Optimal control and Galois theory

Sbornik : Mathematics 204:11 1–15

Matematicheski˘ı Sbornik 204:11 83–98

DOI 10.1070/SM2013v204n11ABEH004352

Optimal control and Galois theory M. I. Zelikin, D. D. Kiselev and L. V. Lokutsievskiy Abstract. An important role is played in the solution of a class of optimal control problems by a certain special polynomial of degree 2(n − 1) with integer coefficients. The linear independence of a family of k roots of this polynomial over the field Q implies the existence of a solution of the original problem with optimal control in the form of an irrational winding of a k-dimensional Clifford torus, which is passed in finite time. In the paper, we prove that for n 6 15 one can take an arbitrary positive integer not exceeding [n/2] for k. The apparatus developed in the paper is applied to the systems of Chebyshev-Hermite polynomials and generalized Chebyshev-Laguerre polynomials. It is proved that for such polynomials of degree 2m every subsystem of [(m + 1)/2] roots with pairwise distinct squares is linearly independent over the field Q. Bibliography: 11 titles. Keywords: Pontryagin’s maximum principle, Lie algebra, dense winding, Galois group, orthogonal polynomials.

§ 1. Introduction Let us consider a smooth 2n-dimensional symplectic manifold M 2n . Suppose that a (2n − 1)-dimensional stratified submanifold S ⊂ M partitions M into finitely S many open domains Ω1 , . . . , Ωk ; M = Ωi . We consider a continuous Hamiltonian H(q, p) : M → R whose restriction Hi = H|Ωi to every set Ωi defines a smooth function C ∞ extendible to a neighbourhood of the set Ωi . One of the main objectives of optimal control theory is the construction of optimal synthesis, that is, a phase portrait of extremals passing through every initial point x0 . Pontryagin’s maximum principle reduces this problem to the study of a Hamiltonian system of differential equations with discontinuous right-hand side, which has discontinuities at the points x0 ∈ S for which the maximum is attained for several different values of control. At these points, the standard existence and uniqueness theorems cannot be applied. The existence problems for a generalized solution are settled by considering an auxiliary differential equation with set-valued right-hand side. The right-hand side of this equation contains the convex closure of all the limit points of the phase velocities as the points in a neighbourhood tend to x0 . The uniqueness of a solution passing through x0 fails to hold fundamentally. The main role is played here by the problem of describing (using the function Hi ) the corresponding integral funnel, that is, the cone of solutions coming to the point x0 AMS 2010 Mathematics Subject Classification. Primary 49J21; Secondary 49J15, 49K21.

c 2013 Russian Academy of Sciences (DoM), London Mathematical Society, Turpion Ltd. ⃝

2

M. I. Zelikin, D. D. Kiselev and L. V. Lokutsievskiy

or going out of the point. The order of the singularity is determined by the number of multiple Poisson brackets of the functions Hi that vanish at x0 (see [1]). For the one-dimensional control or, equivalently, for a discontinuity surface of codimension one with a first-order singularity, the integral funnel consists of three extremals in the typical situation; two of these are placed at the different sides of the discontinuity surface and the third extremal goes along the surface. The situation in general position was found and studied for a second-order singularity by Kupka [2] and Zelikin-Borisov [3]. The extremals come to the point x0 (and go away from this point) in finite time, with countably many intersections with the nonsmoothness surface of H (the chattering behaviour). In the case of multidimensional control (in which the control set is a sphere), the second-order singularities were studied by Milyutin [4], Zelikin-Borisov [3], and Zelikin-Lokutsievskiy-Hildebrand [1]. When approaching the point x0 , the control makes countably many revolutions along some circle during a finite time interval. The corresponding trajectory x(t) arrives at the point x0 , passing an analogue of a logarithmic spiral during a finite time interval. Sometimes combinations of these solutions belonging to distinct two-dimensional subspaces can occur. § 2. Irrational winding of a torus In this section, we study a class of problems in which, on the one hand, the conjugation of singular optimal trajectories with nonsingular ones is unavoidable and, on the other hand, the regular conjugation is impossible by [1], Theorem 5.1. Let us consider the optimization problem Z

+∞

⟨x, Cx⟩ dt → min

J(x) =

(2.1)

0

on the trajectories of the controlled system x(q) = u, x

|u| 6 1, (k)

(0) =

x0k

x ∈ V,

u∈U =V;

for k 6 q − 1.

Here V stands for a Euclidean space with inner product ⟨ · , · ⟩ and C is a symmetric bilinear form. The function x(t) is assumed to be absolutely continuous together with its 2q − 1 derivatives. The control u(t) ∈ L1 (0; +∞) is a measurable function. Let λ1 , . . . , λs be the eigenvalues of the form C and let V1 , . . . , Vs be the corresponding eigensubspaces. Obviously, if at least one of the eigenvalues is negative, then the minimum in the problem is equal to −∞. Therefore, we assume that C is nonnegative definite. In this case, the problem in question has a solution, and this solution is unique1 (see [3]). If dim ker C ̸= 0, then the value of the functional J(x) is preserved under the projection V → V / ker C. Therefore, the optimal trajectories of the problem with dim ker C ̸= 0 can be found trivially from the solutions of the similar problem in V / ker C by taking the preimage. The following theorem can be proved in a similar way (for details, see [1]). 1 For one-dimensional control a proof is presented in [3]. However, this proof can be rewritten verbatim for the case of multidimensional control in a ball.

Optimal control and Galois theory

3

Theorem 1. If L = span(x00 , x01 , . . . , x0q−1 ) is contained in some eigensubspace Vj , then the optimal trajectory of the problem (2.1) remains in L all the time. It can readily be seen that, if the matrix C is positive definite (as was said above, only these problems are of interest), then the problem has precisely one singular extremal x = u = 0, and its global order is equal to2 q. Thus, if q is even, then, by Theorem 5.1 in [1], a regular conjugation of a nonsingular trajectory with the singular trajectory x = u = 0 is not optimal. In this section, we find explicit solutions of the problem (2.1) (belonging to the integral funnel of the point x = 0) for which the optimal control moves along a dense winding of a Clifford torus embedded in the sphere |u| = 1. Let us first formulate a theorem proved earlier in [1]3 , which looks at trajectories that do not leave an eigensubspace Vj of the form C. We consider every two-dimensional plane L in Vj as the complex plane4 C. Theorem 2. In every two-dimensional plane L ⊆ Vj containing 0 there are [q/2] pairs of optimal trajectories5 of (2.1) of the form x(t) = Ak t2q exp{±iαk ln |t|}

(q)

,

u(t) = exp{±iαk ln |t|},

(2.2)

where the αk ∈ R+ are distinct and can be found from the conditions Im Pq (α) = 0,

(−1)q+1 Re Pq (α) > 0,

(2.3)

and the coefficients Ak ∈ R+ are found from the conditions (−1)q+1 Re Pq (αk ) = 1/Ak . Here Pq (α) = (2q + iα)((2q − 1) + iα) · · · (1 + iα). (2.4) If q is odd, then L also contains another trajectory of the form x = tq /q! and u = 1. We note that, in the solutions (2.2), the control makes countably many revolutions along the circle L ∩ {|u| = 1} as t → 0 during a finite time interval. The trajectory x(t) itself is a logarithmic spiral that is also passed along in finite time. In some cases, linear combinations of solutions (2.2) give new optimal solutions of (2.1), in which the control moves along a torus embedded in the sphere |u| = 1 rather than along a circle. To find all trajectories of this kind, let us use the Lie group of symmetries of the problem and distinguish families of trajectories that do not go beyond a single orbit. To this end, we write out the system of the maximum principle. We set x1 = x, x′1 = x2 , . . . , x′q−1 = xq . Then the Hamiltonian (the Pontryagin function) is of the form H = −λ0 ⟨Cx1 , x1 ⟩ + ⟨p1 , x2 ⟩ + · · · + ⟨pq , u⟩ 2 To be more precise, the global order is the compactly supported sequence (0, . . . , 0, dim V, 0, . . . ), where dim V stands at the qth place (for details, see [1]). 3 The formulation of the theorem in [1] contains a misprint, although the proof is correct. 4 The complex structure on L is chosen to be compatible with the Euclidean structure in Rq . 5 Up to a rotation and a time shift.

4


(p1 , . . . , pq stand for the conjugate variables). It can readily be seen that λ0 ̸= 0. Setting λ0 = 1/2, we obtain p′1 = Cx1 ,

p′2 = −p1 ,

...,

p′q = −pq−1 ,

u=

pq . |pq |

The form of this system is drastically simplified if we write zk = (−1)q−k pq−k+1 ,

zq+k = Cxk

for k 6 q.

(2.5)

Then x = C −1 zq+1 and z1′ = z2 ,

z2′ = z3 ,

...,

′ z2q = Cu,

u = (−1)q+1

z1 . |z1 |

(2.6)

This system has the following symmetries. 1. The group G1 = SO(V1 ) × SO(V2 ) × · · · × SO(Vs ) ⊆ SO(V ), where {Vi }si=1 are the eigensubspaces of C introduced above, acts on zk and u by simultaneous rotations and takes velocity vectors (and thus also solutions) of (2.6) to themselves. 2. The group G2 = R+ acts by scaling, namely, for λ ∈ R+ we have zk 7→ λ2q−k+1 zk . In fact, the velocity vector of the system (2.6) is multiplied by λ under this action. However, the integral curves are still taken to integral curves (and only the speed of the motion along the curves becomes λ times larger). Theorem 3. Consider an arbitrary family of two-dimensional planes Lm ⊆ Vjm , m = 1, . . . , N , where the Vjm are some different eigensubspaces of the form C .6 If a family of eigenvalues λj1 , . . . , λjN of the form C satisfies the condition Pq (αj1 ) Pq (αj2 ) Pq (αjN ) = = ··· = =µ λ j1 λj2 λ jN for some distinct αjm , then every trajectory of the form z1 =

N X

bm t2q exp{±iαjm ln |t|}ym ,

m=1

u=µ

N X

(2.7) bm exp{±iαjm ln |t|}ym

m=1

is optimal in problem (2.1) for any choice of the unit vectors ym ∈ Lm and nonzero PN numbers bm ∈ R, provided that µ2 m=1 b2m = 1. Moreover, all possible optimal trajectories (up to time shift) of problem (2.1) that do not go beyond some fixed orbit of the group G = G1 × G2 are described in this way. We note that the control of the solutions (2.7) moves along a winding of the Clifford torus (L1 × · · · × LN ) ∩ {|u| = 1} and passes this winding completely7 in finite time. The trajectory x(t) by itself is the corresponding logarithmic spiral which is also passed during a finite time interval. 6 The case N = 1 is not excluded. For N = 1 every eigenvalue λ of the form C fits, provided j that dim Vj > 2. 7 To be more precise, the half of the winding related to the positive direction of time.


5

Proof of Theorem 3. Since problem (2.1) is convex, it follows that the maximum principle is also a sufficient condition for optimality. Thus, every trajectory of the maximum principle coming to the origin is optimal. Without loss of generality we may assume that a basis in V agrees with the decomposition V = V1 ⊕ · · · ⊕ Vs , that is, z = (w1 , . . . , ws ), where z ∈ V and wj ∈ Vj . The action of the group G on zk defines a representation M 2q ϕ : G = R+ × SO(V1 ) × SO(V2 ) × · · · × SO(Vs ) → GL V , j=1 2q−k+1

πk ϕ(λ, M1 , . . . , Ms )zk = λ

(M1 wk1 , . . . , Ms wks ), where

λ ∈ R+ , Mj ∈ SO(Vj )

(here π denotes the operator of projection to the kth component of the direct sum L2q k j=1 V ). By the existence and uniqueness theorem for ordinary differential equations, an integral curve of the system (2.6) does not leave a fixed orbit of the Lie group G if and only if the velocity vector of the system (2.6) is tangent to the orbit at the initial point z = (z1 , . . . , z2q ). This means that q+1 Cz1 z2 , . . . , z2q , (−1) ∈ dϕ(Te G). |z1 | Here Te G stands for the tangent space to the Lie group G, that is, the Lie algebra R ⊕ so(V1 ) ⊕ · · · ⊕ so(Vs ), where so(Vj ) stands for the Lie algebra of the skew-symmetric forms on Vj . Evaluating dϕ, we see that for some a ∈ R and mj ∈ so(Vj ) we have 

z2 z3 ...



    2qz1 mz1   (2q − 1)z2   mz2    +    = a    ... , ...   Cz 1 (−1)q+1 z2q mz2q |z1 |

(2.8)

where m = (m1 , . . . , ms ). We note that a ̸= 0 because otherwise equation (2.8) has the trivial solution zk = 0 only. Replacing m by a · m, we see that (−1)q+1

z1 = C −1 (2q + m)((2q − 1) + m) · · · (1 + m)z1 = C −1 Q(m)z1 . (2.9) a2q |z1 |

Since one can also act by the scaling group on z1 , it remains to find the matrices m ∈ so(V1 ) ⊕ · · · ⊕ so(Vs ) such that the operator on the right-hand side has a real eigenvalue of sign (−1)q+1 . Thus, let z = (w1 , . . . , ws ) be an eigenvector of the operator C −1 Q(m) = C −1 (2q + m)((2q − 1) + m) · · · (1 + m) with eigenvalue µ of sign (−1)q+1 . Then Q(mj )wj = (2q + mj )((2q − 1) + mj ) · · · (1 + mj )wj = µλj wj .

(2.10)

Thus, either wj = 0 or wj is an eigenvector of the operator Q(mj ). Let us study the eigenvalues of the operator in the previous equation. Using the theorem on the

6


reduction of a skew-symmetric  0 −β1j   0   mj =  0  .  ..   0 0

form, we may assume that8 β1j 0 0 0 .. .

0 0 0 −β2j .. .

0 0 β2j 0 .. .

... ... ... ... .. .

0 0 0 0 .. .

0 0

0 0

0 0

... ...

0 −βrj

 0 0  0  0 . ..  .  βj  r

0

Since eigenvalues of a block-diagonal matrix are eigenvalues of the blocks, it follows that the problem of finding the eigenvalues is reduced to evaluating the eigenvalues of a matrix of rank two. We look at an eigenvector wj belonging to some two-dimensional plane Lkj ⊆ Vj corresponding to the kth block in mj . Using the complex structure in Lkj we shall treat 1 ∈ C as the identity (2 × 2) matrix and i ∈ C as the (2 × 2) matrix with 1 and −1 on the skew diagonal. Then it follows from (2.10) that Q(iβkj )wj = µλj wj and therefore the imaginary part of Q(iβkj ) vanishes. Since Q(iα) coincides with Pq (α) in (2.4), equation (2.10) in the plane Lkj , with regard to the signs, can be represented in the form of the system (2.3) with α = βkj . Every factor in formula (2.4) is a complex number whose argument increases monotonically from 0 to π/2 as α grows from 0 to +∞. Hence, the argument of the product of the factors increases monotonically from 0 to πq. Here, if q is even, then the product crosses the negative part of the real axis [q/2] times and, if q is odd, then it crosses the positive part of the real axis [q/2] times. Suppose now that z = (w1 , . . . , ws ) is a family of vectors wj each of which is either an eigenvector of the operator Q(mj ) or zero. Obviously, z is an eigenvector of the original operator C −1 Q(m) if and only if µ=

Pq (αj1 ) Pq (αj2 ) Pq (αjN ) = = ··· = , λj1 λ j2 λjN

where j1 , . . . , jN are the indices of the nonzero vectors wj . If we write wjm = bm ym , where ym are vectors with unit length and bm are nonzero numbers, then we obtain the trajectories mentioned in the statement of the theorem (the condition for bm is obtained by substituting z1 directly into (2.6)). On every optimal trajectory (2.7), the control ranges over a winding of the Clifford torus TN = (L1 ×· · ·×LN )∩{|u| = 1} and passes to the singular regime x = u = 0 in finite time. The torus TN is embedded in the sphere S n−1 = {|u| = 1} ⊆ V of unit radius. Moreover, if the values αj1 , . . . , αjN are incommensurable over Q, then the winding of TN thus obtained is everywhere dense. As was proved in [1], families of optimal trajectories with control in the form of an irrational dense winding of a torus occur in problem (2.1) for q = 4. An [q/2] affirmative answer to the problem of linear independence of the roots {α2j−1 }j=1 8 The last two-dimensional block in the matrix m is replaced by the one-dimensional zero if j r = dim Vj is odd.


7

of the polynomial (2.4) over the field of rational numbers for an arbitrary q would enable one to obtain the above picture with irrational winding for an arbitrary k-dimensional torus. We note that the coefficients of the polynomial Pq (α) have a very special structure. For this reason, we dare to make the following conjecture. [q/2]

Conjecture 1. The roots {α2j−1 }j=1 of the polynomial Im Pq (α) are linearly independent over the field of rational numbers Q. The next section is devoted to the proof of Conjecture 1 for some particular values of q. § 3. Proof of linear independence Q2n 3.1. Thus, let us consider the polynomial Pn (x) = j=1 (x + j). As was proved in [1], Theorem 6.2, the following system has precisely [n/2] solutions: 1 Im Pn (ix) = 0, x (−1)n+1 Re Pn (ix) > 0.

(3.1) (3.2)

In the proof of the same theorem, it was shown that the polynomials Fn (x) := 1 x Im Pn (ix) and Gn (x) := Re Pn (ix) are separable. It follows immediately from the definition of the polynomials Fn (x) and Gn (x) that there are polynomials fn (x), gn (x) ∈ Z[x] such that Fn (x) = fn (x2 ),

Gn (x) = gn (x2 ).

(3.3)

Since Fn (x) and Gn (x) are separable, it follows immediately from (3.3) that the polynomials fn (x) and gn (x) are separable. The polynomial fn (x) has degree n − 1 and the polynomial gn (x) has degree n. Lemma 1. Let a(x) ∈ k[x] be a separable polynomial of degree n−1, where n ∈ Z>5 . Let Galk (a) ∼ = Sn−1 or Galk (a) ∼ = An−1 . In this case, every [n/2] roots of the polynomial A(x) := a(x2 ) whose squares are pairwise distinct are linearly independent over k. Proof. We choose an arbitrary system of [n/2] roots of the polynomial A(x) such that the squares of any two of its elements are distinct. The set of roots of the polynomial A(x) has structure {±vi }n−1 i=1 . Up to indexing of roots, we can assume without loss of generality that the system v1 , v2 , . . . , v[n/2]

(3.4)

has been chosen. Let the polynomial A(x) be reducible over the field k. We then choose, in an arbitrary way, an irreducible component A1 (x) ∈ k[x]. Since A1 (x) is irreducible over k, it follows that the Galois group Galk (A1 ) acts on the roots of A1 (x) transitively. Since the set {vi2 }n−1 i=1 exhausts the system of roots of the polynomial a(x), it follows from the condition on the group Galk (a) that, first, both the elements vi and −vi cannot simultaneously be roots of the polynomial A1 (x) for any

8


index i = 1, . . . , n − 1 and, second, there is a uniquely defined family of signs {γi }n−1 i=1 ⊂ {−1, 1} such that the polynomial A1 (x) is decomposed over its splitting field into linear factors of the form A1 (x) =

n−1 Y

(x − γi vi ).

i=1

In particular, Galk (A) = Galk (A1 ) ∼ = Sn−1 . If {vij }rj=1 is now a maximal linearly independent subsystem of the system (3.4), then either r = [n/2] or vir+1 = α1 vi1 + · · · + αr vir ,

(3.5)

for a uniquely defined family {αi }ri=1 of elements of the field k (r > 1, because the polynomial a(x) is irreducible over k, by assumption). However, there is an element σ ∈ Galk (A) such that viσj = vij

∀ j = 1, . . . , r,

viσr+1 = vin .

Thus, vir+1 = vin , which contradicts the separability of the polynomial A(x). Suppose now that s ∈ Z>2 and that, for any t = 1, . . . , s, the polynomial A(x) At (x) := Qt−1 2 2 i=1 (x − vi ) is irreducible over the field kt := k(v1 , . . . , vt−1 ) (for t = 1 we set A1 (x) := A(x), k1 := k) and the polynomial As+1 (x) is reducible over the field ks+1 . In this case, every subsystem of s roots of the polynomial A(x) such that the squares of any two elements of the subsystem are distinct is linearly independent over k. Indeed, otherwise, we would arrive at a contradiction to the choice of the number s. If s > [n/2], then the proof is complete. If s < [n/2] and the group Galks+1 (as+1 ), where a(x) at (x) := Qt−1 , a1 (x) := a(x), t = 1, . . . , s, 2 i=1 (x − vi ) is l-transitive for

n l> − s, 2

(3.6)

then, arguing as in the case of s = 1, we find an element σ ∈ Galks+1 (as+1 ) which preserves the chosen linearly independent subsystem of roots in the system (3.4) and transposes the element vn with some other element of (3.4) that can be expressed over ks+1 using the current linearly independent subsystem. We arrive again at a contradiction to the condition that the polynomial in question is separable. It remains now to estimate the transitive degree of the action of the group Galks+1 (as+1 ) on the roots of the polynomial as+1 (x). We write Kt+1 := k(v12 , . . . , vt2 ),

K1 := k.

It can readily be seen that the extension ks+1 /Ks+1 is a Galois extension with the elementary Abelian Galois 2-group. We note that Ks+1 ⊆ ks+1 ∩ Kn ⊆ ks+1 ,


9

which implies, in particular, that the extension ks+1 ∩ Kn /Ks+1 is a Galois extension. Therefore, Gal(Kn /ks+1 ∩ Kn ) is a normal subgroup of Gal(Kn /Ks+1 ). However, since the splitting field of the polynomial as+1 (x) over ks+1 is the composite Kn · ks+1 , it follows that there is an isomorphism Galks+1 (as+1 ) = Gal(Kn · ks+1 /ks+1 ) ∼ = Gal(Kn /ks+1 ∩ Kn ). This means that, up to isomorphism, the group Galks+1 (as+1 ) is a normal subgroup of Gal(Kn /Ks+1 ) and the quotient group by this subgroup is an elementary Abelian 2-group; here, in turn, the group Gal(Kn /Ks+1 ) is isomorphic either to Sn−1−s or to An−1−s . Since the alternating group of degree > 3 has no proper normal subgroups for which the quotient group is a 2-group, it follows that the transitive degree of the form Galks+1 (as+1 ) (as a group of permutations of the roots of the polynomial as+1 (x)) is no less than n − 1 − s − 2 = n − s − 3. Thus, in the worst case, the transitivity degree of Galks ∩Kn (as ) is equal to n − s − 3. Recall that we need a bound (3.6). Therefore, we are to establish the validity of the estimate n − s 6 n − s − 3. (3.7) 2 However, this bound holds for n ∈ Z>5 . This completes the proof of Lemma 1. Remark 1. Lemma 1 shows in fact that, to obtain a solution of the problem of linear independence over Q for numbers satisfying the system (3.1), (3.2), it is sufficient that the Galois group of the polynomial fn (x) over Q act on the roots l-transitively (the only condition is that, when joining another root of the polynomial fn (x2 ), the transitivity degree reduces by at most one), where l > [n/2]. However, if [n/2] > 5, then, by the well-known Jordan theorem (see [5], Ch. 1, § 3.3), only Sn−1 and An−1 can be the desired groups. For positive integers n satisfying the reverse inequality (which is only possible if n 6 11), the Galois group of the polynomial fn−1 (x) over Q turns out to be isomorphic to Sn−1 . This fact is proved below in § 3.3. Therefore, in any case, the proof of linear independence using the considerations of Lemma 1 can be used (at least in the case we are interested in) only if the Galois group of the polynomial a(x) over the field k is isomorphic to either Sn−1 or An−1 . 3.2. On the linear independence of roots of some orthogonal polynomials. Let us present several interesting examples using Lemma 1. All these examples are related in some way to orthogonal polynomials. Definition 1. The polynomials n L(α) n (x) := (−1)

n X n + α (−x)j j=0

n−j

j!

defined for all real values of α ∈ R are referred to as generalized Chebyshev-Laguerre polynomials. The following result holds for generalized Chebyshev-Laguerre polynomials.

10


Theorem 4. Let α ∈ Q \ Z0 . Then for all positive integers n, except possibly for finitely many of them, every system of (α) [(n + 1)/2] roots of the polynomial Ln (x2 ) whose squares are pairwise distinct is linearly independent over Q. Proof. By the Hajir theorem (see [6], Theorem 1.3), for every fixed r ∈ Z>0 , all (1−n−r) polynomials Ln (x), except possibly for finitely many of them, have a Galois group over Q that contains An as a subgroup. By the Filaseta-Lam-Hajir theorem (see [7], Theorem 1.1), for every fixed (α) α ∈ Q \ Z 12, the polynomials H2n (x) and x1 H2(n+1) (x) have the following property: every subsystem of [(n + 1)/2] roots, of either of these polynomials, for which the squares of any two elements of the subsystem are distinct is linearly independent over Q. Proof. Let Kn0 (x) and Kn1 (x) be polynomials of degree n with integer coefficients such that H2n (x) = Kn0 (x) and H2(n+1) (x) = xKn1 (x). In this case, by Schur’s results (see [5], Ch. 5, § 4.2), for n > 12, the Galois groups of the polynomials Kn0 (x) and Kn1 (x) are isomorphic to Sn . It remains to apply Lemma 1. This completes the proof of the theorem. 3.3. Winding of a torus. By Lemma 1 and by the results of § 2, for the existence of an optimal solution of problem (2.1) in the form of a dense winding of a k-dimensional torus for every k 6 [n/2], it suffices to show that the Galois group of the polynomial fn (x) in (3.3) over Q is isomorphic to either Sn or An for n > 5. We note that this result was established in § 2 for n = 4 only (this is the minimal dimension for which the optimal control is not one-dimensional a priori). We show (using the computer algebra system Maple V) that GalQ (fn ) ∼ = Sn−1 for n = 5, . . . , 15. Let us consider the case of n = 6. Under this assumption, one can carry out a proof of the isomorphism GalQ (f6 ) ∼ = S5 , practically without using Maple V. Lemma 2. The polynomial f6 (x) is irreducible over Q. Proof. It follows immediately from the definition of the polynomials fn (x) and gn (x) given in (3.3) that the following recurrent dependence holds: ( fn+1 (x) = −x + (2n + 1)(2n + 2) fn (x) + (4n + 3)gn (x), (3.8) gn+1 (x) = −x + (2n + 1)(2n + 2) gn (x) − (4n + 3)xfn (x). Using the formula (3.8) one can evaluate (rather quickly, even by hand) the reduction of the polynomial f6 (x) modulo 29. After reducing this reduction to unitary form, we obtain f 6 (x) = x5 + 10x4 + 5x3 + 4x2 + 3x + 13 ∈ F29 [x].

(3.9)


11

The irreducibility of the polynomial (3.9) over the field F29 can be established, for 2 example, by evaluating the greatest common divisor (x29 − x, f 6 (x)) = 1 or by applying Berlekamp’s algorithm. This implies the irreducibility of the polynomial f6 (x) over Q because the leading coefficient of the polynomial f6 (x) is coprime to 29. We must now prove that the discriminant of the polynomial f6 (x) is divisible by 563 taken to an odd degree. This is perhaps the only point at which we use computer evaluations. Lemma 3. The multiplicity of the prime 563 in the discriminant of the polynomial f6 (x) is odd. Proof. The evaluation of the discriminant of the polynomial f6 (x) by Maple V gives the expression D(f6 ) = 222 · 314 · 54 · 74 · 1312 · 563 · 466267 · 33137953 · 18669593, which immediately implies the desired statement. Let us introduce the following important definition. Definition 2. Let us consider an exact sequence of finite groups 1

/N

/G

/ F = Gal(K/k)

ϕ

/ 1.

(3.10)

If there is a homomorphism ω : Gal(k sep /k) → G such that ϕ◦ω is the canonical epimorphism restricting the automorphisms of the field k sep over k to automorphisms of the group F , then one says that the embedding problem for the extension (3.10) is soluble. We need the following remarkable result. Theorem 6 (Lur’e). Let p be a prime congruent to 1 modulo 4. Let f (x) ∈ k[x] be an irreducible separable polynomial of degree p over a field k whose characteristic p differs from 2. Suppose further that the problem of embedding the extension k( D(f ))/k in an extension with cyclic Galois group of order (p−1)2 is not soluble. Then the group Galk (f ) is not soluble either. For the proof of this theorem, see [8]. Theorem 7. The Galois group of the polynomial f6 (x) over the field Q is isomorphic to S5 . Proof. By Lemma 2, the polynomial f6 (x) of degree 5 is irreducible over Q. If we prove that the group GalQ (f6 ) is not soluble, then, by Lemma 3, we shall have GalQ (f6 ) ∼ = S5 . By Theorem 6, it suffices to prove that the embedding problem related to the exact sequence 1

/ Z2

/ Z4

ϕ

p / Gal Q( D(f6 ))/Q

(3.11)

is not soluble. As is well known, for problem (3.11) to be insoluble, it is necessary and sufficient that the discriminant D(f6 ) not be representable in the form of two squares of elements of Q. For the proof, see [9], Ch. 1, § 1.

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Thus, if the discriminant D(f6 ) can be represented as a sum of two squares of elements of Q, then it follows from the well-known properties of the Hilbert symbol 563, −1 that the norm residue must be equal to one. However, 563 563, −1 = −1 563 because

−1 563

= (−1)(563−1)/2 = −1,

which contradicts the assumption. This completes the proof of the theorem. For n ∈ {5, 7, 8, 9, 10, 11, 12, 13, 14, 15} we use the following well-known result. Proposition 1. Let a polynomial f (x) ∈ Z[x] of degree n be given. Let p1 , p2 , p3 be primes not dividing the discriminant D(f ) and the leading coefficient of the polynomial f (x). Let the polynomial f (x) be irreducible modulo p1 , decomposable modulo p2 into a product of a linear factor and an irreducible factor of degree n − 1, and decomposable modulo p3 into n − 2 linear factors and a quadratic factor. Then GalQ (f ) ∼ = Sn . The proof of this proposition follows immediately from the classical Dedekind theorem (see [5], Ch. 5, § 4.3, Theorem 5) and also from the well-known fact about the generators of the symmetric group (see [5], Ch. 5, § 4.2, Proposition 1). A much more important fact is that, if these primes exist, then there are infinitely many of them by the Chebotarev density theorem (see [10], Theorem 1.1). Moreover, it follows from the Chebotarev theorem that the most difficult point is to find an appropriate prime of type p3 , because, under the assumption that an isomorphism GalQ (f ) ∼ = Sn exists, there are only N/(2(n − 3)!) numbers of type p3 among every N primes for sufficiently large N , whereas the fractions of numbers of the types p1 and p2 are 1/(n − 1) and 1/(n − 2), respectively (see [11], Proposition 4.1.10). Let us present a table (composed with the help of Maple V) of the least primes of the types p1 , p2 , p3 that are suitable, according to Proposition 1, to prove the isomorphism GalQ (fn ) ∼ = Sn−1 : n 5 6 7 8 9 10 11 12 13 14 15

p1 23 29 37 37 83 79 61 67 59 67 101

p2 19 23 73 41 67 67 103 73 103 163 79

p3 17 19 > 200 11 13 > 200 17 17 19 > 200 > 200


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It can be seen from the table that, using Proposition 1 and Lemma 1, one can justify the existence of a dense winding of an [n/2]-dimensional torus as an optimal solution of the control problem (3.1) for n ∈ {5, . . . , 15} \ {7, 10, 14, 15}. The case n = 7 can be justified by performing the corresponding operation of evaluating the Galois group of the polynomial f7 (x) in Maple V. Let us consider the case n = 10. Proposition 2. The group GalQ (f10 ) is isomorphic to S9 . Proof. Let us consider the reductions of the polynomial f10 (x) modulo the numbers 29, 37, 43, 67, 79, 157. Using the recurrence formulae (3.8) and Berlekamp’s algorithm, one can readily establish by applying Dedekind’s theorem (see [5], Ch. 5, § 4.3, Theorem 5) that under the canonical identification with a subgroup of S9 the group GalQ (f10 ) contains substitutions with the following cycle structure: a 3-cycle, a 7-cycle, the product of independent cycles of lengths 4 and 5, an 8-cycle, a 9-cycle, a 6-cycle (the cycle structures are given in the order corresponding to the family of integers chosen for reduction). Raising the product of cycles of lengths 4 and 5 to the fourth and fifth powers, we obtain the existence of 5-cycles and 4-cycles, respectively, in the group GalQ (f10 ). In particular, the group GalQ (f10 ) acts on the roots of the polynomial f10 (x) 7-transitively. Indeed, the stabilizer of k roots contains a (9 − k)-cycle for k = 0, . . . , 6 and at the same time is the symmetric group of degree 9 − k. Since every 7-transitive group of permutations of degree 9 necessarily contains A9 as a subgroup, this completes the proof of the proposition. Let us consider the cases when n ∈ {14, 15}. It suffices to prove that for either n there is a prime for which the reduction of the polynomial fn (x) modulo this prime decomposes into n − 2 linear factors and a single quadratic factor. In the case of n = 14, we make the reduction modulo 41; we then see that the polynomial f14 (x) is decomposed into the product of nine linear factors, a quadratic factor, and a cubic factor. In particular, GalQ (f14 ) has a permutation of the cycle type (3, 2). Then the cube of this permutation gives the desired transposition. By the Chebotarev density theorem (see [10], Theorem 1.1), there is a desired prime such that the reduction modulo this prime gives this very cycle decomposition. In the case of n = 15, we make a reduction modulo 67; then we see that the polynomial f15 (x) is decomposed into the product of two linear factors, a factor of degree 10, and a factor of degree 3. In particular, GalQ (f15 ) has a substitution of the cycle type (10, 3). Then the group GalQ (f15 ) contains a 10-cycle. Since 10 = 5 · 2, a desired transposition exists. Thus, we have proved the following result. Theorem 8. For every n = 4, . . . , 15 the corresponding optimal control problem (3.1) (for a suitable choice of the matrix C according to Theorem 3) has an optimal solution whose control ranges over a dense winding of a k-dimensional torus (k 6 [n/2]). This winding (to be more precise, the half of this winding that corresponds to the positive direction of time) is passed entirely in finite time. The optimal solution is a logarithmic spiral which is also passed in finite time.

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Remark 2. Proposition 2, together with Jordan’s theorem, shows that, using a rather small family of primes, one can hope to prove the existence of an embedding An ,→ GalQ (fn−1 ) for a more representative family of numbers n. In [11], rather fast ways to define the cycle structure are presented that use reduction modulo a prime p and have time complexity of the order of O(n4 + n2 log p). Unfortunately, the authors could not prove even the irreducibility over Q of the polynomials fn (x) for an infinite range of positive integers n. Nevertheless, the investigation of the cases n 6 15 in the present paper, and also the fact that ‘almost all’ polynomials of degree m with rational coefficients have Galois group isomorphic to Sm , makes it possible to express the following conjecture. Conjecture 2. For any positive integer n the Galois group of the polynomial fn (x) is isomorphic to Sn−1 .

Bibliography [1] M. I. Zelikin, L. V. Lokutsievskiy and R. Hildebrand, “Geometry of neighborhoods of singular trajectories in problems with multidimensional control”, Mathematical control theory and differential equations, A collection of papers dedicated to the 90th birthday of academician Evgenii Frolovich Mishchenko, Tr. Mat. Inst. Steklova, vol. 277, MAIK, Moscow 2012, pp. 74–90; English transl. in Proc. Steklov Inst. Math. 277 (2012), 67–83. [2] I. Kupka, “Generic properties of extremals in optimal control problems”, Differential geometric control theory (Houghton, MI 1982), Progr. Math., vol. 27, Birkh¨ auser, Boston, MA 1983, pp. 310–315. [3] M. I. Zelikin and V. F. Borisov, Theory of chattering control with applications to astronautics, robotics, economics, and engineering, Systems Control Found. Appl., Birkh¨ auser, Boston, MA 1994. [4] A. A. Milyutin. A. E. Ilyutovich, N. P. Osmolovskii and S. V. Chukanov, Optimal control in linear systems, Nauka, Moscow 1993. (Russian) [5] A. I. Kostrikin, Introduction to algebra. Pt. III. Fundamental structures of algebra, Fizmatlit, Moscow 2001. (Russian) [6] F. Hajir, “Algebraic properties of a family of generalized Laguerre polynomials”, Canad. J. Math. 61:3 (2009), 583–603. [7] F. Hajir, “On the Galois group of generalized Laguerre polynomials”, J. Th´eor. Nombres Bordeaux 17:2 (2005), 517–525. [8] B. B. Lur’e, “A criterion for unsolvability by radicals of an equation of prime degree”, Dokl. Ross. Akad. Nauk 388 (2003), 447–448; English transl. in Dokl. Math. 67:1 (2003), 42–43. [9] V. V. Ishkhanov, B. B. Lur’e and D. K. Faddeev, The embedding problem in Galois theory, Nauka, Moscow 1990; English transl., Transl. Math. Monogr., vol. 165, Amer. Math. Soc., Providence, RI 1997. [10] N. V. Durov, “Computation of the Galois group of a polynomial with rational coefficients. I”, Questions of the theory of representations of algebras and groups, 11, Zap. Nauchn. Semin. S.-Peterburg. Otdel. Mat. Inst. Steklov., vol. 319, Nauka, St.-Petersburg Branch, St.-Petersburg 2004, pp. 117–198; English transl. in J. Math. Sci. (N. Y.) 134:6 (2006), 2511–2548.


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[11] N. V. Durov, “Computation of the Galois group of a polynomial with rational coefficients. II”, Questions of the theory of representations of algebras and groups, 12, Zap. Nauchn. Semin. S.-Peterburg. Otdel. Mat. Inst. Steklov., vol. 321, Nauka, St.-Petersburg Branch., St.-Petersburg 2005, pp. 90–135; English transl. in J. Math. Sci. (N. Y.) 136:3 (2006), 3880–3907. Mikhail I. Zelikin Moscow State University, Faculty of Mechanics and Mathematics E-mail: [email protected] Denis D. Kiselev Moscow State University, Faculty of Mechanics and Mathematics E-mail: [email protected] Lev V. Lokutsievskiy Moscow State University, Faculty of Mechanics and Mathematics E-mail: [email protected]

Received 17/JAN/13 and 9/APR/13 Translated by A. SHTERN