Locally but not globally symmetric manifold isometrically embedded in $\mathbb{R}^3$?

Question

I'm trying to get a feeling for the distinction between locally symmetric and globally symmetric Riemannian manifolds.

So I'm wondering, is there a nice example of a locally symmetric manifold which is isometrically embedded in $\mathbb{R}^3$ and is not globally symmetric?

One example is an open unit disc, which is not complete hence not globally symmetric for a rather stupid reason.

Can you give me an example for a Riemannian manifold which is

locally symmetric, but not globally symmetric
isometrically embedded in $\mathbb{R}^3$, so I can 'draw it' or at least 'imagine its shape'
and complete?

There are no such examples. – Moishe Kohan Sep 12 '17 at 12:36 — Moishe Kohan, Sep 12 '17 at 12:36
@Moishe: Could you elaborate? – Jason DeVito - on hiatus Sep 12 '17 at 17:25 — Jason DeVito - on hiatus, Sep 12 '17 at 17:25
@JasonDeVito: See the answer. – Moishe Kohan Sep 13 '17 at 18:52 — Moishe Kohan, Sep 13 '17 at 18:52

score 2 · Accepted Answer · answered Sep 12 '17 at 20:03

First of all, by "isometrically embedded", I assume isometrically embedded with smoothness at least $C^2$. Then:

Every if a locally symmetric complete connected Riemannian surface embeds isometrically into $E^3$, then this surface is globally symmetric (apart maybe from the Moebius band).

A proof is case-by-case analysis.

Let's start with the case of negative curvature. Hilbert proved that a complete simply-connected surface of constant negative curvature (also known as a hyperbolic plane) does not admit an isometric immersion in $E^3$. Since every complete surface of constant negative curvature has universal cover isometric to a hyperbolic plane, it follows that none of them admits an isometric immersion into $E^3$.
Next, consider surfaces of constant positive curvature: up to rescaling, there are only two, the round sphere (globally symmetric) and projective plane (also globally symmetric), but also does not embed (topologically) into $E^3$.
Lastly, consider the zero curvature case. There are 4 of these: (a) flat plane (globally symmetric), product cylinder (globally symmetric), Moebius band (only locally symmetric), flat torus (globally symmetric no matter which flat metric you put on it - nice exercise in Euclidean geometry), the Klein bottle (does not embed in $E^3$ for topological reasons).

This leaves out the Moebius band $M$. For topological reasons, there are no proper embeddings $M\to E^3$. You do not require properness, but completeness. There is a slim chance that such an embedding of the Moebius band exists, but then it will be quite nasty as you approach its ideal boundary (since you want completeness but not cannot have properness).

One last thing: isometric embeddings into Euclidean spaces (which do exist if the codimension is high enough) will tell you nothing about geometry of locally symmetric spaces. The latter should be studied either intrinsically or by considering actions of their fundamental groups on globally symmetric spaces. My suggestion is to take a closer look at the complete flat and hyperbolic metrics on the Moebius band from this viewpoint to get started.

I had forgotten that a locally symmetric, complete, simply connected manifold is globally symmetric (and thus, homogeneous). Thanks for adding the details! — Jason DeVito - on hiatus, Sep 13 '17 at 20:32

Locally but not globally symmetric manifold isometrically embedded in $\mathbb{R}^3$?

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