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"](http://arxiv.org/pdf/math/9201276v1.pdf)


"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.



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


