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?

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    $\begingroup$ Do you know about complex QR decomposition? $\endgroup$ Jun 28, 2019 at 14:13
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    $\begingroup$ Polar Decomposition $\endgroup$
    – AHusain
    Jun 28, 2019 at 14:18
  • $\begingroup$ Thank you. Why does this imply that the "$R$" part of the $QR$ decomposition is real-valued? @MoisheKohan $\endgroup$ Jun 28, 2019 at 14:40
  • $\begingroup$ Thanks @AHusain I will exploit that and come back later. $\endgroup$ Jun 28, 2019 at 14:43
  • $\begingroup$ @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. $\endgroup$ Jun 28, 2019 at 14:52

1 Answer 1


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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