Exercise 24, § 1
Lie Groups and Lie Algebras, VI
We retain the notation and assumptions of the preceding exercise.
a) Let , and let be the orbit of under . Show that the following conditions are equivalent:
(i) .
(ii) or for all .
(If (ii) is not satisfied, construct an element such that , and hence .)
b) Let be a non-empty saturated subset of . Show that contains an element satisfying conditions (i) and (ii) above. (Take to be an element of of minimal length.)
c) Assume that is irreducible. Let be a chamber of , let be the corresponding basis, and let
be the highest root of . Let be the set of such that . Let be a dominant weight. Show that conditions (i) and (ii) of a) are equivalent to each of the following conditions:
² This exercise, hitherto unpublished, was communicated to us by R. Steinberg.
(iii) .
(iv) There exists such that is equal to the corresponding fundamental weight.
A weight satisfying these conditions is called minuscule. Show that every non-empty saturated subset of contains 0 or a minuscule weight.
Solution
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