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.



Gauge groupoid of Lorentz group & complexification
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
$$
\mathrm{SO}^+(1,3)\_{\mathbb C}\rightrightarrows \mathrm{SO}^+(1,3)
$$


(Note that I was, however, unsuccessful in using the universal complexification property to establish the wanted isomorphism.)


