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No. Let $x$ and $y$ be connected by an edge and let's use the graph distance as our metric.
At $x$, connect paths of length $n$ for each $n\in\mathbb N$. At $y$, do the same, but also connect an infinite path.

answered Oct 6 '14 at 17:15

Bjørn Kjos-Hanssen

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23

So it turns out my earlier intuition was incorrect, and one can leverage the order properties of ${\bf R}$ to show:
Theorem. Let $X, Y$ be real Hilbert spaces, with the dimension of $X$ at least two, and let $f: X \to Y$ be a function such that $\|f(x)-f(y)\|=\|x-y\|$ whenever $\|x-y\|$ is a natural number. (We do not assume $f$ to be continuous.) Then $...

22

In genus 2, every surface has a symmetry, namely a hyperelliptic involution. In higher genus, generic surfaces will not have any symmetries. If a surface has a non-trivial symmetry group, then the quotient by the symmetry group will be an orbifold, and the moduli space of hyperbolic structures on this orbifold will have strictly smaller dimension. Hence ...

21

About embeddings, I don't know, but there is an isometric immersion of $\mathrm{SO}(3)$ with its bi-invariant metric into $\mathbb{R}^7$.
To see this, consider the natural representation $\rho_3:\mathrm{SO}(3)\to\mathrm{SO}\big({\mathcal{H}}_3\bigr)$, where $\mathcal{H}_3$ is the $7$-dimensional space consisting of the harmonic cubic polynomials on $\mathrm{...

answered Mar 25 '14 at 1:24

Robert Bryant

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19

There is a 5-point example $\ X := \{x\ y\ a\ b\ c\},\ $ with (symmetric) metrics $\ d\ $ as follows:
$$d(x\ y) = d(a\ b) = 1$$
$$d(x\ a) = d(y\ b) = 2$$
$$d(x\ b) = d(y\ a) = 3$$
$$d(x\ c) = d(y\ c) = 6$$
$$d(a\ c) = 5\qquad\qquad d(b\ c) = 4$$
Oviously, $\ B_r(x)\ $ and $\ B_r(y)\ $ are isometric for every $\ r>0,\ $ while there is no isometry $\ f:...

answered Oct 6 '14 at 22:03

Włodzimierz Holsztyński

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The answer is ``no'', the pointwise condition is not enough. The example exists in dimension 1 already and can be generalized and made arbitrary weird for all dimensions.
Consider a smooth function $v(x)$ such that
$\bullet$ it is positive for $x> 0$
$\bullet\bullet$ it is odd (i.e. $v(x)=-v(-x)$). In particular it vanishes at $x=0$.
$\bullet\...

16

No. Consider, for example, $M=\mathrm{SL}(3,\mathbb{R})/\mathrm{SO(3)}$ endowed with its $\mathrm{SL}(3,\mathbb{R})$-invariant Riemannian metric $g$ (which is unique up to a positive constant multiple). This is an irreducible symmetric space of noncompact type. The identity component of the isometry group of $(M,g)$ is $\mathrm{SL}(3,\mathbb{R})$ itself (...

answered Jul 14 '14 at 18:11

Robert Bryant

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15

Your guess seems to be true. If a map $f$ is not surjective then $f$ can be considered as a continuous map from $S^n$ to $R^n$. Hence there exist two opposite points on $S^n$ which maps to the same point by Borsuk-Ulam theorem.

15

As others have pointed out, it's not hard to show that any function $F(\kappa_1,\kappa_2)$ that is intrinsic to the surface metric must be a function of $K = \kappa_1\kappa_2$, so that settles what one might call the 'lowest-order' case. However, there are certainly higher-order versions. For example, the expression $|\nabla K|^2$ is an intrinsic invariant,...

answered Jun 6 '16 at 11:35

Robert Bryant

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Not exactly 'tweetable', but perhaps the identity (1) may help, if all you want to do is avoid the Euler-Lagrange equations. For simplicity, assume that $M^n$ is oriented. (One can write the identity (1) below as an identity on densities, so the orientability hypothesis is not essential, but I'll leave that detail for the interested.)
If $g$ is a metric ...

answered Feb 15 '17 at 15:16

Robert Bryant

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13

Yes, there are global obstructions. Consider $M= S^1 \times \mathbb R$ and $\phi$ acting by an irrational rotation on $S^1$ and by multiplication by $2$ on $\mathbb R$. There are no finite orbits, but also no invariant metrics: The function that takes a point $P$ on $S^1$ to the length of a vector pointing along $\mathbb R$ at $(P,0)$ is a function on $S^1$ ...

12

Yes. Check out Kobayashi, Transformation Groups in Differential Geometry, theorem 4.1 page 16, and example 2.5 page 8. The automorphisms of a pseudo-Riemannian manifold form a Lie group, as do the automorphisms of a conformal pseudo-Riemannian manifold (in dimension 3 or more), and the automorphisms of a projective connection.
The basic idea of the proof is ...

12

I'll just point out an answer based on somewhat different criteria than explicitly knowing features of the hyperbolic metric: As is well-known, every oriented, compact hyperbolic surface $C$ is canonically a Riemann surface of genus $g\ge2$.
As Ian points out, if $C$ is hyperelliptic, it always has a symmetry, the hyperelliptic involution $\iota:C\to C$, ...

answered Oct 30 '15 at 11:16

Robert Bryant

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12

Let $\mu=(\delta_A+\delta_B)/2$. Then the claim of Arnold - Krylov is the weak convergence of the convolutions $\mu^{*n}*\delta_x$ to the rotation invariant probability measure on the sphere (where $\mu^{*n}$ is the $n$-th convolution power of the probability measure $\mu$ on the group of rotations). A general answer to this question had been given by ...

11

In other words, you are interested in the spaces where isometric subsets are congruent.
If the metric space is locally compact and intrinsic and simply connected then you get only spheres, Euclidean spaces and hyperbolic spaces.
If not simply-conected, then in addition you get real projective spaces.
If the metric is not intrinsic you get discrete spaces ...

11

This phenomenon is to be expected: the point is to recall the classification of isometries and that indirect isometries almost always have fixed points, while direct ones almost always have no fixed points.
The fact that there are two isometries is not relevant to the boundedness of the dynamic: let us look at $U=T_2\circ T_1$. This is a random isometry, ...

answered Jul 12 '13 at 20:25

Benoît Kloeckner

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11

I assume your random $M_i$s are from $O(3)$, not $SO(3)$. In fact, I suspect that the answer depends on whether $M_1M_2$ has eigenvalue $1$ or eigenvalue $-1$.
If I did not make a mistake, when you expand $p_{2i}$, you will get essentially
$$\sum_{k=0}^{i-1} U^k x_1+ V\left(\sum_{k=1}^{i-1} U^i\right)^T x_2$$
Here $U=M_2M_1$ and $V$ is also an orthogonal ...

11

The answer is no if $M$ is assumed to be complete:
Let $x, y \in M$ such that for all $r > 0$ the balls $B_r(x)$ and $B_r(y)$ are isometric via
$$f_r : B_r(x) \to B_r(y).$$
Set $R := \inf \{ r \mid B_r(x) = M\}$ (here $R = \infty$ if $M$ is unbounded).
First I claim that
For all $0 < r < R$ we have $f_r(x) = y$:
Assume on the contrary that $...

10

Question (b) (and, therefore, also Question (a)) was answered affirmatively (for all torsion-free nilpotent groups) in 1998 by R.G.Möller and N.Seifter [Europe J. Comb. 19, 597-602] (see Theorem 4.1(1)).
The answer to Question (a) was rediscovered in 2007 by A.A.Ryabchenko [Siberian Math. J. 48 (2007) 919–922] (see the proof of Theorem 2). This short ...

answered Mar 24 '15 at 3:01

Dave Witte Morris

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9

Terry Tao's argument generalizes to show that $f$ must be an isometry whenever $X$ has dimension greater than 1 and $Y$ is strictly convex. We start by proving a series of lemmas:
Lemma 1: Let $x,y\in X$, and let $a,b\geq0$ be such that $a+b\geq\|x-y\|$ and $|a-b|\leq\|x-y\|$. Then there exists $z\in X$ such that $\|x-z\|=a$ and $\|y-z\|=b$.
Proof: Let $...

8

the nine matrix elements of $SO(3)$ represent a vector in $R^9$, see Isometric Embedding for Homogeneous Compact 3-Manifolds (1996).

answered Mar 24 '14 at 21:26

Carlo Beenakker

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8

Let me add to this nice answer of Ian Agol, that the isometry group is always finite, and contains at most $84(g-1)$ elements, if we count orientation-preserving (conformal) isometries, where $g\geq 2$ is the genus (Hurwitz, Math Ann, 41 (1893)). For an exposition in English, see the book by Tsuji, Potential theory in modern function theory.
EDIT. Let me ...

answered Oct 30 '15 at 2:38

Alexandre Eremenko

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8

There is a discontinuous map $f\colon\mathbb{S}^2\to\mathbb{R}^2$ such that $d_xf$ is defined and isometric for almost all $x$. (If you want a continuous one then I am sure the answer is "no")
To construct such $f$ do the following:
Start with a sequence of finer and finer subdivision $(K_n)$ of $\mathbb{S}^2$ into polygons; say next subdivision divedes ...

answered Mar 17 '17 at 21:05

Anton Petrunin

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7

This is the answer to the original question (Not the one which is posted now).
Look in two papers, mine and the paper of Enrico Le Donne.
You are looking for spaces which admit length-preserving embedding into Hilbert space.
In my paper I prove that a compact length spaces which (roughly) admit a length-preserving map into Euclidean $m$-space has to be ...

answered Jan 19 '14 at 19:01

Anton Petrunin

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7

You should look at Andrew Hick's thesis,
Andrew Hicks, Group actions and the topology of nonnegatively curved $4$-manifolds, Illinois Journal of Mathematics. Volume 41, Issue 3 (1997), 421-437.
A corollary to his Theorem 2 shows that for a positively curved metric on $S^2\times S^2$ with $\delta$ pinching, the size of the isometry group is bounded above ...

7

A surfaces of constant curvature $K$ admit number of local embedding into $\mathbb{E}^3$ as the surfaces of revolution. Direct calculations show that any pair $k_1$ and $k_2$ such that $K=k_1\cdot k_2$ appear this way.
So, "yes", any $P(x,y)=F(x\cdot y)$ for some $F$.

answered Jun 3 '16 at 15:49

Anton Petrunin

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7

The answer is 'no'.
For example, let $H$ be the (5-dimensional) space of symmetric, traceless $3$-by-$3$ matrices, where the Hilbert inner product is $\langle x,y\rangle = \tfrac12\mathrm{tr}(xy)$. Let $A\subset S$ be the subset consisting of those matrices of unit norm for which $0$ is an eigenvalue. Then $A$ is connected and $\mathrm{SO}(3)$ acts ...

answered Jan 3 '17 at 16:08

Robert Bryant

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7

The Mazur-Ulam theorem is used in Bader, Uri; Furman, Alex; Gelander, Tsachik; Monod, Nicolas Property (T) and rigidity for actions on Banach spaces. Acta Math. 198 (2007), no. 1, 57–105 in order to simplify the theory of isometric actions on Banach spaces. In this connection it is worth mentioning that the question on an analogue of the Mazur-Ulam theorem ...

6

Let $k$ denote the value of the constant sectional curvature of $M^m$ and let $X_k$ denote the unique simply-connected complete $m$-dimensional manifold of the constant curvature $k$. Then, since $M$ is simply-connected, there exists a map $dev: M^m\to X_k$, called a developing map of the metric on $M$, which is a local isometry (it is not, in general 1-1). ...

6

If you restrict sets A, B to be simply points, then you are asking for spaces which have the property that for all points A,B\in X there exists an isometry T from X onto X so that T(A)=B (am I understanding you correctly?). Such spaces are called transitive. It is known that if a finite dimensional space is transitive then it is isometric to a Euclidean ...

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