Truncations of Haar distributed matrices and bivariate Brownian bridge

Transcription

1 Truncations of Haar distributed matrices and bivariate Brownian bridge C. Donati-Martin Vienne, April 2011 Joint work with Alain Rouault (Versailles) 0-0

2 G. Chapuy (2007) : σ uniform on S n. Define for p, q n X (n) p,q = #{1 i p, σ(i) q}. A suitable normalisation of the sequence of two-parameters processes converge in distribution to the bivariate Brownian bridge. X (n) ns, nt Note that X (n) p,q = Tr(Σ p,q Σ p,q) where Σ p,q is the truncated matrix of size p q of Σ, the permutation matrix associated with σ. Similar result when Σ replaced by a Haar distributed matrix?

3 Main result: Let U be a Haar unitary, resp. orthogonal, matrix in U(n), resp. in O(n). We consider, for p n and q n, the upper-left p q submatrix V p,q. Set T p,q = Tr(V p,q Vp,q) = U i,j 2. (1) and i p,j q Y (n) p,q = T p,q ET p,q. We define a sequence of two-parameter processes W (n) by ( ) W (n) := Y (n) ns, nt, s, t [0, 1]. W (n) D([0, 1] 2 ) the set of functions on [0, 1] 2 which are right continuous and with limits in all orthants, endowed with the Skorohod topology.

4 Theorem 1 The process W (n) converges in distribution in D([0, 1] 2 ) to a tied-down Brownian bridge 2 β W ( ) where W ( ) is a centered continuous Gaussian process on [0, 1] 2 of covariance E[W ( ) (s, t)w ( ) (s, t )] = (s s ss )(t t tt ), β = 2 in the unitary case and β = 1 in the orthogonal case. Some related works: - d Aristotile, Diaconis and Newman : ns U ii, s [0, 1] i=1 n law MB(C) - Silverstein: q fixed n1/2 ( ns i=1 U iq 2 s), s [0, 1] n law BB

5 The covariance function (unitary case) From the invariance of the Haar measure, we have: For j k, and if i k, j l, From these relations, and E U i,j 2 = 1 n, E U i,j 4 = E ( U i,j 2 U i,k 2) = 2 n(n + 1) 1 n(n + 1) E ( U i,j 2 U k,l 2) = 1 n 2 1. E(T p,q ) = pq n. V ar(t p,q ) = pq n2 n(p + q) + pq n 2 (n 2 1) if p/n s, q/n t. n s(1 s)t(1 t)

6 Proof of Theorem 1: - tightness of the distributions of W (n) in D([0, 1] 2 ): need to estimate moments of increments of the process W (n) over rectangles. - convergence of the finite dimensional distributions : for (s i, t i ) i k [0, 1] 2, prove that (W (n) s i,t i, i k) (law) Gaussian vector, or equivalently for (a i ) i k R, X (n) := k i=1 a i W (n) s i,t i (law) N(0, i,j a i a j Γ((s i, t i ); (s j, t j ))) To prove the convergence of X (n), show that κ r (X (n) ) 0 for r 3 where κ r denotes the cumulant of order r.

7 Weingarten functions (Collins) a) Unitary case : for any choice of indices i l, j m in {1,...,n}, E ( U i1 j 1...U ik j k Ū.. ) i 1j 1.Ūi k,j k = δ p 1 i δ p 2 j Wg U(n) (p 1, p 2 ) p 1,p 2 M U 2k where M U 2k is the set of pairings of [2k] = {1,...,k, 1,..., k} pairing one element of [k] and one element of [ k]. Wg U(n) is the Weingarten matrix, the pseudo inverse of a Gram matrix. b) Orthogonal case: E ( O i1 j 1...O ik j k O...O ) i 1j 1 i kj k = δ p1 i δ p 2 j Wg O(n) (p 1, p 2 ) p 1,p 2 M 2k where M 2k is the set of pairings of [2k] = {1,...,k, 1,..., k}, Wg O(n) is the orthogonal Weingarten matrix.

8 Another formulation: unitary case E ( U i1 j 1...U ik,j k Ū...Ū ) i 1j 1 i k,j k = δα i δβ j Wg(n, βα 1 ) α,β S k where δ α i = 1 if i(s) = i(α(s)) for every s k and 0 otherwise. In particular, if π S k, Wg(n, π) = E(U 11...U kk Ū 1π(1)...Ū kπ(k) ). S k is the set of permutations of [k].

9 Convergence of the finite dimensional distributions Note that T p,q = Tr(D 1 UD 2 U ) where D 1 = diag(i p, 0 n p ), D 2 = diag(i q, 0 n q ). The asymptotic vanishing cumulants of X (n) := k i=1 a iw (n) s i,t i follows from: Proposition 1 Unitary case (Mingo, Sniady, Speicher) Let D = (D 1,...D k ) and D = (D 1,...D k) be two families of deterministic matrices of size n. We set, for 1 i r, X i = Tr(D i UD ī U ). Then, κ r (X 1,...,X r ) = α,β S r C βα 1,A Tr α ( D) Tr β 1(D) (2) A where in the second sum, A P(r) is such that βα 1 A and A β α = 1 r, (3) and C σ,a are the relative cumulants of the unitary Weingarten function.

10 Moreover, if the sequence {D, D} n has a limit distribution, then for r 3, lim n κ r(x 1,...,X r ) = 0. Asymptotics: C σ,a = O(n 2r #(σ)+2#(a) ) Tr α ( D) = O(n #(α) ), Tr β 1(D) = O(n #(β) ) βα 1 A and A β α = 1 r = κ r = O(n 2 r ) From the proposition, we easily get κ r (X (n) ) 0, r 3.

11 The orthogonal case: Description of M 2k For g S 2k, we define η(g) = k (g(i) g(ī)) M 2k. i=1 H k = {g S 2k ; gγ = γg} oùγ = k (iī) i=1 η(g) = η(g ) h H, g = g h. M 2k S 2k /H k E ( O i1 j 1...O ik j k O...O ) 1 i 1j 1 i kj k = H k 2 where Wg is invariant on the H k \S 2k /H k. δ η(g1) i δ η(g 2) j g 1,g 2 S 2k Wg(g 1 1 g 2)

12 Parametrisation of S 2k /H k : For ǫ = (ǫ 1,...,ǫ k ) { 1, 1} k, we define τ ǫ by τ ǫ = For π S k, t π est défini par: i,ǫ i = 1 (iī) H k. t π (i) = i; t π (ī) = π(i). Each class of S 2k /H k contains exactely one permutation for a particular pair (ǫ, π). g ǫ,π = τ ǫ t π

13 Proposition 2 Let D = (D 1,...D k ) and D = (D 1,...D k) be two families of deterministic and symmetric matrices of size n. We set, for 1 i r, X i = Tr(D i OD ī O ). Then, κ r (X 1,...,X r ) = α,β S r,ǫ { 1,1} r λ α,β,r C σ,a Tr α ( D) Tr β 1(D) A where - σ S r is a function of α, β, ǫ satisfying t α 1τ ǫ t β t σ in H k \S 2k /H k - in the second sum, A P(r) is such that σ A and A β α = 1 r, (4) - C σ,a are the relative cumulants of the orthogonal Weingarten function.

14 Asymptotics C σ,a = O(n 2r #(σ)+2#(a) ) Tr α ( D) = O(n #(α) ), Tr β 1(D) = O(n #(β) ) t α 1τ ǫ t β = τ ǫ t σ h, h H k σ A and A β α = 1 r = κ r = O(n 1 )

15 Lemma 1 For A, B P(r) we have #(A) + #(B) r + #(A B). Moreover, if there exists a block A i of A and B j of B such that #(A i B j ) = l, then, Therefore, #(A) + #(B) r l #(A B). #(α) + #(β) r + #(α β), (5) #(A) + #(α β) r + 1, (6) #(A) #(σ). (7) It is enough to study the cases (worst cases): #(α) + #(β) = r + #(α β) #(α) + #(β) = r 1 + #(α β) For example, if i is a fixed point of α and β, then i is a fixed point of σ and in this case (since A α β = 1 r ), #(A) < #(σ).