Exercise 19, § 1

Lie Groups and Lie Algebras, I, 77

(a) Let $L$ be an associative algebra over $K$ . Given $k$ elements $x_1, \ldots, x_k$ of $L$ , we write:

f_k(x_1, \ldots, x_k) = \sum_{\sigma \in S_k} x_{\sigma(1)} \cdots x_{\sigma(k)}.

In the algebra $L \otimes_K K[T_1, \ldots, T_k]$ , where the $T_i$ are indeterminates, $f_k(x_1, \ldots, x_k)$ is the coefficient of $T_1 \ldots T_k$ in $(x_1 T_1 + \cdots + x_k T_k)^k$ ; it is also the coefficient of $T_1 \ldots T_k$ in:

\sum_{j=0}^{k-1} (x_1 T_1 + \cdots + x_{k-1} T_{k-1})^j x_k T_k (x_1 T_1 + \cdots + x_{k-1} T_{k-1})^{k-j-1}.

Given two elements $x, y$ in $L$ , we define $g_{i, k-i}(x, y)$ as the coefficient of $T_1^i T_2^{k-i}$ in $(x T_1 + y T_2)^k$ . Show that

i! (k-i)! g_{i, k-i}(x, y) = f_k(t_1, \ldots, t_k)

where $t_h = x$ for $h \leq i$ and $t_h = y$ for $h \geq i+1$ .

(b) If $L$ is considered as a Lie algebra over $K$ , then in $\mathscr{L}_K(L)$ :

(\mathrm{ad}\ x)^n = \sum_{i=0}^n (-1)^{n-i} \binom{n}{i} L_x^i R_x^{n-i},

where $L_x$ (resp. $R_x$ ) is the multiplication $y \mapsto xy$ (resp. $y \mapsto yx$ ).

Deduce that, if $K$ is of characteristic $p$ ( $p$ prime), then:

(1)

(\mathrm{ad}\ x)^p . y = (\mathrm{ad}\ x^p) . y

(2)

(\mathrm{ad}\ x)^{p-1} . y = \sum_{i=0}^{p-1} x^i y x^{p-1-i}.

Deduce from (2) that:

f_{p-1}(\operatorname{ad} x_1, \ldots, \operatorname{ad} x_{p-1}) \cdot y = f_p(x_1, \ldots, x_{p-1}, y)

and:

g_{i, p-1-i}(\operatorname{ad} x, \operatorname{ad} y) \cdot y = (p - i) g_{i, p-i}(x, y).

Conclude that for any two elements $x, y$ of $L$ :

(3)

(x + y)^p = x^p + y^p + \Lambda_p(x, y)

where:

\Lambda_p(x, y) = \sum_{i=1}^{p-1} (p - i)^{-1} g_{i, p-1-i}(\operatorname{ad} x, \operatorname{ad} y) \cdot y

belongs to the Lie subalgebra of $L$ generated by $x$ and $y$ ("Jacobson's formulae").

Solution

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