles a banana. Bananaman resembles Superman in having a kryptonite style weakness to mouldy bananas, and a Fortress of Solitude style building at the North Pole, made out of a giant banana.

During early board meetings, the designers thought of having Bananagirl accompany the series. The girl would have been called Margaret Wimp, and be the "sister" of Eric. This idea was scrapped later in production, because the concept of two children being related without parents would be too far-fetched for children to understand; however, the idea was revived for a Beano comic strip.

In the 1991 Dandy Annual, Bananaman's origin was changed to that of being a normal Earth baby in a maternity hospital, who obtained his powers after unintentionally eating a banana in which General Blight had hidden a stolen supply of 'Saturnium', and accidentally left it next to Eric. However, later issues referred to the first origin as the real one.

Other characters
Bananaman initially faced a different pastiche supervillain each week, who were often lampoons of the kind of single issue, uncreatively named villains that heroes fought during the Silver Age, or tips of the hat to famous supervillains. Bananaman's arch enemy is General Blight, a parody of Adolf Hitler and generic criminal mastermind who in later strips largely replaced the criminal of the week.

Other villains included mad scientist Doctor Gloom, Bananaman's evil fruit counterpart Appleman, the mischief making Weatherman and dessert fiend Captain Cream.

Eric's punk-style shaved head was replaced by a more typical 1980s-style haircut, Bananaman gained a talking crow sidekick called simply Crow, and Bananaman became so stupid he often forgot how to fly or to use the door. Eventually, Bananaman even began to go to school despite being an adult.

Bananaman is allied with Chief O'Reilly, a stereotyped Irish policeman (apparently in homage to Batman's James Gordon or the equally stereotyped Chief O'Hara in the 1960s Batman television series). He used to wear an Indian feather headdress as a visual pun on Chief, and in later strips wore a hat with a flashing blue light on the top.

Chief works in a police station shaped like a giant police helmet, which frequently has to be rebuilt after Bananaman accidentally destroys it. O'Reilly rings up Eric to get him to talk to Bananaman, presumably thinking Eric is Bananaman's assistant of some kind, as in the cartoon series it is made clear that the Chief is not aware of Eric's being the superhero.

Television cartoon series
From 1983 to 1986, the BBC aired a cartoon series based on Bananaman and featuring the voices of the members of The Goodies. It was produced by 101 Productions. Parts of the character were changed for the series: he was now called Eric Twinge, had a distinctive banana-shaped hairstyle rather than punk stubble, and had a love interest (only when transformed) in the form of Fiona, a newsreader based on Selina Scott and also a possible homage to Lois Lane.

Graeme Garden (incorrectly credited as Greame Garden on some episodes) voiced the characters of Bananaman, General Blight and Maurice of The Heavy Mob, Bill Oddie voiced the characters of Crow, Chief O'Reilly, Doctor Gloom and the Weatherman, and Tim Brooke-Taylor voiced the characters of Eric, King Zorg of the Nerks, Eddie the Gent, Auntie, and Appleman, as well as narrating the episodes.

Jill Shilling voiced Fiona and any additional female characters, including Eric's cousin Samantha (but not her mother Auntie). The programme lasted for forty episodes between 3 October 1983 and 15 April 1986.

Bananaman was aired in the United States by the Nickelodeon cable network, as a companion piece to Danger Mouse, but Bananaman never came close to reaching that series' American popularity. The show also aired during the Australian Broadcasting Corporation's (ABC) after-school timeslot, and is considered one of the Classic ABC shows.

In 1997, some episodes of Bananaman were used on the cartoon series The Pepe and Paco Show, created by Henson International Television.

Some of these episodes would eventually reappear in print form in The Dandy in 1998, coinciding with the BBC repeating the series that year, and were reprinted in the comic in the spring of 2007, now promoting the DVD. Each episode was roughly five minutes long. Phrases from the show, "twenty big men" and "ever alert for the call to action", are still used in the comic today.

On 22 February 2021, Fox Entertainment announced that they would be producing a new Bananaman series with Bento Box Entertainment.

Film adaptation
In March 2014, it was announced that DC Thomson, in conjunction with Elstree Studio Productions, would be producing a movie on Bananaman, with a release date in 2015. In May 2014, DC Thomson unveiled the first teaser poster for the film. By September 2015, the official website stated "coming soon" instead of 2015. In September 2015, it was announced that the movie was in the early stages.

In January 2016, the Bananaman musical's page on Facebook posted that the movie adaptation is now in development, saying "This fruitiest of superheroes is experiencing a revival elsewhere – Bananaman The Movie is also in development". However, a release date was not mentioned.

On 8 June 2016, the now newly formed Beano Studios issued a press release. The release noted that Beano Studios was formed to bring their properties to life through television, film and live performances based upon present projects which were being worked on. "Beano Studios is currently also exploring plans to take Beano characters to the bigger screens and stages worldwide."

By June 2017, the official site had been removed.

Musical production
Near the beginning of January 2016, it was reported that Bananaman would be turned into a musical for West End. An industry launch took place on 2 February 2016, showcasing the musical. The musical ran from the end of 2017 to the beginning of 2018, at the Southwark Playhouse in London. It is unknown if it will be produced worldwide.

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

title: Bananaman


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User:Mukti2

Mukti means Freedom in Sanskrit! I love this Encyclopedia of FREEDOM!!!

title: User:Mukti2


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Linear algebraic group

In mathematics, a linear algebraic group is a subgroup of the group of invertible $$n\times n$$ matrices (under matrix multiplication) that is defined by polynomial equations. An example is the orthogonal group, defined by the relation $$M^TM = I_n$$ where $$M^T$$ is the transpose of $$M$$.

Many Lie groups can be viewed as linear algebraic groups over the field of real or complex numbers. (For example, every compact Lie group can be regarded as a linear algebraic group over R (necessarily R-anisotropic and reductive), as can many noncompact groups such as the simple Lie group SL(n,R).) The simple Lie groups were classified by Wilhelm Killing and Élie Cartan in the 1880s and 1890s. At that time, no special use was made of the fact that the group structure can be defined by polynomials, that is, that these are algebraic groups. The founders of the theory of algebraic groups include Maurer, Chevalley, and. In the 1950s, Armand Borel constructed much of the theory of algebraic groups as it exists today.

One of the first uses for the theory was to define the Chevalley groups.

Examples
For a positive integer $$n$$, the general linear group $$GL(n)$$ over a field $$k$$, consisting of all invertible $$n\times n$$ matrices, is a linear algebraic group over $$k$$. It contains the subgroups
 * $$U \subset B \subset GL(n)$$

consisting of matrices of the form, resp.,
 * $$\left ( \begin{array}{cccc} 1 & * & \dots & * \\ 0 & 1 & \ddots & \vdots \\ \vdots & \ddots & \ddots & * \\ 0 & \dots & 0 & 1\end{array} \right )$$ and $$\left ( \begin{array}{cccc} * & * & \dots & * \\ 0 & * & \ddots & \vdots \\ \vdots & \ddots & \ddots & * \\ 0 & \dots & 0 & *\end{array} \right )$$.

The group $$U$$ is an example of a unipotent linear algebraic group, the group $$B$$ is an example of a solvable algebraic group called the Borel subgroup of $$GL(n)$$. It is a consequence of the Lie-Kolchin theorem that any connected solvable subgroup of $$\mathrm{GL}(n)$$ is conjugated into $$B$$. Any unipotent subgroup can be conjugated into $$U$$.

Another algebraic subgroup of $$\mathrm{GL}(n)$$ is the special linear group $$\mathrm{SL}(n)$$ of matrices with determinant 1.

The group $$\mathrm{GL}(1)$$ is called the multiplicative group, usually denoted by $$ \mathbf G_{\mathrm m}$$. The group of $$k$$-points $$\mathbf G_{\mathrm m}(k)$$ is the multiplicative group $$k^*$$ of nonzero elements of the field $$k$$. The additive group $$\mathbf G_{\mathrm a}$$, whose $$k$$-points are isomorphic to the additive group of $$k$$, can also be expressed as a matrix group, for example as the subgroup $$U$$ in $$\mathrm{GL}(2)$$ :
 * $$\begin{pmatrix}

1 & * \\ 0 & 1 \end{pmatrix}.$$

These two basic examples of commutative linear algebraic groups, the multiplicative and additive groups, behave very differently in terms of their linear representations (as algebraic groups). Every representation of the multiplicative group $$\mathbf G_{\mathrm m}$$ is a direct sum of irreducible representations. (Its irreducible representations all have dimension 1, of the form $$x \mapsto x^n$$ for an integer $$n$$.) By contrast, the only irreducible representation of the additive group $$\mathbf G_{\mathrm a}$$ is the trivial representation. So every representation of $$\mathbf G_{\mathrm a}$$ (such as the 2-dimensional representation above) is an iterated extension of trivial representations, not a direct sum (unless the representation is trivial). The structure theory of linear algebraic groups analyzes any linear algebraic group in terms of these two basic groups and their generalizations, tori and unipotent groups, as discussed below.

Definitions
For an algebraically closed field k, much of the structure of an algebraic variety X over k is encoded in its set X(k) of k-rational points, which allows an elementary definition of a linear algebraic group. First, define a function from the abstract group GL(n,k) to k to be regular if it can be written as a polynomial in the entries of an n×n matrix A and in 1/det(A), where det is the determinant. Then a linear algebraic group G over an algebraically closed field k is a subgroup G(k) of the abstract group GL(n,k) for some natural number n such that G(k) is defined by the vanishing of some set of regular functions.

For an arbitrary field k, algebraic varieties over k are defined as a special case of schemes over k. In that language, a linear algebraic group G over a field k is a smooth closed subgroup scheme of GL(n) over k for some natural number n. In particular, G is defined by the vanishing of some set of regular functions on GL(n) over k, and these functions must have the property that for every commutative k-algebra R, G(R) is a subgroup of the abstract group GL(n,R). (Thus an algebraic group G over k is not just the abstract group G(k), but rather the whole family of groups G(R) for commutative k-algebras R; this is the philosophy of describing a scheme by its functor of points.)

In either language, one has the notion of a homomorphism of linear algebraic groups. For example, when k is algebraically closed, a homomorphism from G ⊂ GL(m) to H ⊂ GL(n) is a homomorphism of abstract groups G(k) → H(k) which is defined by regular functions on G. This makes the linear algebraic groups over k into a category. In particular, this defines what it means for two linear algebraic groups to be isomorphic.

In the language of schemes, a linear algebraic group G over a field k is in particular a group scheme over k, meaning a scheme over k together with a k-point 1 ∈ G(k) and morphisms
 * $$m\colon G \times_k G \to G, \; i\colon G \to G$$

over k which satisfy the usual axioms for the multiplication and inverse maps in a group (associativity, identity, inverses). A linear algebraic group is also smooth and of finite type over k, and it is affine (as a scheme). Conversely, every affine group scheme G of finite type over a field k has a faithful representation into GL(n) over k for some n. An example is the embedding of the additive group Ga into GL(2), as mentioned above. As a result, one can think of linear algebraic groups either as matrix groups or, more abstractly, as smooth affine group schemes over a field. (Some authors use "linear algebraic group" to mean any affine group scheme of finite type over a field.)

For a full understanding of linear algebraic groups, one has to consider more general (non-smooth) group schemes. For example, let k be an algebraically closed field of characteristic p > 0. Then the homomorphism f: Gm → Gm defined by x ↦ xp induces an isomorphism of abstract groups k* → k*, but f is not an isomorphism of algebraic groups (because x1/p is not a regular function). In the language of group schemes, there is a clearer reason why f is not an isomorphism: f is surjective, but it has nontrivial kernel, namely the group scheme μp of pth roots of unity. This issue does not arise in characteristic zero. Indeed, every group scheme of finite type over a field k of characteristic zero is smooth over k. A group scheme of finite type over any field k is smooth over k if and only if it is geometrically reduced, meaning that the base change $$G_{\overline k}$$ is reduced, where $$\overline k$$ is an algebraic closure of k.

Since an affine scheme X is determined by its ring O(X) of regular functions, an affine group scheme G over a field k is determined by the ring O(G) with its structure of a Hopf algebra (coming from the multiplication and inverse maps on G). This gives an equivalence of categories (reversing arrows) between affine group schemes over k and commutative Hopf algebras over k. For example, the Hopf algebra corresponding to the multiplicative group Gm = GL(1) is the Laurent polynomial ring k[x, x−1], with comultiplication given by
 * $$x \mapsto x \otimes x.$$

Basic notions
For a linear algebraic group G over a field k, the identity component Go (the connected component containing the point 1) is a normal subgroup of finite index. So there is a group extension
 * $$1 \to G^\circ \to G \to F \to 1, $$

where F is a finite algebraic group. (For k algebraically closed, F can be identified with an abstract finite group.) Because of this, the study of algebraic groups mostly focuses on connected groups.

Various notions from abstract group theory can be extended to linear algebraic groups. It is straightforward to define what it means for a linear algebraic group to be commutative, nilpotent, or solvable, by analogy with the definitions in abstract group theory. For example, a linear algebraic group is solvable if it has a composition series of linear algebraic subgroups such that the quotient groups are commutative. Also, the normalizer, the center, and the centralizer of a closed subgroup H of a linear algebraic group G are naturally viewed as closed subgroup schemes of G. If they are smooth over k, then they are linear algebraic groups as defined above.

One may ask to what extent the properties of a connected linear algebraic group G over a field k are determined by the abstract group G(k). A useful result in this direction is that if the field k is perfect (for example, of characteristic zero), or if G is reductive (as defined below), then G is unirational over k. Therefore, if in addition k is infinite, the group G(k) is Zariski dense in G. For example, under the assumptions mentioned, G is commutative, nilpotent, or solvable if and only if G(k) has the corresponding property.

The assumption of connectedness cannot be omitted in these results. For example, let G be the group μ3 ⊂ GL(1) of cube roots of unity over the rational numbers Q. Then G is a linear algebraic group over Q for which G(Q) = 1 is not Zariski dense in G, because $$G(\overline {\mathbf Q})$$ is a group of order 3.

Over an algebraically closed field, there is a stronger result about algebraic groups as algebraic varieties: every connected linear algebraic group over an algebraically closed field is a rational variety.

The Lie algebra of an algebraic group
The Lie algebra $$\mathfrak g$$ of an algebraic group G can be defined in several equivalent ways: as the tangent space T1(G) at the identity element 1 ∈ G(k), or as the space of left-invariant derivations. If k is algebraically closed, a derivation D: O(G) → O(G) over k of the coordinate ring of G is left-invariant if
 * $$D \lambda_x = \lambda_x D$$

for every x in G(k), where λx: O(G) → O(G) is induced by left multiplication by x. For an arbitrary field k, left invariance of a derivation is defined as an analogous equality of two linear maps O(G) → O(G) ⊗O(G). The Lie bracket of two derivations is defined by [D1, D2] =D1D2 − D2D1.

The passage from G to $$\mathfrak g$$ is thus a process of differentiation. For an element x ∈ G(k), the derivative at 1 ∈ G(k) of the conjugation map G → G, g ↦ xgx−1, is an automorphism of $$\mathfrak g$$, giving the adjoint representation:
 * $$\operatorname{Ad}\colon G \to \operatorname{Aut}(\mathfrak g).$$

Over a field of characteristic zero, a connected subgroup H of a linear algebraic group G is uniquely determined by its Lie algebra $$\mathfrak h \subset \mathfrak g$$. But not every Lie subalgebra of $$\mathfrak g$$ corresponds to an algebraic subgroup of G, as one sees in the example of the torus G = (Gm)2 over C. In positive characteristic, there can be many different connected subgroups of a group G with the same Lie algebra (again, the torus G = (Gm)2 provides examples). For these reasons, although the Lie algebra of an algebraic group is important, the structure theory of algebraic groups requires more global tools.

Semisimple and unipotent elements
For an algebraically closed field k, a matrix g in GL(n,k) is called semisimple if it is diagonalizable, and unipotent if the matrix g − 1 is nilpotent. Equivalently, g is unipotent if all eigenvalues of g are equal to 1. The Jordan canonical form for matrices implies that every element g of GL(n,k) can be written uniquely as a product g = gssgu such that gss is semisimple, gu is unipotent, and gss and gu commute with each other.

For any field k, an element g of GL(n,k) is said to be semisimple if it becomes diagonalizable over the algebraic closure of k. If the field k is perfect, then the semisimple and unipotent parts of g also lie in GL(n,k). Finally, for any linear algebraic group G ⊂ GL(n) over a field k, define a k-point of G to be semisimple or unipotent if it is semisimple or unipotent in GL(n,k). (These properties are in fact independent of the choice of a faithful representation of G.) If the field k is perfect, then the semisimple and unipotent parts of a k-point of G are automatically in G. That is (the Jordan decomposition): every element g of G(k) can be written uniquely as a product g = gssgu in G(k) such that gss is semisimple, gu is unipotent, and gss and gu commute with each other. This reduces the problem of describing the conjugacy classes in G(k) to the semisimple and unipotent cases.

Tori
A torus over an algebraically closed field k means a group isomorphic to (Gm)n, the product of n copies of the multiplicative group over k, for some natural number n. For a linear algebraic group G, a maximal torus in G means a torus in G that is not contained in any bigger torus. For example, the group of diagonal matrices in GL(n) over k is a maximal torus in GL(n), isomorphic to (Gm)n. A basic result of the theory is that any two maximal tori in a group G over an algebraically closed field k are conjugate by some element of G(k). The rank of G means the dimension of any maximal torus.

For an arbitrary field k, a torus T over k means a linear algebraic group over k whose base change $$T_{\overline k}$$ to the algebraic closure of k is isomorphic to (Gm)n over $$\overline k$$, for some natural number n. A split torus over k means a group isomorphic to (Gm)n over k for some n. An example of a non-split torus over the real numbers R is
 * $$T=\{(x,y)\in A^2_{\mathbf{R}}: x^2+y^2=1\},$$

with group structure given by the formula for multiplying complex numbers x+iy. Here T is a torus of dimension 1 over R. It is not split, because its group of real points T(R) is the circle group, which is not isomorphic even as an abstract group to Gm(R) = R*.

Every point of a torus over a field k is semisimple. Conversely, if G is a connected linear algebraic group such that every element of $$G(\overline k)$$ is semisimple, then G is a torus.

For a linear algebraic group G over a general field k, one cannot expect all maximal tori in G over k to be conjugate by elements of G(k). For example, both the multiplicative group Gm and the circle group T above occur as maximal tori in SL(2) over R. However, it is always true that any two maximal split tori in G over k (meaning split tori in G that are not contained in a bigger split torus) are conjugate by some element of G(k). As a result, it makes sense to define the  k-rank or split rank of a group G over k as the dimension of any maximal split torus in G over k.

For any maximal torus T in a linear algebraic group G over a field k, Grothendieck showed that $$T_{\overline k}$$ is a maximal torus in $$G_{\overline k}$$. It follows that any two maximal tori in G over a field k have the same dimension, although they need not be isomorphic.

Unipotent groups
Let Un be the group of upper-triangular matrices in GL(n) with diagonal entries equal to 1, over a field k. A group scheme over a field k (for example, a linear algebraic group) is called unipotent if it is isomorphic to a closed subgroup scheme of Un for some n. It is straightforward to check that the group Un is nilpotent. As a result, every unipotent group scheme is nilpotent.

A linear algebraic group G over a field k is unipotent if and only if every element of $$G(\overline{k})$$ is unipotent.

The group Bn of upper-triangular matrices in GL(n) is a semidirect product
 * $$B_n = T_n \ltimes U_n,$$

where Tn is the diagonal torus (Gm)n. More generally, every connected solvable linear algebraic group is a semidirect product of a torus with a unipotent group, T ⋉ U.

A smooth connected unipotent group over a perfect field k (for example, an algebraically closed field) has a composition series with all quotient groups isomorphic to the additive group Ga.

Borel subgroups
The Borel subgroups are important for the structure theory of linear algebraic groups. For a linear algebraic group G over an algebraically closed field k, a Borel subgroup of G means a maximal smooth connected solvable subgroup. For example, one Borel subgroup of GL(n) is the subgroup B of upper-triangular matrices (all entries below the diagonal are zero).

A basic result of the theory is that any two Borel subgroups of a connected group G over an algebraically closed field k are conjugate by some element of G(k). (A standard proof uses the Borel fixed-point theorem: for a connected solvable group G acting on a proper variety X over an algebraically closed field k, there is a k-point in X which is fixed by the action of G.) The conjugacy of Borel subgroups in GL(n) amounts to the Lie–Kolchin theorem: every smooth connected solvable subgroup of GL(n) is conjugate to a subgroup of the upper-triangular subgroup in GL(n).

For an arbitrary field k, a Borel subgroup B of G is defined to be a subgroup over k such that, over an algebraic closure $$\overline k$$ of k, $$B_{\overline k}$$is a Borel subgroup of $$G_{\overline k}$$. Thus G may or may not have a Borel subgroup over k.

For a closed subgroup scheme H of G, the quotient space G/H is a smooth quasi-projective scheme over k. A smooth subgroup P of a connected group G is called parabolic if G/P is projective over k (or equivalently, proper over k). An important property of Borel subgroups B is that G/B is a projective variety, called the flag variety of G. That is, Borel subgroups are parabolic subgroups. More precisely, for k algebraically closed, the Borel subgroups are exactly the minimal parabolic subgroups of G; conversely, every subgroup containing a Borel subgroup is parabolic. So one can list all parabolic subgroups of G (up to conjugation by G(k)) by listing all the linear algebraic subgroups of G that contain a fixed Borel subgroup. For example, the subgroups P ⊂ GL(3) over k that contain the Borel subgroup B of upper-triangular matrices are B itself, the whole group GL(3), and the intermediate subgroups
 * $$\left \{ \begin{bmatrix}

* & * & * \\ 0 & * & * \\ 0 & * & * \end{bmatrix} \right \}$$ and $$\left \{ \begin{bmatrix} * & * & * \\ * & * & * \\ 0 & 0 & * \end{bmatrix} \right \}.$$ The corresponding projective homogeneous varieties GL(3)/P are (respectively): the flag manifold of all chains of linear subspaces
 * $$0\subset V_1\subset V_2\subset A^3_k$$

with Vi of dimension i; a point; the projective space P2 of lines (1-dimensional linear subspaces) in A3; and the dual projective space P2 of planes in A3.

Semisimple and reductive groups
A connected linear algebraic group G over an algebraically closed field is called semisimple if every smooth connected solvable normal subgroup of G is trivial. More generally, a connected linear algebraic group G over an algebraically closed field is called reductive if every smooth connected unipotent normal subgroup of G is trivial. (Some authors do not require reductive groups to be connected.) A semisimple group is reductive. A group G over an arbitrary field k is called semisimple or reductive if $$G_{\overline k}$$ is semisimple or reductive. For example, the group SL(n) of n × n matrices with determinant 1 over any field k is semisimple, whereas a nontrivial torus is reductive but not semisimple. Likewise, GL(n) is reductive but not semisimple (because its center Gm is a nontrivial smooth connected solvable normal subgroup).

Every compact connected Lie group has a complexification, which is a complex reductive algebraic group. In fact, this construction gives a one-to-one correspondence between compact connected Lie groups and complex reductive groups, up to isomorphism.

A linear algebraic group G over a field k is called simple (or k-simple) if it is semisimple, nontrivial, and every smooth connected normal subgroup of G over k is trivial or equal to G. (Some authors call this property "almost simple".) This differs slightly from the terminology for abstract groups, in that a simple algebraic group may have nontrivial center (although the center must be finite). For example, for any integer n at least 2 and any field k, the group SL(n) over k is simple, and its center is the group scheme μn of nth roots of unity.

Every connected linear algebraic group G over a perfect field k is (in a unique way) an extension of a reductive group R by a smooth connected unipotent group U, called the unipotent radical of G:
 * $$1\to U\to G\to R\to 1.$$

If k has characteristic zero, then one has the more precise Levi decomposition: every connected linear algebraic group G over k is a semidirect product $$R\ltimes U$$ of a reductive group by a unipotent group.

Classification of reductive groups
Reductive groups include the most important linear algebraic groups in practice, such as the classical groups: GL(n), SL(n), the orthogonal groups SO(n) and the symplectic groups Sp(2n). On the other hand, the definition of reductive groups is quite "negative", and it is not clear that one can expect to say much about them. Remarkably, Claude Chevalley gave a complete classification of the reductive groups over an algebraically closed field: they are determined by root data. In particular, simple groups over an algebraically closed field k are classified (up to quotients by finite central subgroup schemes) by their Dynkin diagrams. It is striking that this classification is independent of the characteristic of k. For example, the exceptional Lie groups G2, F4, E6, E7, and E8 can be defined in any characteristic (and even as group schemes over Z). The classification of finite simple groups says that most finite simple groups arise as the group of k-points of a simple algebraic group over a finite field k, or as minor variants of that construction.

Every reductive group over a field is the quotient by a finite central subgroup scheme of the product of a torus and some simple groups. For example,
 * $$GL(n)\cong (G_m\times SL(n))/\mu_n.$$

For an arbitrary field k, a reductive group G is called split if it contains a split maximal torus over k (that is, a split torus in G which remains maximal over an algebraic closure of k). For example, GL(n) is a split reductive group over any field k. Chevalley showed that the classification of split reductive groups is the same over any field. By contrast, the classification of arbitrary reductive groups can be hard, depending on the base field. For example, every nondegenerate quadratic form q over a field k determines a reductive group SO(q), and every central simple algebra A over k determines a reductive group SL1(A). As a result, the problem of classifying reductive groups over k essentially includes the problem of classifying all quadratic forms over k or all central simple algebras over k. These problems are easy for k algebraically closed, and they are understood for some other fields such as number fields, but for arbitrary fields there are many open questions.

Representation theory
One reason for the importance of reductive groups comes from representation theory. Every irreducible representation of a unipotent group is trivial. More generally, for any linear algebraic group G written as an extension
 * $$1\to U\to G\to R\to 1$$

with U unipotent and R reductive, every irreducible representation of G factors through R. This focuses attention on the representation theory of reductive groups. (To be clear, the representations considered here are representations of G as an algebraic group. Thus, for a group G over a field k, the representations are on k-vector spaces, and the action of G is given by regular functions. It is an important but different problem to classify continuous representations of the group G(R) for a real reductive group G, or similar problems over other fields.)

Chevalley showed that the irreducible representations of a split reductive group over a field k are finite-dimensional, and they are indexed by dominant weights. This is the same as what happens in the representation theory of compact connected Lie groups, or the finite-dimensional representation theory of complex semisimple Lie algebras. For k of characteristic zero, all these theories are essentially equivalent. In particular, every representation of a reductive group G over a field of characteristic zero is a direct sum of irreducible representations, and if G is split, the characters of the irreducible representations are given by the Weyl character formula. The Borel–Weil theorem gives a geometric construction of the irreducible representations of a reductive group G in characteristic zero, as spaces of sections of line bundles over the flag manifold G/B.

The representation theory of reductive groups (other than tori) over a field of positive characteristic p is less well understood. In this situation, a representation need not be a direct sum of irreducible representations. And although irreducible representations are indexed by dominant weights, the dimensions and characters of the irreducible representations are known only in some cases. determined these characters (proving Lusztig's conjecture) when the characteristic p is sufficiently large compared to the Coxeter number of the group. For small primes p, there is not even a precise conjecture.

Group actions and geometric invariant theory
An action of a linear algebraic group G on a variety (or scheme) X over a field k is a morphism
 * $$G \times_k X \to X$$

that satisfies the axioms of a group action. As in other types of group theory, it is important to study group actions, since groups arise naturally as symmetries of geometric objects.

Part of the theory of group actions is geometric invariant theory, which aims to construct a quotient variety X/G, describing the set of orbits of a linear algebraic group G on X as an algebraic variety. Various complications arise. For example, if X is an affine variety, then one can try to construct X/G as Spec of the ring of invariants O(X)G. However, Masayoshi Nagata showed that the ring of invariants need not be finitely generated as a k-algebra (and so Spec of the ring is a scheme but not a variety), a negative answer to Hilbert's 14th problem. In the positive direction, the ring of invariants is finitely generated if G is reductive, by Haboush's theorem, proved in characteristic zero by Hilbert and Nagata.

Geometric invariant theory involves further subtleties when a reductive group G acts on a projective variety X. In particular, the theory defines open subsets of "stable" and "semistable" points in X, with the quotient morphism only defined on the set of semistable points.

Related notions
Linear algebraic groups admit variants in several directions. Dropping the existence of the inverse map $$i\colon G \to G$$, one obtains the notion of a linear algebraic monoid.

Lie groups
For a linear algebraic group G over the real numbers R, the group of real points G(R) is a Lie group, essentially because real polynomials, which describe the multiplication on G, are smooth functions. Likewise, for a linear algebraic group G over C, G(C) is a complex Lie group. Much of the theory of algebraic groups was developed by analogy with Lie groups.

There are several reasons why a Lie group may not have the structure of a linear algebraic group over R.

 * A Lie group with an infinite group of components G/Go cannot be realized as a linear algebraic group.
 * An algebraic group G over R may be connected as an algebraic group while the Lie group G(R) is not connected, and likewise for simply connected groups. For example, the algebraic group SL(2) is simply connected over any field, whereas the Lie group SL(2,R) has fundamental group isomorphic to the integers Z. The double cover H of SL(2,R), known as the metaplectic group, is a Lie group that cannot be viewed as a linear algebraic group over R. More strongly, H has no faithful finite-dimensional representation.
 * Anatoly Maltsev showed that every simply connected nilpotent Lie group can be viewed as a unipotent algebraic group G over R in a unique way. (As a variety, G is isomorphic to affine space of some dimension over R.) By contrast, there are simply connected solvable Lie groups that cannot be viewed as real algebraic groups. For example, the universal cover H of the semidirect product S1 ⋉ R2 has center isomorphic to Z, which is not a linear algebraic group, and so H cannot be viewed as a linear algebraic group over R.

Abelian varieties
Algebraic groups which are not affine behave very differently. In particular, a smooth connected group scheme which is a projective variety over a field is called an abelian variety. In contrast to linear algebraic groups, every abelian variety is commutative. Nonetheless, abelian varieties have a rich theory. Even the case of elliptic curves (abelian varieties of dimension 1) is central to number theory, with applications including the proof of Fermat's Last Theorem.

Tannakian categories
The finite-dimensional representations of an algebraic group G, together with the tensor product of representations, form a tannakian category RepG. In fact, tannakian categories with a "fiber functor" over a field are equivalent to affine group schemes. (Every affine group scheme over a field k is pro-algebraic in the sense that it is an inverse limit of affine group schemes of finite type over k. ) For example, the Mumford–Tate group and the motivic Galois group are constructed using this formalism. Certain properties of a (pro-)algebraic group G can be read from its category of representations. For example, over a field of characteristic zero, RepG is a semisimple category if and only if the identity component of G is pro-reductive.

title: Linear algebraic group


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Talk:Newberry Caldera

Geothermal power
 The Geysers of California, the world's largest producer of geothermal power. More output than Iceland's? This is worth an entry! Wetman 09:32, 10 Dec 2003 (UTC)

Proposed merge
Well, this seems like such an obvious merge that I was about to be bold and do it right now. The articles share both their images, along with much text.

However, these two articles have been separate since the day they were created on 10 December 2003, so it seems worthwhile to be cautious and ask first. I think "Newberry Caldera" (which is an unoffical name anyway) and "Newberry Crater" (the official name for the caldera) should both just be redirects to a single article at Newberry Volcano, which is the official name now for the entire shield volcano including the caldera and everything else. Please see http://geonames.usgs.gov, and search for "Newberry" in Oregon, or try this direct link if it works:

If I don't hear objections, I'll merge within a couple of days. A single article at Newberry Volcano could easily be built into a GA and eventually an FA. Thanks. --Seattle Skier (talk) 08:08, 7 May 2007 (UTC)

 * I fully support the merger. It seems like a "no brainer." --Burntnickel 11:17, 7 May 2007 (UTC)

 * Support the merge, although you may wish to ping User:Maveric149, who wrote the original articles and is still an active editor --- he may remember why he started the two separate articles. Thanks! hike395 05:17, 8 May 2007 (UTC)

 * I thought about doing that when I added the merge tags, and I'm not sure why I didn't. I have now notified him.  Thanks for reminding me to do so.  --Seattle Skier (talk) 09:06, 8 May 2007 (UTC)

 * Truth be told; I created separate articles mostly due to the 4KB per record limit on my PDA when I drafted them on a field study of the Lassen area. My new PDA has a 32 KB per record limit and oddly enough, most of my articles have ended of that long before I upload them. :) So, if a merge can be done in an elegant way, then please, go ahead. --mav 13:16, 16 May 2007 (UTC)

 * Shotgun merge complete. Clean-up help welcome. :) --mav

title: Talk:Newberry Caldera