§ 3. Enveloping algebra of the free Lie algebra

In this paragraph, $A(X) = A_K(X)$ denotes the free associative algebra Libas(X) of the set $X$ over the ring $K$ (Algebra, Chapter III, § 2, no. 7, Definition 2). $X$ is identified with its canonical image in $A(X)$ ; recall that the $K$ -module $A(X)$ admits as basis the free monoid $Mo(X)$ derived from $X$ ; $A^+(X)$ denotes the submodule of $A(X)$ generated by the non-empty words.

1. Enveloping algebra of $L(X)$

Theorem 1

Let* $\alpha : L(X) \to A(X)$ be the unique Lie algebra homomorphism extending the canonical injection of $X$ into $A(X)$ (§ 2, no. 2, Proposition 1). Let $\sigma : L(X) \to U(L(X))$ be the canonical mapping of $L(X)$ into its enveloping algebra and let $\beta : U(L(X)) \to A(X)$ be the unique unital algebra homomorphism such that $\beta \circ \sigma = \alpha$ (Chapter I, § 2, no. 1, Proposition 1). Then: (a) $\alpha$ is injective and $\alpha(L(X))$ is a direct factor submodule of $A(X)$ . (b) $\beta$ is bijective.

Cites 4 · Cited by 4

Cites Proposition 1, I § 2, Proposition 1, II § 2, Corollary 3, I § 2, Corollary, II § 2

Cited by Remark, Theorem 2, II § 6. The Hausdorff series, Proposition 3, VIII § 4

Let B be a unital K-algebra and $\phi$ a mapping of X into B; by Proposition 1 of § 2, no. 2, there exists a Lie algebra homomorphism $\psi : L(X) \to B$ such that $\psi | X = \phi$ ; by Proposition 1 of Chapter I, § 2, no. 1, there exists a unital algebra homomorphism $\theta : U(L(X)) \to B$ such that $\theta \circ \sigma = \psi$ and hence such that $(\theta \circ \sigma) | X = \phi$ . As $\sigma(X)$ generates the unital algebra $U(L(X))$ , the homomorphism $\theta$ is the unique unital algebra homomorphism satisfying the latter condition. This shows that the ordered pair $(U(L(X)), \sigma | X)$ is a solution of the same universal mapping problem as $A(X)$ ; taking $\phi$ to be the canonical injection of X into $A(X)$ , we deduce that $\beta$ is an isomorphism, which proves (b).

Finally, as $L(X)$ is a free K-module (§ 2, no. 11, Corollary to Theorem 1), $\sigma$ is injective and $\sigma(L(X))$ is a direct factor submodule of $U(L(X))$ (Chapter I, § 2, no. 7, Corollary 3 to Theorem 1). By (b), this proves (a).

Corollary 1

There exists on the algebra $A(X)$ a unique coproduct making $A(X)$ into a bigebra such that the elements of X are primitive. Further, $\beta$ is an isomorphism of the bigebra $U(L(X))$ onto $A(X)$ with this bigebra structure.

Cited by 2

Cited by Corollary 2, Theorem 1, II § 6

This follows from assertion (b) of the theorem and the fact that X generates the unital algebra $A(X)$ .

Henceforth $A(X)$ is given this bigebra structure and $L(X)$ is identified with its image under $\alpha$ , that is with the Lie subalgebra of $A(X)$ generated by X.

Corollary 2

If $K$ is a field of characteristic 0, $L(X)$ is the Lie algebra of primitive elements of $A(X)$ .

Cites 2 · Cited by 2

Cites Corollary, II § 1, Corollary 1

Cited by Remark, Theorem 1, II § 6

This follows from Corollary 1 and the Corollary to Proposition 9 of § 1, no. 5.

Remark

(1) Let $K'$ be a commutative ring containing K. If $A(X)$ , $L(X)$ and $L_{K'}(X)$ are identified with subsets of $A_{K'}(X)$ , we deduce from part (a) of Theorem 1 the relation

L(X) = L_{K'}(X) \cap A(X).

(2) Corollary 2 to Theorem 1 remains valid if it only assumes that the additive group of the ring K is torsion-free. For suppose first that $K = \mathbf{Z}$ ; every primitive element of $A(X)$ is a primitive element of $A_q(X)$ and hence is in $L_q(X) \cap A(X) = L(X)$ (Corollary 2 and formula (1)). In the general case, K is flat over $\mathbf{Z}$ and we apply Remark 2 of § 1, no. 2 and Proposition 3 of § 2, no. 5.

(3) Let $\Delta$ be a commutative monoid, $\phi_0$ a mapping of $X$ into $\Delta$ and $\phi : Mo(X) \to \Delta$ the homomorphism of the associated monoid; if $A(X)$ is given the graduation $(A^\delta(X))_{\delta \in \Delta}$ defined in Algebra, Chapter III, § 3, no. 1, Example 3 and $L(X)$ the graduation $(L^\delta(X))_{\delta \in \Delta}$ defined in § 2, no. 6, we have immediately, for $\delta \in \Delta$ , $L^\delta(X) \subset L(X) \cap A^\delta(X)$ . As $L$ is the sum of the $L^\delta(X)$ for $\delta \in \Delta$ , and the sum of the $L(X) \cap A^\delta(X)$ for $\delta \in \Delta$ is direct, this implies

L^\delta(X) = L(X) \cap A^\delta(X).

(4) Let $A$ be a unital associative algebra and $t = (t_i)_{i \in I}$ a family of elements of $A$ . We have a diagram

\begin{array}{ccc} L(I) & \xrightarrow{f_t} & A \\ i \downarrow & & \downarrow g_t \\ A(I) \end{array}

where $i$ is the canonical injection, $f_t$ is the Lie algebra homomorphism defined by $t$ and $g_t$ is the unital algebra homomorphism such that $g_t(i) = t_i$ for $i \in I$ . The diagram is commutative for $g_t \circ i$ and $f_t$ coincide on $I$ . It follows that if $P \in L(I)$ , the element $P((t_i)_{i \in I})$ defined in § 2, no. 4, coincides with the element $P((t_i)_{i \in I})$ defined in Algebra, Chapter III, § 2, no. 8, Example 2.

Cites 6

Cites Theorem 1, Proposition 3, II § 2, Corollary 2, A III § 3. Graded algebras, II § 2. Free Lie algebras, A III § 2. Examples of algebras

2. Projector of $A^+(X)$ onto $L(X)$

Let $\pi$ be the linear mapping of $A^+(X)$ into $L(X)$ defined by

\pi(x_1 \ldots x_n) = (\operatorname{ad}(x_1) \circ \cdots \circ \operatorname{ad}(x_{n-1}))(x_n)

for $n > 0, x_1, \ldots, x_n$ in $X$ .

Proposition 1

(a) The restriction $\pi_0$ of $\pi$ to $L(X)$ is a derivation of $L(X)$ . (b) *For every integer $n \geq 1$ and all $u$ in $L^n(X)$ , $\pi(u) = n.u$ .

Cites 1 · Cited by 1

Cites A III § 10. Derivations

Cited by Corollary

(a) Let $E$ be the endomorphism algebra of the module $L(X)$ and $\theta$ the homomorphism of $A(X)$ into $E$ such that $\theta(x) = \operatorname{ad} x$ for all $x \in X$ . The restriction of $\theta$ to $L(X)$ is a Lie algebra homomorphism of $L(X)$ into $E$ , which coincides on $X$ with the adjoint representation of $L(X)$ , whence

\theta(u).v = [u, v] \quad \text{for } u, v \text{ in } L(X).

Let $a$ be in $A(X)$ and $b$ in $A^+(X)$ ; then

\pi(a.b) = \theta(a).\pi(b).

It suffices to consider the case $a = x_1 \ldots x_p, \ b = x_{p+1} \ldots x_{p+q}$ with $p \geq 0, q \geq 1$ and $x_1, \ldots, x_{p+q}$ in $X$ ; but then (5) follows immediately from (3) since $\theta(x) = \operatorname{ad} x$ for $x \in X$ .

Let $u$ and $v$ be in $L(X)$ ; by (4) and (5),

\pi_0([u, v]) = \pi(uv - vu) = \theta(u) \cdot \pi(v) - \theta(v) \cdot \pi(u) = [u, \pi(v)] - [v, \pi(u)] = [u, \pi_0(v)] + [\pi_0(u), v],

hence $\pi_0$ is a derivation of $L(X)$ .

(b) Let $\pi_1$ be the endomorphism of the module $L(X)$ which coincides on $L^n(X)$ with multiplication by the integer $n \geq 1$ . The formula $[L^n(X), L^m(X)] \subset L^{n+m}(X)$ shows that $\pi_1$ is a derivation (Algebra, Chapter III, § 10, no. 3, Example 6). The derivation $\pi_1 - \pi_0$ of $L(X)$ is zero on $X$ and, as $X$ generates $L(X)$ , $\pi_0 = \pi_1$ , whence (b).

Corollary

Suppose that K is a Q-algebra. Let P be the linear mapping of $A^+(X)$ into itself such that

Cites 1 · Cited by 1

Cites Proposition 1

Cited by Theorem 2, II § 6

P(x_1 \ldots x_n) = \frac{1}{n} (\operatorname{ad} x_1 \circ \cdots \circ \operatorname{ad} x_{n-1})(x_n)

for $n \geq 1$ and $x_1, \ldots, x_n$ in X. Then P is a projector of $A^+(X)$ onto $L(X)$ .

The image of P is contained in $L(X)$ . Further, for all $n \geq 1$ and all $u$ in $L^n(X)$ , $P(u) = \frac{1}{n} \pi(u)$ , whence $P(u) = u$ by Proposition 1. As

L(X) = \sum_{n \geq 1} L^n(X),

we see that the restriction of P to $L(X)$ is the identity.

Remark

Suppose that K is a field of characteristic zero and let Q be the projector of $A(X) = U(L(X))$ onto $L(X)$ associated with the canonical graduation of $U(L(X))$ , cf. § 1, no. 5. For $\alpha \in \mathbf{N}^{(X)}$ , $Q(A^\alpha(X)) \subset L^\alpha(X)$ . For it suffices to verify that the image and the kernel of Q are graded submodules of $A(X)$ with the graduation of type $\mathbf{N}^{(X)}$ . This is obvious for the image, which is equal to $L(X)$ . On the other hand, let $n$ be an integer $\geq 1$ . The vector subspace of $A(X)$ generated by the $y^n$ , where $y \in L(X)$ , is equal to the vector subspace of $A(X)$ generated by the $\sum_{\sigma \in S_n} y_{\sigma(1)} y_{\sigma(2)} \cdots y_{\sigma(n)}$ , where $y_1, y_2, \ldots, y_n$ are homogeneous elements of $L(X)$ ; then this subspace is a graded submodule of $A(X)$ .

(Note that, if Card(X) $\geq 2$ , the projectors P and Q do not coincide on $A^+(X)$ . For let $x, y$ be in X with $x \neq y$ and write

z = x[x, y] + [x, y]x = x^2y - yx^2.

Then $Q(z) = 0$ and $P(z) = \frac{1}{3}[x, [x, y]] \neq 0$ , cf. § 2, no. 10, Example and no. 11, Theorem 1.)

Cites 2

Cites II § 1. Enveloping bigebra of a Lie algebra, II § 2. Free Lie algebras

3. Dimension of the homogeneous components of l(x)

Let X be a set, $\alpha$ an element of $\mathbf{N}^{(X)}$ and d an integer > 0. We write $d|\alpha$ if there exists $\beta \in \mathbf{N}^{(X)}$ such that $\alpha = d\beta$ . The element $\beta$ , which is unique, is then denoted by $\alpha/d$ .

Lemma 1

Let n be an integer > 0, $T_1, \ldots, T_n$ indeterminates and $u_1, \ldots, u_n$ elements of $\mathbf{Z}$ . Let $(c(\alpha))_{\alpha \in \mathbf{N}^n - \{0\}}$ be a family of elements of $\mathbf{Z}$ such that

1 - \sum_{i=1}^n u_i T_i = \prod_{\alpha \neq 0} (1 - T^\alpha)^{c(\alpha)}.

Cites 1 · Cited by 1

Cites A IV § 6. Symmetric functions

Cited by Theorem 2

For all $\alpha \in \mathbf{N}^n - \{0\}$ ,

c(\alpha) = \frac{1}{|\alpha|} \sum_{d|\alpha} \mu(d) \frac{(|\alpha|/d)!}{(\alpha/d)!} u^{\alpha/d}

where $\mu$ is the Möbius function (Appendix).

Formula (7) is equivalent, on taking logarithms on both sides (Algebra, Chapter IV, § 6, no. 9) to:

\log \left( 1 - \sum_{i=1}^n u_i T_i \right) = \sum_{\alpha \neq 0} c(\alpha) \log (1 - T^\alpha).

Now

-\log \left( 1 - \sum_{i=1}^n u_i T_i \right) = \sum_{j \geq 1} \frac{1}{j} \left( \sum_{i=1}^n u_i T_i \right)^j

= \sum_{j \geq 1} \frac{1}{j} \sum_{|\beta|=j} \frac{|\beta|!}{\beta!} u^\beta T^\beta

= \sum_{|\beta|>0} \frac{1}{|\beta|} \frac{|\beta|!}{\beta!} u^\beta T^\beta.

On the other hand

-\sum_{\alpha \neq 0} c(\alpha) \log (1 - T^\alpha) = \sum_{|\alpha|>0, k \geq 1} \frac{1}{k} c(\alpha) T^{k\alpha}

= \sum_{|\beta|>0, k|\beta} \frac{1}{k} c\left( \frac{\beta}{k} \right) T^\beta.

Hence (7) is equivalent to

\sum_{k|\beta} \left| \frac{\beta}{k} \right| c\left( \frac{\beta}{k} \right) = \frac{|\beta|!}{\beta!} u^\beta \quad \text{for all } \beta \in \mathbf{N}^n - \{0\}.

Let $\Lambda$ be the set of $(\lambda_1, \lambda_2, \ldots, \lambda_n) \in \mathbf{N}^n - \{0\}$ such that the g.c.d. of $\lambda_1, \lambda_2, \ldots, \lambda_n$ is equal to 1. Every element of $\mathbf{N}^n - \{0\}$ can be written uniquely in the form $m\lambda$ , where $m$ is an integer $\geqslant 1$ and $\lambda \in \Lambda$ . Condition (12) is equivalent to

\sum_{k|m} \left| \frac{m\lambda}{k} \right| c\left( \frac{m\lambda}{k} \right) = \frac{(m|\lambda|)!}{(m\lambda)!} u^{m\lambda} \quad \text{for all } \lambda \in \Lambda \text{ and all } m \geqslant 1.

By the Möbius inversion formula (Appendix), condition (13) is equivalent to

|m\lambda| c(m\lambda) = \sum_{d|m} \mu(d) \frac{\left| \frac{m\lambda}{d} \right|!}{\left( \frac{m\lambda}{d} \right)!} u^{\frac{m\lambda}{d}}

for all $\lambda \in \Lambda$ and all $m \geqslant 1$ .

Theorem 2

Let $X$ be a finite set and $n = \mathrm{Card}(X)$ .

Cites 6 · Cited by 1

Cites II § 2. Free Lie algebras, A III § 3. Graded algebras, Corollary, II § 2, Corollary 3, I § 2, Theorem 1, Lemma 1

Cited by Remark

(a) For every integer $r \geqslant 1$ , the $K$ -module $L^r(X)$ is free of rank

c(r) = \frac{1}{r} \sum_{d|r} \mu(d) n^{r/d},

where $\mu$ is the Möbius function.

(b) For all $\alpha \in \mathbf{N}^X - \{0\}$ , the $K$ -module $L^\alpha(X)$ (§ 2, no. 6) is free of rank

c(\alpha) = \frac{1}{|\alpha|} \sum_{d|\alpha} \mu(d) \frac{(|\alpha|/d)!}{(\alpha/d)!}.

We already know that the modules $L^r(X)$ , where $r \in \mathbf{N}$ , and $L^\alpha(X)$ , where $\alpha \in \mathbf{N}^X$ , are free (§ 2, no. 11, Corollary to Theorem 1). Consider the multi-gradation $(A^\alpha(X))_{\alpha \in \mathbf{N}^X}$ of $A(X)$ defined by the canonical homomorphism $\phi$ of $Mo(X)$ into $\mathbf{N}^X$ (Algebra, Chapter III, § 3, no. 1, Example 3); then $A^\alpha(X) \cap L(X) = L^\alpha(X)$ by Remark 3 of no. 1. For $\alpha \in \mathbf{N}^X$ , the $K$ -module $A^\alpha(X)$ admits as basis the set of words in which each letter $x$ of $X$ appears $\alpha(x)$ times. Let $d(\alpha)$ be the number of these words, that is the rank of $A^\alpha(X)$ ; we shall calculate in two different ways the formal power series

P((T_x)_{x \in X}) \in \mathbf{Z}[[((T_x)_{x \in X})]]

defined by

P(T) = \sum_{\alpha \in \mathbf{N}^X} d(\alpha) T^\alpha.

(1) We have

P(T) = \sum_{m \in Mo(X)} T^{\phi(m)} = \sum_{r=0}^\infty \sum_{x_1, \ldots, x_r} T_{x_1} \cdots T_{x_r} = \sum_{r=0}^\infty \left( \sum_{x \in X} T_x \right)^r

whence

P(T) = \left( 1 - \sum_{x \in X} T_x \right)^{-1}.

(2) For all $\alpha \in \mathbf{N}^X - \{0\}$ , let $(e_{\alpha,j})_{1 \leq j \leq c(\alpha)}$ be a basis of $L^\alpha(X)$ and give the set $I$ of ordered pairs $(\alpha, j)$ such that $\alpha \in \mathbf{N}^X - \{0\}$ and $1 \leq j \leq c(\alpha)$ a total ordering. By Theorem 1 of no. 1 and the Poincaré–Birkhoff–Witt Theorem (Chapter I, § 2, no. 7, Corollary 3 to Theorem 1), the elements

y_m = \prod_{(\alpha, j) \in I} (e_{\alpha, j})^{m(\alpha, j)},

where the index

m

runs through

\mathbf{N}^{(I)}

, form a basis of

A(X)

. Each

y_m

is of multidegree

\sum_{(\alpha, j) \in I} m(\alpha, j) \alpha

. Let

u(m)

denote this multidegree. It follows that

P(T) = \sum_{m \in \mathbf{N}^{(I)}} T^{u(m)} = \sum_{m \in \mathbf{N}^{(I)}} \prod_{(\alpha, j) \in I} T^{m(\alpha, j)\alpha} = \prod_{(\alpha, j) \in I} \sum_{r=0}^\infty T^{r\alpha} = \prod_{(\alpha, j) \in I} (1 - T^\alpha)^{-1},

whence finally

P(T) = \prod_{\alpha \in \mathbf{N}^X - \{0\}} (1 - T^\alpha)^{-c(\alpha)}.

Comparing (18) and (19), we obtain

1 - \sum_{x \in X} T_x = \prod_{\alpha \in \mathbf{N}^X - \{0\}} (1 - T^\alpha)^{c(\alpha)}.

Lemma 1 then gives (b).

If we now substitute the same indeterminate $U$ for the $T_x$ for $x \in X$ in formula (20), we obtain

1 - nU = \prod_{\alpha \in \mathbf{N}^X - \{0\}} (1 - U^{|\alpha|})^{c(\alpha)} = \prod_{r > 0} (1 - U^r)^{c(r)}.

Applying Lemma 1 again, we deduce (a).

Example

We have

\begin{align*} c(1) &= n, & c(2) &= \frac{1}{2}(n^2 - n), & c(3) &= \frac{1}{3}(n^3 - n), \\ c(4) &= \frac{1}{4}(n^4 - n^2), & c(5) &= \frac{1}{5}(n^5 - n), & c(6) &= \frac{1}{6}(n^6 - n^3 - n^2 + n). \end{align*}

Remark

Let $X$ be a set and let $\alpha \in \mathbf{N}^{(X)}$ ; the rank of the free $K$ -module $L^\alpha(X)$ is also given by formula (16). This follows immediately from Theorem 2 (b) and Proposition 4 of § 2, no. 6.

Cites 2

Cites Proposition 4, II § 2, Theorem 2

Exercises

The letter $X$ denotes a set.

See the exercises for § 3.