Exercise 12, § 11
Tag 01K5, permanent. Cite it as [Bourbaki, Tag 01K5].
Lie Groups and Lie Algebras, VIII, 263
Bourbaki marks this one as harder than the rest.
Assume that is semi-simple. Let and let (resp. ) be the semi-simple (resp. nilpotent) component of . Let (resp. ) be the commutant of (resp. ) in .
a) Show that is a nilpotent element of the semi-simple algebra , and that the commutant of in is equal to . Deduce that dim dim if , i.e. if is not semi-simple.
b) Show that dim rk() if and only if is a principal nilpotent element of .
c) Put G = Aut. Show that, for all , there exists such that . (Show that, if , there exists an -triplet in of which the first component is , and deduce a homomorphism ; take to be the image under of a suitable diagonal element of ) Deduce that belongs to the closure of in the Zariski topology.
d) Show that, if is not semi-simple, does not belong to the closure of in the Zariski topology (use the inequality dim dim , cf. ).
For more details on Exercises 6 to 11, see: B. KOSTANT, The principal threedimensional subgroup and the Betti numbers of a complex simple Lie group,
Amer. J. Math., Vol. LXXXI (1959), pp. 973-1032. e) Assume that is algebraically closed. Prove the equivalence of the following properties:
(i) is semi-simple;
(ii) is closed in in the Zariski topology.
(The implication (ii) (i) follows from ). If (i) is satisfied, and if is in the closure of , Exerc. 15 of §8 shows that the semi-simple component of belongs to , so belongs to the closure of ; conclude by applying d) to and
f) Assume that is algebraically closed. Let be the set of elements such that for every invariant polynomial function on ; we have if and only if the semi-simple component of is G-conjugate to §8, Exerc. 17). Show that is the union of a finite number of orbits of G, and that this number is , where is the rank of . Only one of these orbits is closed: that of ; only one is open in that consisting of the elements such that dim rk().
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