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 $x_{k+1}=Ax_k$ , 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

is part of a certain group, specifically, consider some field mathbf{F}

and define the Special Linear group by

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})

mathsf{SO}(2,mathbf{R})subset mathsf{SL}(2,mathbf{R})

. To see why this is needed, consider A_1

and

both with determinant

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

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.

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

such that

, 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

, $A=USigma V^{top}$ . Differently put, when we use a polar decomposition of

, 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

-dimensional matrix, the determinant requirement translates to ad-bc=1

. Under the following (invertible) linear change of coordinates

becomes

, 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

(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

-dimensional example we can explicitly compute the eigenvalues of any Ain mathsf{SL}(2,mathbf{R})

(plug

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

within the context of a discrete linear dynamical system $x_{k+1}=Ax_k$ ,

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

, defined by all $big(theta,(u,v)big)in mathbf{R}/2pimathbf{Z}times mathbf{D}_o$ satisfying

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

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

(standard rotation matrix parametrized by theta

) plus a symmetric positive definite

, defined by:

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

As a last step, given an approximation widehat{A}:=widetilde{U}

, can we find a control gain

such that

is (efficiently) generated by

. 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

was already a rotation matrix, then K=0

indeed.

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

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

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