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Approximating Heisenberg Geodesics

Math - Heisenberg09
Written by Anton
Tuesday, 23 June 2009 17:38
A Riemannian approximation of the Heisenberg group is given by declaring the following frame field orthonormal in $\mathbb{R}^3$ : $X_s = \frac{\partial}{\partial x} - y \frac{s}{2} \frac{\partial}{\partial t}, \;\; Y_s = \frac{\partial}{\partial x} - y \frac{s}{2} \frac{\partial}{\partial t} \;\; T_s = \sqrt{1-s} \frac{\partial}{\partial t}$ (note that the $T_s$ has changed from the earlier versions). There are many ways to approximate horisontal curves in $\mathbb{H}$ by geodesics in the approximations $\mathbb{H}_s$ . Here's one approximation, going from a sphere in Euclidean space to the apple sphere in Heisenberg space, by way of apple sets in the intermediate spaces:
Last Updated on Tuesday, 23 June 2009 18:10
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HeisenbergGeometry0.1.nb

Math - Heisenberg09
Written by Anton
Thursday, 11 June 2009 14:41
The following notebook contains the Heisenberg geometry routines developed during spring 09. It will be the starting point for this summer's project on Heisenberg geometry visualization. If you make updates, please read the instructions in the notebook. HeisenbergGeometry0.1.nb

More on Metric Spheres of Riemannian Approximations

Math - Heisenberg09
Written by Anton
Monday, 25 May 2009 13:15
(in progress) Since the spaces $\mathbb{H}_s$ are Lie groups, it suffices to analyze their metric properties at the origin. Recall that the geodesic equations are: $x\prime\prime+k y\prime = 0$ $y\prime\prime - k x\prime = 0$ $k\prime=0$ $k = \frac{s}{(1-s)^2}\left(z\prime +\frac{s}{2}(x\prime y - y\prime x)\right)$ Solving these starting at the origin and assuming unit speed, $x(t) = a (\cos(kt) -1) + b \sin(kt)$ $y(t) = a \sin(kt) - b(\cos(kt) - 1)$ $z(t) = \frac{s}{k}t + \frac{s}{2} \left(a^2+b^2\right) \sin (k t)$ For reference, $x\prime(0)=k b$ $y\prime(0) = k a$ $z\prime(0) = \frac{s}{k} + k \frac{s}{2}(a^2+b^2)$
Last Updated on Monday, 25 May 2009 13:40

Riemannian Approximation - Lie groups and metric spheres

Math - Heisenberg09
Written by Anton
Friday, 22 May 2009 17:38
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.
Last Updated on Monday, 25 May 2009 23:51
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Riemannian Approximation

Math - Heisenberg09
Written by Anton
Friday, 15 May 2009 04:51
The fields on the Heisenberg group $\mathbb H$ that gives it its sub-Riemannian and metric structure is given by $X = \frac{\partial}{\partial x} - \frac{1}{2}y \frac{\partial}{\partial t}, Y = \frac{\partial}{\partial y} + \frac{1}{2}x \frac{\partial}{\partial t}$ These are declared orthonormal, and only lines whose velocities are given by $aX + bY$ have a (finite) speed. One may view this setup as a degenerate Riemannian structure, where a third vector field (say, $s \frac{\partial}{\partial t}$ ) has degenerated as s was sent to zero. The Riemannian approximation I find natural interpolates between Euclidean and Heisenberg geometries, and has not yet been explored. Here are the frame fields: $X_s = \frac{\partial}{\partial x} - \frac{s}{2} y \frac{\partial}{\partial t}, Y_s = \frac{\partial}{\partial y} + \frac{s}{2} x \frac{\partial}{\partial t}, T_s = (1-s) \frac{\partial}{\partial t}$ At s=0, this is the standard Euclidean frame field, and at s=1 it degenerates to the Heisenberg structure. In between, it gives Riemannian structures on $\mathbb R^3$ with properties that degenerate as s goes to 1. As with any Riemannian structure, geodesics through a given point with assigned direction exist and are unique. They can be found by solving the geodesic equation $\nabla_{\gamma\prime} \gamma\prime = 0$ for the Levi-Civita connection corresponding to the orthonormal frame field. Given a curve $\gamma = (x,y,z)$ and some calculations, the condition on $\gamma$ becomes: $(1-s)^2 x\prime\prime + s y\prime ( z\prime + \frac{s}{2}( x\prime y - y\prime x) )=0$ $(1-s)^2 y\prime \prime - s x\prime ( z\prime + \frac{s}{2}( x\prime y - y\prime x) )=0$ $( z\prime + \frac{s}{2}( x\prime y - y\prime x) )\prime =0$ These equations are easily solved. The following animations show, as s changes, geodesics leaving the origin with fixed initial velocity. The length of the geodesics changes with s since otherwise they quickly become too small to see.
Last Updated on Friday, 22 May 2009 17:36
Read more... [Riemannian Approximation]

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