# What do radially symmetric functions on a Riemannian symmetric space look like?

Let $M$ be a Riemannian manifold with isometry group $G$. We call a smooth function (on $M$, or on an appropriate neighborhood of $x_0$) radially symmetric about $x_0 \in M$ if it is invariant under the isotropy subgroup (=stabilizer) $K$ of $x_0$ in $G$.

Every point $x_0$ has a normal neighborhood $V$: one which is contained in the image of the exponential $\exp : T_{x_0}M \to M$ and it allows to define the radius $r$ on $V$, locally it is the geodesic distance to $x_0$ and it is invariant under $K$.

Take a geodesic ball $B(x_0, \epsilon)$ contained in a normal neighborhood of $x_0$. It is stable by $K$. Every function of $r$ on this ball is radially symmetric about $x_0$.

Question: If $M$ is a (Riemannian) symmetric space, is any radially symmetric function locally a function of $r$?

It would suffice to show that $K$ acts transitively on sufficiently small geodesic balls $B(x_0, \delta)$. This is the case if and only if $M$ is isotropic, but I cannot assume isotropicity.

A general Riemannian manifold may have trivial isometry group (see Isometry Group of a Manifold) so this certainly cannot hold in general. But a symmetric space has so many isometries that it is credible.

I'm trying to understand $K$-invariant eigenfunctions of $G$-invariant differential operators, which is where this question arises. Notably, the Selberg eigenfunction principle.

I am only familiar with compact symmetric spaces, so this answer will address those.

The group $K$ will act transitively if your symmetric space has rank $1$. That is, if it is a sphere or projective space. I don't know of any classification-free proof but from the classification, they look like $S^{n-1} = SO(n)/SO(n-1)$, $\mathbb{C}P^{n-1} = U(n)/U(n-1)U(1)$, $\mathbb{H}P^{n-1}=Sp(n)/Sp(n-1)Sp(1)$, or $\mathbb{O}P^2 = F_4/Spin(9)$.

If the symmetric space has rank bigger than 1, then in general, $K$ will not act transitively. (I actually don't know of any example where $K$ does act transitively and the rank is larger than $1$, but haven't thought too much about it.)

For example, a general real Grassmannian of oriented $k$-planes in $\mathbb{R}^n$ has the form $SO(n)/SO(k)SO(n-k)$. This has dimension $k(n-k)$, so the relevant sphere around $x_0$ has dimension $k(n-k)-1$. If $k\geq 2$ and $n-k\geq 2$ (so we're talking about rank $\geq 2$) then the sphere has dimension larger than $k - 1$ and $(n-k) - 1$, so the usual action of $SO(k)SO(n-k)$ won't be transitive.

There are, of course, other weird actions inolving $SO$s which are transitive on spheres. But actually, all of the weird ones require working on $Spin(n)$ instead of $SO(n)$, so I guess this means that Grassmannians never have $K$ acting transitively on spheres.

I would bet that this same is true of complex and quaternionic Grassmannians. But of course, there are many others, e.g. $\mathbf{G}_2/SO(4)$.

I'm not sure what this means for your understanding of $K$-invariant eigenfunctions of $G$-invariant differential operators, but hopefully it gives you something to think about.