Central Limit Theorem for Local Linear Statistics in Classical Compact Groups and Related Combinatorial Identities
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1 arxiv:math/ v [math.pr] 3 Aug 999 Central Limit Theorem for Local Linear Statistics in Classical Compact Groups and Related Combinatorial Identities Alexander Soshnikov California Institute of Technology Department of Mathematics Sloan Pasadena, CA 95 USA and University of California, Davis Department of Mathematics Davis, CA 9566, USA Abstract We discuss CLT for the global and local linear statistics of random matrices from classical compact groups. The main part of our proofs are certain combinatorial identities much in the spirit of works by Kac and Spohn. Introduction Let M be a unitary matrix chosen at random with respect to the Haar measure on the unitary group Un. We denote the eigenvalues of M by {expi θ j } n j=, where π θ, θ,..., θ n < π. The joint distribution of the eigenvalues called the Weyl measure is absolutely continuous with respect to the Lebegue measure n j= dθ j on the n-dimensional tori and its density
2 is given by P Un θ,..., θ n = π n n! j<k n expi θ j expi θ k. see [We]. Throughout the paper we will be interested in the global and local linear statistics n S n f = fθ j,. j= S n gl n = n gl n θ j,.3 j= L n L n, n 0. The optimal conditions on f, g for our purposes are k= ˆfk k <,.4 where fx = ĝt t dt <,.5 k= gx = π ˆfk e ikx, ĝt e itx dt However in order to simplify the exposition we will always assume that f has a continuous derivative on a unit circle f C S and g is a Schwartz function g fjr. Let us denote by E n the mathematical expectation with respect to Haar measure. We start with the formulation of the result which is essentially due to C. Andréief [A], for a modern day reference see [TW]and also [Dy].
3 Proposition E n expts n f detid + e tf K n = detid + e tf Q n,.6 where e tf is a multiplicaiton operator and K n, Q n : L S L S are the integral operators with the kernels K n x, y = π sin n x y,.7 sin x y Q n x, y = n j=0 e ijx e ijy.8 π π Remark. K n, Q n are unitary equivalent to each other and are the operators of a finite rank. In particular, Q n is just a projection operator on the first n harmonic functions of the unit circle. One of the ingredients of the proof of the proposition is the following chain of the equalities p Un θ,..., θ n = n! detei j θ k j,k n dete i j θ k j,k n = Q n! det n θ j, θ k = K n! det n θ j, θ k j,k n j,k n.9 Remark allows us to rewrite the Fredholm determinants in.6 as the Toeplitz determinant with the symbol exp t f : E n exp t n j= fθ j = D n expt f π = det exp tfx exp ij kx dx π 0 j,k n.0 3
4 The asymptotics of.0 for large n is given by the Strong Szego Limit Theorem: D n expt f = exp tn ˆf0 + t +. k ˆfk + 0 see [Sz] and [K], [H], [De], [F-H], [G-I], [Wid], [Wid], [McC-W], [Ba-W], [Jo], [Bo], [Bo-S], [Me], [So], [Wie], [D] for further developments. In probabilistic terms. claims that ES n f = n π fθdθ + 0 π π actually the remainder term is zero, and the centralized random variable n j= fθ j = E n n j= fθ j converges in distribution to the normal law N0, k ˆfk. Our first goal is to establish a similar result for the local linear statistics. Theorem. Let g JR, L n +, n 0. Then E n n j= gl n θ j = n π L gxdx, and the centralized random variable n j= gl n θ j E n j= gl nθ j converges in distribution to the normal law N0, π + ĝt t dt. We give a combinatorial proof which holds both in the local and global cases. In some sense our approach is close to the heuristic arguments in [I-D]. We start with L n 4
5 Lemma. Let C l,n f be the l-th cumulant of S n f. Then C l,n f k +...+k l =0 l l m = l, l,...,l m l +...+l m i= l +...+l m i ˆfk... ˆfk l l m= l! l!... l m! n max0, k i max0, l i= k i const l m m l i= l +l k i, k i,..., i= k k l = 0 k k l >n l +l k i, k i,..., i= k ˆfk... ˆfk l. Remark One can see that for sufficiently smooth f the r.h.s. of. goes to zero as n. Remark 3 An analogous result to lemma was established in [Spo] for the determinantal random point field with the sine kernel see also Remark 4 below. The proof of Lemma will be given in. At this state we observe that it implies Lemma The limit of C l,n f, l > exists as n and is equal to k +...+k l =0 ˆfk... ˆfk l Gk,..., k l +G k,..., k l, where G is the piece-wise linear continuous function defined by Gk,..., k l := σ S l max 0, l i= l m= i= m m l l m = l, l,...,l m l +l l +...+l m k σi, k σi,..., i= l!... l m! k σi..3 5
6 Proof of Lemma After opening the brackets in. we observe that the coefficient in front of n is equal to { l m l! m l m=!...l m! =, l =.4 0, l > l l m = l, l,i=,...,m Indeed, the generating function of these coefficients is equal to log + e z = z. Now CLT for n j= fθ j follows from Main Combinatorial Lemma Let k,..., k l be arbitrary real numbers such that their sum equals zero. Let Gk,..., k l be defined as in.3. Then { k = k if l = Gk,..., k l = 0 if l >. We will prove the lemma in 3. Remark 4 A similar combinatorial lemma was stated by Spohn in [Spo]. He studied a time-dependent motion of a system of infinite number of particles governed by the equations dλ j t = i j λ i λ j dt + db j t, where {b j t} + j= -independent standard brownian motions, and the initial distribution of particles is given by determinantal random point field with sin πx y the sine kernel. However, no correct proof of the combinatorial πx y result was given there. For completeness we give a proof of Spohn s lemma independently from the proof of our Main Combinatorial Lemma in 3. Assuming the combinatorial part is done we can quickly finish the proof of Theorem. The formula for the mathematical expectation is trivial. Rewriting. for the higher cumulants of n j= gl n θ j we see that the limit of the l-th cumulant is given by π l ĝt... ĝt l Gt,..., t l + G t,..., t l dt...dt l 6
7 where the integral is over the hyperplane t t l = 0. Theorem is proven. Remark 5 Our method also gives an elementary combinatorial proof of Szegö theorem. for f C S and sufficiently small complex t. It is different from the one suggested by Kac in [K] where the Taylor expansion of D n tg as a function of t was calculated and then a so- called Kac-Spitzer combinatorial lemma was employed to confirm.. Remark 6 Results similar to Theorem have been established for other random matrix models in [Spo], [Jo3], [KKP], [Ba], [B-F], [SSo], [SSo], [BM-K]. The rest of the paper is organized as follows. We prove Lemma in and Main Combinatorial Lemma in 3. The result analogous to Theorem for orthogonal and symplectic groups is established in 4. The author would like to thank Ya. Sinai, P.Diaconis, K. Johanson and A. Khorunzhy for useful discussions. The work was partially supported by the Euler stipend from the German Mathematical Society. Proof of Lemma We start with calculating the moments of S n f. Le us remember that k- point correlation function of the eigenvalues of random unitary matrix is given by n! ρ n,k θ,..., θ k = p Un θ,..., θ n dθ k+...dθ n n k! T n k = det K n θ i, θ j = det Q n θ i, θ j i,j k i,j k. The N-th moment of S n f is equal to E n n i = fθ i... n fθ in, i N = where the indices i,..., i N range independently from to n, and in particular can coincide. Let M = {M,..., M r } be a partition of the set {,,..., N} into subsets determined by coinciding indices among i,..., i N : 7
8 M = {j,..., j s },..., M r = {j r,..., j r s r }, r i= M i = {,,..., N}, s i = M i, i =,...r. Then E n S n f N = over all partitions M E n l l... l r f s θ l... f sr θ lr. Let us consider a typical term in. corresponding to a partition M. E n f s θ l... f sr θ lr = l... l r.3 T r f s x... f sr x r ρ n,r x,..., x r dx...dx r By definition of the determinant and. ρ n,r x,..., x r = σ S r σ r Q n x i, x σi. Writing the permutation σ S r as a product of cyclic permutations we have ρ n,r x,..., x r = over all cyclic permutations of K α over partitions K of {,...r} p α j= Q n x t α j i= q p α α=, x α σt.4 j where {,..., r} = q K α, K α = {t α,..., tα p α }, α =,..., q, p α = K α. Substituting.4 into.3 we arrive at the expression that has the following form :. over partitions M={M,...,M r} of {,...,N} over partitions K={K,...,K q} of {,...,r} To interchange the order of summation we construct a new partition P = {P,..., P q } of {,,...N} as follows: P i = j Ki M j, i =,..., q. Then 8
9 {M j } j Ki gives a partition of P i that we denote by P i. We have E n Sn f N = T t i t i j= over partitions P={P,...,P q} of {,...,N} q i= f P i, x... f P i,t i x ti t i Q n x j, x σj dx...dx ti. over partitions P i of P i :P i ={P i,,...,p i,ti } over cyclic permutations σ S ti.5 We remind that the moments are expressed in terms of cumulants as m N = C P... C Pk. over partitions P={P,...,P k } Comparing the last formula with.5 we arrive at C l,n f = partitions f R x... f Rm x m T m P={R,...,R m} of {,...,l} m m Q n x j, x σj = l m= cyclic permutations j= σ S n m l! over ordered collections l,...,l m: m l i=l,l i f l x... f lm x m m! T m m m Q n x j, x j+ dx...dx m = l m= j= l,..., l m : l l m = l, l i,i=,...,m l!...l m! m!.6 9
10 m m l! l!... l m! f l x... f lm x m T m m Q n x j, x j+ dx...dx m. j=.7 Since Q n x, y = n j=0 e ijx y we can rewrite.7 as C l,n f = n s =0 l m=... l,..., l m : l +...+l m=l, l i n s m=0 m m l! l!...l m! f l sm + s f l s + s... f lm s m + s m. Writing down the Fourier coefficients of the powers of f as the convolutions of the Fourier coefficients of f f l sm + s = ˆfk... ˆfk l, k,..., k l : k +...+k l =s s m f l s + s = k l+,..., k l : k l k l =s s f lm s m + s m = ˆfk l +... ˆfk l, ˆfk lm +... ˆfk lm, k lm +,..., k lm : k lm k lm =s m s m we obtain C l,n f = k +...+k l =0 ˆfk... ˆfk l l m m= l! #{u : 0 u n, 0 u + l!...l m! l k i n,..., 0 u + l +...+l m m k i n }. l,..., l m : l +...+l m=l, l i.8 0
11 The last factor in.8 is equal to n max 0, l l +...+l m k i,..., k i max 0, l k i,..., l +...+l m k i.9 if the expression in.9 is nonnegative or zero otherwise. Lemma is proven. 3 Proof of the Main Combinatorial Lemma First we show that Gk,..., k l is a linear combination of terms k i k is. Then we compute the coefficient in front of every such term and show it to be equal to zero. Assume l >. Consider a partition P = {P,..., P m } of the set {,,..., l}. Let us denote v = j P k j,..., v m = j P m k j. The expression for G can be transformed into l Gk,..., k l = m= P={P,...,P m} m m τ S m max0, v τ, v τ + v τ,..., v τ + v τ v τm. 3. In [R-S] Rudnick and Sarnak, following the ideas of [K] and [Spi] see also [B], [An], used the following identity for the set of real numbers v,...v m with zero sum: max0, v τ, v τ + v τ,..., v τ + v τ +...v τm m τ S m = F!m F! v l 4 F {,...m}, l F F,F C 3.
12 The last formula gives us Gk,..., k l = 4 F! l m= i j F P j k i P={P,...,P m} F F {,...m}, F,F C m F m F!. 3.3 Le us denote by A the subset j F P j of {,,..., l}. Then {P j } j F defines a partition of A, and {P j } j F C a partition of A C = {,,..., l} \ A. We change now the order of summation in 3.3: first we sum over all nonempty subsets A of {,,..., l} and then over all partitions of A and A C : Gk,..., k l = 4 A {,..., l}, over partitions A,A C U={U,...,U r} of A U U! k i. over partitions i A U of A C Finally we note that U={U,...U r} U U! = A! r! t!... t r! r!, A r= U U! t,..., t r : r i= t i= A,t i r 3.4 the expression we already considered in.4. Indeed, there are exactly A! t!... t r! different partitions of A such that {t r!,..., t r } = { U,..., U r }. If A this sum is zero. If A =, then A C = l A and the second factor in 3.4 equals zero by the same argument. Now we turn to a combinatorial lemma first formulated in [Spo]. Let us denote by α = α,..., α l, β = β,..., β l vectors with entries α j {0, }. We consider a lexicographic order on the set of such vectors: α < β
13 iff α j β j, j =,..., l and at least for one j 0 α j0 < β j0. Following [Spo] we call such nonzero vectors branches and a set T of ordered branches T = {α,..., α m }, α < α <... < α m, T = m < l, a tree. We denote by Tl the set of all trees formed by a l-dimensional vectors branches. A combinatorial sum in question is Uk,..., k l = T max0, α k α T 3.5 T Tl Here we used the notation α k = l j= α j k j. We call max0, α k α T the maximum of the tree T. For a warm-up we prove Proposition Uk,..., k l + U k,..., k l = Gk,..., k l, k l+ + G k,..., k l, k l+, where k l+ = k k... k l. Remark 7 Once the proposition is proven we see of course that Uk,..., k l + U k,..., k l is zero for l. Proof of Proposition In the above notations Gk,..., k l, k l+ = T Tl+ T max0, α k α T, T where k = k,..., k l, k l+, and the sum is over all trees T Tl + such that the largest branch of T, α T is less than D =,,...,. Similarly, we can write Uk,..., k l = T Tl+ T max0, α k α T, where the sum is over the trees T Tl + such that the l + th coordinate of α T is zero. We define a rotation on the set of all trees such that α T D : Wα, α,..., α T = α α, α 3 α,..., α T α, D α. Since l+ j= k j = 0, we observe that max0, α k α T + max0, α k α T = max 0, α k α WT + max0, α k α WT
14 The last equality implies Uk,..., k l + U k,..., k l = T Tl+ = = T max0, α k α T + max0, α k α T T Tl+ T T T p=0 max0, α k α W p T + max0, α k α W p T T Tl + α T D + α k α T T max0, α k α T T = Gk,..., k l+ + G k,..., k l+ Here we used that for any T with α T D there exist a unique T with T α l+ = 0 and 0 p < T such that T = W p T. Proposition Uk,..., k l = 0 if l. 3.7 We proceed by induction. It is easy to check the case l =. Let us assume that the proposition is true for some l. Consider Uk 0, k,..., k l. Since U is a symmetric function we may assume k 0 k...k l. The continuity of U implies that it is enough to check 3.7 for nondegenerate vectors k 0, k,..., k l. Therefore we may assume that the coordinates k,...k l are linearly indepdendent over the integers. Fix such k,...k l and consider U as a piecewise linear function of y = k 0, Uy, k = Uy, k, k,..., k l. Our first claim is that Uy, k is zero for all negative y. To show this we write Uy, k = T max0, α y, k α T = T Tl+ T Tl + : T α =0 + T Tl + : T T α =, α =0 4 + T Tl + : T T α =, α =
15 We denote the three subsums by U, U, U 3. The first subsum is equal to T max0, α k α T, the second T Tl T Tl T max0, α k α T + max0, y, and by the induction assumptions both are zero. Now we split the third subsum in two. Consider the smallest branch α T such that the first coordinate of α is, denote this branch by α and denote the preceding may be empty branch by α. We write U 3 = U 3, + U 3,, where in U 3, T T the summation is over T Tl +, such that α =, α = and α α >, 0,..., 0, and in U 3, the summation is over all other trees from U 3. We establish a one-to-one correspondence between U 3, and U 3, : for any tree T with α α >, 0,..., 0 we construct T = {α,..., α, α +, 0,..., 0, α,..., α T }. Clearly, T = T +, therefore T max0, α y, k α T = T max0, α y, k α T, and U 3, and U 3, cancel each other. Now we assume that y is nonnegative and 0 y k < k <... < k l. As we already noted Uy, k,..., k l is a piecewise linear continuous function. We claim that it can change its slope only at y = 0. Indeed, Uy, k,..., k l can change its slope only at the points of degeneracy of y, k,..., k l, where α 0 y + α k = α 0 y + α 0 k and the coordinates of α 0, α, α 0, α take values zero and one. Because k is a non-degenerate vector we must have y +α k = α k or α k = y +α k. Since the tree T contains both branches, α and 0, α only if α α, the only solution for nonnegative vector y, k must be y = 0, α = α. We will finish the proof of the proposition if we show that Uy, k = 0 for sufficiently small positive y. We again write U = U + U + U 3 as before. Then U = 0 by inductive assumption and U 3 is zero for sufficiently small y U 3, and U 3, still cancel each other. We can write the second subsum U as T max0, α k α T + y = T Tl T Tl T max0, α k α T + y T Tl T 3.8 5
16 the last sum includes empty tree. The first term in 3.8 is zero by inductive assumption and the second is also zero since T Tl T = l l m = l +, l i m m l +! l!... l m! = 0. Proposition is proven. 4 Orthogonal and symplectic groups. We start with the orthogonal case. The eigenvalues of matrix M SOn can be arranged in pairs expiθ, exp iθ,..., expiθ n, exp iθ n, 0 θ, θ,..., θ n < π. Consider the normalized Haar measure on SOn. The probability distribution of the eigenvalues is defined by its density see [We]: P SOn θ,..., θ n = cosθ i cosθ j 4. π i<j< n The k-point correlation functions are given by see [So] ρ n,k θ,..., θ k = det K n θ + i, θ j i,j n 4. where K n x, + y = K n x, y + K n x, y = π sin n x y sin + sin x y n x+y sin x+y. 4.3 In [D-S] and [Jo] Diaconis-Shahshahani and Johansson studied asymptotic properties of linear statistics n j= fθ j where for simplicity we may assume that f is real even trigonometric polynomial, fθ = m k= a kl ikθ + 6
17 l ikθ, a k = ˆfk, k =,,..., m. As before we denote the linear statistics by S n f. Then S n f = Trace m k= a km k. It was shown that n E n exp t fθ j = j= exp t m k t ˆfk m + k ˆfk + 0 k= which implies the convergence in distribution of n j= fθ j to the normal law m N m + k ˆfk, k ˆfk. k= Actually 4.4 holds under much weaker conditions it is enough to assume f C +α [0, π], α > 0. Remark 8 Similarly to the unitary case 4.4 is equivalent to the large n asymptotics result for some determinants, this time Hankel determinants see [Jo], [Jo]. Our combinatorial approach allows to prove CLT for all f C [0, π] as well as to study the local linear statistics n j= gl n θ j θ, 0 < θ < π.in particular we establish Theorem Let g be a Schwartz function, L n +, Ln 0 and 0 < θ < π. n Then E n n j= gl n θ j θ = n gxdx+ 0, L n π and the centralized random variable n j= gl n θ j θ E n n j= gl n θ j θ converges in distribution to the normal law N0, π ĝt t dt. Theorem also holds for SOn + and Spn. Let M SOn +. Then one of the eigenvalues of M is and the other n eigenvalues can be arranged in pairs as before. The density of the eigenvalues is equal to n P SOn+ θ,..., θ n = cosθ i cosθ j sin θi. π i<j n k= k= The formula for the k-point correlation function is i= 4.5 ρ n,k θ,..., θ k = det K n θ i, θ j i,j=,...,k 4.6 7
18 where Kn x, y = K nx, y K n x, y = sin nx y π sin sinnx + y x y sin x+y. 4.7 The analogue of 4.4 reads n E n+ exp t fθ j = exp t j= m + k t ˆfk + k= m k ˆfk + 0. k= 4.8 In the symplectic case M Spn the n eigenvalues again can be arranged in pairs expi θ i, exp i θ,..., expi θ n, exp i θ n, 0 θ, θ,..., θ n < π, their density is equal to n P Spn θ,..., θ n = π i<j n cosθ i cosθ j and the formula for k-point correlation function is n sin θ i, 4.9 ρ n,k θ,..., θ k = det K n+ θ i, θ j i,j=,...k. 4.0 i= The analogue of 4.4 reads n E n expt fθ j = j= exp t m + k t ˆfk + k= 4. m k ˆfk + 0. k= We will prove Theorem for SOn. The proofs for SOn + and Spn are almost identical. 8
19 Proof of Theorem The arguments from imply that it is enough to prove Lemma 3 Let C l,n f be the l-th cumulant of n j= fθ j, l. Then C l,n f ˆfk i... ˆfk l Gk,..., k l k +...+k l =0 + G k,..., k l const l 4. k k l = 0 k k l >n k ˆfk... ˆfk l + const l ˆfk... ˆfk l k k l >n We start with the formula.6 which holds for general determinantal random point fields: l C l,n f = m l! m= l l m = l, l i l!... l m! m m K n x j, x j+ + K n x j, x j+ dx...dx m j= we always assume x m+ = x. = l m= l l m = l l i m m [0,π] m f l x... f lm x m m i= ǫ i [0,π] l! l!... l m! ǫ =± ǫ =± [0,π] m f l x... f lm x m... ǫ m=± 4.3 m K n x j, ǫ j x j+ dx... dx m j= Each term in the last sum with m i= ǫ i = is equal to m m f l x... f lm x m K n x j, x j+ dǫ i x i = j= f l x m... f lm x m [0,π] m i= m K n x j, x j+ dx... dx m j= 9
20 we use the fact that fx is even. Combining these terms together we obtain the same expression as for C l,n n j= fθ j in the case of Un, which gives vanishing contribution if l >. Finally we claim that the contribution from the terms with m i= ǫ i = can be bounded from above by const l ˆfk... ˆfk l. Indeed, the integral k k l >n [0,π] m f l x... f lm x m can be rewritten as m n s = n... n s m= n m π j= n s j = n e is jx j ǫ j x j+ dx...dx m f l s ǫ m s m f l s ǫ s... f lm s m ǫ m s m Consider the euclidian basis {e j } m j= in Rm and define f j = e j ǫ j e j, ǫ 0 = ǫ m. The vectors {f j } m j= form a basis in R m iff m j= ǫ j =. Then for any m-tuple t,..., t m there exists the only m-tuple s,..., s m such that t j = s j ǫ j s j, j =,..., m. We write f l j tj = ˆfkl +...+l j +... ˆfk l +...+l j, where the sum is over k i such that l +...+l j l +...+l j + k i = t j. When we plug this into 4.3 we obtain a linear combination of ˆfk... ˆfk m 4.4 It is easy to see that for k k m n the coefficient with the term 4.4 is equal to m l m= l l m = l, l i m m l! l!... l m! = 0 For k k m > n the coefficient is bounded from above by some constant. This finished the proof of Lemma 3. Similar to we obtain the proof of Theorem by applying the lemma to n j= gl n θ j θ. 0
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