Let $G$ be a real Lie group. Then the complexification $G\_\mathbb{C}$ of $G$ is the unique complex Lie group equipped with a map $φ:G\to G\_\mathbb{C}$ such that any map $G\to H$ where $H$ is a 
complex Lie group, extends to a holomorphic map $G\_\mathbb{C}\to H$. If $\mathfrak{g}$ and 
$\mathfrak{g}\_\mathbb{C}$ are the respective Lie algebras, $\mathfrak{g}\_\mathbb{C}≅\mathfrak{g}⊗\_R \mathbb{C}$.


The algebra of para-complex numbers is defined by
$C = \mathbb{R} + e\mathbb{R}$ , $e^2=1$. The Para-complex structure in a vector space$ V $ is:


$K : V \to V$; with $K^2 = 1$. such that
$V = V^+ + V^-$ ;$ dimV^+ = dimV^-$. 
Para-complexication of $(V; K)$ is $V^\mathbb{C}
:= V\otimes C$.


So, In a same method, We can define para-complexification of Lie Groups .


So, My questions are 1) can we say the para-complexification of a lie group is equal to its complexification ? . If not, the paracomplexification of a lie group is unique?. For instance what is the paracomplexification of $U(n)$,


Also we know that the complexification of $U(n)$ is $GL(n, \mathbb{C})$



Para-Complexification of Lie Groups
Include $\mathfrak{g} \to \mathfrak{g}[e]$ by $A \mapsto A-Ae$, and call the image $\mathfrak{g}^{(1,0)}$. 
Include $\mathfrak{g} \to \mathfrak{g}[e]$ by $A \mapsto A+Ae$, and call the image $\mathfrak{g}^{(0,1)}$. 
Linear algebra: we can write every element of $\mathfrak{g}[e]$ uniquely as a sum of a $(1,0)$ with a $(0,1)$, so that $\mathfrak{g}[e]=\mathfrak{g}\oplus\mathfrak{g}$. To be precise, $A+Be=P+Q$ where $P=(A-B)/2-(A-B)e/2$ and $Q=(A+B)/2+(A+B)e/2$.
We can also write the elements of the form $A+0e$ as $\mathfrak{g}[e]\_{\mathbb{R}}$, the real points.
If $G$ is a Lie group, then let $G[e]=G \times G$, so that we can say that $G[e]$ has Lie algebra canonically isomorphic to $\mathfrak{g}[e]$, so that the induced Lie algebra morphism $\mathfrak{g} \to \mathfrak{g}[e]$ is $A \mapsto A+0e$. A paracomplex Lie group is a Lie group with biinvariant splitting of its tangent bundle, $\mathfrak{h}=\mathfrak{h}\_1 \oplus \mathfrak{h}\_2$, and isomorphism of the two Lie algebras $\mathfrak{h}\_1=\mathfrak{h}\_2$, i.e. biinvariant isomorphism $\mathfrak{h}=\mathfrak{g}[e]$ for some Lie algebra $\mathfrak{g}$. If $\phi \colon G \to H$ is a morphism of Lie groups, with $H$ a paracomplex Lie group, then we define the associated Lie algebra morphism $\phi \colon \mathfrak{g} \to \mathfrak{h}$, and extend it uniquely to a Lie algebra morphism $\phi \colon \mathfrak{g}[e] \to \mathfrak{h}$ by $e$-linearity. I will have to think about the group morphisms. But it should be easy for morphisms $\phi \colon G \to H[e]$: you should extend to $\phi \colon G[e] \to H[e]$ by writing the original morphism as $\phi \colon G \to H \times H$, say $\phi=\left(\phi\_1,\phi\_2\right)$ and letting $\phi\left(g\_1,g\_2\right)=\left(\phi\_1\left(g\_1\right),\phi\_2\left(g\_2\right)\right)$.


