Explicit homeomorphism $SL(2,\mathbb{C})\cong SU(2)\times\mathbb{R}^3$.

covering-spacesspin-geometry

Take a look at Gallier, pg. 207. There is a homeomorphism
$$
SO^+(p,q)\cong SO(p)\times SO(q)\times \mathbb{R}^{pq}.
$$
Hence the universal cover of $SO^+(p,q)$ is
$$
\text{Spin}(p)\times\text{Spin}(q)\times\mathbb{R}^{pq}.
$$
In particular, since we know that
\begin{gather}
\text{Spin}(1,3)\cong SL(2,\mathbb{C})
\\
\text{Spin}(3)\cong SU(2)
\end{gather}
The universal property of universal covers implies that there is a homeomorphism
$$
SL(2,\mathbb{C})\cong SU(2)\times\mathbb{R}^3,
$$
endowing $SU(2)\times\mathbb{R}^3$ with a unique group structure. Can this map be made explicit?

Best Answer

I'll post as answer so I can write easily. The point is that $G=GL(n,C)$ has two groups $U=U(n)$ and $T=T(n)$=upper triangular with real positive entries in the diagonal and $U\cap T=1$. This fact easily inplies that $$ut=u't'\Longrightarrow u=u', t=t'$$ (because $ut=u't'$ implies $(u')^{-1}u=t't^{-1}\in U\cap T$)

So, the (restriction of the) multiplication gives an injective map $$U\times T\to G$$ Now you can hit the problem with diff.geom machinery to see that it is also surjective, but my comment was that a look at Gramd-Schmidt ortonormalization methods is in fact a (quite elementry) proof that this map is surjective.

Being said that, now consider $S=SLn$, and the restriction of those maps. If a matrix $A$ has determinant 1 and $A=ut$ then $\det(u)\det(t)=1$. But $u\in U$ implies $\det(u)=e^{i\theta}\in S^1$ and $t\in T$ implies $\det(t)=x\in R_{>0}$. But $e^{i\theta}x=1$ implies $e^{i\theta}=1$ and $x=1$, so $u\in SU$ and $t\in T\cap SL$. As a consequence, $SL(n,C)$ is homeomorphic to $SU(n)\times (T\cap SL(n,C))$.

Notice $T\cap SL(n,C)\cong R_{>0}^{n-1}\times C^{n(n-1)/2}$ because now you need an upper triangular matriz with real positive entries in the diagonal but whose product is 1. For $A\in SL(2,C)$, you have a unique decomposition $A=ut$ with $u\in SU(2)$ and $t=\left(\begin{array}{cc}r&z\\0&r^{-1}\end{array}\right)$ with $r>0$ real and $z\in C$.

Comming back to the explicit homeomorphisms, in one direction is matrix multiplication, in the other direction is Gramd-Schmidt formula for the $u$ factor, and "$u^{-1}A$" for the $t$ factor.

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