Let $N$ be a homogeneous space. Therefore we find a Liegroup $G$ and a isotropy-subgroup $K$ of $G$, such that we can identify $N = G/K$. Then we have a canonical action $l\colon G \times G/K \to G/K$ on $G/K$, given by left-multiplication. 


We now identify the cotangentbundle $T^\*N = G \times\_K \mathfrak{k}^\circ$, where $\mathfrak{k}^\circ$ is the annihilator of $\mathfrak{k}$ in $\mathfrak{g}^\*$ and $T(T^\*N) = G \times\_K (\mathfrak{k}^° \times \mathfrak{g} \times \mathfrak{k}^°)$. We write the elements as $[g,\alpha] \in G \times\_K \mathfrak{k}^°$ and $[g,\alpha, v, \beta] \in G \times\_K (\mathfrak{k}^° \times \mathfrak{g} \times \mathfrak{k}^°)$.


Then the lifted action $\lambda \colon G \times T^\*N \to T^\*N$ is $\lambda\_g([h,\alpha]) =[gh, \alpha]$ and the differential should be $\lambda\_{h\*[g,\alpha]}([g, \alpha, v, \beta])=[hg,\alpha, l\_{h\*}v, \beta]$


Now we can define a left-invariant metric on $\mathfrak{g}$.
If we now assume that $K$ is compact, we find a $Ad\_K$-invariant non-deg bilinearform $\langle \cdot, \cdot \rangle\_1$ on $\mathfrak{g}$ by averaging the left-invariant metric w.r.t. $K$. Since $K$ is still compact, we find also a $Ad^\*\_K$-invariant metric $\langle \cdot,\cdot \rangle\_2$ on $\mathfrak{g}^\*$.


Now we define a riemannian metric on $T^\*N$:


$\langle [g,\alpha, v ,\beta], [g,\alpha, w, \gamma] \rangle=\langle v,w\rangle\_1 + \langle \beta , \gamma \rangle\_2$. 


If I made no mistakes, this should give us a left-invariant riemannian metric.


My question is now:
Assuming $G/K$ is a homogeneous space, can I always find some left-invariant metric on $T^\*(G/K)$ for the lifted action $\lambda$? Even if the isotropy-group isn't compact?


EDIT: Maybe I should explain, why I'm asking: I'm trying to understand, if the set of $G$-orbits with maximal dimension in $T^\*N$ is open and dense in $T^\*N$. Szeghy showed that if $G$ acts isometrically on a connected, semi-riemannian manifold, then it is true for the $G$-orbits in $M$.



Existence of left-invariant metric on the cotangentbundle of homogeneous spaces?
No. The simplest example is $G = \mathrm{SL}(2,\mathbb{R})$ acting on $\mathbb{RP}^1 = G/P$, where $P$ is the (noncompact) subgroup of upper triangular matrices. It's easy to show that $G$ cannot not preserve any Riemannian metric on $T^\*(G/P)$: If it did preserve a metric $g$, then, since $G$ preserves the zero section, it would preserve a metric on the zero section, and hence $G$ would have to preserve a Riemannian metric on $G/P$ itself, but it is too large for that.


More generally, the same argument applies whenever the action of $G$ on $G/K$ is effective and the representation of $K$ on $T\_{eK}G/K$ does not preserve a positive definite metric: A $G$-invariant metric on $T^\*(G/K)$ would induce a $G$-invariant metric on the zero section, and hence a $G$-invariant metric on $G/K$, contradicting the hypothesis.


