A geometrical point of view on linearized beta-deformations
AA geometrical point of view on linearizedbeta-deformations
Andrei Mikhailov † and Segundo P. Mili´an Instituto de F´ısica Te´orica, Universidade Estadual PaulistaR. Dr. Bento Teobaldo Ferraz 271, Bloco II – Barra FundaCEP:01140-070 – S˜ao Paulo, Brasil
Abstract
It is known that the supermultiplet of beta-deformations of N = 4supersymmetric Yang-Mills theory can be described in terms of theexterior product of two adjoint representations of the superconformalalgebra. We present a super-geometrical interpretation of this fact,by evaluating the deforming operator on some special coherent statesin the space of supersingletons. We also discuss generalization of thisapproach to other finite-dimensional deformations of the N = 4 su-persymmetric Yang-Mills theory. † on leave from Institute for Theoretical and Experimental Physics,ul. Bol. Cheremushkinskaya, 25, Moscow 117259, Russia a r X i v : . [ h e p - t h ] N ov ontents GL (4 | , R ) transformations on twistors . . . . . . . 82.4 Solutions of free classical super-Maxwell equations from twistorspace . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 δ ξZ (1) ,Z (2) ( Z ) . . . . . . . . . . . . . . . . . . . . . 144.2 Orientation of L rd and discontinuity of δ ξZ (1) ,Z (2) . . . . . . . . . 154.3 Symmetry of δ ξZ (1) ,Z (2) ( Z ) . . . . . . . . . . . . . . . . . . . . . 164.4 The case when ξ ∧ ξ corresponds to zero deformation . . . . . 174.5 Deformation of the form (cid:82) φψψ . . . . . . . . . . . . . . . . . 19 AdS/CFT correspondence has been under vigorous study for more than 20years, bringing important results. Historically the first application was tocompare the BPS (or “protected”) states of the N = 4 supersymmetric Yang-Mills theory to the solutions of Type IIB supergravity (SUGRA) linearizednear the “background solution” AdS × S . Both are representations of Liesuper-algebra psu (2 , | .Understanding, from various points of view, the structure of the representa-tions appearing in AdS/CFT would be useful. In particular, it is needed forconstructing the string worldsheet massless vertex operators, along the linesof [3, 4, 5, 6]. (The observation of [3] was that, to construct the vertex, it isenough to know the structure of the representation of psu (2 , |
4) in whichthe corresponding state transforms.) And an efficient construction of mass-less vertex operators would, in turn, allow the computation of the SUGRAscattering amplitudes on
AdS × S in the pure spinor formalism [7].In this paper we will suggest a point of view on the finite-dimensional represenations. The idea is to probe them by evaluating the correspondingdeformations of the free Yang-Mills action on certain solutions of free fieldequations. The result is some generalized function of the parameters of thosefree solutions. This generalized function is manifestly supersymmetric andhas a nice super-geometrical interpretation. We discuss mostly the case ofthe so-called beta-deformation [8]. The structure of the corresponding rep-resentation is known from the supergravity analysis [9]. Here we make thisstructure transparent on the field theory side. Then we apply our method toother finite-dimensional representations. Under the assumption (unproven)that they are irreducible, we identify them with particular Young diagramms.We will now proceed to describing our results. Deformations transforming in finite-dimensional representations
Con-sider single trace deformations of N = 4 supersymmetric Yang-Mills theory(SYM): S SYM (cid:55)→ S SYM + ε (cid:90) d xU (1)where U is a single-trace operator. According to AdS/CFT [1], they corre-spond to some deformations of the Type IIB superstring theory on AdS × S .Let us restrict ourselves to infinitesimal deformations, i.e. compute only tothe first order in the deformation parameter ε . Besides ε , there are two other finite dimensional representations of Lie superalgebras are not necessarily semisimple Beta-deformations were discovered in [8], but their structure as representations of psu (2 , |
4) was not studied. It was computed using AdS/CFT in [9]. g Y M and the number of colors N . In this paper, we will consider N being very large. As for the g Y M , thereare two opposite limits: the limit g Y M = 0 of free N = 4 super Yang-Millstheory (SYM) and the strong coupling limit. In the weak coupling limit wecan do perturbative calculations in SYM, and in the strong coupling limitwe can use the superstring theory on the classical supergravity (SUGRA)background AdS × S .Both SYM theory and superstring on AdS × S are invariant underthe superconformal group P SU (2 , | U of Eq.(1) of the form [4]: (cid:90) d x tr(Φ + i Φ ) n +4 + c.c. (2)where Φ , . . . , Φ are the scalar fields of the N = 4-super-Yang-Mills theory.Consider the linear space of all deformations obtained from Eq. (2) by actingwith all possible polynomials of generators of psu (2 , | E free n +4 . E n +4 Most of finite-dimensional deformations of the free theory combine intoinfinite-dimensional representations in the quantum interacting theory. But E free n +4 stays finite-dimensional. We will call it just E n +4 . This follows fromthe finiteness of the operator tr(Φ + i Φ ) n +4 and its descendants. Finite-ness implies that the conformal transformations of this operators are samein quantum theory as in classical theory (no countertems ⇒ no anomalousdimension). Therefore, the subspace generated by acting on Eq. (2) withthe conformal generators is the same in classical and quantum theory. Inparticular, it is finite-dimensional. To generate the full representation, it re-mains to act with supersymmetry and superconformal generators. However,those are nilpotent modulo elements of su (2 , ⊕ su (4), and therefore cannotchange the property of the representation being finite-dimensional.We will now argue that E n +4 is isomorphic to the representation E free n +4 which is generated by acting with psu (2 , |
4) on Eq. (2) in the free N = 4super-Yang-Mills theory. Indeed, let us consider the chiral primary represen-tation , which is defined as the infinite-dimensional representation generatedby the local operator tr(Φ + i Φ ) n +4 (0) by acting on it with psu (2 , | F n +4 . Similarly, there is a F n +4 F free n +4 which is defined in the same way as F n +4 , but in the freetheory. We start by pointing out that F n +4 is isomorphic to F free n +4 , becauseboth are highest weight unitary irreducible representations of psu (2 , |
4) withthe same highest weight tr(Φ + i Φ ) n +4 (0). On the other hand, our finite-dimensional representations E n +4 and E free n +4 are fully determined by F n +4 and F free n +4 , respectively, in the following way. Elements of E n +4 are of the form (cid:82) d x ρ ( x ) O ( x ) where O is an element of F n +4 (a local operator), but in-serted at the point x instead of 0. When we act on (cid:82) d x ρ ( x ) O ( x ) with anelement of psu (2 , | O ( x ), i.e. by the structure of F n +4 . The isomorphism F n +4 (cid:39) −→ F free n +4 impliesa map O (cid:55)→ O free commuting with the action of psu (2 , | E n +4 (cid:39) −→ E free n +4 .Still, it is not clear to us if E n +4 is irreducible or not. If it is irreducible,then considerations of [4] and Section 5 of this paper suggest that it shouldcorrespond to the supersymmetric Young diagramms (see Section 5 and [4]and references there): (3) Beta-deformation
In this letter we will mostly consider a particular ex-ample of a finite-dimensional representation: the linearized β -deformation[8, 9]. In some sense it corresponds to n = −
1, but the construction isdifferent. If we literally take n = − psu (2 , | (cid:90) d x (cid:15) αβ Q [ − , , ] α Q [ − , , ] β tr(Φ + i Φ ) + c.c. (4)where Q [ − , , ] α are supersymmetry generators with weight (cid:2) − , , (cid:3) underthe elements Φ ∂∂ Φ − Φ ∂∂ Φ , Φ ∂∂ Φ − Φ ∂∂ Φ and Φ ∂∂ Φ − Φ ∂∂ Φ of the su (4)Cartan subalgebra. This deformation is preserved by Q [ − , , ] α and Q [ , − , − ]˙ α .It starts with the cubic term, but also has quartic terms. In the language4f N = 1 superspace, the cubic terms come from δW (cid:48)(cid:48) ψψ and quartic termscome from | ( W + δW ) (cid:48) | terms, where W is the superpotential of the N = 4Yang-Mills in the N = 1 notations, and δW its Leigh-Strassler deformation.Notice that the quartic terms are proportional to g YM . In the free fieldlimit the quartic terms are absent, but the cubic terms survive and define adeformation of the free Yang-Mills theory.It was found on the AdS side that the supermultiplet of β -deformationsis related to the wedge product of two copies of the adjoint representation of g = psu (2 , |
4) [9]: ( g ∧ g ) g (5)This is an irreducible representation. The subscript 0 imposes the constraintof zero internal commutator; the origin of this constraing on the AdS sidewas explained in [10, 11, 12]. The geometrical origin of g ∧ g is immediatelyvisible on the AdS side; the corresponding vertex operator is essentially thewedge product of two global symmetry currents on the string worldsheet. Butit is not immediately obvious why such structure would turn up on the fieldtheory side. In the free field limit the deforming operator U is some quibicexpression in elementary fields. It was shown in [13] that these expressionstransform in the representation given by Eq. (5). In this letter we will presenta geometrical construction making this result more transparent. We will usetwistor methods. First in Section 2 we will review the formalism of Harmonicsuperspace, and how it describes classical solutions of free theory. We thenintroduce in Section 3 some special coherent states, which are δ -functionswith support on α -planes. (The ideas of Sections 2 and 3 are well known toexperts, but we could not find a reference with explicit formulas suitable forour needs.) In Section 4 we consider the evaluation of the beta-deformationterm (cid:82) d x U in Eq. (1) on the formal sum of coherent states, also knownas “perturbiner”. For our approach it is essential to work in space-timesignature 2 + 2. In this case the evaluation gives a well-defined generalizedfunction of the parameters of our coherent states. We explicitly describe thisgeneralized function and study its properties. In Section 5 we conjecture asimilar description for the deformations of the form Eq. (2) with n ≥ [16, 17, 18]. Notice that twistor string We would like to thank the Referee for pointing out to us these references CP | , while for our considerationshere real twistors RP | are enough ( cp. Section 3.2 of [14]).Our approach can be interpreted as classifying the possible deformations ofthe action by looking at their effect on the scattering amplitudes. In fact,Eqs. (53) and (91) can be interpreted as the deformation of the k -pointscattering amplitude at the minimal value of k for which the deformation isnonzero. Conceptually similar approach was used in [19] for classification ofcounterterms in supergravity. We will work with the space-time signature 2 + 2. With this signature thesuperconformal group is
P SL (4 | , R ). Following [20] we will think aboutthe four-dimensional N = 4 superspace as a Grassmannian manifold M =Gr(2 | , R | ), which parametrizes subspaces R | ⊂ R | .We will consider the twistor space T = R | ; elements of T are vectorsparametrized by even λ ˙ α , µ α and odd ζ I : T λ, µZ = ( µ , µ , λ ˙1 , λ ˙2 , ζ , ζ , ζ , ζ ) T (6)Here ( . . . ) T means transposition; we think of Z as a “column-vector”, butwrite it as a row in Eq. (6) for typographic reasons. Points of M are 2 | L ⊂ T . They can be parametrized by coordinates x, θ, ˜ θ, u : basis of L ⊥ ⊂ T (cid:48) : ( 1 0 x x θ θ θ θ )( 0 1 x x θ θ θ θ )( ˜ θ ˜ θ ˜ θ ˙11 ˜ θ ˙21 u u u u )( ˜ θ ˜ θ ˜ θ ˙12 ˜ θ ˙22 u u u u ) (7)In these coordinates the condition Z ∈ L is that each row-vector from the setof four vectors (7) has zero scalar product with the column-vector Z . Thesecoordinates are redundant. The redundancy can be described as some vector although with our choice of signature we cannot directly use complex geometry; wewill replace complex analyticity with some polynomiality condition. L , for example: A iα (cid:18) ∂∂ ˜ θ iα + x α ˙1 ∂∂ ˜ θ ˙1 i + x α ˙2 ∂∂ ˜ θ ˙2 i + θ αI ∂∂u Ii (cid:19) (8) A KL (cid:32) ˜ θ KA ∂∂ ˜ θ LA + ˜ θ ˙ AK ∂∂ ˜ θ ˙ AL + u IK ∂∂u IL (cid:33) (9)for any constant A iα and A KL . This can be used to put ˜ θ iα = 0:˜ θ = ˜ θ = ˜ θ = ˜ θ = 0 (10)To summarize: twistors = T = R | (11) space-time = M = Gr(2 | , T ) (12)A special role is played by the 2 | M which will be denoted S M . The fiber of S M over L ⊂ M is L itself. S M Also, PT will denote the projective super-twistor space, i.e. Z modulorescaling: Z (cid:39) κZ . PT For any vector bundle V , we define two line bundles, which we call Ber V and | Ber | V . Sections of Ber V are functions of the bases of the fiber V , Ber | Ber | satisfying the property: σ ( { m JI e J } ) = SDet m σ ( { e I } ) (13)where for a supermatrix m = (cid:18) A BC D (cid:19) :SDet m = det( A − BD − C ) (det D ) − (14)Similarly, section of | Ber | V satisfy: σ ( { m JI e J } ) = sign(det A ) SDet m σ ( { e I } ) (15)7he characteristic property of | Ber | V is that a section of | Ber | V defines anoperation of integration along the fiber : functions on V with compact supporton each fiber V (cid:82) −→ (cid:20) functionson the base (cid:21) (16)It seems logical to call sections of Ber V “integral forms on the fiber”, andsection of | Ber | V “volume elements on the fiber”.We will denote (Ber V ) − the line bundle dual to Ber V , i.e. same as(Ber V ) (cid:48) , and | Ber V| − = | Ber V| (cid:48) . Ber − GL (4 | , R ) transformations on twistors The twistor vector space T should be considered a fundamental representa-tion of GL (4 | , R ). We have: GL (4 | / center = P GL (4 |
4) =
P SL (4 | , R ) (cid:111) U (1) R (17)The u (1) R acts as follows: RZ = (cid:18) − × × × × (cid:19) Z (18)In terms of the parametrization (6): R = − λ ˙ α ∂∂λ ˙ α − µ α ∂∂µ α + ζ I ∂∂ζ I (19) In the free field limit a classical solution of the super-Yang-Mills theory withgauge group U ( N ) can be obtained as an N × N matrix whose entries areclassical solutions of the super-Maxwell equations. For our purpose it isenough to consider free solutions of the form: (cid:20) solution ofsuper-Maxwell equations (cid:21) × (cid:20) N × N matrix (cid:21) (20)8n this Section we will explain the correspondence between solutions of freeclassical super-Maxwell equations on M and sections of ( | Ber | S M ) − whichare polynomial in the coordinates u Ii defined in Eq. (7). The choice of coordi-nates u Ii is somewhat arbitrary, but the statement of polynomial dependencedoes not depend on it; in particular this requirement is P SL (4 | , R ) (cid:111) U (1) R -invariant.Notice that | Ber |S M = ( | Ber | ( T / S M )) − = | Ber | ( T / S M ) (cid:48) = | Ber |S ⊥ M . Thosefour vectors listed in Eq. (7) are the basis of S ⊥ M .Suppose that we are given a section (here Γ( line bundle ) denotes the spaceof sections): σ ∈ Γ (cid:0) ( | Ber | S M ) − (cid:1) (21)which is a polynomial in u . The condition that σ is annihilated by thevector fields defined in Eq. (9) with A KL satisfying A KK = 0 implies that itcan depend on ˜ θ and u only through the expressions ˜ θ ˙ α [1 u I , u I [1 u J and ˜ θ ˙ α [1 ˜ θ ˙ β .Since we require σ to be a polynomial in u , it must be then a polynomial ofthese three expressions. Moreover, σ should have charge 2 under rescaling of(˜ θ, u ), therefore this polynomial is actually a linear function: σ = ˜ θ ˙ α [1 ˜ θ ˙ β F +˙ α ˙ β ( x, θ ) + ˜ θ ˙ α [1 u I Ψ ˙ αI ( x, θ ) + u I [1 u J Φ IJ ( x, θ ) (22)Consider the condition that σ does not change when we add to first andsecond vectors of the basis of Eq. (7) a linear combination of the third andfourth vectors. It implies that σ should be annihilated by the following vectorfields V iα : V iα σ = 0 (23) where V iα = ˜ θ ˙ βi ∂∂x α ˙ β + u Ii ∂∂θ αI (24)In the rest of this Section, we will first prove that Eq. (23) implies the9ollowing equations: (cid:15) ˙ γ ˙ β ∂∂x α ˙ γ F +˙ α ˙ β ( x, θ ) = 0 , (25) ∂∂θ αI F +˙ α ˙ β ( x, θ ) = ∂∂x α ( ˙ β Ψ ˙ α ) I ( x, θ ) , (26) (cid:15) ˙ α ˙ β ∂∂x α ˙ α Ψ ˙ βI ( x, θ ) = 0 , (27) ∂∂θ α ( I Ψ J ) ˙ α ( x, θ ) = 0 , (28) ∂∂θ α [ I Ψ J ] ˙ α ( x, θ ) = ∂∂x α ˙ α Φ IJ ( x, θ ) , (29) ∂∂θ αP Φ IJ ( x, θ ) = ∂∂θ α [ P Φ IJ ] ( x, θ ) , (30)and then show that Eqs. (25)- (30) imply that the θ -expansion of F +˙ α ˙ β ( x, θ ) , Ψ J ˙ α ( x, θ )and Φ IJ ( x, θ ) can be expressed through solutions of free super-Maxwell the-ory. Proof that Eq. (23) implies Eqns. (25) - (30) Consider the expansionof the left hand side of Eq. (23) in powers of ˜ θ . • At the zeroth order in ˜ θ , the term that contributes is u Pm u I [1 u J ∂∂θ αP Φ IJ ( x, θ ) . (31)Its vanishing implies Eq. (30) • At the linear order in ˜ θ , we have − u Im ˜ θ ˙ α [1 u J ∂∂θ αI Ψ ˙ αJ ( x, θ ) + ˜ θ ˙ βm u I [1 u J ∂ α ˙ β Φ IJ ( x, θ ) , (32)where ∂ α ˙ β denotes ∂∂x α ˙ β . Notice: u Ii u Jj = (cid:15) ij u [ I [1 u J ]2] + u ( I ( i u J ) j ) , (33)Therefore the terms of Eq. (32) can be written as: u I [1 u J ˜ θ ˙ αm (cid:18) − ∂∂θ α [ I Ψ J ] ˙ α ( x, θ ) + ∂ α ˙ α Φ IJ ( x, θ ) (cid:19) − u I (1 u J ˜ θ ˙ αm ∂∂θ α ( I Ψ J ) ˙ α ( x, θ ) . (34)The vanishing of this expression implies Eqs. (29) and (28).10 The terms of the second order in ˜ θ are u Im ˜ θ ˙ α [1 ˜ θ ˙ β ∂∂θ αI F +˙ α ˙ β ( x, θ ) + ˜ θ ˙ βm ˜ θ ˙ α [1 u I ∂ α ˙ β Ψ ˙ αI ( x, θ ) , (35)using ˜ θ ˙ βm ˜ θ ˙ αk = (cid:15) ˙ β ˙ α ˜ θ [1( m ˜ θ k ) + (cid:15) mk ˜ θ ( ˙ β [1 ˜ θ ˙ α )2] , (36)Eqn. (35) can be written as u Im ˜ θ ˙ α [1 ˜ θ ˙ β (cid:18) ∂∂θ αI F +˙ α ˙ β ( x, θ ) − ∂ α ( ˙ β Ψ ˙ α ) I (cid:19) + ˜ θ [1(1 ˜ θ u Im (cid:16) (cid:15) ˙ β ˙ α ∂ α ˙ β Ψ ˙ αI ( x, θ ) (cid:17) , (37)This is only zero when both terms vanish separately, implying Eqs.(26) and (27). • At third order in ˜ θ , the term that contributes is: T ˙ γ ˙ β ˙ α ˜ θ ˙ γm ˜ θ ˙ β [1 ˜ θ ˙ α , (38)where T ˙ γ ˙ β ˙ α = ∂ α ˙ γ F +˙ α ˙ β ( x, θ ). The vanishing of this expression impliesthat T ˙ γ ˙ β ˙ α must be totally symmetric, and therefore Eq. (25).This concludes the derivation of the constraints (25)- (30) on the superfields F +˙ α ˙ β ( x, θ ) , Ψ J ˙ α ( x, θ ) and Φ IJ ( x, θ ). Solution to Eqns. (25) - (30) Now we will prove that the componentsof the θ -expansion of F +˙ α ˙ β ( x, θ ) , Ψ J ˙ α ( x, θ ) and Φ IJ ( x, θ ) can be expressedthrough solutions of free super-Maxwell theory.Eq. (30) implies: ∂ γK ∂ βL ∂ Mα Φ IJ ( x, θ ) = ∂ γ [ K ∂ | β | L ∂ M | α | Φ IJ ] ( x, θ ) = 0 , (39)which implies that the expansion of Φ IJ ( x, θ ) in powers of θ terminates atthe second order: Φ IJ ( x, θ ) = ϕ IJ ( x ) + (cid:15) IJP Q θ αP ˜ ψ Qα ( x ) + 12 (cid:15) IJP Q θ αP θ βQ ˜ f αβ ( x ) . (40)11qns. (28), (29) determine Ψ ˙ αJ ( x, θ ). The solution only exists when ∂ [ α | ˙ α ˜ ψ Q | β ] = 0 and ∂ [ α | ˙ α ˜ f | β ] γ = 0 . (41)which are Dirac and Maxwell equations. It is given by the following expres-sion: Ψ J ˙ α ( x, θ ) = ψ J ˙ α ( x ) + θ αI ∂ α ˙ α ϕ JI ( x ) + 12 (cid:15) JIP Q θ αI θ βP ∂ ( α | ˙ α ˜ ψ Q | β ) ( x )+ 13! (cid:15) JIP Q θ αI θ βP θ γQ ∂ ( α | ˙ α ˜ f | βγ ) ( x ) , (42)Eq. (26) determines F +˙ α ˙ β ( x, θ ), but the solution only exists when besidesEqs. (41) also: (cid:3) ϕ IJ = 0 , (43)It is given by: F +˙ α ˙ β ( x, θ ) = f ˙ α ˙ β ( x ) + θ Jρ ∂ ρ ˙ β ψ J ˙ α ( x ) + 12 θ Jρ θ αI ∂ ( ρ | ˙ β ∂ | α ) ˙ α ϕ IJ ( x )+ 13! (cid:15) JIP Q θ Jρ θ αI θ βP ∂ ( ρ | ˙ β | ∂ α | ˙ α | ˜ ψ Qβ ) ( x )+ 14! (cid:15) JIP Q θ Jρ θ αI θ βP θ γQ ∂ ( ρ | ˙ β | ∂ α | ˙ α | ˜ f βγ ) ( x ) . (44)To conclude, σ is given by Eqn. (22) where F +˙ α ˙ β ( x, θ ) , Ψ ˙ βJ ( x, θ ) and Φ IJ ( x, θ )are given by Eqns. (44), (42) and (40), respectively. We will now discuss a special family of solutions of free equations, whichform an orbit of
P SL (4 | Z ∈ T we define a generalized function section of ( | Ber | S M ) − (see Eq. (21)), which we call δ Z . By definition δ Z ( X ), for a fixed 2 | X ⊂ T , is the delta-function of the condition that Z ∈ X . Moreprecisely (here Γ( line bundle ) denotes the space of sections): δ Z ( X )12 Z ∈ Γ ( | Ber | S M ) − such that ∀ s ∈ Γ ( | Ber | S M ) : (cid:90) Z ∈ T f ( Z ) (cid:104) s ( X ) , δ Z ( X ) (cid:105) = (cid:90) Z ∈ X s ( X ) f ( Z ) , (45)Here the integration on the left hand side uses the canonical measure on T ,and the integration measure on the right hand side is the integration alongthe fiber of S M defined by the volume element s (Section 2.2). In coordinates(6), (7): δ Z ( X ) = ( u I ζ I + ˜ θ ˙ α λ ˙ α )( u J ζ J + ˜ θ ˙ α λ ˙ α ) ×× δ ( µ + x α λ ˙ α + θ I ζ I ) δ ( µ + x α λ ˙ α + θ I ζ I ) (46)So defined δ Z ( X ), as a function of X ∈ M (or rather a section of the linebundle ( | Ber | S ) − over M ), encodes a solution of N = 4 super-Maxwellequations of motion as described in Section 2.To determine the corresponding scalar, electromagnetic and spinor fields,we write δ Z ( X ) as a sum of expressions with definite values of the R charge,which corresponds to twice the power of ζ :˜ θ ˙ α ˜ θ ˙ β f ˙ α ˙ β : ˜ θ ˙ α ˜ θ ˙ β λ ˙ α λ ˙ β δ ( µ + xλ ) (47) u J [1 ˜ θ ˙ α ψ ˙ αJ : ˜ θ ˙ α [1 λ ˙ α u J ζ J δ ( µ + xλ ) (48) u I u J ϕ IJ : u I u J ζ I ζ J δ ( µ + xλ ) (49) u [ I u J θ K ] α ˜ ψ IJKα : u [ I u J θ K ] α ζ I ζ J ζ K ∂ α δ ( µ + xλ ) (50) u [ I u J θ Kα θ L ] β ˜ f αβ : u [ I u J θ Kα θ L ] β ζ I ζ J ζ K ζ L ∂ α ∂ β δ ( µ + xλ ) (51) We want to characterize a deforming operator (cid:82) d x U in Eq. (1) by itsvalue on free solutions. In particular, in this paper we are interested inthe case of beta-deformation, which corresponds to deforming the actionby an expression cubic in the elementary fields [13]. Let us evaluate thecorresponding U on the formal sum of three coherent states: (cid:16) ε δ Z (1) ( X ) + ε δ Z (2) ( X ) + ε δ Z (3) ( X ) (cid:17) M (52)13ith bosonic nilpotent coefficients ε , ε , ε , where M is some N × N -matrix(see Eq. (20)). As we explained in Section 1, linearized beta-deformationsare parametrized by B ∈ ( g ∧ g ) / g . We consider the case when B is adecomposable tensor, i.e. is of the form B = ξ ∧ ξ where ξ is an odd element ξ ∧ ξ of g . The condition of zero internal commutator (the subindex 0 in ( g ∧ g ) )means that ξ should be nilpotent. We claim that the result of evaluation of (cid:82) d x U on the free solution given by Eq. (52) is: (cid:90) d x U = ε ε ε δ ξZ (1) ,Z (2) ( Z (3) ) tr M (53)where δ ξZ (1) ,Z (2) ( Z (3) ) is defined by Eq. (57). In fact, our Eq. (57) works forany nilpotent ξ ∈ Π T e P SL (4 | , R ); this includes odd elements of psl (4 | , R )as well as linear combinations of even elements of psl (4 | , R ) with Grassmannodd coefficients .(At the same time, on the string theory side the integrated vertex operatorof the worldsheet sigma-model for this particular B is [9]: U AdS = (cid:90) j a ξ a ∧ j b ξ b (54)Here j a is the Noether current on the worldsheet, a one-form.) δ ξZ (1) ,Z (2) ( Z ) Let us pick two points in PT : [ Z (1) ] and [ Z (2) ]. Let ξ ∈ g be an oddnilpotent element of g = psl (4 | , R ). We can represent ξ as a 4 | × | “Nilpotent” means satisfying ε = ε = ε = 0. To the best of our knowledge, the ideato use nilpotent coefficient was first suggested in [21]. The authors of [21] constructed asolution of classical field equations, which they called “perturbiner”. Our use of nilpotentcoefficients is slightly different; we take a solution of the free field equations in the form ofEq. (52) and evaluate some (cid:82) d x U on it. We need the nilpotence of coefficients to avoidconsidering the square of delta-function. Since we work with coherent states, we are into supergeometry/supermanifolds; wehave to use a “pool” of constant Grassmann odd parameters [22]. Strictly speaking, it is not appropriate to think about “points of a supermanifold”. Itis better to say that, for an arbitrary supermanifold S , we pick two arbitary morphisms Z (1) : S → PT and Z (2) : S → PT [23]. For our purposes, it is enough to take S = R | K for large enough K . Technically, we just allow all our twistors to depend on K constantGrassmann odd parameters. ξ . Since ξ is nilpotent as an element of psl (4 | , R ), the square of this matrix is proportional to the unit matrix: ξ = c (55)where c is some number. We assume that ξ is nondegenerate, in the followingsense: either c (cid:54) = 0, or if c = 0 then Ker ξ = Im ξ . Let us consider a 2 | L generated by Z (1) , Z (2) , ξZ (1) and ξZ (2) : L = R Z (1) + R Z (2) + R ξZ (1) + R ξZ (2) (56)Let us define δ ξZ (1) ,Z (2) ( Z ), essentially a delta-function of Z , in the followingway. For any test function f ∈ C ∞ ( T ): δ ξZ (1) ,Z (2) (cid:90) Z ∈ R | δ ξZ (1) ,Z (2) ( Z ) f ( Z ) == (cid:90) R da ∧ da ∂∂ψ ∂∂ψ f ( a Z (1) + a Z (2) + ψ ξZ (1) + ψ ξZ (2) ) (57)The integration on the left hand side uses the canonical measure on R | (the“ S ” in “ P SL (4 | L canonically defined by ξ as described in Appendix A.Eq. (57) implies that δ ξZ (1) ,Z (2) ( Z ) is a linear function of ξ ⊗ ξ . It hasweight zero in Z and also in both Z (1) and Z (2) ; in other words, for κ ∈ R : δ ξZ (1) ,Z (2) ( κZ ) = δ ξκZ (1) ,Z (2) ( Z ) = δ ξZ (1) ,κZ (2) ( Z ) = δ ξZ (1) ,Z (2) ( Z ) (58)Equivalently, we can define δ ξZ (1) ,Z (2) ( Z ) as follows: δ ξZ (1) ,Z (2) ( Z ) = (59)= (cid:90) R da ∧ da ∂∂ψ ∂∂ψ δ (4 | ( Z − a Z (1) − a Z (2) − ψ ξZ (1) − ψ ξZ (2) ) L rd and discontinuity of δ ξZ (1) ,Z (2) In order to integrate an integral form: da ∧ da ∂∂ψ ∂∂ψ f ( a Z (1) + a Z (2) + ψ ξZ (1) + ψ ξZ (2) ) (60)15ver a supermanifold L (Eq. (56)), we need an orientation of its body L rd = R . This R is generated by Z (1) and Z (2) . Therefore, we need toknow which basis has positive orientation: { Z (1) , Z (2) } or { Z (2) , Z (1) } ? Acomparison with explicit calculation in Section 4.5 shows that the orienta-tion is determined by the sign of (cid:15) ˙ α ˙ β λ (1) ˙ α λ (2) ˙ β . This sounds like a contradic-tion, because (cid:15) ˙ α ˙ β λ (1) ˙ α λ (2) ˙ β is not conformally invariant. However, the sign of (cid:15) ˙ α ˙ β λ (1) ˙ α λ (2) ˙ β is invariant under infinitesimal conformal transformations.Explicit calculation in Section 4.5 shows that δ ξZ (1) ,Z (2) ( Z ), as a function of Z (1) and Z (2) , has a discontinuity when (cid:15) ˙ α ˙ β λ (1) ˙ α λ (2) ˙ β = 0. The limit at thediscontinuity is proportional to δ (2) ( λ − a λ (1) − a λ (2) ), but the sign in frontof δ (2) ( λ − a λ (1) − a λ (2) ) depends on whether we are approaching from (cid:15) ˙ α ˙ β λ (1) ˙ α λ (2) ˙ β > (cid:15) ˙ α ˙ β λ (1) ˙ α λ (2) ˙ β < R , ∪ ∞ = Gr(2 ,
4) (Section 2). For theYang-Mills action to be conformally invariant, the scalar field should trans-form as a section of (Ber S ) − , and spinors as sections of S ⊗ (Ber S ) − (chiral)and S ⊥ ⊗ (Ber S ) − (antichiral), where S is the tautological vector bundle ofGr(2 ,
4) (like S M of Section 2.1 but without “super-”). However, the δ Z ofSection 3 gives sections of ( . . . ) ⊗ | Ber | − instead of ( . . . ) ⊗ Ber − (Section2.2). We would be able to cast sections of | Ber | − as sections of Ber − , if weprovided an orientation of S . But this is only possible locally. In fact, wecan provide orientation of the fiber of S over all x except at infinity. This isacceptable for our computation, because the integral (cid:82) d x U in Eq. (1) issupported on a single point x ∈ R , determined by the intersection of theplanes µ (1) − λ (1) x = 0 and µ (2) − λ (2) x = 0. But when we consider the casewhen λ (1) → λ (2) , the intersection point x goes to ∞ , where the orientationis undefined, essentially because || x || = det x can be positive or negativedepending on how x → ∞ . This leads to the discontinuity of δ ξZ (1) ,Z (2) ( Z ). δ ξZ (1) ,Z (2) ( Z ) The defining Eq. (57) implies that δ ξZ (1) ,Z (2) ( Z ) is symmetric under the ex-change Z (1) ↔ Z (2) . In fact it is symmetric under arbitrary permutations of16 , Z (1) , Z (2) . This can be proven as follows: δ (4 | ( Z − a Z (1) − a Z (2) − ψ ξZ (1) − ψ ξZ (2) ) = (61)= sign( a ) δ (4 | (cid:18) a + ψ ξ Z − Z (1) − a a + ψ ξ Z (2) − a + ψ ξ ψ ξZ (2) (cid:19) == sign( a ) × (62) × δ (4 | (cid:16) − Z (1) + a − Z − ( a − a − a − ψ ψ c ) Z (2) − a − ψ ξZ − (63) − ( a − ψ − a − a ψ ) ξZ (2) (cid:17) The change of variables: ˜ a = a − (64)˜ a = − a − a + a − ψ ψ c (65)˜ ψ = − a − ψ (66)˜ ψ = a − a ψ − a − ψ (67)in the integral (cid:82) R da ∧ da ∂∂ψ ∂∂ψ is equivalent to the exchange Z ↔ Z . ξ ∧ ξ corresponds to zero deforma-tion Suppose that ξ ∧ ξ belongs to the denominator of Eq. (5): ξ ∧ ξ = ∆( ζ ) = (cid:88) C ab t a ∧ [ t b , ζ ] (68)for some ζ ∈ psl ( | ), and C ab the inverse of the Killing metric. We knowfrom [9] that such ξ ∧ ξ should correspond to zero deformations. We will nowprove that for such ξ Eq. (57) indeed evaluates to zero.Let us enumerate the basis vectors of T : { e , e , e , e , f , f , f , f } where e are even and f are odd. Without loss of generality, let us assume that Z (1) = e and Z (2) = e . (In other words, we choose a basis so that the firsttwo elements are Z (1) and Z (2) .) The sign factor sign( a ) is compensated by the possible difference of sign between (cid:15) ˙ α ˙ β λ (1) ˙ α λ (2) ˙ β and (cid:15) ˙ α ˙ β λ ˙ α λ (2) ˙ β . This sign determines the orientation of R which is neededto integrate da ∧ da . ζ is even: ζ = e a ⊗ e ∨ b − δ ba (69)where e ∨ , f ∨ are elements of the dual basis and = (cid:80) c e c ⊗ e ∨ c . In this case:∆( ζ ) = (cid:88) c =1 (cid:18) e a ⊗ e ∨ c − δ ca (cid:19) ∧ (cid:18) e c ⊗ e ∨ b − δ bc (cid:19) + odd ∧ odd (70)The odd ∧ odd terms do not contribute to the computation. Also, the sub-tractions do not contribute to the computation, because the integral of Eq.(59): (cid:90) R da ∧ da ∂∂ψ ∂∂ψ f ( a Z (1) + a Z (2) + ψ ξ Z (1) + ψ ξ Z (2) ) (71)is zero when either ξ or ξ is proportional to (for example, when ξ = α ,the dependence on ψ can be removed by shifting a (cid:55)→ a − ψ α ). Let us picka pair of fermionic constants (cid:15) and η from the pool[22]; we are evaluating: (cid:90) R da ∧ da ∂∂ψ ∂∂ψ (cid:88) c =1 f (cid:16) a e + a e ++ ψ ( (cid:15)δ c e a + ηδ b e c ) + ψ ( (cid:15)δ c e a + ηδ b e c ) (cid:17) (72)This integral is equal to zero. In fact, each term in (cid:80) c =1 vanishes separately.Indeed, when c is 3 or 4 it is proportional to η = 0. If c is 1 or 2, we canshift the integration variable a or a to eliminate the dependence on η ; thenthe integral becomes proportional to (cid:15) = 0.Now consider the case when ζ is odd: ζ = f a ⊗ e ∨ b (73)∆( ζ ) = ( f a ⊗ e ∨ c ) ∧ (cid:18) e c ⊗ e ∨ b − δ bc (cid:19) + (cid:18) f a ⊗ f ∨ c − δ ca (cid:19) ∧ ( f c ⊗ e ∨ b )(74)18gain, the second term does not contribute to the computation. The contri-bution of the first term is: (cid:90) R da ∧ da ∂∂ψ ∂∂ψ (cid:88) c f (cid:16) a e + a e + (75)+ ψ ( δ c f a + ηδ b e c ) + ψ ( δ c f a + ηδ b e c ) (cid:17) The terms with c equal 3 or 4 are zero, being proportional to η = 0. When c is 1 or 2, we can eliminate the dependence on η by shifting a , and then weare left with, for example, when c = 1: (cid:90) R da ∧ da ∂∂ψ ∂∂ψ (cid:88) c f (cid:16) a e + a e + ψ f a (cid:17) (76)This integral is zero, because the integrand does not depend on ψ . (cid:82) φψψ To prove our claim that the evaluation of the deforming operator on the freesolution given by Eq. (52) is indeed δ ξZ (1) ,Z (2) ( Z ), we will consider a particularcase when the test function f ( Z ) of Eq. (57) is: f ( Z ) = A IK ( λ, µ ) (cid:15) IKP Q ζ P ζ Q (77)Eq. (53) implies that evaluation of (cid:82) Z ∈ R | δ ξZ (1) ,Z (2) ( Z ) f ( Z ) with such f ( Z )corresponds to the substitution instead of the term proportional to ε in Eq.(52) of the following solution of the linearized equations of motion ( cp. Eq.(49)): ϕ IK ( x ) = (cid:90) d λd µ A IK ( λ, µ ) δ ( µ + xλ ) (78) ψ = ˜ ψ = F + = F − = 0 (79)Let us choose both ξ and ξ as “rotations of S ”, i.e. :( ξ ∧ ξ ) IK JL = B IJKL (80)with other components of ξ ∧ ξ all zero.19q. (57) implies that (cid:82) Z ∈ R | δ ξZ (1) ,Z (2) ( Z ) f ( Z ) equals to: (cid:90) R A P Q ( a (1) λ (1) + a (2) λ (2) , a (1) µ (1) + a (2) µ (2) ) (cid:15) P QRS B IJRS ζ (1) I ζ (2) J da (1) ∧ da (2) (81)The same result is obtained by evaluating the deformation term in the La-grangian [13], using the formulas of Appendix B: (cid:90) d x(cid:15) IJKL ϕ IJ ( x ) B P QKL ψ (1) P ˙ α ( x ) ψ (2) Q ˙ β ( x ) (cid:15) ˙ α ˙ β (82) where ψ ( A ) ˙ αK ( x ) = λ ( A ) ˙ α ζ ( A ) K δ ( µ ( A ) + xλ ( A ) ) (83)This establishes the contact with the description of the supermultiplet onthe field theory side obtained in [13]; matching of other states follows fromapplying supersymmetry . Fourier transform into the usual momentum space
We will use nowthe notations of [14]. Following the prescription in [14], let us substitute inEq. (78): A IK ( λ, µ ) = δ ( λ − λ (0) ) exp (cid:16) iµ ˜ λ (0) (cid:17) ζ (0) I ζ (0) K (84)and then multiply by the Fourier transform factor exp (cid:16) − iµ (1) ˜ λ (1) − iµ (2) ˜ λ (2) (cid:17) and integrate over µ (1) and µ (2) . Then Eq. (81) gives: (cid:90) da ∧ da δ (cid:0) λ (0) − a (1) λ (1) − a (2) λ (2) (cid:1) × (85) × δ (cid:16) ˜ λ (1) − a (1) ˜ λ (0) (cid:17) δ (cid:16) ˜ λ (2) − a (2) ˜ λ (0) (cid:17) ζ (0) P ζ (0) Q (cid:15) P QRS B IJRS ζ (1) I ζ (2) J Using Appendix B, this is equal to:1 (cid:104) λ (1) , λ (2) (cid:105) δ (cid:18) ˜ λ (1) − (cid:104) λ (2) , λ (0) (cid:105)(cid:104) λ (2) , λ (1) (cid:105) ˜ λ (0) (cid:19) δ (cid:18) ˜ λ (2) − (cid:104) λ (1) , λ (0) (cid:105)(cid:104) λ (1) , λ (2) (cid:105) ˜ λ (0) (cid:19) ×× ζ (0) P ζ (0) Q (cid:15) P QRS B IJRS ζ (1) I ζ (2) J (86) Vanishing of deformations corresponding to the denominator in Eq. (5) was not provenin [13]. It follows from Section 4.4 that such deformations would vanish when evaluatedon (52). This should imply that the deformation is acually zero, although we do not havea rigorous proof. p ( i ) α ˙ α =˜ λ ( i ) α λ ( i ) ˙ α . Eq. (86) can be interpreted as a deformation of the three-pointscattering amplitude. It is a generalized function with support on ˜ λ (1) , ˜ λ (2) and ˜ λ (0) being all collinear to each other. It is likely that the expressions δ ξZ ,Z ( Z ) will serve as building block forevaluation of deformations corresponding to other finite-dimensional repre-sentations. It was conjectured in [4] that deformations of the form: (cid:90) d x ρ ( x ) tr(Φ + i Φ ) ( x ) , (cid:90) d x ρ ( x ) tr(Φ + i Φ ) ( x ) , (cid:90) d x ρ ( x ) tr(Φ + i Φ ) ( x ) , . . . (87)where ρ n are some special polynomial functions described in [4] generatefinite-dimensional representations of g corresponding to supersymmetric Youngdiagramms: , , , . . . (88)and their transposed. An element of such a representation is a super-tracelesstensor B with 2( n +2) lower indices and 2( n +2) upper indices. The symmetrytype of the lower indices is determined by the lower portion of the Youngdiagramms, and of the upper indices by the upper portion. For example, theYoung diagramm (“case n = − B abcd symmetric in ab and antysymmetric in cd , while correspondsto the tensors of the form B abcd antisymmetric in ab and symmetric in cd .When all indices a, b, c, d are fermionic, gives B IJKL = B JIKL = − B IJLK (seeEq. (82)).Consider the tensor B = ξ ⊗ n +2) where ξ is an odd nilpotent element of g . Let us apply the Young symmetrizer corresponding to:2189)We conjecture that the evaluation of the deformation on the coherent state(as in Eq. (20)): (cid:0) ε δ Z ( X ) + ε δ Z ( X ) + . . . + ε n +4 δ Z n +4 ( X ) (cid:1) M (90)gives δ ξZ ,...,Z n +4 tr M n +4 where: δ ξZ ,...,Z n +4 = δ ξZ ,Z ( Z ) δ ξZ ,Z ( Z ) · · · δ ξZ n +2 ,Z n +3 ( Z n +4 ) (91)We leave the verification of this statement for future work. The symmetryof δ ξZ ,...,Z n under permutations of Z i can be proven as in Section 4.3.Let us prove that Eq. (91) depends on ξ ∧ · · · ∧ ξ only through its Youngprojector. To start, notice that each δ ξZ n ,Z n +1 ( Z n +2 ) depends on ξ ∧ ξ via theprojector ξ ⊗ ξ . (In other words, the lower through ξ ( a [ c ξ b ) d ] .) This followsfrom the fact that δ ξZ n ,Z n +1 ( Z n +2 ) = 0 when ξ is degenerate, dim im ξ = 1(this follows from the definition Eq. (57)). This implies that δ ξZ ,...,Z n +4 depends on ξ ∧ · · · ∧ ξ only through: ξ ( a [ c ξ b ) d ] ξ ( a [ c ξ b ) d ] · · · ξ ( a n +2 [ c n +2 ξ b n +2 ) d n +2 ] (92)But this is not all, there are more projectors. Let us consider the special case n = 0. Let us write δ ξZ ,Z ,Z ,Z as δ ξZ ,Z ( Z ) δ ξZ ,Z ( Z ). The indices c and c both contract with Z , therefore they enter symmetrized. The indices d and d also enter symmetrized, because they both contract with Z . Therefore,the ξ ∧ ξ ∧ ξ ∧ ξ enters only through the projector to . In general, ξ ∧ n +2) enters only through the projector of Eq. (89).The second diagramm of Eq. (3) can be probed by replaying our con-struction with T replaced with Π T (the twistor space with flipped statistics).22 cknowledgments The authors are partially supported by the FAPESP grant 2014/18634-9“Dualidade Gravitac , ˜ao/Teoria de Gauge”. The work of A.M. was also sup-ported in part by the RFBR grant 15-01-99504 “String theory and inte-grable systems”. The work of S.P.M was supported by the CNPq grant154704/2014-8. A Measure defined by an odd linear operator
A nondegenerate odd nilpotent linear operator ξ on an n | n -dimensional linearspace defines an integration measure, in the following way. Given any n -tupleof even vectors Z , . . . , Z n we define the integral of an arbitrary function f as follows: (cid:90) f = (cid:90) R n | n da · · · da n ∂∂ψ · · · ∂∂ψ n f (cid:0) a i Z i + ψ i ξZ i (cid:1) (93)In fact, it is not strictly necessary that ξ is nilpotent; it is enough that thesquare of ξ be proportional to a unit matrix: ξ = c (94)where c is some number. All we have to prove that the definition given byEq. (93) is independent of the choice of Z , . . . , Z n . Suppose that we choosea different n -tuple: (cid:101) Z i = Z i + (cid:15) ji ξZ j (95)We replace Z with (cid:101) Z in Eq. (93); the integrand becomes: f (cid:0) a i Z i + ψ i ξZ i (cid:1) = f (cid:0) a i Z i + a i (cid:15) ji ξZ j + ψ i ξZ i + ψ i (cid:15) ji cZ j (cid:1) = (96)= f (cid:0) a i Z i + ( ψ i + a j (cid:15) ij ) ξZ i + ( ψ i + a j (cid:15) ij ) (cid:15) ki cZ k − a j (cid:15) kj (cid:15) ik cZ i (cid:1) The difference between Eq. (96) and the integrand of Eq. (93) can be undoneby the change of variables (cid:101) ψ i = ψ i + a j (cid:15) ij , (cid:101) a i = a i − a j (cid:15) kj (cid:15) ik c . (Notice that thedeterminant of the change from a to (cid:101) a is 1, since (cid:15) ij are fermionic.)23 Some integrals
We denote: (cid:104) λ , λ (cid:105) = (cid:15) ˙ α ˙ β λ α λ β , (cid:104) µ , µ (cid:105) = (cid:15) αβ µ α µ β (97)To integrate a 2-form da ∧ da , we need an orientation of the ( a , a )-plane.We orient it as ( a , a ) if (cid:104) λ , λ (cid:105) > a , a ) otherwize. Then we get: (cid:104) λ , λ (cid:105) (cid:90) da ∧ da δ ( λ − a λ − a λ ) = 1 (98)Therefore: (cid:90) d x δ ( µ + xλ ) (cid:104) λ , λ (cid:105) δ ( µ + xλ ) δ ( µ + xλ ) == (cid:104) λ , λ (cid:105) (cid:90) da ∧ da δ ( λ − a λ − a λ ) ×× (cid:90) d x δ ( µ + xλ ) δ ( µ + xλ ) δ ( µ + xλ ) = (99)= (cid:90) da ∧ da δ ( λ − a λ − a λ ) δ ( µ − a µ − a µ ) = (100)= (cid:104) λ , λ (cid:105) δ (cid:16) (cid:104) λ , λ (cid:105) µ − (cid:104) λ, λ (cid:105) µ + (cid:104) λ, λ (cid:105) µ (cid:17) = (101)= (cid:104) λ , λ (cid:105) |(cid:104) µ , µ (cid:105)| δ (cid:16) (cid:104) λ , λ (cid:105)(cid:104) µ, µ (cid:105) − (cid:104) µ , µ (cid:105)(cid:104) λ, λ (cid:105) (cid:17) ×× δ (cid:16) (cid:104) λ , λ (cid:105)(cid:104) µ, µ (cid:105) − (cid:104) µ , µ (cid:105)(cid:104) λ, λ (cid:105) (cid:17) (102) References [1] O. Aharony, S. S. Gubser, J. M. Maldacena, H. Ooguri, and Y. Oz,
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