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Does the isometry group determine the Riemannian metric?
Suppose $G \subset \text{Iso}(M)$ is a Lie group acting smoothly on a (pseudo-)Riemannian manifold $(M, g)$. Then $G$ induces a Lie algebra of Killing vectors on $M$. In this paper by Goenner and ...
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7
votes
1
answer
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The surjectivity of the exponential map for the isometry group
Little is known on general conditions guaranteeing that the exponential map between a Lie algebra and an associated Lie group is surjective.
Let $M$ be a noncompact connected Riemann manifold, and $G$...
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