geodesics on $G/K$ which are not the orbits of a 1-parameter subgroup of $G$

Question

Let $G$ be Lie group and $K \subset G$ a closed subgroup, such that there exists a $v \in T(G/K)$ whose isotropy-group $G_v$ is discrete (so iff $\dim G_v =0$). Lets assume $g$ acts properly on $T(G/K)$

Now we endow $G/K$ with a $G$-invariant riemannian metric $< \cdot, \cdot >$ such that there exists a geodesic, which is not the orbit of a 1-parameter subgroup of $G$.

So my question is:

Assuming we have

1. $v_1 \in TM$ such that $\dim G_{v_1} =0$
2. $v_2 \in TM$ such that the geodesic induced by $v_2$ is not the orbit of a 1-parameter subgroup of $G$.

Can we now always find an elment $v_0 \in TM$, such that $\dim G_{v_0}=0$ and the geodesic is not the orbit of a 1-parameter subgroup of $G$?

Edit 2: I'm asking, since Paternain and Spatzier say in their paper "New examples of manifolds with completely integrable geodesic flows"

"Suppose the left invariant metric on $G/K$ has a geodesic which is not the orbit of a 1-parameter subgroup of $G$. Then a.e. the isotropy of $\hat{G}$ has dimension zero", where $\hat{G}$ is the action generated by $G$ and the geodesic flow. But to conclude, that the isotropy group of $\hat{G}$ at some point $w \in TM$ has dimension zero, we need to find $w \in TM$, such that $w$ is not the orbit of a 1-parameter subgroup of $G$ and furthermore, the isotropy group of $G$ at $w$ has to have dimension zero. (They also assumed that there is a $v \in TM$, such that $\dim G_v=0$)

Edit:

If we assume that $G$ acts properly on $T(G/K)$, we could try to show that $M_2$ is open, hence $T(G/K)\setminus M_2$ is closed.

Let $l \colon G \times G/K \to G/K$ be the left-multiplication and $l_{g*} \colon T(G/K) \to T(G/K)$ the action on $T(G/K)$ by derivations. Denote by $\exp$ the riemann-exponential and by $e^{\xi}$ the lie exponential.

Take $v_n \in M_2$, such that $v_n \to v_0$. Since $v_n \in M_2$, we find $\xi_n \in \mathfrak{g}$ such that $g_n(t).v_n=l_{e^{t\xi_n}*}(v_n) = \exp_{x_n}(tv_n)$. Since $\exp$ is continuous, we get $\exp_{x_n}(tv_n) \to \exp_{x_0}(tv_0)$.

Now we fix $t$ and so we got a sequence $v_n \to v_0$ and $g_n(t).v_n \to w_0(t) \in T_{y_0}M$

Using the properness of the action on $TM$, we have a subsequence $g_{n_j}(t) \to g_{*t}$ which is convergent and so $e^{t\xi_{n_j}}.v_{n_j} \to g_{*t}.v_0 = \exp_{x_0}(tv_0)$ (for fixed $t$). If I could now be sure, that $g_{*t}$ forms a 1-Parameter subgroup of $G$, I would be done.

Answer 1

Perhaps I'm misunderstanding your question, but what about the following example?

Let $G = \mathrm{SO}(3)$ and let $K=\{e\}$ be the identity subgroup. Then $G/K = \mathrm{SO}(3)$ and $G_v = K$ for all $v\in T\mathrm{SO}(3)$.

Now let $g$ be a metric on $\mathrm{SO}(3)$ that is invariant under the left action of $G$ on itself, but is not right invariant. In fact, for definiteness, let's suppose that the identity component of the isometry group of $(\mathrm{SO}(3),g)$ is $G$ itself acting on the left (this will be the case for the generic left-invariant metric on $G$).

Then nearly all of the geodesics for $g$ are not orbits of $1$-parameter subgroups of $G$. In fact, you have to use elliptic functions to describe them.

(This is equivalent to the problem of the rotating rigid body in $3$-dimensions. See, for example, Arnold "Mathematical Methods of Classical Mechanics" for a thorough discussion.)