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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 $GL(n, \mathbb{C})$

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What would it mean for a Lie group $G$ to admit a para-complex structure? Is there a notion of integrability analogous to that required for an almost complex structure to be complex? –  Peter Crooks Oct 15 '13 at 4:46
You can find here digital.csic.es/bitstream/10261/15773/1/RockyCFG.PDF . –  Hassan Jolany Oct 15 '13 at 9:02
I think the paracomplexification is $G \times G$. –  Ben McKay Oct 15 '13 at 14:07
Thanks Ben, can you explain with details as an answer –  Hassan Jolany Oct 15 '13 at 14:19
An almost-paracomplex manifold is a smooth real even-dimensional manifold with a paracomplex structure (an endomorphism defined as above) defined on its tangent bundle, and morphisms are smooth maps preserving it. I can't remember if there is a notion of integrability. –  Paul Reynolds Oct 15 '13 at 20:11

1 Answer 1

up vote 1 down vote accepted

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 $\phi\left(g_1,g_2\right)=\left(\phi_1\left(g_1\right),\phi_2\left(g_2\right)\right)$.

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This is likely to show that the paracomplexification of a simply connected Lie group $G$ is the diagonal embedding of $G$ into $G\times G$. In general, if $G$ is a connected Lie group, say $G=\tilde{G}/Z$ with $\tilde{G}$ its universal covering, doesn't it follow that the complexification is the diagonal embedding of $G$ into $(\tilde{G}\times\tilde{G})/Z$, where $Z$ is diagonally embedded? –  YCor Oct 18 '13 at 18:13

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