# Gauge groupoid of Lorentz group & complexification

I'm learning about Lie groupoids and was inspired (by Mackenzie's book) to consider the following problem.

Consider first a principal bundle $$P\xrightarrow G M$$; we can construct the quotient manifold $$\Omega=\frac{P\times P}{G},$$ obtained as the orbit space of the pair action $$g(u_2,u_1)=(gu_2,gu_1)$$. In this way, we obtain a Lie groupoid (called the gauge groupoid and denoted $$\Omega\rightrightarrows M$$) with the source and target maps $$s,t\colon \Omega\rightarrow M$$ given by $$s[u_2,u_1]=u_1$$ and $$t[u_2,u_1]=u_2$$, and partial multiplication $$[u_3,u_2'][u_2,u_1]=[u_3,u_1]$$ iff $$u_2'=u_2$$.

In particular, the universal cover $$\mathrm{SU}(2)\rightarrow \mathrm{SO}(3)$$, given by the adjoint representation, may be observed as a principal bundle $${\mathrm{SU}(2)}\xrightarrow{\mathbb Z_2}{\mathrm{SO}(3)}$$. The map $$\mathrm{SU}(2)\times \mathrm{SU}(2)\rightarrow \mathrm{SO}(4)$$, given by $$(p,q)\mapsto (x\mapsto pxq^{-1})$$, where $$p,q,x$$ are viewed as quaternions, induces the isomorphism of Lie groups $$\frac{\mathrm{SU}(2)\times \mathrm{SU}(2)}{\mathbb Z_2}\cong \mathrm{SO}(4),$$ so we actually obtain a Lie groupoid $$\mathrm{SO}(4)\rightrightarrows \mathrm{SO}(3)$$, with the source and target fibres diffeomorphic to $$\mathrm{SU}(2)$$. Additionally, for any $$x\in \mathrm{SO}(3)$$, the vertex group $$s^{-1}(x)\cap t^{-1}(x)\cong\mathbb Z_2$$, indicating that $$\mathrm{SO}(4)\rightrightarrows \mathrm{SO}(3)$$ should be isomorphic to the fundamental groupoid of $$\mathrm{SO}(3)$$.

Now consider the (physically more juicy) universal cover $$\mathrm{SL}(2,\mathbb C)\xrightarrow{\mathbb Z_2}{\mathrm{SO}^+(1,3)}$$, where $$\mathrm{SO}^+(1,3)$$ denotes the component of the Lorentz group $$\mathrm O(1,3)$$ which contains the identity matrix -- this is a Lie subgroup (the proper ortochronous Lorentz group). By the general construction, we obtain a gauge groupoid $$\frac{\mathrm{SL}(2,\mathbb C)\times \mathrm{SL}(2,\mathbb C)}{\mathbb Z_2}\rightrightarrows \mathrm{SO}^+(1,3).$$ We would again like to identify the space of arrows with a known space, but it is now 12-dimensional. The only sensible candidate that comes to my mind is the complexification $$\mathrm{SO}^+(1,3)_{\mathbb C}$$, but I don't know where to start confirming whether this is the case. So the question is: $$\frac{\mathrm{SL}(2,\mathbb C)\times \mathrm{SL}(2,\mathbb C)}{\mathbb Z_2}\stackrel{?}\cong\mathrm{SO}^+(1,3)_{\mathbb C}.$$

Lastly, if anyone has any intuition regarding the obtained gauge groupoid over $$\mathrm{SO}^+(1,3)$$, feel free to make a hand-wavey explanation of what this object could physically represent.

After inspecting the general construction of the complexification of a Lie group, I've arrived to a positive answer to the question of whether $$\mathrm{SO}^+(1,3)_{\mathbb C}$$ is isomorphic to $$\frac{\mathrm{SL}(2,\mathbb C)\times \mathrm{SL}(2,\mathbb C)}{\mathbb Z_2}$$. The argument goes as follows.
Given a Lie group $$G$$, denote by $$\pi\colon\widetilde G\rightarrow G$$ the universal covering projection and by $$\widetilde G_{\mathbb C}$$ the (unique up to iso.) simply connected complex Lie group with Lie algebra $$\mathfrak g_{\mathbb C}=\mathfrak g\otimes \mathbb C$$. Let $$\phi\colon \widetilde G\rightarrow \widetilde G_{\mathbb C}$$ be the unique homomorphism (by Lie's second theorem) such that $$\mathrm d\phi_e$$ is the canonical inclusion $$\mathfrak g\hookrightarrow \mathfrak g_{\mathbb C}$$. The complexification $$G_{\mathbb C}$$ of $$G$$ is constructed abstractly as $$G_{\mathbb C}=\frac{\widetilde G_{\mathbb C}}{\phi(K)^*},$$ where $$K=\ker\pi$$ is the fundamental group of $$G$$, and $$\phi(K)^*$$ is the smallest closed normal subgroup of $$\widetilde G_{\mathbb C}$$ which contains $$\phi(K)$$ (by inspecting the adjoint representation of $$\widetilde G_{\mathbb C}$$, one can show that $$\phi(K)$$ is in the centre of $$\widetilde G_{\mathbb C}$$, hence so is $$\phi(K)^*$$).
In our case, the following isomorphism of Lie algebras is crucial: $$\mathfrak{sl}(2,\mathbb C)_{\mathbb C}\cong \mathfrak{sl}(2,\mathbb C)\oplus \mathfrak{sl}(2,\mathbb C).$$ Note that since $$\mathfrak{so}(1,3)\cong \mathfrak{sl}(2,\mathbb C)$$, there holds (up to an isomorphism) $$\widetilde{\mathrm{SO}^+(1,3)}_{\mathbb C}=\mathrm{SL}(2,\mathbb C)\times \mathrm{SL}(2,\mathbb C).$$ Since $$\ker \pi=\{I,-I\}$$ is the fundamental group of $$\mathrm{SO}^+(1,3)$$, and $$-I=\exp(\mathrm{diag}(i\pi,-i\pi))$$, we have (using the appropriate identifications) $$\phi(\{I,-I\})=\{(I,I),(-I,-I)\},$$ which is already closed and normal in $$\mathrm{SL}(2,\mathbb C)\times \mathrm{SL}(2,\mathbb C)$$.
We may now indeed conclude that the gauge groupoid of the Lorentz group is $$\mathrm{SO}^+(1,3)_{\mathbb C}\rightrightarrows \mathrm{SO}^+(1,3)$$