---

Integrable Discrete Linear Systems

arXiv:hep-th/9109054v1 27 Sep 1991August 16, 2018SISSA-ISAS 107/91/EPIntegrable Discrete Linear Systemsand One-Matrix Random ModelsL.BonoraInternational School for Advanced Studies (SISSA/ISAS), StradaCostiera 11, I-34014 Trieste, Italy and INFN, sezione di TriesteM. MartelliniDipartimento di Fisica, Universit`a di Milano, I-20133 Milano,Italy and INFN, sezione di Pavia, I-27100 Pavia, ItalyandC.S.XiongInternational School for Advanced Studies(SISSA/ISAS), Strada Costiera 11, I-34014 Trieste, ItalyABSTRACTIn this paper we analyze one–matrix models by means of the associated discretelinear systems.

We see that the consistency conditions of the discrete linear systemlead to the Virasoro constraints. The linear system is endowed with gauge

invariances. We show that invariance under time–independent gauge trans-formations entails the integrability of the model, while the double scaling limit isconnected with a time-dependent gauge transformation.

We derive the continuumversion of the discrete linear system, we prove that the partition function is actu-ally the τ–function of the KdV hierarchy and that the linear system completelydetermines the Virasoro constraints.2

1. IntroductionIn the study of two dimensional quantum gravity, one of the main approachesis provided by the random matrix models.

In this context the most interestingand calculable model is the one Hermitian matrix model, which has been studiedintensively [1] (for reviews, see [2]). At the discrete level, it is an integrable system[3], and its partition function satisfies certain constraints, which belong to the Borelsubalgebra of the c=1 non–twisted Virasoro algebra[4].

After taking the doublescaling limit the system is expected to become a KdV system restricted by twistedVirasoro constraints [5,6]. However, there is still a gap between the discrete leveland its continuum version: first, the meaning of the Virasoro constraints is notclear; secondly, the identification of the partition function with the τ–function aswell as the appearance of the twisting has not been justified yet.In this paper we will try to shed light upon these points by analyzing anauxiliary tool in one-matrix models: the relevant associated discrete linear systems.We will show in fact, in section 2, that a one–matrix model is equivalent to a certaindiscrete linear system (DLS) in which the string equation appears as a compatibilitycondition.

In section 3 we will derive the discrete Virasoro conditions from theconsistency conditions of the DLS. In section 4 we examine the “time-independent”gauge symmetry of the DLS and show that it is equivalent to the integrability ofthe system.In section 5 another kind of gauge symmetry is considered whichincludes in particular rescalings of the couplings, and leads naturally to the doublescaling limit.

In the last two sections we examine the continuum limit of the DLSand study its properties. In section 6 we determine the continuum analog of theDLS and find its consistency conditions.

Finally, using the latter exactly as in thediscrete case, in section 7 we determine the continuum Virasoro constraints. Ourresult is that the partition function is actually described by a KdV–τ function, andthe constraints are the twisted ones.

This result had only been conjectured up tonow.1

2.One–Matrix ModelIn this section we review the main results concerning the one–matrix modeland introduce the associated discrete linear system (DLS).The partition function of the one- matrix model is defined by [7]ZN(t) =ZdMe−T rV (M) =ZNYi=1dλi∆2(λ) exp(−NXi=1V (λi))(2.1)with the potentialV (M) =∞Xr=1trMrA powerful tool to study this model is the use of orthogonal polynomials. Theyare normalized as followsPn(λ) = λn + .

. .

(2.2)and satisfy the orthogonal and recursion relationsZdλe−V (λ)Pn(λ)Pm(λ) = δn,mhn(t)(2.3)λPn(λ) = Pn+1(λ) + SnPn(λ) + RnPn−1(λ)(2.4)It is convenient to introduce another set of the polynomialsξn(λ) ≡1√hne−12V (λ)Pn(λ)Then eqs. (2.3) and (2.4) becomeZdλξn(t, λ)ξm(t, λ) = δn,m(2.5a)2

λξn =pRn+1ξn+1 + Snξn +pRnξn−1(2.5b)where the function Rn(t) are defined as the ratioRn ≡hnhn−1,n ≥1.One can show thatSn = −∂∂t1ln hn(t),n ≥0(2.6a)or more generally,∂∂trln hn(t) = −(Qr)nnn ≥0(2.6b)where Q is the Jacobi matrixQnm ≡Zdλξm(t, λ)λξn(t, λ)=pRn+1δn,m−1 + Snδn,m +pRnδn,m+1(2.7)In the same way, we introduce the matrix PPnm ≡Zdλξm(λ) ∂∂λξn(λ)(2.8)Using the orthogonal polynomials, we can perform the integrations in (2.1) andobtainZN(t) = N!h0h1h2 . .

. hN−1(2.9a)∂∂trln ZN(t) = −N−1Xn=0Qrnn ≡−TrQr,r ≥1(2.9b)Next we introduce the discrete linear system alluded to in the introduction.

Letus denote by ξ the column vector with components ξ0, ξ1, ξ2, . .

. , use the recursion3

relations and differentiate the orthogonality relations with respect to tr: we arriveat the following discrete linear system (DLS) of equationsQξ = λξ(2.10a)∂∂trξ = Qraξ(2.10b)∂∂λξ = Pξ(2.11a)P =∞Xk=2ktkQk−1a(2.11b)where the dependence on t and λ has been understood. Here and throughout thepaper we adopt the notation(Qra)nm ≡12(Qr)nm,m < n0,m=n−12(Qr)nm,m > n(2.12)The product in the above equations is of course the matrix product.The consistency conditions for this linear system give rise to the discrete KdVhierarchy[8]∂Q∂tr= [Qra, Q](2.13)and to the so-called string equation[Q, P] = 1(2.14)All the integrability and criticality properties of the matrix model are encodedin the DLS.

Showing this is the aim of our paper.4

It is interesting to answer the question: to what extent is the correspondencebetween discrete linear systems and one–matrix models one to one? There cer-tainly are linear systems that do not correspond to matrix models, however if weimpose the matrix Q to have the Jacobi form (2.7), the correspondence is one toone.

Indeed let us start from the infinite column vector ξ with orthonormalizedcomponents ξ0, ξ1, ξ2, ... as in eq. (2.5a), and write the systemQξ = λξ(2.10a’)∂∂trξ = Qraξ(2.10b’)Then we can reconstruct the partition function from∂∂trln ZN(t) = −TrQr,r ≥1If we define now∂∂λξ = Pξwe have the consistency condition∂P∂tr= [Qra, P](2.11a)This admit the only solutionP =∞Xk=2ktkQk−1a(2.11b)using simply eqs.

(2.10a’) and (2.10b’) beside the orthonormality conditions.To end this section let us pause a bit to make a comment on the DLS. As wewill see, it is justified to consider eq.

(2.10a) as a discrete version of the Schr¨odingerequation where λ plays the role of the spectral parameter; while eq. (2.10b) shows5

the KdV–type of (isospectral) deformations of the discrete Schr¨odinger equationwe will be discussing later on. It is well known that, typically, the Liouville theoryis characterized by a Schr¨odinger equation (actually by two of them, one for eachchirality).Even though we will be using throughout the paper only the linearsystem (2.10-11), it is interesting to notice that we can push this analogy furtherby introducing a discrete version of the Drinfeld-Sokolov linear system.

This isdone as follows.Introduce the two matrices Q1 and Q2 = (Q1)t, t meaning transposition, by(Q1)nm =pXnδn,m +sRnXn−1δn,m+1(2.15)and the block matrix and vectorQ = Q1−1−λQ2!,Ξ = ξ1ξ2! (2.16)Then the discrete Drinfeld-Sokolov linear system isQ Ξ = 0(2.17)This implies in particular Q2Q1ξ1 = λξ1, and we recover eq.

(2.10a), provided Q =Q2Q1. The latter condition allows us to uniquely determine the Xn’s introducedabove in terms of the Sn’s and Rn’s.

One findsX0 = S0,Xn =YnYn−1,n ≥1(2.18)where, introducing for the sake of homogeneity the notation Ri = Rii−1,Y0 = S0Y1 = S1S0 −R10· · ·Yn =X0≤k≤n2(−1)kXi1

We can go further. Let us introduce the sl2 generatorsH = 100−1!,E+ = 0100!,E−= 0010!while 1 will denote the identity 2 × 2 matrix.

Next defineδ = Q1 + Q22,∆= δ1π = Q1 −Q22,Π = πHE+ = E+ + λE−,A = Π −E+(2.20)Then the discrete Drinfeld-Sokolov linear system can be written(∆+ A) Ξ = 0(2.17′)We notice that E+ is the sum of the step operators corresponding to the simpleroots of the affine sl2 algebra, provided we identify the spectral parameter withthe loop parameter. In this sense A can be thought of as the discrete analogue ofan sl2 loop algebra connection.3.

Virasoro constraints from DLSAn important piece of information for the matrix model is contained in theso-called Virasoro constraints[4,5,6]LnZN(t) = 0,n ≥−1(3.1a)whereLn =∞Xk=1ktk∂∂tk−1−2N ∂∂tn+n−1Xk=1∂2∂tk∂tn−k+ N2δn0,(3.1b)[Ln, Lm] = (n −m)Ln+m. (3.1c)They completely determine the possible perturbations.7

In this section we show that the Virasoro constraints result from the consistencyconditions of the linear system (2.10-11), eqs. (2.13-14).To this end we rewrite the string equation (2.14) in the following form∞Xk=2ktk∂∂tk−1Q = −1,(3.2)where we have used the KdV equations (2.13).

Eq. (3.2) implies thatℓSn = −1,n ≥0;ℓ≡∞Xk=2ktk∂∂tk−1(3.3)which, by (2.6a), can be re–expressed asℓ∂∂t1ln hn = 1,∀n ≥0since the operator ℓcommutes with∂∂t1.

After integrating over t1, and using theformula (2.6b), we get∞Xk=1ktk(Qk−1)nn + α = 0,∀n ≥0. (3.4)At first sight the integration constant α seems to depend on t2, t3, ..., but using thediscrete KdV hierarchy and the string equation one can prove that α is actually aconstant.

After summation over n, from (2.9b) we obtain(ℓ−N(t1 −α))ZN(t) = 0We see that we can absorb α by a redefinition of t1. So finally we get(∞Xk=2ktk∂∂tk−1−Nt1)ZN(t) = 0(3.5)which is nothing but the L−1 constraint.8

We remark that choosing even potentials would imply Sn ≡0; therefore eq. (3.3)would be meaningless and would forbid us to recover the L−1 Virasoro condition.We will see later on that in the continuum limit this obstruction is removed.In order to derive the other Virasoro constraints, we introduce the quantitiesB(r)n≡pRn−1 .

. .

Rn−rPn,n−r,r ≥0. (3.6)Due to the string equation (2.14), one finds for the above symbols the recursionrelationsB(r+1)n+1−B(r+1)n= (Sn−r −Sn)B(r)n+ Rn−rB(r−1)n−Rn−1B(r−1)n−1+ δr,0The first few ones are as followsB(0)n= 0,B(1)n= nB(2)n= S0 + S1 + .

. .

+ Sn−2 −(n −1)Sn−1B(3)n=n−3Xi=0S2i +n−4Xi=0SiSi+1 −n−4Xi=0Si(Sn−2 + Sn−1)−Sn−3Sn−1 + 2n−3Xi=0Ri −(n −2)Rn−2. .

. .

. .

(3.7)On the other hand, from the KdV equations, it is easy to see that∂∂t1TrQ = ∂∂t1N−1Xi=0Si = −RN−1. (3.8)Furthermore, since Qr is a symmetric matrix, from the eqs.

(2.11), (3.4) with α = 0,9

one finds that∞Xk=1ktkQ(k−1)nl=Pnl,n > l0,n=l−Pnl,n < l.(3.9)Therefore, for any positive integer r, one obtains∞Xk=1ktkQ(k+r)nn=∞Xk=1n+r+1Xl=n−r−1(ktkQ(k−1))nl(Qr+1)ln= B(r+1)n+r+1 + B(r+1)n+ . .

..(3.10)Then, after summation over n, the above equations give the Lr–constraint. Forinstance, for the first three cases, it is easy to check that∞Xk=1ktkQknn =∞Xk=1ktkQk−1n,n−1Qn−1,n +∞Xk=1ktkQk−1nn Qnn +∞Xk=1ktkQk−1n,n+1Qn+1,n= B(1)n+ B(1)n+1 = 2n + 1∞Xk=1ktkQk+1nn= B(1)n+1(Sn + Sn+1) + B(1)n (Sn + Sn−1) + B(2)n+ B(2)n+2= 2[S0 + S1 + .

. .

+ Sn−1 + (n + 1)Sn]∞Xk=1ktkQk+2nn= B(3)n+3 + B(3)n+ B(2)n (Sn + Sn−1 + Sn−2) + B(2)n+2(Sn + Sn+1 + Sn+2)+ B(1)n (Rn + Rn−1 + Rn−2 + S2n−1 + SnSn−1 + S2n)+ B(1)n+1(Rn + Rn−1 + Rn+1 + S2n+1 + SnSn+1 + S2n),which after summing over n, and making use of (2.9b), (3.7) and (3.8), become the10

Virasoro constraintsLnZN(t) = 0,n = 0, 1, 2(3.11a)Ln =∞Xk=1ktk∂∂tk+n−2N ∂∂tn+n−1Xk=1∂2∂tk∂tn−k+ N2δn,0. (3.11b)The Virasoro algebraic structure (3.1c) ensures that the higher order constraintsare also true.4.

Gauge Symmetry and IntegrabilityAs we anticipated in the introduction the discrete linear system (2.10) is char-acterized by gauge symmetries which, on the one hand, ensure integrability and,on the other hand, allow us to envisage the double scaling limit as a singular gaugetransformation. This section and the following one are devoted to studying theproperties of these gauge tranformations.Let us consider the following transformation (at fixed tk’s)ξ −→ˆξ = G−1ξ,Q −→ˆQ = G−1QG(4.1)where G is a unitary matrix.

If the transformed Jacobi matrix ˆQ has the samestructure as Q, i.e. only the diagonal line and the first two off–diagonal lines arenon–zero, and, moreover, ifˆQˆξ = λˆξ∂∂tr ˆξ = ˆQraˆξ(4.2)then, we say that our linear system is gauge invariant.11

Let us examine these transformations more closely by considering the infinites-imal transformationG = 1 + εg.Then, the invariance requires the matrix g to satisfy the equationsˆQ = Q + ε[Q, g](4.3a)∂∂trg = [Qra, g] −[Qr, g]a. (4.3b)A non–trivial solution isg =XkbkQka,(4.4)where bk’s are time–independent constants.

By abuse of language we will call thisa “time–independent gauge transformation”.Let us consider the case when only bk is nonzero. ThenδQ = ˆQ −Q = εbk[Q, Qka] = −εbk∂∂tkQ.

(4.5)This corresponds to the transformation tk →tk −εbk, which can be rephrased bysaying that the tuning of the time parameters is realized by means of the gaugetransformation (4.4).This transformation has remarkable properties. On the one hand it can beconsidered as the discrete version of the conformal transformations, on the otherhand it leads to the integrability of the linear system (and consequently to that ofthe one–matrix model).12

Let us consider in detail the latter claim. We can think of δQ given by eq.

(4.5)as originating from a Poisson bracket in the following sense:δQ ≡ε{Ar, Q}(4.6)That is{Ar, Q} ≡[Qra, Q]. (4.7)where Ar represents a Hamiltonian to be determined.

For the Hamiltonians wemake the ansatzHr ≡1r∞Xn=0Qrnnr = 1, 2, . .

. (4.8)and corresponding to the choice Ar = Hr, Hr−1, Hr−2, ..., we obtain different Pois-son brackets.

More explicitly we write{Hr−k+1, Q}k = [Qr0, Q],(4.9)In the following we will study in detail only the cases k = 1, 2, 3. Comparing theLHS with the RHS which can be explicitly calculated, we can obtain the Poissonbrackets for Ri and Si.

Of course we have no a priori guarantee that the bracketsso obtained satisfy the Jacobi identity. This has to be checked a posteriori.While explicitly working out the Poisson brackets one realizes that there aretwo distint regimes according to whether Si = 0 or ̸= 0.i) First regime, i.e.

Si = 0. We find two meaningful Poisson brackets:{Ri, Rj}1 = RiRj(δj,i+1 −δi,j+1)(4.10)and{Ri, Rj}3 =RiRj(Ri + Rj)(δj,i+1 −δi,j+1)13

+ RjRj−1Rj−2δj,i+2 −RiRi−1Ri−2δi,j+2(4.11)while{Ri, Rj}2 ≡0ii) Second regime, i.e. Si ̸= 0.

We have two Poisson brackets:{Ri, Rj}1 =RiRj(δj,i+1 −δi,j+1)(4.12a){Ri, Sj}1 =RiSj(δj,i+1 −δi,j)(4.12b){Si, Sj}1 =Riδj,i+1 −Rjδi,j+1. (4, 12c)and{Ri, Rj}2 =2RiRj(Siδi,j−1 −Sjδi,j+1)(4.13a){Ri, Sj}2 =RiRj(δi,j−1 + δi,j) −RiRj+1(δi,j+1 + δi,j+2)+ RiS2j (δi,j −δi,j+1)(4.13b){Si, Sj}2 =(Si + Sj)(Rjδi,j−1 −Riδi,j+1)(4.13c)For k = 3 eq.

(4.9) does not define a consistent bracket.Let us conclude this section with a few remarks. Due to eq.

(4.7) one wouldexpect all the Hamiltonians to commute{Hn, Hm} = 0. (4.14)for any meaningful Poisson structure.

However due to the subtleties connectedwith traces of infinite matrices, one should verify this property starting from thePoisson brackets (4.9) and (4.10). We have done it for the first few cases.14

Moreover, from the definition of the Poisson brackets, we know that{Hr+2, Q}1 = {Hr, Q}3in the first regime and{Hr+1, Q}1 = {Hr, Q}2in the second, namely the two Poisson structures are compatible with each other.That is to say, in both regimes the linear system has infinitely many conservedquantities and possesses a bi–Hamiltonian structure.Finally the Poisson brackets (4.10) and (4.11) are the same we come acrossin the lattice version of the Liouville model [9]. In that case Ri plays the roleof a lattice deformation of the classical Virasoro generators.

However one shouldremember that in the first regime we cannot impose the string equation (see remarkafter eq.(3.5)). So we find a remarkable similarity of structures in the Liouvillemodel on the lattice and in the one-matrix model unconstrained by the stringequation.

Strictly speaking, the Poisson structures of the one-matrix model arethose of the second regime. It is an interesting open problem what field theory onthe lattice they correspond to.5.

Reparametrization and Time-Dependent Gauge TransformationsThe DLS (2.10) is form invariant under reparametrization of the tk couplings.That isQ(˜t)ξ(˜t) = λξ(˜t),∂∂˜trξ(˜t) = (Qra)(˜t)ξ(˜t)where ˜t = (˜t1, ˜t2, ...) and ˜tk is a smooth functions of the tk’s.This invariance is a formal one. However, by combining gauge and reparametriza-tion transformations, we can obtain significant symmetries of the system.

Let us15

consider the transformations( ˜tk = tk + ε(k −n)tk−n,∀k > n,n ≥−1˜tn = tn −2Nε,n ≥1ˆξ(˜t) = G−1(˜t)ξ(˜t),ˆQ(˜t) = G−1(˜t)Q(˜t)G(˜t)(5.1)In this kind of setup it is possible to find G = 1 + εg so thatˆQ(˜t) = Q(t)(5.2)and the linear system becomesQ(t)ˆξ(˜t) = λˆξ(˜t)∂∂tr ˆξ(˜t) = Qra(t)ˆξ(˜t)(5.3)We refer to these as time-dependent gauge transformations.Let us consider two examplesg = P,n = −1g = QP,n = 0(5.4)In these two cases eq. (5.2) is satisfied (the linear system is invariant but remark thatξ is not invariant for n = 0) and this fact leads straightforwardly to the L−1 andL0 Virasoro constraints.

We could as well obtain the other Virasoro constraints,but in these cases the matrix g has a complicated form and will not be writtendown here.16

6. The Double Scaling LimitThe purpose of this section is to recover a continuum version of the DLS (2.10)in the double scaling limit.

In general we will exploit the idea that the doublescaling limit is mimicked by a finite version of the transformation (5.1) abovewhen n = 0:tr −→γrtr,∀r(6.1)where γ is a finite constant. We recall that under this transformation Q remainsinvariant.

In other words, the double scaling limit is connected with a singularcase of a symmetry operation on our DLS.Let us consider a k-th order critical point and define, as usual, the continuumvariablesx ≡nβ,R(x) ≡Rn,ξ(x) ≡ξn.Moreover we setǫ ≡( 1β )12k+1,˜t0 = (1 −nβ )β2k2k+1,∂≡∂∂˜t0,(6.2)The double scaling limit corresponds toβ →∞,N →∞,˜t0 fixedFor large n ∼N one hasx = 1 −ǫ2k˜t0,R(x) = 1 + ǫ2u(˜t0)(6.3)where u(˜t0) is the specific heat.17

The latter ansatz requires a comment.Let us consider the string equation(2.14), suitably rescaled[Q, ¯P] = 1β ,¯P =∞Xr=2r¯trQr−1a,tr = β¯tr(6.4)where ¯tr are renormalized coupling constants (see below). This equation estab-lishes strong restrictions between the limiting expressions of Q and ¯P.

With thesimplifying assumption t2r+1 = 0 ∀r, the above ansatz is correct, as is well known;if we switch on the odd interactions the analysis is more complicated, the aboveansatz is still correct but we have not been able to exclude other solutions. In thefollowing we will stick to the case of even potentials and to (6.3).Let us now write down a few expansions which will be useful in the following.It is easy to see thatRn∓1 = R(x ∓1β ) = 1 + ǫ2u(˜t0 ± ǫ)ξn∓1 = ξ(x ∓1β ) = e±ǫ∂ξ(˜t0).Then, using Taylor expansion, we have the following formulasRn+b = 1 + ǫ2u −bǫ3u′ + b22 ǫ4u′′ −b36 ǫ5u′′′ + .

. .

(6.5a)qYi=0Rn+b+i = 1 + (q + 1)ǫ2u −12(q + 1)(2b + q)ǫ3u′+ ǫ4qXi=0(b + i)22u′′ + (q + 1)q2u2−ǫ5qXi=0(b + i)36u′′′ + (2b + q)(q + 1)q2uu′+ . .

. (6.5b)18

qYi=0pRn+b+i = 1 + q + 12ǫ2u −(q + 1)(2b + q)4ǫ3u′+ ǫ4qXi=0(b + i)24u′′ + (q + 1)(q −1)8u2−ǫ5qXi=0(b + i)312u′′′ + (2b + q)(q + 1)(q −1)8uu′+ . .

. (6.5c)Using these formulas one can deriveQ = 2 + ǫ2(∂2 + u) + O(ǫ3),(6.6a)−Qa = ǫ∂+ 16ǫ3(∂3 + 3u∂+ 32u′) + 18ǫ4(2u′∂+ u′′)+ 18ǫ5 115∂5 + 23u∂3 + u′∂2 + u′′∂−u2∂−uu′ + 13u′′′+ .

. .

,(6.6b)−Q2a = 2ǫ∂+ 43ǫ3(∂3 + 32u∂+ 34u′) + 12ǫ4(2u′∂+ u′′)+ ǫ5 415∂5 + 43u∂3 + 2u′∂2 + 32u′′∂+ 512u′′′+ . .

. ,(6.6c)−Q3a = 6ǫ∂+ ǫ3(5∂3 + 9u∂+ 92u′) + 92ǫ4(2u′∂+ u′′) + ǫ524110∂5+ 15u∂3 + 452 u′∂2 + 372 u′′∂+ 92u2∂+ 92uu′ + 112 u′′′+ .

. .

,(6.6d)−Q4a = 12ǫ∂+ 16ǫ3(∂3 + 32u∂+ 34u′) + 6ǫ4(2u′∂+ u′′) + 2ǫ5245 ∂5+ 16u∂3 + 24u′∂2 + 19u′′∂+ 6u2∂+ 6uu′ + 112 u′′′+ . .

. ,(6.6e)−Q5a = 30ǫ∂+ ǫ3(45∂3 + 75u∂+ 752 u′) + 75ǫ4(2u′∂+ u′′) + 54ǫ5(29∂519

+ 90u∂3 + 135u′∂2 + 111u′′∂+ 45u2∂+ 45uu′ + 33u′′′) + . .

. (6.6f)etc.Let us see now some consequences of the above expansions.

First of all letus notice that in the continuum limit the reduction to even potentials does notcontradict the string equation as in the discrete case. We should remember thatthe contradiction is exposed in eq.

(3.3).From the above expansions it is notdifficult to see that in the continuum limit it does not make sense to single out anequation like (3.3), while the LHS of the string equation is replaced by a differentialoperator even if t2r+1 = 0 ∀r. So in the continuum limit on can safely choose aneven potential.Next we consider the continuum limit of (2.10a).

From eq. (6.6a) we see that in aneighbourhood of the critical point λ ∼2.

Therefore we introduce the renormalizedquantities˜λ ≡ǫ−2(λ −2),˜Q ≡∂2 + u(6.7a)∂∂˜λξ = ˜Pξ,˜P = ǫ2P = ǫ1−2k ¯P. (6.7b)Then, the discrete Schr¨odinger equation goes over to its continuum version(∂2 + u)˜ξ(˜t0) = ˜λ˜ξ(˜t0)(6.8)Similarly the string equation becomes[ ˜Q, ˜P] = 1.

(6.9)We are now in a condition to explicitly determine critical points. From eq.

(2.11b)and (6.7b) (recall that we are working with the simplifying assumption of even po-20

tential) we have˜P = ǫ2(2t2Qa + 4t4Q3a + 6t6Q5a + . .

. )= ǫ1−2k(2¯t2Qa + 4¯t4Q3a + 6¯t6Q5a + .

. .

)(6.10)We remarked above that the string equation (6.9) puts severe restrictions not onlyin the discrete case but also in the double scaling limit. From eq.

(6.6) and (6.7),we see in particular that all the operators Qra’s are vanishing in the limit ǫ →0, sothat if one wants the string equation to be satisfied, one must let a certain subsetof bare coupling constants in P go to infinity (DSL). The practical recipe is to lookfor combinations of ¯t2r’s such that all the singular terms in the second expressionof (6.10) vanish.

Let us see a few significant examples.i) k=2. In this case only ¯t2 and ¯t4 are nonzero, and β = ǫ−5.

Then, from(6.6b,d) we see that only if we sett2 = 158 ǫ−5˜t2 = ǫ−5¯t2,t4 = −532ǫ−5˜t2 = ǫ−5¯t4,are we able to eliminate the ǫ−1∂term in ˜P, and we get the known operator˜P = 52˜t2(∂2 + u)32+ + O(ǫ)The string equation becomes−52˜t2R′2[u] = 1where we have introduced the Gelfand–Dickii polynomialsR′k[u] ≡[(∂2 + u)k−12+, ∂2 + u]. (6.11)As is well-known at the critical point ˜tc2 =815, the above string equation is the21

Painlev´e equation of first kind˜t0 = 13u′′ + u2.ii) k=3. Only t2, t4 and t6 are non-vanishing, β = ǫ−7.

According to the aboverecipe we kill all the negative powers of ǫ in ˜P if we putt2 = −10532 ǫ−7˜t3 = ǫ−7¯t2,t4 = 3564ǫ−7˜t3 = ǫ−7¯t4,t6 = −7192ǫ−7˜t3 = ǫ−7¯t6,It is straightforward to show that˜P = 72˜t3(∂2 + u)52+ + O(ǫ)and the string equation becomes−72˜t3R′3[u] = 1.The third critical point is at ˜tc3 = −1635 and the differential equation is˜t0 = −(u3 −12u′2 −uu′′ + 110u(4)).One can proceed in this way and determine higher order critical points. Asis well-known, on a very general ground the form of the operator ˜P must be asfollows˜P =∞Xn=1(n + 12)˜tn ˜Qn+ 12++ O(ǫ).

(6.12)We conjecture that this form is induced by the following coupling redefinitions:t2r = −∞Xn=r(n + 12)˜tnCna(n)r ǫ−(2n+1) ≡∞Xn=rΓnr ˜tn(6.13)22

wherea(n)r= (−1)r+1 n! (r −1)!

(n −r)! (2r)!We checked eq.

(6.13) for the first few cases and foundC1 = 1,C2 = −34,C3 = 58,C4 = −3564, . .

.In general one hasCn = (−1)n+1(2n −1)!!2n−1n! .These factors are determined in such a way as to reproduce the standard KdVhierarchy.It is worth noticing that 1) in eq.

(6.12) and (6.13) we are considering allthe critical points at a time, and 2) the time transformation (6.13) is made ofa reparametrization plus a scale transformation of the type (6.1).What is left for us to do is to analyze the continuum limit of the KdV hierarchy.On the basis of eq. (6.13) one naively has∂∂˜tn= −nXr=1(n + 12)Cnǫ−(2n+1)a(n)r∂∂t2r.

(6.14)So, in particular, on the basis of (2.10) and (6.6) we must have∂∂˜t1ξ =32ǫ−2∂+ (∂2 + u)32+ + O(ǫ)ξ∂∂˜t2ξ =−158 ǫ−4∂+ (∂2 + u)52+ + O(ǫ)ξ∂∂˜t3ξ =3532ǫ−6∂+ (∂2 + u)72+ + O(ǫ)ξetc. These are however naive formulae since {˜tn,n ≥1} is not a complete setof parameters after we take the continuum limit.

To correct this we have to allow23

also for a ∂–dependent term in the RHS of (6.14). This additional term takescare exactly of the divergent terms (in the ǫ →0 limit) in the RHS of the aboveequations.Finally the evolution equations become∂∂˜tn˜ξ = (∂2 + u)n+ 12+˜ξ,n ≥0(6.15)which result in the standard KdV flow∂∂˜tnu = [(∂2 + u)n+ 12+, ∂2 + u],n ≥0(6.16)In eq.

(6.15) ˜ξ is the limit of ξ possibly multiplied by an ǫ–dependent factorwhich may be necessary in order to obtain a finite result.7. The Virasoro Constraints in the Continuum LimitIn the previous section starting from the DLS we have obtained, near criticality,a continuous linear system(∂2 + u)˜ξ = ˜λ˜ξ(7.1a)∂∂˜tn˜ξ = (∂2 + u)n+ 12+˜ξ(7.1b)˜P =∞Xn=1(n + 12)˜tn ˜Qn+ 12+(7.4)whose consistency conditions are[ ˜Q, ˜P] = 1(7.2a)∂∂˜tnu = [(∂2 + u)n+ 12+, ∂2 + u](7.2b)∂∂˜tn˜P = [(∂2 + u)n+ 12+, ˜P].

(7.2c)We want now to recover the Virasoro constraints in this continuous system.The strategy is the same as for the discrete case. We use essentially the string24

equation (7.2a). First of all, as is well known, in the continuum limit the partitionfunction behaves likeln Z =N−1Xk=1(N −k) ln Rk + constantterms→˜t0Z0d˜t′0(˜t0 −˜t′0)u(˜t′0) + O(ǫ) + regularterms⇒∂2 ln ˜Z = u(˜t0)(7.3)where, in taking the continuum limit, we passed to the normalized partition func-tion˜Z ≡Z(˜t)/Z(T)(7.4)the parameter T being a reference point connected with the extremum of integra-tion ˜t0 = 0.Now, using (7.1c), eq.

(7.2a) can be written−∞Xk=1(k + 12)˜tkR′k[u] = 1. (7.5)Integrating once with respect to t0, we obtain∞Xk=1(k + 12)˜tkRk[u] −Rk[0]+ ˜t0 = 0(7.6)where Rk[0] is Rk[u] computed at ˜t0 = 0IfonedirectlyconsidersthecontinuumlimitofthediscreteV irasoroconstraints,itisnohisisactuallythecase.

.25

In order to simplify the next formulas let us introduce the recursion operatorˆφ ≡14∂2 + u + 12u′∂−1and define the recursion relation for the Gelfand–Dickii polynomialsR′n+1 = ˆφR′n = ˆφn∂u. (7.7)Remembering that, on the basis of our conventions, we have∂−1R′k[u] = Rk[u] −Rk[0]eq.

(7.6) can be rewrittenF ≡˜t0 +∞Xk=1(k + 12)˜tk∂−1 ˆφk−1∂u = 0. (7.8a)On the same basis we can write∂−1 ˆφn+1∂F = 0,n ≥−1(7.8b)To obtain these equations we have used only the string equation.

We can aswell envisage eqs. (7.8) as a consequence of a symmetry of the system, precisely asa consequence of the fact that u(˜t0) and the KdV hierarchy are invariant underthe transformations˜tk −→¯tk = ˜tk + ǫ(k −n + 12)˜tk−nu(˜t) −→ˆφn+1 · 1 + u(¯tk).

(7.9)26

The generators associated with (7.9) areL−1 =∞Xk=1(k + 12)˜tk∂∂˜tk−1+ 14ρ˜t20,L0 =∞Xk=0(k + 12)˜tk∂∂˜tk+ 116,Ln =∞Xk=0(k + 12)˜tk∂∂˜tk+n+ ρ4nXk=1∂2∂˜tk−1∂˜tn−k,n ≥1. (7.10)Here, for later purposes, we have introduced a constant ρ (for example, by rescalingall the ˜t’s).Even though we will not use it in the following, it is worth mentioning thatthere is a larger symmetry of the system: the latter is also invariant under thetransformations˜tk −→˜tk + ǫ(7.11a)whose generators are given byVk = ∂∂˜tk,k ≥0.

(7.11b)The generators Vk’s and Ln’s characterize the master symmetry of the KdV hier-archy [9,11]. The corresponding algebra is[Vk, Vl] = 0,k, l ≥0,(7.12a)[Vk, Ln] = (k + 12)Vk+n,k ≥0, k + n ≥0,(7.12b)[V0, L−1] = 12ρ˜t0(7.12c)[Ln, Lm] = (n −m)Ln+m,n, m ≥−1.

(7.12d)27

But let us return to the derivation of the Virasoro constraint. Representingnow (7.8) in terms of the partition function, we get for n = −1, 0∂ ∞Xn=1(n + 12)˜tn∂∂˜tn−1ln ˜Z + 12ρ˜t20= 0∂ ∞Xn=0(n + 12)˜tn∂∂˜tnln ˜Z= 0or, in general,∂lnp˜Zp˜Z= 0,n ≥−1.

(7.13)where, by definition,l0 = L0 −116,ln = Ln,n ̸= 0.The most general solution of (4.28) has the formlnp˜Z = bnp˜Z(7.14)where the bn’s are ˜t0–independent but arbitrary functions of the other parameters.In order to determine them we remark that they must be compatible with thealgebra (7.12). In particular they must satisfy[ln, bm] −[lm, bn] = (n −m)bn+m + 18nδn+m,0.

(7.15)Moreover we remark that (7.12) is a graded algebra provided we define the degreeas followsdeg[˜tk] ≡−k,deg[ ∂∂˜tl] ≡l,deg[ρ] ≡1Therefore, deg[Ln] = n and, from (7.14),deg[bn] = n,n ≥−1(7.16)28

The general form of bn will be a sum of monomials of the following typeMn(p, q1, ..., qa) = const ρp ˜tq1n1...˜tqanawhere p is a real number and q1, q2, ... are nonnegative real numbers (we excludenegative exponents as it is natural to require a smooth limit of bn as any one ofthe couplings vanishes). Next we remember that the parameter ρ appeared onthe scene because of a rescaling ˜tn →√ρ˜tn.

Therefore if we perform the oppositerescaling the dependence of bn on ρ must disappear. This implies the conditionp = 12(q1 + ... + qa).

So the degree of the above monomial would bedeg[Mn(p, q1, ..., qa)] =aXi=1(12 −ni)qi(7.17)Comparing eq. (7.16) with (7.17), we see all the bn’s are zero except perhaps for b0which must be a constant, and b−1, which could depend on ˜t1 and ρ.

We now usethe consistency conditon (7.15). For n = 1 and m = −1 it tells us that52˜t2∂b−1∂˜t1= 2b0 + 18This allows us to conclude thatb0 = −116.and b−1 does not depend on ˜t1.

Next we use again (7.15) for n = 0 and m = −1and conclude thatb−1 = 0Collecting the above results we obtain the Virasoro constraintsLnp˜Z = 0,n ≥−1(7.18)with Ln given by (7.10). Another proof of the same result is given in Appendix.29

Finally we recall that due to (7.2b) and (7.3),p˜Z is a τ-function of the KdVhierarchy.AppendixThis Appendix is devoted to another proof of eq.(7.18). We start from eq.

(7.15)for m = −1 and m = 0 and analyze the ˜t0 dependence. Since all the bn are ˜t0–independent we get, in the first case (m = −1),∂b−1∂˜tn= 0,n ≥1and, in the second case (m = 0),∂b0∂˜tn= 0,n ≥1In conclusion both b0 and b−1 are constant.

Applying now again eq. (7.15) for n = 0and m = −1, we obtainb−1 = 0i.e.eq.

(7.18) is true for n = −1.Since we have shown above thatp˜Z is aτ-function of the KdV hierarchy, we can now apply a theorem of ref. [11] whichasserts that if L−1τ = 0, then Lnτ = 0, ∀n ≥−1.Consequently eq.

(7.18) isproven.References[1] E. Brezin and V. Kazakov, Phys.Lett.236B (90) 144;M. Douglas and S. Shenker, Nucl.Phys.B335 (90) 635;30

D. Gross and A. Migdal, Phys.Rev.Lett.64 (90) 127;T. Banks, M. Douglas, N. Seiberg and S. Shenker, Phys.Lett.238B (90) 279;M. Douglas, Phys.Lett. 238B (90) 176.

[2] V.A.Kazakov, in Proc.Carg´ese workshop in 2d gravity, eds.O.Alvarez,E.Marinari and P.Windey;L.Alvarez-Gaum´e, Helv.Phys.Acta 64 (1991) 361;A.Bilal, CERN-TH.5867/90. [3] E.J.Martinec, “On the Origin of Integrability of Matrix Models”, Chicagopreprint, EFI–90–67.

[4] A. Gerasimov, A. Marshakov, A. Mironov, A. Morozov, and A. Orlov, “Ma-trix Models of Two–Dimensional Gravity and Toda Theory”, ITEP preprint(1990). [5] R. Dijkgraaf, H. Verlinde, and E. Verlinde Nucl.Phys.

B348(1991)435. [6] M. Fukuma, H. Kawai and R. Nakayama, “Continuum Schwinger–DysonEquations and Universal Structures in Two–Dimensional Quantum Gravity”,Tokyo preprint KEK–TH–251 (1990).

[7] E. Brezin, C. Itzykson, G. Parisi, and J.-B. Zuber, Comm.Math.Phys.

59(78) 35;[8] E. Witten, “Two–Dimensional Gravity and Intersection Theory on ModuliSpace”, IASSNS–HEP–90–45 (1990). [9] L.D.Fadeev and L.Takhtajan, Lect.Notes in Phys.

Vol. 246 (Springer, Berlin,1986), 66;O.Babelon, Phys.Lett.

B215 (1988) 523;A.Volkov, Theor.Math.Phys. 74 (1988) 135.

[10] HoSeong La, “Symmetries in Nonperturbative 2-d Quantum Gravity, Penn-sylvania preprint, UPR–0432 T (1990); W.Oevel, “Master symmetries: weak31

action angle structure for hamiltonian and non-hamiltonian systems” Pader-born preprint (1987); J.Goeree, “W-constraints in two dimensional quantumgravity” Utrecht preprint THU-199/90. [11] V.Kac and A.Schwarz, “Geometric Interpretation of Partition Function of2D Gravity” preprint.32

---

출처: arXiv:9109.054 • 원문 보기