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

  • $\begingroup$ 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? $\endgroup$ – Peter Crooks Oct 15 '13 at 4:46
  • $\begingroup$ You can find here digital.csic.es/bitstream/10261/15773/1/RockyCFG.PDF . $\endgroup$ – user21574 Oct 15 '13 at 9:02
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    $\begingroup$ I think the paracomplexification is $G \times G$. $\endgroup$ – Ben McKay Oct 15 '13 at 14:07
  • $\begingroup$ There are several undefined steps here, since you haven't defined what a paracomplex manifold is, or what the analogue of holomorphic map should be. However, in the linear category, it is clear that the paracomplexification of the Lie algebra is given by taking the direct sum of two copies. When your Lie group is compact (e.g., for your example of $U(n)$), you can use Chevalley's theorem to transport the question into the setting of anisotropic reductive groups: see mathoverflow.net/questions/6079/… $\endgroup$ – S. Carnahan Oct 15 '13 at 15:21
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    $\begingroup$ 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. $\endgroup$ – Paul Reynolds Oct 15 '13 at 20:11

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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    $\begingroup$ 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? $\endgroup$ – YCor Oct 18 '13 at 18:13

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