Representing $\mathbb{C} P^3$ as $Sp(2) / (SU(2)U(1))$? I have a question regarding some representations of the three-dimensional complex projective space $\mathbb{C} P^3$.
I am reading the article 'Homogeneous Nearly Kähler manifolds' by Jean-Baptiste Butruille.
In it, he claims that $\mathbb{C} P^3 = G/H$, where $G = Sp(2)$ and $H = SU(2)U(1)$.
Question: I was wondering: what does $SU(2)U(1)$ mean? What group is it? It is clearly not the Cartesian product $SU(2) \times U(1)$ (for it is not a typo in the article).
Also, he writes the homogeneous space $\mathbb{C} P^3 \cong Sp(2)/ U(1)Sp(1)$. Again, I don't understand what the denominator means.
 A: Some authors use juxtaposition $HK$ to mean a Lie subgroup which is locally isomorphic to $H\times K$.
In this case, the notation $Sp(1)U(1)$ is vague, because there are two subgroups of $Sp(2)$ which are locally isomorphic to $Sp(2)\times Sp(1)$.  The first is all elements of the form $\begin{bmatrix} p & 0 \\ 0 & z\end{bmatrix}$ with $p\in Sp(1)$ and $z\in U(1)$, and the second is all elements of the form $\begin{bmatrix} p & 0 \\ 0 & p\end{bmatrix}\cdot \begin{bmatrix}\cos \theta & \sin\theta \\ -\sin\theta & \cos\theta\end{bmatrix}$ with $p\in Sp(1)$.  The first really is a copy of $Sp(1)\times U(1)$, while the second really is a copy of $U(2)$.  These two groups are not isomorphic: the first has an element of order $2$ in its center while the second doesn't.
Now, I claim that the author of the paper really means the first option.  This is because the first has quotient $\mathbb{C}P^3$, while the other quotient is the Grassmanian $G$ of oriented $2$-planes in $\mathbb{R}^5$.  The two spaces have isomorphic cohomology groups but different ring structures.  The claim about the topology is proven, for example, on page 7 in this paper by Kapovitch and Ziller.
For $Sp(2)/(Sp(1)\times U(1))$, the diffeomorphism is induced from the composition $Sp(2)\rightarrow S^7\rightarrow \mathbb{C}P^3$, where the map from $Sp(2)$ to $S^7$ takes a matrix to its second column, and the map $S^7\rightarrow \mathbb{C}P^3$ is the usual Hopf map.  Alternatively, viewing $\mathbb{H}^2$ as $\mathbb{C}^4$, $Sp(2)$ naturally acts transitively on $\mathbb{C}P^3$ with isotropy group consisting of elements of the first form.
For $Sp(2)/U(2)$, here is an argument.  Since $-I\in U(2)$, the map $Sp(2)/U(2)\rightarrow (Sp(2)/\pm I) / (U(2)/\pm I)$ is a covering map.  The group $Sp(2)/\pm I$ is isomorphic to $SO(5)$.  Since $-I\in Sp(1)\subseteq U(2)$, the map $Sp(2)\rightarrow SO(5)$ restricts to a map $SU(2)\rightarrow SU(2)/\pm I\cong SO(3)\subseteq SO(5)$.  There are, up to conjugacy, precisely two $SO(3)$ in $SO(5)$; the first is the standard block embedding, while the second is maximal.  Because the $SO(3)$ must have non-trivial centralizer (coming from the $U(1)$ part of $U(2)$), $SO(3)$ cannot be maximal in $SO(5)$.  Now, it follows that the image of $Sp(1)U(1)$ in $SO(5)$ is the usual block $3\times 3$ and $2\times 2$ matrices.  This is well known to be a Grassmannian, but can also be argued as above:  the map $SO(5)\rightarrow G$ which takes a matrix in $SO(5)$ to the span of the last $2$ columns induces a diffeomorphism.
In any event, it follows that $Sp(2)/U(2)$ is a covering of the Grassmanian.  However, the Grassmannian is simply connected, so it must be diffeomorphic to $Sp(2)/U(2)$.
