Bourbaki

Exercise 9, § 4

Lie Groups and Lie Algebras, I, 92

(a) Verify that the following multiplication tables define two nilpotent Lie algebras g3,g4g_3, g_4 of dimensions 3 and 4:

g3:[x1,x2]=x3,[x1,x3]=[x2,x3]=0 g_3: [x_1, x_2] = x_3, \quad [x_1, x_3] = [x_2, x_3] = 0
g4:[x1,x2]=x3,[x1,x3]=x4,[x1,x4]=[x2,x3]=[x2,x4]=[x3,x4]=0. g_4: [x_1, x_2] = x_3, \quad [x_1, x_3] = x_4, \quad [x_1, x_4] = [x_2, x_3] = [x_2, x_4] = [x_3, x_4] = 0.

(b) Show that g3g_3 is isomorphic to n(3,K)\mathfrak{n}(3, K) and also to sl(2,K)\mathfrak{sl}(2, K) if KK is of characteristic 2.

(c) Let g1g_1 be the unique Lie algebra of dimension 1. Show that the nilpotent Lie algebras of dimension ≤4\leq 4 are given by the following table:

dimension 1: g1;dimension 2: (g1)2;dimension 3: (g1)3,g3;dimension 4: (g1)4,g3×g1,g4. \begin{array}{ll} \text{dimension 1: } & g_1; \\ \text{dimension 2: } & (g_1)^2; \\ \text{dimension 3: } & (g_1)^3, g_3; \\ \text{dimension 4: } & (g_1)^4, g_3 \times g_1, g_4. \end{array}
(Use Exercise 7 (c). If gg is nilpotent and 3-dimensional and dim⁡Dg=1\dim \mathscr{D} g = 1, observe that Dg\mathscr{D} g is contained in the centre of gg (Exercise 3 (a)), whence g=g3g = g_3. If dim⁡g=4\dim g = 4, dim⁡Dg=1\dim \mathscr{D}g = 1, observe that the bracket on gg defines an alternating bilinear form on g/Dgg / \mathscr{D}g, which is necessarily degenerate and hence that there exists a subalgebra hh of gg contained in the centre of gg with dim⁡h=1\dim h = 1, h∩Dg={0}h \cap \mathscr{D}g = \{0\}, whence g=h×g3g = h \times g_3. If dim⁡g=4\dim g = 4, dim⁡Dg=2\dim \mathscr{D}g = 2, Dg\mathscr{D}g is commutative; applying Theorem 1 to the restrictions to Dg\mathscr{D}g of the adgx\mathrm{ad}_g x (x∈gx \in g), show that there exists a commutative ideal hh of gg with h⊃Dgh \supset \mathscr{D}g, dim⁡h=3\dim h = 3; let x∈gx \in g, x∉hx \notin h; choose a basis of hh such that the restriction of adgx\mathrm{ad}_g x to hh has a Jordan matrix with respect to this basis; then g=g4g = g_4.

Solution

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Transcribed from the printed volume. Not translated.