Is $\text{SL}(2,\mathbb{C})$ a Kähler manifold? Let $\text{SL}(2,\mathbb{C})$ be the special linear group of $2 \times 2$ complex matrices with determinant $1$. We know it's a complex lie group. In particular, it's a (non-compact) complex 3-fold. 
My question is:
Can it be endowed with the structure of a Kähler manifold?
 A: A manifold $X$ of dimension $2n$ admits an almost Kähler structure if its principal $GL(2n, \mathbb{R})$-bundle, given by frames in $TM$, can be reduced to $SO(n,\mathbb{R}) \cap GL(n,\mathbb{C}) \cap Sp(n,\mathbb{R})$ (This amounts to find compatible Riemannian, almost-complex and simplectic structures), since $M=G$ is a lie group, $TG = G \times \mathfrak{g}$ its principal bundle is trivial and the reduction just amounts to the inclusion of groups and those are subgroups, we get that there is always an almost käler structure. If we start with an integrable almost-complex structure, then we have a proper Kähler structure.
In plain english: fix any scalar product $g$ in the Lie algebra of $G$ that is compatible with the almost complex structure $J$ ($J$ is an isometry) given by the multiplication by $i$ in $T_eG = \mathfrak g$ and then move it around $G$ via left-translations to get Riemannian metric. Finally, define the symplectic structure by $\omega(x,y) = g(x,Jy)$, we get a kähler structure.
