# How to see that $\mathbb{C}\mathbb{P}^3\cong\mathrm{SO}(5)/\mathrm{U}(2)$?

I stumbled on this ismorphism in the context of twistor fibrations. See for example 'Twistors in Mathematics and Physics' by Bailey and Baston, p.58. Can anybody provide a construction of this isomorphism?

Sketched proof: \begin{align}\mathbb{H}^2/\mathbb{C}^{\times}~\cong~\mathbb{C}\mathbb{P}^3~\stackrel{?}{\cong}~&SO(5)/U(2)\cr ~\cong~& [SPIN(5)/\mathbb{Z}_2]/\{[U(1)\times SU(2)]/\mathbb{Z}_2\}\cr ~\cong~& SPIN(5)/[U(1)\times SU(2)]\cr ~\cong~& U(2,\mathbb{H})/[U(1)\times U(1,\mathbb{H})],\end{align} \tag{1} because \begin{align} SPIN(5)~\cong~& U(2,\mathbb{H}), \cr U(2)~\cong~& [U(1)\times SU(2)]/\mathbb{Z}_2\cr SU(2)~\cong~& U(1,\mathbb{H}), \end{align} \tag{2}
Removing a $$U(1)$$ phase, it is enough to prove
$$\mathbb{H}^2/\mathbb{R}_+~\stackrel{?}{\cong}~U(2,\mathbb{H})/ U(1,\mathbb{H}). \tag{3}$$
\begin{align} &U(2,\mathbb{H})\cr &~\cong~ \left\{ \left.\begin{pmatrix} \alpha a & \beta b \cr \beta c & \alpha d \end{pmatrix}\right| a,b,c,d\in U(1,\mathbb{H}), ~\bar{a}b+\bar{c}d=0,~\alpha,\beta\in\mathbb{R}_+,~ \alpha^2 +\beta^2=1\right\} \end{align}\tag{4}
in the numerator, while the left-hand side of eq. (3) consists of $$\mathbb{H}^2/\mathbb{R}_+~\cong~\left\{ ( \alpha a , \beta b )\in \mathbb{H}^2 \mid a,b\in U(1,\mathbb{H}),~\alpha,\beta\in\mathbb{R}_+,~ \alpha^2 +\beta^2=1\right\}.\tag{5}$$ $$\Box$$