Bourbaki

Exercise 25, § 6

Lie Groups and Lie Algebras, I, 108

Bourbaki marks this one as harder than the rest.

Let K be a field of arbitrary characteristic pp. Let sp(2n,K)sp(2n, K) denote the Lie algebra of the formal symplectic group in 2n2n variables over K (§ 1, Exercise 26 (c)). Show that this Lie algebra, which is identified with a sub-algebra of gl(2n,K)gl(2n, K), has basis consisting of the elements

Hi=Eii−Ei+n,i+n(1≤i≤n), H_i = E_{ii} - E_{i+n, i+n} \quad (1 \leq i \leq n),
Fij=Eij−Ej+n,i+n(1≤i,j≤n,i≠j), F_{ij} = E_{ij} - E_{j+n, i+n} \quad (1 \leq i, j \leq n, i \neq j),
Gij′=Ei,j+n+Ej,i+n,Gij′′=Ei+n,j+Ej+n,i(1≤i<j≤n), G'_{ij} = E_{i, j+n} + E_{j, i+n}, \qquad G''_{ij} = E_{i+n, j} + E_{j+n, i} \quad (1 \leq i < j \leq n),
Ei,i+n and Ei+n,i(1≤i≤n). E_{i, i+n} \text{ and } E_{i+n, i} \quad (1 \leq i \leq n).

(a) Show that, if p≠2p \neq 2 and n≥1n \geq 1, the algebra sp(2n,K)sp(2n, K) is simple (for n=1,sp(2,K)=sl(2,K)n = 1, sp(2, K) = sl(2, K)). (Same method as in Exercise 24.)

(b) If p=2p = 2 and n≥3n \geq 3, show that the Hi,Fij,Gij′H_i, F_{ij}, G'_{ij} and Gij′′G''_{ij} form an ideal a\mathfrak{a} of dimension n(2n−1)n(2n-1); the elements of a\mathfrak{a} of the form

∑i=1nλiHi+∑i≠jαijFij+∑i<jβijGij′+∑i<jγijGij′′ \sum_{i=1}^{n} \lambda_i H_i + \sum_{i \neq j} \alpha_{ij} F_{ij} + \sum_{i < j} \beta_{ij} G'_{ij} + \sum_{i < j} \gamma_{ij} G''_{ij}

such that ∑i=1nλi=0\sum_{i=1}^{n} \lambda_i = 0 form an ideal b⊂a\mathfrak{b} \subset \mathfrak{a} of dimension n(2n−1)−1n(2n-1) - 1 and the multiples of ∑i=1nHi\sum_{i=1}^{n} H_i an ideal c\mathfrak{c}, the centre of sp(2n,K)sp(2n, K); b,c\mathfrak{b}, \mathfrak{c} are the only ideals of a,b=D(sp(2n,K))a, b = \mathscr{D}(\mathfrak{sp}(2n, K)) and b/cb/c is simple. (Same method.) What are the ideals of sp(4,K)\mathfrak{sp}(4, K) when KK is of characteristic 2?

Solution

No solution has been written yet.

Transcribed from the printed volume. Not translated.