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I'm (still) trying to prove that $SL(2,\mathbb{C})$ is the universal covering group the the proper orthochronous Lorentz group $L$. I have completed the following steps.

(1) Prove that the vector space of $2\times 2$ Hermitian matrices $H$ is isomorphic to Minkowski space.

(2) Demonstrate that the action of $SU(2)$ on $H$ by $X\mapsto AXA^{\dagger}$ induces a group homomorphism $SL(2,\mathbb{C})\to L$.

(3) Prove this is 2:1 by observing that every $2\times 2$ complex matrix can be written as $X+iY$ with $X$ and $Y$ Hermitian.

I still need to prove that this map is surjective though. Here I am completely stuck. All the books and internet resources I have found either gloss over it, or state that it's true but don't prove it.

Could someone possibly give me a proper proof, with mathematician's rigour?!

P.S. My own attempts at a proof have fallen down. I tried the following

(a) Derive a formula for the inverse map locally. I can't see any good way to attack this though.

(b) Prove that the associated Lie algebra homomorphism is an isomorphism. I know theoretically this should be possible, but practically it seems a nightmare!

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up vote 4 down vote accepted

You have shown that $SL(2,\mathbb C)/K \to L$ is an injective homomrphism of Lie groups, where $K$ is the kernel of the representation, which is a discrete subgroup of $SL(2,\mathbb C)$. In other words, $SL(2)/K$ is (isomorphic to) a Lie subgroup of $L$. But by dimensionality reasons you are actually done: since $L$ is connected, there is a 1-to-1 correspondence between connected Lie subgroups and subalgebras of $L$'s Lie algebra. This means that any connected Lie subgroup of the same dimension must be $L$ itself.

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That's marvellous - thank you! Could you possibly point me at a reference for your statement about the 1-to-1 correspondence between connected subgroups and subalgebras? P.S. shall accept the answer as soon as the minimum time limit has expired! – Edward Hughes Feb 12 '13 at 16:19
Sure. I think almost any mathematics text on Lie groups will have this. Here's one explicit reference though:… – Eric O. Korman Feb 12 '13 at 16:24
Brilliant - cheers! – Edward Hughes Feb 12 '13 at 16:25

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