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 $$z = 2^{-1}(z + \tau(z)) + i 2^{-1}(i\tau(z) - iz)$$. 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 $$U \subset G$$. 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.