FREE KAC–MOODY GROUPS AND THEIR LIE ALGEBRAS.

Y. Billig  and  A. Pianzola
Dept. of Mathematics & Statistics  Dept. of Mathematical Sciences
University of New Brunswick  University of Alberta
Fredericton, New Brunswick  Edmonton, Alberta
Canada E3B 5A3  Canada T6G 2G1

Abstract. We compute the Lie algebras of continuous right invariant derivations of free Kac–Moody groups.

Introduction.

Lie’s third Theorem, the existence of a Lie group with given (finite dimensional complex) Lie algebra, is one of the central results of the classical theory of transformation groups. Over arbitrary base fields the analytical tools give way to algebraic geometry and questions of rationality.

For infinite dimensional Lie groups the situation is almost completly uncharted except for a few exceptions, one of which is the case of Kac-Moody groups. The constructions of these groups that is most relevant to us is that of Peterson and Kac as presented in [PK]. The spirit of their construction (at times remenisent of Chevalley’s Tohoku paper) is to first attach to the Lie algebra a “free” Kac-Moody group F and then go on to identify those elements of F whose action on a given representation theory (The integrable

Typeset by AMS-TEX

Both authors acknowledge the generous support of NSERC Canada.

representation in this case. But see also [MP] Ch.6) coincide. At the heart of the present paper is the study the Lie algebra that corresponds to F.

The first question to answer is how to attach a Lie algebra to F to begin with. The answer is rather elegant: There is a natural concept of polynomial function on the group F . These form an algebra, P olF , in which the group acts. As in the classical situation, one then defines the Lie algebra to be the derivations of P olF which are (right) invariant under the group action.

With the Lie algebra constructed we next look at its structure, which as it turns out, is a certain completion of a free Lie algebra. The two key ingredients to the proof are the fact that all elements of P olF can be written as series on certain basic polynomial functions (Theorem 2.10), and secondly that this basic polynomial functions form a shuffle algebra (Theorem 3.7)

We would like to most sincerily thank J. Valencia for his great help in putting together this paper.

Notation, conventions, and terminology.

Let I be a non-empty set which will be assumed fixed throughout our discussion. For each commutative ring 𝕂 and each i ∈ I we consider a copy E_{i}(𝕂) := {E_{i}(λ)|λ ∈ 𝕂} of the additive group of 𝕂. Thus for all λ, μ ∈ 𝕂 E_{i}(λ)E_{i}(μ) = E_{i}(λ + μ). Consider the free product

F(𝕂) := ∗_{i∈I} E_{i}(𝕂).

To deal with F(𝕂) we will introduce a fair amount of notation and terminology. For the reader’s convinience we will gather all this information in the present section (even though some of it will not be needed until much later.) This will make back references much more accesible .

Let W be the free multiplicative monoid (words) on I. We denote by 1 the empty word (identity) in W . If w ∈ W, w ≠ 1, we can uniquely write

w = w_{1} ⋯ w_{w} with w > 0, and w_{1}, ⋯, w_{w} ∈ I.  (1.1)

We thus use bold face characters for words, and the corresponding unbold characters for their lengths.

There is also a unique expression

w = \overline{w}_{1}^{n_{1}} ⋯ \overline{w}n\overline{w} ; \overline{w} with \overline{w}_{n} ≠ \overline{w}_{n+1}, and n_{k} > 0.  (1.2)

We set \overline{w} := \overline{w}_{1} ⋯ \overline{w}\overline{w} , \overline{W} = \{\overline{w} : w ∈ W \} (reduced words) and

w! = n_{1}! ⋯ n\overline{w}!.  (1.3)

1.4 The case w = 1 corresponds to w = 0 in (1.1). We will agree that 1 = \overline{1}, that 1! = 1 , and that 1 ∈ \overline{W} .

If N ∈ ℤ_{>0} we will denote the interval {1, ⋯, N} by [N] , and agree that [0] is the empty set. By means of (1.1) we define the degree function deg: W → ℕ^{I} by w ↦ deg_{w} where

deg_{w}(i) = Card{k ∈ [w] : w_{k} = i}.  (1.5)

Let s and r be elements of W . By a placing map of s into r we understand a map σ : [s] → [r] satisfying the following conditions:

PM1. 1 ≤ p < q ≤ s ⟹ σ(p) ≤ σ(q) (i.e. σ is (non-strictly) monotonic) PM2. s_{p} = r_{σ(p)} for all p ∈ [s].

The totality of such maps will be denoted by [s : r]. Given w ∈ W we define W_{w} = {s ∈ W : [s : w] ≠ ∅}

Remark 1.6. Let σ ∈ [s : r]. For n ∈ [r] the set σ^{−1}(n) := {m|σ(m) = n} is either empty or an interval [p, q] for some 1 ≤ p ≤ q ≤ s. We set σ_{n} := Card σ^{−1}(n) , and σ! := ∏_{n=1}^{r} σ_{n}! . Note that [s : \overline{s}] consists of just one element, denote by ρ_{s} , and that ρ_{s}! = s!

1.7 To each s ∈ W we attach the polynomial ring 𝕂 [s] := 𝕂 [T_{1}, …, T_{s}]. Given a placing map σ ∈ [s : r] we define a ring homomorphism ψ_{σ} : 𝕂 [r] → 𝕂 [s] by

ψ_{σ} : T_{n} ↦ ∑_{m∈σ^{−1}(n)}T_{m}.

Note that if σ^{−1}(n) = ∅ then ψ_{σ}(T_{n}) = 0. By convention we will assume that [1 : r] consists of a single placing map σ satisfying σ! = 1 and whose corresponding ψ_{σ} : 𝕂 [r] → 𝕂 [1] := 𝕂 is given by the augmentation homomorphism.

1.8 If σ ∈ [s : r] we define T^{σ} ∈ 𝕂[r] by T^{σ} := ∏_{n=1}^{r} T_{n}^{σ_{n}} . Similarly for w as in (1.3) we define T^{w} ∈ 𝕂 [\overline{w}] by ∏\overline{w} ; k=1 T_{k}^{n_{k}} . If λ = (λ_{1}, …, λ_{r}) ∈ 𝕂^{r} we define λ^{σ} and λ^{w} by replacing T_{n} by λ_{n} in the previous expressions.

Remark 1.9. It is easy to verify that if σ ∈ [s : r] and τ ∈ [r : t] then τ ∘ σ ∈ [s : t] and ψ_{τ∘σ} = ψ_{σ} ∘ ψ_{τ} .

We now single out two special types of placing maps. Let σ ∈ [s : r]

σ is called a lift if σ is injective and  r = s + 1, and
σ is called a drop if σ is surjective and r = s − 1.

Remark 1.10. By Remark 1.9 we see that every σ ∈ [s : r] can be written as a composition of drops followed by lifts.

1.11 For w ∈ W as in (1.1) we let

F(𝕂)^{w} = {E_{w_{1}}(λ_{1}) … E_{w_{w}}(λ_{w})|λ_{1}, …, λ_{w} ∈ 𝕂} ⊂ F(𝕂).

Let e be the identity element of F(𝕂). If x ∈ F(𝕂), x ≠ e, then there exists a unique r ∈ \overline{W} and unique λ_{1}, …, λ_{r} ∈ 𝕂 \\{0\} such that x = E_{r_{1}}(λ_{1}) … E_{r_{r}}(λ_{r}). We call this the reduced expression of x and say that r is the type of x. By convention F(𝕂)^{1} = {e} and e is of type 1 .

Polynomial Functions on F(𝕂) .

2.1 Given N ∈ ℕ we let Pol(𝕂^{N}) be the ring of polynomial functions from 𝕂^{N} into 𝕂 (Bbk Ch. 4). We have a surjective ring homomorphism κ_{N} : 𝕂 [T_{1}, …, T_{N}] → Pol(𝕂^{N}). It is easy to verify that κ_{1} is injective if and only if κ_{N} is injective for all N ∈ ℕ. If these hold we say that 𝕂 is polynomially faithful, and identify Pol(𝕂^{N}) with 𝕂 [T_{1}, …, T_{N}]

Given w ∈ W as in (1.1) we define π_{w} : 𝕂^{w} → F(𝕂)^{w} ⊂ F(𝕂) by

π_{w} : (λ_{1}, …, λ_{w}) ↦ E_{w_{1}}(λ_{1}) … E_{w_{w}}(λ_{w}).

By convention 𝕂^{0} = {0} and π_{1}(0) = e . (See 1.11.)

A function f : F(𝕂) → 𝕂 is said to be a polynomial function if f ∘ π_{w} is polynomial for all w ∈ W : That is f ∘ π_{w} ∈ P ol(𝕂^{w}) for all w ∈ W. The set of polynomial functions forms a ring which will be denoted by Pol F(𝕂).

In this section we will describe the nature of the ring Pol F as an inverse limit (Theorem 2.3), and as a completion (Theorem 2.10).

Consider the ring 𝕂 [W] := Π_{s∈W} 𝕂[s] as well as its subring 𝕂 [F] := \underleftarrow{lim}_{ψ_{σ}} 𝕂[s] . A typical element of 𝕂 [F] is thus of the form F = (F_{r})_{r∈W} with ψ_{σ}(F_{r}) = F_{s} for all r, s ∈ W and σ ∈ [s : r].

Proposition 2.2. Let f ∈ Pol F(𝕂) . For r ∈ W define F_{r} = f ∘ π_{r}. and F = (F_{r})_{r∈W} ∈ 𝕂[W] . Then F ∈ 𝕂[F] .

Proof. We must show that ψ_{σ}F_{r} = F_{s} for all σ ∈ [s : r] . Assume σ is a lift so that r is of the form r = s_{1} ⋯ s_{k}is_{k+1} ⋯ s_{s} . Then for all λ_{1}, ⋯, λ_{s} ∈ 𝕂

ψ_{σ}F_{r}(λ_{1}, ⋯, λ_{s}) := F_{r}(λ_{1}, ⋯, λ_{k−1}, 0, λ_{k}, ⋯, λ_{s})
= f(E_{s_{1}}(λ_{1}) ⋯ E_{s_{s}}(λ_{s}))
= f ∘ π_{s}(λ_{1}, ⋯, λ_{s})
:= F_{s}(λ_{1}, ⋯, λ_{s}).

The case of drops follows along similar consideration. Proposition 2.2 now follows from 1.10

□

Theorem 2.3. (i) There exists a canonical surjective ring homomorphism ω_{𝕂} : 𝕂[F] → Pol F(𝕂).

(ii) For ω to be an isomorphism it is necessary and sufficient that 𝕂 be polynomially faithful.

Proof. Let F = (F_{r})_{r∈W} ∈ 𝕂 [F]. Define ω_{𝕂}(F) := ω(F) : F(𝕂) → 𝕂 as follows: Given x ∈ F(𝕂) let r be its type, E_{r_{1}}(λ_{1}) … E_{r_{r}}(λ_{r}) its reduced expression, and set ω(F)(x) := F_{r}(λ_{1}, …, λ_{r}). We show that ω(F) is polynomial by establishing that

(2.4) ω(F) ∘ π_{r}(λ_{1}, …, λ_{r}) = F_{r}(λ_{1}, …, λ_{r}) for all λ_{1}, …, λ_{r} ∈ 𝕂 and r ∈ W.

We now reason by induction on r . If r = 0 then (2.4) holds by definition (See 2.1). Assume r > 0 and that (2.4) holds for all words of length < r. We distinguish two cases:

Case 1: r is not reduced. Chose k so that r_{k−1} = r_{k} . Let s = r_{1} ⋯ r_{k−1}r_{k+1} ⋯ r_{r} and let σ ∈ [r : s] be the obvious drop. We have

ω(F) ∘ π_{r}(λ_{1}, …, λ_{r}) = ω(F)(E_{r_{1}}(λ_{1}) … E_{r_{r}}(λ_{r}))
= ω(F)(E_{r_{1}}(λ_{1}) … E_{r_{k−1}}(λ_{k−1})E_{r_{k}}(λ_{k}) … E_{r_{r}}(λ_{r}))
= ω(F) ∘ π_{s}(λ_{1}, …, λ_{k−1} + λ_{k}, …, λ_{r})
= F_{s}(λ_{1}, …, λ_{k−1} + λ_{k}, …, λ_{r}) (by induction)
= F_{r}(λ_{1}, …, λ_{k−1}, λ_{k}, …, λ_{r}) (Since ψ_{σ}(F_{s}) = F_{r}).

Case 2: r is reduced If λ_{1}, …, λ_{r} are all nonzero then (2.4) holds by definition: Namely by (2.3). Assume then that λ_{k} = 0 for some 1 ≤ k ≤ r. Let s = r_{1} ⋯ r_{k−1}r_{k+1} ⋯ r_{r}. Let σ ∈ [s : r] be the obvious lift ( see 1.7 and 2.1 if r = 1. ) We have

ω(F) ∘ π_{r}(λ_{1}, …, λ_{r}) = ω(F)(E_{r_{1}}(λ_{1}) … E_{r_{r}}(λ_{r}))
= ω(F)(E_{r_{1}}(λ_{1}) … E_{r_{k−1}}(λ_{k−1})E_{r_{k+1}}(λ_{k+1}) … E_{r_{r}}(λ_{r}))
= ω(F) ∘ π_{s}(λ_{1}, …, λ_{k−1}, λ_{k+1}, …, λ_{r})
= F_{s}(λ_{1}, …, λ_{k−1}, λ_{k+1}, …, λ_{r}) (by induction)
= F_{r}(λ_{1}, …, λ_{k−1}, 0, λ_{k+1}, …, λ_{r}) (Since ψ_{σ}(F_{r}) = F_{s})
= F_{r}(λ_{1}, …, λ_{r}).

With (2.4) established we have that the image of 𝕂[F] under ω does lye inside Pol F(𝕂). That ω is a ring homomorphism is clear.

Next we show that ω is surjective. We will do this explicitly by constructing certain polynomial functions Y_{𝕂}^{a} which “cover” Pol F(𝕂) (Theorem 2.10 below), and then showing that each Y_{𝕂}^{a} has a preimage in 𝕂[F]. These polynomial function will play a critical role throughout the paper.

Let Ass(Q) be the free associative algebra over Q with free generators {x_{i}}_{i∈I}. We attach to w ∈ W written as in (1.2) elements y^{w} and x^{w} of Ass(Q) by

y^{w} := \frac{1}{w!} x_{w_{1}}^{n_{1}} … xn\overline{w} ; \overline{w} =: \frac{1}{w!} x^{w}.

If s = e then we agree that x^{e} = y^{e} = 1. Thus {y^{s}|s ∈ W } is a Q-basis of Ass(Q). For i ∈ I and n ∈ ℕ the expression y^{i^{n}} is ambiguous. We will agree that y^{i^{n}} = y^{(i^{n})} . Note then that y^{i^{n}} = x_{i}^{n}/n! .

The free ℤ-module

D(ℤ) := ⨁_{s∈W}ℤ y^{s}

is a ℤ -subalgebra of Ass(Q) (algebra of divided powers).

If a ∈ W we define Y_{ℤ}^{(a)} ∈ \overline{D}(ℤ)^{∗} and X_{ℚ}^{(a)} ∈ \overline{A}ss(Q)^{∗} (here and elsewhere overbars will denote completions and * duals) by

Y_{ℤ}^{(a)}( ∑_{b∈W} c_{b}y^{b}) = c_{a} and X_{ℚ}^{(a)}( ∑_{b∈W} c_{b}x^{b}) = c_{a}.

Going back to our base ring 𝕂 we consider the ring D(𝕂) = 𝕂 ⊗_{ℤ} D(ℤ) and its formal completion \overline{D(𝕂)}. The functionals Y_{ℤ}^{(a)} extend naturaly to functionals Y_{𝕂}^{(a)} on D(𝕂) and \overline{D(𝕂)}.

Consider the (divided power version of the) Magnus group

M(𝕂) := \{1 + ∑_{b∈W,b≥1} c_{b}y^{b}\} ⊂ \overline{D(𝕂)}.

For each i ∈ I the map ε_{i} : 𝕂 → M given by λ ↦ ∑ λ^{n}y^{i^{n}} is a group homomorphism. By universal nonsense this yields a (unique) group homomorphism ε : F(𝕂) → M(𝕂) satisfying

ε : E_{i_{1}}(λ_{1}) … E_{i_{N}} (λ_{N}) ↦ ε_{i_{1}}(λ_{1}) … ε_{i_{N}} (λ_{N}).

We use this homomorphism to construct the functions Y_{𝕂}^{a} : F(𝕂) → 𝕂 that we need:

Y_{𝕂}^{a} := Y_{𝕂}^{(a)} ∘ ε.  (2.5)

Remark 2.6. If 𝕂 is a ℚ -algebra we can define functions X_{𝕂}^{a} := X_{𝕂}^{(a)} ∘ ε : F(𝕂) → 𝕂 Note that Y_{𝕂}^{a} = a!X_{𝕂}^{a} .

Proposition 2.7. Let a, s ∈ W and let λ = (λ_{1}, …, λ_{s}) ∈ 𝕂^{s}. Then

Y_{𝕂}^{a}(E_{s}(λ)) = ∑_{σ∈[a:s]} \frac{a!λ^{σ}}{σ!} .

In particular Y_{𝕂}^{a} ∘ π_{s} is the polynomial function on 𝕂^{s} defined by ∑_{σ∈[a:s]} \frac{a!T^{σ}}{σ!} , and Y_{𝕂}^{a} is a polynomial function on F(𝕂). (Note that σ! divides w! in ℤ. This allows us to think of \frac{a!}{σ!} as an element of 𝕂 in a natural way.)

Proof.

Y_{𝕂}^{a}(E_{s}(λ)) : = Y_{𝕂}^{(a)}(ε(E_{s}(λ)))
: = Y_{𝕂}^{(a)}( ∏_{i=1}^{s} ( ∑_{n_{i}≥0} λ_{i}^{n_{i}}y^{s_{i}^{n_{i}}} ))

= Y_{𝕂}^{(a)}( ∑_{n∈ℕ^{s}} λ^{n}y^{s_{1}^{n_{1}}} ⋯ y^{s_{s}^{ns}} ).

With each n ∈ ℕ^{s} we associate in a natural way a word w = s_{1}^{n_{1}} ⋯ s_{s}^{n_{s}} ∈ W and a placing σ ∈ [w : s] such that n_{i} = σ_{i} This procedure establishes a bijection between [w : s] and the occurences of y^{w} in the expansion of ε(E_{s}(λ)) . Multiplication in the divided powers algebra D yields

y^{s_{1}^{n_{1}}} ⋯ y^{s_{s}^{ns}} = \frac{w!}{σ!}y^{w}.

When acting on the above infinite sum Y_{𝕂}^{(a)} picks up only summands corresponding to w = a . Thus

Y_{𝕂}^{a}(E_{s}(λ)) = ∑_{σ∈[a:s]} \frac{a!λ^{σ}}{σ!}

□

Corollary 2.8. Let a, s ∈ W . For Y_{𝕂}^{a} to vanish in F(𝕂)^{s} it is necessary and sufficient that [a : s] be empty.

Proof. If [a : s] is empty then Y_{𝕂}^{a} ∘ π_{s} = 0 . This shows that Y_{𝕂}^{a} vanishes in F(𝕂)^{s} .

If [a : s] is not empty then \overline{a} is a subword of s so that F (𝕂)\overline{a} ⊂ F(𝕂)^{s} . But Σ_{σ∈[a:\overline{a}]} \frac{a!}{σ!}T^{σ} = T_{1}^{n_{1}} ⋯ T n\overline{a} ; \overline{a} (see 1.6), which does not vanish at T_{1} = ⋯ = T_{a} = 1 .

□

2.9 Given S ⊂ W and r ∈ W define S_{r} := {a ∈ S: [a : r] ≠ ∅} (note that S_{a} = S\overline{a} ). Note also that 1 ∈ S_{r} for all r ∈ W (See 1.7).) We say that S is summable if it satisfies the following property:

SUM. For all r ∈ \overline{W} the set S_{r} is finite. A family \{c_{a}Y_{𝕂}^{a}|c_{a} ∈ 𝕂\}_{a∈W} is said to be summable if its support {a : c_{a} ≠ 0} is summable. In this case by Corollary 2.8 there exists a unique function

∑_{a∈W} c_{a}Y_{𝕂}^{a} : F(𝕂) → 𝕂

satisfying

(∑ c_{a}Y_{𝕂}^{a})(x) = ∑ c_{a}Y_{𝕂}^{a}(x)

for all x ∈ F(𝕂).

Theorem 2.10. (i) Every element of Pol F(𝕂) is of the form ∑_{a∈W}c_{a}Y_{𝕂}^{a} for some summable family (c_{a}Y_{𝕂}^{a})_{a∈W} .

(ii) For the family of (i) to be unique it is necessary and sufficient that 𝕂 be polynomially faithful.

Proof. We reason by induction on N ∈ ℕ to establish the existence of a summable family (c_{a}(N)Y_{𝕂}^{a})_{a∈W} such that

(a) The restrictions of f and ∑_{a∈W}c_{a}(N)Y_{𝕂}^{a} to F(𝕂)^{r} agree whenever r ∈ \overline{W} satisfyes r ≤ N, .

(b) c_{a}(N) = 0 whenever \overline{a} > N.

(c) c_{a}(M) = c_{a}(N) whenever \overline{a} ≤ M ≤ N.

If N = 0 we set c_{e}(0) = f(e) and c_{a}(0) = 0 for all a ∈ W, a ≠ e.

Assume N > 0, let

g = f − ∑_{a∈W} c_{a}(N − 1)Y_{𝕂}^{a}.

Then g ∈ Pol F(𝕂) and g vanishes on F_{𝕂}^{s} whenever s < N. It follows that for r ∈ \overline{W} with r = N there is no loss of generality in assuming that g_{r} is given by a polynomial p_{r} which is divisible by T_{1} … T_{N} , hence of the form

p_{r} = ∑_{\overline{a}=r} c_{a}(N)T^{a}.

for some coefficients c_{a}(N) ∈ 𝕂. If we now set c_{a}(N) := c_{a}(N − 1) for all a ∈ W with \overline{a} < N and c_{a}(N) = 0 if \overline{a} > N, the family ∑_{a∈W} c_{a}(N)Y^{a} is as desired.

For a ∈ W let c_{a} := c_{a}(\overline{a}). The the family (c_{a}Y_{𝕂}^{a})_{a∈W} is summable while by (a) we get f = ∑_{a∈W}c_{a}Y_{𝕂}^{a}. This finishes the proof of (i). The proof of (ii) will be given bellow.

□

We now finish the proof of Theorem 2.3. Let f ∈ Pol F(𝕂) and write f = ∑_{a∈W}c_{a}Y_{𝕂}^{a} as in 2.10(i). For s ∈ W define F_{s} := ∑_{a∈W}c_{a}Y_{ℤ}^{a} ∘ π_{s} ∈ 𝕂 [s]. (To see that this makes sense note that if Y_{ℤ}^{a} ∘ π_{s} ≠ 0 then [\overline{a} : s] ≠ ∅ (Corollary 2.8), hence [\overline{a} : \overline{s}] ≠ ∅. Then only finitely many c_{a}’s are nonzero because (c_{a}Y_{𝕂}^{a}) is summable.) By Propositions 2.2 and 2.7 we see that F := (F_{s})_{s∈W} ∈ 𝕂 [F]. It is clear that ω(F) = ∑_{a∈W}c_{a}Y_{𝕂}^{a}. The proof of part (i) of Theorem 2.3 is now complete.

As for (ii) if ω is not injective there exists F ∈ 𝕂[F] and r ∈ W with F_{r} ≠ 0 and ω(F) = 0 . Since κ_{r}(F_{r}) = 0 𝕂 is not polynomially faithful (See 2.1).

Conversely if 𝕂 is not polynomially faithfull, there exists a nonzero polynomial p = Σa_{n}T^{n} ∈ 𝕂[T] with p(λ) = 0 for all λ ∈ 𝕂 . Choose i ∈ I and let f = Σa_{n}Y^{i^{n}}. (There is no ambiguity here: Y^{(i^{n})} = (Y^{i})^{n}. ) By Proposition 2.7 there exists F ∈ 𝕂[F], F ≠ 0 with ω(F) = f , while by Corollary 2.8 f vanishes everywhere in Pol F(𝕂) . The uniqueness statement of Theorem 2.10 follows along similar lines. The proofs of both Theorems are now complete. □

Multiplication of Polynomial Functions.

Let a, b ∈ W. By a shuffle of (a, b) we understand a triple (α, β, s) where s ∈ W, α ∈ [a : s], and β ∈ [b : s] are such that

Sh 1 α and β are injective.

Sh 2 α[a] ∩ β[b] = ∅. Sh 3 s = a + b.

Intuitively s is a word made out of a and b where the support of these subwords do not intersect. It is easy to see that if (α′, β′, s′) is another shuffle of (a, b) then α[a] = α′[a] ⟺ β[b] = β′[b]. In this case we have s = s′ and the two shuffles are equal.

There is a natural bijection between the set Sh(a, b) of all shuffles of (a, b) and the set of stictly increasing (hence injective) maps from [a] into [a + b]. Thus

Card(Sh(a, b)) = \frac{(a + b)!}{a!b!} .  (3.1)

Next we introduce an equivalence relation ∼ on Sh(a, b) that will play an important role when dealing with the multiplicative structure of Pol F(𝕂). when 𝕂 is not a ℚ -algebra (See 3.7 bellow.)

ESh (α, β, s) ∼ (α′, β′, s′) ⟺ ρ_{s} ∘ α = ρ_{s′} ∘ α′ and ρ_{s} ∘ β = ρ_{s′} ∘ β′.

We will denote Sh(a, b)/_{∼} by S̃h(a, b). If P = (α, β, s) ∈ Sh(a, b) then its equivalence class will be denoted by ({ \atop α, β, s}). For P = (α, β, s) ∈ Sh(a, b) we set

sh(P) = sh(P̃) = s (see next Lemma) and  (3.2)
P! = P̃! = (ρ_{s} ∘ α)!(ρ_{s} ∘ β)!

Lemma 3.3. Let P = (α, β, s) ∈ Sh(a, b). Then

(i) If (α′, β′, s′) ∈ P̃ then s′ = s , (ii) Card(P̃) = \frac{s!}{P̃!} , (iii) P̃! divides a! b! .

Proof. It is clear that ρ_{s} ∘ α! divides a! , and similarly for β . This shows that (iii) holds.

If s = \overline{s}_{1}^{ℓ_{1}} ⋯ \overline{s}ℓ\overline{s} ; \overline{s} is the reduced expression of s then a = \overline{s}_{1}^{p_{1}} ⋯ \overline{s}p\overline{s} ; \overline{s} and b = \overline{s}_{1}^{q_{1}} ⋯ \overline{s}q\overline{s} ; \overline{s} for some unique p_{1}, …, q\overline{s} ∈ ℤ_{≥0} satisfying p_{h}+q_{h} = ℓ_{h} for all 1 ≤ h ≤ \overline{s}.

The equivalence class of a shuffle is then determined by the sequence p_{1}, …, q\overline{s} . This observation shows that (i) holds. As for (ii) we have

Card({ \atop α, β, s}) = ∏_{k=1}^{\overline{s}} \binom{ℓ_{k}}{p_{k}}

= ∏_{k=1}^{\overline{s}} \frac{ℓ_{k}!}{p_{k}!q_{k}!} = \frac{s!}{(ρ_{s} ∘ α)!(ρ_{s} ∘ β)!} .

□

Next we introduce the shuffle product ([Rtn] is an excellent reference for this.) The algebra Ass(ℚ) possesses a Hopf algebra structure where the coproduct Δ : Ass(ℚ) → Ass(ℚ) ⊗ Ass(ℚ) is the unique homomorphism of associative algebras satisfying

Δ(x_{i}) := x_{i} ⊗ 1 + 1 ⊗ x_{i}.

This coproduct is co-commutative and can be extended to \overline{A}ss(ℚ) . This endows its dual \overline{A}ss(ℚ)^{∗} with a commutative multiplication ♯ given by

X ♯ Y (x) := X ⊗Y (Δ(x)) ∈ ℚ⊗ℚ ≅ ℚ, for all X, Y ∈ \overline{A}ss(ℚ)^{∗}, and x ∈ \overline{A}ss(ℚ) (3.4)

This multiplication is related to the shuffle product. More precisely (See [Rtn] Proposition 1.8)

X_{ℚ}^{(a)} ♯ X_{ℚ}^{(b)} = ∑_{P∈Sh(a,b)} X_{ℚ}^{sh(P)}  (3.5)

The next Lemma says that the E_{s}(λ)′s are group like.

Lemma 3.6. Let s ∈ W . Then Δ(ϵ(E_{s}(λ))) = ϵ(E_{s}(λ))⊗ϵ(E_{s}(λ)) for all λ ∈ 𝕂^{s}

Proof. Since Δ(1) = 1 ⊗ 1 we may assume s > 0 . We now reason by induction on s > 0 . If s = 1 then s = i ∈ I and for λ ∈ 𝕂

Δ(ϵ(E_{i}(λ))) = Δ(exp(λx_{i}))

= exp(Δ(λx_{i}))

= exp(λ(x_{i} ⊗ 1) + λ(1 ⊗ x_{i}))

= exp(λ(x_{i} ⊗ 1))exp(λ(1 ⊗ x_{i}))

= (exp(λx_{i}) ⊗ 1)(1 ⊗ exp(λx_{i}))

= exp(λx_{i}) ⊗ exp(λx_{i})

= ϵ(E_{i}(λ)) ⊗ ϵ(E_{i}(λ)).

For the induction step, assume s = wi, w ∈ W, i ∈ I . Then

Δ(ϵ(E_{w}(λ)E_{i}(λ_{s}))) = Δ(ϵ(E_{w}(λ))Δ(ϵ(E_{i}(λ_{s})))

= (ϵ(E_{w}(λ)) ⊗ ϵ(E_{w}(λ)))(ϵ(E_{i}(λ_{s})) ⊗ ϵ(E_{i}(λ_{s})))

= ϵ(E_{w}(λ)E_{i}(λ_{s})) ⊗ ϵ(E_{w}(λ)E_{i}(λ_{s})).

Theorem 3.7. (Multiplication Formula)

Let Y (𝕂) be the submodule of Pol F(𝕂) spanned by {Y_{𝕂}^{a}}_{a∈W} , and if 𝕂 is a ℚ –algebra, let X(𝕂) be the submodule of Pol F(𝕂) spanned by {X_{𝕂}^{a}}_{a∈W} (see Remark 2.6). Then

(i) Multiplication in X(𝕂) coinsides with the shuffle product:

X_{𝕂}^{a} X_{𝕂}^{b} = ∑_{P∈Sh(a,b)} X_{𝕂}^{sh(P)}.

(ii) Y (𝕂) is a subalgebra of Pol F(𝕂) whose multiplication is given by

Y_{𝕂}^{a} Y_{𝕂}^{b} = ∑_{P̃∈S̃h(a,b)} \frac{a!b!}{P̃!}Y_{𝕂}^{sh(P̃)}

(Here \frac{a! b!}{P!} ∈ ℤ is viewed naturally as an element of 𝕂 . See Lemma 3.3 )

(iii) Multiplication in Pol F(𝕂) is the (algebraically continuous, in the sense of Section 5) extension of the multiplication in Y (𝕂) .

Proof.

X_{ℚ}^{a} X_{ℚ}^{b}(E_{s}(λ)) = X_{ℚ}^{a}(E_{s}(λ)) X_{ℚ}^{b}(E_{s}(λ))

= X_{ℚ}^{(a)}(ϵ(E_{s}(λ))) X_{ℚ}^{(b)}(ϵ(E_{s}(λ)))

= X_{ℚ}^{(a)} ⊗ X_{ℚ}^{(b)}(ϵ(E_{s}(λ)) ⊗ ϵ(E_{s}(λ)))

= X_{ℚ}^{(a)} ⊗ X_{ℚ}^{(b)}(Δ(ϵ(E_{s}(λ))) by Lemma 3.6

= X_{ℚ}^{(a)}♯X_{ℚ}^{(b)}(ϵ(E_{s}(λ))) by (3.4)

= ∑_{P∈Sh(a,b)} X_{ℚ}^{(sh(P))}(ϵ(E_{s}(λ))) by (3.5)

= ∑_{P∈Sh(a,b)} X_{ℚ}^{sh(P)}((E_{s}(λ)))

This shows that (i) holds in the case 𝕂 = ℚ. The general case follows from Proposition 2.7 and the fact that 𝕂[s] ≃ 𝕂 ⊗ ℚ[s]

(ii) By identifying Y (ℤ) inside X(ℚ) via Y_{ℤ}^{w} = w!X_{ℚ}^{w} we get from (i) and Lemma 3.3 that (ii) holds in the case 𝕂 = ℤ. The general case now follows by Proposition 2.7 together with the canonical isomorphism 𝕂[s] ≅ 𝕂 ⊗_{ℤ} ℤ[s] .

(iii) Let f, g ∈ Pol F(𝕂) . By Theorem 2.10 we can write f = Σ_{a∈R}c_{a}Y_{𝕂}^{a} and g = Σ_{b∈T} d_{b}Y_{𝕂}^{b} where R and T are summable set. If P = (α, β, s) ∈ Sh(a, b) then a, b ∈ W_{s}. It follows that the set {(a, b) ∈ R × T: c_{a}d_{b} ≠ 0} ∩ W_{s} × W_{s} is finite. If we now set

h = Σ_{s∈W} ( ∑_{\begin{matrix}a,b∈W \\ P̃∈S̃h(a,b) \\ sh(P̃)=s \end{matrix}} \frac{a!b!}{P̃!}c_{a}d_{b})Y_{𝕂}^{s}

:= Σ_{s∈W} k_{s}Y_{𝕂}^{s}

we see that S := {s ∈ W: k_{s} ≠ 0} is summable, and that h and fg agree on each F^{r} . A different proof of this result is given in Remark 5.12.

Remark 3.8 The sum X(ℤ) := Σ_{a∈W} ℤX_{ℚ}^{a} ⊂ X(ℚ) is direct and closed under multiplication. In fact X(ℤ) is the ℤ –shuffle algebra (part (i) of the last Theorem).

If 𝕂 is polynomially faithful there exists a unique ℤ –linear map

ζ: Y (ℤ) → X(ℤ)
Y_{ℤ}^{a} ↦ a!X_{ℚ}^{a}.

Part (ii) of the last Theorem shows that ζ is a ℤ –algebra homomorphsim.

Action of F(𝕂) on Pol F(𝕂) .

The group F(𝕂) acts on Pol F(𝕂) naturally by left and right translations: For h ∈ F(𝕂) define L_{h} and R_{h} ∈ End_{𝕂}(Pol F(𝕂)) by (L_{h}f)(g) = f(h^{−1}g) and (R_{h}f)(g) = f(gh) for all h, g ∈ F(𝕂) and f ∈ Pol F(𝕂).

The next proposition gives the explicit formula for the action of the generating subgroups of F(𝕂). To this end we introduce the following notation: given a ∈ W and 0 ≤ p ≤ a we let a^{→}^{p} (respectively a^{←}^{p}) be the word obtained by deleting the first (respectively last) p letters of a . In addition if i ∈ I we let a^{→}(i) (resp. a^{←}(i) ) be the number of i ’s at the begining (resp. end) of a .

Proposition 4.1. Let a = \overline{a}_{1}^{n_{1}} ⋯ \overline{a}n\overline{a} ; \overline{a} be the reduced expression of an element a ∈ W. For i ∈ I and λ ∈ IK we have

L_{E_{i}(λ)}Y_{𝕂}^{a} = a^{→}(i) ; ∑_{p=0}\binom{a^{→}(i)}{p} (−λ)^{p}Y_{𝕂}^{a^{→}^{p}}
R_{E_{i}(λ)}Y_{𝕂}^{a} = a^{←}(i) ; ∑_{p=0} \binom{a^{←}(i)}{p} λ^{p}Y_{𝕂}^{a^{←}^{p}}

Proof. Let s ∈ W and let λ = (λ_{1}, …, λ_{s}) ∈ IK^{s}. Set r = si and μ = (λ_{1}, …, λ_{s}, λ). Then

R_{E_{i}(λ)}Y_{𝕂}^{a}(E_{s}(λ)) = Y_{𝕂}^{a}(E_{s}(λ)E_{i}(λ))
= ∑_{τ∈[a:r]} \frac{a!}{τ!} μ^{τ} (by Proposition 2.7).  (4.2)

If i ≠ \overline{a}\overline{a} then there is a natural bijection between the sets [a : r] and [a : s] . If under this τ ∈ [a : r] corresponds to σ ∈ [a : s] then τ! = σ! and μ^{τ} = λ^{σ} .

Thus in this case we obtain that, as desired,

(4.2) = ∑_{σ∈[a:s]} \frac{a!}{σ!} λ^{σ} = Y_{𝕂}^{a}(E_{s}(λ)).

If i = \overline{a}\overline{a} then for any 0 ≤ p ≤ n\overline{a} there exists a natural bijection between the sets {τ ∈ [a : r]: τ_{s+1} = p and [a^{←}^{p} : s]} (see 1.6). If τ ∈ [a : r] corresponds to σ ∈ [a^{←}^{p} : s] under this bijection then

\{ \begin{matrix}τ! = p!σ! and \\ a^{←}^{p}!p−1 ; ∏_{h=0}(n\overline{a} − h) = a! \\ μ^{τ} = λ^{σ} λ^{p}. \end{matrix}  (4.3)

Thus

(4.2) = ∑_{p=0}^{a^{←}(i)} ∑_{\begin{matrix}τ∈[a:r] \\ τ_{s+1}=p \end{matrix}} \frac{a!}{τ!} μ^{τ}

= ∑_{p=0}^{a^{←}(i)} ∑_{σ∈[ a^{←}^{p}:s]} \binom{a^{←}(i)}{p} \frac{a^{←}^{p}!}{σ!} λ^{p}λ^{σ} (by 4.3)

= ∑_{p=0}^{a^{←}(i)} \binom{a^{←}(i)}{p} λ^{p}Y_{𝕂}^{a^{←}^{p}} (Proposition 2.7)

which is as desired. The proof for the left action runs along similar lines.

□

Continuous operators.

Throughout this section we will assume that 𝕂 is polynomially faithful.

Given ∂ ∈ End_{𝕂}(Pol F(𝕂)) and s ∈ W define a family of scalars ∂_{w}^{s} by writting

∂Y_{𝕂}^{s} = ∑_{w∈W} ∂_{w}^{s}Y_{𝕂}^{w}, ∂_{w}^{s} ∈ IK.

in the way prescribed by Theorem 2.10 Lemma 5.1. For ∂ ∈ End_{𝕂}(Pol F(𝕂)) the following two conditions are equivalent

PC1 For every summable S the following two conditions hold: (i) For w ∈ W the set {s ∈ S : ∂_{w}^{s} ≠ 0} is finite (ii) For every w ∈ W the set {u ∈ W_{w} : ∃s ∈ S : ∂_{u}^{s} ≠ 0} is finite. In other words,

the set {w ∈ W : ∃s ∈ S : ∂_{w}^{s} ≠ 0} is summable.

PC2 For all w ∈ W there exists r ∈ W such that

∀u ∈ W_{w} and ∀s ∈ W ∂_{u}^{s} ≠ 0 ⇒ s ∈ W_{r}.

Proof. Suppose PC2 holds for ∂ . Let S be summable and let w ∈ W .Choose r ∈ W as in PC2. Then {s : ∂_{w}^{s} ≠ 0} ∩ S = {s : ∂_{w}^{s} ≠ 0} ∩ S_{r}. Since S_{r} is finite for every r PC1(i) holds.

As for PC1(ii) we have

{u ∈ W_{w} : ∃s ∈ S : ∂_{u}^{s} ≠ 0} = {u ∈ W_{w} : ∃s ∈ S_{r} : ∂_{u}^{s} ≠ 0} = ∪_{s∈S_{r}}{u ∈ W_{w} : ∂_{u}^{s} ≠ 0}.

Since supp ∂(Y_{𝕂}^{s}) := {u ∈ W : ∂_{u}^{s} ≠ 0} is summable, it follows that ∪_{s∈S_{r}}{u ∈ W_{w} : ∂_{u}^{s} ≠ 0} is a finite union of finite sets, hence finite. The set of PC1(ii) is thus summable.

Assume next that PC1 holds. Fix w ∈ W , and consider the set

\{(u, s) ∈ W_{w} × W : ∂_{u}^{s} ≠ 0\} = ∪_{v∈\overline{W}} P_{v},

where

P_{v} := \{(u, s) : u ∈ W_{w}, \overline{s} = v, ∂_{u}^{s} ≠ 0\}.

Let Q_{v} := \{u : ∃s : (u, s) ∈ P_{v}\}. Note that the set R := \{v ∈ \overline{W} : Q_{v} is infinite\} is finite, for otherwise we can construct a summable set violating PC1(ii) as follows: Let {v_{(n)}} be an infinite sequence of distinct elements of R . Construct the sequences {s_{(k)}} and {u_{(k)}} inductively as follows: since Q_{v_{(n)}} is infinite, there exist u_{(n)} ∈ Q_{v_{(n)}}\{u_{(k)}}_{k<n} and s_{(n)} such that (u_{(n)}, s_{(n)}) ∈ P_{v_{(n)}} . The set S = {s_{(n)}}_{n∈ℕ} which is summable (since it contains at most one element for every reduced type) violates PC1(ii).

The set T := \{v ∈ \overline{W} : Q_{v} is finite and non empty\} is also finite: For if v(n) is an infinite sequence of distinct elements of T we can consider an infinite sequence of pairs (u_{(n)}, s_{(n)}) ∈ P_{v_{(n)}} . If an infinite number of different u_{(n)} ’s appear in these pairs then PC1(ii) fails, while if the different u_{(n)} ’s are finite in number then a subsequence of the s_{(n)} ’s violates PC1(i).

Note that PC1(i) implies that

∀ u ∃ q(u) ∈ W such that ∂_{u}^{s} ≠ 0 ⇒ s ∈ W_{q(u)}.

Construct r in such a way that W_{r} contains W_{q(u)} for all u ∈ T and contains W_{u} for all u ∈ R . We claim that PC2 holds. Indeed, if ∂_{u}^{s} ≠ 0 then either \overline{s} ∈ R , in which case s ∈ W\overline{s} ⊂ W_{r} , or \overline{s} ∈ T , in which case s ∈ W_{q(u)} ⊂ W_{r} . The proof of the Lemma is now complete..

5.2 The functions described in this last Lemma will be refered to as precontinuous. ∂ as above will be said to be algebraically continuous if it is precontinuous and in addition satisfyes the following property

AC For every summable set S and family c_{s} ∈ 𝕂 we have

∂( ∑_{s∈S} c_{s}Y_{𝕂}^{s}) = ∑_{w∈W} ( ∑_{s∈S} c_{s}∂_{w}^{s})Y_{𝕂}^{w}.

Remark 5.3 By abuse of notation we will interpret continuity as the following equality:

∂( ∑_{s∈W} c_{s}Y_{𝕂}^{s}) = ∑_{s∈W} c_{s}∂(Y_{𝕂}^{s})

Lemma 5.4. For r ∈ W and f ∈ Pol F(𝕂) the following conditions are equivalent (i) f vanishes in F(𝕂)^{r} (ii) f ∘ π_{r} = 0 (iii) If f = Σc_{a}Y_{𝕂}^{a} then c_{a} = 0 whenever a ∈ W_{r} .

Proof. (i) ⇒ (ii) is obvious.

(ii) ⇒ (iii) In Proposition 2.7 (with s replaced by r ) note that regardless of a different σ ’s produce different monomials T^{σ} ’s. If we now write f as in Theorem 2.10 and take the argument of Corollary 2.8 into account we see that f ∘ π_{r} is a nonzero polynomial in 𝕂[r] whenever suppf ∩ W_{r} ≠ ∅ . Our assumption that 𝕂 be polynomially faithful shows then that (ii) fails whenever (iii) does.

(iii) ⇒ (i) See Corrollary 2.8. □

5.5 The set of functions described in this last Lemma will be denoted by U_{r} . Note that U_{r} = U\overline{r} and U_{rs} ⊂ U_{r} ∩U_{s} . It follows that the sets f +U_{r} form a neighborhood base of f ∈ Pol F(𝕂) of a (linear) topology on Pol F(𝕂) . For this topology continuity of an endomorphism ∂ ∈ End_{𝕂}(Pol F(𝕂)) is equivalent to

TC For all w ∈ W there exists r ∈ W such that ∂(U_{r}) ⊂ U_{w} .

These maps will be called topologycally continuous.

Proposition 5.6. Let ∂ ∈ End_{𝕂}(Pol F(𝕂)) .

(i) If ∂ is algebraically continuous it is topologically continuous. (ii) If ∂ is topologically continuous it is precontinuous.

Proof.

(i) Let ∂ ∈ End_{𝕂}(Pol F(𝕂)) be algebraically continuous. Given w ∈ W choose r as in PC2. Taking Lemma 5.4 into account we see that for s ∈ W ∂Y_{𝕂}^{s} ∉ U_{w} ⇒ ∃u ∈ W_{w}: ∂_{u}^{s} ≠ 0 ⇒ s ∈ W_{r} ⇒ Y_{𝕂}^{s} ∉ U_{r} . Thus Y_{𝕂}^{s} ∈ U_{r} ⇒ ∂Y_{𝕂}^{s} ⊂ U_{w}. That TC holds now follows from AC.

(ii) Let ∂ ∈ End_{𝕂}(Pol F(𝕂)) be topologically continuous and given w ∈ W choose r as in TC. Let u ∈ W_{w} and s ∈ W then

∂_{u}^{s} ≠ 0 ⇒ ∂Y_{𝕂}^{s} ∉ U_{u}
⇒ ∂Y_{𝕂}^{s} ∉ U_{w} ( see 5.4 (iii))
⇒ Y_{𝕂}^{s} ∉ U_{r} (by TC)
⇒ s ∈ W_{r} (Corollary 2.8)
□

Corollary 5.7.. The composition ∂_{1}∂_{2} of two algebraically continuous endomorphisms ∂_{1}, ∂_{2} ∈ End_{𝕂} Pol F(𝕂) is algebraically continuous.

Proof. The Proposition shows that ∂_{1}∂_{2} is precontinuous. That AC holds follows easily from the assumption that AC holds for both ∂_{1} and ∂_{2} .

Remark 5.8. Let Y (𝕂) := ∑_{w∈W} 𝕂Y_{𝕂}^{w} ⊂ Pol F(𝕂) This is a dense subset of Pol F(𝕂) . For 𝕂 -linear maps from Y (𝕂) into Pol F(𝕂) condition AC trivially holds: algebraic continuity and precontinuity are thus equivalent in this case.

Proposition 5.9.. Left and right multiplication by elements of the group (see section 4) are algebraically continuous endomorphisms of Pol(𝕂^{N}).

Proof. We leave this as a (non entirely trivial) exercise.

Proposition 5.10.. Let ∂ ∈ Hom_{𝕂}(Y (𝕂), Pol F(𝕂)) be precontinuous. Then

(i) There exists a unique algebraically continuous 𝕂 -linear endomorphism of Pol F(𝕂) extending ∂.

(ii) If ∂ is a derivation or right or left invariant then so is its extension.

Proof. Taking Theorem 2.10 into account define ∂ ∈ End_{𝕂} Pol F(𝕂) by means of AC. The uniqueness is clear. That ∂ is a derivation follows from the description of the multiplication in Pol F(𝕂) given in Theorem 3.7(iii). The invariance follows easily from the last Proposition. □

As an easy consequence of (i) we get

Corollary 5.11.. If two algebraically continuous endomorphism of Pol F(𝕂) coincide in Y (𝕂) , they are equal.

Remark 5.12. Let f, g, and h be as in the proof of Theorem 3.7(iii). We see that fg and h have the same image in each of the quotient rings Pol F(𝕂)/U_{r}. That 3.7(iii) holds follows from ∩_{r∈W} U_{r} = (0) .

Right invariant derivations and free Lie algebras.

Throughout this section we will assume that 𝕂 is polynomially faithful. Algebraically continuous functions we will simply refer to as continuous. Starting with 6.8 we also assume that 𝕂 is torsion free.

We give the free 𝕂 -module Y (𝕂) a ℤ^{I} grading by declaring Y_{𝕂}^{w} to be homogeneous of degree deg_{w} (See 1.5.) The shuffle product is compatible with this grading making Y_{𝕂} into a ℤ^{I} -graded algebra thereof. This grading extends to Pol F(𝕂) : Every element of Pol F(𝕂) is a sum over a summable set of homogeneous elements (see Theorem 2.10.)

Given ∂ ∈ End_{𝕂}(Pol F(𝕂)) and ω ∈ ℤ^{I} define ∂_{ω} ∈ Hom_{𝕂}(Y (𝕂) , Pol F(𝕂)) by

∂_{ω}(Y_{𝕂}^{s}) = ∑_{w deg (w)= deg (s)+ω} ∂_{w}^{s}Y_{𝕂}^{w}.  (6.1)

Note that this sum is always finite so that ∂_{ω} is always precontinuous. It follows (see Proposition 5.9) that ∂_{ω} extends uniquely to an element, also denoted ∂_{ω} of End_{𝕂}(Pol F(𝕂)) . Note also that ∂_{ω} = 0 unless suppω := {i ∈ I: ω(i) ≠ 0} is finite.

Lemma 6.2. Let ∂ ∈ End_{𝕂}(Pol F(𝕂)) be continuous. Let ω ∈ ℤ^{I} . Then

(i) If ∂ is right-invariant then ∂_{ω} is right-invariant. (ii) If ∂ is a derivation then ∂_{ω} is a derivation.

Proof.

(i) Let i ∈ I , λ ∈ 𝕂 and w ∈ W . By Proposition 4.1 and the fact that both ∂ and R_{h} are continuous we have for h = E_{i}(λ) ,

R_{h}∂(Y_{𝕂}^{s}) = ∑_{w∈W} ∂_{w}^{s} ∑_{p=0}^{w^{←}(i)} λ^{p} ( \begin{matrix}w^{←}(i) \\ p \end{matrix} ) Y_{𝕂}^{\overleftarrow{w}^{p}}

∂(R_{h}Y_{𝕂}^{s}) = ∑_{p=0}^{s^{←}(i)} λ^{p} ( \begin{matrix}s^{←}(i) \\ p \end{matrix} ) ∂(Y_{𝕂}^{\overleftarrow{s}^{p}}).

Our assumption on IK (look at the above as polynomials in one variable evaluated at λ ) together with the right-invariance of ∂ yields the system of equalities

( \begin{matrix}s^{←}(i) \\ p \end{matrix} ) ∂(Y_{𝕂}^{\overleftarrow{s}^{p}}) = ∑_{w∈W} ∂_{w}^{s} ( \begin{matrix}w^{←}(i) \\ p \end{matrix} ) Y_{𝕂}^{\overleftarrow{w}^{p}}

for all p ≥ 0 . Equating now the terms of degree deg(\overleftarrow{s}^{p}) + ω we obtain, taking Theorem 2.10 into account, that

( \begin{matrix}s^{←}(i) \\ p \end{matrix} ) ∂_{ω}(Y_{𝕂}^{\overleftarrow{s}^{p}}) = ∑_{w∈W deg(w)=deg(s)+ω} ∂_{w}^{s} ( \begin{matrix}w^{←}(i) \\ p \end{matrix} ) Y_{𝕂}^{\overleftarrow{w}^{p}},

which is equivalent to the right-invariance of ∂_{ω} .

(ii) Let u, v ∈ S. Equating the terms of degree deg(u) + deg(v) + ω on the left and on the right of ∂(Y_{𝕂}^{u} Y_{𝕂}^{v}) = ∂(Y_{𝕂}^{u}) Y_{𝕂}^{v} + Y_{𝕂}^{u} ∂(Y_{𝕂}^{v}) yields ∂_{ω}(Y_{𝕂}^{u} Y_{𝕂}^{v}) = ∂_{ω}(Y_{𝕂}^{u}) Y_{𝕂}^{v} + Y_{𝕂}^{u} ∂_{ω}(Y_{𝕂}^{v}) . That ∂_{ω} is a derivation now follows from Proposition 5.10.

Proposition 6.3. Let ∂ ∈ End_{𝕂}(Pol F(𝕂)) be continuous and right-invariant. If ∂_{ω} ≠ 0 then ω ∈ ℤ_{−}^{I} := {ω ∈ ℤ^{I} : ω(i) ⩽ 0 for all i ∈ I}

Proof. Suppose ω ∉ ℤ_{−}^{I}. Let us prove that ∂_{ω}(Y_{𝕂}^{s}) = 0 by induction on s . The proof is based on the observation that the only polynomial functions invariant (left or right) under the group action are the constant functions. If s = 0 then Y_{𝕂}^{s} = 1 is a constant function and R_{h}∂_{ω}(1) = ∂_{ω}R_{h}(1) = ∂_{ω}(1) . Thus ∂_{ω}(1) is a constant function which is zero since ω ≠ 0 . Assume now that ∂_{ω}(Y_{𝕂}^{w}) = 0 for all w with w < s. For h ∈ F(𝕂)) we have by Proposition 4.1 that

R_{h}(∂_{ω}Y_{𝕂}^{s}) = ∂_{ω}(R_{h}Y_{𝕂}^{s})

= ∂_{ω}(Y_{𝕂}^{s}) + ∑_{w<s} c_{w}∂_{ω}(Y_{𝕂}^{w}) = ∂_{ω}(Y_{𝕂}^{s}).

It follows that ∂_{ω}Y_{𝕂}^{s} is a constant function. But deg (∂_{ω}Y_{𝕂}^{s}) = ω + deg_{s} ≠ 0. Hence ∂_{ω}Y_{𝕂}^{s} = 0 and since ∂_{ω} is continuous ∂_{ω} = 0 . □

Given i ∈ I define ∂_{i} ∈ End_{𝕂}Y (𝕂) as follows: For s = \overline{s}_{1}^{n_{1}} … \overline{s}n\overline{s} ; \overline{s}

∂_{i}(Y_{𝕂}^{s}) = s^{→}(i)Y s⃗^{1} ; 𝕂.  (6.4)

Note that if 𝕂 is a ℚ -algebra we have

∂_{i}(X_{ℚ}^{s}) = \{ \begin{matrix}Xs⃗^{1} ; ℚ, & if i = s_{1} \\ 0, & if i ≠ s_{1}. \end{matrix}  (6.5)

∂_{i} is clearly precontinuous and its unique continuous extension to End_{𝕂}(Pol F(𝕂)) will also be denoted by ∂_{i} .

Proposition 6.6. ∂_{i} ∈ End_{𝕂}(Pol F(𝕂)) is a continuous homogeneous right-invariant derivation of degree −deg_{i}, Proof. We have just seen that ∂_{i} is continuous. That ∂_{i} is homogeneous of degree −deg_{i} is clear. By Proposition 5.10 to show that ∂_{i} is right invariant it suffices to show that ∂_{i} is right invariant when viewed as an element of End_{𝕂}Y (𝕂) , for which in turn it suffices to show that

R_{h}∂_{i}(Y_{𝕂}^{s}) = ∂_{i}(R_{h}Y_{𝕂}^{s}), for all j ∈ I, λ ∈ 𝕂, h = E_{j}(λ), and s ∈ W.

By taking Proposition 4.1 into account it is easy to see that this holds if s̄ > 1. If s̄ = 1 then the only non trivial case is when i = j and s = i^{n}. We then have

R_{h}∂_{i}(Y_{𝕂}^{s}) = nR_{h}Y s⃗^{1} ; 𝕂 = ∑_{p=0}^{n−1} n ( \begin{matrix}n − 1 \\ p \end{matrix} ) λ^{p}Y s⃗^{p+1} ; 𝕂

= ∑_{p=0}^{n−1} \frac{n!}{p!(n − p − 1)!}λ^{p}Y s⃗^{p+1} ; 𝕂

= ∑_{p=0}^{n} ( \begin{matrix}n \\ p \end{matrix} ) (n − p)λ^{p}Y s⃗^{p+1} ; 𝕂

= ∂_{i}( ∑_{p=0}^{n} ( \begin{matrix}n \\ p \end{matrix} ) λ^{p}Y s⃗^{p} ; 𝕂)

= ∂_{i}(R_{h}Y_{𝕂}^{s}).

Finally, let us prove that ∂_{i} is a derivation of Pol F(𝕂). Again it suffices to show that ∂_{i} is right invariant as an element of End_{𝕂}Y (𝕂). Since Y (𝕂) ≅ 𝕂 ⨂_{ℤ} Y (ℤ) it will suffice to establish the result in the case 𝕂 = ℤ . That this last is the case follows from the commutativity of the diagram in Remark 3.8 commutes, together with the fact that ∂_{i} is a derivation of X(ℤ) ([Rtn] 1.4.3.) □

In what follows D will denote the associative 𝕂 -subalgebra of End_{𝕂}(Pol F(𝕂)) generated by the ∂_{i} ’s. If s = s_{1} ⋯ s_{s} ∈ W we set ∂^{s} := ∂_{s_{s}} ⋯ ∂_{s_{1}} . By convention ∂^{1} = Id. The Lie subalgebra of Der Pol F(𝕂) generated by the ∂_{i} ’s will be denoted by L .

Proposition 6.7. All elements of D are continuous and right invariant.

Proof. Apply Corollary 5.7 and Proposition 6.6. □

Proposition 6.8. If 𝕂 is torsion free then D is a free associative algebra. Furthermore the natural ℤ^{I} grading of D is compatible with that of Y (𝕂) .

Proof. We need to show that the ∂^{s}’s are linearly independent over 𝕂 . The constant term of ( ∑_{s∈W}c_{s}∂^{s})Y_{𝕂}^{w} is c_{w} w! · 1, hence ∑_{s∈W}c_{s}∂^{s} = 0 implies c_{s} = 0 for all s ∈ W. We assigned ∂^{s} = ∂_{s_{s}} … ∂_{s_{1}} degree −deg_{s} . The compatibility is clear.

Corollary 6.9. If 𝕂 is torsion free L is a free Lie algebra, freely generated by the ∂_{i} ’s.

Remark 6.10. When 𝕂 has torsion L is not free.

Remark 6.11 In Proposition 6.7 and its Corollary Pol F(𝕂) may be replaced by Y (𝕂) .

For the remainder of the paper we will assume 𝕂 is torsion free. The algebra D , being free associative, affords a unique coproduct Δ : D → D ⊗ D satysfying

Δ(∂_{i}) = ∂_{i} ⊗ 1 + 1 ⊗ ∂_{i}. Denote by μ : Pol F(𝕂)⊗Pol F(𝕂) → Pol F(𝕂) the multiplication in Pol F(𝕂).

Lemma 6.12. Let ∂ ∈ D, and f, g ∈ Pol F(𝕂) . Then μ(Δ(∂)(f ⊗ g)) = ∂(μ(f ⊗ g)).

Proof. It is sufficient to prove the Lemma for ∂ = ∂^{w}, w ∈ W . This we shall do by induction in w > 0 (The case w = 0 being clear: ∂^{0} = Id .)

If w = 1 then ∂ = ∂_{i} and we have

μ(Δ(∂_{i})(f ⊗ g)) = μ[(∂_{i} ⊗ 1 + 1 ⊗ ∂_{i})(f ⊗ g)]

= μ(∂_{i}f ⊗ g + f ⊗ ∂_{i}g) = (∂_{i}f)g + f(∂_{i}g) = ∂_{i}(fd) = ∂_{i}(μ(f ⊗ g)).

For the inductive step, let ∂^{w} = ∂^{v}∂_{i} and assume the Lemma holds for ∂^{v} . Then

μ(Δ(∂^{v}∂_{i})(f ⊗ g)) = μ[(Δ(∂^{v})Δ(∂_{i}))(f ⊗ g)]

= μ(Δ(∂^{v})(Δ(∂_{i}))(f ⊗ g)) = μ[Δ(∂^{v})(∂_{i}f ⊗ g + f ⊗ ∂_{i}g)]
= ∂^{v}((∂_{i}f)g + f(∂_{i}g)) = ∂^{v}(∂_{i}(fg)) = ∂^{v}∂_{i}μ(f ⊗ g).

Proposition 6.13. Let ∂ ∈ D . Then ∂ ∈ Der_{𝕂}Y (𝕂) if and only if ∂ ∈ L.

Proof. Make D (respectively Y (𝕂) ) into a ℤ -graded algebra by declaring the ∂_{i}’s (respectively Y_{𝕂}^{a} ) to be of degree −1 (respectively a.) These gradings are compatible so that we may assume without loss of generality that ∂ is homogeneous of degree −N . Then

Δ(∂) = ∂ ⊗ 1 + 1 ⊗ ∂ + ∑_{\begin{matrix}a,b \\ a<N,b<N \\ a+b=N \end{matrix}} c_{ab}∂^{a} ⊗ ∂^{b}.

Thus if a + b = N, a < N, b < N we have

Δ(∂)(Y^{a} ⊗ Y^{b}) = a! b! c_{ab}1 ⊗ 1

and hence

μ(Δ(∂)((Y^{a} ⊗ Y^{b})) = a! b! c_{ab} 1.

By the last Lemma

μ(Δ(∂)((Y^{a} ⊗ Y^{b})) = ∂(X^{a} X^{b})
= ∂(Y^{a}) Y^{b} + Y^{a} ∂(Y^{b}) = 0,

Thus c_{ab} = 0 and Δ(∂) = ∂ ⊗ 1 + 1 ⊗ ∂. By Friedrichs’ Theorem ([Rtn] Corollary 4.17 or [Bbk]Ch.2 §3.1) we conclude that ∂ ∈ L .

Proposition 6.14. Let ∂ ∈ End_{𝕂}(Pol F(𝕂)) be continuous and right invariant. If ω ∈ ℤ_{−}^{I}. then ∂_{ω} is a linear combination of the ∂^{s} with deg_{s} = −ω.

Proof. Let ∂_{ω}(Y^{s}) = c_{s} · 1 for s ∈ W with deg_{s} = −ω. An argument similar in spirit to that of Proposition 6.3 shows that ∂_{ω} − ∑_{\begin{matrix}s∈W \\ deg_{s}=−ω \end{matrix}}c_{s}∂^{s} = 0.

□

Consider the following completions of the free associative algebra D generated by operators ∂_{i} :

\overline{D} = ∏_{w∈W} IK∂^{w}

and

\overline{D}_{0} = ⨁_{r∈\overline{W}} ∏_{\begin{matrix}w∈W \\ \overline{w}=r \end{matrix}} IK ∂^{w}.

Note that \overline{D} acts on the space Y (𝕂) = ⊕_{w∈W} IKY^{w} . If ∂ = ∑_{u∈W} d_{u}∂^{u} then

∂(Y^{s}) = ∑_{\begin{matrix}u,w \\ s=uw \end{matrix}} d_{u}c_{s,u}Y^{w}  (6.15)

for some coefficients {c_{s,u}} all but a finite number of which are not zero.

Proposition 6.16. An endomorphism ∂ ∈ \overline{D} can be extended to a continuous endomorphism of Pol F(𝕂) if and only if ∂ ∈ \overline{D}_{0} .

Proof. We mantain the notation of 6.15. If ∂ ∈ \overline{D}_{0} then the set T = \{t ∈ \overline{W} |∃v ∈ W : \overline{v} = t and d_{v} ≠ 0\} is finite. If we set r = tw , where t is the product in any order of the elements of T then with the aid of 6.16 one shows that PC2 holds. Thus ∂ is precontinuous and hence has a continuous extension (Proposition 5.10.)

Conversely if ∂ ∉ \overline{D}_{0} then the set T above is infinite and we can construct an infinite set S ⊂ W such that S contains at most one word of every reduced type and for every s ∈ S d_{s} ≠ 0 and \overline{s} ∈ T . Then the set S is summable and since ∂_{1}^{s} = d_{s}c_{ss} = d_{s}s! ≠ 0 we see that PC1 (i) fails. □

Let \overline{L} be the algebra of Lie series of L (See [Rtn] Ch. 3) Define \overline{L}_{0} = \overline{L} ∩ \overline{D}_{0} .

Theorem 6.17. Let IK be a ring which is torsion free and polinomially faithful. Then there exists a natural isomorphism between \overline{L}_{0} and the Lie algebra of all continuous right-invariant derivations of Pol F(𝕂). Proof. Let ∂ ∈ End_{𝕂}(Pol F(𝕂)) be a continuous right invariant derivation. By Propositions 6.2, 6.3, and 6.14 we can identify ∂ with Σ_{ω∈ℤ_{−}^{I}}∂_{ω} ∈ \overline{L}. Now Proposition 6.13 shows that ∂ ∈ \overline{L}. Since ∂ is continuous the last Proposition shows that ∂ ∈ \overline{D}_{0}

Conversely if ∂ ∈ \overline{L}_{0} then we can think of ∂ as a continuous element of End_{𝕂}(Pol F(𝕂)) (again by the last Proposition.) When we restrict ∂ to Y (𝕂) we obtain a right invariant derivation (Propositions 6.7 and 6.13.) The unique continuous extension \overline{∂} of ∂ to End_{𝕂}(Pol F(𝕂)) is then right invariant derivation (Proposition 5.10.) Since \overline{∂} is continuous ∂ = \overline{∂} (Corollary 5.11). □

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