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

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?

• Do you know about complex QR decomposition? Jun 28, 2019 at 14:13
• Polar Decomposition Jun 28, 2019 at 14:18
• Thank you. Why does this imply that the "$R$" part of the $QR$ decomposition is real-valued? @MoisheKohan Jun 28, 2019 at 14:40
• Thanks @AHusain I will exploit that and come back later. Jun 28, 2019 at 14:43
• @big-lion: I did not say it is real-valued. What is true is that the diagonal part can be taken to be positive: This yields the uniqueness of the decomposition. Jun 28, 2019 at 14:52

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.