Bourbaki

Exercise 18, § 4

Lie Groups and Lie Algebras, I, 95

Let g be a 7-dimensional vector space over K with a basis (ei)1≤i≤7(e_i)_{1 \leq i \leq 7}. An alternating bracket is defined on g by the formulae:

(1)

[ei,ej]=αijei+j(1≤i<j≤7,i+j≤7) [e_i, e_j] = \alpha_{ij} e_{i+j} \quad (1 \leq i < j \leq 7, i + j \leq 7)
and all the other brackets [ei,ej][e_i, e_j] are zero for i<ji < j.

(a) For the Jacobi identity to hold, it is necessary and sufficient that:

(2)

−α23α15+α13α24=0 -\alpha_{23}\alpha_{15} + \alpha_{13}\alpha_{24} = 0
(3)
α12α34−α24α16+α14α25=0. \alpha_{12}\alpha_{34} - \alpha_{24}\alpha_{16} + \alpha_{14}\alpha_{25} = 0.

(b) Suppose henceforth that all the αij\alpha_{ij} are ≠0\neq 0. Show that the ideals C2g,C3g,C4g,C5g,C6g\mathscr{C}^2g, \mathscr{C}^3g, \mathscr{C}^4g, \mathscr{C}^5g, \mathscr{C}^6g admit the following bases: (e3,e4,e5,e6,e7)(e_3, e_4, e_5, e_6, e_7), (e4,e5,e6,e7)(e_4, e_5, e_6, e_7), (e5,e6,e7)(e_5, e_6, e_7), (e6,e7)(e_6, e_7), (e7)(e_7). Show that the centralizer hh of C6g\mathscr{C}^6g admits the basis (e2,e3,e4,e5,e6,e7)(e_2, e_3, e_4, e_5, e_6, e_7).

(c) Let (ei′)1≤i≤7(e'_i)_{1 \leq i \leq 7} be another basis of g such that C6g\mathscr{C}^6g admits the basis (e7′)(e'_7), C5g\mathscr{C}^5g admits the basis (e6′,e7′)(e'_6, e'_7), …\ldots, hh admits the basis (e2′,e3′,e4′,e5′,e6′,e7′)(e'_2, e'_3, e'_4, e'_5, e'_6, e'_7). Show that

[ei′,ej′]=αij′ei+j′+βei+j+1′+γei+j+2′+⋯+λe7′, [e'_i, e'_j] = \alpha'_{ij} e'_{i+j} + \beta e'_{i+j+1} + \gamma e'_{i+j+2} + \cdots + \lambda e'_7,
where
α14′α25′α16′−1α24′−1=α14α25α16−1α24−1. \alpha'_{14}\alpha'_{25}{\alpha'_{16}}^{-1}{\alpha'_{24}}^{-1} = \alpha_{14}\alpha_{25}\alpha_{16}^{-1}\alpha_{24}^{-1}.

(d) Deduce that there exists, if K is infinite, an infinity of non-isomorphic nilpotent Lie algebras of dimension 7 over K and that there exist complex nilpotent Lie algebras which cannot be derived from real nilpotent Lie algebras by extending the base field from R\mathbf{R} to C\mathbf{C}.

Solution

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Transcribed from the printed volume. Not translated.