# Universal Covering Group of $SO(1,3)^{\uparrow}$

I'm trying to prove that $SL(2,\mathbb{C})$ is the universal covering group for the proper orthochronous Lorentz group $SO(1,3)^{\uparrow}$. The standard way goes as follows.

(1) Exhibit a real vector space isomorphism between Minkowski space and the space of $2\times 2$ Hermitian matrices, $H$.

(2) Let $SL(2,\mathbb{C})$ act on $H$ by $X\mapsto AXA^{\dagger}$ and prove this induces a surjective, 2:1 homomorphism from $SL(2,\mathbb{C})$ to $SO(1,3)^{\uparrow}$.

I'm wondering whether there is a better way however. To prove that $SU(2)$ is the universal covering group of $SO(3)$ it suffices to go to the Lie algebra and demonstrate that the adjoint representation is an isomorphism of Lie algebras. Can I do something analogous here?

Here's what I've tried. The Lie algebra of $SO(1,3)$ is $su(2)\oplus su(2)$ which naturally acts on a 4-dimensional complex vector space. The Lie algebra of $SL(2,\mathbb{C})$ is the space of traceless complex matrices, of dimension 6. I can't now see how to proceed.

Maybe this approach doesn't work at all now. Is it just a special property of $SU(2)$ and $SO(3)$ that happens because $SU(2)$ happens to have dimension 3, exactly the right number for an $SO(3)$ action?

Many thanks in advance!

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I'd say it's purely coincidental and follows from identification of vectors and antisymmetric matrices in dimensions $3$. In any case, could you elaborate how does the adjoint rep. of $su(2)$ help us prove that $SU(2)$ is a universal cover of $SO(3)$? –  Marek Feb 12 '13 at 12:56
Certainly. First one shows that the adjoint representation of $su(2)$ yields an isomorphism to $so(3)$, by identifying $su(2)$ with $\mathbb{R}^3$. Next since $su(2)$ is simply connected we know that $ad$ exponentiates to $Ad$, yielding a homomorphism of $SU(2)$ into $SO(3)$. But since $SO(3)$ is connected and the Lie algebras are isomorphic, this must be surjective. To prove it's 2:1 takes a little more work, but not too much! I'm inclined to agree it's coincidental having thought a bit more about it. –  Edward Hughes Feb 12 '13 at 13:37
Thanks, that's a nice argument. The adjoint rep. won't work in general, but finding any isomorphism of Lie algebras will, so I guess that's good enough. –  Marek Feb 12 '13 at 16:18
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## 2 Answers

Maybe this is more of a comment but a general method to show that a Lie group $G$ is isomorphic to $SO^+(p,q)$ is to find a $p+q$ dimensional representation of $G$ that preserves an inner product of signature $(p,q)$. Then if $\dim G = \dim SO(p,q)$ and $G$ is connected, this will give an isomorphism $G/\ker \to SO^+(p,q)$.

In the case of $G = SU(2)$ and $SO(3)$, we need a 3-dimensional rep of $SU(2)$. Since $SU(2)$ is three dimensional, we can try the adjoint representation of $SU(2)$ on its Lie algebra. Since $SU(2)$ is compact any real representation is orthogonal so this maps into $SO(3)$. Then we just need to check that the kernel is $\{\pm 1\}$. Note also here that there is no need to go to the Lie algebra.

For your case of $SL(2,\mathbb C)$ the adjoint representation is 6 dimensional and irreducible since $SL(2,\mathbb C)$ is simple.

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Thanks for your insight. I don't see how you get surjectivity without going to the Lie algebra though. Could you expand on that? Perhaps as an answer to this question I have just raised? –  Edward Hughes Feb 12 '13 at 16:11
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I am not totally sure, but I think you can use the same method. Here $\mathfrak{su}(2)$ has real dimension 3 and $\mathfrak{sl}(2)$ has real dimension 6.

Any matrix in $\mathrm M \in \mathfrak{sl}(2)$ splits into $\mathrm M = \mathrm M_1 + i\mathrm M_2$ where $\mathrm M_1$ and $\mathrm M_2$ are trace zero hermitian matrices. This shows that $\mathfrak{sl}(2) \simeq \mathfrak{su}(2) \oplus \mathfrak{su}(2)$.

So I guess that the adjoint representation should give an isomorphism of Lie algebras.

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I don't quite see how you show that it's exactly the adjoint representation that gives this isomorphism. Could you possibly elaborate? –  Edward Hughes Feb 12 '13 at 13:41
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