I was wondering if the following holds:
Let $G/H$ be a (complete) Riemannian homogeneous manifold (i.e., a homogeneous manifold with a prescribed Riemannian metric on it, and $G$ acts transitively on $G/H$ through isometries). Suppose $\xi_0$ is a totally geodesic submanifold of $G/H$, and let $H_0 = \left\lbrace g \in G | g \cdot \xi_0 = \xi_0 \right\rbrace$ be its isotropy group. Can we say that $H_0$ acts transitively on $\xi_0$?
It seems to be true in simple cases of $\mathbb{R}^n$ and $\mathbb{S}^n$. In the first case, $G$ is the group of rigid motions characterized by the matrices of the type $\left[ \begin{matrix} A & b \\ 0& 1 \end{matrix} \right]$, where $A \in O(n)$ and $b \in \mathbb{R}^n$, with the action given by $\left[ \begin{matrix} A & b \\ 0& 1 \end{matrix} \right] \cdot x = Ax + b$, for $x \in \mathbb{R}^n$. Here, if I consider $\mathbb{R}^{n - 1}$ as a totally geodesic submanifold of $\mathbb{R}^n$, we get its isotropy group to be
$$H_0 = \left\lbrace \left[ \begin{matrix} A & 0 & b \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right] : A \in SO(n-1) \text{ and } b \in \mathbb{R}^{n - 1} \right\rbrace \cup \left\lbrace \left[ \begin{matrix} A & 0 & b \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{matrix} \right] : A \in O^{-}(n-1) \text{ and } b \in \mathbb{R}^{n - 1} \right\rbrace,$$
where $O^{-}(n-1)$ is the set of all orthogonal matrices with determinant negative. From this characterization, it is easy to see that $H_0$ does act transitively on $\xi_0$ (through translations alone!). Similar computations can be done for the case of $\mathbb{S}^n$, where the totally geodesic submanifolds are lower dimensional spheres.
Now, I want to generalize this to arbitrary Riemannian homogeneous manifolds (that need not be embedded in any Euclidean space, like $\mathbb{S}^n$). I have an intuition that this should be true. That is because, just like the case of $\mathbb{R}^n$, there are group elements from $G$ that take one point to another, and then there are elements from $H_0$ and do not take us outside $\xi_0$. Therefore, it seems that by choosing proper element, we will remain inside $\xi_0$ and yet be able to go from one point to another using elements of $H_0$.
Any help in formalizing these ideas (or disproving them) will be highly appreciated!
In general, no, there are totally geodesic subgroups which are not acted on transitively by the group.
To see this, consider the homogeneous spaces $SU(n+1)/SU(n)$, which happens to be diffeomorphic to the sphere $S^{2n+1}$. Here, for definiteness, I'm thinking of the $SU(n)$ as being embedded in $SU(n+1)$ as matrices of the block diagonal form $\operatorname{diag}(1,A)$ with $A\in SU(n)$.
Let $\xi_0$ denote the equatorial $S^{2n}\subseteq S^{2n+1}\subseteq \mathbb{C}^{n+1}$ given by $S^{2n} = \{(z_1,...,z_{n+1})\in S^{2n+1}: Im(z_1) = 0\}$. Then $\xi_0$ is totally geodesic, being the fixed point set of the isometry $\phi:S^{2n+1}\rightarrow S^{2n+1}$ given by $\phi(z_1,...,z_{n+1}) = (\overline{z}_1, z_2,..., z_{n+1}).$
Then, for any $n\geq 1$, the isotropy group $H_0$ does not act transitively on $\xi_0$.
Here is one way to see it. From the explicit description of $SU(n)$ above, it follows that $SU(n)\subseteq H_0$. The subgroup $SU(n)\subseteq SU(n+1)$ is almost maximal (among connected groups): it is contained in $U(n)$, but no larger group. Thus, the identity component of $H_0$ (which acts transitively on $\xi_0$ iff the whole $H_0$ action on $\xi_0$ is transitive) is either $SU(n)$ or $U(n)$. But from the classification of transitive actions on a sphere, there is no transitive action of $SU(n)$ or $U(n)$ on $S^{2n}$.
In the example above, a bi-invariant metric on $SU(n+1)$ does not induce the round metric on $S^{2n+1}$, so one may still ask for a normal homogeneous space where this kind of thing happens. Here is an example.
Consider the homogeneous space $G_2/SU(3)$, which is diffeomorphic to $S^6$. Here, a bi-invariant metric on $G_2$ does induce the round metric on $S^6$. Now, let $\xi_0$ denote any equatorial $S^4$. From the classification of transitive actions on spheres, the only group acting transitively on $S^4$ is (up to covers) $SO(5)$. But there is no $SO(5)\subseteq G_2$.