All Questions
5 questions
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{...
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 ...
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 ...
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 $G$...
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