Group actions and identities for the simple Lie algebra sl 2 (C)

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Group actions and identities for the simple Lie algebra sl 2 (C IMECC - Unicamp September 02, 2011

Beginning: A. Giambruno, A. Regev, Wreath products and PI algebras, J. Pure Appl. Algebra 35 (1985 133-145.

To have a description of G-identities of sl 2 (C, where G is a finite group that acts faithfully on sl 2 (C.

Berele in 2004 in the paper Polynomial identities for 2 2 matrices with finite group actions, J. Algebra 274 (1 (2004 202-214, described bases of the G-identities for the matrix algebra of order two M 2 (C, where G is a finite group acting faithfully on M 2 (C. It is well known that if G is a finite group that acts faithfully on M 2 (C, then it must be one of the groups Z n, D n,. His proofs rely on the concrete basis of the 2-graded identities for M 2 (C found by Di Vincenzo in 1993 and on computations in the group algebras of the corresponding groups.

Our problem is analogous to Berele s problem for the simple three-dimensional Lie algebra sl 2 (C. Fortunately, in this case it is well known that the finite groups that act faithfully on sl 2 (C are the same as those for M 2 (C, and they act on sl 2 (C in the same way that they act on M 2 (C, by conjugation of matrices. We use the concrete form of the 2-graded identities for sl 2 (C, and moreover, some methods and techniques developed by Berele. We exhibit bases of the corresponding G-identities for sl 2 (C.

How we approach the problem:

How we approach the problem: The same way that Berele did. is the easiest case, because of that we discuss it separately. We treat each group family in turn, so we divide the problem in three cases. In the first two cases, where the groups are Z n and D n, we consider computations in the group algebras of the corresponding groups. In the last case, where the groups are we consider sl 2 (C as a G-module.

Notation:

Notation: X = {x 1, x 2, x 3,...} will denote an infinite countable set. G will denote a group. By a G-Lie Algebra we understand a Lie Algebra together with a group G that acts on it. LC X ; G will denote the C-free Lie algebra of symbols g(x, where g G and x X, that we will refer to as G-free Lie algebra.

Definition Given f LC X ; G and a G- Lie algebra A, f is called a G-polynomial identity of A, if for every G-homomorphism ϕ : LC X ; G A, ϕ(f = 0. Remark The set of all G-polynomial identities of A is a G-ideal of LC X ; G, that is invariant under all G-endomorphisms. This set will be denoted by T G (A. The relatively free algebra of the G-Lie algebra A is the algebra LC X ; G. T G (A

G = : Let G = {e, g} = and its action on sl 2 (C be given by: ( a b e c a ( a b = c a and ( a b g c a ( a b = c a.

G = : Let G = {e, g} = and its action on sl 2 (C be given by: ( a b e c a ( a b = c a and ( a b g c a ( a b = c a. G-identity: [e(x + g(x, e(y + g(y] = 0.

The algebra sl 2 (K, where K is an infinite field of characteristic different from 2, admits a natural -grading: sl 2 (K = K(e 11 e 22 (Ke 12 + Ke 21.

The algebra sl 2 (K, where K is an infinite field of characteristic different from 2, admits a natural -grading: sl 2 (K = K(e 11 e 22 (Ke 12 + Ke 21. Theorem (Koshlukov The graded identities for sl 2 (K with the above grading follow from the identity [x 1, x 2 ] = 0 where x 1 and x 2 are even (neutral variables.

The algebra sl 2 (K, where K is an infinite field of characteristic different from 2, admits a natural -grading: sl 2 (K = K(e 11 e 22 (Ke 12 + Ke 21. Theorem (Koshlukov The graded identities for sl 2 (K with the above grading follow from the identity [x 1, x 2 ] = 0 where x 1 and x 2 are even (neutral variables. For all x sl 2 (C define π 0 (x = e(x + g(x. Then it follows from Koshlukov s theorem that all the G-polynomial identities are consequences of:

The algebra sl 2 (K, where K is an infinite field of characteristic different from 2, admits a natural -grading: sl 2 (K = K(e 11 e 22 (Ke 12 + Ke 21. Theorem (Koshlukov The graded identities for sl 2 (K with the above grading follow from the identity [x 1, x 2 ] = 0 where x 1 and x 2 are even (neutral variables. For all x sl 2 (C define π 0 (x = e(x + g(x. Then it follows from Koshlukov s theorem that all the G-polynomial identities are consequences of: [π 0 (x, π 0 (y] = 0.

G = Z/nZ = Z n (n 3 : Let G = Z n be generated by g G.

G = Z/nZ = Z n (n 3 : Let G = Z n be generated by g G. g acts on sl 2 (C by conjugation by a matrix A of order n. ( 1 0 We can assume that A is a diagonal matrix of the form 0 ω where ω is a n-th primitive root of unity. ( ( a b a ω Then g = 1 b c a ωc a,

G = Z/nZ = Z n (n 3 : Let G = Z n be generated by g G. g acts on sl 2 (C by conjugation by a matrix A of order n. ( 1 0 We can assume that A is a diagonal matrix of the form 0 ω where ω is a n-th primitive root of unity. ( ( a b a ω Then g = 1 b c a ωc a CG = n 1 i=0 Ce i, where each e i, i {0,..., n 1}, is a primitive idempotent and e i = (1/n n 1 j=0 ω ji g j. Since G is abelian we can use this decomposition to obtain an equivalence between the G-actions and the G-gradings. sl 2 (C = n 1 i=0 e isl 2 (C. If i 0, 1, n 1, then e i sl 2 (C = 0.,

Denote e n 1 by e 1, then instead of considering the Z n grading on sl 2 (C, we can consider a Z-grading concentrated in { 1, 0, 1}.

Denote e n 1 by e 1, then instead of considering the Z n grading on sl 2 (C, we can consider a Z-grading concentrated in { 1, 0, 1}. Thus we have that ( ( ( a b a 0 a b e 0 =, e c a 0 a 1 c a ( a e 1 c b = a ( 0 b 0 0 =. ( 0 0 c 0, and

Denote e n 1 by e 1, then instead of considering the Z n grading on sl 2 (C, we can consider a Z-grading concentrated in { 1, 0, 1}. Thus we have that ( ( ( a b a 0 a b e 0 =, e c a 0 a 1 c a ( a e 1 c b = a ( 0 b 0 0 =. G-identities: [e i (x 1, e i (x 2 ] = 0, for all i { 1, 0, 1}. ( 0 0 c 0, and

Theorem Let G = Z n, where n 3. Then all the G-identities of sl 2 (C are consequences of the following: 1 e 0 (x + e 1 (x + e 1 (x = x; 2 For α { 1, 0, 1}, [e α (x 1, e α (x 2 ] = 0.

Theorem Let G = Z n, where n 3. Then all the G-identities of sl 2 (C are consequences of the following: 1 e 0 (x + e 1 (x + e 1 (x = x; 2 For α { 1, 0, 1}, [e α (x 1, e α (x 2 ] = 0. To prove this theorem we construct a basis for the relatively free LC X ; G algebra T G (sl 2 (C.

Theorem The following set is a basis for the relatively free algebra LC X ; G T G (sl 2 (C : 1 [e 1 (x i1, e 0 (x m, e 1 (x j1, e 1 (x i2, e 1 (x j2,..., e 1 (x iα, e 1 (x jα ]; 2 [e 1 (x i1, e 0 (x m, e 1 (x j1, e 1 (x i2, e 1 (x j2,..., e 1 (x iα ]; 3 [e 1(x i1, e 0(x m, e 1(x j1, e 1(x i2, e 1(x j2,..., e 1(x iα, e 1(x jα, e 1(x jα+1 ]; 4 [e 1(x j1, e 0(x m ]; 5 e 0(x i1 ; where m 0, α 1 and [e k (x i1, e 0(x m ], for all k { 1, 1} means [e k (x i1, e 0(x (1,..., e 0(x (m ].

G = D n : D n = g, h : g n = h 2 = 1, hgh = g 1.

G = D n : D n = g, h : g n = h 2 = 1, hgh = g 1. It is well known that: If n = 2m, for some m N, then CD n = C C C C M 2 (C... M 2 (C. }{{} m 1 If n = 2m + 1, for some m N, then CD n = C C M 2 (C... M 2 (C. }{{} m

G = D n : D n = g, h : g n = h 2 = 1, hgh = g 1. It is well known that: If n = 2m, for some m N, then CD n = C C C C M 2 (C... M 2 (C. }{{} m 1 If n = 2m + 1, for some m N, then CD n = C C M 2 (C... M 2 (C. }{{} m Let e i be as before, then for each i, he i = e i h. The two copies of C in CD n are generated by the idempotents (1/2(1 he 0 and (1/2(1 + he 0, and for i > 0 there is a copy of M 2 (C with basis given by e i, e i, he i, and he i.

Embedding ( D n into PGL 2 (C, ( we can take g as before, 1 0 0 1 g =, and h as. 0 ω 1 0

Embedding ( D n into PGL 2 (C, ( we can take g as before, 1 0 0 1 g =, and h as. 0 ω 1 0 So the action by h is ( ( a b a c h = c a b a.

As before, if i 2 then each e ±i and he ±i acts as zero on sl 2 (C.

As before, if i 2 then each e ±i and he ±i acts as zero on sl 2 (C. For the remaining elements we have that

As before, if i 2 then each e ±i and he ±i acts as zero on sl 2 (C. For the remaining elements we have that ( ( a b 2a 0 (1 he 0 = c a 0 2a ( a e 1 c ( a e 1 c, ( ( ( b 0 0 a b 0 c =, he a c 0 1 = c a 0 0 ( ( b 0 b a b = and he a 0 0 1 = c a, ( 0 0 b 0.

Theorem Let G = D n, where n 1. Then, all G-identities of sl 2 (C are consequences of the following: (1 he 0 1 (x + e 1 (x + e 1 (x = x; 2 2 For α { 1, 1}, [e α (x 1, e α (x 2 ] = 0; 3 For α { 1, 1}, [he α (x 1, he α (x 2 ] = 0; 4 For α { 1, 1}, [e α (x 1, he α (x 2 ] = 0.

Theorem Let G = D n, where n 1. Then, all G-identities of sl 2 (C are consequences of the following: (1 he 0 1 (x + e 1 (x + e 1 (x = x; 2 2 For α { 1, 1}, [e α (x 1, e α (x 2 ] = 0; 3 For α { 1, 1}, [he α (x 1, he α (x 2 ] = 0; 4 For α { 1, 1}, [e α (x 1, he α (x 2 ] = 0. Again, to prove this theorem we construct a basis for the relatively LC X ; G free algebra T G (sl 2 (C.

Note that this case is very similar to the previous one, the difference being that we have two elements in the components 1 and -1.

Note that this case is very similar to the previous one, the difference being that we have two elements in the components 1 and -1. ( ( a b 2a 0 Component 0: (1 he 0 =. c a 0 2a Component 1: ( a b e 1 = c a ( 0 0 c 0 ( a and he 1 c b = a ( 0 0 b 0. Component -1: ( a b e 1 = c a ( 0 b 0 0 ( a and he 1 c b a = ( 0 c 0 0.

Thus we have a basis that looks similar to the basis in the Z n case for n 3, the differences being:

Thus we have a basis that looks similar to the basis in the Z n case for n 3, the differences being: 1 e 0 is replaced by (1 he 0 ; 2 Wherever e 1 appeared, we have two possibilites: e 1 or he 1 ; 3 Wherever e 1 appeared, we have two possibilites: e 1 or he 1.

Theorem The following set is a basis for the relatively free algebra LC X ; G T G (sl 2 (C : 1 [e 1 (x i1, (1 he 0 (x m, e 1 (x j1, e 1 (x i2, e 1 (x j2,..., e 1 (x iα, e 1 (x jα ]; 2 [e 1 (x i1, (1 he 0 (x m, e 1 (x j1, e 1 (x i2, e 1 (x j2,..., e 1 (x iα ]; 3 [e 1 (x i1, (1 he 0 (x m, e 1 (x j1, e 1 (x i2, e 1 (x j2,..., e 1 (x iα, e 1 (x jα, e 1 (x jα+1 ]; 4 [e 1 (x j1, (1 he 0 (x m ]; 5 (1 he 0 (x i1 ; 6 [e 1 (x i1, (1 he 0 (x m, he 1 (x j1, e 1 (x i2, he 1 (x j2,..., e 1 (x iα, he 1 (x jα ]; 7 [e 1 (x i1, (1 he 0 (x m, he 1 (x j1, e 1 (x i2, he 1 (x j2,..., e 1 (x iα ]; 8 [e 1 (x i1, (1 he 0 (x m, he 1 (x j1, e 1 (x i2, he 1 (x j2,..., e 1 (x iα, he 1 (x jα, he 1 (x jα+1 ]; 9 [he 1 (x j1, (1 he 0 (x m ];

Theorem 10 [he 1 (x i1, (1 he 0 (x m, e 1 (x j1, he 1 (x i2, e 1 (x j2,..., he 1 (x iα, e 1 (x jα ]; 11 [he 1 (x i1, (1 he 0 (x m, e 1 (x j1, he 1 (x i2, e 1 (x j2,..., he 1 (x iα ]; 12 [he 1 (x i1, (1 he 0 (x m, e 1 (x j1, he 1 (x i2, e 1 (x j2,..., he 1 (x iα, e 1 (x jα, e 1 (x jα+1 ]; 13 [e 1 (x i1, (1 he 0 (x m, e 1 (x j1, e 1 (x i2, e 1 (x j2,..., e 1 (x ik, e 1 (x jk, he 1 (x ik+1, e 1 (x jk+1,..., he 1 (x iα, e 1 (x jα ]; 14 [e 1 (x i1, (1 he 0 (x m, e 1 (x j1, e 1 (x i2, e 1 (x j2,..., e 1 (x ik, e 1 (x jk, he 1 (x ik+1, e 1 (x jk+1,..., he 1 (x iα ]; 15 [e 1 (x i1, (1 he 0 (x m, e 1 (x j1, e 1 (x i2, e 1 (x j2,..., e 1 (x ik, e 1 (x jk, he 1 (x ik+1, e 1 (x jk+1,..., he 1 (x iα, e 1 (x jα, e 1 (x jα+1 ]; 16 [e 1 (x i1, (1 he 0 (x m, e 1 (x j1, e 1 (x i2, e 1 (x j2,..., e 1 (x ik, e 1 (x jk, e 1 (x ik+1, he 1 (x jk+1,..., e 1 (x iα, he 1 (x jα ]; 17 [e 1 (x i1, (1 he 0 (x m, e 1 (x j1, e 1 (x i2, e 1 (x j2,..., e 1 (x ik, e 1 (x jk, e 1 (x ik+1, he 1 (x jk+1,..., e 1 (x iα ]; 18 [e 1 (x i1, (1 he 0 (x m, e 1 (x j1, e 1 (x i2, e 1 (x j2,..., e 1 (x ik, e 1 (x jk, e 1 (x ik+1, he 1 (x jk+1,..., e 1 (x iα, he 1 (x jα, he 1 (x jα+1 ]; where m 0, α 1 and (1 he 0 (x m is analogous to e 0 (x m.

G = A 4, A 5 or S 4 : All these cases are similar.

G = A 4, A 5 or S 4 : All these cases are similar. The span of the image of G in the group of automorphisms of sl 2 (C is isomorphic to C sl 3 (C.

G = A 4, A 5 or S 4 : All these cases are similar. The span of the image of G in the group of automorphisms of sl 2 (C is isomorphic to C sl 3 (C. Thus it contains elements ɛ ij, where i, j {1, 2, 3}, that act on sl 2 (C in the following way: ɛ ij v α = δ j,α v i, where v 1 = e 11 e 22, v 2 = e 12 and v 3 = e 21 form a basis for sl 2 (C.

( a b If A = c a we have:

( a b If A = we have: c a Component 0: ( a 0 ɛ 11 (A = 0 a Component 1: ( 0 a ɛ 21 (A = 0 0 ( b 0 = e 0 (A, ɛ 12 (A = 0 b ( 0 b, ɛ 22 (A = 0 0 Component -1: ( ( 0 0 0 0 ɛ 31 (A =, ɛ a 0 32 (A = b 0 ( c 0, ɛ 13 (A = 0 c ( 0 c = e 1 (A and ɛ 23 (A = 0 0 ( 0 0 and ɛ 33 (A = c 0.. = e 1 (A.

Theorem Let G = A 4, A 5, S 4. Then all G-identities of sl 2 (C are consequences of the following: 1 ɛ 11 (x + ɛ 22 (x + ɛ 33 (x = x; 2 For α, β {1, 2, 3}, [ɛ 1α (x, ɛ 1β (y] = 0; 3 For α, β {1, 2, 3}, [ɛ 2α (x, ɛ 2β (y] = 0; 4 For α, β {1, 2, 3}, [ɛ 3α (x, ɛ 3β (y] = 0.

Theorem Let G = A 4, A 5, S 4. Then all G-identities of sl 2 (C are consequences of the following: 1 ɛ 11 (x + ɛ 22 (x + ɛ 33 (x = x; 2 For α, β {1, 2, 3}, [ɛ 1α (x, ɛ 1β (y] = 0; 3 For α, β {1, 2, 3}, [ɛ 2α (x, ɛ 2β (y] = 0; 4 For α, β {1, 2, 3}, [ɛ 3α (x, ɛ 3β (y] = 0. Again, to prove this theorem we construct a basis for the relatively LC X ; G free algebra T G (sl 2 (C.

Again we have a basis that looks similar to the basis in the Z n case for n 3, the differences being:

Again we have a basis that looks similar to the basis in the Z n case for n 3, the differences being: 1 Wherever e 0 appeared, we have three possibilites: e 0 = ɛ 11, ɛ 12, or ɛ 13; 2 Wherever e 1 appeared, we have three possibilites: ɛ 21, e 1 = ɛ 22, or ɛ 23; 3 Wherever e 1 appeared, we have three possibilites: ɛ 31, ɛ 32, or e 1 = ɛ 33.

Again we have a basis that looks similar to the basis in the Z n case for n 3, the differences being: 1 Wherever e 0 appeared, we have three possibilites: e 0 = ɛ 11, ɛ 12, or ɛ 13; 2 Wherever e 1 appeared, we have three possibilites: ɛ 21, e 1 = ɛ 22, or ɛ 23; 3 Wherever e 1 appeared, we have three possibilites: ɛ 31, ɛ 32, or e 1 = ɛ 33. Thus, this basis is worse than in the previous case, and will not be displayed here.

With similar but more complicated calculations we can show that the bases for the G-identities for are the same as those for sl 2 (C.

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