Minimal spheres in rank one symmetric spaces Let $M^n$ be a compact rank one symmetric space endowed with its canonical metric. For a given point $p \in M$, does there exist a positive number $r$ such that the sphere
$$S_r(p) = \{ x \in M: d(x,p)=r \}$$
is minimally embedded?
For instance, I know this is true when $M$ is a round sphere.
 A: Here is the result:  such an $r$ always exists except when $M$ is isometric to $\mathbb{R}P^n$ for $n > 2$.  Moreover, $r$ is unique (under the assumption that $r \leq \operatorname{diam} M$) in the case of $S^n$ with $n\geq 2$, as well as for $\mathbb{C}P^n$ or $\mathbb{H}P^n$ with $n > 2$, and also for $\mathbb{R}P^2$.  On the other hand, for $\mathbb{C}P^2$,$\mathbb{H}P^2$, and $\mathbb{O}P^2$, there are precisely two choices for $r$.  For $\mathbb{R}P^1 = S^1$, there are infinitely many choices for $r$.
Let's prove all this.
The case of $\mathbb{R}P^1= S^1$ is special because for any $r$ smaller than the diameter, $S_r(p)$ consists of $2$-points, which are trivially minimal submanifolds.  We will ignore this manifold in what follows.
Let $M$ denote any compact rank 1 symmetric space of dimension at least $2$.  Note that $M$ is homogeneous:  $M = G/H$ where $(G,H)$ is a symmetric pair of compact Lie groups.
By inspection, in each case, the isotropy action of $H$ on $T_{eH} G/H$ (where $e\in G$ is the identity) is transitive on the unit sphere.  Exponentiating out, we find that for each $r$, $H$ the restriction of the $G$ action to $H$ preserves $S_r(eH)$ and acts transitively.
Let's assume momentarily that $r = \operatorname{diam}(M)$.  In this case, for $M = S^n$, $S_r(p)$ is a point, so not a sphere.  The remaining spaces are of the form $\mathbb{K}P^n$ where $\mathbb{K}\in \{\mathbb{R}, \mathbb{C}, \mathbb{H}, \mathbb{O}\}$; in these cases $S_r(p)$ is diffeomorphic to $\mathbb{K}P^{n-1}$.  This space is totally geodesic, hence minimal.  But it is only diffeomorphic to a sphere when $n=2$.
So, to finish the proof, we need to show that a) for $\mathbb{R}P^n$ with $n\geq 2$, no $r$ smaller than the diameter works and b) for the other spaces, that there is a unique $r$ smaller than the diameter which works.
We may now assume that $0 < r < \operatorname{diam} M$. As mentioned above, for any $p\in M$, $S_r(p)$ is homogeneous under the action by $H$.  According to the main theorem in this paper of Hsiang, $S_{r}(p)$ is minimal iff it is extremal among its orbit type.  Here, extremal means that the $(\dim M - 1)$-volume of $S_r(p)$ is a max or min among all nearby $H$-orbits of the same orbit type.  But all $S_r(p)$ with $r$ smaller than the diameter have the same orbit type.  So, we need to compute the $(\dim M-1)$-volume of $S_r(p)$ as a function of $r$, and determine the extreme values of this function.
When $M$ is a sphere, it easily follows that the extremal $H$-orbits occur precisely when $r$ is half the diameter.  This is why there is a unique $r$-value for spheres.  Using the fact that $\mathbb{R}P^n$ is locally isometric to $S^n$, we see the volume of $S_r(p)$ is a strictly increasing function of $r$, so has no extrema.  This completes task a), and task b) for spheres.
The remaining spaces are all of the form $\mathbb{K}P^n$, as above.  Here, each $S_r(p)$ is naturally the total space of a bundle.  Precisely, it's a bundle of the form $$S^{k-1}\rightarrow S_r(p)\cong S^{(n+1)k-1}\rightarrow \mathbb{K}P^{n-1},$$ where $k:=\dim_\mathbb{R}\mathbb{K}$.
For concreteness, let us assume the metric on $\mathbb{K}P^n$ is normalized to have minimum curvature $1$.  Now, note the induced metric on $S_r(p)$ is not round.  Rather, as seen in Bettiol's answer here we see that the induced metric on $S_r(p)$ has the form
$$g_r=\sin^2 r\, (g_{hor}+\cos^2 r \, g_{ver}),$$ where $g_{hor}+g_{ver}$ is the round metric on $S_r(p)$ and $g_{hor}$ and $g_{ver}$ refer to the horizontal and vertical components relative to the sphere bundle above.
Let's compute the volume of $g_r$.  We can do this using a A version of Fubini's theorem.  Namely, we find that the metric on the fiber $S^{k-1}$ is $\sin^2 r \cos^2 r$ times the round metric.  This has volume $(\sin^2 r \cos^2 r)^{k-1}\cdot vol(S^{k-1}(1))$, where $S^{k-1}(1)$ refers to the round sphere of radius $1$.  The base is a $\mathbb{K}P^{n-1}$ of diameter  $\sin^2(r)/2$, so it has volume $(\sin^2(r)/2)^{k(n-1)}\cdot Vol(\mathbb{K}P^{n-1}(1)$, where $\mathbb{K}P^{n-1}(1)$ indicates the metric is scaled to have minimum curvature $1$.
So, up to multiplicative constant (which doesn't affect the location of extrema), we find $$f(r):=Vol(S_r(p)) = (\sin^2 r \cos^2 r)^{k-1} \cdot (\sin^2(r))^{k(n-1)} = \sin^{2(kn-1)}(x)\cos^{2(k-1)}(x).$$
From here, computing a derivative $f'(r)$ is straight forward.  We get $$f'(r) = 2\cos^{2(k-1)-1}(x) \sin^{2(kn-1)}(x)\Big((kn-1)\cos^2(x) - (k-1)\sin^2(x)\Big).$$
On the interval $(0,\pi/2)$ (which is the relevant interval for $r$ since we've  normalized minimum curvature to $1$), $\sin(r)$ and $\cos(r)$ are both non-zero.  So, $f'(r)$ vanishes only when the factor in parenthesis vanishes.  This factor is $0$ iff $\tan^2(r) = \frac{kn-1}{k-1}$.  Since $\tan^2:(0,\pi/2)\rightarrow (0,\infty)$ is a bijection, there is precisely one $r$ which solves this equation.  Thus, $f$ has a unique critical point on $(0,\pi/2)$.
Of course, critical points don't have to be extrema, but let's verify that they are in this case.  The function $f$, viewed as a map on $[0,\pi/2]$ vanishes on both end points and is obviously positive in the interior.  Thus, it must achieve its absolute max on some interior point, which must therefore be a critical point.  Since there is only one critical point, this must be an extrema of $f$.
