Construction G-Invariant Riemannian Metric

Question

Let $M$ be a smooth manifold and $G$ be a lie group acting transitively on $M$.
I know by Corollary 1.27 of these notes that there to exist a Riemannian metric $g_G$ on $M$ satisfying the in-variance relation $$ (\forall x,y \in M)(\forall g \in G) \,g_G(x,y)=\,g_G(g\cdot x,g\cdot y)? $$ When is this Riemannian metric unique? How can it be contructed explicitly if $M$ is $\mathbb{R}^d$?

Answer 1

First, this is not true as stated. That is, if $G$ is a non-compact Lie group acting transitively on a manifold $M$, there may not be a Riemannian metric on $M$ for which the $G$ action preserves the distance.

For example, consider $G = PGl(2,\mathbb{C})$ acting on the Riemann sphere $S^2$ via Möbius transformations.. Suppose there is a $G$-invariant distance on $S^2$. Let $H\subseteq G$ denote the isotropy group at $p:=(1,0,0)$. That is, for $h\in G$, $h\in H$ iff $h\ast p = p$. For any $\epsilon > 0$, because $G$ preserves distance, $H$ will preserve the set $\{q\in S^2: d(p,q) = \epsilon\}$. For small $\epsilon$, this set should be a circle. However, since $G$ acts $3$-transitively on $S^2$, it follows that $H$ acts transitively on $S^2\setminus \{p\}$, so $H$ cannot preserve any circle.

What is true is that if $G$ is compact, then you can always find a $G$-invariant metric. As Sam mentioned in the comments, this can be achieved via an averaging procedure.

Second, as Sam mentions, there is never a unique invariant metric, because you can always scale things. For many examples, this is the only ambiguity. That is, it is often the case that $M$ admits a unique-up-to-scaling $G$-invariant Riemannian metric. However, it is also common for $M$ to admit a multi-parameter family of $G$-invariant metrics. Here's how to tell the difference.

Suppose $G$ acts on $M$ transitively. Pick your favorite $p\in M$ and let $H = G_p = \{g\in G: G\ast p = p\}$ be the isotropy group at $p$. Then $G/H\cong M$. Writing the Lie algebra of $G$ and $H$ as $\mathfrak{g}$ and $\mathfrak{h}$ respectively, the conjugation action of $H$ on $G$ descends to an action of $H$ on $\mathfrak{g}$ which preserves $\mathfrak{h}$. If one uses, say, a bi-invariant metric on $G$, the resulting inner product on $\mathfrak{g} = T_e G$ becomes $H$-invariant, so we get an orthogonal direct sum decomposition $\mathfrak{g} = \mathfrak{h}\oplus \mathfrak{p}$. (Careful: $\mathfrak{p}$ is a vector space, but is rarely a subalgebra.)

The important proposition here is that $G$-invariant metrics on $M\cong G/H$ are in bijective correspondence with $H$-invariant inner products on $\mathfrak{p}$. (The idea of the proof here is that $\mathfrak{p}$ can be identified with $T_p M$. Given an inner product on $\mathfrak{p}\cong T_p M$, you push it around all of $M$ using $G$. The $H$-invariance guarantees the result is independent of how you push it. Conversely, given a $G$-invariant metric on $M$, restrict it to $T_p M\cong \mathfrak{p}$).

So, we need to count $H$-invariant inner products on $\mathfrak{p}$. Intepreting the $H$ action on $\mathfrak{p}$ as a representation, we get the so-called isotropy representation. Since $H$ is compact (being a closed subset of $G$), this representation splits into a sum of irreducible representations: $\mathfrak{p} = \bigoplus \mathfrak{p}_i$.

Now, as a consequence of this answer, on each $\mathfrak{p}_i$, there is, up to scaling, a unique $H$-invariant inner product. In particular, if $\mathfrak{p}$ is irreducible, then $M$ has a unique-up-to-scaling $G$-invariant metric. On the other hand, if $\mathfrak{p}$ is reducible, you can scale an invariant inner product on each one individually, giving rise to more $G$-invariant metrics on $M$.