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The generalized no-ghost theorem for N=2 SUSY critical

arXiv:hep-th/9111047v1 23 Nov 1991EFI 91-65November 1991The generalized no-ghost theorem for N=2 SUSY criticalstrings 1Jadwiga Bie´nkowskaDepartment of Physics andEnrico Fermi Institutethe University of ChicagoChicago, Il 60637AbstractWe prove the no-ghost theorem for the N=2 SUSY strings in (2,2) dimen-sional flat Minkowski space. We propose a generalization of this theoremfor an arbitrary geometry of the N=2 SUSY string theory taking advan-tage of the N=4 SCA generators present in this model.

Physical statesare found to be the highest weight states of the N=4 SCA.1work supported in part by DOE grant DE-FG02-90ER40560.

The N=2 SUSY critical string theory with (2,2) real signature seems to be aninteresting model to test duality properties of string theories [1]. It is particularlysimple because the only degree of freedom in the model is believed to be a scalar fieldplaying the role of the K¨ahler potential deformations.

Even though there is no reasonto suppose that there are other physical states present in this model there was up todate no explicit no-ghost theorem proven.In this letter we present such a proof for the flat Minkowski space with (2,2) realsignature. The proof is a direct generalization of the modern version of the no-ghosttheorem by Thorn [6] which uses the BRST quantisation prescription [10].

We presentthe line of the proof for the N=2 critical string theory pointing out the differencesbetween the N=0 [6] and N=2 case.The N=2 string theory has one advantage over cases with lower N. It possesesan additional symmetry generated by the spectral flow [2].In 4 real dimensions(critical N=2 strings) the spectral flow generators are dimension 1 operators andcan be identified as currents of the local SU(2) (in Euclidean space) or SU(1,1) (inMinkowski type space) groups [3]. The enlarged set of generators of the N=2 theoryincludes all the N=4 SCA generators [4], with the gauge symmetry generators beingonly the ones belonging to the N=2 SCA subset.

Using this larger set of the operatorswe can look at the states in this model as states from the Verma module of the N=4SCA. We show that gauging away the N=2 SUSY is enough to prove that the onlyphysical states are the highest weight states of the N=4 SCA.

In the Euclidean modelthe only physical state is the identity, while in the Minkowski case there are enoughstates left in the physical subspace to make the model interesting.The critical N=2 SUSY string lives in the two dimensional complex space withtwo complex bosons aµ(∗)nand fermions bµ(∗)nµ = 1, 2 [5]. The free field representationof the N=2 SCA is:Ln=−Xsa∗n−s · as +Xs(n2 −s)b∗n−s · bsGn=√2Xsb∗n−s · asG∗n =√2Xsbn−s · a∗sTn=−12Xsb∗n−s · bs(1)where [aµr , anus ] = −rηµνδr+s, {bµr , bnus } = −ηµνδr+s.

Dots signify a scalar productwith respect to the ηµν Minkowski ( diag(1, −1) ) or Euclidean ( diag(1, 1)) hermitian1

metric. The generators obey the N=2 SCA comutation relations:[Ln, Lm]=(n −m)Ln+m + 12kn(n2 −1)δn+m{Gr, G∗s}=2Lr+s + 2(r −s)Tr+s + 12k(4r2 −1)δr+s[Lm, Gr]=(m2 −r)Gm+r[Lm, G∗r] = (m2 −r)G∗m+r[Tm, Gr]=12Gm+r[Tm, G∗r] = −12G∗m+r[Lm, Tn]=−nTm+r[Tm, Tn] = 12kmδn+m(2)and all other commutators are zero .

The moding of bµ(∗)r, G(∗)roperators is governedby the spectral flow parameter η [2],[5] and we will restrict our attention to one sector,let us say the NS with r ∈Z + 12. The physical states are defined by the conditionsLn|s⟩=Gr|s⟩= G∗r|s⟩= Tn|s⟩= 0 for n, r > 0 L0|s⟩= T0|s⟩= 0(3)In the BRST formulation of the theory for each gauge constraint one introducesa pair of ghosts with a statistics opposite to that of the constraint [5],[10].ForN=2 there are four pairs of ghosts, associated with Ln, Tn, Gn, G∗n gauge constraintsrespectively, satisfying the commutation relations {cn, ¯cm} = δm+n, {tn, ¯tm} = δm+n,[γn, ¯γm] = δm+n, [γ∗n, ¯γ∗m] = δm+n.

The BRST operator Q expressed in terms of theseoperators isQ=Lnc−n + Gpγ−p + G∗pγ∗−p + Tnt−n −12(m −n)c−mc−n¯cn+m −2γ−pγ∗−q¯cp+q−2(p −q)γ−pγ∗−q¯tp+q + nc−mt−n¯tm+n + (m2 −p)c−m(γ−p¯γm+p + γ∗−p¯γ∗m+p)+12t−m(γ−p¯γm+p −γ∗−p¯γ∗m+p)(4)Q2 = 0 for two complex bosons and fermions [5].We are set to begin the proof of the no ghost theorem for the Minkowski signatureN=2 SUSY critical string theory. We follow closely the line of the proof proposed byThorn [6].

In Euclidean space the L0|state⟩= 0 condition trivially forces the onlyphysical state to be the identity. To fix the gauge conditions ( 3) we introduce thegauge fixing operators A+n , B+n , A∗−n , B∗−nwhere the notation (+, −) is unambiguousin the (1,1) signature and A±(∗)n(B±(∗)r) refer to the free boson (fermion) operatorsprojected on the lightlike directions.Following the line of the proof [6] we wantto construct the Hilbert space of this model using gauge generators, gauge fixing2

conditions and states which are simultanously annihilated by all of them. We considerthe statesLλ1−1.

. .

Lλl−lGg1−1 . .

. Ggr−rG∗g∗1−1 .

. .

G∗g∗r∗−r∗T τ1−1 . .

. T τn−nA+α1−1.

. .

A+αa−a B+β1−1 . .

. B+βb−b A∗−α∗1−1.

. .

A∗−α∗a∗−a∗B∗−β∗1−1. .

. B∗−β∗b∗−b∗|t⟩(5)where Ln|t⟩= Gr|t⟩= G∗r|t⟩= Tn|t⟩= A∗+n |t⟩= B∗+r |t⟩= A−n |t⟩= B−r |t⟩= 0 forn, r > 0 and L0|t⟩= h|t⟩.

The ”mixed” (+, −) gauge fixing conditions are needed toassure the linear independence of the vectors ( 5), as we explain below.The first step is to prove the linear independence of the vectors having the form( 5) at each level N. First of all we notice that the number of generators in the basis( 5) is the same as the number of generators in the Hilbert space spanned by the freeboson and free fermion operators acting on the vacuum with a momentum p |0, p⟩,so at each level the number of vectors from the set ( 5) is the same as the number ofvectors from the free field representation of the Hilbert space which clearly forms alinearly independent set (a basis of the vector space at each mass level).What remains to be proven then is that at each level N we can express any vectorfrom the basis of A±n , A∗±m , B±r , B∗±sby the vectors from ( 5). This is enough sinceat each level N we have a finite dimensional vector space.The proof follows byinduction.

For N =12 the check is trivial since the G−12, G∗−12 generators take therole of B−−12, B∗+−12 operators. At level N = 1 we check by straightforward algebrathat any vector from the free field basis can be expressed by vectors from the set( 5).

In particular at this level it becames clear that we need the mixed gauge fixingconditions as above to ensure the linear independence of vectors ( 5). We requirethat both A±0 are different from zero.

The case when either of them is equal zero iseasy to prove since the physical state condition L0|s⟩= 0 automatically requires thatthere are no mass excitations different from 0. The p = 0 case has to be addressedseparately and we discuss it at the end of the main line of the proof.Having checked the validity of our statement at level 1, we assume that it istrue for any vector at level n < N, i.e.

states from free field basis at mass leveln can be expressed in terms of vectors ( 5).Any state from the free field basisat level N which is a product of several A±(∗)n(B±(∗)r) operators can be expressedin the basis ( 5) by the induction assumption and using the commutation relationswith the gauge generators. The B∗+−N+ 12|h⟩, B−−N+ 12|h⟩states are expressed easily byG−N+ 12|h⟩= A−0 B∗+−N+ 12|h⟩+ other states from the basis,3

G∗−N+ 12|h⟩= A∗+0 B−−N+ 12|h⟩+other states from the basis. The states A∗+−N|h⟩, A−−N|h⟩can also be expressed in this basis as follows.

Consider the expansion of the stateG−pG∗−N+p|h⟩(12 < p < N −12 and p ∈Z + 12 ) in the free fields basis.Wesee that it contains the states (A∗−0 A+−N + A∗+0 A−−N)|h⟩plus other states includingthe product of the several operators from the free field basis. By induction we canexpress these states in the (5) basis with one exception.There is the possibilitythat doing so we will cancel the A∗+0 A−−N)|h⟩term.

The cancellation occurs fromthe state 2A−0 A∗+0 B−p−NB∗+−p|h⟩if we try to express the B∗+−p operator using the G−pgauge generator and then commute the B−−N+p to the right.To solve this prob-lem we realize that T−N = −12PN−12p= 12 (B∗+−pB−p−N + B∗+−pB−p−N)|h⟩and so the statePN−12p= 12 B∗+−pB−p−N can be expressed in the basis as shown above.Then it followsfrom (PN−12p= 12 G−pG∗−N+p + 4A−0 A∗+0 T−N)|h⟩= −2NA∗+0 A−−N + other states wherenow all other states are from the basis (5), that we can express A−−N in this ba-sis.The remaining state A∗+−N|h⟩can then be expressed by expanding L−N|h⟩=A−0 A∗+−N|h⟩+ other states where all other states can be expressed in the basis ( 5).This completes the proof of the linear independence of the vectors from the set ( 5).The proof of the no-ghost theorem is now relatively simple if we follow Thorn’sapproach [6].We consider the BRST Hilbert space built as follows from ghostsand vectors ( 5). We introduce the N=2 SCA generators which include the ghostsoperators [5]:{Q, ¯cn}=Ln + Lghn = Ln{Q, ¯tn} = Tn + T ghn= Tn[Q, ¯γp]=Gp + Gghp = GphQ, ¯γ∗pi= G∗p + G∗ghp= G∗p(6)We define the ghost vacuum as ¯c0|0⟩ghost = ¯t0|0⟩ghost = 0.

Then our BRST basisis formed by the statesLλ1−1. .

. Lλl−l¯c¯κ1−1 .

. .

¯c¯κ¯k−¯kGg1−1 . .

. Ggr−r¯γ ¯χ1−1 .

. .

¯γ ¯χ¯s−¯sG∗g∗1−1. .

. G∗g∗r∗−r∗¯γ∗¯χ∗1−1.

. .

¯γ∗¯χ∗¯s∗−¯s∗T τ1−1 . .

. T τn−n¯t¯ν1−1 .

. .

¯t¯ν ¯m−¯mA+α1−1. .

. A+αa−a cκ1−1 .

. .

cκk−kB+β1−1 . .

. B+βb−b γχ1−1 .

. .

γχs−sA∗−α∗1−1. .

. A∗−α∗a∗−a∗tκ∗1−1 .

. .

tκ∗k∗−k∗B∗−β∗1−1. .

. B∗−β∗b∗−b∗γ∗χ∗1−1 .

. .

γ∗χ∗s∗−s∗|t⟩gh(7)where |t⟩gh means now a tensor product of the ghost vacuum and previously defined|t⟩states from the physical sector. The above vectors form a linearly independentset since adding ghosts does not destroy the linear independence of the basis ( 5).

A4

more useful for the no-ghost theorem is the basis in which we symmetrise the bosonicoperators and antisymmetrise the fermionic ones. Using the N=2 SCA commutationrules ( 2) and the ghosts commutation properties we can do it without spoiling thelinear independence of the basis ( 7).

We introduce shorthand notation to refer tothis symmetrised basisLλ{−l}¯c¯κ[−¯k] Gg[−r] ¯γ ¯χ{−¯s} G∗g∗[−r∗] ¯γ∗¯χ∗{−¯s∗} T τ{−n} ¯t¯ν[−¯m]A+α{−a}cκ[−k] B+β[−b] γχ{−s} A∗−α∗{−a∗} tκ∗[−k∗] B∗−β∗[−b∗] γ∗χ∗{−s∗} |t⟩gh(8)where the {}, [ ] means symmetrization or antisymmetrization in the indices enclosed.Following the proof by Thorn [6] we can easily verify that the specific symmetrycombination of the first pair L−n, ¯c−m can be expressed by other states with a lowernumber of L’s and a pure gauge. Working down the number of L’s we can express anystate with a symmetry associated to the Young tableau1+¯κ| |λby other statesand a pure gauge .

Thus we are left with the states of the symmetry ¯κ| |λ+1only. Further requirement that these states be physical ( annihilated by Q) drives usto the conclusion that the symmetry properties exclude any state with (λ, ¯κ) ̸= (0, 0)to be physical (for a more detailed discussion see [6]).This procedure can be repeated smoothly for all other pairs of gauge generatorsand antighosts.

We are left with the basis formed by the gauge fixing operators andghosts A+α{−a}cκ[−k] B+β[−b] γχ{−s} A∗−α∗{−a∗} tκ∗[−k∗] B∗−β∗[−b∗] γ∗χ∗{−s∗} |t⟩gh. We can easily see fromthe commutation relationshQ, A+−mi=mc−nA+n−m + m√2γ∗−nB+n−mhQ, A∗−−mi=mc−nA∗−n−m + m√2γ−nB∗−n−m{Q, B+−m}=(m −n2)c−nB+n−m + 12t−nB+n−m +√2γ−nA+n−m{Q, B∗−−m}=(m −n2)c−nB∗−n−m + 12t−nB∗−n−m +√2γ−nA∗−n−m(9)that the procedure of removing specific symmetry combinations of gauge fixing con-ditions and ghosts works when A±0 ̸= 0.

We can remove, in the standard way [6],the first pair A+α{−a} cκ[−k]. Then by the BRST charge Q acting on the vector formedfrom the basis B+β[−b] γχ{−s} B∗−β∗[−b∗] γ∗χ∗{−s∗} A∗−α∗{−a∗} tκ∗[−k∗] |t⟩gh we will reproduce, due tothe commutation relations ( 9), the states containing operators A+−n, c−m[6].

By the5

previous step, since we consider the states annihilated by Q, states containing any ofthese operators have to be a pure gauge. The procedure of removing the states with(β, χ) ̸= (0, 0) and (β∗, χ∗) ̸= (0, 0) from this basis follows.We are left with the states formed from the basis A∗−α∗{−a∗} tκ∗[−k∗] |t⟩gh.

We requirethat for a linear combination of the above states |s⟩Q|s⟩= 0.Looking at thecommutation relations ( 9) and [Q, t−n] = mcm−nt−m −2(m −2n)γm−nγ∗−m we seethat it is enough to consider only the γ∗−p ghost conservation number. The stateswith the non zero γ∗−p ghost have to cancel among themselves and starting from thestate with the highest t−n excitation we see that the only physical state Q|phys⟩= 0is the one with no t ghost.

Then it is easy to see that the states built only from A∗−−nexcitations are not physical. This completes the proof for A± ̸= 0.For A+ = 0 or A−= 0 the physical state condition L0|phys⟩= 0 automaticallyexcludes any excitation so we are left with the physical states subspace |0, p⟩wherepµp∗µ = 0.The pµ = 0 case is special, as we know from the N=0,1 SUSY case [7].

We haveto check directly which states with zero L0, T0 eigenvalue are physical. In this casec1|0⟩gh ̸= 0 (L−1|0⟩= 0), γ 12|0⟩gh ̸= 0 (G−12|0⟩= 0), γ∗12|0⟩gh ̸= 0 (G∗−12|0⟩= 0) andthere are several nontrivial possibilities for the states with non zero ghost numbersto be physical L0|s⟩= 0.

We investigated all possible states built up from the aboveghosts and we found that there are in this case additional non zero ghost numberphysical states(aµ−1c1 +√2bµ−12γ∗12)|0, pµ = 0⟩(a∗µ−1c1 +√2b∗µ−12γ 12)|0, pµ = 0⟩(10)This completes the proof for the no-ghost theorem for the N=2 SUSY criticalstrings in flat space.We generalize the no-ghost theorem for the case of an arbitrary geometry of theN=2 SUSY string theory in (2,2) signature real dimension space by taking advantageof the larger set of generators existing in these theories due to the spectral flow [2], [3].In the case of the (4,0) real signature space the group is enlarged by the spectral flowgenerators T +, T −which can be identified with the generators of the SU(2) group.The full set of generators is described by the N=4 SCA operators [3], [4]. One cancheck that in the free field representation ( 1) the N=2 SCA algebra is completed tothe N=4 by adding the generators.T +n =Xsb∗1n−sb∗2sT −n = −Xsb1n−sb2s(11)6

It is important to have a hermitian conjugate basis for the operators (T +n )† = T −−nsince this unambiguously defines the commutation relations in the algebra.Theadditional generators of the N=4 SCA and their commutation rules (non-zero ones)arehT +n , G∗ri=√2Xt(b∗2n+r−ta∗1t −b∗1n+r−ta∗1t ) = Bn+rhT −n , Gri=−√2Xt(b2n+r−ta1t −b1n+r−ta1t) = −B∗n+rhT +n , T −mi=2Tn+m + 12knδn+mhTn, T ±mi= ±T ±m+nhT +m, B∗ri=−Gr+mhT −m, Bri= G∗r+m{Br, B∗s}=2Lr+s + 2(r −s)Tr+s + 12k(4r2 −1)δr+s[Tm, Br]=12Br+m[Tm, B∗r] = −12B∗r+m{Br, Gs}=2(r −s)T +s+r{B∗r, G∗s} = −2(r −s)T −r+s(12)which is equivalent to the N=4 SCA with the SU(2) subgroup discussed in [4] withGr =1√2(G1r + ¯G2r), G∗r =1√2(G2r + ¯G1r), Br =1√2(−G1r + ¯G2r), B∗r =1√2(G2r −¯G1r). Theabove defined operators are hermitian in the sense (Br)† = B∗−r, (Gr)† = G∗−r.

It isthen straightforward to find the N=4 SCA in the case of Minkowski signature space.The algebra has, as could be expected, the SU(1,1) subgroup and the commutationsrelations for the generators T ±n and B(∗)rdefined as in ( 11, 12):hT +n , T −mi=−2Tn+m + 12knδn+mhT +m, B∗ri=Gr+mhT −m, Bri= −G∗r+m{Br, B∗s}=−(2Lr+s + 2(r −s)Tr+s) −12k(4r2 −1)δr+s(13)where we have listed only the commutators which are different from case in (12) .The algebra we obtain in this case is considerably different from the Euclideanspace algebra, since it contains as a subgroup the noncompact SU(1,1). The SU(1,1)group is the part of the rotation group unbroken by the background.

Typical physicalstates of the theory belong to finite dimensional representations of SU(1,1). Fortu-nately the representations of this group were classified some time ago [8].Thereare three basic types of the SU(1,1) representations classified by the eigenvaluesof the quadratic Casimir operator C = −gabT a0 T b0, where gab = diag(−1, −1, 1)7

C|0, j, m⟩= −j(j + 1)|0, j, m⟩, and the discrete values of the compact directionoperator T 30 |0, j, m⟩= T0|0, j, m⟩= m|0, j, m⟩. The ground states of the N=4 SCAtheory ( 13) belong to the irreducible representations of the global SU(1,1) algebraand we introduced notation |0, j, m⟩where 0 refers to the state zero excitation of theN=4 SCA.

There are following irreducible (non necessarily unitary) SU(1,1) repre-sentations: continuous ones1)C0j , j = −12 + iκ and m = 0, ±1, ±2 . .

.2)C12j , j = −12 + iκ and m = ±12, ±32 . .

.3)Ej, j ∈R, j ̸∈Z, 2j ̸∈Z m = 0, ±1, ±2, . .

. or m = ±12, ±32, .

. .and there is no highest state anihilated either by T +0or T −0 ; infinite discreterepresentations:4)D+j , j = −12, −1, −32 .

. .

and m = −j, −j + 1, . .

.5)D−j , j = −12, −1, −32 . .

. and m = j, j −1, .

. .and finite discrete representations;6)Fj, j = 12, 1, .

. .

and m = −j, −j + 1, . .

. , j −1, jFor the D+j representation there exists a highest state |s⟩+ = |0, j −j⟩such thatT −0 |s⟩+ = 0 and for the D−j representation there is a state |s⟩−= |0, j, j⟩such thatT +0 |s⟩−= 0.

These remarks are useful in the generalization of the no-ghost theoremwe present below.The space of all states in the theory we consider is a tensor product of the statesof the Verma module of the N=4 SCA, obeying the commutation relations ( 2, 12, 13)(SU(1,1) case), and some other states which are the trivial representation of the N=4SCA. The goal of our proof is to show that requirement that the states obey the N=2SCA gauge conditions ( 3) is enough to gauge away all excitations from the N=4 SCAVerma module.

This leaves the physical states space of the theory containing onlythe highest weight states of the N=4 SCA.We consider a state from a Verma module built on a state belonging to one ofthe global representations of the SU(1,1) algebra. The states built from the orderedproduct of the N=4 SCA operators acting on the highest weight state:Lλ1−1.

. .

Lλl−lGg1−1 . .

. Ggr−rG∗g∗1−1 .

. .

G∗g∗r∗−r∗T τ1−1 . .

. T τn−nT +α1−1.

. .

T +αa−a Bβ1−1 . .

. Bβb−bT∗−α∗1−1.

. .

T−α∗a∗−a∗B∗β∗1−1 . .

. B∗β∗b∗−b∗|0, j, m⟩(14)form a good basis for the Verma module, in the sense that there are no linear combi-nations of them equaling zero (of course there are linear combinations of states with8

norm equal to zero). The ordering of the operators ( last four T +−n, B−p, T −−n, B∗−p ) isimportant for the no-ghost proof and should be set respectively to the representationof the |0, j, m⟩state.

For the continuous representations C12j , C0j , Ej it does not mat-ter, for the D−j representations the positive charge generators T +m, Bn should cometo the left as in ( 14) and for the D+j representations the negative charge operatorsT −m, B∗n come first from the left.The no ghost theorem follows in much the same way as for flat space. Howeverthere are some differences which we point out below.

We also restrict our discussionto the case when the ground state belongs to D−j or continuous representation, butthe D+j case is proven analogously.To gauge away the N=2 SUSY we use the BRST prescription with the BRSTcharge defined by ( 4). The full BRST Hilbert space takes the familiar formLλ{−l}¯c¯κ[−¯k] Gg[−r] ¯γ ¯χ{−¯s} G∗g∗[−r∗] ¯γ∗¯χ∗{−¯s∗} T τ{−n} ¯t¯ν[−¯m]T +α{−a}cκ[−k] Bβ[−b] γχ{−s} T −α∗{−a∗} tκ∗[−k∗] B∗β∗[−b∗] γ∗χ∗{−s∗} |0, j, m⟩gh(15)Adding ghosts and the (anti) symmetrization procedure does not spoil the linearindependence of the basis ( 14).

But unlike in the previous case the subspace HTBbuilt from the vectors T +α{−a}cκ[−k] Bβ[−b] γχ{−s} T −α∗{−a∗} tκ∗[−k∗] B∗β∗[−b∗] γ∗χ∗{−s∗} |0, j, m⟩gh is notQ invariant.There exist vectors |s⟩∈HTB such that Q|s⟩̸∈HTB.This is theconsequence of the commutation relations of these generators with the BRST chargehQ, T +−mi=mc−nT +n−m + T +n−mt−n −γ∗−nBn−mhQ, T −−mi=mc−nT −n−m −T −n−mt−n + γ−nB∗n−m{Q, B−m}=−(m + n2)Bn−mc−n −12Bn−mt−n −2(n + m)γ−nT +n−m{Q, B∗−m}=−(m + n2)B∗n−mc−n + 12B∗−n−mt−n + (n + m)γ−nT −n−m(16)and the N=4 SCA ( 12), ( 13).As in the previous case, we would like to express the states with a symmetry1+¯κ| |λin the Lλ{l} and ¯c¯κ[¯k] operators successively by the states with the lowernumber of L−n’s and pure gauge terms and eventually eliminate them from the phys-ical states space. In the case of the basis ( 15) there will be some states with the samenumber λ of L−n representing the above symmetry state in our induction procedure[6], which did not happen previously.

The unwanted L−n generators will come from9

the commutation relations ( 16) and a subsequent commutation of the B, B∗gen-erators. We are getting one L−n generator at the expense of exchanging two B, B∗pair for the ghost.

For each finite N level space we can subsequently work down thenumber of L−n-producing generators eliminating them after a finite number of steps,and so our procedure of eliminating the symmetry combination1+¯κ| |λsur-vives. Then we consider how the states with the remaining symmetry¯κ| |λ+1can be made physical and we come to the conclusion that the only possibility is that(λ, ¯κ) = (0, 0).

Working further with the basis of the states without the L’s or ¯c’swe would again encounter the L state reproduced when we try to express the com-binationg| |¯χ+1by other states and pure gauge terms. But by the first step,and physical state condition, any such state has to be pure gauge and the procedurefollows all the way thus taking care of all the states containing the gauge generatorsand antighosts.The next difference showing up in this case is that the role of the A+0 , A∗−0op-erators (which are C-numbers in flat Minkowski space ) is played by the T ±0 as wesee from ( 16).

Since the ground states in this theory belong to the SU(1,1) infiniterepresentations, this step of the proof also survives. To check it we need the basisordered as was discussed before.Looking at the commutator[Q, T +(α++1)−{a}c(κ−1)−[k]]=XiT+α+1{−1 .

. .

ˆT+α+i−i. .

. T +α+a−a} iT +0 c−icκ−1−[k] ++other states(17)we see that the state with the symmetryκ| |α+1can be expressed by otherstates and a pure gauge if there always exists a state at zero level |0, j, m −1⟩suchthat |0, j, m⟩= T +0 |0, j, m −1⟩.

This statement is always true in the continuous andD−j representation since we ordered our basis to assure that the states containingT +n , cn, Bm, γm operators can be always removed. We can also remove in this way thestates built from the T −n , tn, B∗m, γ∗m operators as long as the state |0, j, m⟩is differentfrom the highest weight state of the D−j global representation |s⟩−.

For this statethere does not exists |˜s⟩such that T −0 |˜s⟩= |s⟩−. We have to consider states builton the highest state |s⟩−separately.

Any other state from the basis ( 16), which10

has |0, j, m⟩some other state from the D−j or a continuous representation is eithernonphysical or a pure gauge.We have to check directly which linear combinations of T −α∗{−a∗} tκ∗[−k∗] B∗β∗[−b∗] γ∗χ∗{−s∗} |s⟩−states can be physical e.i. annihilated by Q.

When Q acts on the linear product ofthe above states we can realise that the terms (. .

. )T −0 c−n|s⟩−= (.

. .

)c−n|0, j, j −1⟩have to cancel by themselves, by the ghost conservation number and orthogonality ofthe states in the SU(1,1) representation D−j . This excludes the possibility of the T −−nexcitations in the physical state.

Let us look next at the commutators of Q with theremaining generators. From the rules ( 16), [Q, t−n] = mcm−nt−m−2(m−2n)γm−nγ∗−mand [Q, γ∗−n] = ( 32m−n)c−mγ∗m−n−12t−mγ∗m−n we see that the only terms containing theγ−p ghost come from the commutators with the t−n ghost and therefore have to cancelby themselves.

Starting from the state containing t−n with the highest n we can easilysee that the only possibility is that there are no t−n ghost excitations at all. We arefinally left with the linear combination of states B∗β∗[−b∗] γ∗χ∗{−s∗} |s⟩−, and by looking atthe [Q, B∗−p] commutators, we see the states of the form T −0 γ∗−p|s⟩−have to cancel eachother.

Starting from the state with highest excitation number of B∗−n and workingdown the B∗−p’s number we see that there is no possibility to cancel the last one ([Q, γ∗p]does not contain the T −0 ) and so the only solution to the physical state condition isthe one where β∗= 0. It is straightforward to see that the remaining states built outof the γ∗−p ghosts cannot be physical for nonzero ghost excitation.

For the |0, j, m⟩be-longing to the finite dimension representations Fj we can gauge away any state fromthe basis ( 15) with m > −j as above. We have to consider only the linear combina-tion of states T −α{−a} cκ[−k] Bβ[−b] γχ{−s} T −α∗{−a∗} tκ∗[−k∗] B∗β∗[−b∗] γ∗χ∗{−s∗} |0, j, −j⟩.

We requirethat Q acting on such states is zero. From the commutators ( 16) we see that states[.

. .

]T +0 c−n|0, j, −j⟩have to cancel among themselves and so α = 0. We can furthereliminate any B−p excitations realising that [.

. .

]T +0 c−n|0, j, −j⟩states have to canceleach other. The remaining states T −α∗{−a∗} tκ∗[−k∗] B∗β∗[−b∗] γ∗χ∗{−s∗} |0, j, −j⟩are gauged awayas in equation ( 17) since we can always express |0, j, −j⟩= T −0 |0, j, −j + 1⟩.

Thiscompletes the proof that the states from the Verma module of the N=4 SCA are eithernonphysical or pure gauge in the N=2 SUSY theory. The only possible physical statesare the highest weight states of the N=4 SCA which fall into the representations ofthe global SU(1,1) group.The physical state constraints ( 3) could tell us more about which of these statesare physical.

We can think of the Hilbert space of this theory as a product of the11

SU(1,1) Kac-Moody algebra representation and the space which is trivial with respectto it. The L0|state⟩= 0 condition does not provide additional information.

Fromthe Sugawara construction for the SU(1,1) group [9] and the value of the centralcharge c=6 [3] we see that LSU(1,1)0|0, j, m⟩= −j(j + 1)|0, j, m⟩and easilly we get−j(j + 1) < 0. Since the SU(1,1) trivial part of the space remains undeterminedwe can not say which combination of states from the global SU(1,1) representationsare allowed.

The charge of the state is that of the SU(1,1) representation and therequirement T0|state⟩= 0 means that we should consider only the SU(1,1) stateswith zero charge. This excludes the discrete unitary representations D±j , C12j , andnoninteger m Ej representations since we have m ̸= 0 in these cases.In the proof presented the Verma modules built from the trivial (j = 0) repre-sentation of the global SU(1,1) were not considered.

In such a case the presentedabove line of the proof breaks down since both operators T ±0 annihilate the groundstate of the theory. This situation is similar to the zero momentum case in Minkowskispace but here we are unable to determine, from general arguments, what is the L0eigenvalue of the |0, j = 0, 0⟩state since we can not say much about its SU(1,1) trivialpart.I would like to thank Emil Martinec for bringing this problem to my attentionand for many enlightening discussions.

I thank also Peter Freund and Tohru Eguchifor helpful remarks. This work is submitted in partial fulfillment of the requirementsfor a Ph.D. degree in physics at the University of Chicago.References[1]H.Ooguri and C.Vafa, Geometry of N=2 strings EFI91/05, HUTP-91/A003.Mod.

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