A SHORT CLASSIFICATION OF FINITE SUBGROUPS OF SL(3,C) 0 ...

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2 YASUSHI GOMI, IKU NAKAMURA, KEN-ICHI SHINODA1. Types (A)-(D)Lemma 1.1. If V is a reducible G-module, then G is type (A) or (B).Proof. Clear from the definition.Lemma 1.2. If G is imprimitive, then G is type (C) or (D).Proof. In the definition 0.2, it is only possible that n = 3 and dim V k = 1. Then itis clear that G is type (C) or (D).Lemma 1.3. Let N be a normal subgroup of G of type (A). Then V is a reducibleG-module.Proof. Let V = ⊕ n i=1 V i be the decomposition of V into N-modules with differenteigenvalues. Then g(V i )=V i for any g ∈ G because the eigenvalues of N on V i aredifferent.Lemma 1.4. Let N be a normal subgroup of G of type (B). Then V is a reducibleG-module.Proof. In this case N has a unique eigenvector v in V , which is again an eigenvectorof G because N is normal. Hence V is G-reducible.2. Types (E)-(G)Theorem 2.1. If G is a primitive subgroup of SL(V ) with an imprimitive normalsubgroup N of G, then G is of type (E), (F) or (G).For the proof of the theorem we start with the following observation.Since N is imprimitive, there is an N-invariant decomposition V = V 1 ⊕ V 2 ⊕ V 3 ,that is, σ(V i ) ⊂ V j for σ ∈ N, anyi ans some j. Then we have a homomorphismφ V·: N → S 3 defined by N(V i )=V φ(i) . Since N is imprimitive, the image φ V·(N)is at least of order 3. Hence there exists an element T ∈ N of order 3 such thatT (V i mod 3 ) ⊂ V i+1 mod 3 for any i mod 3. (We write this simply as T (V i ) ⊂ V i+1for any i.) Choosing a basis v of V i , we define v i = T i−1 (v).Suppose that V = V 1 ⊕V 2 ⊕V 3 is the unique N-invariant decomposition of V . SinceN is normal, the decomposition V = V 1 ⊕ V 2 ⊕ V 3 is G-invariant. It follows from theuniqueness of the decomposition that V i is G-invariant for any i. This contradictsthat V is G-irreducible. Hence V has at least another N-invariant decompositionV = U 1 ⊕ U 2 ⊕ U 3 .Thus the following is a key to the proof of the theorem.Lemma 2.2. If N is imprimitive and if there are at least two N-invariant decompsitionsof V into 3 one-dimensional subspaces, then there are exactly four N-invariantdecompositions of V . The group N is idenitified with the Heisenberg group G(3) ofdegree 3, or an extension of G(3) by the involution R defined below, while the fourN-invariant decompositions of V are identified with the four singular Hesse cubics.Proof. There are two casesCase 1. φ U·(N) is of order 3 for any N-invariant V = U 1 ⊕ U 2 ⊕ U 3 .Case 2. φ U·(N) =S 3 for some N-invariant V = U 1 ⊕ U 2 ⊕ U 3 .
FINITE SUBGROUS 3Since φ U·(N) for V = U 1 ⊕ U 2 ⊕ U 3 is of order 3, there exists an element S ∈ Nof order 3 such that S(U i ) ⊂ U i+1 for any i mod 3. There are two cases whereT (U i )=U i for any i, orT (U i ) ⊂ U i+1 for any i mod 3.Case (i). T (U i )=U i for any i, S(V j )=V j for any jCase (ii). T (U i )=U i for any i, S(V j )=V j+1 for any jCase (iii). T (U i ) ⊂ U i+1 for any i mod 3, S(V j )=V j for any j,Case (iv). T (U i ) ⊂ U i+1 for any i mod 3, S(V j )=V j+1 for any j.Here we note Cases (ii) and (iii) are easily reduced to Case (i).We consider Case (1i). Then U i is one of the eigenspaces of T , hence V =U 1 ⊕ U 2 ⊕ U 3 is unique. Then we may assume S(v i )=ω i−1 v i because N ⊂ SL(3).We note [S, T ]=ω. Hence N contains a subgroup generated by S and T , which isthe Heisenberg group G(3) of degree 3. In this case the family of Hesse cubicsv 3 1 + v 3 2 + v 3 3 − 3µv 1 v 2 v 3 =0is invariant under G(3), whence all the singular Hesse cubics for µ = ∞, orµ 3 =1are invariant under G(3). We note [S, T ]=ω.Here is a sublemma:Sublemma Let σ ∈ N. Assume σ(V i ) = V i and σ(U i ) = U i for any i. Thenσ = ω k · id. for some k.Proof. Let σ(v i )=a i v i for any i. IfT (U i )=U i for any i, then U i is an eigenspace ofT , which is spanned by v 1 +ω k v 2 +ω 2k v 3 . It follows from σ(U i )=U i that a 1 = a 2 =a 3 .IfT (U i )=U i+1 for any i and a 3 ≠ a 1 ,a 2 , then we have V 3 = ker(σ − a 3 ), whichis one of the eigenspaces of σ. Hence we may assume V 3 = U 3 by renumbering U i ifnecessary. Since T (U i )=U i+1 , one sees U i = T i (U 3 )=T i (V 3 )=V i , a contradiction.Thus a 1 = a 2 = a 3 = ω k .Now we prove N = G(3). Let σ ∈ N. By the assumptions of Case (1i) we mayassume σ(V i )=V i , σ(U i )=U i for any i. Hence by Sublemma, σ = ω k ∈ G(3).Hence N = G(3). As is well known V is G(3)-irreducible. It follows that thereare exactly four G(3)-invariant decompositions U· of V , for any of which the imageφ U·(N) is order 3.Next we consider Case (1iv). In this case by Sublemma we see that S = ω k · Tfor some k. Hence if always Case (iv) occurs, then we derive a contradiction. Forthis, let σ ∈ N. By the assumption of Case 1 we may assume σ(V i )=V i andσ(U i )=U i for any i. By Sublemma σ = ω k and N = 〈T,ω〉. Hence N is abelian,which contradicts that V is N-irreducible.Hence there is an N-invariant decomposition V = U 1 ′ ⊕ U 2 ′ ⊕ U 3 ′ and an elementS ′ ∈ N such that one of the cases (i)-(iii) occurs. In this case we see N = G(3) asabove and there are exactly four G(3)-invariant decompositions U· of V .Next we consider Case (2i). As in Case (1i) we see [S, T ]=ω, G(3) = 〈S, T 〉. Bythe assumption of Case 2, we may assume there is R ∈ N such that R(V 1 )=V 2 ,R(V 2 )=V 1 and R(V 3 )=V 3 . Let τ = R 2 . Since φ U·(N) isS 3 , S k τ keeps any V i andsome U j invariant for some k, where we may assume j = 1. Now we choose S 2k R forR. ThusR(v 1 )=av 2 , R(v 2 )=bv 1 , R(v 3 )=cv 3 and R(v 1 +v 2 +v 3 )=d(v 1 +v 2 +v 3 )for some a, b, c, d. It follows a = b = c = d = −1. We note φ U·(R) =(2, 3) ∈ S 3 .
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A SHORT CLASSIFICATION OF FINITE SUBGROUPS OF SL(3,C) 0 ...

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