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Riemannian Approximation - Lie groups and metric spheres

Math - Heisenberg09

Friday, 22 May 2009 17:38 | Author: Anton

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 ( Monday, 25 May 2009 23:51 )

Naan

Food - Breads

Tuesday, 19 May 2009 15:30 | Author: Anton

Indian flat bread that puffs up in oven. From "1000 Indian Recipes".

Ingredients:

2 tsp yeast
1 tsp sugar
1/4 cup warm water
1/2 cup yogurt
2 Tbsp oil
2 cups flour
1/4 tsp salt
1/4 cup melted butter

Instructions:

1. Dissolve yeast and sugar in water, after 5 minutes mix in yogurt and oil.
2. Mix flour and salt, slowly mix in (1), and form into a ball.
3. Let rise until doubled, 3-4 hours.
4. Split dough into 12 balls.
5. Take each ball, coat with flour and form into 7-8 inch triangle (equilateral?).
6. Place 3-4 triangles on baking tray, lightly baste with water to prevent drying out.
7. Broil 4-5 inches below heating element: about 1 minute until brown spots start to appear, then about 30 seconds turned over until golden.

Hummus

Food - Hors d'Oeuvres and Such

Sunday, 17 May 2009 21:25 | Author: Anton

Blend:

4 minced garlic cloves
2 15-oz cans of chickpeas, drained and rinsed
2/3 cup tahini (sesame seed paste)
1/3 cup lemon juice
1/2 cup water
1/4 cup olive oil
1/2 tsp salt

Last Updated ( Sunday, 17 May 2009 21:34 )

Friday, 15 May 2009 02:01 | Author: Anton

Riemannian Approximation

Math - Heisenberg09

Friday, 15 May 2009 04:51 | Author: Anton

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 ( Friday, 22 May 2009 17:36 )

Riemannian Approximation - Lie groups and metric spheres

Naan

Hummus

Riemannian Approximation

The Heisenberg 3-Sphere