66 H.2. Let 2 g be <strong>subregular</strong> <strong>nilpotent</strong>. Let 2 C0 be p{regular, i.e., with 0 < h + ; _i
H.3. So far we havestudied only the categories C with 2 C 0 0. In order to get all possible C we should also look at the 2 C0 with h + ; _ 0 i = p. Weknow 0 0 from H.1 that there exist <strong>in</strong> many cases 2 C0 with C = C 0, but there are also cases where this does not hold. When we try to extend the theory for weights <strong>in</strong> C 0 0 to the rema<strong>in</strong><strong>in</strong>g weights <strong>in</strong> C0, then we encounter two major problems: The pro<strong>of</strong> <strong>in</strong> D.3 that certa<strong>in</strong> modules are simple will not work for all 2 C0, and there is not an easy de nition <strong>of</strong> a <strong>in</strong>variant as <strong>in</strong> D.5. The rst problem occurs only for relatively few and we are go<strong>in</strong>g to ignore them here. De nition: We say that a weight 2 C0 is nice if for all simple roots , all <strong>subregular</strong> 2 g with (p ) = 0, and all w 2 W with w ,1 simple and hw( + ); _ i > 0 the module Z (w ; ) is simple. Note that Lemma D.3.a says that each 2 C 0 0 is nice. I hope that all 2 C0 are nice <strong>in</strong> good characteristic; what I can prove is this: Lemma: Let 2 C0 with h + ; _ 0 i = p. a) In case R is <strong>of</strong> type E8, F4, orG2, assume that p>h+1. If there exists a simple root with $ m<strong>in</strong>uscule such that y + p$ 2 C 0 0 , then is nice. b) If there exists for each simple root with h + ; _ i > 0 a weight <strong>in</strong> the same facet as with h + ; _ i =1,then is nice. c) Suppose that R is not <strong>of</strong> type G2. Ifh + ; _ i > 0 for all simple roots , then is nice. Pro<strong>of</strong> : Consider , , and w as <strong>in</strong> the de nition above. a) Set 0 = y + p$ .Wehave Z (w ; ) ' Z (wy ,1 0 ; ): (1) Recall from H.1 that y permutes [f, 0g. Wehave therefore (wy ,1 ) ,1 = y w ,1 2 [f, 0g: So the right hand side <strong>in</strong> (1) is simple by Lemma D.4.a (<strong>in</strong> case y w ,1 = , 0) or by Lemma D.3.a (<strong>in</strong> case y w ,1 simple) provided we know <strong>in</strong> the second case that hwy ,1 ( 0 + ); _ i > 0. However, one checks easily that hwy ,1 ( 0 + ); _ i = hw( + ); _ i + ph$ ;y w ,1 _ i > 0 us<strong>in</strong>g wy ,1 ( 0 + )=w( + )+pwy ,1 $ . b) Under these assumptions Lemma D.2 holds with F (J) replaced by the facet <strong>of</strong> . Therefore the pro<strong>of</strong> <strong>of</strong> Lemma D.3.a works <strong>in</strong> this case. c) Note that p = h + ; _ 0 i and 2 C0 imply p h ; _ 0 i = h , 1. If R is <strong>of</strong> type A1, then = 0 and w =1. Then Z ( ; )= Z ( )isa Ste<strong>in</strong>berg module, hence irreducible. 67