Exercise 3, § 6

Lie Groups and Lie Algebras, V

Bourbaki marks this one as harder than the rest.

Let K be a commutative ring, E a free K-module with basis $(e_1, \ldots, e_l)$ , and $f_1, \ldots, f_l$ elements of the dual of E. Put $a_{ij} = f_i(e_j)$ . If $1 \leq i \leq l$ , let $s_i$ be the pseudo-reflection $s_{e_i, f_i}$ (§ 2, Exerc. 1). Then

s_i(e_j) = e_j - a_{ij} e_i.

Put $c = s_1 \ldots s_l$ , and $z_i = c(e_i)$ .

a) For $1 \leq i, k \leq l$ , put

y_i^k = s_1 \ldots s_k(e_i) \quad \text{and} \quad y_i = s_1 \ldots s_{i-1}(e_i) = y_i^{i-1}.

Thus $y_i^0 = e_i$ and $y_i^l = z_i$ .

Show that

y_i^{k-1} - y_i^k = a_{ki} y_k.

Deduce the formulas

c_i = y_i + \sum_{k < i} a_{ki} y_k

z_i = y_i - \sum_{k \geq i} a_{ki} y_k.

b) Let C be the matrix of $c$ with respect to the basis $(e_i)$ . Let $U = (u_{ij})$ and $V = (v_{ij})$ be the matrices defined by

u_{ij} = \begin{cases} a_{ij} & \text{if } i < j \\ 0 & \text{otherwise,} \end{cases} \quad v_{ij} = \begin{cases} 0 & \text{if } i < j \\ a_{ij} & \text{otherwise.} \end{cases}

The matrix $I + U$ is invertible with determinant 1. Show that

C = (I - V)(I + U)^{-1}.

Deduce that

\det(\lambda I - C) = \det((\lambda - 1)I + V + \lambda U),

in other words

\det(\lambda I - C) = \left| \begin{array}{cccccc} (\lambda - 1) + a_{11} & \lambda a_{12} & \lambda a_{13} & \cdots & \lambda a_{1l} \\ a_{21} & \lambda(\lambda - 1) + a_{22} & \lambda a_{23} & \cdots & \lambda a_{2l} \\ a_{31} & a_{32} & (\lambda - 1) + a_{33} & \cdots & \lambda a_{3l} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ a_{l1} & a_{l2} & a_{l3} & \cdots & (\lambda - 1) + a_{ll} \end{array} \right|.

c) Let $\Gamma = (I, S)$ be the graph whose set of vertices is $I = [1, l]$ and whose set $S$ of arrows is the set of subsets $\{i, j\}$ of $I$ with 2 elements such that $a_{ij} \neq 0$ or $a_{ji} \neq 0$ . If $\alpha = \{i, j\}$ belongs to $S$ , denote by $\sigma_\alpha$ the transposition of $i$ and $j$ (considered as an element of the symmetric group $S_I$ ), and put $a_\alpha = -a_{ij}a_{ji}$ .

Let $\mathscr{B}$ be the set of subsets of $S$ consisting of the arrows whose vertices are disjoint. If $X \in \mathscr{B}$ , denote by $C(X)$ the set of $i \in I$ that are not vertices of any arrow of $X$ , and put $\sigma_X = \prod_{\alpha \in X} \sigma_\alpha, a_X = \prod_{\alpha \in X} a_\alpha$ .

Let $\sigma \in S_I$ , and let $d_\sigma$ be the term corresponding to $\sigma$ in the expansion of the determinant of $(\lambda - 1)I + V + \lambda U$ . Show that, if $\sigma$ is of the form $\sigma_X$ , with $X \in \mathscr{B}$ , then

d_\sigma = a_X \lambda^{\operatorname{Card}(X)} \prod_{i \in C(X)} (\lambda - 1 + a_{ii}).

Assume now that $\Gamma$ is a forest (Chap. IV, Appendix, no. 3). Show that, if $\sigma \in S_l$ is not of the form $\sigma_X$ with $X \in \mathscr{B}$ , then $d_\sigma = 0$ . Deduce the formula

\det(\lambda - c) = \sum_{X \in \mathscr{B}} a_X \lambda^{\operatorname{Card}(X)} \prod_{i \in C(X)} (\lambda - 1 + a_{ii}).

d) Consider the polynomial

P(X) = \left| \begin{array}{cccc} X & a_{12} & \ldots & a_{1l} \\ a_{21} & X & \ldots & a_{2l} \\ \vdots & \vdots & \ddots & \vdots \\ a_{l1} & a_{l2} & \ldots & X \end{array} \right|.

To the assumptions in c), we add that $a_{ii} = 1$ for all $i$ . Show that in that case

\det(\lambda^2 - c) = \lambda^l P(\lambda + \lambda^{-1}).

EXERCISES

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