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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 $\ma …
Robert Bryant's user avatar
22 votes

Isometric embedding of SO(3) into an euclidean space

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:\mat …
Robert Bryant's user avatar
7 votes
Accepted

Homogeneous subsets of the sphere

More generally, if $G$ is a connected compact Lie group acting irreducibly by isometries on a (finite-dimensional) Hilbert space $H$, then each of the orbits of $G$ in the unit sphere $S\subset H$ will …
Robert Bryant's user avatar
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 …
Robert Bryant's user avatar
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 cann …
Robert Bryant's user avatar
13 votes

Isometry group of a left-invariant Riemannian metric on $\mathrm{SU}(2)$

Up to a scalar multiple, it is the standard round sphere, so the group of isometries of $g$ is $\mathrm{O}(4)$. …
Robert Bryant's user avatar
19 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 identit …
Robert Bryant's user avatar
16 votes
Accepted

The surjectivity of the exponential map for the isometry group

You'll run into the same problem with the conformal group in this example, because the space of conformal transformations in this example is the same as the space of isometries. …
Robert Bryant's user avatar
2 votes
Accepted

Are all symmetries of the Dirichlet functional isometries?

The answer to your question is "yes, every symmetry (in the sense you have specified) is an isometric immersion". To see why, first note that, if $(M,g)$ is a compact Riemannian $m$-manifold and $h$ …
Robert Bryant's user avatar
13 votes

Isometry group of a compact hyperbolic surface

Then the group of orientation preserving isometries of $C$ will be an extension of $\Gamma$ by a $\mathbb{Z}_2$ (because of the hyperelliptic involution). … The orientation-preserving isometries of $C$ are exactly the elements of $\mathrm{PSL}(3,\mathbb{C})$ that preserve the quartic $Q$. …
Robert Bryant's user avatar
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 on …
Robert Bryant's user avatar