The general form of the *-commutator on the Grassman algebra

23 Pages·0.24 MB·English

by  I. V. Tyutin

FIAN/TD/01-04
The general form of the ∗–commutator on the Grassman algebra
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0
0
2 I.V.Tyutin ∗

n
a
J
I.E.Tamm Department of Theoretical Physics,

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1 P.N.Lebedev Physical Institute,

117924, Leninsky Prospect 53, Moscow, Russia.
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v
8
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Abstract
0
1
0 We study the general form of the ∗–commutator treated as a deformation of the
/ Poisson bracket on the Grassman algebra. We show that, up to a similarity transfor-
h
t mation, there are other deformations of the Poisson bracket in addition to the Moyal
-
p commutator (one at even and one at odd n, n is the number of the generators of the
e
Grassman algebra) which are not reduced to the Moyal commutator by a similarity
h
: transformation.
v
i
X
r
a
∗
E-mail: [email protected]
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1 Introduction
As is well known, the major hopes for the construction of quantum mechanics on nontrivial
manifolds are connected with the so–called geometric or deformation quantizations ([1] –
[5]). The functions on phase space are put into correspondence with the operators and the
productoftheoperatorsandtheircommutatoraredescribedbyanassociative∗–productand
∗–commutator offunctions, whichrepresent deformationsofausual“pointwise” productand
the Poisson bracket. On even manifolds, at least locally, the ∗–commutator coincides with
usual commutator in algebra with the associative ∗–product. It is interesting to find out,
what is the situation in the case, when the phase space is a supermanifold. In present paper
we investigate a general form of the deformation of the nonsingular Poisson bracket on the
Grassmanalgebra. Weshow thatontheGrassmanalgebrainadditiontothe∗–commutators
which are equivalent to the Moyal commutator [6], there are also other deformations which
are not reduced to the Moyal commutator by a similarity transformation.
The paper is organized as follows. In Sect. 2 we formulate the problem. In Sect. 3 the
solution of the Jacobi identity considered as an equation to the ∗–commutator is found in
the lowest approximation in deformation parameter. In Sect. 4 the higher deformations are
considered and the main result of the paper is formulated. In two Appendices solutions of
the equations considered in the main text are presented.
Notations and conventions.
ξα, α = 1,...,n are odd anticommuting generators of the Grassman algebra:
∂
ε(ξα) = 1, ξαξβ +ξβξα = 0, ∂ ≡ ,
α ∂ξα
ε(A) denotes the Grassman parity of A;
[ξα]0 ≡ 1, [ξα]k ≡ ξα1···ξαk,1 ≤ k ≤ n, [ξα]k = 0,k > n,
[∂ ]0 ≡ 1, [∂ ]k ≡ ∂ ···∂ ,1 ≤ k ≤ n, [∂ ]k = 0,k > n,
α α α1 αk α
←− ←− ←−
[∂ ]k ≡ ∂ ··· ∂ ,
α α1 αk
T ≡ T , T = −T , i = 1,...,k −1.
...[α]k... ...α1...αk... ...αiαi+1... ...αi+1αi...
2 Formulation of the problem
We consider the Grassman algebra G over the field K = C or R, with the generators ξα,
K
α = 1,2,...,n. The generic element f of the algebra (a function of the generators) is
n 1
f ≡ f(ξ) = f [ξα]k, f ≡ f = const, f ∈ K.
k! [α]k [α]0 0 [α]k
kX=0
Let the nonsingular Poisson bracket
←−
{f ,f }(ξ) ≡ [f ,f ] (ξ) ≡ c (ξ|f ,f ) = f (ξ)∂ ωαβ(ξ)∂ f (ξ) (1)
1 2 1 2 ∗0 0 1 2 1 α β 2
3
be given on the Grassman algebra G where ωαβ(ξ) is the symplectic metric, the even
K
symmetric tensor function satisfying the Jacobi identity:
ωβα = ωαβ, ωαδ∂ ωβγ +ωγδ∂ ωαβ +ωβδ∂ ωγα = 0.
δ δ δ
We are interested in the general form of the ∗–commutator, i.e. the bilinear mapping
G ×G → G :
K K K
f (ξ) ≡ [f ,f ] (ξ) = C(ξ|f ,f ),
3 1 2 ∗ 1 2
with the following properties:
i) The mapping is even
ε(C(ξ|f ,f )) = ε(f )+ε(f ).
1 2 1 2
ii) Antisymmetry
C(ξ|f ,f ) = −(−1)ε(f1)ε(f2)C(ξ|f ,f ).
2 1 1 2
iii) The Jacobi identity
(−1)ε(f1)ε(f3)[f ,[f ,f ] ] +cycle(1,2,3) = 0, ∀f ,f ,f , (2)
1 2 3 ∗ ∗ 1 2 3
or, equivalently,
(−1)ε(f1)ε(f3)C(ξ|f ,f )+cycle(1,2,3) = 0, ∀f ,f ,f , (3)
1 23 1 2 3
where we denote f (ξ) ≡ C(ξ|f ,f ).
ij i j
iv) It is supposed, that the ∗–commutator is defined in terms of the series in the defor-
mation parameter h¯2,
[f ,f ] (ξ) = h¯2k[f ,f ] (ξ) = h¯2kc (ξ|f ,f ), (4)
1 2 ∗ 1 2 ∗k k 1 2
kX=0 kX=0
where [f ,f ] = c (ξ|f ,f ) coincides with the Poisson bracket (1) (the boundary condition
1 2 ∗0 0 1 2
or the correspondence principle). Differently, we want to establish the general form of a
deformation of the Poisson bracket on the Grassman algebra. The deformation parameter is
denoted by h¯2 and, without loss of generality, it is possible to consider h¯ as real and positive
number.
We treat condition (2) (or, equivalently, (3)) as an equation on possible structure of the
∗–commutator. Note that C (or, equivalently, the ∗ –commutator),
T T
C (ξ|f ,f ) = T−1C(ξ|Tf ,Tf ), (5)
T 1 2 1 2
T isanonsingularevenlinearmappingG → G ,satisfiesthepropertiesi)-iii),ifC(ξ|f ,f )
K K 1 2
satisfies these properties. However, the boundary condition can change. In general, the
operator T can be represented as
T = T′(1+h¯2t +...), T′ = T| .
1 ¯h=0
Let T′ = T be the operator responsible for the change of generators:
η
T ξα=ηα(ξ), T−1ξα=T ξα=ζα(ξ), T f(ξ) = f(η(ξ)), T−1f(ξ) = T f(ξ) = f(ζ(ξ)), (6)
η η ζ η η ζ
4
where ζα(ξ) is inverse change, ηα(ζ(ξ)) = ξα, ε(ηα) = ε(ζα) = 1. The ∗ –commutator obey
Tη
the property iv) but now with modified Poisson bracket:
←−
∂ ∂
ωαβ(ξ) −→ ωαβ(ξ) = ξα ωα′β′(ζ(ξ)) ξβ.
Tη ∂ζα′ ∂ζβ′
We say, that the ∗ –commutator and ∗–commutator are related by a similarity transforma-
T
tion, or call them equivalent. The ∗–commutator obtained from the Poisson bracket by a
similarity transformation is a trivial deformation of the Poisson bracket.
We assume, that after appropriate similarity transformation the Poisson bracket is of
canonical form
←−
{f ,f }(ξ) ≡ [f ,f ] (ξ) ≡ c (ξ|f ,f ) = f (ξ)∂ λ ∂ f (ξ), λ2 = 1, (7)
1 2 1 2 ∗0 0 1 2 1 α α α 2 α
where λα = 1 in the case of the Grassman algebra over complex numbers (GC) and λα = ±1
in the case of the Grassman algebra over real numbers (GR).
Eq. (2) (or equivalent eq. (3)) will be solved in terms of expansion in deformation
parameter h¯2. It is obviously satisfied in the zeroth approximation. In the first order one
gets
(−1)ε(f1)ε(f3)[f ,[f ,f ] ] +(−1)ε(f1)ε(f3)[f ,[f ,f ] ] +cycle(1,2,3) = 0, (8)
1 2 3 ∗1 ∗0 1 2 3 ∗0 ∗1
or, equivalently,
(−1)ε(f1)ε(f3)[f ,c (f ,f )] (ξ)+(−1)ε(f1)ε(f3)c (ξ|f ,[f ,f ] )+cycle(1,2,3) = 0. (9)
1 1 2 3 ∗0 1 1 2 3 ∗0
Besides, the conditions i) and ii) should be satisfied, or
c (ξ|f ,f ) = −(−1)ε(f1)ε(f2)c (ξ|f ,f ), ε(c (ξ|f ,f )) = ε(f )+ε(f ).
1 2 1 1 1 2 1 1 2 1 2
The obvious solution of equation (8) or (9) is a bilinear functional c , obtained from the
1triv
Poisson bracket by the similarity transformation Tf = T(ξ|f) = (1 +h¯t + O(h¯2))f(ξ) =
1
f(ξ)+h¯t (ξ|f)+O(h¯2), ε(Tf) = ε(t f) = ε(f):
1 1
c (ξ|f ,f ) = [f ,t f ] (ξ)−t (ξ|[f ,f ] )+[t f ,f ] (ξ). (10)
1triv 1 2 1 1 2 ∗0 1 1 2 ∗0 1 1 2 ∗0
Consider the algebra B = B of even k–linear functionals Φ (ξ|f ,...,f ) ∈ B (k–
k k 1 k k
⊕
P
linear mappings (G×)k → G), ε(Φ (ξ|f ,...,f )) = ε(f ) + ··· + ε(f ) (mod 2), k ≥ 1,
k 1 k 1 k
B = K, with the property
0
Φ (ξ|f ,...,f ,f ,f ,...,f ) = −(−1)ε(fi)ε(fi+1)Φ (ξ|f ,...,f ,f ,f ,...,f ).
k q i−1 i i+1 n k q i−1 i+1 i n
The product (◦) in this algebra (a mapping B ×B → B ) is defined as follows:
k l k+l
1
Φ ◦Φ (ξ|f ,...,f ) = sign(P)Φ (ξ|f ,...,f )Φ (ξ|f ,...,f ),
k l 1 k+l (k +l)! k i1 ik l ik+1 ik+l
XP
wherethesumistakenoverallpermutationsP ofnumbers1,2,...,k+l; sign(P)istheparity
of the permutation P calculated by the following rule: the permutation of the neighboring
5
indexes i and i gives the sign factor (−1)1+ε(fip)ε(fiq). In this algebra there is a natural
p q
grading g: g(Φ ) = k, turning the algebra B into the graded algebra, g(B ) = k, and the
k k
linear operator d , the differential coboundary Chevalley operator B → B , g(d ) = 1,
S k k+1 H
acting according to the rule:
i−1
k+1 i+1+ε(fi) ε(fl)
dSΦk(f1,...,fk+1) = (−1) lP=1 fi(ξ)∗0 Φk(ξ|f1,...,fi−1,fi+1,...,fk+1)+
Xi=1
+ (−1)j+1+ε(fj)(ε(fi+1+···+ε(fj−1))Φ (ξ|f ,...,f ,f ∗ f ,f ,...,f ,f ,...,f ),
k 1 i−1 i 0 j i+1 j−1 j+1 k+1
1≤i<Xj≤k+1
d B = 0.
S 0
It is easy to prove that
d2 = 0, d (Φ ◦Φ ) = (d Φ )◦Φ +(−1)g(Φk)Φ ◦d Φ .
S S k l S k l k S l
Eqs. (9) and (10) can be rewritten in terms of the Chevalley operator d as
S
d c (ξ|f ,f ,f ) = 0,
S 1 1 2 3
c (ξ|f ,f ) = d t (ξ|f ,f ),
1triv 1 2 S 1 1 2
i.e. it means that c belongs to the second Chevalley cogomology group, while the first
1
order trivial deformations of the Poisson bracket are coboundaries, i.e. they belong to the
zeroth Chevalley cogomology. Thus, the problem of description of all solutions of eq. (9) can
be now formulated as a problem of calculation of the second Chevalley cogomology group
H2(d ,B).
S
3 The first order deformation
It is convenient to use the momentum representation of the ∗–commutator (any bilinear
functional can be represented in such a form)
n ←−
c (ξ|f ,f ) = f (ξ) [∂ ]kc[α]k|[β]l(ξ)[∂ ]lf (ξ).
1 1 2 1 α 1 β 2
kX,l=0
The properties i) and ii) give:
ε(c[α]k|[β]l) = k +l(mod 2), c[β]l|[α]k = −(−1)klc[α]k|[β]l. (11)
1 1 1
6
The equation for the coefficient functions c[α]k|[β]l(ξ) follows from eq. (9)
1
←− n ←−
f (ξ)∂ λ ∂ f (ξ) [∂ ]kc[α]k|[β]l(ξ)[∂ ]lf (ξ) +
1 γ γ γ(cid:18) 2 α 1 β 3 (cid:19)
k,Pl=0
n ←− ←−
+(−1)ε(f2)ε(f3) f (ξ) [∂ ]kc[α]k|[β]l(ξ)[∂ ]lf (ξ) ∂ λ ∂ f (ξ)−
(cid:18) 1 α 1 β 3 (cid:19) γ γ γ 2
k,Pl=0
n ←− ←−
− f (ξ) [∂ ]kc[α]k|[β]l(ξ)[∂ ]lf (ξ) ∂ λ ∂ f (ξ)+
(cid:18) 1 α 1 β 2 (cid:19) γ γ γ 3
k,Pl=0 (12)
n ←− ←−
+f (ξ) [∂ ]kc[α]k|[β]l(ξ)[∂ ]l f (ξ)∂ λ ∂ f (ξ) +
1 α 1 β (cid:18) 2 γ γ γ 3 (cid:19)
k,Pl=0
←− n ←−
+(−1)ε(f2)ε(f3) f (ξ)∂ λ ∂ f (ξ) [∂ ]kc[α]k|[β]l(ξ)[∂ ]lf (ξ)−
(cid:18) 1 γ γ γ 3 (cid:19) α 1 β 2
k,Pl=0
←− n ←−
− f (ξ)∂ λ ∂ f (ξ) [∂ ]kc[α]k|[β]l(ξ)[∂ ]lf (ξ) = 0.
(cid:18) 1 γ γ γ 2 (cid:19) α 1 β 3
k,Pl=0
We also write the similarity transformation in the momentum representation
n
t (ξ|f) = t[α]k(ξ)[∂ ]kf(ξ), ε(t[α]k) = k(mod 2),
1 1 α 1
kX=0
←− n
c (ξ|f ,f ) = f (ξ)∂ λ ∂ t[α]k(ξ)[∂ ]kf (ξ) −
1triv 1 2 1 γ γ γ(cid:18) 1 α 2 (cid:19)
kP=0 (13)
n ←− n ←−
− t[α]k(ξ)[∂ ]k f (ξ)∂ λ ∂ f (ξ) + t[α]k(ξ)[∂ ]kf (ξ) ∂ λ ∂ f (ξ).
1 α (cid:18) 1 γ γ γ 2 (cid:19) (cid:18) 1 α 1 (cid:19) γ γ γ 2
kP=0 kP=0
We solve eq. (12) studying the coefficients in front of different degrees of the derivatives
of the functions f . From now on it is assumed that n ≥ 2.
i
S t e p 0 Consider the factor in eq. (12) in front of f (ξ) (without derivatives):
1
n ←−
(−1)ε(f2)ε(f3) c0|[β]l(ξ)[∂ ]lf (ξ) ∂ λ ∂ f (ξ)−
(cid:18) 1 β 3 (cid:19) γ γ γ 2
Xl=0
n ←− n ←−
− c0|[β]l(ξ)[∂ ]lf (ξ ∂ λ ∂ f (ξ)+ c0|[β]l(ξ)[∂ ]l f (ξ)∂ λ ∂ f (ξ) = 0.
(cid:18) 1 β 2 (cid:19) γ γ γ 3 1 β (cid:18) 2 γ γ γ 3 (cid:19)
Xl=0 Xl=0
0|0
From antisymmetry condition (11) it follows that c (ξ) = 0.
1
0a The coefficient in front of ∂αf2(ξ) gives:
n ←− ←−
(−1)lc0|[β]l(ξ)∂ λ [∂ ]lf (ξ)−c0|α(ξ)∂ λ ∂ f (ξ) = 0.
1 α α β 3 1 γ γ γ 3
Xl=1
From these equations it follows (see Appendix 1) that:
0|β 0|α 0|α
λ ∂ c (ξ)+λ ∂ c (ξ) = 0 =⇒ c (ξ) = λ ∂ c ξ, ε(c ) = 0, (14)
α α 1 β β 1 1 α α 10 10
∂ c0|[β]l(ξ)[∂ ]lf (ξ) = 0 =⇒ c0|[β]l(ξ) = const, l ≥ 2.
α 1 β 3 1
0b Thecoefficient infrontof∂α∂βf2(ξ),f3(ξ) = exp(ξγpγ),pαξβ+ξβpα = pαpβ+pβpα =
0, provides:
λ c0|[γ]l−1α[p ]l−1p −λ c0|[γ]l−1β[p ]l−1p = 0, l ≥ 2. (15)
α 1 γ β β 1 γ α
7
The general solution of eq. (15) is (see Appendix 1):
c0|[α]l = 0, 2 ≤ l ≤ n−1, c0|[α]n = c ε[α]n, c = const, ε(c ) = n(mod 2).
1 1 1n 1n 1n
The contribution from the similarity transformation to the factor c0|[α]k arises only from
1
the term t0 in the expression (13) and is equal to
1
c0|α = −λ ∂ t0(ξ), c0|[α]k = 0, k ≥ 2.
1triv α α 1 1triv
Performing the similarity transformation with t0(ξ) = c (ξ), we obtain the expression for
1 10
∗–commutator with c0|[α]k = c[α]k|0 = 0, k ≤ n−1, and
1 1
c0|[α]n = c ε[α]n, c[α]n|0 = −c ε[α]n, c = const, ε(c ) = n(mod 2).
1 1n 1 1n 1n 1n
Note,thatinthecaseoftheGrassmanalgebrasoverCorRthecoefficientc canbedifferent
1n
from zero only at even n. Furthermore, if one requires, that the functions f(ξ) = const ∗–
commute with any function, then c = 0.
1n
S t e p 1 The coefficient in eq. (12) in front of ∂ f (ξ) gives:
γ 1
n ←− n
λ f (ξ) (−1)k[∂ ]k∂ c[α]k|[β]l(ξ)[∂ ]lf (ξ)− ∂ cγ|[β]l(ξ)[∂ ]lf (ξ)λ ∂ f (ξ)+
γ 2 α γ 1 β 3 α 1 β 2 α α 3
k,Pl=1 lP=1
n
+(−1)ε(f2)λ ∂ f (ξ) ∂ cγ|[β]l(ξ)[∂ ]lf (ξ)−2c ε[β]n[∂ ]nf (ξ)λ ∂ f (ξ)+
α α 2 α 1 β 3 1n β 2 γ γ 3
lP=1 (16)
n
+(−1)ε(f2)2λ ∂ f (ξ)c ε[β]n[∂ ]nf (ξ)− cγ|[β]l[∂ ]l ∂ f (ξ)λ ∂ f (ξ) −
γ γ 2 1n β 3 (cid:20) 1 β (cid:18) α 2 α α 3 (cid:19)
lP=0
−cγ|[β]l[∂ ]l∂ f (ξ)λ ∂ f (ξ)−(−1)l(1+ε(f2))cγ|[β]l∂ f (ξ)λ [∂ ]l∂ f (ξ) = 0.
1 β α 2 α α 3 1 α 2 α β α 3 (cid:21)
1a The coefficient in front of ∂αf2(ξ), [∂γ]kf3(ξ) provides:
λ ∂ cβ|[γ]k(ξ)+λ ∂ cα|[γ]k(ξ)+δ λ ∂ cβ|α(ξ)+δ 2c λ δ ε[γ]n = 0, k = 1,...,n.
α α 1 β β 1 1k γ γ 1 kn 1n α αβ
1a1 k = 1
λ ∂ cβ|γ(ξ)+λ ∂ cα|γ(ξ)+λ ∂ cβ|α(ξ) = 0 =⇒ cα|β(ξ) = λ ∂ cβ(ξ)+λ ∂ cα(ξ), (17)
α α 1 β β 1 γ γ 1 1 α α 1 β β 1
with some function cα(ξ) (see Appendix 1); here it was taken into account that from anti-
1
symmetry properties (11) of the coefficients c[α]k|[β]l it follows that cα|β = cβ|α.
1 1 1
1a2 2 ≤ k ≤ n−1
λ ∂ cβ|[γ]k(ξ)+λ ∂ cα|[γ]k(ξ) = 0, =⇒ cα|[β]k(ξ) = λ ∂ c[β]k(ξ), k = 2,...,n−1, (18)
α α 1 β β 1 1 α α 1
with some functions c[β]k(ξ) (see Appendix 1).
1
1an k = n
λ ∂ cβ|[γ]n(ξ)+λ ∂ cα|[γ]n(ξ) = λ c δ ε[γ]n =⇒ (19)
α α 1 β β 1 α 1n αβ
1
cα|[β]n(ξ) = (λ ∂ c′ (ξ)− ξαc )ε[β]n,
1 α α 1n 2 1n
8
with some function c′ (ξ). Perform the similarity transformation with t0(ξ) = 0, t[α]k(ξ) =
1n 1 1
−c[α]k(ξ), k = 1,...,n−1, t[α]n = −c′ (ξ)ε[α]n. It does not change the coefficients c0|[α]k and
1 1 1n
reduces cα|[β]k to the form
1
1
cα|[β]k(ξ) = 0, k ≤ n−1, cα|[β]n(ξ) = − ξαc ε[β]n.
1 1 2 1n
1b The coefficients in front of [∂α]kf2(ξ)[∂β]lf3(ξ), k,l ≥ 2, give:
←−
λ f (ξ)[∂ ]k∂ c[α]k|[β]l(ξ)[∂ ]lf (ξ)−
γ 2 α γ 1 β 3 (20)
−δ (−1)(l−1)ε(f2)Ck−1qξγc ε[α]k−1[β]l−1λ ∂ [∂ ]k−1f (ξ)∂ [∂ ]l−1f (ξ) = 0.
k+l,n+2 n 1n σ σ α 2 σ β 3
It is easy to see that second term in eq. (20) is identically equal to zero, so from eq. (20) it
follows
c[α]k|[β]l(ξ) = c[α]k|[β]l = const, ∀ k,l ≥ 2.
1 1
For n = 2 we have c[α]2|[β]2 = c(2)ε[α]2ε[β]2. Antisymmetry condition (11) gives c(2) =
1 1 1
(2)
−c = 0. Thus one obtains for n = 2
1
[f ,f ] (ξ) = d t (ξ|f ,f )+
1 2 ∗1 S 1 1 2
←− ←− (21)
+c f (ξ)(1− 1 ∂ ξα)ε[β]2[∂ ]2f (ξ)f (ξ)[∂ ]2ε[α]2(1− 1ξβ∂ )f (ξ) .
12(cid:18) 1 2 α β 2 1 α 2 β 2 (cid:19)
In what follows n ≥ 3.
S t e p 2 The coefficients in eq. (12) in front of ∂ ∂ f (ξ) and derivatives of f (ξ)
α β 1 2
and f (ξ) of orders ≥ 2 give:
3
←− ←− ←− ←−
k λ f (ξ)∂ [∂ ]k−1cβ[γ]k−1|[δ]l −λ f (ξ)∂ [∂ ]k−1cα[γ]k−1|[δ]l [∂ ]lf (ξ)+
(cid:18) α 2 α γ 1 β 2 β γ 1 (cid:19) δ 3
←−
+lf (ξ)[∂ ]k c[γ]k|β[δ]l−1λ ∂ [∂ ]l−1f (ξ)−c[γ]k|α[δ]l−1λ ∂ [∂ ]l−1f (ξ) − (22)
2 γ (cid:18) 1 α α γ 3 1 β β γ 3 (cid:19)
←− ←−
−2(−1)k+klCk−1 λ f (ξ)∂ [∂ ]k−1cαβ|[γ]k−1[δ]l−1∂ [∂ ]l−1f (ξ) = 0, k,l ≥ 2.
k+l−2 σ 2 σ γ 1 σ δ 3
2a f2(ξ) = ξγξσ, f3(ξ) = exp(ξδpδ):
δ λ cασ|[δ]l −δ λ cβσ|[δ]l −δ λ cαγ|[δ]l +δ λ cβγ|[δ]l [p ]l =
(cid:18) γβ β 1 γα α 1 σβ β 1 σα α 1 (cid:19) δ
(23)
= l δ cγσ|β[δ]l−1 −δ cγσ|α[δ]l−1 −δ cαβ|σ[δ]l−1 +δ cαβ|γ[δ]l−1 λ p [p ]l−1.
(cid:18) αδ 1 βδ 1 γδ 1 σδ 1 (cid:19) δ δ δ
The general solution of eq. (23) is (see Appendix 1):
cαβ|[δ]l[p ]l = (δ aβ[δ]l−1 −δ aα[δ]l−1)λ p [p ]l−1, l ≥ 2, (24)
1 δ αδ 1 βδ 1 δ δ δ
where tensor aα1···αl = a[α]l is totally antisymmetric. Note that according to formula (24),
1 1
cαβ|[δ]n = 0.
1
Perform the similarity transformation with t[α]k(ξ) = (2/k)a[α]k, k ≥ 2. According to eq.
1 1
(13) one gets
←− ←−
c (ξ|f ,f ) = −f (ξ)∂ ∂ (δ aβ[δ]l−1 −δ aα[δ]l−1)λ ∂ [∂ ]l−1f (ξ)+...,
1triv 1 2 1 α β αδ 1 βδ 1 δ δ δ 2
Xl=2
9
where dots mean the terms containing the third and higher derivatives of the function f (ξ).
1
This transformation cancels the contribution to the ∗–commutator containing the terms
cαβ|[δ]l, 2 ≤ l < n, i.e., one may put
1
cαβ|[δ]l = 0, 2 ≤ l ≤ n.
1
S t e p 3 Now eq. (22) is simplified because the third term in the left hand side is
equal to zero (including the cases k +l ≥ n+2):
←− ←− ←− ←−
k λ f (ξ)∂ [∂ ]k−1cβ[γ]k−1|[δ]l −λ f (ξ)∂ [∂ ]k−1cα[γ]k−1|[δ]l [∂ ]lf (ξ)+
(cid:18) α 2 α γ 1 β 2 β γ 1 (cid:19) δ 3
(25)
←−
+lf (ξ)[∂ ]k c[γ]k|β[δ]l−1λ ∂ [∂ ]l−1f (ξ)−c[γ]k|α[δ]l−1λ ∂ [∂ ]l−1f (ξ) = 0, k,l ≥ 2.
2 γ (cid:18) 1 α α δ 3 1 β β δ 3 (cid:19)
3a The coefficients in front of [∂γ]nf2(ξ), f3(ξ) = exp(ξδpδ).
Introducing the notation c[α]k|[β]n = c[α]kε[β]n, c[α]n|[β]l = −(−1)lnc[β]lε[α]n and using the
1 1 1 1
relation
←− ←− 1 ←−
∂ [∂ ]n−1εβ[γ]n−1 = δ [∂ ]nε[γ]n,
α γ αβ γ
n
we obtain ( the first term in eq. (25) is identically equal to zero)
cα[δ]l−1λ p [p ]l−1 = cβ[δ]l−1λ p [p ]l−1. (26)
1 β β δ 1 α α δ
From eq. (26) it follows that (see Appendix 1) c[α]k = 0, k < n, i.e.
1
c[α]k|[β]n = c[α]n|[β]l = 0, k,l < n.
1 1
For k = l = n one has:
c[α]n|[β]n = ε[α]nc(n)ε[β]n, ε(c(n)) = 0, c(n) = −(−1)nc(n), (27)
1 1 1 1 1
the last equality in (27) follows from antisymmetry condition (11) and that means that the
(n)
coefficients c can be nonzero only for odd n.
1 ←−
It is useful to point out that the operator [∂ ]nε[α]nc(n)ε[β]n[∂ ]n can be also represented
α 1 β
as
←− ←− n
[∂ ]nε[α]nc(n)ε[β]n[∂ ]n = c′(n) ∂ λ ∂ , c′(n) = (−1)pn!λ ···λ c(n). (28)
α 1 β 1 (cid:18) α α α(cid:19) 1 1 n 1
3b k = l = 3 (f2(ξ) = ξα3ξα2ξα1, f3(ξ) = ξβ1ξβ2ξβ3).
δ λ cβα2α3|β1β2β3 +δ λ cβα3α1|β1β2β3 +δ λ cβα1α2|β1β2β3−
αα1 α 1 αα2 α 1 αα3 α 1
−δ λ cαα2α3|β1β2β3 −δ λ cαα3α1|β1β2β3 −δ λ cαα1α2|β1β2β3+
βα1 β 1 βα2 β 1 βα3 β 1 (29)
+δ λ cα1α2α3|ββ2β3 +δ λ cα1α2α3|ββ3β1 +δ λ cα1α2α3|ββ1β2−
αβ1 α 1 αβ2 α 1 αβ3 α 1
−δ λ cα1α2α3|αβ2β3 −δ λ cα1α2α3|αβ3β1 −δ λ cα1α2α3|αβ1β2 = 0.
ββ1 β 1 ββ2 β 1 ββ3 β 1
The general solution of eq. (29) is (see Appendix 1)
cα1α2α3|β1β2β3 = λα1λα2λα3Σ δ δ δ +δ δ δ +δ δ δ −
1 n(n−1)(n−2)(cid:18) α1β1 α2β2 α3β3 α1β3 α2β1 α3β2 α1β2 α2β3 α3β1
(30)
−δ δ δ −δ δ δ −δ δ δ
α1β2 α2β1 α3β3 α1β3 α2β2 α3β1 α1β1 α2β3 α3β2(cid:19)
10
or
←− 6Σ ←− 3
[∂ ]3c[α]3|[β]3[∂ ]3 = ∂ λ ∂ . (31)
α 1 β n(n−1)(n−2)(cid:18) α α α(cid:19)
(Compare with expression (28)).
3c We finally consider the remaining part of eq. (25). Choosing f2(ξ) = exp(qγξγ),
f (ξ) = exp(ξδp ), q qβ +qβq = p q +p q = q ξβ +ξβq = 0, we have:
3 δ α α α β β α α α
k λ q [q ]k−1cβ[γ]k−1|[δ]l −λ q [q ]k−1cα[γ]k−1|[δ]l [p ]l+
(cid:18) α α γ 1 β β γ 1 (cid:19) δ
(32)
+l[q ]k c[γ]k|β[δ]l−1λ p [p ]l−1 −c[γ]k|α[δ]l−1λ p [p ]l−1 = 0, 3 ≤ k,l < n, k +l ≥ 7.
γ (cid:18) 1 α α δ 1 β β δ (cid:19)
It follows from eq. (32) that (see Appendix 1):
c[α]k|[β]l = 0, k 6= l, k +l 6= n,
1
c[α]k|[β]n−k = c(k)ε[α]k[β]n−k, k,n−k ≥ 4, 2k 6= n, (33)
1 1
(k) (k) (n−k)
ε(c ) = n(mod 2), c = −c .
1 1 1
(k)
Since c are numbers they can be nonzero only for even n.
1
S t e p 4 The coefficients in eq. (12) in front of ∂ ∂ ∂ f (ξ) and derivatives of
α1 α2 α3 1
f (ξ) and f (ξ) of orders ≥ 3.
2 3
The first three terms in eq. (12) are equal to zero. The other three give an equation of
the following general structure:
c[β]3|[γ]k+l−2[∂]kf [∂]lf +c[β]k|[γ]l+1[∂]kf [∂]lf +c[β]k+1|[γ]l[∂]kf [∂]lf = 0. (34)
1 2 3 1 2 3 1 2 3
4a k = l−1, 4 ≤ l < n, 2l 6= n, f2(ξ) = exp(qβξβ).
In this case, the first two terms in eq. (34) are equal to zero and we obtain:
(δ cα2α3[β]l−2|[γ]l +δ cα3α1[β]l−2|[γ]l +δ cα1α2[β]l−2|[γ]l)λ q [q ]l−2 = 0. (35)
α1β 1 α2β 1 α3β 1 β β β
The general solution of this equation is (see Appendix 1)
c[β]l|[γ]l = 0.
1
Thus, the structure of all coefficients c[α]k|[β]l is determined for odd n.
1
4b n = 2m is even, k = m−1, l = m, f2(ξ) = exp(qβξβ), f3(ξ) = exp(ξγpγ).
Eq. (34) gives (with already established structure of the coefficients c[α]k|[β]2m−k, k 6= m,
1
see formulas (33)):
3Cm−2 c(3)εα1α2α3[β]m−2[γ]m−1λ q [q ]m−2p [p ]m−1+
2m−3 1 δ δ β δ γ
+(−1)mCm2+1c1(m−1)(cid:18)εα1α2[β]m−1[γ]m−1δα3γ +cycle(1,2,3)(cid:19)[qβ]m−1λγpγ[pγ]m−1− (36)
−C2 cα1α2[β]m−2|[γ]mδ +cycle(1,2,3) λ q [q ]m−2[p ]m = 0.
m(cid:18) 1 α3β (cid:19) β β β γ
It follows from eq. (36) (see Appendix 1)
c[α]m|[β]m = c(m)ε[α]m[β]m.
1 1