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Exterior Differentials in Superspace and Poisson Brackets Published by Institute of Physics Publishing for SISSA/ISAS Received: Accepted: Exterior Differentials in Superspace and Poisson Brackets JHEP00(2002)000 Dmitrij V. Soroka and Vyacheslav A. Soroka Kharkov Institute of Physics and Technology, 61108 Kharkov, Ukraine E-mail: [email protected], [email protected] Abstract: It is shown that two definitions for an exterior differential in superspace, giving the same exterior calculus, yet lead to different results when applied to the Poisson bracket. A prescription for the transition with the help of these exterior differentials from the given Poisson bracket of definite Grassmann parity to another bracket is introduced. It is also indicated that the resulting bracket leads to generalization of the Schouten-Nijenhuis bracket for the cases of superspace and brackets of diverse Grassmann parities. It is shown that in the case of the Grassmann-odd exterior differential the resulting bracket is the bracket given on exterior forms. The above-mentioned transition with the use of the odd exterior differential applied to the linear even/odd Poisson brackets, that correspond to semi-simple Lie groups, results, respectively, in also linear odd/even brackets which are naturally connected with the Lie superalgebra. The latter contains the BRST and anti- arXiv:hep-th/0211280v3 24 Mar 2003 BRST charges and can be used for calculation of the BRST operator cohomology. Keywords: BRST Quantization, BRST Symmetry, Superspaces, Differential and Algebraic Geometry.. Dedicated to the memory of Aleksandr Ivanovich Soroka and Raisa Artemovna Soroka (Gelukh) c SISSA/ISAS 2018 http://jhep.sissa.it/JOURNAL/JHEP3.tar.gz Contents 1. Introduction: two definitions of an exterior differential in superspace 1 2. Poisson brackets related with the exterior differentials 3 3. Linear Poisson brackets related with semi-simple Lie groups 6 JHEP00(2002)000 4. Lie superalgebra for the BRST and anti-BRST charges 8 5. Conclusion 10 1. Introduction: two definitions of an exterior differential in superspace There exist two possibilities to define an exterior differential in superspace with coordinates a a z having Grassmann parities g(z ) ≡ ga and satisfying the permutation relations zazb = (−1)gagb zbza. The first one realized when we set the Grassmann parity of the exterior differential d0 = a d0z ∂za to be equal to zero a g(d0z )= ga, a where ∂za ≡ ∂/∂z . In this case the symmetry property of an exterior product of two differentials is a b gagb+1 b a d0z ∧ d0z = (−1) d0z ∧ d0z (1.1) a b and a permutation relation of the exterior differential d0z with the coordinate z has the form a b gagb b a d0z z = (−1) z d0z . Note that relation (1.1) can be rewritten in the following form ga a b (ga+1)(gb+1) gb b a (−1) d0z ∧ d0z = (−1) (−1) d0z ∧ d0z . (1.2) By defining an exterior product of a differential p-form 0 ap a1 0 0 Φ = d0z ∧···∧ d0z φa1...ap , g(φa1...ap )= ga1 + · · · + gap – 1 – and a q-form 0 bq b1 0 0 Ψ = d0z ∧···∧ d0z ψb1...bq , g(ψb1...bq )= gb1 + · · · + gbq as 0 0 pq bq b1 ap a1 0 0 Φ ∧ Ψ = (−1) d0z ∧···∧ d0z ∧ d0z ∧···∧ d0z φa1...ap ψb1...bq , we obtain the following symmetry property of this product 0 0 pq 0 0 Φ ∧ Ψ = (−1) Ψ ∧ Φ . JHEP00(2002)000 By setting the exterior differential of the p-form Φ0 as follows 0 ap a1 0 p ap a1 b 0 b d0Φ = d0 ∧ d0z ∧···∧ d0z φa1...ap = (−1) d0z ∧···∧ d0z ∧ d0z ∂z φa1...ap , (1.3) we obtain the Leibnitz rule for the exterior differential of the exterior product of two forms 0 0 0 0 p 0 0 d0(Φ ∧ Ψ ) = (d0Φ ) ∧ Ψ + (−1) Φ ∧ (d0Ψ ). Note that very often another definition for the exterior differential is adopted 0 a1 ap b 0 b d0Φ = d0z ∧···∧ d0z ∧ d0z ∂z φap...a1 which differs from (1.3) with the absence of the grading factor (−1)p and leads to the following form of the Leibnitz rule 0 0 q 0 0 0 0 d0(Φ ∧ Ψ ) = (−1) (d0Φ ) ∧ Ψ +Φ ∧ (d0Ψ ). Another definition for the exterior differential in superspace arises when the Grassmann a parity of the exterior differential d1 = d1z ∂za is chosen to be equal to unit a g(d1z )= ga + 1. Then the symmetry property of the exterior product for two differentials is defined as1 a b (ga+1)(gb+1) b a d1z ∧˜d1z = (−1) d1z ∧˜d1z (1.4) and a rule for the permutation of such a differential with the coordinate zb has to be a b (ga+1)gb b a d1z z = (−1) z d1z . Relation (1.4) can be represented in the form ga a b gagb+1 gb b a (−1) d1z ∧˜d1z = (−1) (−1) d1z ∧˜d1z . (1.5) If the exterior product of the p-form 1 ap ˜ ˜ a1 1 1 Φ = d1z ∧··· ∧d1z φa1...ap , g(φa1...ap )= ga1 + · · · + gap 1In this case we use another notation ∧˜ for the exterior multiplication. – 2 – and q-form 1 bq ˜ ˜ b1 1 1 Ψ = d1z ∧··· ∧d1z ψb1...bq , g(ψb1...bq )= gb1 + · · · + gbq is defined in the following way 1 1 p(q+g +···+g ) b b a a 1 1 ˜ b1 bq q ˜ ˜ 1 ˜ p ˜ ˜ 1 Φ ∧Ψ = (−1) d1z ∧··· ∧d1z ∧d1z ∧··· ∧d1z φa1...ap ψb1...bq , then the symmetry property of this product is Φ1∧˜Ψ1 = (−1)pqΨ1∧˜Φ1. JHEP00(2002)000 Let us define in this case the exterior differential of the p-form Φ1 in the following form 1 ap a1 1 p+ga +···+gap ap a1 b 1 ˜ ˜ ˜ 1 ˜ ˜ ˜ b d1Φ = d1∧d1z ∧··· ∧d1z φa1...ap = (−1) d1z ∧··· ∧d1z ∧d1z ∂z φa1...ap . Then the Leibnitz rule for the exterior differential of the exterior product of a p-form Φ1 and a q-form Ψ1 will be 1 1 1 1 p 1 1 d1(Φ ∧˜Ψ ) = (d1Φ )∧˜Ψ + (−1) Φ ∧˜(d1Ψ ). Note that in this case the symmetry properties of the exterior product (1.4) coincide with the ones for the usual Grassmann product of two differentials for the coordinate za a b (ga+1)(gb+1) b a d1z d1z = (−1) d1z d1z . (1.6) The equivalence of the exterior calculi obtained with the use of the above mentioned different definitions for the exterior differential can be established as a result of the direct verification by taking into account relations (1.1) and (1.5) and by putting the following relations between coefficients of the corresponding p-forms Φ0 and Φ1 [p/2] ga2k 0 kP=1 1 φa1...ap = (−1) φa1...ap , where [p/2] denotes a whole part of the quantity p/2. Thus, we proved that two definitions for the exterior differential, differed with the Grassmann parities, result in the same exterior calculus. 2. Poisson brackets related with the exterior differentials Now we show that application of these differentials leads, however, to the different results under construction from a given Poisson bracket with a Grassmann parity ǫ = 0, 1 of another one. A Poisson bracket, having a Grassmann parity ǫ, written in arbitrary non-canonical variables za ← → ab {A, B}ǫ = A ∂ za ωǫ (z) ∂ zb B (2.1) – 3 – has the following main properties: g({A, B}ǫ) ≡ gA + gB + ǫ (mod 2), (gA+ǫ)(gB+ǫ) {A, B}ǫ = −(−1) {B, A}ǫ, (−1)(gA+ǫ)(gC +ǫ){A, {B,C} } = 0, X ǫ ǫ (ABC) ab which lead to the corresponding relations for the matrix ωǫ JHEP00(2002)000 ab g ωǫ ≡ ga + gb + ǫ (mod 2), (2.2) ab (ga+ǫ)(gb+ǫ) ba ωǫ = −(−1) ωǫ , (2.3) (ga+ǫ)(gc+ǫ) ad bc (−1) ω ∂ d ω = 0, (2.4) X ǫ z ǫ (abc) where gA ≡ g(A) and a sum with a symbol (abc) under it designates a summation over cyclic permutations of a, b and c. The Hamilton equations for the phase variables za, which correspond to a Hamiltonian Hǫ (g(Hǫ)= ǫ), a dz a ab = {z ,H } = ω ∂ b H (2.5) dt ǫ ǫ ǫ z ǫ can be represented in the form a dz ab ab ∂(dζ Hǫ) def a b = = ωǫ ∂z Hǫ ≡ ωǫ b (z , dζ Hǫ)ǫ+ζ , dt ∂(dζ z ) where dζ (ζ = 0, 1) is one of the exterior differentials d0 or d1. By taking the exterior differential dζ from the Hamilton equations (2.5), we obtain a d(dζ z ) ab ∂(dζ Hǫ) ζ(ga+ǫ) ab def a b = = (dζ ωǫ ) b + (−1) ωǫ ∂z (dζ Hǫ) (dζ z , dζ Hǫ)ǫ+ζ . dt ∂(dζ z ) As a result of two last equations we have by definition the following binary composition a a a for functions F and G of the variables z and their differentials dζ z ≡ yζ ← → ← → ← → ab ζ(g +ǫ) ab c ab a a a a c (F, G)ǫ+ζ = F ∂ z ωǫ ∂ b +(−1) ∂ y ωǫ ∂ zb + ∂ y yζ ∂z ωǫ ∂ b G. (2.6) h yζ ζ ζ yζ i ab In consequence of the grading properties (2.2) for the matrix ωǫ this composition has the Grassmann parity ǫ + ζ g[(F, G)ǫ+ζ ] ≡ gF + gG + ǫ + ζ (mod 2). ab By using the symmetry property (2.3) of ωǫ , we can establish the symmetry of the composition (2.6) (gF +ǫ+ζ)(gG+ǫ+ζ) (F, G)ǫ+ζ = −(−1) (G, F )ǫ+ζ . – 4 – ab At last, taking into account relations (2.3) and (2.4) for the matrix ωǫ , we come to the Jacobi identities for this composition (−1)(gE +ǫ+ζ)(gG+ǫ+ζ)(E, (F, G) ) = 0.
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