Remark

Tag 0AE3, permanent. Cite it as [Bourbaki, Tag 0AE3].

Lie Groups and Lie Algebras, chapter III, § 5. Formal calculations in Lie groups

(1) We write condition (ii) of Proposition 2 for t an integer $\geqslant 0$ :

\begin{align*} 0 &= \psi_0(x) \\ x &= \psi_0(x) + \psi_1(x) \\ x^{[2]} &= \psi_0(x) + 2\psi_1(x) + \psi_2(x) \\ &\ldots \ldots \ldots \ldots \ldots \ldots \ldots \end{align*}

These formulae suffice to determine the $\psi_i$ .

(2) We see that $\psi_0(x) = 0, \psi_1(x) = x, \psi_2(x) = x^{[2]} - 2x$ ,

x^{[-1]} = \sum_{i=1}^\infty (-1)^i \psi_i(x).

(3) The above expression for $\psi_2$ and formula (6) prove that

\psi_2(x) \equiv \mathrm{B}(x, x) \mod \deg 3.

Using Proposition 2, (i) and (ii), we see that

x^{[t]} \equiv tx + \binom{t}{2} \mathrm{B}(x, x) \mod \deg 3.

(4) Let $\psi_{p,m}(x)$ and $h_{t,m}(x)$ denote the homogeneous components of $\psi_p(x)$ and $x^{[t]}$ of degree m. Then $\psi_{p,m} = 0$ for $m < p$ . On the other hand, Proposition 2 (ii) gives

h_{t,m}(x) = \sum_{p \leq m} \frac{t(t-1)\ldots(t-p+1)}{p!} \psi_{p,m}(x)

that is

h_{t,m}(x) = \sum_{1 \leq r \leq m} t^r \phi_{r,m}(x)

where the

\phi_{r,m}

are homogeneous polynomial mappings of degree m of

\mathbf{K}^n

into

\mathbf{K}^n

. In particular, by (23),

\phi_{1,m}(x) = \sum_{p \leq m} \frac{(-1)^{p-1}}{p} \psi_{p,m}(x)

\phi_{m,m}(x) = \frac{1}{m!} \psi_{m,m}(x).

(5) If $\mathbf{K}$ is of characteristic $> 0$ , the results of nos. 1 and 2 remain true, provided $e_\alpha$ , in no. 2, is defined by $\langle e_\alpha, \sum_\beta \lambda_\beta x^\beta \rangle = \lambda_\alpha$ . In no. 3, if the $\psi_j$ are defined as above, the argument again proves that $x^{[t]} = \sum_{i=1}^\infty \binom{t}{i} \psi_i(x)$ if $t \in \mathbf{N}$ .

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

Cited by

Nothing.

Cites

Proposition 2