About division by 1

Alain Lascoux^{∗}

Abstract

The Euclidean division of two formal series in one variable produces a sequence of series that we obtain explicitly, remarking that the case where one of the two initial series is 1 is sufficiently generic. As an application, we define a Wronskian of symmetric functions.

The Euclidean division of two polynomials P (z), Q(z), in one variable z, of consecutive degrees, produces a sequence of linear factors (the successive quotients), and a sequence of successive remainders, both families being symmetric functions in the roots of P and Q separately.

Euclidean division can also be applied to formal series in z, but it never stops in the generic case, leaving time enough to observe the law of the coefficients appearing in the process.

Moreover, since the quotient of two formal series is also a formal series, it does not make much difference if we suppose that one of the two initial series is 1. This renders the division of series simpler than that of polynomials; in fact the latter could be obtained from the former.

By formal series we mean a unitary series

f(z) = 1 + c_{1}z + c_{2}z^{2} + · · · .

We shall moreover formally factorize it

f(z) = σ_{z}(𝔸) := ∏_{a∈𝔸}(1 − za)^{−1} = ∑_{i=0}^{∞} z^{i} S_{i}(𝔸) ,

the alphabet 𝔸 being supposed to be an infinite set of indeterminates, or of complex numbers, the coefficients S_{i}(𝔸) being called the complete functions in 𝔸.

∗text written during the Conference Applications of the Macdonald Polynomials, at the Newton Institute in April 2001.

Given two series, dividing f_{−1}(z) = σ_{z}(𝔸) by f_{0}(z) = σ_{z}(𝔹) means finding the unique coefficients α, β such that

(σ_{z}(𝔸) − (1 + αz) σ_{z}(𝔹)) \frac{1}{β} z^{−2}  (1)

is a unitary series f_{1}(z) = σ_{z}(ℂ).

Dividing in turn f_{0}(z) by f_{1}(z), one obtains f_{2}(z), and iterating one gets from the initial pair (f_{−1}, f_{0}) an infinite sequence of series f_{−1}, f_{0}, f_{1}, f_{2}, f_{3}, …

However, all the equations

f_{k−1}(z) = (1 + α_{k} z) f_{k}(z) + β_{k} z^{2} f_{k+1}(z)  (2)

can be divided by f_{0}(z) = σ_{z}(𝔹). If the k-th remainder for the pair (σ_{z}(𝔸), σ_{z}(𝔹)) is σ_{z}(ℂ), then the k-th remainder for the pair (σ_{z}(𝔸)/σ_{z}(𝔹) , 1) is

σ_{z}(ℂ − 𝔹) = ∑_{i=0}^{∞} z^{i} S_{i}(ℂ − 𝔹) := \frac{∏_{b∈𝔹}(1 − zb)}{∏_{c∈ℂ}(1 − zc)} .

Indeed, already at the first step, one sees that the equation

σ_{z}(𝔸) = (1 + zS_{1}(𝔸 − 𝔹)) σ_{z}(𝔹) + S_{2}(𝔸 − 𝔹) z^{2} f_{1}(z)  (3)

is equivalent to

σ_{z}(𝔸 − 𝔹) = (1 + zS_{1}(𝔸 − 𝔹)) 1 + S_{2}(𝔸 − 𝔹) z^{2} f_{1}(z)/σ_{z}(𝔹)  (4)

since the two coefficients are functions of 𝔸 − 𝔹.

Of course equation (4) expands into

σ_{z}(𝔸 − 𝔹) = (1 + zS_{1}(𝔸 − 𝔹)) + S_{2}(𝔸 − 𝔹) z^{2} ∑_{i} z^{i} \frac{S_{2+i}(𝔸 − 𝔹)}{S_{2}(𝔸 − 𝔹)} (5)

We shall see that the other remainders can be as easily written. We just need to recall the definition of a Schur function S_{λ}(𝔸 − 𝔹), where λ = [λ_{1} ≥ λ_{2} ≥ ⋯ ≥ λ_{ℓ} ≥ 0] is a partition :

S_{λ}(𝔸 − 𝔹) = det(S_{λ_{i}+j−i}(𝔸 − 𝔹))_{1≤i,j≤ℓ} .

Then Theorem 1 The k-th remainder in the Euclidean division of σ_{z}(𝔸) by 1 is

f_{k}(z) = S_{(k+1)^{k}}(𝔸)^{−1} ∑_{i=0}^{∞} z^{i} S_{k+1+i,(k+1)^{k−1}}(𝔸) .  (6)

The k-th remainder of the division of σ_{z}(𝔸) by σ_{z}(𝔹) is

f_{k}(z) = S_{(k+1)^{k}}(𝔸 − 𝔹)^{−1} σ_{z}(𝔹) ∑_{i=0}^{∞} z^{i} S_{k+1+i,(k+1)^{k−1}}(𝔸 − 𝔹) . (7)

Proof. Merging all equations (2) together, one can characterize in the first case f_{k}(z) as the unique series such that there exists scalars α_{1}^{k}, …, α_{k−1}^{k}, β_{1}^{k}, …, β_{k}^{k}, γ^{k} :

z^{2k}γ^{k} f_{k}(z) = (1 + α_{1}^{k}z + ⋯ + α_{k−1}^{k}z^{k−1}) σ_{z}(𝔸) − (1 + β_{1}^{k}z + ⋯ + β_{k}^{k}z^{k}) 1 . (8)

Defining

σ_{y}(𝔸 ± \frac{1}{z}) = σ_{y}(𝔸)σ_{y}(±\frac{1}{z})) = σ_{y}(𝔸)(1 − y/z)^{∓1} ,

I claim that

z^{k−1}S_{(k+1)^{k−1}}(𝔸 − \frac{1}{z}) σ_{z}(𝔸) + (−z)^{k}S_{k^{k}}(𝔸 + \frac{1}{z})  (9)

is equal to z^{2k}S_{(k+1)^{k}}(𝔸) modulo terms of higher degree in z.

Indeed, z^{k−1}S_{(k+1)^{k−1}}(𝔸 − 1/z) σ_{z}(𝔸) can be written as the k × k determinant

|\begin{matrix}z^{k−1}σ_{z}(𝔸) & ⋯ & σ_{z}(𝔸) \\ S_{k}(𝔸) & ⋯ & S_{2k−1}(𝔸) \\ ⋮ & & ⋮ \\ S_{2}(𝔸) & ⋯ & S_{k+1}(𝔸) \end{matrix}|

because, subtracting from each column, except the first, 1/z times the preceding column and using S_{j}(𝔸 − 1/z) = S_{j}(𝔸) − S_{j−1}(𝔸)/z, j ∈ ℤ, the determinant factorizes into z^{k−1}σ_{z}(𝔸) S_{(k+1)^{k−1}}(𝔸 − 1/z).

The coefficients of z^{k+1}, …, z^{2k−1} are the functions S_{2,(k+1)^{k−1}}(𝔸), …, S_{k,(k+1)^{k−1}}(𝔸) which are zero, having two identical rows in their determinantal expression.

The coefficients of z^{0}, …, z^{k} are

S_{1−k,(k+1)^{k−1}}(𝔸), …, S_{1,(k+1)^{k−1}}(𝔸)

which, after permuting their first row with the others, are recognized to be

S_{k^{k−1},0}(𝔸), …, S_{k^{k−1},k}(𝔸) ,

up to the sign (−1)^{k−1}.

These terms sum up to (−z)^{k}S_{k^{k}}(𝔸 + 1/z), and the required series f_{k}(z) is obtained by dividing by (−1)^{k−1}S_{k^{k−1}}(𝔸). QED

The first defining equations for the f_{i}’s are, writing S_{j} for S_{j}(𝔸) :

z^{2}S_{2} f_{1} = σ_{z}(𝔸) − (1 + zS_{1}) = σ_{z}(𝔸) − zS_{1}(𝔸 + \frac{1}{z}) ,

− z^{4}S_{33}/S_{2} f_{2} = (1 − zS_{3}/S_{2})σ_{z}(A) − (1 + zS_{21}/S_{2} + z^{2}S_{22}/S_{2}) =

= −S_{3}(𝔸 − \frac{1}{z})σ_{z}(𝔸)/S_{2} − S_{22}(𝔸 + \frac{1}{z})/S_{2} ,

z^{6}S_{444}/S_{33} f_{3} = (1−zS_{43}/S_{33}+z^{2}S_{44}/S_{33}) σ_{z}(𝔸)−(1+zS_{331}/S_{33}+z^{2}S_{332}/S_{33}

+ z^{3}S_{333}/S_{33}) = z^{2}S_{44}(𝔸 − \frac{1}{z}) σ_{z}(𝔸)/S_{33} − z^{3}S_{333}(𝔸 + \frac{1}{z})/S_{33}

Equation (9) can be understood as giving the Padé approximant of degree [k, k − 1] of the series σ_{z}(𝔸) :

σ_{z}(𝔸) = (−1)^{k−1} \frac{z S_{k^{k}}(𝔸 + 1/z)}{S_{(k+1)^{k−1}}(𝔸 − 1/z)} + z^{2k} \frac{S_{(k+1)^{k}}(𝔸)}{z^{k−1}S_{(k+1)^{k−1}}(𝔸 − 1/z)} f_{k}(z)(10)

Sylvester treated the Euclidean division by a different method, using summations on subsets of roots (cf. [LP]). We shall for our part interpret now the Euclidean division as producing a sequence of alphabets from a given one. This time, it is more convenient to divide 1 by a series, and put f_{−1} = 1 = σ_{z}(0), f_{0} = σ_{z}(𝔸) = σ_{z}(𝔸^{0}), …, f_{k} = σ_{z}(𝔸^{k}), . . ..

From (7) one has

σ_{z}(𝔸^{k} − 𝔸^{0}) = S_{(k+1)^{k}}(0 − 𝔸^{0})^{−1} ∑_{i=0}^{∞} z^{i} S_{k+1+i,(k+1)^{k−1}}(0 − 𝔸^{0})

that is,

S_{k^{k+1}}(𝔸) σ_{z}(𝔸^{k}) = σ_{z}(𝔸) ∑_{i=0}^{∞} z^{i} S_{k+1+i,(k+1)^{k−1}}(−𝔸^{0})  (11)

Proposition 2 The successive remainders σ_{z}(𝔸^{k}) in the division of 1 by σ_{z}(𝔸) satisfy

S_{k^{k+1}}(𝔸) σ_{z}(𝔸^{k}) = ∑_{i=0}^{∞} z^{i} S_{k+i,k^{k}}(𝔸)  (12)

Proof. Instead of having recourse to determinants, and using relations between minors, let us invoke symmetrizing operators. Suppose the cardinality of 𝔸 to be finite, 𝔸 = {a_{1}, …, a_{N}}, before letting it tend to infinity. Let π_{ω} be the maximal isobaric divided difference, that is, the operator such that the image of a^{λ} := a_{1}^{λ_{1}} ⋯ a_{N}^{λ_{N}} is S_{λ}(𝔸), for any λ ∈ ℕ^{N}. The series on the right of (12) is the image under π_{ω} of

(a_{1} ⋯ a_{k+1})^{k}/(1 − za_{1})

and also of

(a_{1} ⋯ a_{k+1})^{k}/(1 − za_{1}) ⋯ (1 − za_{k+1}) ,

which can be written

\frac{(a_{1} ⋯ a_{k+1})^{k} (1 − za_{k+2}) ⋯ (1 − za_{N})}{(1 − za_{1}) ⋯ (1 − za_{N})} .

The denominator is symmetrical in a_{1}, …, a_{N}, and thus commutes with π_{ω}. As for the numerator, the only monomials giving a non-zero contribution have exponents

\underbrace{k, …, k}_{k+1} \underbrace{1, …, 1}_{i} \underbrace{0, …, 0}_{N−k−1−i} , 0 ≤ i ≤ N − k − 1 .

Therefore

∑_{i=0}^{∞} S_{k+i,k^{k}}(𝔸) = ∑(−z)^{i}S_{k^{k+1},1^{i}}(𝔸) σ_{z}(𝔸) ,

which gives (11) since S_{λ}(𝔸) = (−1)^{|λ|}S_{λ^{∼}}(−𝔸), where λ^{∼} is the conjugate to the partition λ. QED

The sequences 𝔸, 𝔸^{1}, 𝔸^{2}, … have been considered by B.Leclerc, to whom the following notion of a Wronskian of complete symmetric functions is due (there are more general Wronskians associated to any set of symmetric functions, and any alphabet).

Let n be an integer, k_{1}, …, k_{n} ∈ ℕ, and 𝔸 be an alphabet. Then the Wronskian W (S_{k_{1}}, …, S_{k_{n}}; 𝔸) is the determinant

det|S_{k_{j}−i}(𝔸_{i})|_{0≤i≤n−1,1≤j≤n} ,

where 𝔸^{0} := 𝔸, 𝔸^{1}, 𝔸^{2} … is the sequence of alphabets obtained in the Euclidean division of 1 by σ_{z}(𝔸).

As an application of his study of relations between minors [Bl], B. Leclerc obtained in an unpublished note :

Theorem 3 Let n be an integer, K = [k_{1}, …, k_{n}] ∈ ℕ^{n}, 𝔸 be an alphabet. Then

W (S_{k_{1}}, …, S_{k_{n}}; 𝔸) = S_{K+[n−1,... ,0]}(𝔸)/S_{(n−1)^{n}}(𝔸) .  (13)

Proof. To simplify the notation, we shall take n = 4, K[α, β, γ, δ]. Then the Wronskian, in terms of functions of 𝔸 only, is, according to (12),

|\begin{matrix}S_{α000} & S_{β000} & S_{γ000} & S_{δ000} \\ S_{α100} & S_{β100} & S_{γ100} & S_{δ100} \\ S_{α220} & S_{β220} & S_{γ220} & S_{δ220} \\ S_{α333} & S_{β333} & S_{γ333} & S_{δ333} \end{matrix}|  (14)

up to division of the second, third, last row by S_{11}, S_{222}, S_{3333} respectively. All the entries of this determinant are 4 × 4 minors of the matrix :

column index [\begin{matrix}0 & 1 & 2 & 3 & 4 & 5 & α & β & γ & δ \\ S_{0} & S_{1} & S_{2} & S_{3} & S_{4} & S_{5} & S_{α+3} & S_{β+3} & S_{γ+3} & S_{δ+3} \\ . & S_{0} & S_{1} & S_{2} & S_{3} & S_{4} & S_{α+2} & S_{β+2} & S_{γ+2} & S_{δ+2} \\ . & . & S_{0} & S_{1} & S_{2} & S_{3} & S_{α+1} & S_{β+1} & S_{γ+1} & S_{δ+1} \\ . & . & . & S_{0} & S_{1} & S_{2} & S_{α} & S_{β} & S_{γ} & S_{δ} \end{matrix}] (15)

Designating a minor by the sequence indexing its columns, the determinant to study is

|\begin{matrix}[012α] & [012β] & [012γ] & [012δ] \\ [013α] & [013β] & [013γ] & [013δ] \\ [034α] & [034β] & [034γ] & [034δ] \\ [345α] & [345β] & [345γ] & [345δ] \end{matrix}|

and according to Bazin (cf. [Bl]), factorizes into

[0123] [0134] [0345] [αβγδ]

that is, more explicitly, into the product

S_{0000}(𝔸) S_{1100}(𝔸) S_{2220}(𝔸) S_{3333}(𝔸) .

Reintroducing the missing factor S_{11}(𝔸), S_{222}(𝔸), we obtain the theorem. QED

As we have said, Padé approximants, formal orthogonal polynomials, continued fraction expansion of formal series are all related to Euclidean division.

As a last example, we illustrate how to write a continued fraction expansion of a series, which one can find in the work of Wronski, Chebyschef or Stieltjes.

Proposition 4 Given an alphabet 𝔸, then \frac{1}{z} σ_{1/z}(𝔸) =

\frac{1}{z + S_{1}(0 − 𝔸^{0}) + \frac{S_{2}(0 − 𝔸^{0})}{z + S_{1}(𝔸^{0} − 𝔸^{1}) + \frac{S_{2}(𝔸^{0} − 𝔸^{1})}{z + S_{1}(𝔸^{1} − 𝔸^{2}) + \frac{S_{2}(𝔸^{1} − 𝔸^{2})}{...}}}}

Proof. The validity of such expansion amounts to the recursions (3)

σ_{1/z}(𝔸^{k}) = (1 + \frac{1}{z}S_{1}(𝔸^{k−1} − 𝔸^{k}))σ_{1/z}(𝔸^{k}) + S_{2}(𝔸^{k−1} − 𝔸^{k}) \frac{1}{z^{2}} σ_{1/z}(𝔸^{k+1})

with which we began this text.

References

[Bl] B. Leclerc, On identities satisfied by minor of a matrix, Advances in Maths. 100 (1993)101–132.

[LP] A. Lascoux, P. Pragacz, Sylvester’s formulas for Euclidean division and Schur multifunctions, Max Planck Institut Preprints (2001).

C.N.R.S., Institut Gaspard Monge Université de Marne-la-Vallée,

5 Bd Descartes, Champs sur Marne, 77454 Marne La Vallée Cedex 2 FRANCE

                                    Alain.Lascoux@univ-mlv.fr
                           http://phalanstere.univ-mlv.fr/∼al