Curvilinear coordinates from orbits

[mnb96]

Hello,

I have the following problem where I have two groups of transformations R_\alpha (rotation) and S_\lambda (scaling) acting on the plane, so that the orbits of any arbitrary point x=(x₀,y₀) under the actions of S_\lambda and R_\alpha are known (in the former case they are straight lines from the origin; in the latter case they are concentric circles with center in 0).

From this information, how can I "build" a system of curvilinear coordinates, where the coordinates are exactly the parameters (α,λ) of the transformations?

PS: I know that the answer leads to the log-polar coordinates, but I need a procedure to arrive at it.

 

[quasar987]

The rotation group is diffeomorphic to the circle S^1, while the scaling group is diffeomorphic to the real line. It is clear that if you take the standard coordinates θ on S^1 and r on R^1, then the map

(-pi,pi) x (0,infty)-->R x S --->R^2
(θ,r)------------->(R_θ,S_r)--->(R_θ\bulletS_r\bullet(1,0)

is a diffeomorphism onto R²\{the x<=0 ray} commonly referred to as "the polar coordinates on R^2"! ("\bullet" stands for the group action)

But there is no Log anywhere in this construction so perhaps this is not what you are after?

 

[mnb96]

Hi quasar987,

thanks! your explanation was pretty clear. I simply didn't think of making such observations in terms of diffeomorphisms. And by the way you are right, the log-polar is not important in this case. I was just a bit confused.

One more thing I would be very interested to know. If I replace the rotation and scaling groups with two different and not well-known groups of transformations, can we do the same? What is the procedure to find a system of curvilinear coordinates in the plane, when we know the orbits of the points under the action of the two groups?

 

[quasar987]

Well, in general, no, the analogue of the map we created for general group is not going to be a diffeomorphism. For instance, if instead of S^1 and R^1, you take two R^1's (both acting in the same way on R²), then we see fairly easily that the differential will be at every point non surjective. Similarly if we take two S^1's. The key for the above example to work was that the orbits are transverse.