Random matrices with external source and KP τ functionsDong Wang Citation: Journal of Mathematical Physics 50, 073506 (2009); doi: 10.1063/1.3167802 View online: http://dx.doi.org/10.1063/1.3167802 View Table of Contents: http://scitation.aip.org/content/aip/journal/jmp/50/7?ver=pdfcov Published by the AIP Publishing Articles you may be interested in Kerov's interlacing sequences and random matrices J. Math. Phys. 54, 113302 (2013); 10.1063/1.4830024 A block variational procedure for the iterative diagonalization of non-Hermitian random-phase approximationmatrices J. Chem. Phys. 136, 034111 (2012); 10.1063/1.3677667 On universality of bulk local regime of the Hermitian sample covariance matrices J. Math. Phys. 51, 103516 (2010); 10.1063/1.3498377 Asymptotic independence of the extreme eigenvalues of Gaussian unitary ensemble J. Math. Phys. 51, 023514 (2010); 10.1063/1.3290968 Moments of ratios of characteristic polynomials of a certain class of random matrices J. Math. Phys. 50, 043518 (2009); 10.1063/1.3119483

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Random matrices with external source and KP � functionsDong Wanga�

Centre de Recherches Mathématiques, Université de Montréal, C. P. 6128, Succ. CentreVille, Montréal, Québec H3C 3J7, Canada

�Received 13 March 2009; accepted 12 June 2009; published online 13 July 2009�

In this paper we prove that the partition function in the random matrix model withexternal source is a KP � function. © 2009 American Institute of Physics.�DOI: 10.1063/1.3167802�

I. INTRODUCTION

Let A�Hn�n be an n�n Hermitian matrix and d��x�=w�x�dx be a measure on R with allmoments finite. Then we define the partition function,

Zn�A� = �M�Hn�n

eTr�AM�d��M� , �1�

where d��M� denote a unitary invariant measure on Hn�n such that if eigenvalues of M are�1 , . . . ,�n,

d��M� = �i=1

n

w��i�dM . �2�

Due to the unitary invariance, we assume A=diag�a1 , . . . ,an�, ai�R without loss of generality. Weconsider Zn�A� as a function of eigenvalues of A and find a KP �Kadomtsev-Petviashvili� �function property of it.

Zn�A� arises in the random matrix model with external source.7,8,30,31 Let A�Hn�n be an n�n Hermitian matrix, and V�x� be a function defined on R, such that e−V�x� decays sufficientlyfast. We consider the ensemble of n�n Hermitian matrices with the probability density function,

P�M� =1

ZnV�A;R�

e−Tr�V�M�−AM�, �3�

where the normalization constant ZnV�A ;R�, also called the partition function, is defined as

ZnV�A;R� = �

M�Hn�ne−Tr�V�M�−AM�dM . �4�

The ensemble is called the random matrix model with external source, and we can easily identifyZn

V�A ;R� as the Zn�A� in �1� with w�x�=e−V�x�. In the study of gap probability,25,5,3,6,2 we need toconsider also

a�Electronic mail: wangdong@crm.umontreal.ca.

JOURNAL OF MATHEMATICAL PHYSICS 50, 073506 �2009�

50, 073506-10022-2488/2009/50�7�/073506/10/$25.00 © 2009 American Institute of Physics

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ZnV�A;E� = � M�Hn�n

all eigenvalues of M�E

e−Tr�V�M�−AM�dM , �5�

such that the gap probability that all eigenvalues of M are in E, a subset of R, isZn

V�A ;E� /ZnV�A ;R�. Similarly we identify Zn

V�A ;E� as the Zn�A� with w�x�=e−V�x��E�x�, where �E isthe indicator function.

The random matrix model with external source was introduced in the 1990s �Refs. 7 and 8� asa generalization of the standard matrix model, i.e., the A=0 case, which is first proposed byWigner in the 1950s.19 However, a special form of the random matrix model with external sourcehas been studied by statisticians since the 1920s, under the name of Wishart ensemble, one of themost important models in multivariate statistics.24,13

Consider N independent, identically distributed complex samples x1 , . . . ,xN, all of which aren�1 column vectors, and we further assume that the sample vectors xi’s are Gaussian with mean0 and covariance matrix �, which is a fixed n�n positively defined Hermitian matrix. If we putxi’s into an n�N rectangular matrix X= �x1 : . . . :xN�, then the sample covariance matrix S= �1 /N�XX† is an n�n positively defined Hermitian matrix matrix. If we assume N�n, then theprobability density function of S is

P�S� =1

Zn,NWis���

e−N Tr��−1S��det S�N−n, �6�

where the normalization constant Zn,NWis���, analogous to the partition function in �4�, is

Zn,NWis��� = � S�Hn�n

S is positively defined

e−N Tr��−1S��det S�N−ndS , �7�

which is the Zn�−N�−1� with w�x�=xN−n��0,���x�. To study the distribution of the eigenvalues of S,we also need partition functions such as the Zn

V�A ;E� in �5�. See, e.g., Refs. 4, 12, 21, and 20.The main result in this paper is that the partition function Zn�A� is a KP � function. To make

the statement precise, we denote power sums of eigenvalues of A

tk =1

k�i=1

n

aik, k = 1,2, . . . . �8�

Since Zn�A� is a symmetric function in a1 , . . . ,an, it can be regarded as a function of t= �t1 , t2 , . . .�. We should be cautious that tk’s are not independent among one another, so that Zn�A�cannot be written as a function of tk’s in a unique way. However, we will eliminate the ambiguity

in the following way. First, we define explicitly a function Zn�t� in �26�, and then identify

Zn�A� = Zn�t��tk=1/k�i=1n ai

k. �9�

Thus the rigorous statement is the following.

Theorem 1: Zn�t� defined by (26) is a KP � function in variables t1 , t2 , . . ..

Since Zn�t� is a KP � function, it satisfies Hirota bilinear equations,9 which are equivalent toa series of partial differential equations �PDEs�. For example,

�4

�t14 + 3

�2

�t22 − 4

�2

�t1 � t3log Zn�t� + 6 �2

�t12 log Zn�t�2

= 0, �10�

which corresponds to the first nontrivial Hirota bilinear equation.

073506-2 Dong Wang J. Math. Phys. 50, 073506 �2009�

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In Refs. 16 and 32, it is proven that the HCIZ �Harish-Chandra-Itzykson-Zuber� integral is aToda � function. This result is closely related to Theorem 1, since the partition function Zn�A� isusually evaluated by the HCIZ formula,7 although we do not use it in this paper, and KP �functions are closely related to Toda � functions.

Zn�A� is also very similar to the partition function of the generalized Kontsevich model�GKM� ZN

�V��M� �following the notation in Ref. 17�, which is also proven to be a KP � function,with the negative power sums of eigenvalues of M as the KP flow parameters. See Refs. 17, 15,23, and 22, for details. The definition of ZN

�V��M� in the GKM is more complicated, and there is acrucial difference between ZN

�V��M� and Zn�A�: For different dimensions N, ZN�V��M� are specializa-

tions of the same � function, however, Zn�t� are different � functions when the dimensions n are

different. For example, for different n, the Virasoro constraints for Zn�t� are not identical, and wecan no longer find the Virasoro constraints by the trick “taking the large N limit,” which works forthe GKM.

Zn�A� was studied as a multicomponent KP � function by Adler and van Moerbeke et al. InRef. 1 it is proven that the determinant of a moment matrix for several weights, after addingdeformation parameters, is a multicomponent KP � function. By de Bruijn’s formula �Lemma 3�,we find that Zn�A� is a special case of these determinants of moment matrices. Especially, wheneigenvalues of A have only two distinct values, Zn�A� is a three-component KP � function, and isdetailed in Ref. 2. For the nondegenerate case, i.e., eigenvalues a1 , . . . ,an of A are distinct, Zn�A�is a �1+n�-component KP � function.

The intriguing fact is that our KP � function structure is independent of this multicomponentKP � function structure. In particular, our set of KP flow parameters t is none of these sets of KPflow parameters in Ref. 1.

Remark 1: With the help of Virasoro constraints, it is shown in Ref. 2 that for quadratic V,Zn

V�A ;E� in (5) satisfies a PDE, which is derived from one Hirota bilinear equation of the multi-

component KP � function. We can also find Virasoro constraints for Zn�t� as the KP � function andget new PDEs satisfied by Zn

V�A ;E�. It will be done in a forthcoming paper.29

In Sec. II we summarize necessary preliminaries. Then in Sec. III we give the definition of

Zn�t� and prove Theorem 1.

II. BOSON-FERMION CORRESPONDENCE AND KP � FUNCTIONS

The definition of KP � functions follows that in Ref. 14, and all materials on symmetricfunctions are from Ref. 27.

KP � functions can be defined through representations of the Heisenberg algebra, an associa-tive algebra. Over any field K with characteristic 0, such as R or C, the Heisenberg algebra H isgenerated by �hk�k�Z\�0�, satisfying

�hk,hl� = kk,−l. �11�

We can construct a representation of H over the so-called boson Fock space, which isK�t1 , t2 , . . .�, the space of polynomials with infinitely many variables. hk’s �k−1� act as multi-plication operators, and hk’s �k�1� act as derivations,

hk → − kt−k for k − 1

�

�tkfor k � 1. � �12�

This representation of H is called the Boson representation.On the other hand, H has another representation over the so-called fermion Fock space. To

define the fermion Fock space, we take an infinite dimensional vector space V with basis �vi�i�Z.The fermion Fock space � is composed of semi-infinite forms, spanned by the basis vi0,i−1,i−2,. . .,which are defined as

073506-3 Random matrices and KP � functions J. Math. Phys. 50, 073506 �2009�

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vi0,i−1,i−2,. . . = vi0∧ vi−1

∧ vi−2∧ . . . , �13�

with i0 , i−1 , . . . strictly decreasing and i−k=−k for k sufficiently large.We call the semi-infinite form v0,−1,−2,. . .=v0∧v−1∧v−2∧ . . . the vacuum, following the physical

terminology. Later in this paper, we use form to mean a semi-infinite form, unless otherwiseclaimed.

Formally, we define the action of hi on V by

hk�vl� = vl−k �14�

and get the induced action on � by

hk�vi0,i−1,i−2,¯� = �j�0

vi0∧ vi−1

∧ ¯ ∧ vi−j+1∧ vi−j−k ∧ vi−j−1

∧ ¯ . �15�

Although the action of hk on V is not consistent with the commutation relation �11� of H, theaction of hk on � defines a representation of H. This is the fermion representation of H.

We can observe that the boson representation and the fermion representation of H are equiva-lent. The correspondence � between K�t1 , t2 , . . .� and � is

��1� = vacuum,

��ktk� = h−k�vacuum� ,

��f�t1,2t2, . . .�� = f�h−1,h−2, . . .��vacuum� , �16�

where k�0 and f is a polynomial. Since �hk ,hl�=0 for k , l�Z−, the polynomial of operatorsf�h−1 ,h−2 , . . .� is well defined. Although it is not difficult to check the validity of the correspon-dence �, the images of monomials on the boson Fock space become messy combinations of thebasis of �. It is an interesting question what the preimage of vi0,i−1,i−2,. . . is. The answer is nontrivialand can be best formulated in notations of symmetric functions.

Proposition 1: Let i0= 0 , i−1= 1−1 , i−2= 2−2 , . . . with 0� 1� 2� . . ., such that� 0 , 1 , . . .�= is a partition, then �−1�vi0,i−1,i−2,. . .�= s �t�, where s is a polynomial in the defini-tion of Schur functions by power sums: If we regard �tk� as power sums defined in (8), then for anyn the Schur functions s satisfies

s �a1, . . . ,an� = s �t� . �17�

Since every symmetric function can be written uniquely as a polynomial of power sums, s iswell defined.

To define KP � functions, we need the concept of decomposability of forms. We call v��decomposable, if and only if

v = u0 ∧ u−1 ∧ u−2 ∧ . . . , �18�

where ui’s are linear combinations of vi’s. Now we are ready to give the definition.Definition 1: f�t� is a KP � function if and only if f =�−1�v�, where v is a decomposable form.

Here we note that the Zn�t� that we are going to study is not a polynomial in t1 , t2 , . . ., butrather a series. We can generalize the correspondence � such that it maps “convergent” series f�t�into a form that is an infinite sum of vi0,i−1,i−2,. . .. Then we define that f�t� is a KP � function if and

only if ��f� is a decomposable form. Thus Definition 1 is also applicable to Zn�t�. For the rigorousstatement of convergence condition, see Ref. 26.

073506-4 Dong Wang J. Math. Phys. 50, 073506 �2009�

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III. PROOF OF THEOREM 1

In order to define Zn�t�, we first expand Zn�A� in Schur polynomials. By the Weyl integralformula, we have

Zn�A� =1

Cn� ¯� ����2�

U�n�eTr�AU diag���U−1�dUd���1� . . . d���n� , �19�

where Cn is a constant and dU is the Haar measure over U�n�. Then we use the identity18

�U�n�

eTr�AU diag���U−1�dU = �k=0

�1

k! � ⊢k

l� �n

C �l1, . . . ,ln�C ��1, . . . ,�n�C �1, . . . ,1�

, �20�

where ’s are partitions of k, l� � is the length of , and C ’s are constant multiples of Schurpolynomials s , with the normalization11

� ⊢k

C �x1, . . . ,xn� = �x1 + . . . + xn�k. �21�

Furthermore, we know27,11

C =k!

H� �s and s �1, . . . ,1� =

�n�

H� �, �22�

where H� � is the hook length product of and �n� is the extended Pochhammer symbol: if = � 1 , 2 , . . . , l�, then

�n� = �i=1

l

�j=1

i

�n − i + j� . �23�

Thus if we denote

G =� ¯� �2���s ��1, . . . ,�n�d���1� ¯ d���n� , �24�

we have

Zn�A� =1

Cn� ¯� ����2�

k=0

�

� ⊢k

l� �n

s �a1, . . . ,an�s ��1, . . . ,�n��n�

d���1� . . . d���n�

=1

Cn�k=0

�

� ⊢k

l� �n

1

�n�

G s �a1, . . . ,an� , �25�

and we define

Zn�t� =1

Cn�k=0

�

� ⊢k

l� �n

1

�n�

G s �t� , �26�

where s is the polynomial defined in �17�. Then it is clear that �9� holds.

To determine whether Zn�t� is a KP function, we turn to the fermion representation and mapit as

073506-5 Random matrices and KP � functions J. Math. Phys. 50, 073506 �2009�

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��Zn�t�� =1

CnV, with V = �

k=0

�

� ⊢k

l� �n

1

�n�

G v , �27�

where for = � 0 , 1 , . . . , l�, we denote v =v0+ 0,−1+ 1,. . .,−l+ l,−l−1,. . ..

To prove that Zn�t� is a KP � function is equivalent to prove that V is a decomposable form.We have a simple criterion �Plücker relations� for decomposability of forms, and first introducetwo kinds of linear operators vi∧ �v� and ıvi

��v� on any v��,

vi ∧ �vi0∧ vi−1

∧ ¯� = vi ∧ vi0∧ vi−1

∧ ¯ , �28�

ıvi��vi0

∧ vi−1∧ ¯� = ��− 1� jvi0

∧ ¯ ∧ vi−j+1∧ vi−j−1

∧ ¯ if vi−j= vi

0 otherwise.� �29�

Here we note that vi∧ �v� and ıvi��v� are not in �: they are in �+ and �−, respectively. �+ is

spanned by forms vi0,i−1,¯+ =vi0

∧vi−1∧¯, such that i0 , i−1 , . . . are strictly decreasing and i−k=−k

+1 for k sufficiently large; �− is spanned by forms vi0,i−1,¯− =vi0

∧vi−1∧¯, such that i0 , i−1 , . . . are

strictly decreasing and i−k=−k−1 for k sufficiently large.Now we can state the criterion.14

Proposition 2: v�� is decomposable, if and only if the tensor of forms in �+ � �− ,

�i=−�

�

vi ∧ �v� � ıvi��v� = 0. �30�

For notational simplicity, for a partition = � 0 , . . . , l�, we denote v + and v

−, analogous to v ,

v + = v1+ 0,0+ 1,. . .,−l+1+ l,−l,. . ., v

− = v−1+ 0,−2+ 1,. . .,−l−1+ l,−l−2,. . . . �31�

For a partition and i an integer, we define partitions + i and − i. First, = � 0 , . . . , l�corresponds to a v

−��−, whose subscript �−1+ 0 ,−2+ 1 , . . . ,−l−1+ l ,−l−2, . . .� is a decreas-ing sequence of integers. If i� �−1+ 0 ,−2+ 1 , . . . ,−l−1+ l ,−l−2, . . .�, then we say + i is notwell defined; otherwise we can arrange elements in �−1+ 0 ,−2+ 1 , . . . ,−l−1+ l ,−l−2, . . .�� �i� into a decreasing sequence which is the subscript of a form v ���, and we define + i= �. Symmetrically, we can define − i: If i� �1+ 0 ,0+ 1 , . . . ,−l+1+ l ,−l , . . .�, then − i isnot well defined; otherwise we can arrange �1+ 0 ,0+ 1 , . . . ,−l+1+ l ,−l , . . .� \ �i� into a decreas-ing sequence, which is the subscript of a v ���, and we define − i= �.

Now we consider the tensor of forms �i=−�� vi∧ �V� � ıvi

��V���+ � �− and compute its coeffi-cients of every v�

+� v�

− term. By book keeping, we get the result.Lemma 1: For any two partitions � and �, the coefficient of the v�

+� v�

− term of V is

c��,�� = �i�Z;

�−i and �+i are both well defined;

l��−i�n, l��+i�n

sgn+��,i�sgn−��,i��n��−i�n��+i

G�−iG�+i, �32�

where sgn+�� , i�= �−1� j if v�+ =va0,a−1,. . . and i=a−j; sgn−�� , i�= �−1� j� if v�

− =vb0,b−1,. . . and b−j�� i�b−j�+1.

To simplify c�� ,��, we first prove the following.Lemma 2: For any partitions � and �, and integer i such that �− i and �+ i are both well

defined, we have for any n�Z+

�n��−i�n��+i = n�n + 1���n − 1��. �33�

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Proof: Let �= ��0 ,�1 , . . . ,�q�, �= ��0 ,�1 , . . . ,�r�, and �j=0,1 , . . .�

a−j = �� j − j + 1 for j q

− j + 1 for j � q ,� �34�

b−j = �� j − j − 1 for j r

− j − 1 for j � r .� �35�

We assume that i=a−k and b−l� i�b−l−1, then we have

�n��−i = �j=0

k−1

�p=n−j

n+a−j−1

p �j=k+1

�

�p=n−j+1

n+a−j−1

p , �36�

�n��+i = �j=0

l

�p=n−j

n+b−j−1

p �p=n−l−1

n+i−1

p �j=l+1

�

�p=n−j−1

n+b−j−1

p . �37�

Here we take the convention that �p�p�=1 if p�� p�.

On the other hand, we have

�n + 1�� = �j=0

k−1

�p=n−j+1

n+a−j−1

p �p=n−k−1

n+a−k−1

p �j=k+1

�

�p=n−j+1

n+a−j−1

p , �38�

�n − 1�� = �j=0

l

�p=n−j−1

n+b−j−1

p �j=l+1

�

�p=n−j−1

n+b−j−1

p . �39�

Thus, we get

�n��−i�n��+i

�n + 1���n − 1��

=� j=0

k−1�n − j��n−l−1

n+i−1p

� j=0

l�n − j − 1��n−k+1

n+a−k−1p

. �40�

Notice that a−k= i, we can verify that

�n��−i�n��+i

�n + 1���n − 1��

= n , �41�

and prove the lemma. �

From this result we observe that if both the �− i and �+ i are well defined, the conditionmax�l��− i� , l��+ i��n is equivalent to max�l���−1, l���+1�n. Then we have for l���n+1 and l���n−1,

c��,�� =1

n�n + 1���n − 1���

i�Z;

�−i and �+i are

both well defined;

max�l���−1,l���+1�n

sgn+��,i�sgn−��,i�G�−iG�+i, �42�

and later in this section we assume max�l���−1, l���+1�n. We find that for all but finitely manyi�Z, either G�−i or G�+i is not well defined, and we can write �42� as a finite summation,

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c��,�� =1

n�n + 1���n − 1���j=0

n

sgn+��,a−j�sgn−��,a−j�G�−a−jG�+a−j

, �43�

where a−j is given by �34�, and we assume G�+a−j=0 if �+a−j is not well defined.

Now we recall the determinantal formula for Schur polynomials28 that if = � 0 , . . . , l� withln, then

s �x1, . . . ,xn� =

�x1

n−1+ 0 x2n−1+ 0 . . . xn

n−1+ 0

x1n−2+ 1 x2

n−2+ 1 . . . xnn−2+ 1

] ] . . . ]

x1n−l−1+ l x2

n−l−1+ l . . . xnn−l−1+ l

x1n−l−2 x2

n−l−2 . . . xnn−l−2

] ] . . . ]

1 1 . . . 1

���x1, . . . ,xn�

. �44�

Therefore

����2s ��1, . . . ,�n� = ������1

n−1+ 0 . . . �nn−1+ 0

�1n−2+ 1 . . . �n

n−2+ 1

] . . . ]

1 . . . 1� . �45�

To simplify the integrals in �42�, we need another formula.10

Lemma 3: �de Bruijn’s� For any f0 , . . . , fn−1 ,g0 , . . . ,gn−1�L2�R�,

� ¯� d�1 . . . d�n�fn−1��1� fn−1��2� . . . fn−1��n�fn−2��1� fn−2��2� . . . fn−2��n�

] ] . . . ]

f0��1� f0��2� . . . f0��n���

gn−1��1� gn−1��2� . . . gn−1��n�gn−2��1� gn−2��2� . . . gn−2��n�

] ] . . . ]

g0��1� g0��2� . . . g0��n��

= n ! det� f i�x�gj�x�dx0i, jn−1

. �46�

Now we denote

Gi =� xid��x� , �47�

and by �45� and Lemma 3, for 0 jn we have �a−j is given by �34��

073506-8 Dong Wang J. Math. Phys. 50, 073506 �2009�

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G�−a−j=� ¯� ����2s�−a−j

��1, . . . ,�n�d���1� . . . d���n�

= n !�Gn−1+a0

Gn−1+a0+1 . . . Gn−1+a0+n−1

Gn−1+a−1Gn−1+a−1+1 . . . Gn−1+a−1+n−1

] ] . . . ]

Gn−1+a−jGn−1+a−j+1 . . . Gn−1+a−j+n−1

] ] . . . ]

Gn−1+a−nGn−1+a−n+1 . . . Gn−1+a−n+n−1

� , �48�

where ˆ means the entry is deleted. For �+a−j, we denote v�+ =bb0,b−1,. . ., and similarly have that if

�+a−j is well defined and l��+a−j�n, then

sgn−��,a−j�G�+a−j= sgn−��,a−j�� ¯� ����2s�+a−j

��1, . . . ,�n�d���1� . . . d���n�

= n ! �Gn−1+a−j

Gn−1+a−j+1 . . . Gn−1+a−j+n−1

Gn−1+b0Gn−1+b0+1 . . . Gn−1+b0+n−1

Gn−1+b−1Gn−1+b−1+1 . . . Gn−1+b−1+n−1

] ] . . . ]

Gn−1+b−n+2Gn−1+b−n+2+1 . . . Gn−1+b−n+2+n−1

� . �49�

Here we notice that the n�n matrix in �48� is constructed from an �n+1��n matrix with the�j+1�th row eliminated, and the first row in the matrix in �49� is the same as the deleted row in theconstruction of the matrix in �48�. Thus we have a determinantal formula,

�j=0

n

sgn+��,a−j�sgn−��,a−j�G�−a−jG�+a−j

= �n!�2�B0 B1 . . . Bn−1

Gn−1+b0Gn−1+b0+1 . . . Gn−1+b0+n−1

Gn−1+b−1Gn−1+b−1+1 . . . Gn−1+b−1+n−1

] ] . . . ]

Gn−1+b−n+2Gn−1+b−n+2+1 . . . Gn−1+b−n+2+n−1

� ,

�50�

where all rows except for the first one are the same as those in the matrix in �49�, and the first row�B0 ,B1 , . . . ,Bn−1� is

�B0,B1, . . . ,Bn−1� = �j=0

n

sgn+��,a−j�G�−a−j�Gn−1+a−j

,Gn−1+a−j+1, . . . ,Gn−1+a−j+n−1� . �51�

Notice that sgn+�� ,a−j�= �−1� j, we find that �51� is equivalent to that Bk is the determinant of a�n+1�� �n+1� matrix �k=0,1 , . . . ,n−1�,

Bk = �Gn−1+a0+k Gn−1+a0

Gn−1+a0+1 . . . Gn−1+a0+n−1

Gn−1+a−1+k Gn−1+a−1Gn−1+a−1+1 . . . Gn−1+a−1+n−1

] ] ] . . . ]

Gn−1+a−n+k Gn−1+a−nGn−1+a−n+1 . . . Gn−1+a−n+n−1

� . �52�

Since in the matrix, the first column is identical to the �k+2�th column, we get Bk=0. Therefore� j=0

n sgn+�� ,a−j�sgn−�� ,a−j�G�−a−jG�+a−j

=0, which means that c�� ,��=0 by �43�.

073506-9 Random matrices and KP � functions J. Math. Phys. 50, 073506 �2009�

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Summing up these results, we get the conclusion that V is a decomposable form, and by �27�prove Theorem 1.

ACKNOWLEDGMENTS

I would like to thank M. Adler, J. Baik, J. Harnad, A. Yu. Orlov, and P. Zinn-Justin for helpfulcomments, and the referee for suggestions on the literature about GKM.

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