MATRIX INTEGRALS AND MAP ENUMERATION 2

Transcription

1 MATRIX ITEGRALS AD MAP EUMERATIO 2 IVA CORWI Abstract. We prove the generating function formula for one face maps and for plane diagrams using techniques from Random Matrix Theory and orthogonal polynomials. Much of this material is from the article Matrix Integrals and Map Enumeration by A. Zvonkin, as well as from the maps book by Lando and Zvonkin.. Matrix Integrals () A function F : H H is unitary invariant if F(U HU) = F(H) for every unitary matrix U (i.e., U = U ). Such a function can then be considered as just a function of Λ, the diagonal matrix of eigenvalues λ,...,λ). Proposition.. If F is a unitary invariant function on H then F(H)dµ(H) = c F(Λ) (λ i λ j ) 2 dµ(λ ) dµ(λ ) H i<j for some constant c depending only on the dimension of matrices. Here the matrix Λ is the diagonal eigenvalue matrix, and dµ(h) is the Gaussian measure defined on H via the quadratic form 2 tr(h2 ), and dµ(λ) is standard Gaussian measure on the line. (2) It is worth noting that the product term is just the square of the Vandermonde determinant and so 2 λ λ 2 λ (λ i λ j ) 2 = λ 2 λ 2 2 λ 2 i<j λ λ2 λ (3) We will, later, use orthogonal polynomials to determine the constant c. 2. Generating function for one face maps () Recall the connection between one face maps and matrix integrals: Theorem 2.. Let T n () be T n () := tr(h 2n ) = tr(h 2n )dµ(h). H Then n/2 n/2 ( ) g T n () = ǫ g (n) k+ 2g = k+ ǫ g (n) 2, g=0 where ǫ g (n) is the number of label one-face maps of genus g with n edges. Date: October 26, Mathematics Subject Classification. 82C22, 60K35. Key words and phrases. Asymmetric Simple Exclusion Process, Interacting Particle Systems, Last Passage Percolation. g=0

2 MATRIX ITEGRALS AD MAP EUMERATIO 2 2 (2) Let T(,s) = + 2s + 2s n= T n () (2n )!! sn. Then we can determine what the ǫ g (n) are from the following result Theorem 2.2. The generating function T(,s) is equal to ( ) + s T(,s) =. s From this we have that the numbers ǫ g (n) satisfy the recurrence relation (n + 2)ǫ g (n + ) = (4n + 2)ǫ g (n) + (4n 3 n)ǫ g (n ) where ǫ g (0) = for g = 0 and 0 otherwise. (3) We prove this in two steps, the first uses random matrices and the second is purely combinatoric. (a) We have the following result which we prove using random matrix theory: Lemma 2.3. The function t(,n) = is a polynomial in n of degree. Proof. H tr H 2n dµ(h) = c R tr(λ 2n ) T n() (2n )!! i<j (λ i λ j ) 2 dµ(λ i ). Replace tr(λ 2n ) by λ using symmetry. Expand the Vandermonde and integrate over every λ i for i. We are left with a polynomial in λ of degree 2n+2 2 with coefficients constant for fixed. Moreover, every monomial is of the form λ 2n+2k for 0 k. Integrating such a monomial against Gaussian measure gives (2n + 2k )!! (2n )!! which is a polynomial of degree k in n. (b) We can interprete T n () as the number of gluing of the 2n-gon which agree with possible colors of the vertices in (not more than) colors. Then letting T n () be the number of gluings which agree with colors with precisely colors, we have T n () = L= ( L ) T n (L). (c) Observe that T 0 () = T () = = T 2 () = 0 since the resulting graph of a 2n-gon on a surface has at most n + vertices. By the previous lemma and the formula above, one sees by induction that t(,n) = T n () (2n )!! is also a polynomial in n of degree. We know the roots of this polynomial are 0,,2,..., 2. So ( ) n t(n,) = A ( )!. i=

3 MATRIX ITEGRALS AD MAP EUMERATIO 2 3 Substituting this gives T n () = (2n )!! L= A L ( n L )( L ) (L )!. We can find the constant by considering the leading term (coefficient of n+ in T n ()) and using the fact that it coincides with the Catalan numbers A n+ (2n )!! (n + )! n! = Cat n = (2n)! n!(n + )!, thus we find A n+ = 2 n /n! and T n () = (2n )!! ( )( ) n 2 L. L L L= It is easy now to check that this gives the desired generating function. 3. Solving for c () Recall the formula for integrating a unitary invariant function against the Gaussian measure on H. If we consider the function F(Λ) =, then we have = dµ(h) = c (λ i λ j ) 2 dµ(λ i ). H R So if we can evaluate that integral, we can find c. (2) Introduce the Hermite polynomials: Listing a few we have i<j i= H n (x) = ( ) n e x2 /2 dn dx n(e x2 /2 ). H 0 (x) = H (x) = x H 2 (x) = x 2 H 3 (x) = x 3 3x H 4 (X) = x 4 6x (a) H is a polynomial in x of degree n; (b) The leading coefficient of H n equals ; (c) The polynomials H 0,H,...,H n form an orthogonal basis in the polynomials of degree n with respect to the standard Gaussian measure on R and the scalar product (P,Q) = R PQdµ(λ); (d) H n uniquely satisfies the previous property; (e) (H n,h n ) = n(h n,h n ). (3) Using elementary row operations we can rewrite the Vandermonde determinant in terms of the Hermite polynomials: 2 H 0 (λ ) H 0 (λ 2 ) H 0 (λ ) (λ i λ j ) 2 = H (λ ) H (λ 2 ) H (λ ) i<j H (λ ) H (λ 2 ) H (λ )

4 MATRIX ITEGRALS AD MAP EUMERATIO 2 4 When you expect this determinant each term H i (λ j ) appears with at most degree two. Moreover, each monomial has the λ j term occur exactly twice, so we can split it into each pair. By orthogonality the only terms which contribute to the integral are H 2 0 (λ σ())h 2 (λ σ(2)) H 2 (λ σ()) where σ S. Each such monomial contributes the same, and there are! of them. So (λ i λ j ) 2 dµ(λ i ) =!(H 0,H 0 ) (H,H ). R i<j i= Using the recurrence relation for (H n,h n ) = n(h n,h n ) we find that (H n,h n ) = n! and so (λ i λ j ) 2 dµ(λ i ) =!( )! 2!! and hence R i<j c = i=!( )! 2!!. 4. Enumeration of Quadrangulations () We will enumerate labeled quadrangulations, where we define a labeled quadrangulation as a connected graph with n faces, all the face being of degree 4, faces are labeled by numbers,...,n, the edges incident to each face are labeled by numbers,2,3,4 and this labeling corresponds to their cyclic order. Alternatively we can use the dual notion of a diagram which is a connected graph with n vertices, all the vertices being of degree 4, vertices are labeled by numbers,..., n, the edges incident to each vertex are labeled by numbers,2,3,4 and this labeling corresponds to their cyclic order. (2) A coupling will have the same labeling properties as the above definitions, however the underlying graph need not be connected. It will be easier to count couplings than diagrams. (3) Count the number of couplings and diagrams with n = and n = 2. (4) Let C g (n) be the number of couplings with n vertices and Euler characteristic χ = 2 2g (we use Euler characteristic since the coupling may involve disjoint graphs and hence characteristics add). Let D g (n) be the number of diagrams with n vertices and genus g (it is alright to use genus since the graph is connected). By Taylor expansion and the Wick formula Theorem 4.. (5) Similarly define Z(t,) := { } t exp tr H4 = D(t,) := ( t) n C g (n) 2 2g. n! n=0 ( t) n D g (n) 2 2g. n! n= (6) A standard result in enumerative combinatorics shows that these two generating functions, one of general labeled objects and the other for connected labeled objects, are related as follows:

5 MATRIX ITEGRALS AD MAP EUMERATIO 2 5 Lemma 4.2. (7) Factoring out the 2 we have Theorem 4.3. Let E(t, ) be Then, where log Z(t,) = D(t,) E(t,) := log Z(t,) = 2 2 log E(t,) = g 0 E g (t) = n ( ) g 2 E g (t), n! ( t)n D g (n) exp { t } tr H4. and D g (n) is the number of connected diagrams of genus g with n vertices. 5. The free energy of the V = λ 4 ensemble and the planar approximation () Consider the measure dµ V (λ) = e tv (λ) dµ(λ) on the line, where dµ is standard Gaussian measure and Define a scalar product V = V (λ) = λ 4. (P,Q) V = PQdµ V. (2) With respect to this measure define generalized Hermite polynomials P V,0 (λ),p V, (λ),... such that P V,k is a polynomial of degree k with leading term λ k, The P V,k are orthogonal with respect to dµ V. If V = 0 then these are the regular Hermite polynomials. (3) For V (λ) = λ 4, P V,k are even for k even and odd for k odd. (4) As a function of t, let h k (t) = h k = (P k,p k ) V. Just as before, we can replace the Vandermonde determinant with these polynomials and using orthogonality show that t tr H4 dµ(h) = c!h h 2 h 0. H e As we will be taking the log of this, it is very convenient to have this product structure. We need to find these h n (t) though. (5) For every family of orthogonal polynomials we have a three term recurrence, which, due to the parity, takes the form here of Lemma 5.. λp k = P k+ + R k P k for some constant R k = R V,k (t). From this lemma we can relate h k = (P k,λp k ) V = (λp k,p k ) V = (P k+ + R k P k,p k ) V = R k h k,

6 MATRIX ITEGRALS AD MAP EUMERATIO 2 6 and therefore ttr H4 e dµ(h) = c!r R2 2 R h 0. H Finally, for P,Q polynomials, integration by parts gives (P,Q ) V = PQ e 2 λ2 tv (λ) dλ = (P,Q) V + (P,(λ + tv )Q) V. sqrt2π Using these tools we have kh k = (P k,p k ) V and factoring out h k we get = (P k,p k) V + (P k,(λ + 4tλ 3 )P k ) V = (λp k,p k ) V + 4t(λP k,λ 2 P k ) V = h k + 4t(P k + R k P k 2,λ 2 P k ) V = h k + 4t [ (λp k,λp k ) V + R k (λ 2 P k 2,P k ) V ] = h k + 4t(h k+ + R 2 k h k + R k h k ) = h k (R k + 4tR k (R k+ + R k + R k )), k = R k + 4tR k (R k+ + R k + R k ), which is the discrete Painlevé I equation. (6) We are interested in E 0 (t) which is the first order term in the asymptotics of E(t,) as goes to infinity: E 0 (t) = lim 2 log e t tr H4 dµ(h). H This corresponds to the generating function for planar diagrams (labeled gluings of squares). So if we substitute t/ in for t everywhere and divide everything by our discrete Painlevé I equation becomes k = r k(t)( + 4t(r k+ (t) + r k (t) + r k (t))) with r k (t) = R k (t)/. Taking k and to infinity together so that k/ x [0,], we get (non-rigorously) x = r( + 2tr) R where r = r(x,t) = lim x (t) n. We non-rigorously assume that this ratio tends to a continuous function. (7) Take x = and solve for r(,t) = α(t) as r(,t) = α(t) = t 24t = n 0( ) n (2t) n Cat n. ow recall that e t tr H4 dµ(h) =!R R2 2 R h 0. c H Here c can be intepreted as the same integral with t = 0. Therefore, taking logs we have 2 log e t tr H4 dµ(h) = ( k ) log R k(t) H R k (0) log h 0(t) h 0 (0). k=

7 MATRIX ITEGRALS AD MAP EUMERATIO 2 7 Taking the limit as for t fixed gives ( ) r(x,t) E 0 (t) = ( x)log dx. x 0 We can change this into an integral over r instead of x since dx/dr = + 2rt which is positive for t and r positive. Hence we get E 0 (t) = α 0 ( r 2tr)( + 24tr)log( + 2tr)dr = 2 log α + ( α)(α 9). 24 Using the series for α(t) we get n (2n )! t n E 0 (t) = ( 2) (n + 2)! n! = ( 2) n Cat n 2n(n + 2) tn. m= (8) Therefore the number of plane diagrams of n vertices (or dually n faces) is given by n= n (2n )! D 0 (n) = (2) (n + 2)!. 6. The free energy for invariant ensembles See Asymptotics of the Partition Function for Random Matrices via Riemann-Hilbert Techniques, and Applications to Graphical Enumeration by.m. Ercolani and K.D. T-R McLaughlin. I. Corwin, Courant Institute of the Mathematical Sciences, ew York University, 25 Mercer Street, ew York, Y 002, USA address: corwin@cims.nyu.edu