Bourbaki

Exercise 7, § 3

Lie Groups and Lie Algebras, V

Let q(ξ1,…,ξn)=∑i,jaijξiξjq(\xi_1, \ldots, \xi_n) = \sum_{i,j} a_{ij} \xi_i \xi_j (aij=ajia_{ij} = a_{ji}) be a degenerate positive quadratic form on Rn\mathbf{R}^n, such that aij⩽0a_{ij} \leqslant 0 for i≠ji \neq j. Assume that there is no partition I∪JI \cup J of {1,2,…,n}\{1, 2, \ldots, n\} such that aij=0a_{ij} = 0 for i∈I,j∈Ji \in I, j \in J.

a) Show that, if we put ξi=0\xi_i = 0, we obtain a non-degenerate positive quadratic form with respect to ξ1,…,ξi−1,ξi+1,…,ξn\xi_1, \ldots, \xi_{i-1}, \xi_{i+1}, \ldots, \xi_n.

b) Show that aii>0a_{ii} > 0 for all ii. (Let (ζ1,…,ζn)(\zeta_1, \ldots, \zeta_n) be an element of the kernel of qq, with ζ1>0,…,ζn>0\zeta_1 > 0, \ldots, \zeta_n > 0. Use the equality a1iζ1+⋯+aniζn=0a_{1i} \zeta_1 + \cdots + a_{ni} \zeta_n = 0.)

c) Show that if we replace one of the aija_{ij} by some aij′<aija'_{ij} < a_{ij}, the new form is non-positive. (Use the equality ∑i,jaijζiζj=0\sum_{i,j} a_{ij} \zeta_i \zeta_j = 0.)

Solution

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