2
$\begingroup$

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.

$\endgroup$

1 Answer 1

0
$\begingroup$

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.)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.