KP nuevo takasaki

8/14/2019 KP nuevo takasaki

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arXiv:0912

.4867v1

[math-ph]24Dec2009

-EXPANSION OF KP HIERARCHY:

RECURSIVE CONSTRUCTION OF SOLUTIONS

KANEHISA TAKASAKI AND TAKASHI TAKEBE

Abstract. The -dependent KP hierarchy is a formulation of the KP hierar-

chy that depends on the Planck constant and reduces to the dispersionlessKP hierarchy as 0. A recursive construction of its solutions on the

basis of a Riemann-Hilbert problem for the pair (L,M) of Lax and Orlov-Schulman operators is presented. The Riemann-Hilbert problem is convertedto a set of recursion relations for the coefficients Xn of an -expansion of

the operator X = X0 + X1 + 2X2 + for which the dressing operatorW is expressed in the exponential form W = exp(X/). Given the lowest

order term X0, one can solve the recursion relations to obtain the higherorder terms. The wave function associated with W turns out to have

the WKB form = exp(S/), and the coefficients Sn of the -expansion

S = S0 + S1 + 2S2 + , too, are determined by a set of recursion rela-tions. This WKB form is used to show that the associated tau function has

an -expansion of the form log = 2F0 + 1F1 + F2 + .

0. Introduction

The KP hierarchy can be completely solved by several methods. The most clas-sical methods are based on Grassmann manifolds [SS], [SW], fermions and vertexoperators [DJKM] and factorisation of microdifferential operators [Mu]. Unfor-

tunately, those methods are not very suited for a quasi-classical (-dependent,where is the Planck constant) formulation [TT2] of the KP hierarchy.

The -dependent formulation of the KP hierarchy was introduced to study thedispersionless KP hierarchy [KG], [Kr], [TT1] as a classical limit (i.e., the lowestorder of the -expansion) of the KP hierarchy. This point of view turned out to bevery useful for understanding various features of the dispersionless KP hierarchysuch as Lax equations, Hirota equations, infinite dimensional symmetries, etc., inthe light of the KP hierarchy. In this paper, we return to the -dependent KPhierarchy itself, and consider all orders of the -expansion.

We first address the issue of solving a Riemann-Hilbert problem for the pair(L, M) of Lax and Orlov-Schulman operators [OS]. This is a kind of quantisationof a Riemann-Hilbert problem that solves the dispersionless KP hierarchy [TT1].Though the Riemann-Hilbert problem for the full KP hierarchy was formulated

in our previous work [TT2], we did not consider the existence of its solution in ageneral setting. In this paper, we settle this issue by an -expansion of the dressingoperator W, which is assumed to have the exponential form W = exp(X/) with anoperator X of negative order. Roughly speaking, the coefficients Xn, n = 0, 1, 2, . . .,of the -expansion of X are shown to be determined recursively from the lowestorder term X0 (in other words, from a solution of the dispersionless KP hierarchy).

Date: 24 December 2009.

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2 KANEHISA TAKASAKI AND TAKASHI TAKEBE

We next convert this result to the language of the wave function . This, too,is to answer a problem overlooked in our previous paper [TT2]. Namely, giventhe dressing operator in the exponential form W = exp(X/), we show that theassociated wave function has the WKB form = exp(S/) with a phase functionS expanded into nonnegative powers of. This is genuinely a problem of calculusof microdifferential operators rather than that of the KP hierarchy. A simplestexample such as X = x()1 demonstrates that this problem is by no meanstrivial. Borrowing an idea from Aokis exponential calculus of microdifferentialoperators [A], we show that dressing operators of the form W = exp(X/) and wavefunctions of the form = exp(S/) are determined from each other by a set ofrecursion relations for the coefficients of their -expansion. More precisely, we needmany auxiliary quantities other than X and S, for which we can derive a large setof recursion relations. Thus our construction is essentially recursive. Consequently,the wave function of the solution of the aforementioned Riemann-Hilbert problem,too, are recursively determined by the -expansion.

Having the

-expansion of the wave function, we can readily derive an

-expansionof the tau function as stated in our previous work [TT2]. This -expansion is ageneralisation of the genus expansion of partition functions in string theories andrandom matrices [D], [Kr], [Mo], [dFGZ].

This paper is organised as follows. Section 1 is a review of the -dependent formu-lation of the KP hierarchy. Relevant Riemann-Hilbert problems are also reviewedhere. Section 2 presents the recursive solution of the Riemann-Hilbert problem. Atechnical clue is the Campbell-Hausdorff formula, details of which are collected inAppendix A. The construction of solution is illustrated for the case of the Kont-sevich model [AvM] in Appendix B. Section 3 deals with the -expansion of thewave function. Aokis exponential calculus is also briefly reviewed here. Section 4mentions the -expansion of the tau function. Section 5 is devoted to concludingremarks.

Acknowledgements. The authors are grateful to Professor Akihiro Tsuchiya fordrawing our attention to this subject and to Professor Masatoshi Noumi for in-structing how to use Mathematica in algebra of microdifferential operators. Com-putations in Appendix B were done with the aid of one of his Mathematica pro-grammes. This work is partly supported by Grant-in-Aid for Scientific Research No.19540179 from the Japan Society for the Promotion of Science. TT is partly sup-ported by the grant of the State University Higher School of Economics, Russia,for the Individual Research Project 09-01-0047 (2009).

1. -dependent KP hierarchy: review

In this section we recall several facts on the KP hierarchy depending on a formal

parameter

in [TT2], 1.7.The -dependent KP hierarchy is defined by the Lax representation

(1.1) L

tn= [Bn, L], Bn = (L

n)0, n = 1, 2, . . . ,

where the Lax operator L is a microdifferential operator of the form

(1.2) L = +n=1

un+1(, x , t)()n, =

x,


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-EXPANSION OF KP 3

and ( )0 stands for the projection onto a differential operator dropping negativepowers of . The coefficients un(, x , t) of L are assumed to be formally regularwith respect to . This means that they have an asymptotic expansion of the form

un(, x , t) =

m=0 mu

(m)n (x, t) as 0.

We need two kinds of order of microdifferential operators: one is the ordinaryorder,

(1.3) ord

an,m(x, t)nm

def= max{m | an,m(x, t) = 0},

and the other is the -order defined by

(1.4) ord

an,m(x, t)nm

def= max{m n | an,m(x, t) = 0}.

In particular, ord = 1, ord = 1, ord = 0. For example, the conditionwhich we imposed on the coefficients un(, x , t) can be restated as ord

(L) = 0.The principal symbol(resp. the symbol of order l) of a microdifferential operator

A = an,m(x, t)nm with respect to the -order is(A)

def=

mn=ord(A)

an,m(x, t)m(1.5)

(resp. l (A)def=

mn=l

an,m(x, t)m).(1.6)

When it is clear from the context, we sometimes use instead ofl .

Remark 1.1. This order coincides with the order of an microdifferential operatorif we formally replace with 1t0 , where t0 is an extra variable. In fact, naivelyextending (1.1) to n = 0, we can introduce the time variable t0 on which nothingdepends. See also [KR].

As in the usual KP theory, the Lax operator L is expressed by a dressing operator

W:(1.7) L = Ad W() = W()W1

The dressing operator W should have a specific form:

W = exp(1X(,x,t,))()()/,(1.8)

X(,x,t,) =k=1

k(, x , t)()k,(1.9)

ord(X(,x,t,)) = ord () = 0,(1.10)

and () is a constant with respect to x and t. (In [TT2] we did not introduce ,which will be necessary in Section 2.)

The Orlov-Schulman operatorM is defined by

(1.11) M = Ad

Wexp1(t,)

x = W

n=1

ntn()n1 + x

W1

where (t,) =

n=1 tn()n. It is easy to see that M has a form

(1.12) M =n=1

ntnLn1 + x + ()L1 +

n=1

vn(, x , t)Ln1,

and satisfies


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ord(M) = 0; the canonical commutation relation: [L, M] = ; the same Lax equations as L:

(1.13) M

tn= [Bn, M], n = 1, 2, . . . .

Remark 1.2. If an operator M of the form (1.12) satisfies the Lax equations (1.13)and the canonical commutation relation [L, M] = with the Lax operator L of theKP hierarchy, then () in the expansion (1.12) does not depend on any tn nor onx. In fact, expanding the canonical commutation relation, we have

+

x()1 + (lower order terms) = ,

which implies x = 0. Similarly, from (1.13) followstn

= 0 with the help of (1.1)

and [Ln, M] = nLn1.

The following proposition (Proposition 1.7.11 of[TT2]) is a dispersionful coun-terpart of the theorem for the dispersionless KP hierarchy found earlier (Proposition7 of [TT1]; cf. Proposition 1.4 below).

Proposition 1.3. (i) Suppose that operators f(, x, ), g(, x,), L and Msatisfy the following conditions:

ord f = ord g = 0, [f, g] = ; L is of the form (1.2) and M is of the form (1.12), [L, M] = ; f(, M , L) and g(, M , L) are differential operators:

(1.14) (f(, M , L))


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-EXPANSION OF KP 5

The principal symbol of the Orlov-Schulman operator is

(1.19) M =

n=1

ntnLn1 + x + 0L1 +

n=1

v0,nLn1, v0,n :=

(vn),

which is equal to

(1.20) M = exp

ad{,} X0

exp

ad{,} 0 log

exp

ad{,} (t, )

x,

where (t, ) =

n=1 tnn. The series M satisfies the canonical commutation

relation with L, {L, M} = 1 and the Lax type equations:

(1.21)M

tn= {Bn, M}, n = 1, 2, . . . .

The Riemann-Hilbert type construction of the solution is essentially the same asProposition 1.3. (We do not need to assume the canonical commutation relation{L, M} = 1.)

Proposition 1.4. (i) Suppose that functions f0(x, ), g0(x, ), L and M satisfythe following conditions:

{f0, g0} = 1; L is of the form (1.15) and M is of the form (1.19) f0(M, L) and g0(M, L) do not contain negative powers of

(1.22) (f0(M, L))


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are both differential operators (cf. Proposition 1.3). Let us expand X and withrespect to the -order as follows:

X(,x,t,) =n=0

nXn(x,t,), Xn(x,t,) =

k=1

n,k(x, t)()k,(2.2)

() =n=0

nn,(2.3)

where n,k and n do not depend on and hence k in (1.9) is expanded ask =

n=0

nn,k.Assume that the solution of the dispersionless KP hierarchy corresponding to

((f), (g)) is given. In other words, assume that a symbol X0 =

k=1 0,k(x, t)k

and a constant 0 are given such that

(f)(L, M) = exp

ad{,} X0

exp

ad{,} 0 log

exp

ad{,} (t, )

(f)(x, )

(g)(L, M) = expad{,}

X0 expad

{,}

0log expad

{,}(t, )(g)(x, )

do not contain negative powers of :

(2.4)

(f)(L, M)


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-EXPANSION OF KP 7

Expand P(i1) and Q(i1) with respect to the -order as

P(i1) = P(i1)0 + P

(i1)1 + +

kP(i1)k + (2.11)

Q(i1) = Q(i1)0 + Q

(i1)1 + +

kQ(i1)k + .(2.12)

(ord P(i1)k = ord

Q(i1)k = 0.)

(Step 2) Put P0 := (P(i1)

0 ), Q0 := (Q

(i1)0 ), P

(i1)i :=

(P(i1)i ),

Q(i1)i :=

(Q(i1)i ) and define a constant i and a series Xi(x,t,) =

k=1 i,k(x, t)k by

(2.13) i log + Xi :=

Q0

P(i1)i

P0

Q(i1)i

1

d.

The integral constant of the indefinite integral is fixed so that the righthand side agrees with the left hand side.

(Step 3) Define a series Xi(x,t,) = k=1 i,k(x, t)k byXi = X

i

1

2{(X0), X

i } +

p=1

K2p(ad{,}((X0)))

2pXi ,

Xi := i log + Xi(x, ) exp(ad{,} (X0))(i log ).

(2.14)

Here K2p is determined by the generating function

(2.15)z

ez 1= 1

z

2+

p=1

K2pz2p,

or K2p = B2p/(2p)!, where B2ps are the Bernoulli numbers. (Step 4) The operator Xi(x,t,) is defined as the operator with the prin-

cipal symbol Xi:

(2.16) Xi =k=1

i,k(x, t)()k.

The main theorem is the following:

Theorem 2.1. Assume that X0 and 0 satisfy (2.4) and construct Xis and isby the above procedure recursively. Then X and defined by (2.2) satisfy (1.14).Namely W = exp(X/)()/ is a dressing operator of the -dependent KP hier-archy.

The rest of this section is the proof of Theorem 2.1 by induction.Let us denote the known part of X and by X(i1) and (i1) as in (2.8)

and, as intermediate objects, consider P(i1)

and Q(i1)

defined by (2.9) and (2.10),which are expanded as (2.11) and (2.12).IfXand are expanded as (2.2) and (2.3), the dressing operator W = exp(X/) exp( log()/)

is factored as follows by the Campbell-Hausdorff theorem:

(2.17) W = expi1(i log() + Xi) +

iX>i

exp1X(i1)

exp

1(i1) log()

,


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where ord(i log() + Xi(x,)) = 0, ord(X>i) 0 and the principal symbol

of i log() + Xi(x,) is defined by

(2.18) (i log() + Xi)(x, )

=n=1

(ad{,} (X0))

n1

n!(Xi) + exp

ad{,}

(X0)

(i log ).

Note that the only log term in (2.18) is i log and the rest is sum of negative

powers of . The principal symbol of Xi is recovered from Xi by the formula

(Xi) = (Xi)

1

2{(X0),

(Xi)} +

p=1

K2p(ad{,}((X0)))

2p(Xi),

(Xi) := (Xi)(x, ) exp(ad{,}

(X0))(i log )

=n=1

(ad{,} (X0))n1

n!(Xi).

(2.19)

Here coefficients K2p are defined by (2.15). This inversion relation is the origin of(2.14). (Note that the principal symbol determines the operator Xi, since it is ahomogeneous term in the expansion (2.2).) We prove formulae (2.17) and (2.19) inAppendix A.

The factorisation (2.17) implies

P = Ad

expi1(i log() + Xi) +

iX>i

P(i1)

=P(i1) + i1[(i log() + Xi) + X>i, P(i1)] + (terms of-order < i).

Thus, substituting the expansion (2.11) in the step 1, we have

P =P(i1)0 + P

(i1)1 + +

iP(i1)i +

+ i1[i log() + Xi, P(i1)

0 ]

+ (terms of-order < i).

(2.20)

Comparing this with the -expansion of P (2.5), we can express Pis in terms of

P(i1)j , Xi and i as follows:

Pj = P(i1)j (j = 0, . . . , i 1),(2.21)

0(Pi) = 0(P(i1)i + h

1[i log() + Xi, P(i1)0 ]).(2.22)

Similar equations for Q are obtained in the same way. The first equations (2.21)show that the terms of -order greater than i in (2.5) are already fixed byX0, . . . , X i1 and 0, . . . , i1, which justifies the inductive procedure. That isto say, we are assuming that X0, . . . , X i1 and 0, . . . , i1 have been already de-

termined so that Pj = P(i1)j and Qj = Q

(i1)j for j = 0, . . . , i 1 are differential

operators.The operator Xi and constant i should be chosen so that the right hand side

of (2.22) and the corresponding expression for Q are differential operators. Taking


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-EXPANSION OF KP 9

equations P(i1)

0 = P0 and Q(i1)0 = Q0 into account, we define

P(i)i := P

(i1)i +

1[i log() + Xi, P0],

Q(i)i := Q

(i1)i +

1[i log() + Xi, Q0].(2.23)

Then the condition for Xi and i is written in the following form of equations forsymbols:

(2.24) (0 (P(i)i ))1 = 0, (

0 (Q(i)i ))1 = 0.

(The parts of -order less than 1 should be determined in the next step of the

induction.) To simplify notations, we denote the symbols 0 (P(i)i ),

0 (P

(i1)i ) and

so on by the corresponding calligraphic letters as P(i)i , P

(i1)i etc. By this notation

we can rewrite the equations (2.24) in the following form:

(2.25)(P

(i)i )1 = 0, P

(i)i := P

(i1)i + {i log + Xi, P0} = 0,

(

Q

(i)

i )1 = 0,

Q

(i)

i := Q

(i1)

i + {i log +

Xi, Q0} = 0.The above definitions of P

(i)i and Q

(i)i are written in the matrix form:

(2.26)

P0x

P0

Q0x

Q0

(i log + Xi)

x(i log + Xi)

=

P

(i)i P

(i1)i

Q(i)i Q

(i1)i

.

Recall that operators P(i1) and Q(i1) are defined by acting adjoint operation tothe canonically commuting pair (f, g) in (2.9), (2.10) and (2.7). Hence they alsosatisfy the canonical commutation relation: [P(i1), Q(i1)] = . The principalsymbol of this relation gives

{P(i1)0 , Q

(i1)0 } = {P0, Q0} = 1,

which means that the determinant of the matrix in the left hand side of (2.26) isequal to 1. Hence its inverse matrix is easily computed and we have

(2.27)

(i log + Xi)

x(i log + Xi)

=

Q0

P0

Q0x

P0x

P

(i)i P

(i1)i

Q(i)i Q

(i1)i

.

We are assuming that P0 and Q0 do not contain negative powers of and we are

searching for i log + Xi such that P(i)i and Q

(i)i are series of without negative

powers. Since i is constant with respect to x, the left hand side of (2.27) containonly negative powers of . Thus taking the negative power parts of the both hand

sides in (2.27), we have

(2.28)

(i log + Xi)

x(i log + Xi)

=

Q0

P(i1)i

P0

Q(i1)i

1

Q0x

P(i1)i

P0x

Q(i1)i

1

.

This is the equation which determines i and Xi.


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-EXPANSION OF KP 11

Since the time variables tn do not play any role in this section, we set them tozero. As the factor ()/ in (1.8) becomes a constant factor z when it is appliedto exz/, we also omit it here.

Let A(, x, ) =

n an(, x)()n be a microdifferential operator. The total

symbol of A is a power series of defined by

(3.4) tot(A)(, x , ) :=n

an(, x)n.

Actually, this is the factor which appears when the operator A is applied to exz/:

(3.5) Aexz/ = tot(A)(, x , z)exz/.

Using this terminology, what we show in this section is that a operator of the formeX/ has a total symbol of the form eS/ and that an operator with total symboleS/ has a form eX/. Exactly speaking, the main results in this section are thefollowing two propositions.

Proposition 3.1. Let X = X(, x,) be a microdifferential operator such thatord X = 1 and ord X = 0. Then the total symbol of eX/ has such a form as

(3.6) tot(exp(1X(, x,))) = eS(,x,)/,

where S(, x , ) is a power series of 1 without non-negative powers of and hasan-expansion

(3.7) S(, x , ) =n=0

nSn(x, ).

Moreover, the coefficient Sn is determined by X0, . . . , X n in the -expansion(2.2) of X =

n=0

nXn.

Explicitly, Sn is determined as follows:

(Step 0) Assume that X0, . . . , X n are given. Let Xi(x, ) be the total symboltot(Xi(x,)).

(Step 1) Define Y(l)k,m(x, y, , ) and S

(l)(x, ) by the following recursion re-lations:

Y(l)k,1 = 0(3.8)

S(0)m = 0,(3.9)

Y(l)

0,m(x, y, , ) = l,0Xm(x, )(3.10)

for l 0, m = 0, . . . , n,

(3.11) Y(l)k+1,m(x, y, , )

=1

k + 1

yY(l)k,m1(x, y, , ) + 0ll10mm

Y(l)k,m(x, y, , )yS

(ll)mm(y, )

for k 0, and

(3.12) S(l+1)m (x, ) =1

l + 1

l+mk=0

Y(l)k,m(x,x,,).


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-EXPANSION OF KP 13

(We shall show that Y(l)k,m,j = 0 for l + k > j .) Schematically this procedure

goes as follows:

Y(l)k,m,1 = l,0k,0Sm,1

Y(l)k,m,1(m

< m) Y(l)k,m,2 (k, l = 0) Y

(0)0,m,2 Sm,2

Y(l)k,m,1, Y

(l)k,m,2(m

< m) Y(l)k,m,3 (k, l = 0) Y

(0)0,m,3 Sm,3

...

(Step 2) Xn(x, ) =

j=1 Y(0)

0,n,j(x,x,,). (The infinite sum is the homo-

geneous expansion in terms of powers of .)

Combining these propositions with the results in Section 2, we can, in principle,make a recursion formula for Sn (n = 0, 1, 2, . . . ) of the wave function of the solution

of the KP hierarchy corresponding to the quantised canonical transformation (f, g)as follows: let S0, . . . , S i1 be given.

(1) By Proposition 3.2 we have X0, . . . , X i1.(2) We have a recursion formula for Xi by Theorem 2.1.(3) Proposition 3.1 gives a formula for Si.

If we take the factor ()/ into account, this process becomes a little bit compli-cated, but essentially the same.

The rest of this section is devoted to the proof of Proposition 3.1 and Proposi-tion 3.2.

These statements might seem obvious, but since the multiplication in the defi-nition of

(3.19) eX/ =n=0

X(, x,)n

nn!

is non-commutative, while the multiplication of total symbols in the series

(3.20) eS/ =n=0

S(, x , )n

nn!

is commutative, it needs to be proved. In fact, computation of the simplest exampleof X = x()1 would show how complicated the formula can be:

tot(ex()1/) =

n=0

1

n!ntot(X

n)

=1 + 11!

x1 + 12!2

(x22 x3)

+1

3!3(x33 3x24 + 32x5)

+1

4!4(x44 6x35 + 152x26 153x7) +

=exp1

x1

x3

2+

x5

2

5x7

8+

.


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It is not obvious, at least to authors, why there is no more negative powers of in the last expression, which can be obtained by calculating the logarithm of theprevious expression.

To avoid confusion, the commutative multiplication of total symbols a(, x , )and b(, x , ) as power series is denoted by a(, x , ) b(, x , ) and the non-commutativemultiplication corresponding to the operator product is denoted by a(, x , ) b(, x , ). Recall that the latter multiplication is expressed (or defined) as follows:

a(, x , ) b(, x , ) = eya(, x , )b(, y , )|y=x,=

=n=0

n

n!n a(, x , )

ny b(, y , )|y=x,=.

(3.21)

(See, for example, [S], [A] or [KR].) The order of an operator corresponding tosymbol a(, x , ) is denoted by ord a(, x , ), which is the same as the order ofa(, x , ) as a power series of . The -order is the same as that of the operator:

ord

x = ord

= 0, ord

= 1.The main idea of proof of propositions is due to Aoki [A], where exponential cal-

culus of pseudodifferential operators is considered. He considered analytic symbolsof exponential type, while our symbols are formal ones. Therefore we have only toconfirm that those symbols make sense as formal series.

First, we prove the following lemma.

Lemma 3.3. Leta(, x , ) and b(, x , ) be two symbols such that ord a(, x , ) =M, ord a(, x , ) = 0, ord b(, x , ) = N, ord b(, x , ) = 0, ord p(, x , ) =ord q(, x , ) = 0, ordp(, x , ) = ord q(, x , ) = 0.

Then there exists a symbol c (ord c = N + M, ord c = 0), r (ord r = 0,ord r = 0) such that

(3.22)

a(

, x , )e

p(,x,)/

b(

, x , )e

q(,x,)/= c(

, x , )e

r(,x,)/

.In the proof of Proposition 3.1 and Proposition 3.2, we use the construction ofc

and r in the proof of Lemma 3.3.

Proof. Following [A], we introduce a parameter t and consider

(3.23) (t) = (t; , x , y , , ) := etya(, x , )b(, y , )e

p(,x,)+q(,y,)

/.

If we set t = 1, y = x and = , this reduces to the operator product of (3.21).The series (t) satisfies a differential equation with respect to t:

(3.24) t = y, (0) = a(, x , )b(, y , )e

p(,x,)+q(,y,)

/.

We construct its solution in the following form:

(3.25)

(t) = (t)ew(t)/,

(t) = (t; , x , y , , ) =n=0

ntn,

w(t) = w(t; , x, y, , ) =k=0

wktk.


15/28


16/28


depending on a parameter s. The total symbol of E(s) is defined as

(3.34) E(s;

, x , ) =

k=0

sk

kk! X

(k)

(

, x , ), X

(0)

= 1, X

(k+1)

= X X

(k)

.

Taking the logarithm (as a function, not as an operator) of this, we can defineS(s) = S(s;, x , ) by

(3.35) E(s; , x , ) = e1S(s;,x,).

What we are to prove is that S(s), constructed as a series, makes sense at s = 1and formally regular with respect to .

Differentiating (3.35), we have

(3.36) X(, x , ) E(s; , x , ) =S

seS(s;,x,)/.

By Lemma 3.3 (a X, b 1, p 0, q S) and the technique in its proof, we

can rewrite the left hand side as follows:(3.37) X(, x , ) E(s; , x , ) = Y(s;, x, x, , )eS(s;,x,)/,

where Y(s;, x , y , , ) =

k=0 Yk and Yk(s; , x , y , , ) are defined by

Yk+1(s; , x , y , , )

=1

k + 1(yYk(s; , x , y , , ) + Yk(s; , x, y, , )yS(s; , y , )),

Y0(s;, x , y , , ) = X(, x , ).

(3.38)

Yk(s) corresponds to k in the proof of Lemma 3.3, while wk there corresponds tok,0S(s). On the other hand, substituting (3.37) into the left hand side of (3.36),we have

(3.39)S

s (s; , x , ) = Y(s; , x, x, , ).

We rewrite the system (3.38) and (3.39) in terms of expansion ofS(s;, x , ) andYk(s; , x , y , , ) in powers of s and :

S(s; , x , ) =l=0

S(l)(, x , )sl =l=0

n=0

S(l)n (x, )nsl,

Yk(s; , x , y , , ) =l=0

Y(l)k (, x , y , , )s

l =l=0

n=0

Y(l)k,n(x, y, , )

nsl,

(3.40)

The coefficient ofnsl in (3.38) is

(3.41) Y(l)k+1,n(x, y, , )

=1

k + 1

yY(l)k,n1(x, y, , ) +

l+l=ln+n=n

Y(l)k,n(x, y, , )yS

(l)n (y, )

,

(Y(l)k,1 = 0) and

(3.42) Y(l)

0,n(x, y, , ) = l,0Xn(x, ),


17/28


18/28


Let us prove that Y(l)k,ns determined above satisfy

Y(l)k,n = 0, if k > l + n,(3.49)

ord Y(l)k,n k l 1.(3.50)

(We define that ord 0 = .) In particular, the sum in (3.43) is well-defined and

(3.51) ord S(l+1)n l 1.

If n = 1, both (3.49) and (3.50) are obvious. Fix n0 0 and assume that wehave proved (3.49) and (3.50) for n < n0 and all (l, k).

When l = 0, (3.49) and (3.50) are true for all k because of (3.46) and (3.45),respectively. Fix l0 0 and assume that we have prove (3.49) and (3.50) for l l0and all k. As a result (3.51) is true for l l0.

The recursion relation (3.41) implies (3.49) and (3.50) for l = l0 + 1 and allk > 0. In fact, if k + 1 > (l0 + 1) + n, then k > (l0 + 1) + (n 1) and k > l + n

which guarantees that Y

(l0+1)

k,n1 = 0 and Y

(l)

k,n = 0 in the recursion relation (3.41)for l = l0 + 1 by the induction hypothesis. The estimate (3.50) is easy to check forl0 + 1. (Recall once again that lowers the order by one.)

For k = 0, (3.49) is void and (3.50) is true because of (3.42) and ord Xn 1.The step l = l0 + 1 being proved, the induction proceeds with respect to l and

consequently with respect to n.

In summary we have constructed Y(s; , x , y , , ) and S(s; , x , ) satisfying

(3.37) and (3.39). Thanks to (3.51), Sn(x, ) =

l=0 S(l)n (x, ) is meaningful as

a power series of . Thus Proposition 3.1 is proved.

Proof of Proposition 3.2. We reverse the order of the previous proof. Namely, givenS(, x , ), we shall construct X(, x , ) such that the corresponding S(1;, x , ) inthe above proof coincides with it.

Suppose we have such X(, x , ). Then the above procedure determine Y(l)k,n and

S(l)n . We expand them as follows:

S(, x , ) =n=0

Sn(x, )n, Sn(x, ) =

j=1

Sn,j(x, ),

X(, x , ) =n=0

Xn(x, )n, Xn(x, ) =

j=1

Xn,j(x, ),

S(s;, x , ) =l=0

n=0

S(l)n (x, )nsl, S(l)n (x, ) =

j=1

S(l)n,j(x, ),

Yk(s; , x , y , , ) Y(l)k,n(x, y, , )

=l=0

n=0

Y(l)k,n(x, y, , )

nsl, =j=1

Y(l)k,n,j(x, y, , ).

Here terms with index j are homogeneous terms of degree j with respect to and.

At the end of this proof we shall determine Xn by (3.42),

(3.52) Xn(x, ) = Y(0)

0,n (x, y, , ).


19/28

-EXPANSION OF KP 19

(In particular, Y(0)

0,n (x, y, , ) should not depend on y and .) For this purpose,

Y(0)

0,n should be determined by

(3.53) Y(0)

0,n (x, y, , ) = Sn(x, )

(l,k)=(0,0)l,k0

1l + 1

Y(l)k,n(x,x,,)

because of (3.43) and Sn(x, ) =

l=1 S(l)n (x, ).

Since ord Y(l)k,n should be less than l k (cf. (3.50)), we expect

(3.54) Y(l)k,n,1 = 0

for (l, k) = (0, 0). Hence picking up homogeneous terms of degree 1 with respectto , the following equation should hold:

(3.55) Y(0)

0,n,1 = Sn,1

All Y(l)k,n,1 are determined by the above two conditions, (3.54) and (3.55). Note also

that

(3.56) Y(l)

0,n,j = 0 for l = 0

because Y0 should not depend on s because of (3.42).

Having determined initial conditions in this way, we shall determine Y(l)k,n,j in-

ductively. To this end we rewrite the recursion relation (3.41) by (3.43) and pickup homogeneous terms of degree j:

(3.57) Y(l)k+1,n,j(x, y, , ) =

1

k + 1

yY

(l)k,n1,j1(x, y, , )+

+

l+l=l,l1j+j=j1,n+n=n

0k

1

l Y

(l)

k,n,j(x, y, , )yY

(l1)

k,n,j(x,x,,)

(As before, terms like Y(l)k,1,j1 appearing the above equation for n = 0 can be

ignored.)

Fix n0 0 and assume that Y(l)k,0,j , . . . , Y

(l)k,n01,j

are determined for all (l , k, j).

(1) First we determine Y(l)k,n0,1

for all (l, k) by (3.55) and (3.54).

(2) Fix j0 2 and assume that Y(l)k,n0,j

are determined for j = 1, . . . , j0 1 and

all (l, k). (The above step is for j0 = 2.)Since all the quantities in the right hand side of the recursion relation

(3.57) with j = j0 are known by the induction hypothesis, we can determine

Y(l)k,n0,j0 for l = 0, 1, 2, . . . and k = 1, 2, . . . .

(3) Together with (3.56), Y(l)

0,n0,j0= 0 for l = 1, 2, . . . , we have determined all

Y(l)k,n0,j0

except for the case (l, k) = (0, 0).

(4) It follows from (3.57), (3.55) and (3.56) by induction that for all Y(l)k,n0,j

determined in (1), (2) and (3),

(3.58) Y(l)k,n,j = 0 for l + k + 1 > j.


20/28


This corresponds to ord Y(l)k,n0

l k 1 (3.50) in the proof of Proposi-tion 3.1.

(5) We determine Y

(0)

0,n0,j0 by

(3.59) Y(0)

0,n0,j0= Sn0,j0

(l,k)=(0,0)

l,k0

1

l + 1Y

(l)k,n0,j0

(x,x,,),

which is the homogeneous part of degree j0 in (3.53). The sum in theright hand side is finite thanks to (3.58).

(6) The induction with respect to j proceeds by incrementing j0.

Thus all Y(l)k,n0,j

are determined and Xn0 is determined by Xn0(x, ) =

j=1 Y(0)

0,n0,j

(cf. (3.52)), which completes the proof of Proposition 3.2.

4. Asymptotics of the tau function

In this section we derive an expansion

(4.1) log (, t) =n=0

n2Fn(t)

of the tau function from the expansion (3.7) of the S-function. Note that we havesuppressed the variable x, which is understood to be absorbed in t1.

Let us recall the fundamental relation [DJKM]

(4.2) (t; z) =(t [z1])

(t)e

1(t,z),

where [z1] = (1/z, 1/2z2, 1/3z3, . . . ) and (t, z) =

n=1 tnzn. This implies that

(4.3) 1S(t; z) =

eD(z) 1

log (t)

where S(t; z) = S(t; z) (t, z) and D(z) =

j=1

zjj

tj . Differentiating this with

respect to z, we have

1

zS(t; z) = D(z)eD(z) log (t)

= D(z)(log w(t; z) + log (t)),(4.4)

where D(z) := z D(z) =

j=1 zj1

tj. Hence

(4.5) D(z)log (t) = 1

z+ D(z)

S(t; z)

Multiplying zn to this equation and taking the residue, we obtain a system ofdifferential equations

(4.6) tn

log (t) = 1 Res zn

z

+ D(z)

S(t; z) dz, n = 1, 2, . . .

which is known to be integrable [DJKM]. We can thus define the tau function (t),up to multiplication (t) c(t) by a nonzero constant c, as a solution of (4.5).

By substituting the -expansions

(4.7) log (t) =n=0

n2Fn(t), S(t; z) =

n=0

nSn(t; z),


21/28


22/28


Special solutions stemming from string theory and random matrices [Mo], [dFGZ](e.g. the Kontsevich model) can admit a more efficient approach such as the methodof Eynard and Orantin [EO]. Those methods are based on a quite different principle.In the method of Eynard and Orantin, it is the so called loop equation for correla-tion functions of random matrices. The loop equations amount to constraints onthe tau function. Eynard and Orantins topological recursion relations determinea solution of those constraints rather than of an underlying integrable hierarchy; itis somewhat surprising that a solution of those constraints gives a tau function.

Lastly, let us mention that the results of this paper can be extended to the Todahierarchy. That case will be treated in a forthcoming paper.

Appendix A. Proof of formulae (2.17) and (2.19)

In this appendix we prove the factorisation ofW (2.17) and an auxiliary formula(2.19).

The main tool in this appendix is the Campbell-Hausdorff theorem:

(A.1) exp(X)exp(Y) = exp

n=0

cn(X, Y)

,

where cn(X, Y) is determined recursively:

c1(X, Y) = X+ Y,

cn+1(X, Y) =1

n + 1

1

2[X Y, cn]+

+

p1,2pn

K2p

(k1,...,k2p)k1++k2p=n

[ck1 , [ , [ck2p , X+ Y] ]]

.

(A.2)

The coefficients K2p are defined by (2.15). See, for example, [B].First we prove

(A.3) exp(1X(x,t, ))

= expi1Xi + (terms of-order < i + 1)

exp

1X(i1)

,

where the principal symbol of Xi is

(A.4) (Xi) :=n=1

(ad{,} (X0))n1

n!(Xi),

as is defined in (2.19). For simplicity, let us denote

(A.5) A :=1

X(i1) =1

i1

j=0

jXj , B :=

1

j=i

jXj .

Note that A + B = X/ and ord A 1, ord B i + 1. We prove the followingby induction:

(A.6) Cn := cn(A + B, A) =(ad A)n1

n!(B) + (terms of-order < i + 1).

This is obvious for n = 1 since C1 = (A+B)+ (A) = B. Assume that (A.6) is truefor n = 1, . . . , N . This means, in particular, ord Cn ord

B 0 (1 n N),


23/28

-EXPANSION OF KP 23

which implies that for any operator Z ord[Cn, Z] is less than ord Z by more thanone. Hence the term of the highest -order in the recursive definition (A.2) is thefirst term. More precisely, it is decomposed as

1

N + 1

1

2[(A + B) (A), CN] =

1

N + 1[A, CN] +

1

2(N + 1)[B, CN],

and the first term in the right hand side has the highest -order. By the inductionhypothesis and ord A 1, we have

1

N + 1[A, CN] =

1

N + 1

A,

(ad A)N1

N!(B) + (terms of order < i + 1)

=(ad A)N

(N + 1)!(B) + (terms of-order < i + 1).

(A.7)

This proves (A.6) for all n. Taking its symbol of order i + 1, we have

(A.8) (cn(A + B, A)) = (ad{,}

(A))n1

n!(B),

which gives the terms of (A.4). Substituting this into the Campbell-Hausdorffformula (A.1), we have (A.3).

By factorisation (A.3), we can factorise W = exp(X/)() as follows ((i1) :=i1j=0

jj):

exp(1X(x,t,))()

=expi1Xi + (terms of-order < i + 1)

exp

1X(i1)

expi1i log() + (terms of-order < i + 1)

exp1(i1) log()=exp

i1Xi + (terms of-order < i + 1)

exp

ead(1X(i1))

i1i log() + (terms of-order < i + 1)

exp1X(i1)

exp

1(i1) log()

.

(A.9)

Since the symbol of order i+1 ofead(1X(i1))

i1i log()

is ead{,}X0(i log ),

(A.9) is rewritten as (2.17) by using the Campbell-Hausdorff formula (A.1) onceagain.

In order to recover Xi from Xi (or Xi), we have only to invert the definition(A.4). In the definition (A.4) we substitute ad{,}(

(X0)) in the equation

et 1

t=

n=1

tn1

n!.

Hence substitution t = ad{,}((X0)) in its inverse

t

et 1= 1

t

2+

p=1

K2pt2p


24/28


gives the inverse map Xi Xi. Here the coefficients K2p are defined in (2.15).Hence equation (2.19):

(Xi) = (Xi)

12

{(X0), (Xi)} +

p=1

K2p(ad{,}((X0)))

2p(Xi)

gives the symbol of Xi.

Appendix B. Example (Kontsevich model)

According to Adler and van Moerbeke [AvM], the solution of the KP hierarchyarising in the Witten-Kontsevich theorem [Ko] satisfies

(B.1) (L2)1 = 0,

1

2ML1

1

4L2 L

1

= 0.

This corresponds to the case where

(B.2) f(x, ) = ()2, g(x, ) = 12

x()1 4

()2 + .

Let us apply our procedure to this case. We fix all the time variables tn to0, which means that we restrict ourselves to the so-called small phase space intopological string theory [D]. (The first time variable t1 can be re-introduced byshifting x to t1 + x.)

To begin with, let us determine the leading terms of X and , which are theinitial data for our procedure. The dispersionless limit of (L, M) satisfies

(B.3) (L2)1 = 0,

1

2ML1 L

1

= 0.

Since L has the form (1.15), L2 should be a second order polynomial of : L2 =2 + u(x). When tn = 0, M has the form

M = x + 0L1 +

n=1

v0,nLn1.

(See (1.19).) Using this and L = (1 + u(x)2)1/2, we have

1

2ML1 L = +

x2

u

2

1 +

02

2 + O(3).

Hence, due to the second equation of (B.3), we have u(x) = x, 0 = 0 and,consequently,

(B.4) P0 := L2 = 2 + x, Q0 :=

1

2ML1 L = .

Combining them, we have the following expression for M:

M =2L(L ) = 2(2 + x) 22

1 + x21/2

=x n=2

2

1/2

n

xn2n+2

=x +x2

42

x3

84+

5x4

646

7x5

1288+

21x6

51210+ O(12)

(B.5)


25/28

-EXPANSION OF KP 25

On the other hand, from the expression M = exp(ad{,} X0)x follows

M =

n=0

ad{,} X0

n! x

=x 0,12 20,2

3

+

30,3

0,10,1

2

4 + (40,4 20,2

0,1)

5 + ,

(B.6)

where 0,k are the coefficients in the expansion ofX0 (2.16). Comparing (B.5) and(B.6), we can determine 0,k inductively and hence X0 is determined:

X0 = x2

4()1 +

x3

48()3

x4

384()5

+x6

6144()9

x7

61440()11 + .

Having determined X0 and 0, we can start the algorithm discussed in Section 2.In the step 1 for i = 1 we define P(0) and Q(0) by (2.9) and (2.10):

P(0) = ()2 + x +

2()1

x

4()3 +

2

8()4 +

5x2

32()5

412x

96()6 +

233

96

3x3

32

()7 +

3012x2

384()8

+

833x

48+

53x4

1024

()9 +

1914

192

5432x3

512

()10

+ 87833x2

1536

119x5

4096 ()11 + ,

Q(0) = ()

4()2 +

3x

8()4

32

8()5

29x2

64()6

+1572x

96()7 +

493

32+

x3

2

()8

5192x2

128()9

+

43453x

384

1077x4

2048

()10 +

13394

128+

39612x3

512

()11

+ .

We extract terms (symbols) of-order 0 from the -expansion of them:

P0(x, ) = 2 + x, Q0(x, ) = ,

and those of-order 1:

P(0)1 (x, ) =

1

21

x

43 +

5x2

325

3x3

327 +

53x4

10249

119x5

409611 +

Q(0)1 (x, ) =

1

42 +

3x

84

29x2

646 +

x3

28

1077x4

204810 + .


26/28


27/28


28/28


[Mu] Mulase, M., Complete integrability of the Kadomtsev-Petviashvili equation. Adv. inMath. 54 (1984), 5766.

[OS] Orlov, A. Yu. and Schulman, E. I., Additional symmetries for integrable equations and

conformal algebra representation, Lett. Math. Phys. 12 (1986), 171179; Orlov, A. Yu.,Vertex operators, -problems, symmetries, variational indentities and Hamiltonian for-malism for 2 + 1 integrable systems, in: Plasma Theory and Nonlinear and TurbulentProcesses in Physics (World Scientific, Singapore, 1988); Grinevich, P. G., and Orlov,

A. Yu., Virasoro action on Riemann surfaces, Grassmannians, det j and Segal Wilson function, in: Problems of Modern Quantum Field Theory (Springer-Verlag, 1989).

[S] Schapira, P., Microdifferential systems in the complex domain, Grundlehren der mathe-matischen Wissenschaften 269, Springer-Verlag, Berlin-New York, (1985)

[SS] Sato, M., and Sato, Y., Soliton equations as dynamical systems on infinite dimensional

Grassmann manifold, in Nonlinear Partial Differential Equations in Applied Science;Proceedings of t he U.S.-Japan Seminar, Tokyo, 1982, Lect. Notes in Num. Anal. 5

(1982), 259271.

[SW] Segal, G., and Wilson, G., Loop groups and equations of KdV type. Inst. Hautes Etudes

Sci. Publ. Math. 61 (1985), 565.[TT1] Takasaki, K., and Takebe, T., SDiff(2) KP hierarchy, Int. J. Mod. Phys. A7S1B (1992),

889-922.[TT2] Takasaki, K., and Takebe, T., Integrable Hierarchies and Dispersionless Limit, Rev.

Math. Phys. 7 (1995), 743-803.

Graduate School of Human and Environmental Studies, Kyoto University, Yoshida,

Sakyo, Kyoto, 606-8501, Japan

E-mail address: [email protected]

Faculty of Mathematics, State University Higher School of Economics, VavilovaStreet, 7, Moscow, 117312, Russia

E-mail address: [email protected]

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