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 2-dimensional linear discrete-time dynamical system x_{k+1}=Ax_k, which preserves the volume of the cube [-1,1]^2 under the dynamics, i.e. mathrm{Vol}([-1,1]^2)=mathrm{Vol}(A^k[-1,1]^2) for any kin mathbf{Z}.

Formally put, this means that A is part of a certain group, specifically, consider some field mathbf{F} and define the Special Linear group by

     mathsf{SL}(n,mathbf{F}):={ Ain mathsf{GL}(n,mathbf{F});|; mathrm{det}(A)=1 }.

Now, assume that we are only interested in matrices Ain mathsf{SL}(2,mathbf{R}) such that the cube [-1,1]^2 remains bounded under the dynamics, i.e., lim_{kto infty}A^k[-1,1]^2 is bounded. In this scenario we restrict our attention to mathsf{SO}(2,mathbf{R})subset mathsf{SL}(2,mathbf{R}). To see why this is needed, consider A_1 and A_2 both with determinant 1:

   A_1 = left[   begin{array}{ll}   2 & 0    0 & frac{1}{2}    end{array}right], quad  A_2 = left[   begin{array}{ll}   1 & 2    0 & 1    end{array}right],

If we look at the image of [-1,1]^2 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.

Linear maps 

To have a somewhat interesting problem, assume we are given x_{k+1}=(A+BK)x_k with Ain mathsf{SL} while it is our task to find a K such that A+BKin mathsf{SO}, hence, find feedback which not only preserves the volume, but keeps any set of initial conditions (say [-1,1]^2) bounded over time.

Towards a solution, we might ask, given any Ain mathsf{SL}(2,mathbf{R}) what is the nearest (in some norm) rotation matrix? This turns out to be a classic question, starting from orthogonal matrices, the solution to mathcal{P}:min_{widehat{A}in mathsf{O}(n,mathbf{R})}|A-widehat{A}|_F^2 is given by widehat{A}=UV^{top}, where (U,V) follows from the SVD of A, A=USigma V^{top}. Differently put, when we use a polar decomposition of A, A=widetilde{U}P, then the solution is given by widetilde{U}. See this note for a quick multiplier proof. An interesting sidenote, problem mathcal{P} is well-defined since mathsf{O}(n,mathbf{R}) is compact. To see this, recall that for any Qin mathsf{O}(n,mathbf{R}) we have |Q|_F = sqrt{n}, hence the set is bounded, plus mathsf{O}(n,mathbf{R}) is the pre-image of I_n under varphi:Amapsto A^{top}A, Ain mathsf{GL}(n,mathbf{R}), hence the set is closed as well.

This is great, however, we would like to optimize over mathsf{SO}subset mathsf{O} 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 mathsf{SL}(2,mathbf{R}) and largely follow (DK1999), for such a 2-dimensional matrix, the determinant requirement translates to ad-bc=1. Under the following (invertible) linear change of coordinates

 left[   begin{array}{l}   a     d     b     c     end{array}right]  =  left[   begin{array}{llll}   1 & 1 & 0 & 0    1 & -1 & 0 & 0 0 & 0 & 1 & 1    0 & 0 & 1 & -1   end{array}right] left[   begin{array}{l}   p     q     r     s     end{array}right]

mathrm{det}(A)=1 becomes p^2+s^2=q^2+r^2+1, i.e., for any pair (q,r)in mathbf{R}^2 the point (p,s) runs over a circle with radius (q^2+r^2+1)^{1/2}. Hence, we obtain the diffeomorphism mathsf{SL}(2,mathbf{R})simeq mathbf{S}^1 times mathbf{R}^2. We can use however a more pretty diffeomorphism of the sphere and an open set in mathbf{R}^2. To that end, use the diffeomorphism:

     (theta,(u,v)) mapsto frac{1}{sqrt{1-u^2-v^2}} left[ begin{array}{ll}     cos(theta)+u & -sin(theta)+v      sin(theta)+v & cos(theta)-u      end{array}right] in mathsf{SL}(2,mathbf{R}),

for theta in mathbf{R}/2pi mathbf{Z} (formal way of declaring that theta should not be any real number) and the pair (u,v) being part of mathbf{D}_o:={(u,v)in mathbf{R}^2;|; u^2+v^2<1} (the open unit-disk). To see this, recall that the determinant is not a linear operator. Since we have a 2-dimensional example we can explicitly compute the eigenvalues of any Ain mathsf{SL}(2,mathbf{R}) (plug ad-bc=1 into the characteristic equation) and obtain:

     lambda_{1,2} = frac{mathrm{Tr}(A)pmsqrt{mathrm{Tr}(A)^2-4}}{2},quad mathrm{Tr}(A) = frac{2cos(theta)}{sqrt{1-u^2-v^2}}.

At this point, the only demand on the eigenvalues is that lambda_1 cdot lambda_2 =1. Therefore, when we would consider A within the context of a discrete linear dynamical system x_{k+1}=Ax_k, 0 is either a saddle-point, or we speak of a marginally stable system (|lambda_{1,2}|=1). We can visualize the set of all marginally stable Ain mathsf{SL}, called M, defined by all big(theta,(u,v)big)in mathbf{R}/2pimathbf{Z}times mathbf{D}_o satisfying

     left|frac{cos(theta)}{sqrt{1-u^2-v^2}}pm left(frac{1-u^2-v^2-sin(theta)^2}{1-u^2-v^2} right)^{1/2}  right|=1

Once we visualize theta using the equivalence relation (instead of merely as the interval [0,2pi)), then this set M lives in the solid corresponding to the Torus mathbf{T}^2. Note that M resembles two adjoined croissants. This set is however larger than mathsf{SO} since we still contain matrices like A_2 from before.

To find close (in Frobenius-norm) rotation matrices, given any Ain M, we observe that we can do the Polar decomposition (A=widetilde{U}P) explicitly in big(theta,(u,v)big) coordinates, given any Ain mathsf{SL}(2,mathbf{R}) we get widetilde{U}in mathsf{SO}(2,mathbf{R}) (standard rotation matrix parametrized by theta) plus a symmetric positive definite P, defined by:

     P = frac{1}{sqrt{1-u^2-v^2}} left[ begin{array}{ll}  1+cos(theta) u + sin(theta)v & cos(theta)v-sin(theta)u   cos(theta)v-sin(theta)u & 1-sin(theta)v-cos(theta)u   end{array}right].
Croissants with jam 

It turns out that all these widetilde{U} 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 A in big(theta,(u,v)big), which is the crux.

As a last step, given an approximation widehat{A}:=widetilde{U} to A, can we find a control gain K such that widehat{A}-A is (efficiently) generated by BK. If we assume that Bin mathsf{GL}(2,mathbf{R}), the solution is given by Kbig(B,theta,(u,v)big)=B^{-1}widetilde{U}(I_2-P). So, if A was already a rotation matrix, then K=0 indeed.

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