The Campbell-Hausdor theorem

Transcription

1 M.M. Schipper The Campbell-Hausdor theorem Bachelor's thesis, June 26, 2014 Supervisor: Dr. L.D.J. Taelman Mathematisch Instituut, Universiteit Leiden

2 Contents 1 Introduction 3 2 Associative and Lie algebras Denitions Free associative and free Lie algebras Enveloping algebra Completions and the exponential and logarithmic map The Campbell-Hausdor theorem Specialisations Proof of the Campbell-Hausdor theorem Campbell-Hausdor for matrices Convergence of matrix series Matrix exponential and logarithmic map Commuting matrices 23 2

3 1 Introduction The map exp: C C is a group homomorphism. That is, for all x, y C we have exp(x + y) = exp(x) exp(y). Additionally, one can dene exp(a) for a complex n n matrix A by the usual power series expansion A i exp(a) = i!. Let A and B be complex n n matrices. If we impose certain criteria on A and B, there exists a matrix C such that i=0 exp(c) = exp(a) exp(b). In general, C is not equal to A + B, but we can take C = log(exp(a) exp(b)) where log is dened by the usual power series expansion. We compute exp(a) exp(b) = 1 + A + B + AB + A2 2! + B2 + higher-order terms. 2! We take the logarithm to nd the lowest-order terms of C. We have C = A + B + 1 (AB BA) (A2 B + AB 2 + BA 2 + B 2 A 2ABA 2BAB) + higher-order terms = A + B [A, B] + ([A, [A, B]] [B, [A, B]]) + higher-order terms, 2 12 where [A, B] denotes the commutator AB BA. The Campbell-Hausdor theorem tells us that C is a series of which all terms are linear combinations of iterated commutators in A and B. The series itself does not depend on n, A or B. More naturally: C is an element of the free Lie algebra on {A, B}. The formula for C is even unique. See theorem 3.1 for the exact statement of the Campbell-Hausdor theorem. Campbell, Baker and Hausdor created the rst proof in In 1968 Eichler produced a totally dierent, purely algebraic proof. He shows by induction on n that all terms of order n are iterated commutators. Eichler proved the theorem in the context of matrices. In this thesis we will present Eichlers proof in the language of free Lie algebras. If A and B commute all terms of C, except for A + B, equal zero. In this case we have exp(a + B) = exp(a) exp(b). Less trivially, there exist non-commuting matrices A, B where 1 1 [A, B] + ([A, [A, B]] [B, [A, B]]) + higher-order terms = and thus exp(a + B) = exp(a) exp(b). 3

4 2 Associative and Lie algebras 2.1 Denitions Let k be a eld. Denition 2.1. An associative algebra over k is a k-vector space A with a k-bilinear map : A A A and an element 1 A A such that for all a, b, c A we have (A1) (a b) c = a (b c); (A2) 1 A a = a 1 A = a. We will use A instead of (A,, 1 A ) to denote an associative algebra. For x, y A we will use xy to indicate x y. Let A and A be associative k-algebras. A map f : A A is called a homomorphism of associative algebras over k if it is k-linear, respects and f(1 A ) = 1 A. Example 2.2. The vector space Mat n (k) of n n matrices over k with matrix multiplication forms an associative k-algebra with unit id Matn(k). Denition 2.3. A Lie algebra over k is a k-vector space L with a k-bilinear map [, ]: L L L such that for all x, y, z L we have (L1) [x, x] = 0; (L2) [x, [y, z]] + [z, [x, y]] + [y, [z, x]] = 0. We will often denote a Lie algebra (L, [, ]) simply by L. Let L and L be Lie algebras over k. We will call a map f : L L a homomorphism of Lie algebras over k if f is k-linear and it respects [, ]. Lemma 2.4. Let L be a Lie algebra over k with x, y L. We have [x, y] = [y, x]. Proof. Since [, ] is k-bilinear, we have [x + y, x + y] = [x, x] + [x, y] + [y, x] + [y, y]. By (L1) we have [x, x] = [y, y] = [x + y, x + y] = 0. Hence [x, y] = [y, x]. Example 2.5. Let A be an associative k-algebra. Then A together with the operation [, ]: A A A: (a, b) ab ba is a Lie algebra over k. The only non-trivial thing to check is (L2), which follows from the associativity of the product in A. 4

5 2.2 Free associative and free Lie algebras Let T be a set. Denition 2.6. A free associative algebra on T over k is a pair (A, f) with A an associative algebra over k and f : T A a map such that for all associative k-algebras B and for all maps g : T B there is a unique k-algebra homomorphism h: A B such that the diagram commutes. T g f Proposition 2.7. Let (A, f) and (A, f ) be free associative algebras on T over k. There is a unique isomorphism of associative k-algebras h: A A with h f = f. Proof. By the universal property of A there is a unique k-algebra homomorphism h: A A such that h f = f. We will show that h is an isomorphism. By the universal property of A there is a homomorphism of associative k-algebras h : A A such that h f = f. We will show that h and h are mutually inverse to each other. The map h h: A A is a k-algebra homomorphism with h h f = h f = f. The homomorphism id A has this property as well. According to the universal property of A we have h h = id A. Similarly one obtains h h = id A. Thus h is the unique isomorphism as described. Since a free associative algebra on T over k is uniquely unique, it is allowable to speak of the free associative algebra on T over k. Denition 2.8. Let i, j N. Regard T i as the set of words of length i. Let Ass i T be the vector space over k with basis T i. The dimension of Ass i T is T i. Concatenation denes an associative map : T i T j T i+j. It extends uniquely to a k-bilinear map Ass i T Assj T Assi+j T. Let A B h Ass T = i 0 Ass i T. Varying i and j, the system of maps : Ass i T Assj T Assi+j T gives a map : Ass T Ass T Ass T. This map is k-bilinear, associative and has a unit element, the empty word. Hence Ass T is an associative algebra over k. The set T can be regarded as a subset of Ass T, by identifying T with the set of words of length 1. Proposition 2.9. Let χ: T Ass T be the inclusion. Then (Ass T, χ) is the free associative algebra on T over k. 5

6 Proof. Let A be an associative k-algebra and let g : T A be a map. Let h: Ass T A be the k-linear map which sends a word w = x 1 x 2 x n with n N and x i T to g(x 1 )g(x 2 ) g(x n ). The map h is a k-algebra homomorphism with h χ = g. Let h : Ass T A be a homomorphism of associative k-algebras with h χ = g. For all x T we have h (x) = h (χ(x)) = g(x). Hence such a map is xed on T. Since Ass T is generated by T as an associative algebra, h is uniquely determined by its images on T. In conclusion h: Ass T A is the unique k-algebra homomorphism with h χ = g. The associative algebra Ass T can be regarded as the set of polynomials over k in non-commuting variables in the set T. We will now discuss the free Lie algebra. Denition A free Lie algebra on T over k is a pair (L, f) with L a Lie algebra over k and f : T L a map such that for all maps g : T M with M a Lie algebra over k, there is a unique map h: L M a homomorphism of Lie algebras over k such that the diagram commutes. T g f Theorem Let (L, f) and (L, f ) be free Lie algebras on T over k. There is a unique isomorphism of Lie algebras h: L L with h f = f. 2. A free Lie algebra on T over k exists. 3. Let (L, f) be a free Lie algebra. Then the map f is injective. 4. Let (L, f) be a free Lie algebra. We have L = i 1 Li such that L 1 is the k-vector space with basis T and for all for all i 2 we have L M h i 1 L i = [L j, L i j ], where [L i, L i j ] denotes te k-vector space generated by {[a, b]: a L j, b L i j }. Proof. Since a free Lie algebra is dened by a universal property, it can be proven that it is uniquely unique analogously to the free associative algebra. In [2, II.2.2, def.1] a Lie algebra L(T ) is constructed. In [2, II.2.2, prop.1] a map φ: T L(T ) is given and it is proven that the pair (L(T ), φ) is a free Lie algebra. The map φ is injective, see [2, II.2.2, after cor.2]. By the uniqueness of the free Lie algebra we have f = h φ for a certain isomorphism h. Hence f is injective. The free Lie algebra (L(T ), φ) has a vector space grading as described, see [2, II.2.6, eq.12 and above eq.12]. The vector space L has such a grading as well, since it is isomorphic to L(T ) as a Lie algebra over k. 6

7 Since a free Lie algebra on T over k is uniquely unique, we will speak of the free Lie algebra on T over k. We will denote its Lie algebra by L T and we will identify T with its image in L T. Notice that the vector spaces L i in thm are uniquely determined. It is possible to give an explicit basis H i of each L i T, but we will not do this. See [2, II.2.10, def.2 and prop.11] for a recursive construction of the sets H i. In [2, II.2.11, thm.1] it is proven that i 1 H i forms a basis of L T. Example Let T = {x, y} with x y. The construction of the rst four sets H i gives: H 1 = {x, y} H 2 = {[x, y]} H 3 = {[x, [x, y]], [y, [x, y]]} H 4 = {[x, [x, [x, y]]], [y, [x, [x, y]]], [y, [y, [x, y]]]}. For instance the element [x, [y, [x, y]]] L 4 T can be expressed in the basis H 4 as follows: [x, [y, [x, y]]] = [x, [[x, y], y]] by lemma 2.4 = [y, [x, [x, y]]] + [[x, y], [y, x]] by (L2) = [y, [x, [x, y]]] by (L1) and lemma 2.4. Remark Ass T is a Lie algebra over k with Lie bracket [a, b] = ab ba for a, b Ass T. (See example 2.5.) By the universal property of L T there exists a unique Lie homomorphism h: L T Ass T with h(x) = x for all x L T. The map h expands the nested Lie brackets into iterated commutators. For example for x, y L T we have h([x[x, y]]) = x(xy yx) (xy yx)x. Note that h(l i T ) Assi T injective. for all i. At the end of the next section we will see that h is 2.3 Enveloping algebra Let L be a Lie algebra over k. Denition An enveloping algebra of L is a pair (U, f) with U an associative k- algebra and f : L U a homomorphism of Lie algebras over k such that for all associative k-algebras A and for all Lie homomorphisms g : L A there is a unique homomorphism of associative k-algebras h: U A such that the diagram commutes. L g f U A h 7

8 Theorem Let (U, f) and (U, f ) be enveloping algebras of L. There is a unique isomorphism of associative k-algebras h: U U with h f = f. 2. An enveloping algebra of L exists. 3. Let (U, f) be an enveloping algebra of L. Then f is injective. Proof. The proof of the uniqueness is analogous to the proof of the uniqueness of the free associative algebra. In [1, III.1, after def.1.1] a pair (U L, ɛ) is constructed. It is proven that this pair is an enveloping algebra of L, see [1, III.1, thm.1.2]. One should read between the lines to nd the injectivity of the map ɛ. We will now clarify this. The objects T 1 L, T L and I are dened in the text mentioned. The map ɛ is a composition of three maps, of which the rst two are clearly injective. For all elements z in the ideal I of T L, we have z / T 1 L. Hence for all x, y L = T 1 L T L with x y the classes of x and y in the quotient T L /I are not the same. Hence ɛ is injective. By the uniqueness of an enveloping algebra, it follows that f is injective. Since an enveloping algebra of L is uniquely unique, we will speak of the enveloping algebra of L. We will denote it by (U L, f). Example Let L be a Lie algebra over k with [, ] the zero map. Such a Lie algebra is called an abelian Lie algebra. Let (e i ) i I be a basis of L as k-vector space. We will show that U L is the polynomial ring on variables X i for i I. The Lie algebra L is identied with the space of monomials via f : L k[x i : i I], e i X i. The map f is a Lie algebra homomorphism since k[x i : i I] is a commutative ring. Let A be an associative algebra and let g : L A be a Lie algebra homomorphism. Since the g(e i ) commute, there is a unique algebra homomorphism h: k[x i : i I] A such that X i g(e i ). Let h: L T Ass T be as in remark Theorem The pair (U LT, f T ) is the free associative algebra on T over k; there is a unique algebra isomorphism ψ : Ass T U LT such that ψ χ = f T. Proof. Since h: L T Ass T is a Lie homomorphism, there is a unique k-algebra homomorphism φ: U LT Ass T such that φ f = h, by the universal property of U LT. χ T L T U LT Ass T h φ ψ f 8

9 According to the universal property of Ass T there is a unique homomorphism of associative algebras ψ : Ass T U LT with ψ χ = f T. We obtain φ ψ χ = φ f T = h T = χ. By the universal property of Ass T we have φ ψ = id AssT. Furthermore we nd ψ φ f T = ψ h T = ψ χ = f T. Since ψ φ can be interpreted as a Lie homomorphism, the universal property of L T gives ψ φ f = f. Then we have ψ φ = id ULT by the universal property of U LT. Hence ψ and φ are mutually inverse bijections. So there is a unique isomorphism of associative algebras ψ : Ass T U LT such that the diagram χ T Ass T f T ψ U LT is commutative. Since f is injective, h = φ f is. We will use the injectivity of this map in the proof of the Campbell-Hausdor theorem. 2.4 Completions and the exponential and logarithmic map We aim to dene an exponential and a logarithmic map by the usual power series expansions on the free associative algebra on T over k. These series need not be elements of Ass T. In this section T is supposed to be a nite set. Denition We dene the completions of the free Lie and free associative algebra to be ˆL T := L i T, and ˆ Ass T := Ass i T respectively. The k-bilinear maps [, ] and extend to k-bilinear maps on ˆL T ˆL T and Ass ˆ T Ass ˆ T. This turns ˆL T and Ass ˆ T into a Lie algebra and an associative algebra respectively. Furthermore dene ˆm T := i=0 Ass i T. 9

10 This is the ideal of Ass ˆ T generated by T. We will denote an element (f i ) i Ass ˆ T by i f i. In this way we regard Ass ˆ T as the set of power series over k in non-commuting variables in T. For a Ass ˆ T, we will use a i to indicate the homogeneous term of degree i of a. If a is not the zero element, dene ord(a) := min{i N: (a) i 0}. Otherwise, let ord(a) =. Denition The injective Lie homomorphism h: L T Ass T extends to a homomorphism of Lie algebras h: ˆL T ˆm T, f i h(f i ). By remark 2.13 we have h(f i ) Ass i T. Hence h(f i) Assi T = ˆm T, so h is well dened. The map h: ˆL T ˆm T is an injective Lie homomorphism since h: L T Ass T is. We will now dene an exponential and a logarithmic function by the usual formulas on subspaces of Ass ˆ T. To avoid division by zero, we assume char(k) = 0. Denition Dene the maps exp: ˆm T 1 + ˆm T and log: 1 + ˆm T ˆm T by the following formulas: a i exp(a) = i! and where a ˆm T. log(1 + a) = i=0 ( 1) i+1 a i We will verify that exp is well dened. Let a ˆm T with α = ord(a). We nd that ord(a i ) = αi i. Therefore we have ( n ) a i (exp(a)) n = Ass n T i! i=0 n for all n N. Hence exp(a) Ass ˆ T. Since (exp(a)) 0 = 1, exp is well dened. In the same way we can verify that log is well dened. Lemma exp and log are mutually inverse bijections. Proof. Let a ˆm T. It is known that log and exp are each others inverses in the power series ring Q[[X]]. Since k is an extension eld of Q, there is a map of Q-algebras ψ : Q[[X]] Ass ˆ T which transforms X into a. The power series λ i X i is mapped to λi a i, which lies in Ass ˆ T since ord(a i ) i. If we apply ψ to the equality exp(log(1 + X)) = 1 + X, we get exp(log(1 + a)) = 1 + a. In the same way we nd that exp is a right inverse of log. Remark Let x, y T. Since 1+ ˆm T is closed under multiplication exp(x) exp(y) 1 + ˆm T. By lemma 2.21 there is a unique z ˆm T such that exp(z) = exp(x) exp(y). i 10

11 3 The Campbell-Hausdor theorem In the previous chapter we have introduced all objects and maps we need to formulate the Campbell-Hausdor theorem. In chapter 3, let T be the set {x, y} with x y. The ground eld of the free associative and free Lie algebra is supposed to be of characteristic zero. Let h: ˆL T ˆm T be as in denition We will now formulate the main theorem of this thesis: Theorem 3.1. Let z = log(exp(x) exp(y)) ˆm T. Then we have z h(ˆl T ). We will use the whole of chapter 3 to present Eichlers proof of the Campbell-Hausdor theorem. (See [3] and [4, 7.7].) The map h: ˆL T ˆm T is an injective Lie homomorphism. So we can identify h( L ˆ T ) with L ˆ T. We will call an element in ˆm T Lie if it lies in L ˆ T. We can regard the quotient space ˆm T /ˆL T. For a, b ˆm T we will use a b to indicate that the classes of a and b are the same in ˆm T /ˆL T. So we aim to prove that z 0. Since z ˆm T, it can be written uniquely as z = F n (3.1) n=1 with F n Ass n T. We will prove by induction that F n 0 for all n. Since F n Ass n T and since L n T Assn T, we then nd that F n L n T and it follows that 3.1 Specialisations z = n=1 F n n 1 L n T = ˆL T. To prove the theorem, we need to substitute other elements for x, y in the polynomials F i. We will now make this rigorous. Denition 3.2. Let A be an associative algebra with a, b A. Let ψ be the unique homomorphism of associative algebras ψ : Ass T A which maps x, y to a, b. (We use the universal property of Ass T.) Dene for each F Ass T the element F (a, b) by F (a, b) := ψ(f ) A. Let S be a set and let a, b substitute its elements. Ass ˆ S. Since Ass ˆ S is an associative algebra, we can Lemma 3.3. Let F Ass T. Let α = ord(a), β = ord(b) and ζ = ord(f ). Then ord(f (a, b)) ζ min{α, β}. 11

12 Proof. If F 0, the statement is clearly true. Otherwise, the shortest word w of F has length ζ. If w = u ζ where u = x if α = min{α, β} and u = y if β = min{α, β}, we nd ord(w(a, b)) = ζ min{α, β}. This is the shortest word possible in F (a, b). Hence ord(f (a, b)) ζ min{α, β}. We have seen how to do substitutions in non-commutative polynomials. We can also do substitutions in power series with non-commuting variables in T. That is, we will dene how to do substitutions in elements of Ass ˆ T. Remember a i denotes the homogeneous component of a, where a Ass ˆ T. Denition 3.4. Let ψ be as in denition 3.2 with Ass ˆ S substituted for A and assume a, b ˆm T. Let ˆψ : Ass ˆ T Ass ˆ S, f i ψ(f i ). By lemma 3.3 the map ˆψ is well dened. Dene f(a, b) by f(a, b) := ˆψ(f) for f i=0 i=0 ˆ Ass T. Lemma 3.5. We have log(exp(a) exp(b)) = F i (a, b). Proof. By denition of F i (a, b), ˆψ and z we obtain F i (a, b) = ( ) ψ(f i ) = ˆψ F i = ˆψ(log(exp(x) exp(y))). Since ψ is an algebra homomorphism we have ˆψ(log(exp(x) exp(y))) = = ψ(log(exp(x) exp(y)) j ) log(exp(ψ(x)) exp(ψ(y))) j = log(exp(a) exp(b)). If we substitute Lie power series in a Lie polynomial, we expect to nd a Lie power series. This turns out to be the case. Lemma 3.6. For a, b ˆ L T ˆm T and F L T Ass T we have F (a, b) ˆL T. Proof. Let ψ be as in denition 3.2 with Ass ˆ T substituted for A. Let i: ˆL T Ass ˆ T be the inclusion. Let ψ be the unique homomorphism of Lie algebras ψ : L T ˆL T which sends x, y to a, b. (We use the universal property of L T.) 12

13 T L T Ass T ˆL T ψ i ψ ˆ Ass T Notice that ψ is a Lie homomorphism if we regard Ass T and Ass ˆ T as Lie algebras. The Lie algebra homomorphism ψ LT and i ψ from L T to Ass ˆ T are equal if we restrict them to T. So by the universal property of L T the square commutes. Hence F (a, b) = ψ(f ) = (i ψ )(F ) = ψ (F ) ˆL T. 3.2 Proof of the Campbell-Hausdor theorem We will prove by induction on n that F n 0 for all n N 1. By writing down the terms of z containing words of length 1 and 2 one nds F 1 = x+y and F 2 = 1 2 (xy yx). Since we identify h(li T ) and Li T, the terms F 1 and F 2 lie in L 1 T and L 2 T respectively. In conclusion we have F 1, F 2 0. Let N N >2 and assume all polynomials F n with 1 n < N are Lie. We will use the rest of this chapter to prove that F N is a Lie polynomial. The rst step is to derive the following equation: Lemma 3.7. Let a, b, c Ass 1 T. We have F N(a, b)+f N (a+b, c) F N (a, b+c)+f N (b, c). Proof. The associativity of the product in Ass ˆ T gives ( ) ( ) exp(a) exp(b) exp(c) = exp(a) exp(b) exp(c). Simple as it is, this step is crucial in Eichlers proof. Since exp and log are mutually inverse and since a, b, c ˆm T we can apply lemma 3.5 to obtain exp F j (a, b) exp(c) = exp(a) exp F j (b, c). Note that F j(a, b) and F j(b, c) lie in ˆm T. Taking the logarithm and applying lemma 3.5 a second time gives F j (a, b), c = a, F j (b, c) ˆm T. (3.2) F i Let G N be the homogeneous term of degree N of the left side of equation 3.2. By lemma 3.3 one nds that the terms with i > N do not contribute to G N. Hence we have N G N = F j (a, b), c. (3.3) F i F i N 13

14 We will now discuss which j, depending on i, contribute to G N. We need the following two lemmas. Lemma 3.8. Let A, B ˆm T, let i N 2, let n N 1. Let ɛ i n Assi T. For all F Ass i T we have (F (A + ɛ, B)) n = (F (A, B)) n. Proof. We will give a proof by induction on i. Assume ɛ 0. Step 1. Let F Ass 2 T. Then F is a linear combination of the words x2, xy, yx and y 2. We have ord(ɛa) = ord(aɛ) n + 1 and ord(ɛ 2 ) 2n. So we get ((A + ɛ) 2 ) n = (A 2 + Aɛ + ɛa + (ɛ) 2 ) n = (A 2 ) n. In the same way we obtain ((A + ɛ)(b)) n = (AB) n. The statement follows for i = 2. Step 2. Let I N >2 and assume the statement is true for all 2 i < I. It is sucient to prove the statement for monomials in Ass I T. Let F AssI T be a monomial. Then F = Hx or F = Hy for a certain H Ass I 1 T. We will prove the statement for the rst case. Using lemma 3.3 we nd ord(h(a + ɛ, B)ɛ) n + 1. Since (A) 0 = 0 and by the induction hypothesis we obtain Hence (F (A + ɛ, B)) n = (F (A, B)) n. (H(A + ɛ, B)(A + ɛ)) n = (H(A + ɛ, B)A) n n 1 = (H(A, B)) j (A) n j j=0 = (H(A, B)A) n. Lemma 3.9. Let A, B ˆm T and let ɛ i 2 Assi T. Let n N. For all F Assn T we have (F (A + ɛ, B)) n = (F (A, B)) n. Proof. We will prove this lemma by induction on n. Assume ɛ 0. Step 1. For n = 0 and for all F Ass 0 T we have F (A + ɛ, B) = F = F (A, B). Step 2. Let N N and assume the statement holds for all n N with n < N. It is sucient to prove the statement for monomials in Ass N T. Let F AssN T be a monomial. Then F = Hx or F = Hy for a certain H Ass N 1 T. We will prove the statement for the rst case. Using lemma 3.3, one nds ord(h(a + ɛ, B)ɛ) N + 1. Hence (H(A + ɛ, B)(A + ɛ)) N = (H(A + ɛ, B)A) N. Since (A) 0 = 0 and since ord(h(a + ɛ, B)) N 1 we have (H(A + ɛ, B)A) N = (H(A + ɛ, B)) N 1 (A) 1. By the induction assumption the right side equals (H(A, B)) N 1 (A) 1 = (H(A, B)A) N. It follows that (F (A + ɛ, B)) N = (F (A, B)) N. 14

15 We will now go back to equation 3.3. For i = 1 we have F 1 F j (a, b), c = F j (a, b) + c N N = F N (a, b) since F j (a, b) lies in Ass j T and since c is homogeneous of degree 1 N. For 2 i N 1 we use lemma 3.8 to obtain N 1 N 1 N 1 F j (a, b), c = F j (a, b), c. i=2 F i For i = N we use lemma 3.9 to nd F N F j (a, b), c Hence N 1 G N = F N (a, b) + i=2 N F i N i=2 F i = (F N (F 1 (a, b), c)) N = F N (F 1 (a, b), c). N 1 F j (a, b), c N N + F N (a + b, c). We will now determine the class of G N in ˆm T /ˆL T. This is where we use the induction assumption. For 2 j < N we have F j L T by this assumption. Since a, b Ass 1 T = L1 T we can apply lemma 3.6 to obtain F j (a, b) L T. Applying the induction assumption and lemma 3.6 a second time, we nd N 1 i=2 F i( N 1 F j(a, b), c) L T. In conclusion G N F N (a, b) + F N (a + b, c). We can do the same computation for the right side of equation 3.2 to obtain the important result F N (a, b) + F N (a + b, c) F N (a, b + c) + F N (b, c). (3.4) This nishes the proof of lemma 3.7. We will now substitute several variables for a, b and c in equation 3.4 to nally obtain F N 0. Before we start substituting, we will derive some useful facts. Lemma Let A, B ˆm T with AB = BA. We have exp(a) exp(b) = exp(a + B). Proof. Since A and B commute, we can apply Newton's binomial theorem to nd (A + B) i i A i j B j exp(a + B) = = i! j!(i j)!. Furthermore we have i=0 exp(a) exp(b) = i=0 j=0 i=0 j=0 A i B j. i!j! The coecient of A i j B j 1 with 0 j i is j!(i j)! for both exp(a+b) and exp(a) exp(b). Since the terms of both series are equal, we get exp(a+b) = exp(a) exp(b) 1+ ˆm T. 15

16 We will use the following three facts while performing substitutions. Lemma Let r, s k. We have 1. F N (ra, sa) = F N (a, 0) = F N (0, a) = F N (ra, rb) = r N F N (a, b). Proof. Since ra and sa commute, we have exp(ra) exp(sa) = exp(ra + sa) by lemma We take the logarithm and use lemma 3.5 to get F i (ra, sa) = F 1 (ra, sa). Because a lies in Ass 1 T, the polynomial F i(ra, sa) is homogeneous of degree i. By the uniqueness of the power series in Ass ˆ T we have F i (ra, sa) = 0 for all i > 1. Hence F N (ra, sa) = 0. If we take r = 1 and s = 0 we nd F N (a, 0) = 0 and analogously F N (0, a) = 0. Since F N is homogeneous of degree N, we have F N (ra, rb) = r N F N (a, b). We will now start substituting in equating 3.4. Bear in mind that we aim to derive F N 0. We will rst derive a relation between F N (a, b) and F N (b, a). Lemma We have F N (a, b) ( 1) N+1 F N (b, a). Proof. First of all, substitute b for c in equation 3.4. (Note that b Ass 1 T.) We nd F N (a, b) + F N (a + b, b) F N (a, 0) + F N (b, b) 0, where the last equivalence holds by fact 2 and 1 of lemma Therefore we have F N (a, b) F N (a + b, b). (3.5) Next, substitute a for b in equation 3.4 and use fact 1 and 2 of lemma 3.11 to nd F N (a, a) + F N (0, c) F N (a, a + c) + F N ( a, c) 0. Then, replacing a, c by a, b respectively gives F N (a, b) F N ( a, a + b). (3.6) 16

17 With equation 3.5 and 3.6 we can nd a relation between F N (a, b) and F N (b, a). We have F N (a, b) F N (a + b, b) F N ( a b, a) F N ( b, a) ( 1) N F N (b, a), where the last equivalence follows from lemma F N (a, b) and F N (b, a) is Hence the relation between F N (a, b) ( 1) N+1 F N (b, a). (3.7) Equation 3.4 has more to oer. Lemma We have (1 2 1 N )F N (a, b) (1 + ( 1) N )2 N F N (a, a + b). Proof. Now we substitute b/2 for c. (Note that b/2 Ass 1 T.) This gives F N (a, b) + F N (a + b, b/2) F N (a, b/2) + F N (b, b/2) F N (a, b/2), where the last equivalence follows from lemma So we have F N (a, b) F N (a, b/2) F N (a + b, b/2). (3.8) Next, substituting b/2 for a in equation 3.4 and using lemma gives F N ( b/2, b) + F N (b/2, c) F N ( b/2, b + c) + F N (b, c) We replace b, c by a, b respectively, to obtain F N (b/2, c). F N (a, b) F N (a/2, b) F N ( a/2, a + b). (3.9) With equation 3.8 and 3.9 we can pass from polynomials in A, B to polynomials in A/2, B/2. This enables us to nd a relation between F N (a, b) and itself. We will use 3.9 to rewrite the two terms on the right side of 3.8. F N (a, b/2) F N (a/2, b/2) F N ( a/2, a + b/2) F N (a/2, b/2) + F N (a/2, a/2 + b/2) 2 N (F N (a, b) + F N (a, a + b)), 17

18 where the last two equivalences follow from 3.6 and fact 3 of lemma respectively. The second term of 3.8 is rewritten by 3.9 as follows: F N (a + b, b/2) F N (a/2 + b/2, b/2) F N ( a/2 b/2, a + b/2) F N (a/2, b/2) + F N (a/2 + b/2, a/2) 2 N ( F N (a, b) + F N (a + b, a)), using 3.5, 3.6 and lemma So 3.8 becomes F N (a, b) 2 1 N F N (a, b) + 2 N F N (a, a + b) 2 N F N (a + b, a). We use 3.7 to simplify this equation to (1 2 1 N )F N (a, b) (1 + ( 1) N )2 N F N (a, a + b). (3.10) If N is odd, division by N gives the result F N (a, b) 0. (We use the fact that N > 1.) If N is even, we need to do a little work to derive the same. Lemma If N is even, we have F N (a, b) 0. Proof. We replace b by b a in equation 3.10, obtaining Applying 3.6 on the left side of 3.11 gives (1 2 1 N )F N (a, b a) 2 1 N F N (a, b). (3.11) F N ( a, b) 21 N N F N(a, b). (3.12) This equation allows us to nd a relation between F N (a, b) and itself. We substitute a for a in equation 3.12 and use the equation itself to obtain Since N > 2, we have 21 N F N (a, b) N F N( a, b) ( ) 2 1 N N F N (a, b). ( ) 2 1 N 2 1. (3.13) Thus we have F N (a, b) N Finally, we substitute x, y for a, b respectively (which is possible since x, y Ass 1 T ), to nd the desired result. F N = F N (x, y) = F N (a, b) 0. Eichler has used slightly dierent substitutions in his proof. This `freedom' suggests there is a shorter way to obtain F N 0. 18

19 4 Campbell-Hausdor for matrices We will show that the Campbell-Hausdor theorem in the context of matrices is implied by theorem 3.1. We need to think about convergence of series in Mat n (C) before we start substituting matrices for the variables of the free associative algebra. 4.1 Convergence of matrix series Denition 4.1. Let be the map : Mat n (C) R 0, (a ij ) i,j a ij 2 where a ij is the norm of a ij on C. The map is a norm on Mat n (C) and (Mat n (C), ) is isomorphic to R 2n2 with the euclidean norm. The norm is submultiplicative: Lemma 4.2. We have AB A B for all A, B Mat n (C). Proof. If A = (a ij ) i,j and B = (b ij ) i,j, then we have by the triangle inequality and the multiplicative property of the norm on C n (AB) ij = a il b lj l=1 n a il b lj. By the Cauchy-Schwarz inequality we get n a il b lj n a il 2 n b lj 2. It follows that l=1 l=1 l=1 AB = (AB) ij 2 i,j i = A B. n a il 2 j l=1 l=1 i,j n b lj 2 l=1 19

20 Let T = {x, y} and take the ground eld k for Ass T and L T equal to C. Let F Ass T and let A, B Mat n (C). Then F (A, B) Mat n (C) makes sense. (See denition 3.2.) In this chapter we will denote an element f Ass ˆ T by i f i where each f i is a monomial or by a w w where we sum over all words w with a w C. Lemma 4.3. Let i f i Ass ˆ T. Assume i f i( A, B ) converges absolutely in R, then i f i(a, B) converges in Mat n (C). Proof. Since i f i( A, B ) converges in R, the sequence of partial sums ( n ) f i ( A, B ) i=0 is Cauchy. Let ɛ R >0. Let N N be such that for all n, m N >N with n m we have n n f i ( A, B ) = f i ( A, B ) < ɛ. i=m+1 i=m+1 Then we have by the triangle inequality and the submultiplicative property of on Mat n (C) the inequality n n f i (A, B) f i ( A, B ). i=m+1 i=m+1 We use the fact that each f i is a monomial. For n, m N >N with n m the last term is smaller than ɛ. Hence ( i f i(a, B)) n is a Cauchy sequence in Mat n (C). By the completeness, the sequence converges. 4.2 Matrix exponential and logarithmic map n ( 1) i+1 λ i Remark 4.4. The series exp(λ) = i=0 λi i! and log(1 + λ) = i are absolutely convergent for all λ R and for all λ ( 1, 1) respectively. By lemma 4.3 i=0 Ai i! converges and ( 1) i+1 A i i converges if A < 1 with A Mat n (C). Denition 4.5. Dene the matrix exponential and logarithmic map to be exp: Mat n (C) Mat n (C), A log : D Mat n (C), 1 + A with D = {1 + A Mat n (C): ( 1) i+1 A i i i=0 A i i!, ( 1) i+1 A i converges}. i 20

21 Lemma 4.6. Let A, B Mat n (C). 1. If A and B commute, we have exp(a + B) = exp(a) exp(b). 2. exp(a) is invertible with inverse exp( A). Proof. The rst statement is proven as in lemma Since A and A commute we have exp(a) exp( A) = exp(0) = 1. Similarly exp( A) is a left inverse of exp(a). Remark 4.7. By lemma 4.6 the codomain of the matrix-exponential map can be taken equal to the unit group of the domain (analogous to the exponential map on C). We shall denote the unit group of Mat n (C) by GL n (C). The map exp: C C is a group homomorphism, whereas exp: Mat n (C) GL n (C) is not if n > 1. Let A, B Mat n (C). Denition 4.8. Let ψ : Ass T Mat n (C) be the map as in denition 3.2 which replaces x, y by A, B. Dene { ˆψ : f i Ass ˆ T such that } f i (A, B) converges Mat n (C), f i ψ(f i ). i i i i Lemma 4.9. Let i f i = log(exp(x)) and i g i = exp(log(1 + x)). 1. If i f i( A, B ) converges absolutely we have log(exp(a)) = A. 2. If i g i( A, B ) converges absolutely we have exp(log(1 + A)) = 1 + A. 3. If exp( A ) 1 < 1 the series i f i( A, B ) converges absolutely. 4. If A < 1 the series i g i( A, B ) converges absolutely. Proof. If i f i( A, B ) converges absolutely, the terms of i f i(a, B) can be interchanged. Hence we have ( ) ˆψ f i = f i (A, B) = log(exp(a)). i i We apply ˆψ to the equality log(exp(x)) = x (see lemma 2.21) to nd log(exp(a)) = A. The second statement is proven analogously. Let h = h w w ˆm T with h w 0 for all words w. We can write h i log(1 h) = i = a w w, ( 1) i+1 h i log(1 + h) = = b w w. i By the triangle inequality and since a w 0 we have b w a w for all w. So if aw w( A, B ) converges, the series w b ww( A, B ) converges absolutely. We can take h equal to exp(x). If we assume exp( A ) 1 < 1, it follows that w b ww( A, B ) converges absolutely. The fourth statement is derived similarly starting with exp( log(1 x)) = a w w and exp(log(1 + x)) = b w w. 21

22 Let A, B Mat n (C) with A, B < log( 2). Lemma For i f i = log(exp(x) exp(y)), the series i f i( A, B ) converges absolutely. Proof. The proof is analogous to the proof of lemma where we take h equal to exp(x) exp(y) and use the fact that exp( A ) exp( B ) 1 < exp(log( 2)) 2 1 = 1. Remark By lemma 4.10 the series i f i(a, B) converges and its terms can be interchanged. Therefore we can dene the matrix C := log(exp(a) exp(b)) = i f i(a, B). By lemma 4.9, C is the unique matrix with exp( C ) 1 < 1 and exp(c) = exp(a) exp(b). Theorem The matrix C is a series of which all terms are linear combinations of iterated commutators in A and B. Proof. As in lemma 3.5 we can prove C = n=1 F n(a, B) where F n is as in equation 3.1. The last equality in this lemma holds since it is allowed to interchange the terms of log(exp(a) exp(b)). Each F i (A, B) is a linear combination of iterated commutators in A and B by theorem

23 5 Commuting matrices Let A, B Mat n (C). If A and B commute, that is [A, B] = 0, we have exp(a + B) = exp(a) exp(b). In general, the converse implication does not hold. Lemma 5.1. There exists an n N and matrices A, B Mat n (C) such that [A, B] 0 and exp(a + B) = exp(a) exp(b). Proof. Take n = 2, A = ( ) πi and B = ( ) πi Because A is diagonal, we compute ( ) exp(0) 0 exp(a) = = 1 0 exp(2πi) Matn(C). We can compute exp(b) using the Jordan Normal Form of B. eigenvalues of B, this form is diagonal. We have ( ) 0 0 B = Q Q 1 = Q A Q 1 0 2πi Since 0 and 2πi are for a certain matrix Q GL n (C). Hence exp(b) = Q exp(a) Q 1 = 1 Matn(C). In the same way one nds exp(a + B) = 1 Matn(C). In conclusion we have exp(a + B) = exp(a) exp(b), but ( ) ( ) πi AB = 0 4π 2 0 4π 2 = BA. However, for a certain subspace of Mat n (C) the converse implication is true. Denition 5.2. Dene for all l N with 1 l n the subspace T n,l := {X = (x ij ) i,j Mat n (C): x ij = 0 if j < i + l}. T n,l is the subspace of matrices where all non-zero entries lie above l-th diagonal. For example, for A T 4,2 the matrix has the following shape 0 0 A = We will often denote T n,1 simply by T n. This is the subspace of upper triangular matrices with zeros on the main diagonal. 23

24 Remark 5.3. We have the inclusions of subspaces T n = T n,1 T n,2... T n,n = {0}. For all l, m N with 1 l, m n we have T n,l T n,m = {AB : A T n,l, B T n,m } T n,l+m. (5.1) Proposition 5.4. The maps exp: T n 1 + T n and log: 1 + T n T n are mutually inverse bijections. Proof. Let A T n. By remark 5.3 we have exp(a) = 1 + n 1 Ai i! 1 + T n. Since A is nilpotent, log(a) is well dened and log(a) T n. For i f i = exp(log(x)), the series i f i( A, B ) converges absolutely since it concerns a nite sum. By lemma we have exp(log(a)) = A. We will now prove that exp: T n 1 + T n is injective. Let A, B T n with exp(a) = exp(b). I will prove by induction that A B T n,i for all 1 i n. We clearly have A B T n,1. Let I N and assume A B T n,i for all 1 i < I. Since exp(a) = exp(b) we have Since A B = n 1 j=2 B j A j. j! B j A j = B(B j 1 A j 1 ) (A B)A j 1, we nd that B j A j T n,i if B j 1 A j 1 T n,i 1 using the induction assumption and remark 5.3. Since B A T n,i 1, it follows inductively that B j A j T n,i for all 2 j n 1. Hence we obtain A B T n,n = {0}. Since A = B, the map exp is injective. It then follows that exp and log are mutually inverse to each other. Theorem 5.5. Let A, B T n. We have [A, B] = 0 if and only if exp(a + B) = exp(a) exp(b). Proof. We will prove the implication from right to left. Assume exp(a+b) = exp(a) exp(b). We claim that [A, B] T n,i for all 1 i n. We will prove this by induction on i. Step 1. For i = 1 we have [A, B] T n,2 T n,1 by remark 5.3. Step 2. Let I N, suppose the claim is true for all 1 i < I. Let F j be as in equation 3.1. Applying equation 5.1 j 1 times, we obtain F j (A, B) T n,j. Hence for all j n we have F j (A, B) = 0. According to the proof of theorem 4.12 we have exp(a) exp(b) = exp(a + B + F 2 (A, B) + F 3 (A, B) F n 1 (A, B)). Since exp(a + B) = exp(a) exp(b) and by proposition 5.4 one nds 0 = F 2 (A, B) + F 3 (A, B) F n 1 (A, B). For j > 2, we have F j (A, B) T n,i by equation 5.1 and the induction assumption. In conclusion, we have F 3 (A, B) F n 1 (A, B) T n,i. If F 2 (A, B) T n,i there is a j with j < i + I and (F 2 (A, B)) ij 0. We then have F 2 (A, B) + F 3 (A, B) F n (A, B) 0. This is a contradiction. Hence [A, B] = 2F 2 (A, B) T n,i. It follows that [A, B] T n,n = {0}. 24

25 Bibliography [1] Jean-Pierre Serre. Lie algebras and Lie groups. W.A. Benjamin, Inc., New York- Amsterdam, [2] Nicolas Bourbaki. Lie groups and Lie algebras, Part I. Addison-Wesley, Massachusetts, [3] M. Eichler. A new proof of the Baker-Campbell-Hausdor formula. Journal of the Mathematical Society of Japan. 20:23-25, [4] John Stillwell. Naive Lie Theory. Springer, New York,