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.


1 Answer 1


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.)


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