The expansion of Wronskian Hermite polynomials in the Hermite basis

We express Wronskian Hermite polynomials in the Hermite basis and obtain an explicit formula for the coefficients. From this we deduce an upper bound for the modulus of the roots in the case of partitions of length 2. We also derive a general upper bound for the modulus of the real and purely imaginary roots. These bounds are very useful in the study of irreducibility of Wronskian Hermite polynomials.


Introduction
Let {H n (x)} n≥0 be the classical Hermite polynomials, solutions to the equation y (x)−2xy (x)+2ny(x) = 0. In this paper we study the Wronskian of such polynomials: if n 1 < n 2 < . . . < n r is a sequence of nonnegative integers, we can define the Wronskian Wr[H n1 (x), H n2 (x), . . . , H nr (x)] as the determinant .
Wronskians of Hermite polynomials appear in the study of rational potentials admitted by the Schrödinger operator L = − ∂ 2 ∂x 2 +V (x). Oblomkov ([10]) characterized rational potentials of monodromy-free Schrödinger operators that grow as x 2 at infinity. In this case, the potentials have the form From (1) the zeros of Wr[H n1 (x), H n2 (x), . . . , H nr (x)] are precisely the poles of the potential. Because of this relationship it is important to understand the geometry of the zeros of the Wronskian ( [4], [6]). Unfortunately not much is known about the set of zeros. Veselov (see [6]) conjectured that all the zeros are simple, except possibly at the origin. This conjecture is known to be true in a few special cases, but in general it is still open. In contrast, for the Hermite polynomials H n (x) it is well known that the zeros are real and simple.
Then He λ (x) is a monic polynomial of degree n. Furthermore, He λ (x) has up to scaling the same set of zeros as the Wronskian of {H n1 (x), H n2 (x), . . . , H nr (x)}.
Recently, Bonneux, Dunning and Stevens ([1]), following earlier work ( [2]), found an explicit formula for the coefficients of He λ (x) in terms of the characters of the symmetric group. Theorem 4.2). Let λ n. Then where H(λ) := n! χ λ (1) . Our main contribution in this paper is to establish a dual version of Theorem 1, where we determine the coefficients of He λ (x) in the Hermite basis.
Our interest in Theorem 2 stems from the need to find good bounds for the modulus of the roots, proportional to √ n. For Hermite polynomials, Szegő proved the following.
We made no effort to optimize the constant in Corollary 4, as the correct bound is likely close to 2 √ λ 1 . One can also obtain bounds for the real and purely imaginary roots from Theorem 2.
Corollary 5. Let λ n. If z is a real or purely imaginary root of He λ (x) then |z| ≤ x n , where x n is the largest root of He n (x).
However, Corollary 5 does not give the full picture. By exploiting the Schrödinger equation it is possible to obtain a better bound. Proposition 6. Let λ = (λ 1 , λ 2 , . . . , λ r ). If z is a real root of He λ (x) then |z| ≤ x λ1+r−1 , where x λ1+r−1 is the largest root of He λ1+r−1 (x).
This result is already apparent in the work of García-Ferrero and Gómez-Ullate ( [8], Section 2). However, the proof in [8] is done under the additional assumption of semi-degeneracy, which is so far unproven for Wronskian Hermite polynomials. Therefore, we include a proof of Proposition 6.
These bounds are very useful in the study of irreducibility of He λ (x). We will treat this topic in a forthcoming paper.
The rest of this note is organized as follows. In Section 2, we fix the notation and state the auxiliary results that we will need. In Section 3, we prove Theorem 2. In Section 4, we derive the two corollaries. Finally, in Section 5, we prove Proposition 6.

Notation and auxiliary results
In this section we define the notation that we use throughout the paper. We also state several results that will be needed for the proof.

2.1.
Partitions and the symmetric group. If n ≥ 0 is an integer, a partition λ of n, denoted λ n, is a sequence of non-negative integers λ 1 ≥ λ 2 ≥ . . . ≥ λ r such that r i=1 λ i = n. We denote |λ| = n and call (λ) := r the length of the partition λ. We say that λ i are the parts of the partition.
Note that we allow parts of size 0. This will simplify many of our statements and proofs. We shall frequently use the notation (λ 1 , λ 2 , . . . , λ r ) for λ. Sometimes we will also use the notation µ = 1 r1 2 r2 3 r3 . . ., meaning that the partition µ has r i parts of size i.
The degree sequence of the partition λ is defined as n λ := (λ r , λ r−1 + 1, . . . , This can be represented as a collection of unit squares arranged in rows, with the i-th row having λ i squares. For example, The conjugate partition λ is obtained from λ by transposing the Ferrers diagram: λ j is the largest index i such that λ i ≥ j. If λ and µ are two partitions, a semistandard Young tableau of shape λ and type µ is a filling of the Ferrers diagram of λ with the numbers 1, 2, . . . , (µ) such that the number i appears µ i times, the numbers weakly increase along rows, and strictly increase along columns. The Kostka number K λµ is the number of semistandard Young tableaux of shape λ and type µ.
Clearly K λµ ≥ 0. Furthermore, K λµ > 0 if and only if λ dominates µ, written λ µ, that is when |λ| = |µ| and We denote by χ λ the irreducible character of S n associated to λ. Let F λ := χ λ (1) be the degree of the irreducible representation. Then this is given by the formula (see [7], (4.11)): 2.2. Schur polynomials. Let x 1 , x 2 , . . . , x k be k variables. We let e i (x 1 , . . . , x k ) be the elementary symmetric polynomials: By convention e i (x 1 , . . . , x k ) = 0 when i > k. Similarly, we let h i (x 1 , . . . , x k ) be the complete symmetric polynomials: Finally, we let p i (x 1 , . . . , x k ) be the power-sum polynomials: We are not going to write the variables if they are clear from the context. For our purposes we will also use the convention e 0 = h 0 = p 0 = 1.
If λ = (λ 1 , λ 2 , . . . , λ r ) is a partition, we define e λ = e λ1 e λ2 · · · e λr , Now assume that k ≥ r. By adding parts of size 0, we can further assume that r = k. In this case we define W λ (x 1 , . . . , Then W λ is an alternating polynomial, and hence it is divisible by the Vandermonde determinant W 0 (x 1 , . . . , We call s λ the Schur polynomial for λ. Then s λ is a symmetric polynomial, and is defined for any partition λ with (λ) ≤ k. In the case (λ) > k we define s λ (x 1 , . . . , x k ) := 0. We will need the following consequence of Pieri's rule (see [7], Appendix A.1).
Theorem 7. For any partition µ we have where the sum is taken over all partitions λ with non-zero parts.
We shall also need the following (see [7], Lecture 4, (4.10), with C i the conjugacy class of µ). . However, Turán only proved an inequality for |Im z| for a decomposition in the base of classical Hermite polynomials. For convenience, we explain how to adapt the proof to obtain Theorem 9.
Proof. Let x k,1 ≤ x k,2 ≤ . . . ≤ x k,k be the roots of He k (x).
If z is a complex number, we define D(z) = min |z − x k,i |, where the minimum is taken over all 1 ≤ i ≤ k ≤ n. The main step of the proof is showing that if z is a root of P (x) and z is not a root of He n (x). The inequality (4) follows verbatim from Turán's proof by replacing |y| with D(z) everywhere (see also [9] for an exposition of the proof). Therefore we shall not reproduce the proof here.
Using (4) we will show the inequality in the theorem. If z is a root of He n (x) then the inequality is trivially true, as x n,n ≥ 0. So we may assume that z is not a root of He n (x). As the roots of Hermite polynomials interlace, we have x n,1 ≤ x k,i ≤ x n,n . The roots of Hermite polynomials are also symmetric around the origin, i.e. x n,1 = −x n,n . Therefore the root with largest absolute value among x k,i is x n,n . Then Therefore D(z) ≥ |z| − x n,n , and the inequality follows from (4).

Proof of the main result
We start by writing the base-change formula.
Proof. We use the fact that Replacing every power of x in the expression of He λ (x) gives We know that a j = (−1) j H(λ) 2 j j!(n−2j)! χ λ (2 j 1 n−2j ) from Theorem 1. Replacing a j in Lemma 10 gives Let S λ k be the sum k j=0 (−1) j k j χ λ (2 j 1 n−2j ). The somewhat surprising fact is that this sum can be evaluated.
Lemma 11. Let λ n and 0 ≤ k ≤ n 2 . Set µ k := 2 k 1 n−2k . Then S λ k = 2 k K λ µ k . Proof. By adding parts of size 0, we may assume without lack of generality that λ has length n, and that we have n variables x 1 , . . . , x n . Set µ j := 2 j 1 n−2j , 0 ≤ j ≤ k, and define M λ as the monomial x λ1+n−1 . . x λn n . From Theorem 8, we know that χ λ (µ j ) is the coefficient of M λ in the polynomial p µj (x 1 , . . . , x n )W 0 (x 1 , . . . , x n ). Therefore S λ k must be the coefficient of M λ in the polynomial where the last equality follows from the binomial theorem.
However, p 2 1 − p 2 = 2e 2 . Furthermore, p 1 = e 1 . Hence the above is the polynomial 2 k e n−2k when (ρ) ≤ n, and s ρ = 0 otherwise, where in the last sum the appropiate number of zero parts were added to ρ such that W ρ is defined. Recall that we need the coefficient of M λ . We argue that this can only come from ρ = λ. Let ρ be a partition such that the determinant W ρ = det[x ρj +n−j i ] contains the monomial M λ . Assume for a contradiction that ρ = λ. By comparing powers of each x i , there must exist a permutation σ such that Let i be minimal such that σ(i) = i. Then σ(i) > i. Let j > i such that σ(j) = i. As ρ i ≥ ρ σ(i) we have Therefore the coefficient of M λ comes from ρ = λ only, and so it is 2 k K λ µ k . This finishes the proof.
Theorem 2 now follows directly by replacing S λ k in (5):

An upper bound for the modulus of the roots
In this section we derive the bounds on the absolute value of the roots. We again write He λ (x) = n 2 k=0 b k He n−2k (x). Proof of Corollary 4. The plan is to the compute the coefficients b k exactly and then use Theorem 9.
This finishes the proof.
Turán's theorem is modelled after Walsh's theorem ( [13], see also [9]), which gives a similar bound, but in terms of the expansion in the usual basis {x n } n≥0 . However, applying Walsh's theorem to the coefficients in Theorem 1 only gives a bound of the order O(n √ n). The bound does not change for partitions of length 2: for example, for He n,2 (x) it is possible to compute the character values exactly and show that the upper bound in Walsh's theorem is at least Ω(n √ n). Hence changing to the Hermite basis and relying on Theorem 2 is necessary.
Let us now look at the real and purely imaginary roots of He λ (x).
Proof of Corollary 5. Let x k,1 ≤ x k,2 ≤ . . . ≤ x k,k be the roots of He k (x). We show that He λ (x) has no real root in the interval (x n,n , +∞). The roots of Hermite polynomials interlace, so x n,1 ≤ x k,i ≤ x n,n for all 1 ≤ i ≤ k ≤ n. Therefore if z ∈ R is greater than x n,n , then He k (z) > 0 for all k ≤ n. Furthermore, b k ≥ 0 for all k and b 0 = 1. Hence He λ (z) > 0, so z is not a root.
Proposition 12 is equivalent to Proposition 6. This follows from Definition 1, the fact that He n (x) is a rescaling of H n (x), and the fact that the roots of Hermite polynomials interlace.
The proof relies on the following simple observation.
Then the statement follows from Observation 13. Now assume r ≥ 3 and the induction hypothesis holds. Define From the induction hypothesis it follows that ψ nr−2 (x), ψ nr−1 (x) and ψ nr (x) do not vanish in I R . Then from the definition they are repeatedly differentiable in I R . In this situation, Crum ([5]) showed that for i ∈ {n r−1 , n r }, ψ i verifies the Schrödinger equation where V (x) = x 2 − 2 ∂ 2 ∂x 2 log ψ nr−2 (x). Crum proved this in the case when n 1 , n 2 , . . . , n r−2 are consecutive integers starting from 0, and only for the interval (0, 1) with boundary conditions. However, the proof of (8) remains valid for a sequence n 1 < n 2 < . . . < n r−2 of non-consecutive integers, in a neighborhood of x where the Wronskians do not vanish.