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4 votes
1 answer
255 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
1 answer
129 views

$\operatorname{Hess}r$ is scalar matrix $\implies$ $M$ is isometric to the space form

I'm trying to prove the rigidity part of a theorem in my paper, which requires the use of the classical Hessian comparison theorem's rigidity part: $$\DeclareMathOperator\sn{sn}\operatorname{Hess}r=\...
HeroZhang001'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
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
6 votes
0 answers
149 views

What is the minimum $n$ for which $\Bbb H^3$ can be isometrically embedded in $\Bbb R^n$ as a bounded set?

Consider the hyperbolic $3$-space $\Bbb H^3$ (i.e., the unique, simply-connected, $3$-dimensional complete Riemannian manifold with a constant negative sectional curvature equal to $-1$). The Nash ...
Random's user avatar
  • 1,097
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
9 votes
1 answer
334 views

Can a knotted sphere isometrically embed into $\mathbb R^3$?

All smooth simple closed curves in $\mathbb R^3$ (knotted or not) can be isometrically embedded into $\mathbb R^2$ as a circle of equal arclength. The situation for knotted spheres seems more ...
Christian Bueno's user avatar
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
15 votes
1 answer
2k views

Uniform distribution of points on Riemannian manifolds

Recently, I came across a beautiful paper by Arnol'd and Krylov (Uniform distribution of points on a sphere...) that contains the following theorem: Theorem: Let A and B be two rotations of the ...
José Navarro's user avatar
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
1 vote
0 answers
187 views

Does there exist an isometry between a regular polygon and a circle?

In order to define the question in a meaningful fashion, I am referring to a smooth manifold $\mathcal{M}$ within an $\epsilon$-neighborhood of a regular polygon $\mathcal{P}$ satisfying $$\max\{\|x-p\...
Talmsmen's user avatar
  • 547
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
16 votes
2 answers
967 views

Tweetable way to see Riemannian isometries are harmonic?

$\newcommand{\al}{\alpha}$ $\newcommand{\euc}{\mathcal{e}}$ $\newcommand{\Cof}{\operatorname{Cof}}$ $\newcommand{\Det}{\operatorname{Det}}$ Smooth Riemannian isometries are harmonic. Can one conclude ...
Asaf Shachar's user avatar
  • 6,741
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
3 votes
1 answer
454 views

Pogorelov's rigidity theorem vs Cohn-Vossen rigidity theorem

There is the following rigidity theorem of Cohn-Vossen as stated on p. 86 of these lecture notes: http://www.math.brown.edu/~deigen/chern.pdf Any isometry between two closed smooth convex surfaces (...
asv's user avatar
  • 21.8k
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
517 views

Local isometry implies covering map: nonempty boundary case [closed]

The following theorem is well known in the literature: Let $M$ and $N$ be riemannian manifolds and let $f : M \to N$ be a local isometry. If $M$ is complete and $N$ is connected, then $f$ is a ...
Eduardo Longa's user avatar
4 votes
0 answers
101 views

Isometries of fiber bundles

Let $F\to S\overset{\pi}{\to} B$ a Riemannian submersion with totally geodesic fibers. Question: How much information about the isometries of $S$ we have if we know the isometries of $F$ and $B$? For ...
Dinisaur's user avatar
  • 223
6 votes
1 answer
185 views

Cohn-Vossen rigidity theorem in hyperbolic space

There is the following rigidity theorem of Cohn-Vossen as stated on p. 86 of these lecture notes: http://www.math.brown.edu/~deigen/chern.pdf Any isometry between two closed smooth convex surfaces in ...
asv's user avatar
  • 21.8k
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
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
14 votes
2 answers
506 views

Do curvature differences obstruct a.e orientation-preserving isometries?

Is there an example of a pair $M,N$ of connected, oriented equidimensional Riemannian manifolds with the following properties: $M$ is everywhere non-flat, $N$ is flat. There exist a map $f:M \to N$ ...
Asaf Shachar's user avatar
  • 6,741
11 votes
1 answer
486 views

About Palais' remark that an isometry of Riemannian manifolds does not induce an isometry of the Hilbert manifolds of curves

In the paper ``Morse theory on Hilbert manifolds'' (1963), on page 326, Richard Palais makes a remark that if $\phi\colon V \to W$ is an isometry (of submanifolds of $\mathbb{R}^n$), then this does ...
Jaap Eldering's user avatar
1 vote
1 answer
142 views

How does this orthogonality follow from the map being an isometry?

This is a step of a proof in the book Variational Problems in Geometry by Seiki Nishikawa. I will ignore the background and change some of the statements and notations for simplicity. Let $(M,g)$ ...
trisct's user avatar
  • 283
21 votes
1 answer
1k views

A differentiable isometry is smooth?

I posted this question in MSE but got no response (even after giving a bounty), so I am trying here. Let $M,N$ be smooth $d$-dimensional Riemannian manifolds. Suppose $f:M \to N$ is a differentiable ...
Asaf Shachar's user avatar
  • 6,741
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
9 votes
2 answers
499 views

There is no arcwise isometry from a high dimensional manifold into a low dimensional manifold

$\newcommand{\al}{\alpha}$ $\newcommand{\ga}{\gamma}$ $\newcommand{\e}{\epsilon}$ Let $X,Y$ be Riemannian manifolds, such that $\dim(X) > \dim(Y)$. I am trying to prove the following statement (...
Asaf Shachar's user avatar
  • 6,741
16 votes
1 answer
601 views

If all balls around two points are isometric... -- manifold version

This question is a natural follow-up of this other question, asked earlier today by wspin. Let's say that a metric space $(X,d)$ has two poles if: there are two distinct points $x$, $y$ such that ...
Marco Golla's user avatar
  • 10.9k
1 vote
2 answers
383 views

Conformal harmonic maps in high dimensions are scaled isometries

This is a cross-post from MSE (where I got no answer). It is well-known that conformal maps between $2$-dimensional Riemannian manifolds are harmonic. I discovered lately that in dimension $d>2$, ...
Asaf Shachar's user avatar
  • 6,741
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
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
7 votes
1 answer
373 views

Are metric isometries smooth at the boundary?

Let $M,N$ be smooth Riemannian manifolds with boundary (In particular, we assume the boundaries are smooth). Suppose we have a map $\phi:M \to N$ which satisfies the following properties: $$(1) \, \,...
Asaf Shachar's user avatar
  • 6,741
6 votes
2 answers
208 views

Are all symmetries of the Dirichlet functional isometries?

This is a cross-post from MSE (no answer there). Let $M,N$ be oriented $d$-dimensional Riemannian manifolds, $M$ compact*, and let $f:M \to N$ be smooth. Consider the Dirichlet energy functional: $...
Asaf Shachar's user avatar
  • 6,741
8 votes
1 answer
320 views

Does nonexpanding map between manifolds decrease volume?

(This question is a special case of a question I asked at SE, which got no answer there) Let $M,N$ be diffeomorphic connected compact Riemannian manifolds, and let $f:M \to N$ be a surjective ...
Asaf Shachar's user avatar
  • 6,741
3 votes
1 answer
1k views

Isometry groups of Riemannian submersions with totally geodesic fibers

Suppose $F\to M\stackrel{\pi}{\to} B$ is a Riemannian submersion with totally geodesic fibers, all manifolds compact. In general, unless $M=B\times F$ is a Riemannian product, the isometry groups of ...
Renato G. Bettiol's user avatar
4 votes
1 answer
194 views

How isometric action on Riemannian manifold acts on cut locus

Assume that $M$ is a simply connected closed Riemannian manifold with no boundary and nonnegative sectional curvaure Assume that ${\bf Z}_n=(g),\ n\geq 3$ acts on $M$ isometrically. Then if $gx=x$, i....
Hee Kwon Lee's user avatar
  • 1,100
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
1 vote
1 answer
285 views

Obstruction to the existence of global isometries on a constant-curvature Riemannian manifold

Let $M$ be an $m$-dimensional simply connected Riemannian manifold that is not geodesically complete. Suppose $M$ has constant sectional curvature. Because the curvature is constant, locally $M$ ...
Josh Burby'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
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
1 vote
1 answer
207 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
7 votes
0 answers
669 views

Homometric $\Rightarrow$ isometric?

Suppose you know that there is a mapping between two Riemmanian manifolds $M_1$ and $M_2$ such that, for each $x_1 \in M_1$, the (codimension-1) measure of the set of points at distance $d$ from $x_1$ ...
Joseph O'Rourke's user avatar