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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
2 votes
0 answers
207 views

Minkowski isometries

Consider theorem 1.7 from chapter III of 'Elementary differential geometry' by O'Neill. It says that: Theorem 1.7: If $\phi$ is an isometry of $E^3 $, then there exists a unique translation $T$ and a ...
Soennecken's user avatar
1 vote
0 answers
92 views

Instantaneous rotation field in relation to a developable surface

I have a ruled surface, let it be given by $\Sigma: U \subset \mathbb{R}^2 \rightarrow \mathbb{R}^3$ parametrized by $(u,v)$ with the rulings along the $u$-lines. Now, let $X: U \subset \mathbb{R}^2 \...
RWien's user avatar
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1 vote
0 answers
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Mapping to distorted constant Gauss curvature surfaces of revolution

There are three questions here. We imagine a flexible membrane that is scrolled out so as to straighten it. 1) How can we find a surface isometrically mapped from a surface of constant negative Gauss ...
Narasimham's user avatar
1 vote
0 answers
561 views

Bending Beltrami Pseudosphere

The Beltrami Pseudosphere $$[x = a \sin p \cos t , y= - a ( \cos p + \log \tan p/2 ) , z= b+ a \sin p \sin t \; ], (.1 <p<\pi/2), (0< t< 2 \pi), \; (b>a) $$ is bent to a non-...
Narasimham's user avatar
1 vote
1 answer
345 views

What is general expression for the moment map of a Kaehler Hamiltonian G-manifold

A Kaehler Hamiltonian G-manifold is a Kaehler manifold with a Hamiltonian G-action, i.e., a G-action generated by a moment map. In particular, the Killing vector fields which generate the G-action are ...
Mtheorist's user avatar
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0 votes
0 answers
1k views

Surface locally isometric to a sphere.

If for any two points p,q in a regular, compact surface $S\subseteq R^3$, there exists an isometry $f:S\rightarrow S$ s.t. f(p)=q. How to prove that $S$ is locally isometric to the sphere?
diego0's user avatar
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