Good evening to everyone.
I'm having a problem in the following setting:
If I'm having a homogeneous manifold $M=G/K$, where $K \subset G$ is a closed subgroup, I can always find a $G$-invariant riemannian metric on $G/K$. Using this metric, there is for each $(x,v) \in TM$ a unique geodesic, denoted by $\gamma_v(t) \colon I \to M$, such that $\gamma_v(0)=x$ and $\dot{\gamma}_v(0)=v$.
Now we can define an action $\phi \colon \mathbb{R} \times TM \to TM$, by the geodesic flow, namely for $(x,v) \in TM$ and $t \in \mathbb{R}$, we map $(t,(x,v)) \mapsto (\gamma_v(t), \dot{\gamma}_v(t))$. Since we are having a homogeneous space, our riemannian manifold is geodesically complete.
My question now is: Is it a proper action?
So, taking a sequence $(x_n,v_n) \in TM$ and $t_n \in \mathbb{R}$, such that $x_n \to x_0 \in M$ and $v_n \to v_0 \in TM$ and $\left(\gamma_{v_n}(t_n) ,\dot\gamma_{v_n}(t_n)\right) \to (y_0,w_0)$, I need to show, that there exists a convergent subsequence of $t_n$ with $t_{n_j} \to t_0$ and $(\gamma_{v_0}(t_0), \dot{\gamma}_{v_0}(t_0)) = (y_0, w_0)$. If I can find a convergent subsequence, then the last equality follows directly.
I thought, I could maybe take the sequence $\{t_n\}_n$ and then $\sigma_n := (\gamma_{v_0}(t_n), \dot\gamma_{v_0}(t_n))$ and somehow conclude, that $\sigma_n \to (y_0,w_0)$
1) Would that action maybe be proper, if our homogenous space is compact?
Edit:
I think it is important that we have a homogeneous space, not only for the completeness, but also in some other facts (which I'm not sure which one). So, an example would be $S^1\subset \mathbb{R}^2$ with the standard-metric. Then taking $(x_n,v_n) =((1,0),(0,2)) \in T_{(1,0)}S^1$ for all $n$ and $t_n =2\pi n$, then $t_n \cdot (x_n,v_n) =((1,0),(0,2))$ but there is no convergent subsequence of $\{t_n\}$.