# 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. 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). 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, 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 $z=2^{-1}(z+\tau (z))+i2^{-1}(i\tau (z)-iz)$ . 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 $U\subset G$ . 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 $U$ ; the conjugates of $B$  are called Borel subgroups.