The Campbell-Hausdor theorem

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M.M. Schipper

The Campbell-Hausdor theorem

Bachelor's thesis, June 26, 2014 Supervisor: Dr. L.D.J. Taelman

Mathematisch Instituut, Universiteit Leiden

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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

exp(A) =

∞

X

i=0

Aⁱ 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

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

2! + higher-order terms.

We take the logarithm to nd the lowest-order terms of C. We have C = A + B + 1

2(AB − BA) + 1

12(A²B + AB²+ BA²+ B²A − 2ABA − 2BAB) + higher-order terms

= A + B + 1

2[A, B] + 1

12([A, [A, B]] − [B, [A, B]]) + higher-order terms,

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 1906. In 1968 Eichler pro- duced 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

2[A, B] + 1

12([A, [A, B]] − [B, [A, B]]) + higher-order terms = 0 and thus exp(A + B) = exp(A) exp(B).

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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 1A∈ A such that for all a, b, c ∈ A we have

(A1) (a · b) · c = a · (b · c);

(A2) 1A· a = a · 1_A= a.

We will use A instead of (A, ·, 1A) to denote an associative algebra. For x, y ∈ A we will use xy to indicate x · y. Let A and A⁰ be associative k-algebras. A map f : A → A⁰ is called a homomorphism of associative algebras over k if it is k-linear, respects · and f (1A) = 1A⁰.

Example 2.2. The vector space Matn(k)of n × n matrices over k with matrix multiplication forms an associative k-algebra with unit idMatn(k).

Denition 2.3. A Lie algebra over k is a k-vector space L with a k-bilinear map [−, −] : L × L → Lsuch 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 L⁰be Lie algebras over k. We will call a map f : L → L⁰ 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) 7→ 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.

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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

T A

B

f

g h

commutes.

Proposition 2.7. Let (A, f) and (A⁰, f⁰)be free associative algebras on T over k. There is a unique isomorphism of associative k-algebras h: A → A⁰ with h ◦ f = f⁰.

Proof. By the universal property of A there is a unique k-algebra homomorphism h: A → A⁰ such that h ◦ f = f⁰. We will show that h is an isomorphism.

By the universal property of A⁰ there is a homomorphism of associative k-algebras h⁰: A⁰→ Asuch that h⁰◦ f⁰= f. We will show that h and h⁰are mutually inverse to each other. The map h⁰◦ h : A → Ais a k-algebra homomorphism with h⁰◦ h ◦ f = h⁰◦ f⁰= f.

The homomorphism idA has this property as well. According to the universal property of A we have h⁰ ◦ h = id_A. Similarly one obtains h ◦ h⁰ = id_A⁰. 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 Tⁱ as the set of words of length i. Let Assⁱ_T be the vector space over k with basis Tⁱ. The dimension of Assⁱ_T is |T |ⁱ.

Concatenation denes an associative map ·: Tⁱ× T^j → T^i+j. It extends uniquely to a k-bilinear map · AssⁱT× Ass^j_T → Ass^i+j_T . Let

Ass_T =M

i≥0

Assⁱ_T.

Varying i and j, the system of maps ·: Assⁱ_T× Ass^j_T → Ass^i+j_T gives a map ·: AssT × Ass_T → AssT. 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 AssT, by identifying T with the set of words of length 1.

Proposition 2.9. Let χ: T ,→ AssT be the inclusion. Then (AssT, χ) is the free associative algebra on T over k.

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Proof. Let A be an associative k-algebra and let g : T → A be a map. Let h: AssT → A be the k-linear map which sends a word w = x1x2· · · x_n with n ∈ N and xi ∈ T to g(x₁)g(x₂) · · · g(x_n). The map h is a k-algebra homomorphism with h ◦ χ = g.

Let h⁰: Ass_T → A be a homomorphism of associative k-algebras with h⁰◦ χ = g. For all x ∈ T we have h⁰(x) = h⁰(χ(x)) = g(x). Hence such a map is xed on T . Since AssT

is generated by T as an associative algebra, h⁰is uniquely determined by its images on T . In conclusion h: AssT → Ais the unique k-algebra homomorphism with h ◦ χ = g.

The associative algebra AssT 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 2.10. 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

T L

M

f

g h

commutes.

Theorem 2.11. 1. Let (L, f) and (L⁰, f⁰) be free Lie algebras on T over k. There is a unique isomorphism of Lie algebras h: L → L⁰ with h ◦ f = f⁰.

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 = L_i≥1Lⁱ such that L¹ is the k-vector space with basis T and for all for all i ≥ 2 we have

Lⁱ =

i−1

X

j=1

[L^j, L^i−j],

where [Lⁱ, 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.

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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 LT and we will identify T with its image in LT.

Notice that the vector spaces Lⁱ in thm. 2.11.4 are uniquely determined. It is possible to give an explicit basis Hi of each Lⁱ_T, but we will not do this. See [2, II.2.10, def.2 and prop.11] for a recursive construction of the sets Hi. In [2, II.2.11, thm.1] it is proven that S

i≥1H_i forms a basis of LT.

Example 2.12. Let T = {x, y} with x 6= y. The construction of the rst four sets Hi

gives:

H₁= {x, y}

H₂= {[x, y]}

H3= {[x, [x, y]], [y, [x, y]]}

H4= {[x, [x, [x, y]]], [y, [x, [x, y]]], [y, [y, [x, y]]]}.

For instance the element [x, [y, [x, y]]] ∈ L⁴T can be expressed in the basis H4 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 2.13. AssT is a Lie algebra over k with Lie bracket [a, b] = ab − ba for a, b ∈ AssT. (See example 2.5.) By the universal property of LT there exists a unique Lie homomorphism h: LT → Ass_T with h(x) = x for all x ∈ LT. The map h expands the nested Lie brackets into iterated commutators. For example for x, y ∈ LT we have

h([x[x, y]]) = x(xy − yx) − (xy − yx)x.

Note that h(Lⁱ_T) ⊆ Assⁱ_T for all i. At the end of the next section we will see that h is injective.

2.3 Enveloping algebra

Let L be a Lie algebra over k.

Denition 2.14. 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

L U

A

f

g h

commutes.

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Theorem 2.15. 1. Let (U, f) and (U⁰, f⁰) be enveloping algebras of L. There is a unique isomorphism of associative k-algebras h: U → U⁰ with h ◦ f = f⁰.

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 (UL, ) 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 . We will now clarify this. The objects T_L¹, TLand I are dened in the text mentioned. The map is a composition of three maps, of which the rst two are clearly injective. For all elements z in the ideal I of TL, we have z /∈ T_L¹. Hence for all x, y ∈ L = T_L¹ ⊆ T_L with x 6= y the classes of x and y in the quotient TL/I are not the same. Hence 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 (UL, f ).

Example 2.16. Let L be a Lie algebra over k with [−, −] the zero map. Such a Lie algebra is called an abelian Lie algebra. Let (ei)_i∈I be a basis of L as k-vector space.

We will show that ULis the polynomial ring on variables Xifor i ∈ I. The Lie algebra Lis identied with the space of monomials via

f : L → k[Xi: i ∈ I], ei 7→ X_i.

The map f is a Lie algebra homomorphism since k[Xi: 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(ei) commute, there is a unique algebra homomorphism h: k[Xi: i ∈ I] → A such that Xi 7→ g(e_i).

Let h: LT → Ass_T be as in remark 2.13.

Theorem 2.17. The pair (ULT, f |T)is the free associative algebra on T over k; there is a unique algebra isomorphism ψ : AssT → U_L_T such that ψ ◦ χ = f|T.

Proof. Since h: LT → Ass_T is a Lie homomorphism, there is a unique k-algebra homomorphism φ: ULT → Ass_T such that φ ◦ f = h, by the universal property of ULT.

T LT ULT

AssT χ

f h

φ ψ

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According to the universal property of AssT there is a unique homomorphism of associative algebras ψ : AssT → U_L_T with ψ ◦ χ = f|T. We obtain

φ ◦ ψ ◦ χ = φ ◦ f |_T = h|_T = χ.

By the universal property of AssT we have φ ◦ ψ = idAssT. 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 LT gives ψ ◦ φ ◦ f = f. Then we have ψ ◦ φ = idU_LT by the universal property of ULT. Hence ψ and φ are mutually inverse bijections. So there is a unique isomorphism of associative algebras ψ : AssT → U_L_T such that the diagram

T ULT

AssT f |T

χ ∼

ψ

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 expan- sions on the free associative algebra on T over k. These series need not be elements of AssT. In this section T is supposed to be a nite set.

Denition 2.18. We dene the completions of the free Lie and free associative algebra to be

LˆT :=

∞

Y

i=1

Lⁱ_T, and

Assˆ _T :=

∞

Y

i=0

Assⁱ_T

respectively. The k-bilinear maps [−, −] and · extend to k-bilinear maps on ˆLT × ˆL_T and Âss_T × Âss_T. This turns ˆLT and Âss_T into a Lie algebra and an associative algebra respectively. Furthermore dene

ˆ m_T :=

∞

Y

i=1

Assⁱ_T .

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This is the ideal of ˆAss_T generated by T .

We will denote an element (fi)i ∈ ÂssT by P_ifi. In this way we regard ÂssT as the set of power series over k in non-commuting variables in T . For a ∈ ÂssT, we will use ai

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 6= 0}. Otherwise, let ord(a) = ∞.

Denition 2.19. The injective Lie homomorphism h: LT → Ass_T extends to a homomorphism of Lie algebras

h : ˆL_T → ˆm_T,

∞

X

i=1

f_i 7→

∞

X

i=1

h(f_i).

By remark 2.13 we have h(fi) ∈ Assⁱ_T. Hence P^∞_i=1h(f_i) ∈Q∞

i=1Assⁱ_T = ˆm_T, so h is well dened. The map h: ˆLT → ˆm_T is an injective Lie homomorphism since h: LT → Ass_T is.

We will now dene an exponential and a logarithmic function by the usual formulas on subspaces of ˆAss_T. To avoid division by zero, we assume char(k) = 0.

Denition 2.20. Dene the maps exp: ˆmT → 1 + ˆmT and log: 1 + ˆmT → ˆmT by the following formulas:

exp(a) =

∞

X

i=0

aⁱ i!

and

log(1 + a) =

∞

X

i=1

(−1)ⁱ⁺¹aⁱ i where a ∈ ˆm_T.

We will verify that exp is well dened. Let a ∈ ˆmT with α = ord(a). We nd that ord(aⁱ) = αi ≥ i. Therefore we have

(exp(a))n=

n

X

i=0

aⁱ i!

!

n

∈ Assⁿ_T

for all n ∈ N. Hence exp(a) ∈ ˆAssT. Since (exp(a))0 = 1, exp is well dened. In the same way we can verify that log is well dened.

Lemma 2.21. exp and log are mutually inverse bijections.

Proof. Let a ∈ ˆmT. 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]] → ˆAss_T which transforms X into a. The power series P λiXⁱ is mapped to P λ_iaⁱ, which lies in ˆAssT since ord(aⁱ) ≥ 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 2.22. Let x, y ∈ T . Since 1+ ˆmT 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).

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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 6= y. The ground eld of the free associative and free Lie algebra is supposed to be of characteristic zero. Let h: ˆLT → ˆm_T be as in denition 2.19. We will now formulate the main theorem of this thesis:

Theorem 3.1. Let z = log(exp(x) exp(y)) ∈ ˆmT. Then we have z ∈ h(ˆLT).

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: ˆLT → ˆm_T is an injective Lie homomorphism. So we can identify h( ˆL_T) with ˆL_T. We will call an element in ˆm_T Lie if it lies in ˆL_T. We can regard the quotient space ˆmT/ˆLT. For a, b ∈ ˆmT 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 =

∞

X

n=1

F_n (3.1)

with Fn∈ Assⁿ_T. We will prove by induction that Fn≡ 0 for all n. Since Fn∈ Assⁿ_T and since LⁿT ⊆ Assⁿ_T, we then nd that Fn∈ Lⁿ_T and it follows that

z =

∞

X

n=1

Fn∈ Y

n≥1

Lⁿ_T = ˆLT.

3.1 Specialisations

To prove the theorem, we need to substitute other elements for x, y in the polynomials Fi. 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 : AssT → A which maps x, y to a, b. (We use the universal property of AssT.) Dene for each F ∈ AssT the element F (a, b) by F (a, b) := ψ(F ) ∈ A.

Let S be a set and let a, b ∈ ˆAssS. Since ˆAssS is an associative algebra, we can substitute its elements.

Lemma 3.3. Let F ∈ AssT. Let α = ord(a), β = ord(b) and ζ = ord(F ). Then ord(F (a, b)) ≥ ζ min{α, β}.

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Proof. If F 6= 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 ˆAssT. Remember ai denotes the homogeneous component of a, where a ∈ ˆAss_T.

Denition 3.4. Let ψ be as in denition 3.2 with ˆAssS substituted for A and assume a, b ∈ ˆm_T. Let

ψ : ˆˆ Ass_T → ˆAss_S,

∞

X

i=0

f_i 7→

∞

X

i=0

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 ∈ ˆAss_T. Lemma 3.5. We have

log(exp(a) exp(b)) =

∞

X

i=1

Fi(a, b).

Proof. By denition of Fi(a, b), ˆψ and z we obtain

∞

X

i=1

F_i(a, b) =

∞

X

i=1

ψ(F_i) = ˆψ

∞

X

i=1

F_i

!

= ˆψ(log(exp(x) exp(y))).

Since ψ is an algebra homomorphism we have ψ(log(exp(x) exp(y))) =ˆ

∞

X

j=1

ψ(log(exp(x) exp(y))_j)

=

∞

X

j=1

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 ∈ ˆLT ⊂ ˆmT and F ∈ LT ⊂ Ass_T we have F (a, b) ∈ ˆLT.

Proof. Let ψ be as in denition 3.2 with ˆAss_T substituted for A. Let i: ˆLT ,→ ˆAss_T be the inclusion. Let ψ⁰ be the unique homomorphism of Lie algebras ψ⁰: LT → ˆLT which sends x, y to a, b. (We use the universal property of LT.)

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T L_T Ass_T

Lˆ_T Assˆ _T

ψ⁰ ψ

i

Notice that ψ is a Lie homomorphism if we regard AssT and ˆAssT as Lie algebras. The Lie algebra homomorphism ψ|LT and i ◦ ψ⁰ from LT to ˆAss_T are equal if we restrict them to T . So by the universal property of LT the square commutes. Hence

F (a, b) = ψ(F ) = (i ◦ ψ⁰)(F ) = ψ⁰(F ) ∈ ˆL_T.

3.2 Proof of the Campbell-Hausdor theorem

We will prove by induction on n that Fn≡ 0 for all n ∈ N≥1.

By writing down the terms of z containing words of length 1 and 2 one nds F1 = x + y and F2 = ¹₂(xy − yx). Since we identify h(LⁱT) and LⁱT, the terms F1 and F2 lie in L¹T

and L²_T respectively. In conclusion we have F1, F₂≡ 0.

Let N ∈ N>2 and assume all polynomials Fnwith 1 ≤ n < N are Lie. We will use the rest of this chapter to prove that FN is a Lie polynomial. The rst step is to derive the following equation:

Lemma 3.7. Let a, b, c ∈ Ass¹T. We have FN(a, b)+F_N(a+b, c) ≡ F_N(a, b+c)+F_N(b, c). Proof. The associativity of the product in ˆAss_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





∞

X

j=1

F_j(a, b)



exp(c) = exp(a) exp





∞

X

j=1

F_j(b, c)



.

Note that P^∞_j=1F_j(a, b)and P^∞_j=1F_j(b, c)lie in ˆm_T. Taking the logarithm and applying lemma 3.5 a second time gives

∞

X

i=1

F_i





∞

X

j=1

F_j(a, b), c



=

∞

X

i=1

F_i



a,

∞

X

j=1

F_j(b, c)



∈ ˆm_T. (3.2) Let GN be the homogeneous term of degree N of the left side of equation 3.2. By lemma 3.3one nds that the terms with i > N do not contribute to GN. Hence we have

G_N =





N

X

i=1

F_i





∞

X

j=1

F_j(a, b), c









N

. (3.3)

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We will now discuss which j, depending on i, contribute to GN. We need the following two lemmas.

Lemma 3.8. Let A, B ∈ ˆm_T, let i ∈ N≥2, let n ∈ N≥1. Let ∈ L_i≥nAssⁱ_T . For all F ∈ Assⁱ_T we have

(F (A + , B))_n= (F (A, B))_n. Proof. We will give a proof by induction on i. Assume 6= 0.

Step 1. Let F ∈ Ass²_T. Then F is a linear combination of the words x², xy, yx and y². We have ord(A) = ord(A) ≥ n + 1 and ord(²) ≥ 2n.So we get

((A + )²)n= (A²+ A + A + ()²)n= (A²)n.

In the same way we obtain ((A + )(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 AssÎT. Let F ∈ AssÎT be a monomial. Then F = Hx or F = Hy for a certain H ∈ AssÎ−1_T . We will prove the statement for the rst case. Using lemma 3.3 we nd ord(H(A + , B)) ≥ n + 1. Since (A)0 = 0 and by the induction hypothesis we obtain

(H(A + , B)(A + ))_n= (H(A + , B)A)_n

=

n−1

X

j=0

(H(A, B))j(A)n−j

= (H(A, B)A)n. Hence (F (A + , B))n= (F (A, B))n.

Lemma 3.9. Let A, B ∈ ˆmT and let ∈ L_i≥2Assⁱ_T. Let n ∈ N. For all F ∈ Assⁿ_T we have

(F (A + , B))_n= (F (A, B))_n.

Proof. We will prove this lemma by induction on n. Assume 6= 0.

Step 1. For n = 0 and for all F ∈ Ass⁰_T we have F (A + , 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^NT. Let F ∈ Ass^NT 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 + , B)) ≥ N + 1. Hence

(H(A + , B)(A + ))_N = (H(A + , B)A)_N. Since (A)0 = 0 and since ord(H(A + , B)) ≥ N − 1 we have

(H(A + , B)A)_N = (H(A + , 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 + , B))N = (F (A, B))_N.

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We will now go back to equation 3.3. For i = 1 we have



F1





∞

X

j=1

Fj(a, b), c









N

=





∞

X

j=1

Fj(a, b) + c





N

= FN(a, b)

since Fj(a, b)lies in Ass^j_T and since c is homogeneous of degree 1 6= N. For 2 ≤ i ≤ N −1 we use lemma 3.8 to obtain





N −1

X

i=2

F_i





∞

X

j=1

F_j(a, b), c









N

=





N −1

X

i=2

F_i





N −1

X

j=1

F_j(a, b), c









N

. For i = N we use lemma 3.9 to nd



FN





∞

X

j=1

Fj(a, b), c









N

= (FN(F1(a, b), c))N = FN(F1(a, b), c).

Hence

G_N = F_N(a, b) +





N −1

X

i=2

F_i





N −1

X

j=1

F_j(a, b), c









N

+ F_N(a + b, c).

We will now determine the class of GN in ˆm_T/ˆL_T. This is where we use the induction assumption. For 2 ≤ j < N we have Fj ∈ L_T by this assumption. Since a, b ∈ Ass¹_T = L¹_T we can apply lemma 3.6 to obtain Fj(a, b) ∈ L_T. Applying the induction assumption and lemma 3.6 a second time, we nd P^{N −1}_i=2 Fi(PN −1

j=1 Fj(a, b), c) ∈ L_T. In conclusion GN ≡ F_N(a, b) + FN(a + b, c).

We can do the same computation for the right side of equation 3.2 to obtain the important result

FN(a, b) + FN(a + b, c) ≡ FN(a, b + c) + FN(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 3.10. 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 exp(A + B) =

∞

X

i=0

(A + B)ⁱ

i! =

∞

X

i=0 i

X

j=0

A^i−jB^j j!(i − j)!. Furthermore we have

exp(A) exp(B) =

∞

X

i=0

∞

X

j=0

AⁱB^j i!j! .

The coecient of A^i−jB^jwith 0 ≤ j ≤ i is_j!(i−j)!¹ 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.

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We will use the following three facts while performing substitutions.

Lemma 3.11. Let r, s ∈ k. We have 1. FN(ra, sa) = 0.

2. FN(a, 0) = FN(0, a) = 0.

3. FN(ra, rb) = r^NF_N(a, b).

Proof. Since ra and sa commute, we have exp(ra) exp(sa) = exp(ra + sa) by lemma 3.10. We take the logarithm and use lemma 3.5 to get

∞

X

i=1

Fi(ra, sa) = F1(ra, sa).

Because a lies in Ass¹_T, the polynomial Fi(ra, sa) is homogeneous of degree i. By the uniqueness of the power series in ˆAssT we have Fi(ra, sa) = 0 for all i > 1. Hence F_N(ra, sa) = 0.

If we take r = 1 and s = 0 we nd FN(a, 0) = 0and analogously FN(0, a) = 0. Since FN is homogeneous of degree N, we have FN(ra, rb) = r^NFN(a, b).

We will now start substituting in equating 3.4. Bear in mind that we aim to derive FN ≡ 0. We will rst derive a relation between FN(a, b)and FN(b, a).

Lemma 3.12. We have

FN(a, b) ≡ (−1)^{N +1}FN(b, a).

Proof. First of all, substitute −b for c in equation 3.4. (Note that −b ∈ Ass¹_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 3.11. Therefore we have FN(a, b) ≡ −FN(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)

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With equation 3.5 and 3.6 we can nd a relation between FN(a, b) and FN(b, a). We have

FN(a, b) ≡ −FN(a + b, −b)

≡ F_N(−a − b, a)

≡ −F_N(−b, −a)

≡ −(−1)^NF_N(b, a),

where the last equivalence follows from lemma 3.11.3. Hence the relation between F_N(a, b)and FN(b, a)is

FN(a, b) ≡ (−1)^{N +1}FN(b, a). (3.7)

Equation 3.4 has more to oer.

Lemma 3.13. We have

(1 − 2^1−N)F_N(a, b) ≡ (1 + (−1)^N)2^−NF_N(a, a + b).

Proof. Now we substitute −b/2 for c. (Note that −b/2 ∈ Ass¹T.) 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 3.11.1. 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 3.11.1 gives

F_N(−b/2, b) + F_N(b/2, c) ≡ F_N(−b/2, b + c) + F_N(b, c)

≡ F_N(b/2, c).

We replace b, c by a, b respectively, to obtain

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 FN(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) + FN(a/2, a/2 + b/2)

≡ 2^−N(FN(a, b) + FN(a, a + b)),

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where the last two equivalences follow from 3.6 and fact 3 of lemma 3.11.3 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 3.11.3. So 3.8 becomes

F_N(a, b) ≡ 2^1−NF_N(a, b) + 2^−NF_N(a, a + b) − 2^−NF_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^−NF_N(a, a + b). (3.10)

If N is odd, division by 1 − 2^1−N gives the result FN(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 3.14. If N is even, we have FN(a, b) ≡ 0. Proof. We replace b by b − a in equation 3.10, obtaining

(1 − 2^1−N)F_N(a, b − a) ≡ 2^1−NF_N(a, b). (3.11) Applying 3.6 on the left side of 3.11 gives

−F_N(−a, b) ≡ 2^1−N

1 − 2^1−NFN(a, b). (3.12) This equation allows us to nd a relation between FN(a, b)and itself. We substitute −a for a in equation 3.12 and use the equation itself to obtain

−F_N(a, b) ≡ 2^1−N

1 − 2^1−NF_N(−a, b)

≡ −

2^1−N 1 − 2^1−N

2

F_N(a, b).

(3.13)

Since N > 2, we have

2^1−N 1 − 2^1−N

2

6= 1.

Thus we have FN(a, b) ≡ 0.

Finally, we substitute x, y for a, b respectively (which is possible since x, y ∈ Ass¹T), to

nd the desired result.

FN = FN(x, y) = FN(a, b) ≡ 0.

Eichler has used slightly dierent substitutions in his proof. This `freedom' suggests there is a shorter way to obtain FN ≡ 0.

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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 Matn(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 7→

sX

i,j

|a_ij|² where |aij|is the norm of aij on C.

The map | · | is a norm on Matn(C) and (Matn(C), | · |) is isomorphic to R²ⁿ² with the euclidean norm. The norm | · | is submultiplicative:

Lemma 4.2. We have |AB| ≤ |A||B| for all A, B ∈ Matn(C).

Proof. If A = (aij)_i,j and B = (bij)_i,j, then we have by the triangle inequality and the multiplicative property of the norm on C

|(AB)_ij| = |

n

X

l=1

ailblj|

≤

n

X

l=1

|a_il||b_lj|.

By the Cauchy-Schwarz inequality we get

n

X

l=1

|a_il||b_lj| ≤ v u u t

n

X

l=1

|a_il|² v u u t

n

X

l=1

|b_lj|². It follows that

|AB| = s

X

i,j

|(AB)_ij|²

≤ v u u t

X

i n

X

l=1

|a_il|²X

j n

X

l=1

|b_lj|²

= |A||B|.

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Let T = {x, y} and take the ground eld k for AssT and LT equal to C. Let F ∈ AssT

and let A, B ∈ Matn(C). Then F (A, B) ∈ Matn(C) makes sense. (See denition 3.2.) In this chapter we will denote an element f ∈ ˆAssT by P_ifiwhere each fi is a monomial or by P aww where we sum over all words w with aw∈C.

Lemma 4.3. Let P_if_i ∈ ˆAss_T. Assume P_if_i(|A|, |B|)converges absolutely in R, then P

ifi(A, B)converges in Matn(C).

Proof. Since P_i|f_i(|A|, |B|)| converges in R, the sequence of partial sums

Xⁿ

i=0

fi(|A|, |B|)

n

is Cauchy. Let ∈ R>0. Let N ∈ N be such that for all n, m ∈ N>N with n ≥ m we

have n

X

i=m+1

f_i(|A|, |B|) =

n

X

i=m+1

f_i(|A|, |B|)

< .

Then we have by the triangle inequality and the submultiplicative property of | · | on Mat_n(C) the inequality

n

X

i=m+1

f_i(A, B)

≤

n

X

i=m+1

f_i(|A|, |B|) .

We use the fact that each fi is a monomial. For n, m ∈ N>N with n ≥ m the last term is smaller than . Hence (P_ifi(A, B))n is a Cauchy sequence in Matn(C). By the completeness, the sequence converges.

4.2 Matrix exponential and logarithmic map

Remark 4.4. The series exp(λ) = P^∞_i=0^λ_i!ⁱ and log(1 + λ) = P^∞_i=1⁽⁻¹⁾ⁱ⁺¹_i ^λⁱ are absolutely convergent for all λ ∈ R and for all λ ∈ (−1, 1) respectively. By lemma 4.3 P∞

i=0Aⁱ

i! converges and P^∞_i=1⁽⁻¹⁾ⁱ⁺¹_i ^Aⁱ converges if |A| < 1 with A ∈ Matn(C).

Denition 4.5. Dene the matrix exponential and logarithmic map to be

exp : Mat_n(C) → Matn(C), A 7→

∞

X

i=0

Aⁱ i! , log : D → Matn(C), 1 + A 7→

∞

X

i=1

(−1)ⁱ⁺¹Aⁱ i with D = {1 + A ∈ Matn(C): P^∞_i=1⁽⁻¹⁾ⁱ⁺¹_i ^Aⁱ converges}.

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Lemma 4.6. Let A, B ∈ Matn(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 3.10. 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 Matn(C) by GLn(C). The map exp: C → C^∗ is a group homomorphism, whereas exp: Matn(C) → GLn(C) is not if n > 1.

Let A, B ∈ Matn(C).

Denition 4.8. Let ψ : AssT → Mat_n(C) be the map as in denition 3.2 which replaces x, y by A, B. Dene

ψ :ˆ n X

i

fi∈ ˆAssT such that X

i

fi(A, B)convergeso

→ Mat_n(C),X

i

fi 7→X

i

ψ(fi).

Lemma 4.9. Let P_if_i= log(exp(x)) and P_ig_i = exp(log(1 + x)). 1. If P_ifi(|A|, |B|) converges absolutely we have log(exp(A)) = A.

2. If P_ig_i(|A|, |B|)converges absolutely we have exp(log(1 + A)) = 1 + A.

3. If | exp(|A|) − 1| < 1 the series P_ifi(|A|, |B|)converges absolutely.

4. If |A| < 1 the series P_ig_i(|A|, |B|)converges absolutely.

Proof. If P_ifi(|A|, |B|) converges absolutely, the terms of P_ifi(A, B) can be interchanged. Hence we have

ψˆ X

i

f_i

=X

i

f_i(A, B) = log(exp(A)).

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 = P hww ∈ ˆm_T with hw ≥ 0for all words w. We can write

− log(1 − h) =

∞

X

i=1

hⁱ i =X

aww,

log(1 + h) =

∞

X

i=1

(−1)ⁱ⁺¹hⁱ

i =X

b_ww.

By the triangle inequality and since aw ≥ 0 we have |bw| ≤ a_w for all w. So if P a_ww(|A|, |B|)converges, the series P_wb_ww(|A|, |B|) converges absolutely.

We can take h equal to exp(x). If we assume | exp(|A|) − 1| < 1, it follows that P

wbww(|A|, |B|)converges absolutely. The fourth statement is derived similarly starting with exp(− log(1 − x)) = P aww and exp(log(1 + x)) = P bww.

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Let A, B ∈ Matn(C) with |A|, |B| < log(√ 2).

Lemma 4.10. For P_if_i = log(exp(x) exp(y)), the series P_if_i(|A|, |B|) converges absolutely.

Proof. The proof is analogous to the proof of lemma 4.9.1 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 4.11. By lemma 4.10 the series P_if_i(A, B) converges and its terms can be interchanged. Therefore we can dene the matrix C := log(exp(A) exp(B)) = P

ifi(A, B). By lemma 4.9, C is the unique matrix with | exp(|C|) − 1| < 1 and exp(C) = exp(A) exp(B).

Theorem 4.12. 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 = P^∞_n=1F_n(A, B) where Fn 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 Fi(A, B)is a linear combination of iterated commutators in A and B by theorem 3.1.

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5 Commuting matrices

Let A, B ∈ Matn(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 ∈ Matn(C) such that [A, B] 6= 0 and exp(A + B) = exp(A) exp(B).

Proof. Take n = 2,

A =0 0 0 2πi

and B =0 1 0 2πi

. Because A is diagonal, we compute

exp(A) =exp(0) 0 0 exp(2πi)

= 1_Mat_n₍_C).

We can compute exp(B) using the Jordan Normal Form of B. Since 0 and 2πi are eigenvalues of B, this form is diagonal. We have

B = Q ·0 0 0 2πi

· Q⁻¹ = Q · A · Q⁻¹

for a certain matrix Q ∈ GLn(C). Hence exp(B) = Q · exp(A) · Q⁻¹ = 1_Mat_n₍_C). In the same way one nds exp(A + B) = 1Matn(C). In conclusion we have exp(A + B) = exp(A) exp(B), but

AB =0 0

0 −4π²

6=0 2πi 0 −4π²

= BA.

However, for a certain subspace of Matn(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): xij = 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 ∈ T4,2 the matrix has the following shape

A =







0 0 ∗ ∗ 0 0 0 ∗ 0 0 0 0 0 0 0 0





 .

We will often denote Tn,1simply by Tn. This is the subspace of upper triangular matrices with zeros on the main diagonal.

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Remark 5.3. We have the inclusions of subspaces Tn= T_n,1⊃ T_n,2⊃ .... ⊃ T_n,n = {0}.

For all l, m ∈ N with 1 ≤ l, m ≤ n we have

T_n,lT_n,m= {AB : A ∈ T_n,l, B ∈ T_n,m} ⊆ T_n,l+m. (5.1) Proposition 5.4. The maps exp: Tn → 1 + T_n and log: 1 + Tn → T_n are mutually inverse bijections.

Proof. Let A ∈ Tn. By remark 5.3 we have exp(A) = 1 + Pⁿ⁻¹_i=1 ^A_i!ⁱ ∈ 1 + T_n. Since A is nilpotent, log(A) is well dened and log(A) ∈ Tn. For P_ifi = exp(log(x)), the series P

if_i(|A|, |B|) converges absolutely since it concerns a nite sum. By lemma 4.9.1 we have exp(log(A)) = A. We will now prove that exp: Tn→ 1 + T_n is injective.

Let A, B ∈ Tn with exp(A) = exp(B). I will prove by induction that A − B ∈ Tn,i for all 1 ≤ i ≤ n. We clearly have A − B ∈ Tn,1. Let I ∈ N and assume A − B ∈ Tn,ifor all 1 ≤ i < I. Since exp(A) = exp(B) we have

A − B =

n−1

X

j=2

B^j− A^j j! . Since

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 ∈ Tn,I−1, it follows inductively that B^j− A^j ∈ T_n,I for all 2 ≤ j ≤ n − 1. Hence we obtain A − B ∈ Tn,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 ∈ Tn. 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] ∈ Tn,i for all 1 ≤ i ≤ n. We will prove this by induction on i.

Step 1. For i = 1 we have [A, B] ∈ Tn,2⊂ T_n,1 by remark 5.3.

Step 2. Let I ∈ N, suppose the claim is true for all 1 ≤ i < I. Let Fj be as in equation 3.1. Applying equation 5.1 j − 1 times, we obtain Fj(A, B) ∈ T_n,j. Hence for all j ≥ n we have Fj(A, B) = 0.According to the proof of theorem 4.12 we have

exp(A) exp(B) = exp(A + B + F₂(A, B) + F₃(A, B) + ... + F_n−1(A, B)).

Since exp(A + B) = exp(A) exp(B) and by proposition 5.4 one nds 0 = F₂(A, B) + F₃(A, B) + ... + F_n−1(A, B).

For j > 2, we have Fj(A, B) ∈ T_n,I by equation 5.1 and the induction assumption. In conclusion, we have F3(A, B) + ... + F_n−1(A, B) ∈ T_n,I. If F2(A, B) 6∈ T_n,I there is a j with j < i + I and (F2(A, B))ij 6= 0. We then have

F₂(A, B) + F₃(A, B) + ... + F_n(A, B) 6= 0.

This is a contradiction. Hence [A, B] = 2F2(A, B) ∈ T_n,I. It follows that [A, B] ∈ Tn,n= {0}.

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Bibliography

[1] Jean-Pierre Serre. Lie algebras and Lie groups. W.A. Benjamin, Inc., New York- Amsterdam, 1965.

[2] Nicolas Bourbaki. Lie groups and Lie algebras, Part I. Addison-Wesley, Mas- sachusetts, 1975.

[3] M. Eichler. A new proof of the Baker-Campbell-Hausdor formula. Journal of the Mathematical Society of Japan. 20:23-25, 1968.

[4] John Stillwell. Naive Lie Theory. Springer, New York, 2008.

The Campbell-Hausdor theorem