An example of simple Lie superalgebra with several invariant bilinear forms

Transcription

1 math-ph/ An example of simple Lie superalgebra with several invariant bilinear forms arxiv:math-ph/011063v 1 Jan 00 S.E.Konstein I.E.Tamm Department of Theoretical Physics, P. N. Lebedev Physical Institute, 11794, Leninsky Prospect 53, Moscow, Russia. Abstract Simple associative superalgebra with independent supertraces is presented. Its commutant is a simple Lie superalgebra which has at least independent invariant bilinear forms. konstein@lpi.ru This work was supported by the Russian Basic Research Foundation, grant and grant

2 1 Introduction In this paper the associative superalgebra A = SH 3 (0) with two-dimensional space of supertraces is presented. It is shown that (i) it is simple, (ii) its commutant A 1 = SH 3 (0), SH 3 (0)} is a simple Lie superalgebra and (iii) A 1 has at least -dimensional space of nondegenerate bilinear invariant forms. Would A 1 be finitedimensional superalgebra having -dimensional space of invariant bilinear forms, it necessarily would not be simple. The superalgebra SH 3 (0) belongs to the class of superalgebras of observables of N-particle Calogero model which is denoted in 8] as SH N(ν), where ν is a coupling constant of this Calogero model 1]. Let x i R, i = 1,..., N. The Hamiltonian of Calogero model after some similarity transformation attains the form H Cal = 1 which coincides with the operator N i=1 x i x i + ν j i x i x j x i (1) H = 1 N { } a 0 i, a 1 i i=1 () on the space of symmetric functions. The annihilation and creation operators a α i (α = 0, 1 correspondingly) acting on the space of complex valued infinitely smooth functions of x 1, x,..., x N are expressed via the differential-difference Dunkl operators D i (x) 4] a α i = 1 (x i + ( 1) α D i (x)), where D i = x i + ν and K ij = K ji are the operators of elementary transpositions N l i 1 x i x l (1 K il ), (3) K ij x i = x j K ij, K ij x k = x k K ij if k i and k j, (4) satisfying as a consequence the following relations: K ij a α i = a α j K ij, K ij a α k = aα k K ij if k i and k j, K ij = 1. (5) The operators a α i satisfy the following commutation relations a α i, a β ] j = ǫ α β δ ij + νδ ij N l=1, l i where ǫ α β = ǫ β α, ǫ 0 1 = 1, which lead 6] to the relations H, a α i ] = ( 1)α a α i, K il νδ i j K ij, and, therefore, a 1 i when acting on the Fock vacuum 0 such that a 0 j 0 = 0 for any j, give all the wavefunctions of the model () and, consequently, of the model (1). In 7] Perelomov

3 operators ], describing the wavefunctions of the model (1) are expressed via these operators a α i. Definition 1. SH N (ν) is the associative superalgebra of all polynomials in the operators K ij and a α i with Z -grading π defined by the formula: π(a α i ) = 1, π(k ij) = 0. The superalgebra SH N (ν) is called the algebra of observables. It is clear that SH N (ν) = SH 1 (0) SH N(ν), where SH 1 (0) is generated by the operators a α = i a α i / N satisfying the commutation relations a α, a β ] = ǫ αβ, while SH N (ν) is generated by the transpositions K ij and by all linear combinations of the form i λ i a α i with coefficients λ i satisfying the condition i λ i = 0. The group algebra CS N ] of the permutation group S N, generated by elementary transpositions K ij is a subalgebra of SH N(ν). Definition. The supertrace on an arbitrary associative superalgebra A is complexvalued linear function str( ) on A which satisfies the condition str(fg) = ( 1) π(f)π(g) str(gf) (6) for every f, g A with definite parity. It is proven in 8] that SH N (ν) has nontrivial supertraces and the dimension of the space of the supertraces on SH N (ν) is equal to the number of partitions of N into the sum of positive odd integers. 1 In particular, SH (ν) has only one supertrace and SH 3 (ν) has two independent supertraces. Since SH N (ν) = SH 1 (0) SH N (ν) and SH 1(0) has one supertrace, it follows that the numbers of supertraces on SH N (ν) and on SH N (ν) are the same. Every supertrace str( ) defines the bilinear form B str : B str (f, g) = str(fg). (7) Since the complete set of null-vectors of every invariant bilinear form constitutes the twosided ideal of the algebra, the supertrace is a good tool for finding such ideals. It should be noticed that the existence of supertraces depends on the choice of Z -grading π, while the existence of an ideals does not depend on the grading. In the well known case of SH (ν) corresponding to the usual two-particle Calogero model the only supertrace defines the bilinear form (7) which degenerates when ν is half-integer 5], hence, the superalgebra SH (ν) has an ideal for these values of ν. When N 3 the situation is more complicated as SH N(ν) has more than one supertrace. In the case of finitedimensional superalgebra this would be sufficient for existence of ideals but for infinitedimensional superalgebras under consideration the existence of ideals is not completely investigated yet. It is shown in 10] that in the superalgebras SH 3(ν) with ν = n ± 1/3 or ν = n + 1/ (n is an integer) there exists one supertrace with degenerated corresponding bilinear form (7) and that for the other values of ν all the bilinear forms corresponding to supertraces are non-degenerate. One can suggest that for the latest case the corresponding superalgebras are simple. Below it is shown that the associative superalgebra SH 3 (0) is simple and that its commutant is simple Lie superalgebra. 1 The analogous result for Calogero model based on arbitrary root system 3] is presented in 11]. 3

4 The paper is organized as follows. In Section the generating elements of associative superalgebra SH 3(0) are presented together with the relations between them. The simplicity of SH 3 (0) is proved in Section 3 and in the last section the simplicity of SH 3 (0), SH 3 (0)} is established. The algebra SH 3(0) In this Section we present the generating elements of associative superalgebra A = SH 3 (0) and relations between them. Let λ = exp(πi/3). As a basis in C S 3 ] let us choose the vectors L k = 1 3 (λk K 1 + K 3 + λ k K 31 ), Q k = 1 3 (1 + λk K 1 K 13 + λ k K 1 K 3 ), L i+3 = L i, Q i+3 = Q i, L ± def = L ±1, Q ± def = Q ±1. Instead of the generating elements a α i let us introduce the vectors x α = a α 1 + λaα + λ a α 3, x def = x 0, x + def = x 1, y α = a α 1 + λ a α + λa α 3, y def = y 0, y + def = y 1, which are derived from a α i by subtracting the center of mass a α = 1 a α i 3. i=1,,3 The Lie algebra sl of inner automorphisms of SH 3(0) is generated by the generators T αβ These generators satisfy the usual commutation relations T αβ = 1 3 (xα y β + x β y α ). (8) T αβ, T γδ ] = ǫ αγ T βδ + ǫ αδ T βγ + ǫ βγ T αδ + ǫ βδ T αγ, (9) and act on generating elements x α and y α as follows: T αβ, x γ] = ǫ αγ x β + ǫ βγ x α, T αβ, y γ] = ǫ αγ y β + ǫ βγ y α (10) leaving the group algebra CS 3 ] invariant: ] T αβ, K ij = 0. Clearly, SH 3 (0) decomposes into the infinite direct sum of finitedimensional irreducible representations of this sl. Let A s (s is a spin) be the direct sum of all s + 1-dimensional irreducible representations in this decomposition. So Evidently, A 0 is a subalgebra of A = SH 3(0). A = A 0 A 1/ A 1... (11) 4

5 As every element of the subspace A s for s > 0 is a linear combinations of the vectors of the form f, T αβ ] where T αβ is defined in (8), every supertrace vanishes on this subspace and has non-trivial values only on the associative subalgebra A 0 SH 3 (0) of all sl -singlets. Let m = 1 4 {xα, y α }. Clearly, m is a singlet under the action of sl (8); m can also be expressed in the form m = 1 (xα y α + 3) or, equivalently, (1) m = 1 (y αx α 3). (13) In these formulas the greek indices are lowered and raised with the help of the antisymmetric tensor ε αβ : a α = β ε αβ a β. The generating elements x α, y α, Q i, L i and the element m satisfy the following relations: where δ 0 = 1 and δ i = 1 if i 0, L i L j = δ i+j Q j, L i Q j = δ i j L j, (14) Q i L j = δ i+j L j, Q i Q j = δ i j Q j, (15) L i x α = y α L i+1, L i y α = x α L i 1, (16) Q i x α = x α Q i+1, Q i y α = y α Q i 1, (17) L i m = ml i, Q i m = mq i, (18) x, x + ] = 0, y, y + ] = 0, y, x + ] = x, y +] = 3, x, y] = x +, y +] = 0, (19) and m, x α ] = 3 (xα ), m, y α ] = 3 (yα ), (0) Corollary 1. There exists Z 3 grading ρ on SH 3 (0) defined by the formulas: ρ(q i ) = 0, ρ(l i ) = i, ρ(x α ) = 1, ρ(y α ) = 1, such that fq i = Q i f for any element f SH 3 (0) with ρ(f) = 0. Particularly, (xα ) 3, Q i ] = (y α ) 3, Q i ] = 0. Corollary. There exists the antiautomorphism τ of associative algebra SH 3 (0): τ(x α ) = y 1 α, τ(y α ) = x 1 α, τ(m) = m, τ(l i ) = L i, τ(q i ) = Q i. Every supertrace on SH 3 (ν) is completely determined by its values on C S 3] SH 3 (ν), i.e., by the values str(1), str(k 1 ) = str(k 3 ) = str(k 31 ) and str(k 1 K 3 ) = str(k 1 K 13 ), 5

6 which are consistent with (6) and admit the extension of the supertrace from C S 3 ] to SH 3(ν) in a unique way, if and only if str(k ij ) = ν( str(1) str(k 1 K 3 )) 8]. It was noticed above that only trivial representations of sl can contribute to any supertrace on SH 3 (0). Evidently, the associative algebra A 0 of sl -singlets contains only the polynomials of m with coefficients in C S 3 ]. To describe the restrictions of all the supertraces on A 0, it is convenient to use the generating functions, which are computed in 10] for any values of ν, and for ν = 0 take the following form: where L i Q i def = str(e /3ξm L i ) = 0, (1) def = str(e /3ξm Q i ) = P i, () P 0 = S 1 + S ( e ξ + e ξ e ξ e ξ), (3) P + = 3 S ( 1 e ξ + e ξ) + S ( e ξ e ξ + 3 ), (4) P = 3 S ( 1 e ξ + e ξ) + S ( e ξ e ξ + 3 ) (5) = exp( 3ξ)(exp(3ξ) + 1). (6) Here S 1, S are arbitrary parameters, specifying the supertrace in the two-dimensional space of supertraces: S 1 = str(1) str(k 1 K 3 ), S = 8 3 str(1) 3 str(k 1K 3 ). 3 The simplicity of SH 3 (0). To prove the simplicity of SH 3 (0) it is sufficient to prove that elements Q i belong to any nonzero ideal I. Let f I, f = f 0 + f 1/ + f , where f s A s. As T αβ, f] I then f s I. Proposition 1. Let I SH 3 (0) be two-sided nonzero ideal. Then there exist nonzero polynomial f(x α, y α ) and a polynomial g(x α, y α ) such that fq 0 + gl 0 I. Indeed, let r I, r 0. Then at least one of the polynomials rq 0, rl 0, rxq 0, rxl 0, ryq 0, ryl 0 has the desired form. Proposition. Let I SH 3 (0) be two-sided nonzero ideal. Then there exist nonzero polynomial f 0 (m) and a polynomial g 0 (m) such that f 0 Q 0 + g 0 L 0 I. Let f(x α, y α ) 0 and h 0 = fq 0 + gl 0 I A s with some polynomial g(x α, y α ). Let h k+1 = ] T 00, h k (7) Evidently, h k I for any k. Then there exists such integer K that h K 0 and h K+1 = 0. The vector h K is the highest vector of some irreducible representation of sl algebra (8) and has the form s ( ) h K = (x) i (y) s i q i (m)q 0 + (x) i (y) s i p i (m)l 0 i=0 6 (8)

7 where some polynomial q n (m) 0. Then F = (y + ) n (x + ) s n h K I. Let F 0 be the singlet part of F. Evidently, F 0 0, F 0 I, and F 0 = f 0 (m)q 0 +g 0 (m)l 0 with nonzero polynomial f 0 (m). Proposition 3. Let I SH 3 (0) be two-sided nonzero ideal. Then there exists nonzero polynomial f 0 (m) such that F 0 = f 0 (m)q 0 I. Let F = f(m)q 0 + g(m)l 0 I, f(m) 0. Then mf(m)q 0 = 1 (mf + Fm) I due to commutational relations. Proposition 4. Let I SH 3 (0) be two-sided nonzero ideal. Then there exist nonzero polynomials f i (m) such that F i = f i (m)q i I for i = 0, ±1. Let F 0 = f 0 (m)q 0 I is nonzero polynomial from Proposition 3. Let us denote the singlet part of arbitrary element F A as (F) 0 Then F 1 = (y + F 0 x) 0 = f 1 (m)q 1 and F = (x + F 0 y) 0 = f (m)q are nonzero and belong to I. Proposition 5. Let I i = {f(m) Cm] : f(m)q i I}. Then I i are nonzero ideals in Cm]. This proposition follows trivially from Proposition 4. Hence there exist polynomials ϕ i (m) = m k i +... such that I i = ϕ i (m) Cm]. Proposition 6. The polynomials ϕ i (m) generating the ideals I i satisfy the relations ϕ 0 (m) = ϕ 0 ( m), (9) ϕ 1 (m) = ±ϕ ( m). (30) Indeed, ϕ 0 ( m)q 0 = L 0 ϕ 0 (m)q 0 L 0 I, hence ϕ 0 ( m) I 0 and all the roots of ϕ 0 (m) are the roots of ϕ 0 ( m). Further, L 1 ϕ 1 (m)q 1 L = ϕ 1 ( m)q and, as a consequence, ϕ 1 ( m) = p(m)ϕ (m) with some nonzero polynomial p(m). Analogously, ϕ ( m) = q(m)ϕ 1 (m) with some nonzero polynomial q(m). So p(m)q( m) = 1, and ϕ ( m) = ±ϕ 1 (m). Now we can prove the following theorem: Theorem 1. The associative superalgebra A = SH 3(0) is simple. We will prove that arbitrary nonzero ideal in A contains the element 1 A. Let I be nonzero ideal in A and ϕ i (m) generate the ideals I i described above. So ϕ i (m)q i I and hence y 3 ϕ i (m)q i (x + ) 3 I. Consider the sl -singlet part of this element which also belongs to I and can be computed with the help of (0): So P(m)ϕ i (m + 9/) I i and (y 3 ϕ i (m)q i (x + ) 3 ) 0 = P(m)ϕ i (m + 9/)Q i, I (31) where P(m) = const(m + 3/)(m + 3)(m + 9/) 0. (3) P(m)ϕ i (m + 9/) = p i (m)ϕ i (m) with some polynomials p i (m). Since ϕ 0 (m) is even polynomial, the real part of the rightest root of ϕ 0 (m) is non negative, and so this rightest root is not a root of polynomial (m + 3/)(m+3)(m+9/)ϕ 0 (m+3/). So ϕ 0 (m) has no roots and ϕ 0 (m) = 1. One can apply the same consideration to the function ϕ(m) = ϕ 1 (m)ϕ (m) which is even due to Proposition 6 and satisfies the equation P(m) ϕ(m + 9/) = p 1 (m)p (m)ϕ(m). So ϕ 1 (m)ϕ (m) = 1 and ϕ 1 (m) = ϕ (m) = 1. In such a way, Q i I and 1 = Q 0 + Q 1 + Q I. 7

8 4 Lie superalgebra A L Now let us consider Lie superalgebra A = A L which consists of elements of associative algebra A = SH 3 (0) with operation f, g} = fg ( 1)π(f)π(g) gf. This Lie superalgebra has an ideal A 1 = A, A} with the following evident properties (i) codima 1 =, (ii) A 1/ A 1... A 1. Let Z be the center of SH 3(0). Obviously, Z A 0 and so every element of Z is a polynomial of m with coefficients in C S 3 ]. Proposition 7. Z = C. Let f(m) Z. As m is even then f is even element also. As ml i = L i m, the relation f, m] = 0 gives that f(m) = f i (m)q i. Due to equalities (x) 3, Q i ] = 0 and (0) one has (x) 3, f(m)] = (x) 3 (f(m) f(m 9/)) and as a consequence f(m) = f(m 9/). So f(m) does not depend on m, f(m) C S 3 ]. To finish the proof it is sufficient to commute f with x or y. Proposition 8. Z A 1 = {0}. As there exists such supertrace that str(1) 0 and Z = C, it follows from str(z A 1 ) = 0 that Z A 1 = {0}. Proposition 9. Let I A 1 be an ideal and I even = {0}. Then I odd = {0}. Let r I odd. Due to sl invariance one can choose r in the form r = s i=0,1, n=0 ( (x) n (y) s n f ni (m)q i + (x) n (y) s n g ni (m)l i ) with odd value of s. As (x) 3 A 3/ A 1 then (x) 3, r} I even and {(x) 3, r} = 0. So the following relations take place: s i=0,1, n=0 (x) n+3 (y) s n (f ni (m) + f ni (m + 9/))Q i = 0 (33) and so f ni = 0. Further, it follows from 0 = {(x) 3, r} s i=0,1, n=0 ((x)3 + (y) 3 ) (x) n (y) s n g ni (m)l i that g n i(m) = 0 also. Here the sign is used to denote the equality up to polynomials of lesser degrees. Theorem Lie superalgebra A 1 is simple. This Theorem follows from the next two general theorems proven by S.Montgomery in 9]. Theorem 3 Let A be associative simple superalgebra and I A L, A L } be an ideal of A L, A L } such that I A L, A L }. Then I 3 = {0}, where I 1 = I, I}, I = I 1, I 1 }, I 3 = I, I }. Theorem 4. Let A be associative simple superalgebra and I A L, A L } be an ideal of A L, A L } such that I even, I} Z(A) where Z(A) is the center of A. Then I even Z(A). Indeed, let I A 1 be an ideal such that I A 1. Due to Theorem 3 one can consider that I, I} = {0}. Then it follows from Theorem 4 and Proposition 8 that I even = {0}. Further, Proposition 9 gives that I odd = {0}, and so I = {0}. 8

9 Theorem 5 Simple Lie superalgebra A 1 has at least independent bilinear invariant forms. Two-dimensional space of supertraces str( ) on A = SH 3 (0) generates some space of invariant bilinear forms B(u, v) = str(uv) on A 1. This space is also -dimensional. Indeed, let some supertrace str 0 ( ) on A leads to bilinear form B 0 on A 1 which is equal to zero identically. The elements x, y +, xq 1 and y + Q 1 belong to A 1/ A 1. So, 0 = B 0 (x, y + ) B 0 (y +, x) = str 0 (x, y + ]) = 3str 0 (1) and 0 = B 0 (x, y + Q 1 ) B 0 (y +, xq 1 ) = str 0 (x, y + ]Q 1 ) = 3str 0 (Q 1 ). It follows from () that str 0 (1) = 1(S 6 1 3S ) and str 0 (Q 1 ) = 1(S S ). So S 1 = S = 0 and str 0 is equal to zero identically. Acknowledgments The author is very grateful to M. Vasiliev and to D. Leites for useful discussions. References 1] F. Calogero, J. Math. Phys., 10 (1969) 191, 197; ibid, 1 (1971) 419. ] A. M. Perelomov, Teor. Mat. Fiz., 6 (1971) ] M. A.Olshanetsky and A. M. Perelomov, Phys. Rep., 94 (1983) ] C.F.Dunkl, Trans. Am. Math. Soc., 311 (1989) ] M.A. Vasiliev, JETP Letters, 50 (1989) 344; Int. J. Mod. Phys., A6 (1991) ] L. Brink, H. Hansson and M.A. Vasiliev, Phys. Lett., B86 (199) ] L. Brink, H. Hansson, S.E. Konstein and M.A. Vasiliev, Nucl. Phys., B401 (1993) ] S.E. Konstein and M.A. Vasiliev, J. Math. Phys., 37 (1996) 87. 9] S.Montgomery, J. of Algebra, 195 (1997) ] S.E. Konstein, Theor. Math. Phys., 116 (1998); hep-th/ ] S.E. Konstein, math-ph/