complex Lie algebra

In mathematics, a complex Lie algebra is a Lie algebra over the complex numbers.

Given a complex Lie algebra $\mathfrak{g}$ , its conjugate $\overline{\mathfrak g}$ is a complex Lie algebra with the same underlying real vector space but with $i = \sqrt{-1}$ acting as $-i$ instead.{{harvnb|Knapp|2002|loc=Ch. VI, § 9.}} As a real Lie algebra, a complex Lie algebra $\mathfrak{g}$ 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).

Real form

Given a complex Lie algebra $\mathfrak{g}$ , a real Lie algebra $\mathfrak{g}_0$ is said to be a real form of $\mathfrak{g}$ if the complexification $\mathfrak{g}_0 \otimes_{\mathbb{R}}\mathbb{C}$ is isomorphic to $\mathfrak{g}$ .

A real form $\mathfrak{g}_0$ is abelian (resp. nilpotent, solvable, semisimple) if and only if $\mathfrak{g}$ is abelian (resp. nilpotent, solvable, semisimple).{{harvnb|Serre|2001|loc=Ch. II, § 8, Theorem 9.}} On the other hand, a real form $\mathfrak{g}_0$ is simple if and only if either $\mathfrak{g}$ is simple or $\mathfrak{g}$ is of the form $\mathfrak{s} \times \overline{\mathfrak{s}}$ where $\mathfrak{s}, \overline{\mathfrak{s}}$ are simple and are the conjugates of each other.

The existence of a real form in a complex Lie algebra $\mathfrak g$ implies that $\mathfrak g$ is isomorphic to its conjugate; indeed, if $\mathfrak{g} = \mathfrak{g}_0 \otimes_{\mathbb{R}} \mathbb{C} = \mathfrak{g}_0 \oplus i\mathfrak{g}_0$ , then let $\tau : \mathfrak{g} \to \overline{\mathfrak{g}}$ denote the $\mathbb{R}$ -linear isomorphism induced by complex conjugate and then

: $\tau(i(x + iy)) = \tau(ix - y) = -ix- y = -i\tau(x + iy)$ ,

which is to say $\tau$ is in fact a $\mathbb{C}$ -linear isomorphism.

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 $\mathbb{C}$ -linear isomorphism $\tau: \mathfrak{g} \overset{\sim}\to \overline{\mathfrak{g}}$ ; without loss of generality, we can assume it is the identity function on the underlying real vector space. Then define $\mathfrak{g}_0 = \{ z \in \mathfrak{g} | \tau(z) = z \}$ , which is clearly a real Lie algebra. Each element $z$ in $\mathfrak{g}$ can be written uniquely as $z = 2^{-1}(z + \tau(z)) + i 2^{-1}(i\tau(z) - iz)$ . Here, $\tau(i\tau(z) - iz) = -iz + i\tau(z)$ and similarly $\tau$ fixes $z + \tau(z)$ . Hence, $\mathfrak{g} = \mathfrak{g}_0 \oplus i \mathfrak{g}_0$ ; i.e., $\mathfrak{g}_0$ is a real form.

Complex Lie algebra of a complex Lie group

Let $\mathfrak{g}$ be a semisimple complex Lie algebra that is the Lie algebra of a complex Lie group $G$ . Let $\mathfrak{h}$ be a Cartan subalgebra of $\mathfrak{g}$ and $H$ the Lie subgroup corresponding to $\mathfrak{h}$ ; the conjugates of $H$ are called Cartan subgroups.

Suppose there is the decomposition $\mathfrak{g} = \mathfrak{n}^- \oplus \mathfrak{h} \oplus \mathfrak{n}^+$ given by a choice of positive roots. Then the exponential map defines an isomorphism from $\mathfrak{n}^+$ to a closed subgroup $U \subset G$ .{{harvnb|Serre|2001|loc=Ch. VIII, § 4, Theorem 6 (a).}} The Lie subgroup $B \subset G$ corresponding to the Borel subalgebra $\mathfrak{b} = \mathfrak{h} \oplus \mathfrak{n}^+$ is closed and is the semidirect product of $H$ and $U$ ;{{harvnb|Serre|2001|loc=Ch. VIII, § 4, Theorem 6 (b).}} the conjugates of $B$ are called Borel subgroups.

Notes

References

{{Fulton-Harris}}
{{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}}.
{{cite book |first=Jean-Pierre |last=Serre |title=Complex Semisimple Lie Algebras |publisher=Springer |location=Berlin |date=2001 |isbn=3-5406-7827-1}}

Category:Lie algebras