# A Special Group, Volume Preserving Feedback |16 Nov. 2019|tags: math.LA, math.DS

This short note is inspired by the beautiful visualization techniques from Duistermaat and Kolk (DK1999) (themselves apparently inspired by Atiyah).

Let's say we have a -dimensional linear discrete-time dynamical system , which preserves the volume of the cube under the dynamics, i.e. for any .

Formally put, this means that is part of a certain group, specifically, consider some field and define the Special Linear group by Now, assume that we are only interested in matrices such that the cube remains bounded under the dynamics, i.e., is bounded. In this scenario we restrict our attention to . To see why this is needed, consider and both with determinant : If we look at the image of under both these maps (for several iterations), we see that volume is preserved, but also clearly that the set is extending indefinitely in the horizontal direction. To have a somewhat interesting problem, assume we are given with while it is our task to find a such that , hence, find feedback which not only preserves the volume, but keeps any set of initial conditions (say ) bounded over time.

Towards a solution, we might ask, given any what is the nearest (in some norm) rotation matrix? This turns out to be a classic question, starting from orthogonal matrices, the solution to is given by , where follows from the SVD of , . Differently put, when we use a polar decomposition of , , then the solution is given by . See this note for a quick multiplier proof. An interesting sidenote, problem is well-defined since is compact. To see this, recall that for any we have , hence the set is bounded, plus is the pre-image of under , , hence the set is closed as well.

This is great, however, we would like to optimize over instead. To do so, one usually resorts to simply checking the sign - and if necessary - flipping it via finding the closest matrix with a change in determinant sign. We will see that by selecting appropriate coordinates we can find the solution without this checking of signs.

For today we look at and largely follow (DK1999), for such a -dimensional matrix, the determinant requirement translates to . Under the following (invertible) linear change of coordinates  becomes , i.e., for any pair the point runs over a circle with radius . Hence, we obtain the diffeomorphism . We can use however a more pretty diffeomorphism of the sphere and an open set in . To that end, use the diffeomorphism: for (formal way of declaring that should not be any real number) and the pair being part of (the open unit-disk). To see this, recall that the determinant is not a linear operator. Since we have a -dimensional example we can explicitly compute the eigenvalues of any (plug into the characteristic equation) and obtain: At this point, the only demand on the eigenvalues is that . Therefore, when we would consider within the context of a discrete linear dynamical system , is either a saddle-point, or we speak of a marginally stable system ( ). We can visualize the set of all marginally stable , called , defined by all satisfying  Once we visualize using the equivalence relation (instead of merely as the interval ), then this set lives in the solid corresponding to the Torus . Note that resembles two adjoined croissants. This set is however larger than since we still contain matrices like from before.

To find close (in Frobenius-norm) rotation matrices, given any , we observe that we can do the Polar decomposition explicitly in coordinates, given any we get (standard rotation matrix parametrized by ) plus a symmetric positive definite , defined by:  It turns out that all these matrices form the jam filling of the croissants. More interestingly, we see that we can find the closest rotation matrix immediately instead of first finding the closest orthogonal matrix. Of course, this insight was facilitated by the reparametrization of in , which is the crux.

As a last step, given an approximation to , can we find a control gain such that is (efficiently) generated by . If we assume that , the solution is given by . So, if was already a rotation matrix, then indeed.

(DK1999) J.J. Duistermaat and J.A.C. Kolk : ‘‘Lie Groups’’, 1999 Springer.