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

Solution 1:

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.

Solution 2:

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.