Quoting from Wikipedia [article](https://en.wikipedia.org/wiki/Quaternionic_projective_space) on quaternionic projective space:



> 
> Therefore the quotient manifold
>  $$
> \mathbb{HP}^{2}/\mathrm{U}(1)
> $$
>  may be taken, writing $U(1)$ for the circle group. It has been shown that this quotient is the $7$-sphere, a result of Vladimir Arnold from 1996, later rediscovered by Edward Witten and Michael Atiyah.
> 
> 
> 


Does such a situation hold for higher quaternionic projective spaces? More specifically, does there exist a $U(1)$ action on $\mathbb{HP}^n$, and is the quotient a sphere? 



Circle Action on Quaternionic Projective Space
$\mathbb{HP}^n\cong \mathrm{Sp}(n+1)/(\mathrm{Sp}(n)\times \mathrm{Sp}(1))$ is a symmetric space, so every one-parameter subgroup of $\mathrm{Sp}(n+1)$ acts on it. As was noted in the comments, such an action always has at least $n+1$ fixed points. Let me restrict to the action $z\cdot [x\_0:\cdots: x\_n] = [x\_0z:\cdots:x\_nz]$, since this is what was used in the case $n=2$.


For $i\in\{0,\cdots,n\}$, there is a coordinate chart $\phi\_i:\mathbb{C}^{2n}\cong U\_i\subset \mathbb{HP}^n$ defined by
$$
\phi\_i(x\_1,\cdots,x\_n,y\_1,\cdots,y\_n) = [x\_1+jy\_1:\cdots:x\_{i-1}+jy\_{i-1}:1:x\_i+jy\_i:\cdots:x\_n+jy\_n]
$$
where the $1$ is in the $i$-th homogeneous coordinate. Then
\begin{align\*}
\hspace{5pt}&\hspace{-5pt}z\cdot \phi\_i(x\_1,\cdots,x\_n,y\_1,\cdots,y\_n)\\
 &= [(x\_1+jy\_1)z:\cdots:(x\_{i-1}+jy\_{i-1})z:z:(x\_i+jy\_i)z:\cdots:(x\_n+jy\_n)z]\\
&= [z^{-1}(x\_1+jy\_1)z:\cdots:z^{-1}(x\_{i-1}+jy\_{i-1})z:1:z^{-1}(x\_i+jy\_i)z:\cdots:z^{-1}(x\_n+jy\_n)z]\\
&= [x\_1+jy\_1z^2:\cdots:x\_{i-1}+jy\_{i-1}z^2:1:x\_i+jy\_iz^2:\cdots:x\_n+jy\_nz^2]\\
&= \phi\_i(x\_1,\cdots,x\_n,z^2y\_1,\cdots,z^2y\_n)
\end{align\*}
Of course, $-1$ commutes with all quaternions and thus acts by the identity. We can replace $U(1)$ by $U(1)/(\mathbb{Z}/2)$ which amounts to replacing $z^2$ by $z$. (In fact, this action is just that of the automorphisms of $\mathbb{H}$ restricting to the identity on $\mathbb{C}$, which explains the relation to complex conjugation.)


Thus $U\_i$ is stable under the $U(1)$-action, and $U\_i/U(1)\cong \mathbb{C}^n\times (\mathbb{C}^n/U(1))$, where $U(1)$ acts on $\mathbb{C}^n$ by scalar multiplication. Using polar coordinates, one easily sees that $\mathbb{C}^n/U(1)\cong C\mathbb{CP}^{n-1}$ is homeomorphic to the cone on $S^{2n-1}/U(1)\cong \mathbb{CP}^{n-1}$. Hence the quotient $\mathbb{HP}^n/U(1)$ is covered by $n+1$ open sets, each of which is homeomorphic to $\mathbb{C}^n\times C\mathbb{CP}^{n-1}$. When we remove the tip of the cone from any neighbourhood of it, the resulting open set retracts onto $\mathbb{CP}^{n-1}$. Looking at $H\_2$, we see that if the tip of the cone (with some element of $\mathbb{C}^n$ in the first component) has a euclidean neighbourhood, we must have $n = 2$. Conversely, in this case $C\mathbb{CP}^{n-1}\cong CS^2 \cong \mathbb{R}^3$, so $U\_i/U(1)\cong \mathbb{R}^7$ and the quotient is a manifold. In higher dimensions, the quotient is not even an orbifold, but one still understands its singularities.


Here is a proof that for $n=2$ the quotient is homeomorphic to $S^7$: Note that $U\_n = \mathbb{HP}^{n}\setminus \mathbb{HP}^{n-1}$. For $n=0$, the quotient is a point; for $n=1$ it is compact and the union of a point and $\mathbb{C}\times C\mathbb{CP}^0\cong \mathbb{R}^2\times\mathbb{R}\_{\ge 0}$, so it is the one-point compactification of the half-space, which is contractible. Consequently, $\mathbb{HP}^{2}/U(1)\to (\mathbb{HP}^{2}/U(1))/(\mathbb{HP}^{1}/U(1))$ is a homotopy equivalence, and the target is the one-point compactification of $\mathbb{C}^2\times C\mathbb{CP}^1\cong \mathbb{R}^7$ which is $S^7$. So $\mathbb{HP}^2/U(1)$ is a manifold and a homotopy sphere; by the Poincare conjecture, it is homeomorphic to $S^7$.


