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4 votes
1 answer
254 views

Isometry group of a left-invariant Riemannian metric on $\mathrm{SU}(2)$

Recall that \begin{equation} \mathbb{S}^3=\operatorname{SU}(2)=\left\{ \begin{pmatrix} z&w\\ -\bar{w}&\bar{z} \end{pmatrix} ,|z|^2+|w|^2=1 \right\} \end{...
Adterram's user avatar
  • 1,441
10 votes
4 answers
711 views

Palais's and Kobayashi's theorems on automorphism groups of geometric structures

My question concerns two results in the neighborhood of the standard theorem of Myers-Steenrod that isometry groups of Riemannian manifolds are Lie groups. Both appear in the first chapter of ...
Chris Wendl's user avatar
6 votes
0 answers
691 views

Isometries of Compact Semisimple Lie Groups

In this delightful question, the poster mentioned that the isometry group of a compact Lie group $G$, equipped with the metric from the Killing form, is $G\times G/Z(G)$, where $Z(G)$ is the center of ...
Robin Goodfellow's user avatar
7 votes
1 answer
1k views

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 $G$...
Alex M.'s user avatar
  • 5,407
12 votes
1 answer
2k views

Isometry group of pseudo Riemannian manifold always a Lie group? (Myers-Steenrod)

Myers-Steenrod states that the isometry group of a Riemannian manifold is a Lie group. Is that also true for pseudo Riemannian manifolds? I didn't find anything related to that. Cheers
Stephan Meier's user avatar