The planar Chain Rule and the Differential Equation for the planar Logarithms

Transcription

1 arxiv:math/ v1 [math.ra] 17 Feb 2005 The planar Chain Rule an the Differential Equation for the planar Logarithms L. Gerritzen ( ) Abstract A planar monomial is by efinition an isomorphism class of a finite, planar, reuce roote tree. If x enotes the tree with a single vertex, any planar monomial is a non-associative prouct in x relative to m array grafting. A planar power series f(x) over a fiel K in x is an infinite sum of K multiples of planar monomials incluing the unit monomial represente by the empty tree. For every planar power series f(x) there is a universal ifferential f(x) which is a planar power series in x an a planar polynomial in a variable y which is the ifferential of x. We state a planar chain rule an apply it to prove that the erivative (Exp k(x)) is the k ary planar exponential series. A special case of the planar chain rule is prove an it erive that the planar universe series Log k (1+x) of Exp k (x) satisfies the ifferential equation ( (1+x) ) (Log k (1+x)) = 1 where (1+x) is the erivative which when applie to x results in 1+x. Introuction For every planar power series f(x) in one variable x, there is a universal ifferential f(x) which is a planar power series in x an a planar polynomial in a variable y which is the ifferential of x. If g(x) is a planar power series of orer 1, then the substitution of g(x) for x in f(x) gives a power series f(g(x)) an the ifferential f(g(x)) of f(g(x)) can be compute as ϕ g (f) where ϕ g is the substitution homomorphism inuce by g an ϕ g is the homomorphism on the algebra of universal ifferential forms in x an extening ϕ g 1

2 an mapping into g. This formula is calle the planar chain rule. We note an application to the planar exponential series Exp k (x),k N,k 2, see [G1], section 3. One obtains a functional equation for the ifferential ω k of Exp k (x), see section 6. It follows that the erivative (Exp k(x)) is equal to Exp k (x). This gives a conceptual proof of this ifferential relation, while the proof in [G1], section 3, is quite technical. We also prove a special chain rule, if g (x) where g (x) = g(x) is the erivative with respect to x. From it we fin that ( (1+x) ) (Log k (1+x)) = 1, if Log k (1+x) is the k ary planar logarithm, see [G 2], section 5. In section 1 notions about {x, y}- labele finite, planar, reuce roote trees are fixe. Some basic facts about the algebra K{{x,y} of power series in x an polynomials in y over a fiel K are presente in section 2. The universal ifferential : K{{x}} K{{x,y} is introuce in section 4 an the planar chain rule is prove in section 5. The application to the erivative of Exp k (x) relative to x is given in section 6. In section 7 we prove a special rule for the erivative, if the substitution series g(x) satisfies the equation g(x) = 1+g(x) an we show that ( (1+x) ) (Log k (1+x)) = 1. If h n is the homogeneous component of Log k (1+x) of egree n, then h n+1 (x) = ( n)h n(x) for n 1. In [G2], section 6, the homogeneous component h 4 of Log k (1+x) has been compute to be h 4 (x) = ( k 3 3![2] 1 ) (k +1)(k 2)(k 3) x 4 + 4![3]! ( ![2] 2 ) (k 2) (x x 3 +x 3 x) 4![3]! 2

3 ( + ( ![2]! 2 8 4![3]! 1 k 2 2(k+1)(k 2) 2 3![2]! 4![3]! ) (x (x x 2 )+x (x 2 x)+(x x 2 ) x+(x 2 x) x) ( 1 + 3![2]! (k +1) 8 4![3]! ) x 2 x 2 ) (x x x 2 +x x 2 x+x 2 x x). From this it follows that h 3 (x) = ( 1 3 )h 4 (x) = 1 k 4[2] (x x2 +x 2 x) 1 3! h 2 (x) = ( 1 2 )h 3(x) = 1 2 x2, h 1 (x) = h 2 (x) = x. 1 Labele planar roote trees k 2 [2] x 3, Let x,y be ifferent elements; they will be calle variables in the sequel. A {x,y}- labele finite, planar reuce roote tree is a pair S = (T,λ) where T is a finite, planar, reuce roote tree, see [G1], an λ is a map L(T) {x,y} where L(T) enotes the set of leaves of T. Then eg x (T,λ) := #λ 1 (x) is calle the x-egree of (T,λ) which is the number of leaves of T labele with x by λ. Also eg y (T,λ) := #λ 1 (y) is calle the y-egree of (T,λ) which is the number of leaves of T labele with y by λ. Then eg (T,λ) = eg T := #L(T) is calle the total egree of (T,λ). We enote by P(x,y) the set of isomorphism classes of {x,y} labele finite, planar, reuce roote trees. Given S 1,...,S m P(x,y),S i = (T i,λ i ) for 1 i m. Then there is a unique S P(x,y) with the following properties: S = (T,λ) an if ρ T is the root of T, then T ρ T is a planar roote forest whose i- th component is isomorphic with T i. The set of leaves of L(T) is the isjoint union of L(T 1 ),...,L(T n ) an the labeling λ inuces λ i on T i for all 1 i m. 3

4 Werecallthataplanarrooteforestisanorereisjoint systemofplanarroote trees. The construction escribe above gives a unique map m : (P(x,y)) m P(x,y) which is calle the m-ary grafting operation for labele planar roote trees. Let P (x,y) = P(x,y) {1 P } where 1 p is a symbol representing the empty tree. There is a unique extension of the m ary grafting operation to a map m : (P (x,y)) m P (x,y) satisfying the properties: m(1 P,...,1 P ) = 1 P 2(S,1 P ) = 2(1 P,S) = S for all S P (x,y), m (S 1,...,S m ) = m 1 (S 1,...,S i 1,S i+1,...,s m ) if S i = 1 P for all S i P (x,y). It is also refere to as the m-ary grafting operation. 2 Polynomials an Power series Let K be a fiel an V be the K- vectorspace of all K value functions f on P (x,y) such that for all n N the set {S P (x,y) : f(s) 0,or x (S) = n} is finite. For any f V the value f(s) at S P (x,y) will be enote by c S (f). It will be calle the coefficient of f relative to S. For any m N,m 2, we are going to efine a m ary operation m on V by putting f := m (f 1,...,f m ) if c S (f) := c S1 (f 1 )... c Sm (f m ) where the summation is extene over all S 1,...,S m such that S = m (S 1,...,S m ). It is easy to check that m is a K multilinear map V m V. We will write f 1 f 2... f m for m (f 1,...,f m ) sometimes. In this note a K algebra is by efinition a K vectorspace W together with a family ( m ) m 2 of K multilinear maps m : W m W. In the context of opera theory such an algebra can be efine as an algebra over an opera in the category of K linear spaces which is freely generate by {µ m : m 2} where µ m has egree m, see [MSS], Chap. (1.9). It is a tree opera, see [BV]. Trees are inspire by the attempt to obtain a general composite operation from a collection of inecomposable operations, see [BV]. Other authors, see[sv], use the term hyperalgebra to enote this type of algebras. For the general theory of operas an algebras over operas see also [L], [LR], [H], [F]. 4

5 Therefore (V,( m ) m 2) is a K algebra; it is enote by K{{x,y} an will be refere to as the algebra of planar power series in x an polynomials in y. For any S P (x,y) there is a unique f S K{{x,y} such that { 1 : T = S c T (f s ) = 0 : T P (x,y),t S. We ientify f S with S an thus consier P (x,y) as a subset of K{{x,y}. It is easy to check that P (x,y) is a K linearly inepenent subset of K{{x,y} an that the K vectorspace of K{{x,y} generate by P (x,y) is a subalgebra. It is enote by K{{x,y} an is calle the algebra of planar polynomials: in x an y over K. For f K{{x,y} efine or x (f) =,if f = 0 an if then f 0, or x (f) = min{n N : there is S P(x,y) with eg x (S) = n an c S (f) 0.} It is calle the x orer of f Let f x := { 0 : f = 0 ( 1 2 )or (f) : f 0. Then f 1 f 2... f m x = f 1 x f 2 x... f m x m as or(f 1 f 2... f m ) = or(f i ) an, x efines a istance function on K{{x,y} by f,g x := f g x which is calle the x aic istance. Proposition 2.1. Any Cauchysequence in K{{x, y} relative to the x aic istance is convergent. Proof. By stanar arguments. 3 Substitution homomorphisms Proposition 3.1. Let g,h K{{x,y} an let or x (g) 1. Then there is a unique K algebra homomorphism such that ϕ (g,h) = ϕ : K{{x,y} K{{x,y} 5

6 (i) ϕ(x) = g (ii) ϕ(y) = h (iii) ϕ is continuous with respect to the x-aic topology. It is calle substitution homomorphism inuce by ( g, h). Proof. 1) First we efine ϕ (S) for any S P(x,y). If S = (T,λ) where T is a planar roote tree of egree, there is a m ary operation T associate to T, see [G2], Proposition (2.7) Let ϕ (S) := T(Z 1,...,Z m ) where Z i = g if λ(l i ) = x an Z i = h if λ(l i ) = y. Here l i is the i-th leaf of T. 2 )There is a unique K-linear extension of the map ϕ in 1) to a linear map ϕ : K{{x,y} K{{x,y}. 3) If f K{{x,y},f n := c S (f) S where the sum is extene over all S P(x,y) for which eg x (S) = n, then f n is a finite sum an f = f n. n=0 From 1) it follows that or x (ϕ(f)) or x (f) for all f K{{x,y} as or x (g) 1. Thus or x (ϕ(f n )) n for all n. One efines ϕ(f) = ϕ(f n ). The infinite sum is converging, as the sequence of n=0 partial sums is a Cauchysequence with respect to x. Thus a continuous K- linear map ϕ : K{{x,y} K{{x,y} is well efine. One can check that ϕ is an algebra homomorphism because its restriction to K{{x,y} is an algebra homomorphism by the efinition in 1) an 2). Notation: For ϕ = ϕ ( g,h) an f K{{x,y}, we enote ϕ(f) also by f(g,h). Then f = f(x,y) as ϕ(x,y) = i. Also f(g,h) = f(g(x,y),h(x,y)). It is obtaine by replacing x by g an y by h. For any k K we get f(kx,ky) = k m f(x,y), if f is homogeneous of egree m. 4 Universal erivation WeenotebyK{{x}}thecloseunitalK-subalgebraofK{{x,y}generatebyx. 6

7 Proposition 4.1. There is a unique continuous K-linear map such that (i) (x) = y : K{{x}} K{{x,y} (ii) (f 1 f 2... f m ) = m = (f 1... f i 1 (f i ) f i+1... f m ) for all m 2an all f 1,...,f m K{{x,y} The map is calle the universal erivation on K{{x}}. It is customary to enote (f) by f an call it the ifferential of f. Especially y =. Proof. 1) First (S) is efine for all S P(x,y). If S = 1 P, then (1 P ) := 0. If S 1 P,S = (T,λ),λ(l i ) = x for all leaves l 1,...,l m of T,m = eg(t),l(t) = {l 1,...,l m },then m let (S) := (T,λ (i) ), where λ (i) (l j ) = y if j = i an λ (i) (l j ) = x if j i. 2) As P (x) is K-linearly inepenent, the map in 1) has a unique extension to a K linear map 0 : K{x} K{x,y}. 3) There is a unique continuous extension of 0 to a K linear map : K{{x}} K{{x,y}. It is easy to check that property (ii) hols. 5 Chain rule Let g K{{x}},or(g) 1. As a special case of Proposition (3.1) we get a continuous K algebra homomorphism ϕ g = ϕ : K{{x}} K{{x}} such that ϕ(x) = g. Also there is a unique continuous K algebra homomorphism y : K{{x,y} K{{x,y} such that ϕ(x) = ϕ(x) = g ϕ(y) = g 7

8 Proposition 5.1. (ϕ) = ϕ. Proof. 1) Let δ := (ϕ) ϕ. Then δ is a continuous K linear map from K{{x}} into K{{x, y}} which satisfies 2) It is easy to see that δ(x) = 0 δ(f 1... f m ) = m ϕ(f 1 )... ϕ(f i 1 ) δ(fi) ϕ(f i+1 )... ϕ(f m. UsingthisformulaonecanprovebyinuctionontheegreethatforanyS P(x) one gets δ(s) = 0. If S = S 1... S m, then eg x (S i ) < eg x (S) an by inuction hypothesis δ(s i ) = 0,( i?) By the formula above then δ(s) = 0 3) As δ is continuous an K{x} is ense in K{{x}} an δ(f) = 0 by 2) an the linearity of δ, we see that δ = 0. Example: Let Then an g = k x,k K. ϕ(f(x)) = f(kx) (ϕ)ω(x,y) = ω(kx,ky). If f (resp. ω) is homogeneous of egree m, then f(kx) = k m f(x),ω(kx,ky) = k m ω(x,y). 6 The erivative of the exponentials Let now char(k) = 0,k Nk 2, an f = Exp k (x) be the k ary exponential series, see [G1]. Let ω = f be the ifferential of f. Then ω(kx,ky) = f k by the chain rule. If f = f n an f n is the homogeneous part of egree n, then n= ω = f n n=1 8

9 an f n is the homogeneous component ω n of ω of egree n with f 0 = 0. More precisely: eg x (f n ) = n 1, eg y (f n ) = 1. Now by the functional equation satisfie by f, we get ( ) f k (x) = f(kx) = k n f n (x) = where the summation is over all i i k =n k n f n (x) n=0 (i 1,...,i k ) N k f i1 f i2... f ik with i i k = n. Proposition 6.1. k n ω n (x,y) = k fi 1... fi j 1 ω(fi j ) fi j+1... fi k. j=1 i i k =n Proof. Immeiate by property (ii) of Proposition 4.1. There is a unique continuous unital algebra homomorphism ψ : K{{x,y} K{{x}} such that ψ(x) = x,ψ(y) = 1. Recall that ψ(f) = f for all f K{{x}}. Then (f ) : K{{x}} K{{x}} is the erivative relative to x, also enote symbolically by, see [G1] Proposition 6.2. (Exp k(x)) = Exp k (x). Proof. 1) We have to show that the erivative f n = (f n) of the homogeneous component of f(x) = Exp k (x) is equal to f n 1 for n 1. By applying ψ to the formula in Proposition (6.1) we get k n f n (x) = k j=1 i i k =n f i1... f ij 1 f i j f ij f ik. 9

10 2) We will show that f n(x) = f n 1 (x) for n 1 by inuction on n. If n = 1, then it is obviously true as or x (f (1+x)) 2. Let now n 2. As f i j = 0 if i j = 0 an f i j = f ij 1 for n > 0 j 1, we obtain k n f n (x) = k j= yi k =n i j 1 Now for any 1 i k we get i i k =n 1 f i1..., f ij 1 f ij 1 f ij f ik i 1 t...ti k =n i j 1 f i1 f i2... f ik = k n 1 f n 1 (x) f i1... f ij 1 f ij 1 f ij fi k = by ( ) It follows that an k n f n (x) = k kn 1 f n 1 (x) f n (x) = f n 1(x). 7 Special chain rules an logarithm Let h K{{x}}. Then there is a unique algebra homomorphism ϕ (x,h) = ϕ : K{{x,y} K{{x}} such that ϕ(x) = x an ϕ(y) = h Then θ := ϕ o : K{{x}} K{{x}} is a erivation on K{{x}} which means that is K linear, continuous an satisfies the general prouct rule m θ(f 1... f m ) = f 1... f i 1 θ(f i ) f i+1... f m for all f 1,...,f m K{{x}} This erivation θ will be formally enote by h. Be aware that (h )(f) h (f) in general. We enote (f) also by f an call it the erivative of f with respect to x. Proposition 7.1. : Let g K{{x}} such that g = 1+g Then (f(g(x)) = f (g(x))+(x )(f)(g(x)) = ((1+x) (f)(g(x)) 10

11 Proof. :1) First we will prove the formula in case f is a tree T of egree n If n = 1, then T = xan x = 1 while (1+x) (T) = 1+x.Thus f(g(x)) = g(x) (g) = 1+g from which the formula follows. Let now n > 1. Then an n (T) = T (i) where T (i) = T (z 1,,z n ) with z j = { x: j i 1 P : j = i an Also with an x (T) = n T n (T(g(x)) = S i S (i) = T (W 1,...,W n ) w j = { g(x): j i g (x): j = i As g = 1+g, we obtain from the multilinearity of T, that S(i) = T (g,...,g)+ T (i)(g,...,g) From these computations it follows that the proposition hols, if f is a tree 2) From 1) one can exten the result to polynomials by multilinearity. As both sies of the formula are continuous in f the result follows for power series. Let Log k (1+x) be the k ary planar logarithm. see [G2], section 5. Proposition 7.2. : for all ((1+x) (Log k(1+x) = 1 k N, k > 2. 11

12 Proof. Let Then g = Exp k (x) 1 Log k (Exp k (x)) = x From the special chain rule above, we can conclue that (((1+x) )(Log k))exp k (x)) = 1 From this the formula follows as f(g(x)) = 1 only if f(x) = 1. Corollary 7.3. : Let h n be the homogeneous component of Log k (1+x) of egree n. Then h 0 = 0 an for Proof. : h n+1 = nh n n 1 Log k (1+x) = h n =: h n=0 an (1+x) (h n) = h n +nh n, as (s (h n) = n h n. Moreover h n is homogeneous of egree (n 1). Thus h n+1 +nh n = 0 for all because n 1 (1+x) (h) = 1. 12

13 Example 7.4. We give the homogeneous terms h n (x) of Log k (1+x) of egree n 4. Recall that [n] = kn n 1 1 k 1 = k i an that So especially [n]! = n [i]. [2] = k +1, [3] = k 2 +k +1,[3]! = (k +1)(k 2 +k +1) = k 3 +2k 2 +2k +1. For any planar roote tree T in P, let {T} be the sum of all planar trees S in P whose unerlying roote tree is isomorphic to the unerlying roote tree of T. One can show that Exp k (x) an Log k (x) are infinite sums over {T} with T P. Then the homogeneous components h n (x) for n 4 can be written as, see [G2]: h 1 (x) = x h 2 (x) = 1 2 x2 h 3 (x) = 1 k 4 [2] {x x2 } 1 3! k 2 [2] x 3 i=0 h 4 (x) = ( k 3 1 (k +1)(k 2)(k 3![2] 4![3]! 3)) {x4 }+ +( (k 2 3![2] 4![3]! 2)){x x3 } +( 1 3![2]! ![3]! ){x (x x2 )} +( 1 3![2]! (k+1) 4![3]! ) {x 2 x 2 } +( 1 k 2 2(k+1)(k 2) 2 3![2]! Easy computations give 4![3]! )+{x x x 2 } ({x x x2 }) = 3 {x x 2 }+6 {x x x} ({x2 x 2 }) = 2 {x x 2 } ({x (x x2 )}) = 6 {x x 2 } {x x3 } = 2{x 3 }+3 {x x 2 } 13

14 from which one can erive that h 4(x) = 3h 3 (x) Similarly one gets In [DG] Log 2 (x) was compute to be h 3(x) = 2h 2 (x) h 2 (x) = h 1(x) Log 2 (x) = x 1 2 x (1 2 x x x2 x) 1( x (x x2 ) + 4 x (x2 x) + 5 x2 x (x x2 ) x (x2 x) x)+ higher terms. It is obtaine by substituting 2 for k in Log k (x). Open question: In there a proceure to construct h n+1 (x) which is an integral of ( n) h n (x) irectly from h n (x)? The problem arises from the fact that the space of homogeneous polynoms of egree n+1 whose erivative is zero is non- trivial for n 2 References [BV] Boarman, J. M. - Vogt, R. M.: Homotopy invariant algebraic structures on topological spaces, Springer-Verlag 1973 [DG] Drensky, V.- Gerritzen, L.: Nonassociative exponential an logarithm, Journal of Algebra 272 (2004) [F] Fresse, B.: Koszul uality of operas an homology on partition posets, in Homotopy theory an its applications (Evanston, 2002), Contemp. Math. 346 (2004), [G1] Gerritzen, L,: Planar roote trees an non-associative exponential series, Avances in Applie Mathematics 33 (2004) [G2] Gerritzen, L,: Automorphisms of the planar tree power series algebra an the non-associative logartithm, Serica Math. J., 30, Nos. 2-3 (2004), [Hol] Holtkamp, R.: On Hopf algebra structures over operas, preprint, July 2004, ArXiv:math.RA/

15 [L] Loay, J. - L.: La renaissance es operaes, Seninaire Burbaki, Vol. 1994/95, Asterisque No. 237 (1996) Exp. No. 792, 3, [LR3] Loay, J.-L. - Ronco, M.: Algebr e Hopf colibres, c. R. Aca. Sci. Paris, 337. (2003), [MSS] Markl, M.- Shnier, S.- Stasheff, J.: Operas in Algebra, Topology an Physics, in: Math. Surveys Monogr., 2002, Chapter (1.5), p. 52 (see [G1]). [SU] Shestakov, F. P. - Umirbaev, U. U.: Free Akivis algebras, primitive elements an hyperalgebras, J. of Algebra 250 (2002),