17
votes
Accepted
Tweetable way to see Riemannian isometries are harmonic?
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 ...
16
votes
Accepted
Are there some intrinsic invariants of surfaces other than Gaussian curvature?
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 ...
13
votes
Uniform distribution of points on Riemannian manifolds
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 $\...
13
votes
Accepted
What is an example of two Banach spaces $X,Y$ such that $X$ embeds isometrically but not linearly into $Y$?
Yes, if $H$ is a nonseparable Hilbert space then it embeds isometrically into the Arens-Eells space ${\rm AE}(H)$, but not linearly isometrically, or even linearly homeomorphically. See Theorem 5.21 ...
12
votes
Accepted
Realizing mapping classes as isometries?
As you suspect, in general, no.
For example, if $M$ is compact, and $\psi:M\to M$ fixes a metric $g$ on $M$, then the closure of $\{\psi^k\ |\ k\in\mathbb{Z}\ \}$ is a compact abelian subgroup of $\...
10
votes
Accepted
Are two metric spaces isometric if they have the same $\varepsilon$-covering and $\varepsilon$-packing numbers for all $\varepsilon>0$?
Certainly no. Consider metric spaces on $n$ points and all distance 1 and 2. There are $2^{n^2/2+o(n^2)}$ such spaces. But only polynomially many different covering and packing functions.
8
votes
Do curvature differences obstruct a.e orientation-preserving isometries?
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 ...
8
votes
Can a knotted sphere isometrically embed into $\mathbb R^3$?
A spun knot will give a 2-sphere embedded in $\mathbb{R}^4$ whose intrinsic metric embeds into $\mathbb{R}^3$ isometrically as a surface of revolution.
Take a tangle $T$ in $\mathbb{R}^3_+$ with two ...
8
votes
Realizing mapping classes as isometries?
For any compact manifold you need your diffeo to be of finite order in the mapping class group. Here is a proof.
Proof. Suppose $M$ is compact and $\varphi$ is an isometry of $M$ for some metric $g$. ...
8
votes
Accepted
Are polyhedra with equilateral triangular faces rigid?
This depends on how do you define a "polyhedron". If you accept a doubly covered lozenge (two copies of two adjacent equilateral triangles), then no. But under reasonable nondegeneracy ...
7
votes
Accepted
Laplace-Beltrami and the isometry group
For the maximally symmetric case with the second order operator, the proof is very simple.
A second order differential operator acting on scalars can be written in the form
$$ b^{ij} \nabla^2_{ij} +...
7
votes
Are there some intrinsic invariants of surfaces other than Gaussian curvature?
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$...
7
votes
Accepted
Homogeneous subsets of the sphere
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)$. ...
6
votes
Accepted
A differentiable isometry is smooth?
By the nontrivial fact that $f$ is a local homeomorphism, we can assume without loss of generality that $f$ is a bijective homeomorphism. The usual textbook proof of the formula for the differential ...
6
votes
What are the applications of the Mazur-Ulam Theorem?
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 ...
6
votes
Accepted
Pogorelov's rigidity theorem vs Cohn-Vossen rigidity theorem
Below is the answer from the comments. First some terminology.
A convex surface is the boundary of a compact convex body in $\mathbb R^3$.
Each convex surface comes with two metrics: the path-...
6
votes
Accepted
Separable Banach spaces isometric to quotient of a Banach space
The answer is yes.
Following
Dowling, P. N.(1-MMOH); Lennard, C. J.(1-PITT-MS)
Every nonreflexive subspace of L1[0,1] fails the fixed point property.
Proc. Amer. Math. Soc. 125 (1997), no. 2, 443--...
6
votes
Realizing mapping classes as isometries?
To complement Dmitri Panov's answer:
Let $\phi$ be a (smooth) diffeomorphism of a closed manifold. Then $\phi$ preserves a Riemannian metric if and only if $\phi$ is isotopic to a diffeomorphism of ...
6
votes
Accepted
Are two metric spaces isometric if they have the same $\varepsilon$-covering numbers for all $\varepsilon>0$?
No. For $0 < \delta \leq 2$ let $E_\delta$ be the metric space
consisting of three points $A,B,C$ with
$d(A,B) = d(A,C) = 1$ and $d(B,C) = \delta$.
I claim that the $E_\delta$ for $1 \leq \delta \...
5
votes
If $E\oplus_\phi E \cong E\oplus_\psi E,$ does it imply that $\phi= \psi$?
It was proved by E. Behrends in Studia Math. 55, 71-85 (1976) that apart from $E=\mathbb{R}^2$ with the sup norm, which is isometric to $E$ with the $\ell_1$-norm, a Banach space $E$ admits a ...
5
votes
Accepted
If $E\oplus_\phi E \cong E\oplus_\psi E,$ does it imply that $\phi= \psi$?
It is a sketch of the proof in the case of $p\ne q$ (a complete
proof on these lines is rather lengthy). Something similar can be
done in more general case, possibly with some exceptions.
Our plan is ...
5
votes
Accepted
There is no arcwise isometry from a high dimensional manifold into a low dimensional manifold
Your proof is correct, but you need to add words "amost everywhere" at ane more place.
We use Rademacher's theorem and lemma about length of curve, which says that if a curve parametrized by length ...
5
votes
Accepted
Conformal harmonic maps in high dimensions are scaled isometries
This result is well-known in the theory of harmonic morphisms, about which, there is an extensive literature. It is a quite general fact (not depending on the conformally flat case of Euclidean space)...
5
votes
Tweetable way to see Riemannian isometries are harmonic?
Let $u: M\rightarrow N$ be a diffeomorphism. By staring at the Dirichlet energy formula and knowing that integration by parts works just as well here as for the classical case for functions, you know ...
5
votes
Accepted
Fixed points on spherical buildings
Since spherical buildings are CAT(1), we get a fixed point if $$\mathop{\rm rad}S<\tfrac \pi 2.$$
5
votes
Accepted
Does the isometry group determine the Riemannian metric?
I think that you are missing some hypotheses on the action of $G$, otherwise there are trivial counterexamples. For example, if $G\subset\mathrm{Iso}(M,g)$ is the trivial group, then one clearly ...
5
votes
Accepted
Is every 1-Lipschitz homeomorphism $f:X\to X$ from a compact metric space to itself an isometry?
It follows from 1.6.15(1) in "A Course in Metric Geometry" by Burago, Burago, and Ivanov and 1.6.15(2) is a more general statement:
Any distance-noncontracting map from a compact metric ...
5
votes
Unitary versus isometric operators
This is always true. Consider $P(t)=1-U(t)U(t)^*$. Then $T=\{t\in\mathbb R: P(t)=0\}$ is open and closed and $0\in T$. The set is open because if $t_0\in T$, then $\|P(t)\|<1$ for all $t$ ...
4
votes
Norms on $\mathbb{R}^d$ whose linear isometries are the hypercube group
Here is a complete answer in dimension 2.
Among all the norms on $\mathbb{R}^2$ which are invariant under the dihedral group $D_8$, the extremal rays are norms whose unit ball are octagons with the ...
4
votes
Accepted
How isometric action on Riemannian manifold acts on cut locus
It can happen that $g.x\in cut(x)$ for some $x$. This is what happens for $S_{n+1}$ acting by permutation of homogeneous coordinates on $\mathbb{CP}^n$.
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