4
$\begingroup$

I am now hopelessly confused:

Metric

If $u$ and $v$ are two vectors in real ''n''-dimensional vector space $R_n$ with the usual Euclidean norm, both of which have norm less than 1, then we may define an isometric invariant by

$ \delta (u, v) = 2 \frac{\lVert u-v \rVert^2}{(1-\lVert u \rVert^2)(1-\lVert v \rVert^2)} \ $

where $\lVert \cdot \rVert $ denotes the usual Euclidean norm. Then the distance function is

$d(u, v) = \operatorname{arcosh} (1+\delta (u,v)) $

Such a distance function is defined for any two vectors of norm less than one, and makes the set of such vectors into a metric space which is a model of hyperbolic space of constant curvature −1. The model has the conformal property that the angle between two intersecting curves in hyperbolic space is the same as the angle in the model.

Does this quote mean that the Poincare disk model is isometric? (I doubt this very much)

Or are "isometric invariant" and "isometric" just terms that have the sad suggestion that they have a similar meaning but mean something completely different?

What does "isometric invariant" mean?

This question came up when I saw the answer of Behaviour on the question Models of the hyperbolic plane?

Behavior wrote:

Poincaré disk is an isometric model of the hyperbolic plane, and so is the halfplane with the hyperbolic metric. Yes, they share all properties that the hyperbolic plane has. They serve as concrete representatives of the isometry class of complete surfaces of constant curvature −1.

What seems to me an direct contradiction of Hilberts theorem ( https://en.wikipedia.org/wiki/Hilbert%27s_theorem_%28differential_geometry%29 )

can anybody shed a lot of light in this?

$\endgroup$
1
  • $\begingroup$ the word isometric is used in various distinct ways here. There is no surface in $\mathbb R^3$ which, using the inherited metric from the ambient space, has constant curvature $-1,$ is simply connected, complete, and so on. $\endgroup$
    – Will Jagy
    Commented Dec 28, 2014 at 22:20

1 Answer 1

5
$\begingroup$

The concept of isometry requires a Riemannian manifold, and you must be careful not to separate the manifold from its metric.

If you have a manifold $M$ with metric $g$, and a second manifold $N$ with metric $g'$, then the two manifolds are isometric if there exists a diffeomorphism $\phi: M\to N$ that preserves the metric, i.e. $g(v,w) = g'(\phi_*v, \phi_*w)$. The Poincar$\acute{\textrm{e}}$ disk is isometric to the hyperbolic plane in this sense; note that the metric of the disk is not the Euclidean metric (it is given further down in the article you've linked).

A manifold $M$ with metric $g$ can be isometrically embedded in $\mathbb{R}^n$ if there exists a submanifold $N$ of $\mathbb{R}^n$ with the induced Euclidean metric which is isometric to $M$. The hyperbolic plane cannot be isometrically embedded into $\mathbb{R}^2$, but the disk model is not a contradiction, since the metric of the disk model is not the Euclidean metric.

So the two terms do not have completely different meaning; an isometric embedding implies existence of an isometry to a submanifold in Euclidean space with one particular metric.

$\endgroup$
2
  • $\begingroup$ Am I then correct in thinking that for example the Beltrami–Klein model en.wikipedia.org/wiki/Beltrami%E2%80%93Klein_model is also an isometric model ? , makes me wonder will it not be hard to find an non isometric model $\endgroup$
    – Willemien
    Commented Dec 29, 2014 at 19:56
  • $\begingroup$ Yes, every embedding can be made an isometry by pulling back the metric. For these models, the metric has a particularly nice form. $\endgroup$
    – user7530
    Commented Jan 5, 2015 at 17:12

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .