I came across the following statement from Berger's book 'A Panoramic View of Riemannian Geometry":
Assume one has a Riemannian metric g on G and that one wants to push it down to the quotient G/H. By the general nonsense quoted above, one needs g to be invariant under the action of H. More precisely, H is called the isotropy group of G/H and its action is to be considered on TeG by the differentials of the right translations associated to the h ∈ H. A Lie group acting faithfully on a real vector space can leave invariant a positive definite quadratic form if and only if it is compact.
To understand this I examined the example when $G=\mathbb{R}^2$, being the Lie group under addition, with $e=(0, 0)$ the identity element. And $g=dx^2+dy^2$ is the standard Euclidean metric, which is bi-variant since translations are isometric.
Now let $H=\{(0, y)\mid y\in\mathbb{R}\}$ be the $y$-axis, being a subgroup. The quotient space $G/H$ is just the real line $\mathbb{R}$, and the projection $\pi: (x, y)\mapsto x$ is a Riemannian submersion, which pushes down the metric from $\mathbb{R}^2$. But $H$ is not compact.
What is the problem here? What does the statement exactly mean?
A Lie group acting faithfully on a real vector space can leave invariant a positive definite quadratic form if and only if it is compact.
The statement
is false as stated. What is true is:
(a) If $G$ is a compact group acting (linearly and continuously) on a real vector space $V$, then $G$ preserves an inner product on $V$.
(b) If $G$ is a topological group acting (linearly and continuously) on a finite-dimensional vector space $V$ preserving an inner product on $V$, then the image of $G$ in the group of linear automorphisms of $V$ is relatively compact. (Equivalently, $G$ acts on $V$ via uniformly bounded linear operators.)
As a simple example, consider the real Lie group $G\cong {\mathbb R}$. This group embeds in the compact Lie group $T^2=S^1\times S^1$.
Edit. More precisely, there exists an injective continuous homomorphism $h: {\mathbb R}\to T^2$. For instance, $T^2\cong {\mathbb R}^2/{\mathbb Z}^2$. Let $L\subset {\mathbb R}^2$ denote a line (a 1-dimensional linear subspace) with irrational slope, e.g. $$ L=\{(t,t \sqrt{2}): t\in {\mathbb R}\}. $$
Then the restriction of the covering homomorphism ${\mathbb R}^2\to T^2$ to the line $L$ defines an injective embedding $h: {\mathbb R}\to T^2$.
The latter group embeds in $U(2)$ which acts orthogonally on ${\mathbb C}^2$. However, $G$ itself is noncompact. Nevertheless, its image in $GL(2, {\mathbb C})$ is relatively compact.
In the setting of infinite-dimensional representations, the situation becomes more interesting.
Definition. A locally compact topological group $G$ is called unitarizable if every continuous linear representation of $G$ on a (complex) Hilbert space $V$ whose image consists of uniformly bounded operators, preserves some inner product on $V$ (which defines the original same topology on $V$).
Every compact topological group is unitarizable and, more generally, so is every amenable group. It is an open problem due to Dixmier if every unitarizable locally compact group is amenable.