Lemma 1

Tag 0A49, permanent. Cite it as [Bourbaki, Tag 0A49].

Lie Groups and Lie Algebras, chapter II, § 3. Enveloping algebra of the free Lie algebra

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)}.

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 $m \geqslant 1$ .

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

Cited by

Theorem 2, § 3

Cites

Algebra, Chapter IV, § 6, no. 9