Rectangular Young tableaux and the Jacobi ensemble

It has been shown by Pittel and Romik that the random surface associated with a large rectangular Young tableau converges to a deterministic limit. We study the fluctuations from this limit along the edges of the rectangle. We show that in the corner, these fluctuations are gaussian wheras, away from the corner and when the rectangle is a square, the fluctuations are given by the Tracy-Widom distribution. Our method is based on a connection with the Jacobi ensemble.


Statement of the results
From a formal point of view, a rectangular Young tableau of size (m, n) can be defined as an mn-tuple of integers (X 1,1 , . . .X m,n ) satisfying, for all i, j, X i,j < min(X i,j+1 , X i+1,j ) {X 1,1 , . . .X m,n } = {1, 2, . . .mn} We denote by X m,n the set of mn-tuples of this form.
Pittel and Romik [11] studied the limit shape of the surface associated with a random large rectangular Young tableau.Fix a real t > 0, and consider a Young tableau of size (n, ⌊tn⌋), chosen uniformly at random.Then they proved that for all reals (r, s) ∈ [0, 1] 2 , X ⌊rn⌋,⌊stn⌋ /tn 2 converges in law as n → ∞ to a deterministic quantity g(r, s, t), which is the solution of a minimization problem.We want to study the fluctuations of this shape along the edges of the rectangle.First, we state a completely explicit result on the corner: Theorem 1 Let (X 1,1 , . . ., X m,n ) be a uniform random variable on X m,n .Then for n ≤ k ≤ mn − m + 1, Let us compare this result with the following model.Let R be a rectangle of size ((m − 1)(n − 1), m + n − 1) on an integer lattice.Consider the set of paths going from the south-west corner to the north-east corner with only north and east steps.Choose such a path uniformly at random.Then it is an elementary exercise to check that the probability for the n-th north step to be the k-th step is equal to (1).However, we have not been able to find a combinatorial link between this simple model and the corner of a rectangular Young tableau.
A consequence of Theorem 1 is the following: Corollary 1 Fix some real t ∈ R * + .For every n, put m n = ⌊tn⌋ and let (X where G is a standard gaussian random variable.
Our second result only holds when the rectangle is a square and deals with the fluctuations on the edges.
n,n ) be a uniform random variable on X n,n .Then there exists a function r : (0, 1) → R * + such that for t ∈ (0, 1), as where T W has the Tracy-Widom distribution.
Pittel-Romik's result tells us that n −2 EX (n) ⌊tn⌋,n converges to the value of the limit shape at (t, 1), namely (1 + √ 2t − t 2 )/2.Our method consists in studying Young tableaux in a slightly modified framework.As the asymptotic shape only makes sense when the "height" is renormalized, that is, the X i,j are divided by mn, it seems natural to work directly in a continuous framework.The formal setup is the following.
Definition.For a pair of integers (m, n), let Y m,n be the set of mn-tuples We want to study uniform random variables on Y m,n .Denote by ∆(x 1 , . . .x k ) the Vandermonde of the k-tuple (x 1 , . . .x k ): When the rectangle is a square, we get: Theorem 3 Let n be a positive integer and (Y 1,1 , . . ., Y n,n ) be a uniform random variable on the set Y n,n .Then for every k ∈ has a marginal density proportional to For a general rectangle, the result reads: has a marginal density proportional to has a marginal density proportional to Of course, for a diagonal of the form (Y m−k+1,1 , Y m−k+2,2 , . . ., Y m,k ), we get a similar expression as in (i).The densities appearing in these two theorems belong to the general class called the Jacobi ensemble.This ensemble also appears in various models, among others the MANOVA procedure in statistics [10], log-gas theory [6], Wishart matrices and random projections [4].For a detailed account on random matrices, we refer to [1].
Using Theorems 3 and 4 together with known results on the Jacobi ensemble enables us to derive the results stated above.Moreover, the deterministic limit shape can also be recovered this way, see Section 4.An alternative form of Theorem 2 in the continuous setting is as follows: where TW has the Tracy-Widom distribution.
Remark that Y n,n is a convex subset (indeed, a compact polytope) of R n 2 .From this point of view, Corollary 2 can be seen as a result on the projection of the uniform measure on a convex set in high dimension.This is reminiscent of the classical result saying that if (X 1 , . . .X n ) is a random vector distributed according to the uniform measure on the euclidean n-dimensional sphere with radius √ n, then X 1 is asymptotically gaussian.
The remainder of this note is organized as follows.We prove Theorems 3 and 4 in Section 2. Theorem 1 is derived in Section 3. Section 4 is devoted to the proofs of the asymptotic results.Some concluding remarks are made in Section 5

Diagonals of continuous tableaux
We prove here Theorem 3. The proof of Theorem 4 is the same and is omitted.The basic idea is to use a random generation algorithm filling the tableau diagonal by diagonal, using conditional densities.
We begin with a preliminary lemma.Let n ≥ 2 and define by induction the following polynomials: g 1 (x) = 1 and for i ≤ n − 1, Lemma 1 (i) For every i ∈ [1, n], there exists a constant c i such that , there exists a constant c i such that Proof An elementary proof of (i) can be found in Baryshnikov [2].To deduce (ii), we proceed by induction.For m ≤ n, define K m (ε, y 1 , . . .y n−m ) as the integral We want to evaluate On the one hand, using (i) easily gives for some positive constant C(m, n), where the equivalent is uniform over all On the other hand, where the equivalent is uniform over all (n + m)-tuples (r 1 , . . .y 1 , . . .s m ) satisfying where C ′ (m, n) is the constant such that It follows that uniformly over all (n − m)-tuples (x 1 , . . .x n−m ) satisfying (4), Comparing this estimate with (3) yields (ii).We now describe an algorithm generating random elements in Y n,n .A similar algorithm can be used to generate random permutations with a prescribed profile of ascents and descents [9].In the remainder of this section, k and n are the integers in the statement of Theorem 3.For i ∈ [1, n], we denote If (x 1 , . . .x j ) and (y 1 , . . .y j+1 ) are two sequences of reals, we say that they are interlacing in [0, 1] if 0 ≤ y 1 ≤ x 1 ≤ y 2 . . .≤ x j ≤ y j+1 ≤ 1.We denote the event that this interlacing relation is satisfied by Inter((x 1 , . . .x j ), (y 1 , . . .y j+1 ))

Algorithm
• Choose the diagonal D k at random according to the density where • By induction, for i from k down to 2, conditional on D i , choose D i−1 according to the conditional density • By induction, for i from k to n − 1, conditional on D i , choose D i+1 according to the conditional density g i+1 (x 1 , . . ., x i+1 )1 Inter(Di,(x1,...,xi+1)) g i (D i ) • By induction, for i from n to 2n − 2, conditional on D i , choose D i+1 according to the conditional density First, remark that the conditional densities used by the algorithm are indeed probability densities.That is, they are measurable, positive functions and their integral is 1.The latter fact is easy to verify: for instance, by definition of g i , and thus, using (5), we get We claim that the algorithm yields a random element of Y n,n with the uniform measure.Indeed, by construction, the n 2 -tuple generated by the algorithm has a density which is the product of the conditional densities of the diagonals D 1 , D 2 . . .D 2n−1 .Hence this density is given by The expression above is a telescopic product and after simplification, we find that the density is constant on the set Y n,n .This proves our claim.Finally, the density of D k is proportional to according to Lemma 1.This proves Theorem 3.

The law of the corner
One can relate the discrete and the continuous model of Young tableaux.To construct a continuous Young tableau (Y i,j ) of size (m, n) from a discrete Young tableau (X i,j ) of the same size, proceed as follows: • Let (X i,j ) be a uniform random variable on X m,n .
• For every pair (i, j), let k(i, j) be the integer satisfying X i,j = k(i, j).
Then put Y i,j = Z k(i,j) .
Proposition 1 Consider the (mn)-tuple (Y i,j ) constructed as above.Then (i) (Y i,j ) is distributed according to the uniform measure on Y m,n , (ii) For every pair (i, j), The proof of (i) is elementary and (ii) follows from a simple variance computation, using the fact that the density of Z k is As a consequence of Proposition 1, let f i,j be the marginal density of Y i,j .For every 1 ≤ k ≤ mn put p i,j (k) = P(X i,j = k).Using Proposition 1, we get that the density f i,j is equal to where h k (x) is the density of Z k .Thus This way one can deduce the probabilities in the discrete model from the densities in the continuous model.For the case i = 1, j = n, according to Theorem 4 (i), To obtain the desired decomposition, divide both sides of ( 6) by (1 − x) mn−1 and use the change of variables y = x/(1 − x) to get and p i,j (k) = 0 otherwise.This proves Theorem 1.
4 Asymptotic results

Proof of Theorem 2
For every n, let (Y n,n ) be distributed according to the uniform measure on Y n,n .Collins' results on the "soft edge" of the Jacobi ensemble [4] can be translated in our context as follows.For every t ∈ (0, 1), there exists a sequence (s n (t)) and a constant r(t) such that, for every k, as n goes to infinity, Consider the empirical measure General results on Jacobi ensembles apply in this case and we get that µ n (t) converges in distribution, as n goes to infinity, to the deterministic probability measure with density where See for instance the first proposition in [5].Reformulating this result, we get the following.Let r, s ∈ [0, 1] 2 , r ≥ s and put t = 1 − s + r.Then as n tends to infinity, X ⌊rn⌋,⌊sn⌋ converges in law to the Dirac point mass δ g(r,s) where being the inverse of the function f t (y)dy and f t being given by ( 8) and (9).The function g is an alternative fomulation of the limit shape found by Pittel and Romik.Remark however that this result is weaker than Pittel-Romik's, since it only gives the convergence along a diagonal, whereas Pittel and Romik show a uniform convergence on the whole rectangle.In fact, a major weakness of our method is that it only allows us to work on a single diagonal, and not on several diagonals simultaneously.

Concluding remarks
Consider the case when the rectangle is a square.It would be interesting to study the transition between the deterministic regime of X n,n and the fluctuations of order n 4/3 for X ⌊tn⌋,n , as well as the transition between the fluctuations of order n 4/3 for X ⌊tn⌋,n and the fluctuations of order n 3/2 for X 1,n .A natural conjecture is the following: Conjecture 1 Let (a n ) be a nondecreasing sequence with a n → ∞ as n → ∞.Then, up to a multiplicative constant, where T W has the Tracy-Widom distribution.
Let us explain where this conjecture comes from.The function r from Theorem 2 can be computed using [4]: Computing the asymptotics when t → 1 leads to the second part of the conjecture.The first part comes from a link with the random matrix model known as the GUE.For a fixed k, define a family of variables (T Then it follows immediately from Theorem 3 that, as n goes to infinity, the renormalized diagonal (T k,k ) has a limit density proportional to This is the density of the eigenvalues of a random (k, k) matrix from the GUE, and classical results [1] naturally lead to the first part of the conjecture.Proving it would involve an exchange of limits, which does not seem to have been achieved in the literature so far.
There is a link between rectangular Young tableaux and a particle system known as the TASEP, see [12].In this view, a phenomenon of arctic circle arises when the rectangle is a square.In the general case, the arctic curve is no longer a circle but it is still algebraic.To compute an equation of this curve, one has to determine the parameters of the Jacobi ensemble associated with the rectangle using Theorem 4, and then compute the values λ ± for this Jacobi ensemble as in [5].
The horizontal strips of the TASEP diagram where this arctic curve appears correspond to diagonals of the Young tableau.Moreover, the places of the vertical steps inside a horizontal strip correspond to the integers in the diagonal of the Young tableau, and Theorems 3 and 4 tell us that these vertical steps are asymptotically distributed like the eigenvalues of a Jacobi ensemble.A similar result has been established by Johansson and Nordenstam for domino tilings of the Aztec diamond [8], where the Jacobi ensemble is replaced by the GUE.
In the case of a GUE of size n, the fluctuations of the eigenvalues in the bulk are of order log n/n and are asymptotically gaussian [7].It is not clear whether the same behaviour occurs in our context.
A model of corners of Jacobi ensembles was studied recently by Borodin and Gorin [3], who showed a convergence to the gaussian free field.Their model is slightly different from ours, but it would be interesting to know whether their results could be transposed in our case.
Finally, the method used here can be applied to generate at random a standard filling of a general polyomino: compute the conditional densities of the diagonals, which will be polynomials given by multiple integrals, and then use a generating algorithm as in Section 2. Of course, the problem is that for a general polyomino, the corresponding polynomials will not have a simple form as for rectangles.