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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
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
  • 245
6 votes
1 answer
370 views

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 ...
Katerina's user avatar
  • 203
7 votes
2 answers
257 views

Are polyhedra with equilateral triangular faces rigid?

Convex polyhedra are rigid by Cauchy’s theorem. Steffen’s polyhedron is an example of a non-convex polyhedron that is flexible (i.e., non-rigid). However, it appears to have edges of different lengths....
Hussein's user avatar
  • 264
-1 votes
1 answer
244 views

Isometry between Minkowski space and Tangent space in an article by Stefan Waldmann [closed]

In the notes Geometric Wave Equations by Stefan Waldmann at page 70 they have Having a fixed Lorentz metric $g$ on a spacetime manifold $M$ we can now transfer the notions of special relativity, ...
amilton moreira's user avatar
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
8 votes
3 answers
629 views

Realizing mapping classes as isometries?

Let $\phi : M \to M$ be a diffeomorphism. Is there a metric $g$ on $M$ and a diffeomorphism $\psi$ isotopic to $\phi$ so that $\psi$ is an isometry with respect to $g$? I'm guessing the answer is no,...
user101010's user avatar
  • 5,349
2 votes
1 answer
101 views

Characterization of extrinsic distance prevserving embedding (see the definition given!) from low dimensional Euclidean spaces to high dimensions

P.S. I asked the question on MSE more than a week ago, but didn't get any desired answer, so asking here. Let $m < n \in \mathbb{N}$. Let us equip $\mathbb{R}^m, \mathbb{R}^n $ with their ...
Learning math's user avatar
1 vote
0 answers
65 views

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
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
6 votes
1 answer
1k views

Laplace-Beltrami and the isometry group

H$\vphantom{a}$i. Consider the Laplacian on $\mathbb R^n$, $$ \Delta=\partial_i^2 $$ It is easy to prove that the most general differential operator that commutes with rotations and translations is ...
AccidentalFourierTransform'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
  • 1,155
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
17 votes
2 answers
1k views

Are there some intrinsic invariants of surfaces other than Gaussian curvature?

The principal curvatures of a surface is denoted by $\kappa_{1}, \kappa_{2}$. Let $P(x,y)$ be a polynomial with real coefficients. Assume that $P(\kappa_{1}, \kappa_{2})$ is an intrinsically ...
Ali Taghavi's user avatar
2 votes
1 answer
261 views

isometric action on the $n$-sphere

Let $S^n$ be the $n$-sphere. If $n=2k+1$ is odd, then we can identify $S^n$ as a subset of $\mathbb{C}^{k+1}$. We define the $S^1$ action on $S^n$ by multiplication, namely $$ \Psi \colon S^1 \times ...
Olorin's user avatar
  • 501
13 votes
3 answers
2k views

Isometry group of a compact hyperbolic surface

Consider a compact surface $M$ of genus $g \geq 2$ with a metric of constant negative curvature. My question is, is it known under what sorts of sufficient conditions such a metric will have non-...
user82102's user avatar
  • 133
6 votes
1 answer
734 views

Are the Sasaki metrics on tangent and cotangent bundle isomorphic?

Let $(M,g)$ be a Riemannian manifold. Then there is the well-known Sasaki metric that makes $(TM,\hat{g})$ a Riemannian manifold. In a similar way, one can construct a Sasaki metric $\bar{g}$ on the ...
Jaap Eldering's user avatar
13 votes
3 answers
2k views

Is there a global obstruction for a diffeomorphism to be an isometry?

Let $V$ be a finite dimensional vector space. Let us call an automorphism $T:V\rightarrow V$ admissible if there exists an inner product $\langle , \rangle$ on $V$ making $T$ an isometry. We know $T$...
Asaf Shachar's user avatar
  • 6,741
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
25 votes
6 answers
3k views

Isometric embedding of SO(3) into an euclidean space

Consider $SO(3)$ with its bi-invariant metric and $R^n$ the euclidean space of dimension $n$. What is the minimal value of $n$ such that there exists an isometric embedding $f: SO(3) \to R^n$?
Mauro Patrão's user avatar
5 votes
2 answers
2k views

Isometric embeddings of metric spaces in Hilbert spaces

There are plenty of isometric embeddings of metric spaces in Banach spaces. Nevertheless, I have been unable to find any significant result on isometric embeddings into Hilbert spaces. My question is: ...
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
1 vote
1 answer
206 views

Riemann isometry vs Euclidean bi-Lipschitz mapping

Assume that $\gamma$ is a rectifiable Jordan curve in the complex plane of length $2\pi$. Then there exists a Riemann isometry $f$ between $\gamma$ and the unit circle $T$. My question is, does this ...
djoke's user avatar
  • 303
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
  • 1
7 votes
1 answer
497 views

Open problems about CMC hypersurfaces with symmetries?

Recently, Andrews and Li announced a complete classification of CMC ($H=const.$) tori in $S^3$, confirming a conjecture of Pinkall and Sterling. Their main result is that any such torus is ...
Renato G. Bettiol's user avatar
18 votes
1 answer
980 views

Possible isometries of a positively curved $S^2\times S^2$

Just to put things in perspective, recall that the Hopf Conjecture asks whether $S^2\times S^2$ admits a metric of positive sectional curvature. By the work of Hsiang-Kleiner, it is known that, if $S^...
Renato G. Bettiol's user avatar
5 votes
4 answers
2k views

Testing for Riemannian isometry

In most physics situations one gets the metric as a positive definite symmetric matrix in some chosen local coordinate system. Now if on the same space one has two such metrics given as matrices then ...
Anirbit's user avatar
  • 3,541