Identifying the orbit space of the unitary group $U(n)$ in the compact symplectic group $Sp(n)$

Question

Let $Sp(n)$ be the compact symplectic group. Let $U(n)$ the unitary group, and $O(n)$ the orthogonal group. What is $Sp(n)/U(n)$? What is $U(n)/O(n)$?

I obtain that $Sp(1)/U(1)=S^3/U(1)=\mathbb{C}P^1\cong S^2$.

Answer 1

The homogeneous space $Sp(1) / U(1)$ is an example of compact classical symmetric space.

Here, $Sp(n)$ is the group of transformations that preserve the quaternionic structure of $\mathbb{H}^n$, and $U(n)$ is the group of transformations that preserve a given complex structure on $\mathbb{H}^n$ along with the usual inner product $(a, b) \mapsto \Re(a \bar{b})$ on $\mathbb{H}^n$. So, we can view $Sp(n) / U(n)$ as the space (the symmetric space of type CI) of complex structures on $\mathbb{H}^n$ that preserve the inner product.

In the case $n = 1$, like you say we can regard $Sp(1)$ as the group of unit quaternions in $\mathbb{H}$, which topologically is $\mathbb{S}^3$, and the $U(1)$-orbits are precisely the circles parameterized by $t \mapsto u \exp(i t)$. The projection $$Sp(1) \cong \mathbb{S}^3 \to \mathbb{S}^3 / \mathbb{S}^1 \cong Sp(1) / U(1) \cong \mathbb{S}^2$$ is this case is precisely the Hopf Fibration. In fact, we can explicitly identify the complex structures that preserve the inner product on $\mathbb{H}$: They are precisely the maps $$x \mapsto (ai + bj + ck) x, \qquad a^2 + b^2 + c^2 = 1,$$ and by identifying respectively with the points $ai + bj + ck \in \mathbb{H}$ we can identify $Sp(1) / U(1) \cong \mathbb{S}^2$ as the set of imaginary quaternions of unit length.