Given a finite-dimensional Lie Group $G$, is there always a Riemannian manifold $M$, such that $G$ is the group of isometries of $M$?
2026-05-06 03:22:27.1778037747
Is every Lie group the automorphism group of a riemannian manifold?
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As Torsten Hĕrculĕ Cärlemän points out, this question has been answered on MathOverflow:
Can every Lie group be realized as the full isometry group of a Riemannian manifold?