Let $M$ be a differentiable manifold and suppose I have a group action $G \subseteq {\rm Diff}(M)$ where $G$ is a finite-dimensional Lie group (not necessarily compact). Does there exist a theory to find the metric $g$ which has $G$ as isometries (assuming one exists)? For example if I set $M = \mathbb{S}^2 \subset \mathbb{R}^3$ and consider the matrix group $G = SO(3)$ acting in the usual way on $\mathbb{R}^3$ I would like to discover the round metric on $\mathbb{S}^2$ (or the Euclidean metric on $\mathbb{R}^3$).
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4$\begingroup$ If $G$ is compact, there is always a $G$-invariant metric over $M$, by integration over $G$. For General $G$, it should not be true $\endgroup$– Feng HaoCommented Jun 27, 2017 at 1:52
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3$\begingroup$ The isometry group is Lie group that acts properly and conversely, any proper effective action of a Lie group is isometric in some Riemannian metric. The former is proven e.g. in Kobayashi-Nomidzy textbook, vol I, Theorem I.4.7. The latter is due to Palais Theorem 4.3.1 in [On the existence of slices for actions of non-compact Lie groups, Ann. of Math. 73 (1961), 295–323.] See vmm.math.uci.edu/ExistenceOfSlices.pdf. $\endgroup$– Igor BelegradekCommented Jun 27, 2017 at 3:10
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