# 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.