Proposition 2

Tag 0AE2, permanent. Cite it as [Bourbaki, Tag 0AE2].

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

(i) $\psi_j \equiv 0 \mod \deg j$ . (ii) If $t \in \mathbf{K}$ , then

x^{[t]} = \sum_{i=1}^8 \binom{t}{i} \psi_i(x)

where the formal power series on the right is meaningful because of (i).

(We write

\binom{t}{i} = \frac{t(t-1)\ldots(t-i+1)}{i!}

for all $t \in \mathbf{K}$ .)

Assertion (i) is obvious from the definition of the $\psi_j$ .

We prove (ii) for $t$ an integer $\geq 0$ . By (14),

x^{[t]} = \sum_{\alpha(1), \ldots, \alpha(t) \in \mathbf{N}^n} a_{\alpha(1), \ldots, \alpha(t)} x^{\alpha(1) + \ldots + \alpha(t)}.

For $\alpha = (\alpha(1), \ldots, \alpha(t)) \in (\mathbf{N}^n)^t$ , let $\sigma(\alpha)$ denote the set of $j \in \{1, 2, \ldots, t\}$ such that $\alpha(j) \neq 0$ . If, in the sum (17), we group together the terms for which $\sigma(\alpha)$ is the same, then

x^{[t]} = \sum_{\sigma \subset \{1, t\}} h_{t, \sigma}(x)

where

h_{t, \sigma}(x) = \sum_{\sigma(\alpha) = \sigma} a_{\alpha(1), \ldots, \alpha(t)} x^{\alpha(1) + \ldots + \alpha(t)}.

Let $\sigma = \{j_1, j_2, \ldots, j_q\}$ with $j_1 < j_2 < \ldots < j_q$ . In (14) (where $j$ is replaced by $t$ ), we substitute 0 for $x(k)$ for $k \notin \sigma$ ; as 0 is the identity element of $G$ , we obtain the expansion of $x(j_1) \cdot x(j_2) \ldots x(j_q)$ as an integral series about the origin:

x(j_1) \cdot x(j_2) \ldots x(j_q) = \sum_{\sigma(\alpha) \subset \sigma} a_{\alpha(1), \ldots, \alpha(t)} x(j_1)^{\alpha(j_1)} x(j_2)^{\alpha(j_2)} \ldots x(j_q)^{\alpha(j_q)}

and hence, by the definition of $\psi_q$ :

\psi_q(x) = \sum_{\sigma(\alpha) = \sigma} a_{\alpha(1), \ldots, \alpha(t)} x^{\alpha(j_1) + \ldots + \alpha(j_q)}.

By (19) and (20), we see that $h_{t, \sigma}(x) = \psi_{\mathrm{card}\,\sigma}(x)$ . Then (18) implies

x^{[t]} = \sum_{i=0}^t \binom{t}{i} \psi_i(x) = \sum_{i=0}^\infty \binom{t}{i} \psi_i(x).

Then we write $x^{[t]'} = \sum_{i=0}^\infty \binom{t}{i} \psi_i(x)$ for all $t \in \mathbf{K}$ . In the integral series $x^{[t]}$ and $x^{[t]'}$ , each coefficient is a polynomial function of $t$ . For this is obvious for $x^{[t]'}$ . As far as $x^{[t]}$ is concerned, it suffices to prove that, for all $u \in \mathrm{U}(G)$ , the image of $u$ under $x \mapsto x^{[t]}$ is a polynomial function of $t$ . Now, for $u \in \mathrm{U}^m(G)$ , this image is $t^m u$ (§ 4, no. 3, Proposition 7 (iv)).

As $x^{[t]} = x^{[t]'}$ for $t$ an integer $\geq 0$ , it then follows that $x^{[t]} = x^{[t]'}$ for all $t \in \mathbf{K}$ .

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

Cited by

Cites

§ 4, no. 3, Proposition 7