Riemannian Approximations as Lie groups

The Heisenberg group is given by the group law

(x_1,y_1,t_1)+(x_2,y_2,t_2) = \left(x_1+x_2, y_1+y_2, t_1+t_2+\frac{1}{2}( x_1 y_2 - y_1 x_2) \right)

The left action leaves the sub-Riemannian structure, and thus the metric, invariant.

The frame fields for the Riemannian approximations is also invariant under a group law (one for each s):

 (x_1,y_1,t_1)+(x_2,y_2,t_2) = \left(x_1+x_2, y_1+y_2, t_1+t_2-\frac{s}{2}( x_1 y_2 - y_1 x_2) \right)

Note that the negation in the t coordinate. This just means I flipped the t coordinate when writing down the approximations, and I'll go through and fix that.

The approximationg spaces \mathbb{H}_s are then Lie groups in their own right, endowed with a Riemannian structure. These groups are again two-step nilpotent (except for s=0), so in particular not semi-simple. So I guess what we have is an approximation of the Heisenberg group by other Carnot groups very similar to \mathbb{H} and degenerating metrics.

Looking for spheres

How well are the metric properties of \mathbb{H} approximated by the spaces \mathbb{H}_s? Not too well, it turns out.

The unit sphere around the origin is a basic object to analyze, and can (naively, it turns out) be thought of as the image of the unit sphere under the exponential map. That is, I take all the paths leaving the origin at unit speed and see where they end up after unit time.

Here is what seems to happen to the unit sphere at the origin as s varies from 0 to 1:

 

The "spheres" swirl around nicely enough, but they surely don't converge to the "apple" of Heisenberg space. Instead, all the unit-speed geodesics from the origin shrivel up and die, unless they are coming out along the XY plane: any T component to the speed is fatal as s increases. Not very good.

 However, these aren't actually the unit spheres in \mathbb{H}_s.Geodesics are only distance-minimizing for some neighborhood of the identity, and the injectivity radius of the exponential map reduces to 0 as s goes to 1.

Recall that the distance in a Riemannian space is defined as the length of the shortest path joining the two points, defined as the integral of the norm of the derivative (the shortest path might not exist, so we take the infemum). In a Riemannian structure the shortest paths exist and are unique between points that are sufficiently close to each other.

Consider now a point on the Heisenberg sphere. What is the distance from the origin to it in \mathbb{H}_s? At most the s-length of the corresponding Heisenberg geodesic.

One such unit-speed geodesic \gamma is given by

x(t) = \sin(t), y(t) = \cos(t)-1, z(t) = (t-\sin(t))/2

Intererstingly, the derivative \gamma \prime is written as a linear combination of the frame fields the same way for every space \mathbb{H}_s. For any s\neq 1,

\gamma \prime = \cos(t) X + \sin(t) Y + (1-\cos(t))/2 T

In particular, the length of the curve is the same for every space \mathbb{H}_s and, unfortunately, equal to ~1.00551, not 1.

This has several implications:

1. The s-length of a horisontal curve does not converge to its Heisenberg length.

2. The s-distance between two points might not converge to the Heisenberg distance.

3. The "sphere" images above do not show spheres, since shortening \gamma slightly gives a point of distance at most 1 from the origin for every s.This point stays above the XY plane for all s.

However, the velocities of Heisenberg horisontal curves seem to have a constant representation in every approximating space, and the curves themselves have constant length.

 

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Last Updated ( Monday, 25 May 2009 23:51 )