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 $M$.


1 Answer 1


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.

  • $\begingroup$ Thank you very much for your quick response. Since it it not true, I will have to see, if it could be true, that the $G$-orbits in $T^*(G/K)$ with maximal dimension are an open and dense set in $T^*(G/K)$ $\endgroup$
    – Olorin
    Mar 7, 2016 at 18:25
  • $\begingroup$ It can be true, but it is not always true that there are open $G$-orbits in $T^*(G/K)$. It's true in the example that I gave, but there are other examples in which it is false. $\endgroup$ Mar 7, 2016 at 18:27

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