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In mathematics, a '''complex Lie algebra''' is a [[Lie algebra]] over the complex |
In[[mathematics]], a '''complex Lie algebra''' is a [[Lie algebra]] over the [[complex number]]s. |
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Given a complex Lie algebra <math>\mathfrak{g}</math>, its '''conjugate''' <math>\overline{\mathfrak g}</math> is a complex Lie algebra with the same underlying [[real number|real]] [[vector space]] but with <math>i = \sqrt{-1}</math> acting as <math>-i</math> instead.<ref name=Knapp>{{harvnb|Knapp|2002|loc=Ch. VI, § 9.}}</ref> As a real Lie algebra, a complex Lie algebra <math>\mathfrak{g}</math> is trivially [[isomorphic]] to its conjugate. A complex Lie algebra is isomorphic to its conjugate [[if and only if]] it admits a real form (and is said to be defined over the real numbers). |
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== Real form == |
== Real form == |
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Given a complex Lie algebra <math>\mathfrak{g}</math>, a real Lie algebra <math>\mathfrak{g}_0</math> is said to be a [[real form]] of <math>\mathfrak{g}</math> if the [[complexification]] <math>\mathfrak{g}_0 \otimes_{\mathbb{R}}\mathbb{C}</math> is isomorphic to <math>\mathfrak{g}</math>. |
Given a complex Lie algebra <math>\mathfrak{g}</math>, a real Lie algebra <math>\mathfrak{g}_0</math> is said to be a [[real form]] of <math>\mathfrak{g}</math> if the [[complexification]] <math>\mathfrak{g}_0 \otimes_{\mathbb{R}}\mathbb{C}</math> is isomorphic to <math>\mathfrak{g}</math>. |
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A real form <math>\mathfrak{g}_0</math> is abelian (resp. nilpotent, solvable, semisimple) if and only if <math>\mathfrak{g}</math> is abelian (resp. nilpotent, solvable, semisimple).<ref name="Theorem 9">{{harvnb|Serre|loc=Ch. II, § 8, Theorem 9.}}</ref> On the other hand, a real form <math>\mathfrak{g}_0</math> is simple if and only if either <math>\mathfrak{g}</math> is simple or <math>\mathfrak{g}</math> is of the form <math>\mathfrak{s} \times \overline{\mathfrak{s}}</math> where <math>\mathfrak{s}, \overline{\mathfrak{s}}</math> are simple and are |
A real form <math>\mathfrak{g}_0</math> is abelian (resp. nilpotent, solvable, semisimple) if and only if <math>\mathfrak{g}</math> is abelian (resp. nilpotent, solvable, semisimple).<ref name="Theorem 9">{{harvnb|Serre|2001|loc=Ch. II, § 8, Theorem 9.}}</ref> On the other hand, a real form <math>\mathfrak{g}_0</math>is[[simple Lie algebra|simple]] if and only if either <math>\mathfrak{g}</math> is simple or <math>\mathfrak{g}</math> is of the form <math>\mathfrak{s} \times \overline{\mathfrak{s}}</math> where <math>\mathfrak{s}, \overline{\mathfrak{s}}</math> are simple and are the conjugates of each other.<ref name="Theorem 9" /> |
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The existence of a real form in a complex Lie algebra <math>\mathfrak g</math> implies that <math>\mathfrak g</math> is isomorphic to its conjugate;<ref name=Knapp /> indeed, if <math>\mathfrak{g} = \mathfrak{g}_0 \otimes_{\mathbb{R}} \mathbb{C} = \mathfrak{g}_0 \oplus i\mathfrak{g}_0</math>, then let <math>\tau : \mathfrak{g} \to \overline{\mathfrak{g}}</math> denote the <math>\mathbb{R}</math>-linear isomorphism induced by complex conjugate and then |
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:<math>\tau(i(x + iy)) = \tau(ix - y) = -ix- y = -i\tau(x + iy)</math>, |
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which is to say <math>\tau</math> is in fact a <math>\mathbb{C}</math>-linear isomorphism. |
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Conversely,{{clarify|There is a suggestion that the converse direction has a problem. See [https://math.stackexchange.com/questions/4873533/complex-lie-algebra-isomorphic-to-its-conjugate-has-a-real-form]|date=July 2024}} suppose there is a <math>\mathbb{C}</math>-linear isomorphism <math>\tau: \mathfrak{g} \overset{\sim}\to \overline{\mathfrak{g}}</math>; without loss of generality, we can assume it is the identity function on the underlying real vector space. Then define <math>\mathfrak{g}_0 = \{ z \in \mathfrak{g} | \tau(z) = z \}</math>, which is clearly a real Lie algebra. Each element <math>z</math> in <math>\mathfrak{g}</math> can be written uniquely as <math>z = 2^{-1}(z + \tau(z)) + i 2^{-1}(i\tau(z) - iz)</math>. Here, <math>\tau(i\tau(z) - iz) = -iz + i\tau(z)</math> and similarly <math>\tau</math> fixes <math>z + \tau(z)</math>. Hence, <math>\mathfrak{g} = \mathfrak{g}_0 \oplus i \mathfrak{g}_0</math>; i.e., <math>\mathfrak{g}_0</math> is a real form. |
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== Complex Lie algebra of a complex Lie group == |
== Complex Lie algebra of a complex Lie group == |
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Let <math>\mathfrak{g}</math> be a complex Lie algebra that is the Lie algebra of a [[complex Lie group]] <math>G</math>. Let <math>\mathfrak{h}</math> a [[Cartan subalgebra]] of <math>\mathfrak{g}</math> and <math>H</math> the Lie |
Let <math>\mathfrak{g}</math> be a semisimple complex Lie algebra that is the Lie algebra of a [[complex Lie group]] <math>G</math>. Let <math>\mathfrak{h}</math>be a [[Cartan subalgebra]] of <math>\mathfrak{g}</math> and <math>H</math> the Lie subgroup corresponding to <math>\mathfrak{h}</math>; the conjugates of <math>H</math> are called [[Cartan subgroup]]s. |
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Suppose |
Suppose there is the decomposition <math>\mathfrak{g} = \mathfrak{n}^- \oplus \mathfrak{h} \oplus \mathfrak{n}^+</math> given by a choice of positive roots. Then the [[exponential map (Lie theory)|exponential map]] defines an isomorphism from <math>\mathfrak{n}^+</math> to a closed subgroup <math>U\subset G</math>.<ref>{{harvnb|Serre|2001|loc=Ch. VIII, § 4, Theorem 6 (a).}}</ref> The Lie subgroup <math>B \subset G</math> corresponding to the [[Borel subalgebra]] <math>\mathfrak{b} = \mathfrak{h} \oplus \mathfrak{n}^+</math> is closed and is the semidirect product of <math>H</math> and <math>U</math>;<ref>{{harvnb|Serre|2001|loc=Ch. VIII, § 4, Theorem 6 (b).}}</ref> the conjugates of <math>B</math> are called [[Borel subgroup]]s. |
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== |
== Notes == |
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{{reflist}} |
{{reflist}} |
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== References == |
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* {{Fulton-Harris}} |
* {{Fulton-Harris}} |
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* {{cite book|author-link=A. W. Knapp|last=Knapp|first=A. W.|title=Lie groups beyond an introduction|isbn=0-8176-4259-5|publisher=Birkhäuser|series=Progress in Mathematics|volume=120|edition=2nd|year=2002|location=Boston·Basel·Berlin}}. |
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* Jean-Pierre Serre |
* {{cite book |first=Jean-Pierre |last=Serre |title=Complex Semisimple Lie Algebras |publisher=Springer |location=Berlin |date=2001 |isbn=3-5406-7827-1}} |
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{{algebra-stub}} |
{{algebra-stub}} |
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[[Category:Lie algebras]] |
Inmathematics, a complex Lie algebra is a Lie algebra over the complex numbers.
Given a complex Lie algebra , its conjugate
is a complex Lie algebra with the same underlying real vector space but with
acting as
instead.[1] As a real Lie algebra, a complex Lie algebra
is trivially isomorphic to its conjugate. A complex Lie algebra is isomorphic to its conjugate if and only if it admits a real form (and is said to be defined over the real numbers).
Given a complex Lie algebra , a real Lie algebra
is said to be a real formof
if the complexification
is isomorphic to
.
A real form is abelian (resp. nilpotent, solvable, semisimple) if and only if
is abelian (resp. nilpotent, solvable, semisimple).[2] On the other hand, a real form
issimple if and only if either
is simple or
is of the form
where
are simple and are the conjugates of each other.[2]
The existence of a real form in a complex Lie algebra implies that
is isomorphic to its conjugate;[1] indeed, if
, then let
denote the
-linear isomorphism induced by complex conjugate and then
which is to say is in fact a
-linear isomorphism.
Conversely,[clarification needed] suppose there is a -linear isomorphism
; without loss of generality, we can assume it is the identity function on the underlying real vector space. Then define
, which is clearly a real Lie algebra. Each element
in
can be written uniquely as
. Here,
and similarly
fixes
. Hence,
; i.e.,
is a real form.
Let be a semisimple complex Lie algebra that is the Lie algebra of a complex Lie group
. Let
be a Cartan subalgebraof
and
the Lie subgroup corresponding to
; the conjugates of
are called Cartan subgroups.
Suppose there is the decomposition given by a choice of positive roots. Then the exponential map defines an isomorphism from
to a closed subgroup
.[3] The Lie subgroup
corresponding to the Borel subalgebra
is closed and is the semidirect product of
and
;[4] the conjugates of
are called Borel subgroups.
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