Manipulating exponential products

Transcription

1 Manipulating exponential products Instead of working with complicated concatenations of flows like z(t) =e t 9 t (f 8 0 +f 1 +f 2 ) dt...e t 2 t (f 1 0 +f 1 f 2 ) dt e t 1 0 (f 0+f 1 +f 2 ) dt (p) it is desirable to rewrite the solution curve using a minimal number of vector fields f πk that span the tangent space (typically using iterated Lie brackets of the system fields f 0,f 1,...f m ) Coordinates of the first kind z(t) =e b 1(t,u)f π1 +b 2 (t,u)f π1 +b 3 (t,u)f π b n (t,u)f π n(p) Coordinates of the second kind z(t) =e c 1(t,u)f π1 e c 2 (t,u)f π1 e c 3 (t,u)f π3... e c n (t,u)f π n(p) Using the Campbell-Baker-Hausdorff formula, this is possible, but a book-keeping nightmare. Moreover, the CBH formula does not use a basis, but uses linear combinations of all possible iterated Lie brackets. Yet, by the Jacobi identity (and anticommutativity), in ever Lie algebra e.g. [X, [Y,[X, Y ]]] + [Y,[[X, Y ],X]]+[[X, Y ], [X, Y ]]] = 0 }{{} and hence [X, [Y,[X, Y ]]] = [Y,[X, { [X, }} Y ]]] { Plan: Work with bases for (free) Lie algebras. Find useful formulae for the coefficients b k (t, u) orc k (t, u). 12

2 The Chen Fliess series K. T. Chen, 1957: Geometric invariants of curves in R n M. Fliess, 1970s: adaptation to control The formal control system Ṡ = S m i=1 u i X i, S(0) = I on the associative algebra Â(X 1...X m ) of formal power series in the noncommuting indeterminates (letters) X 1,...X m has the unique solution S CF (T,u)= I T t1 tp 1 u i p (t p )...u i 1 (t 1 ) dt 1...dt p 0 } 0 0 {{ } Υ I (T,u) X i1...x ip }{{} X I What is the CF-series good for? For any given control system ẋ = m u i (x)f i (x), x(0) = p, with output y = ϕ(x) i=1 φ(x(t,u)) = I T t1 tp 1 u i p (t p )...u i 1 (t 1 ) dt 1...dt p 0 } 0 0 {{ } Υ I (T,u) (f i1... f ip ϕ)(p) (uniform convergence for small T and IC s on compacta [Sussmann, 1983]) The CF-series was basic tool for deriving many high-order conditions for controllability and optimality. [Hermes, Stefani, Sussmann, Kawski,...] 13

3 Inadequacies of the CF-series The Chen Fliess series is a good starting point, BUT It has too many terms ( 2 when only should do ) Lots of duplication: Repetition in the iterated integrals High-order partial diff operators where only 1 st or low order PDO s should occur It is not geometric: Its character as exponential Lie series is not obvious It is not geometric: Truncations do not correspond to any systems at all (not directly useful for obtaining approximating systems) More desirable alternatives: Expand the series in either of the forms or S CF (T,u)=exp S CF (T,u)= B B B B α B (t, u) B exp (β B (T,u) B) for suitable bases B of the free Lie algebra L(X 1,...X m ). Question: Existence? 14

4 Ree s theorem and exponential Lie series Theorem [Ree, 1957]: A formal power-series I c I X I is an exponential Lie series iff the coefficients satisfy the shuffle relations c I x J = c I c J for all I,J Exercise: The coefficients Υ I (T,u) of the CF-series satisfy the shuffle relations. (Simple induction. Shuffles correspond to pointwise multiplication of integrated integral functionals. Recursive definition of shuffle product corresponds to repeated integration by parts.) Corollary: Either expansion of the CF-series (exp of sum, or productofexp)ispossible. Issues/questions: Need explicit basis for the free Lie algebra Want explicit formulae for the iterated integral coefficients α B (T,u) and/or β B (T,u). 15

5 Shuffle product Combinatorial definition (for words w, z and letters a, b): ( wa) X ( zb)=((wa) X z ) b +(w X ( zb)) a Example: The shuffle product of two words (ab) X (cd) = abcd + a c b d + c abd + a cdb + c a d b + cdab Algebraic definition: On the free associative algebra A = A k (X ) (algebra of polynomials, or words ) over a set X (of noncommuting indeterminates, or letters ) define a co-product Δ: A A A by Δ(a) =1 a + a 1 for a X Define the shuffle product X as the transpose of Δ (on the algebra  = Âk(X ) of formal power series) <v X w, z >= <v w, Δ(z) > 16

6 Shuffles and simplices On permutations algebras Duchamp and Agrachev consider partially commutative and noncommutative shuffles. Illustration: σ σ σ x 2 = σ 21 σ 1 In the case of three letters {1, 2, 3} = σ (12) x 3 = σ 312 σ 132 σ 123 E.g. σ (12) x 3 = {t:0 t 1 t 2 1, 0 t 3 1} For multiplicative integrands f(x, y, z) =f 1 (x) f 2 (y) f 3 (z) (using x, y, z, instead of t 1,t 2,t 3 for better readability): ( 1 ) y ( )dx dy 1 ( )dz = y x ( ) dz dx dy + 1 y z ( ) dx dz dy + 1 z y ( ) dx dy dz

7 CF-coefficients satisfy shuffle-relations Sketch of proof of exercise (by induction on the combined lengths of the coefficients) Υ 1 (t, u) 1 Υ a x 1 (t, u) = Υ a (t, u) = Υ a (t, u) 1 = Υ a (t, u) Υ 1 (t, u) for any letter a X Υ (wa) x (zb) (T,u)= = Υ ((wa) x z)b+(w x (zb))a (T,u) = Υ ((wa) x z)b (T,u)+Υ (w x (zb))a (T,u) = T 0 Υ (wa) x z (t, u) u b (t)dt + T 0 Υ w x (zb) (t, u) u a (t)dt (1) = T 0 (Υ wa (t, u) Υ z (t, u) u b (t)+υ w (t, u) Υ zb (t, u) u a (t)) dt = T 0 ( Υwa (t, u) d dt Υ zb(t, u)+ d dt (Υ wa(t, u)) Υ zb (t, u) ) dt = Υ wa (T,u) Υ zb (T,u) Morale: When working with repeated integrations by parts, omit integrals and similar notational ballast. Instead work purely combinatorially in shuffle algebra. 18

8 Product expansion of CF-series Using Ree s theorem the existence of exponential product expansions of the CF-series is assured. for suitable bases B of L(X 1,...X m ). S CF (T,u)= B B exp (β B (T,u) B) Recall remaining issues/questions: Need explicit basis for the free Lie algebra Want explicit formulae for the iterated integral coefficients α B (T,u) and/or β B (T,u). Using different bases for the free Lie algebra explicit formulae for the dual bases (iterated integral functionals β B (T,u) have been rediscovered several times in different contexts: Schützenberger (1958), Séminaire Dubreil Sussmann (1986), Nonlinear control Melançon and Reutenauer (1989), Combinatorics Grayson and Grossman (1991), Realizations of free nilpotent Lie algebras = See historical slide 19

9 Lazard elimination Theorem [Lazard elimination]: Suppose k a field of scalars, X is a set and c X. Then the free Lie algebra L k (X )overk generated by X is the direct sum of the one-dimensional subspace {λc: λ k} and of a Lie-subalgebra of L k (X ) that is freely generated by the set {(ad j c, b): b X\{c}, j 0}. This principle is at the heart of constructions involving Hall bases for free Lie algebras, for Sussmann s derivation of the exponential product expansion by solving DEs by iteration, and thereby it is closely connected to Zinbiel structures. 20

10 Hall and Lyndon bases Ph. Hall, 1930s, calculus of commutator groups M. Hall, 1950s, first bases for free Lie algebras Lyndon, 1950s, different (?) bases Sǐrsov, 1950s, different (?) bases Viennot, 1970s, only one kind of practical basis AHallsetover a set X is any strictly ordered subset H of the free magma M(X ) (i.e. the set of all parenthesized words, or labelled binary trees) that satisfies X H Suppose a X. Then (w, a) H iff w H, w < a and a<(w, a). Suppose u, v, w, (u, v) H. Then (u, (v, w)) H iff v u (v, w) andu<(u, (v, w)). Original Hall bases as in Bourbaki require that ordering be compatible with the length. Viennot showed that is not necessary. The image of a Hall set under the canonical map ϕ: M(X ) L k (X ) from the free magma into the free Lie algebra is a basis for L k (X ). 21

11 Lie brackets and formal brackets Need to distinguish formal brackets and elements of a Lie algebra. E.g., consider {x, y, (x, y), (y, (x, (x, y)))} H M({x, y}). Then ϕ((x, y)) = [x, y] = [y, x],and ϕ((y, (x, (x, y)))) = [y, [x, [x, y]]] = [x, [y, [x, y]]] (due to anti-commutativity and Jacobi-identity in L k (X )). Consequently, ϕ 1 ([x, [y, [x, y]]]) = ϕ 1 ([y, [x, [x, y]]]) = (y, (x, (x, y))) (x, (y, (x, y))) in M(X ) Similarly, ϕ 1 ([ y, x]) = ϕ 1 ([x, y]) = (x, y) (y, x) in the algebra over M(X ) But coding of iterated integrals depends critically on the factorization of the Hall words, requiring well-defined left and right factors. 22

12 Hall and Lyndon bases, examples Lyndon basis b (((((ab)b)b)b)b) ((((ab)b)b)b) (a((((ab)b)b)b)) (((ab)b)b) ((ab)(((ab)b)b)) (a(((ab)b)b)) (a(a(((ab)b)b))) ((ab)b) ((ab)((ab)b)) (a((ab)((ab)b))) (a((ab)b)) ((ab)(a((ab)b))) (a(a((ab)b))) (a(a(a((ab)b)))) (ab) ((a(ab))(ab)) (a((a(ab))(ab))) (a(ab)) (a(a(ab))) (a(a(a(ab)))) (a(a(a(a(ab))))) b Lyndon words L< [LR] <R Read backwards, a word is Lyndon if it is strictly larger in lexicographical order than any of its cyclic rearrangements Hall words (in narrow sense) a X a H w, z H,u<v u < v If a X,then (ua) H u<aand u<(ua) (u, (vw)) H u, (vw) Hand v u<(vw), u< (u(vw)) 23 Hall basis (as in Bourbaki) ((a(ab))(b(ab))) ((ab)(b(b(ab)))) ((ab)(b(a(ab)))) ((ab)(a(a(ab)))) (b(b(b(b(ab))))) (b(b(b(a(ab))))) (b(b(a(a(ab))))) (b(a(a(a(ab))))) ((ab)(b(ab))) ((ab)(a(ab))) (b(b(b(ab)))) (b(b(a(ab)))) (b(a(a(ab)))) (a(a(a(ab)))) (b(b(ab))) (b(a(ab))) (a(a(ab))) (b(ab)) (a(ab)) (ab) b a

13 Hall-Viennot bases: unique factorization Bases for free Lie algebras proposed by M. Hall, Lyndon, and Siršov were originally considered to be distinct, until Viennot showed that they all arise from a fundamental unique factorization principle. By construction: The restriction of the map ϕ ϕ(a) = a for ; a X,and ϕ((w, z)) = [ϕ(w),ϕ(z)] for w, z M(X ). to any Hall-set H M(X ) is one-to-one by construction [Viennot]. Hence the inverse image ϕ 1 (H) H ϕ( H) L k (X ) is well-defined H M(X )ofanelement Practically speaking: When working with a fixed Hall set there is no need to write down any parentheses! Consequence: Every word w W (X ) factors uniquely into a nonincreasing product of Hall words, i.e. there exist unique H j H, such that w = H 1 H 2...H s and H 1 H 2... H s 24

14 Zinbiel products In the most simple basic control system ẏ 1 = u 1 u 1 1 ẏ 2 = u 2 u 2 1. ẏ 3 = y 1 u 2 many consider as a virtual third control the function u 3 (t) =(u 1 u 2 )(t) def = ( ) t u 0 1(s) ds u 2 (t) The ubiquitous occurrence of this product justifies to call it THE PRODUCT of control theory Exercise: Verify that this product satisfies: (u 1 u 2 + u 2 u 1 )(t) = d (( )( t dt u )) t 0 1(s) ds u 0 2(s) ds and also the three term right Zinbiel identity (u 1 (u 2 u 3 )) (t) = ( t 0 u 1 (s),ds) (( t 0 u 2 (s)ds) u 3 (t)) = ( t 0 ( s 0 u 1 (σ) dσ) u 2 (s)ds) u 3 (t) +( t 0 ( s 0 u 2 (σ) dσ) u 1 (s)ds) u 3 (t) = ((u 1 u 2 ) u 3 )(t)+((u 2 u 1 ) u 3 )(t) 25

15 Zinbiel algebras A (right) Zinbiel algebra is a linear space C (over a field k) thatis endowed with a bilinear product : C C C which satisfies the (right) Zinbiel identity v (w z) =(v w) z +(w v) z for all v, w, z C (2) Example 1:(repeated) L loc ([0, )) with (u 1 u 2 )(t) def =( t 0 u 1 (s) ds) u 2 (t) Example 2: AC loc ([0, )) with (U 1 U 2 )(t) def =( t 0 U 1 (s)u 2(s) ds) Easy exercise: (F (G H))(t) = t 0 F (s) ( d s ds 0 G(σ)H (σ) dσ ) ds = t 0 F (s)g(s) H (s) ds = t 0 ( s 0 F (σ)g (σ) dσ + s 0 G(σ)F (σ) dσ) H (s) ds = ((F G) H)(t)+((F G) H)(t) Also the symmetrized product is pointwise multiplication (U 1 U 2 )+(U 2 U 1 )=(U 1 U 2 ) for U 1,U 2 AC loc,0 26

16 Zinbiel algebras, examples There are many familiar Zinbiel subalgebras of the Zinbiel algebras AC loc,0 and AC loc,0, most notably those of polynomial and of (real, or complex) exponential functions. Typical multiplication rules are x n x m = 1 n+1 xn+m+1 e nt e mt = 1 n e(n+m)t x n x m = m n+m xn+m e nt e mt = m n+m e(n+m)t Exercise: In each case verify directly that the right Zinbiel identity is satisfied. For example: (x m x n ) x k = 1 m+n+2 1 m+1 x m+n+k+2 (x n x m ) x k = 1 m+n+2 1 n+1 x m+n+k+2 x m (x n x k ) = 1 m+1 1 n+1 x m+n+k+2 27

17 Zinbiel products and solving DEs Zinbiel products efficiently encode the solution of time-varying linear differential equations by iteration: The integrated form of the universal control system Ṡ = S Φ, (0) = 1 with Φ= m i=1 u i X i is compactly rewritten using Zinbiel products S =1+S Φ Iteration yields the explicit series expansion S = 1 +(1+S Φ) Φ = 1+Φ+ ((1 + S Φ) Φ) Φ = 1+Φ+(Φ Φ) + (((1 + S Φ) Φ) Φ) Φ = 1+Φ+(Φ Φ) + ((Φ Φ) Φ) + ((((1 + S Φ) Φ) Φ) Φ) Φ. = 1+Φ+(Φ Φ) + ((Φ Φ) Φ) + (((Φ Φ) Φ) Φ)... Corollaries: Coefficients of the CF-series and in the exp.prod.expansion: For a word w W (X ) and a letter a X Υ wa (t, u) = T 0 Υw (s, u) ( d ds Υa (s, u) ) ds = (Υ w (,u) Υ a (,u)) (t) C HK (t, u) =C H (t, u) C K (t, u) if H, K, HK H 28

18 Normal forms for nilpotent systems Zinbiel products provide the most compact way for specifying a normal form for free nilpotent control systems of rank r. The key is to index the coordinates by Hall words H H (r) def = {H H: h r} (rather than by the integers): Normal form for a free system (maximally free Lie algebra) ẋ a = u a if a X x HK = x H x K if H, K, HK H (r) (X ) Example: Normal from for a free nilpotent system (of rank r =5)usinga typical Hall set on the alphabet X = {0, 1} ẋ 0 = u 0 ẋ 1 = u 1 ẋ 01 = x 0 ẋ 1 = x 0 u 1 ẋ 001 = x 0 ẋ 01 = x 2 0 u 1 using ψ 1 (001) = (0(01)) ẋ 101 = x 1 ẋ 01 = x 1 x 0 u 1 using ψ 1 (101) = (1(01)) ẋ 0001 = x 0 ẋ 001 = x 3 0 u 1 using ψ 1 (0001) = (0(0(01))) ẋ 1001 = x 1 ẋ 001 = x 1 x 2 0 u 1 using ψ 1 (1001) = (1(0(01))) ẋ 1101 = x 1 ẋ 101 = x 2 1 x 0 u 1 using ψ 1 (1101) = (1(1(01))) ẋ = x 0 ẋ 001 = x 4 0 u 1 using ψ 1 (00001) = (0(0(0(01)))) ẋ = x 1 ẋ 0001 = x 1 x 3 0 u 1 using ψ 1 (10001) = (1(0(0(01)))) ẋ = x 1 ẋ 1001 = x 2 1 x2 0 u 1 using ψ 1 (11001) = (1(1(0(01)))) ẋ = x 01 ẋ 001 = x 01 x 3 0 u 1 using ψ 1 (01001) = ((01)(0(01))) ẋ = x 01 ẋ 101 = x 01 x 2 1x 0 u 1 using ψ 1 (01101) = ((01)(1(01))) 29

19 Free Zinbiel algebra On A k,0 (X )andâk,0(x) noncommutative polynomials and power series with zero constant term, define a (bilinear, noncommutative, nonassociative) product by w a = wa for any nonempty word w W 0 (X )anda X,and w (z a) =(w z) a +(z w) a for a X,w,z W 0 (X ) Exercise: With this product C k (X ) def =(A k,0 (X ), ) is a Zinbiel algebra that is free in the usual sense: If C is any Zinbiel algebra then any γ: X C extends uniquely to a Zinbiel algebra homomorphism ˆγ : C k (X ) C. Observe: The shuffle product is (the extension to A k (X ) of) the symmetrization of the Zinbiel product (and 1 1 cannot be defined meaningfully) w z + z w = w X z for w, z W 0 (X ) Lots of useful identities (e.g. to hide unwanted factorials) For w A k,0 (X ) define w 1 = λ 1 (w) =w x 1 = w, and inductively for n 1 λ n+1 (w) = w λ n (w) w (n+1) w x (n+1) = w n w = w X w x n = w x n X w Theorem: w w (n 1) = (n 1) w n λ n (w) = (n 1)! w n w x n = n! w n (= nλ n (w)) 30

20 Zinbiel products and dual PBW bases Every Hall-word H W (X ) \X, factors uniquely in the form H 1 H 2 H 3... H s 2 H s 1 H s < a This factorization (due to the connection with Lazard factorization) perfectly matches Sussmann s variation of parameters approach to obtaining the iterated integrals β H (T,u) in the exponential product expansion of the CF-series 1 β H = (β H1 (β H2...(β Hs 1 (β Hs β a ))...)) (...)! Compare to the mix of products fromdifferent algebras in Reutenauer s and Melançon s formula for the dual-pbw bases. In the shuffle algebra (A(X ), X ) the transposes of both the left and right translation by a letter λ a : w aw, andϱ a : w wa are derivations. However: On the Zinbiel algebra, only λ a is a derivation, ϱ a is not. λ a(w z) = (λ aw) z + w (λ az) ϱ a(w z) = w (ϱ az) 31

21 Realization of free Zinbiel algebra Compare the standard realization of polynomial algebras: k[x 1,...X n ] polynomials w/ coeff s in k k[x 1,...x n ] polynomial functions k n k = the subalgebra of Map(k n, k) =k kn generated by the projections x k = π k :(p 1,...,p n ) p k Similarly realize the free Zinbiel algebra as a Zinbiel algebra of timevarying scalars. E.g for an index set X of letters U =AC loc ([0, ), R) π a : U X U, π a ({u b : b X})=u a IIF(X ) Map(U X, U) subalgebra generated by projections π a,a X Theorem[Kawski and Sussmann]: The map ΥΥ(: C(X ) IIF(X ) defined by ΥΥ(: a π a is a Zinbiel algebra isomorphism. Zinbiel algebra surjective homomorphism is rather clear by now. The one-to-one-ness requires a sufficiently rich coefficient ring and some analysis (Nagano s theorem...). 32

22 CF series as identity map The Chen Fliess series of iterated integral functionals corresponds to the identity map under the Zinbiel algebra isomorphism ΥΥ, i.e. it is natural object. id A = w w w Schützenberger, Melançon & R. Combinatorics idâ ΥΥ w w ΥΥ(w) Sussmann Diff Equns proof H exp ([H] S H ) id  ΥΥ H exp ([H] β H ) Hom(A, A) = idâ ΥΥ Â A  AIIF free Zinbiel Zinbiel.algebra iterated integral algebra isomorphism functionals 33

23 Koszul duality and Leibniz operad The Zinbiel operad is dual to the Leibniz operad. ( pre-lie algebra structure, or noncommutative Lie algebra ) (Left) Leibniz identity v (w z)=(v w) z+ v (w z) for all v, w, z (3) Compare / recall: (right) Zinbiel identity v (w z) =(v w) z +(w v) z for all v, w, z (4) and the ( left) chronological identity of Gamkrelidze and Agrachev) x (y z) (x y) z = y (x z) (y x) z x (y z) y (x z) = (x y) z (y x) z sometimes suggestively written as L [x,y] =[L x,l y ] Puzzle: The anti-commutativity of the Lie-brackets appears so natural in control yet algebraically it appears to be only a coincidence. What in control corresponds geometrically to the Zinbiel algebra it must be connections; but they are not much used in controllability and optimality should they? What do they add? For details on Koszul duality of operads see Ginzburg and Kapranov, Duke. J. Math, For details on Leibniz algebras, and their role in cyclic homology, see numerous articles by Loday. 34