GENERATING FUNCTIONS FOR MOMENTS OF THE QUASI–NILPOTENT DT–OPERATOR

KEN DYKEMA AND CATHERINE YAN

Abstract. We prove a recursion formula for generating functions of certain renormalizations of ∗–moments of the DT(δ_{0}, 1)–operator T, involving an operation ⊙ on formal power series and a transformation ℰ that converts ⊙ to usual multiplication. This recursion formula is used to prove that all of these generating functions are rational functions, and to find a few of them explicitly.

Introduction

In combinatorics, one of the most useful methods for studying a sequence is to give its generating functions. The two most common types of generating functions are ordinary generating functions ∑ f(n)x^{n}, and exponential generating functions ∑ f(n)x^{n}/n!. In this paper we study the ∗–moment generating functions — a family of multivariable power series F_{n}— of a particular operator T that arose in the theory of free probability. We prove that F_{n}’s are all rational by applying a linear transformation between these two types of generating functions.

The central object of this paper is the collection of ∗–moments of a particular bounded operator T on Hilbert space, which was constructed in [1] and which is a candidate for an operator without a nontrivial hyperinvariant subspace. (A hyperinvariant subspace of an operator T on a Hilbert space ℋ is a closed subspace ℋ_{0} ⊆ ℋ that is invariant under every operator S that commutes with T, i.e. S(ℋ_{0}) ⊆ ℋ_{0}. It is an open problem whether every operator on Hilbert space that is not a multiple of the identity has a nontrivial hyperinvariant subspace.) The von Neumann algebra generated by T has a unique normalized trace τ, and by the ∗–moments of T we mean the values

M(k_{1}, ℓ_{1}, …, k_{n}, ℓ_{n}) = τ((T^{∗})^{k_{1}}T^{ℓ_{1}} … (T^{∗})^{k_{n}}T^{ℓ_{n}}),

with n ∈ N, k_{1}, …, k_{n}, ℓ_{1}, …, ℓ_{n} ∈ N ∪ {0}. These ∗–moments determine a representation of T on a Hilbert space, (which can be shown to be bounded, see [1]), via the construction of Gelfand, Naimark and Segal, (cf. [2]) and hence they encode all essential properties of the operator. Our effort to understand the ∗–moments is part of an attempt better to understand the operator T.

Date: Jan. 10, 2002.

K.D. supported in part by NSF grant DMS–0070558. He thanks also the Mathematical Sciences Research Institute, where part of this work was carried out. Research at MSRI is supported in part by NSF grant DMS–9701755. C.Y. supported in part by NSF grant DMS-0070574. She is also partly supported by NSF grant DMS-9729992 through the Institute for Advanced Study.

To be precise, T is a DT(δ_{0}, 1)–operator, in the class of DT–operators constructed in [1]. T can be realized as the limit in ∗-moments of strictly upper triangular random matrices having i.i.d. complex gaussian entries. In [1], it was proved that T is quasi–nilpotent, i.e. has spectrum {0}. It was shown that M(k_{1}, ℓ_{1}, …, k_{n}, ℓ_{n}) = 0 if k_{1} + ⋯ + k_{n} ≠ ℓ_{1} + ⋯ + ℓ_{n}, and also a recursion formula for M was proved (cf. [1, Theorem 8.5]). It was conjectured in [1] that

τ(((T^{∗})^{k}T^{k})^{n}) = \frac{n^{nk}}{(nk + 1)!}  (k, n ∈ N).  (1)

In this paper, we will consider the quantities

N(k_{1}, ℓ_{1}, …, k_{n}, ℓ_{n}) = \begin{cases}0 & if \begin{matrix}k_{1} + ⋯ + k_{n} \\ ≠ ℓ_{1} + ⋯ + ℓ_{n} \end{matrix} \\ (m + 1)! M(k_{1}, ℓ_{1}, …, k_{n}, ℓ_{n}) & if \begin{matrix}m := & k_{1} + ⋯ + k_{n} \\ = & ℓ_{1} + ⋯ + ℓ_{n} \end{matrix} \end{cases}

(which must be nonnegative integers by [1, Lemma 2.1]), and their generating functions

F_{n}(z_{1}, w_{1}, …, z_{n}, w_{n}) = ∑_{\begin{matrix}k_{1},...,k_{n}≥0 \\ ℓ_{1},...,ℓ_{n}≥0 \end{matrix}} N(k_{1}, ℓ_{1}, …, k_{n}, ℓ_{n})z_{1}^{k_{1}}w_{1}^{ℓ_{1}} … z_{n}^{k_{n}}w_{n}^{ℓ_{n}}.  (2)

We will prove a recursion formula for the functions F_{n}, involving an operation ⊙ on power series, which can be decsribed as multiplication with reweighting of homogeneous parts. In order to compute these generating functions, we will define a linear transformation of formal power series, called the ℰ–transform, which maps these power series to exponential generating functions of a new variable q, and converts the operation ⊙ to the usual multiplication. This will allow us to prove that F_{n} is a rational function for all n, and to give an algorithm for computing F_{n}.

In §1, the operation ⊙ and the transform ℰ on formal power series are introduced, and several properties are proved. In §2, recursion formulas for N(k_{1}, ℓ_{1}, …, k_{n}, ℓ_{n}) and for the generating functions F_{n} are proved. The latter are shown to be rational and several examples are computed. In §3, we use the method of contour integration to verify the conjecture (1) in the case n = 2.

1. Formal power series

In this section, fix variables x_{1}, x_{2}, … and fix N ∈ N. (In the next section we will use only the case N = 2.) Let Θ = Θ_{N} be the algebra of all formal power series in x_{1}, x_{2}, … of the form

f(x_{1}, …, x_{n}) = ∑_{k_{1},...,k_{n}≥0} c_{k_{1},...,k_{n}}x_{1}^{k_{1}} … x_{n}^{k_{n}}  (3)

for some n ∈ N, and such that c_{k_{1},...,k_{n}} = 0 whenever k_{1} + ⋯ + k_{n} is not divisible by N. Given a formal power series as in (3) and given k ∈ N ∪ {0}, write f^{(k)} for its Nk–homogeneous part:

f^{(k)}(x_{1}, …, x_{n}) = ∑_{k_{1}+···+k_{n}=Nk} c_{k_{1},...,k_{n}}x_{1}^{k_{1}} … x_{n}^{k_{n}}.  (4)

Definition 1.1. Let ⊙ = ⊙_{N} be the binary operation on Θ given by

f ⊙ g = ∑_{k,ℓ≥0} \binom{k + ℓ}{k}f^{(k)}g^{(ℓ)}.

Note that ⊙ is bilinear, commutative and associative. If f_{i} ∈ Θ (i ∈ {1, …, p}) then

(f_{1} ⊙ f_{2} ⊙ ⋯ ⊙ f_{p}) = ∑_{k_{1},...,k_{p}≥0} \binom{k_{1} + ⋯ + k_{p}}{k_{1}, k_{2}, …, k_{p}} ∏_{i=1}^{p} f_{i}^{(k_{i})}.

Let Θ̃ be the set of all formal power series in variables x_{1}, x_{2}, … and the additional variable q.

Definition 1.2. The ℰ–transform ℰ = ℰ_{N} : Θ → Θ̃ is the map given by

ℰf = ∑_{k≥0} \frac{q^{k}}{k!} f^{(k)}.

Note that ℰ is linear.

Proposition 1.3. (i) If f, g ∈ Θ then ℰ(f ⊙ g) = (ℰf)(ℰg).

(ii) If f ∈ Θ and if a ∈ Θ is a homogeneous polynomial of degree N then

\frac{d}{dq}(ℰ(af)) = a ℰf, i.e.

ℰ(af)(x_{1}, x_{2}, …, q) = a ∫_{0}^{q} ℰf(x_{1}, x_{2} …, t)dt.

(iii) If r ∈ N and if u_{1}, …, u_{r} ∈ Θ are distinct homogeneous polynomials of degree N, then

ℰ(\frac{1}{(1 − u_{1})(1 − u_{2}) … (1 − u_{r})}) = ∑_{i=1}^{r} \frac{u_{i}^{r−1}e^{qu_{i}}}{∏_{j≠i}(u_{i} − u_{j})},

which in the case r = 1 means ℰ(1/(1 − u_{1})) = e^{qu_{1}}. (iv) Let r ∈ N and k ∈ {0, 1, …, r − 1}. If u_{1}, …, u_{r}, a_{1}, …, a_{k} ∈ Θ are homoge-

neous polynomials of degree N with u_{1}, …, u_{r} distinct, then

ℰ(\frac{a_{1} … a_{k}}{(1 − u_{1})(1 − u_{2}) … (1 − u_{r})}) = a_{1} … a_{k} ∑_{i=1}^{r} \frac{u_{i}^{r−k−1}e^{qu_{i}}}{∏_{j≠i}(u_{i} − u_{j}).}

Proof. For (i),

ℰ(f ⊙ g) = ∑_{r≥0} \frac{q^{r}}{r!} ∑_{k+ℓ=r} \binom{r}{k}f^{(k)}g^{(ℓ)} = ∑_{k,ℓ≥0} (\frac{q^{k}}{k!} f^{(k)})(\frac{q^{ℓ}}{ℓ!} g^{(ℓ)}) = (ℰf)(ℰg).

4 T-moments

For (ii),

\frac{d}{dq} ℰ(af) = \frac{d}{dq} ℰ( ∑_{k=0}^{∞} af^{(k)}) = a\frac{d}{dq} ∑_{k=0}^{∞} \frac{q^{k+1}}{(k + 1)!}f^{(k)} = a ∑_{k=0}^{∞} \frac{q^{k}}{k!} f^{(k)} = a ℰf.

For (iii),

ℰ(\frac{1}{(1 − u_{1}) … (1 − u_{r})}) = ∑_{k=0}^{∞} \frac{q^{k}}{k!} ∑_{k_{1}+···+k_{p}=k} u_{1}^{k_{1}} … u_{r}^{k_{r}}.

On the other hand,

∑_{i=1}^{r} \frac{u_{i}^{r−1}e^{qu_{i}}}{∏_{j≠i}(u_{i} − u_{j})} = ∑_{k=0}^{∞} \frac{q^{k}}{k!} ∑_{i=1}^{r} \frac{u_{i}^{k+r−1}}{∏_{j≠i}(u_{i} − u_{j}).}

Hence it will suffice to show

∑_{k_{1}+···+k_{r}=k} u_{1}^{k_{1}} … u_{r}^{k_{r}} = ∑_{i=1}^{r} \frac{u_{i}^{k+r−1}}{∏_{j≠i}(u_{i} − u_{j})}  (5)

for all k ≥ 0. We will proceed by induction on r, treating u_{1}, …, u_{r} in (5) as any n distinct elements in a field. The case r = 1 is clear. If r = 2, then we have

∑_{k_{1}+k_{2}=k} u_{1}^{k_{1}}u_{2}^{k_{2}} = u_{2}^{k} ∑_{k_{1}=0}^{k} (\frac{u_{1}}{u_{2}} )^{k_{1}} = u_{2}^{k} \frac{(u_{1}/u_{2})^{k+1} − 1}{(u_{1}/u_{2}) − 1}

= \frac{u_{1}^{k+1} − u_{2}^{k+1}}{u_{1} − u_{2}} = \frac{u_{1}^{k+1}}{u_{1} − u_{2}} + \frac{u_{2}^{k+1}}{u_{2} − u_{1}} ,

as required. Assume r ≥ 3. Using the induction hypothesis, we have

∑_{k_{1}+···+k_{r}=k} u_{1}^{k_{1}} … u_{r}^{k_{r}} = u_{1}^{k} ∑_{k′=0}^{k} ∑_{k_{2}+···+k_{r}=k′} (\frac{u_{2}}{u_{1}} )^{k_{2}}(\frac{u_{3}}{u_{1}} )^{k_{3}} … (\frac{u_{r}}{u_{1}} )^{k_{r}}

= u_{1}^{k} ∑_{k′=0}^{k} ∑_{i=2}^{r} \frac{(u_{i}/u_{1})^{k′+r−2}}{∏_{j≠1,i} ((u_{i}/u_{1}) − (u_{j}/u_{1}))}

= u_{1}^{k} ∑_{i=2}^{r} \frac{((u_{i}/u_{1})^{k+1} − 1)(u_{i}/u_{1})^{r−2}}{((u_{i}/u_{1}) − 1) ∏_{j≠1,i} ((u_{i}/u_{1}) − (u_{j}/u_{1}))}

= ∑_{i=2}^{r} \frac{u_{i}^{k+r−1}}{∏_{j≠i}(u_{i} − u_{j})} − u_{1}^{k+1} ∑_{i=2}^{r} \frac{u_{i}^{r−2}}{∏_{j≠i}(u_{i} − u_{j}).}

We must show

− ∑_{i=2}^{r} \frac{u_{i}^{r−2}}{∏_{j≠i}(u_{i} − u_{j})} = \frac{u_{1}^{r−2}}{∏_{j≠1}(u_{1} − u_{j})}

or, equivalently,

∑_{i=1}^{r} u_{i}^{r−2} ∏_{\begin{matrix}1≤k<ℓ≤r \\ k,ℓ≠i \end{matrix}}(u_{ℓ} − u_{k}) = 0.  (6)

Using the Vandermonde determinant

det (\begin{matrix}1 & 1 & … & 1 \\ u_{1} & u_{2} & … & u_{r−1} \\ ⋮ & ⋮ & ⋮ & ⋮ \\ u_{1}^{r−2} & u_{2}^{r−2} & … & u_{r−1}^{r−2} \end{matrix}) = ∏_{1≤k<ℓ≤r−1}(u_{ℓ} − u_{k})

and expanding the determinant

det (\begin{matrix}u_{1}^{r−2} & u_{2}^{r−2} & … & u_{r}^{r−2} \\ 1 & 1 & … & 1 \\ u_{1} & u_{2} & … & u_{r} \\ ⋮ & ⋮ & ⋮ & ⋮ \\ u_{1}^{r−2} & u_{2}^{r−2} & … & u_{r}^{r−2} \end{matrix})

along the first row, we find the latter is equal to the LHS of (6), but is also clearly equal to zero.

To prove (iv), use (iii) and iterate (ii). □

Remark 1.4. If one knows that h ∈ Θ̃ is in the image of ℰ, then to find ℰ^{−1}(h), simply replace every occurrence of q^{k} by k!. This can sometimes be used to compute f ⊙ g.

We now turn to some cases of f ⊙ g which will be used in §2.

Proposition 1.5. If m ∈ N, k ∈ {0, 1, …, m−1} and if a_{1}, …, a_{k}, u_{1}, …, u_{m}, v ∈ Θ are homogeneous polynomials of degree N such that u_{1}, …, u_{m} are distinct, then

\frac{a_{1} … a_{k}}{(1 − u_{1}) … (1 − u_{m})} ⊙ \frac{1}{(1 − v)} = \frac{a_{1} … a_{k}(1 − v)^{m−k−1}}{(1 − u_{1} − v)(1 − u_{2} − v) … (1 − u_{m} − v)}.

Proof. Using Proposition 1.3, we have

\frac{a_{1} … a_{k}}{(1 − u_{1}) … (1 − u_{m})} ⊙ \frac{1}{(1 − v)}

↦^{ℰ} (a_{1} … a_{k} ∑_{i=1}^{m} \frac{u_{i}^{m−k−1}e^{qu_{i}}}{∏_{j≠i}(u_{i} − u_{j})})e^{qv}

= a_{1} … a_{k} ∑_{i=1}^{m} \frac{((u_{i} + v) − v)^{m−k−1}e^{q(u_{i}+v)}}{∏_{j≠i}(u_{i} − u_{j})}

= a_{1} … a_{k} ∑_{p=0}^{m−k−1}(−1)^{p}\binom{m − k − 1}{p}v^{p} ∑_{i=1}^{m} \frac{(u_{i} + v)^{m−k−p−1}e^{q(u_{i}+v)}}{∏_{j≠i} ((u_{i} + v) − (u_{j} + v))}

↦^{ℰ^{−1}} ∑_{p=0}^{m−k−1}(−1)^{p}\binom{m − k − 1}{p}v^{p} \frac{a_{1} … a_{k}}{∏_{i}(1 − u_{i} − v)}

= \frac{a_{1} … a_{k}(1 − v)^{m−k−1}}{(1 − u_{1} − v)(1 − u_{2} − v) … (1 − u_{m} − v)}.

□

Proposition 1.6. Let m, n ∈ N and let k ∈ {0, 1, …, m − 1}, ℓ ∈ {0, 1, …, n − 1}. Let u_{1}, …, u_{m}, a_{1}, …, a_{k}, v_{1} …, v_{n}, b_{1}, …, b_{ℓ} ∈ Θ be homogeneous polynomials of degree N such that (u_{i} + v_{j})_{1≤i≤m, 1≤j≤n} is a family of mn distinct polynomials. Then

\frac{a_{1} … a_{k}}{(1 − u_{1}) … (1 − u_{m})} ⊙ \frac{b_{1} … b_{ℓ}}{(1 − v_{1}) … (1 − v_{n})} = \frac{a_{1} … a_{k} b_{1} … b_{ℓ} P_{m,n}^{k,ℓ}}{∏_{i,j}(1 − u_{i} − v_{j}),}

where P_{m,n}^{k,ℓ} = P_{m,n}^{k,ℓ}(u_{1}, …, u_{m}, v_{1}, …, v_{n}) is a polynomial in the variables (u_{i})_{1}^{m} and (v_{j})_{1}^{n}, having degree bounded above by mn − k − ℓ − 1.

Proof. Using Proposition 1.3 and Remark 1.4, we have

\frac{a_{1} … a_{k}}{∏_{i}(1 − u_{1})} ⊙ \frac{b_{1} … b_{ℓ}}{∏_{j}(1 − v_{j})}

↦^{ℰ} a_{1} … a_{k} b_{1} … b_{ℓ}( ∑_{i=1}^{m} \frac{u_{i}^{m−k−1}e^{qu_{i}}}{∏_{i′≠i}(u_{i} − u_{i′})})( ∑_{j=1}^{n} \frac{v_{j}^{n−ℓ−1}e^{qv_{i}}}{∏_{j′≠j}(v_{j} − v_{j′})})

↦^{ℰ^{−1}} a_{1} … a_{k} b_{1} … b_{ℓ} ∑_{i,j} \frac{u_{i}^{m−k−1}v_{j}^{n−ℓ−1}}{(1 − u_{i} − v_{j})( ∏_{i′≠i}(u_{i} − u_{i′}))( ∏_{j′≠j}(v_{j} − v_{j′}))}

= (\frac{a_{1} … a_{k} b_{1} … b_{ℓ}}{∏_{i,j}(1 − u_{i} − v_{j})}) \frac{Q}{( ∏_{p<q}(u_{p} − u_{q}))( ∏_{r<s}(v_{r} − v_{s}))},

where Q = Q(u_{1}, …, u_{m}, v_{1}, …, v_{n}) is the polynomial

\begin{aligned}Q = ∑_{i,j}(−1)^{i+j}u_{i}^{m−k−1}v_{j}^{n−ℓ−1} & ( ∏_{(i′,j′)≠(i,j)}(1 − u_{i′} − v_{j′}))· \\ & · ( ∏_{p<q, p≠i, q≠i}(u_{p} − u_{q}))( ∏_{r<s, r≠j, s≠j}(v_{r} − v_{s})). \end{aligned}

(7)

We will show that for every 1 ≤ a < b ≤ m, u_{a} − u_{b} divides Q. It will suffice to show that substituting u_{a} = x and u_{b} = x into the RHS of equation (7) yields zero. This substitution makes all terms of the sum zero except possibly when i ∈ {a, b}. These remaining terms are equal to the common factor

x^{m−k−1}( ∏_{\begin{matrix}i′∉\{a,b\} \\ 1≤j′≤n \end{matrix}}(1 − u_{i′} − v_{j′}))( ∏_{\begin{matrix}p<q \\ p,q∉\{a,b\} \end{matrix}}(u_{p} − u_{q}))·

· ∑_{j=1}^{n}(−1)^{j}v_{j}^{n−ℓ−1}(1 − x − v_{j}) ∏_{j′≠j}(1 − x − v_{j′})^{2}

times the quantity

(−1)^{a} ∏_{\begin{matrix}p<q \\ b∈\{p,q\} \\ a∉\{p,q\} \end{matrix}}(u_{p} − u_{q}) + (−1)^{b} ∏_{\begin{matrix}p<q \\ a∈\{p,q\} \\ b∉\{p,q\} \end{matrix}}(u_{p} − u_{q}).  (8)

However, since a < b, we see that the quantity (8) is equal to

(−1)^{a}( ∏_{p<b, p≠a}(u_{p} − x))( ∏_{q>b}(x − u_{q})) + (−1)^{b}( ∏_{p<a}(u_{p} − x))( ∏_{q>a, q≠b}(x − u_{q}))

=((−1)^{a}(−1)^{b−a−1} + (−1)^{b})( ∏_{p<a}(u_{p} − x))( ∏_{q>a, q≠b}(x − u_{q})) = 0.

By symmetry, we have that v_{c}−v_{d} divides the polynomial Q, for every 1 ≤ c < d ≤ n. Thus

P_{m,n}^{k,ℓ} = \frac{Q}{( ∏_{p<q}(u_{p} − u_{q}))( ∏_{r<s}(v_{r} − v_{s}))}  (9)

is a polynomial in u_{1}, …, u_{m}, v_{1}, …, v_{n}. The upper bound on its degree is easily computed from (9) and the expression (7). □

Proposition 1.5 shows P_{m,1}^{k,0} = (1 − v_{1})^{m−k−1} and P_{1,n}^{0,ℓ} = (1 − u_{1})^{n−ℓ−1}. Here are some other examples.

Examples 1.7.

P_{2,2}^{0,1} = 1 − u_{1}u_{2} − v_{1} − v_{2} + v_{1}v_{2}
P_{2,2}^{1,1} = 2 − u_{1} − u_{2} − v_{1} − v_{2}
P_{2,3}^{1,1} = \begin{matrix}2 − 3u_{1} + u_{1}^{2} − 3u_{2} + 4u_{1}u_{2} − u_{1}^{2}u_{2} + u_{2}^{2} − u_{1}u_{2}^{2} − v_{1} + u_{1}v_{1} + u_{2}v_{1} − u_{1}u_{2}v_{1} \\ − v_{2} + u_{1}v_{2} + u_{2}v_{2} − u_{1}u_{2}v_{2} − v_{3} + u_{1}v_{3} + u_{2}v_{3} − u_{1}u_{2}v_{3} + v_{1}v_{2}v_{3} \end{matrix}

P_{2,3}^{1,2} = \begin{matrix}3 − 3u_{1} + u_{1}^{2} − 3u_{2} + u_{1}u_{2} + u_{2}^{2} − 2v_{1} + u_{1}v_{1} + u_{2}v_{1} − 2v_{2} + u_{1}v_{2} + u_{2}v_{2} \\ + v_{1}v_{2} − 2v_{3}u_{1}v_{3} + u_{2}v_{3} + v_{1}v_{3} + v_{2}v_{3}. \end{matrix}

2. Generating functions

We will now define N(k_{1}, ℓ_{1}, …, k_{n}, ℓ_{n}), but extending the definition found in the introduction to allow k_{j}, ℓ_{j} to be −1. The reason for this extension is to allow a recursion formula to be proved for N(k_{1}, ℓ_{1}, …, k_{n}, ℓ_{n}) that holds for all k_{j}, ℓ_{j} ∈ N ∪ {0}.

Definition 2.1. Let n ∈ N and k_{1}, …, k_{n}, ℓ_{1}, …, ℓ_{n} ∈ N∪{0, −1}. If k_{1} +⋯+k_{n} ≠ ℓ_{1}+⋯+ℓ_{n} then set N(k_{1}, ℓ_{1}, …, k_{n}, ℓ_{n}) = 0. Suppose now k_{1}+⋯+k_{n} = ℓ_{1}+…, ℓ_{n}, and let m = k_{1} + ⋯ + k_{n}. If k_{j} ≥ 0 and ℓ_{j} ≥ 0 for all j, then let

N(k_{1}, ℓ_{1}, …, k_{n}, ℓ_{n}) = (m + 1)!τ((T^{∗})^{k_{1}}T^{ℓ_{1}} … (T^{∗})^{k_{n}}T^{ℓ_{n}}).

In the case n = 1, k_{1} = ℓ_{1} = −1 let N(−1, −1) = 1. In all other cases, (where n ≥ 2 and k_{j} = −1 or ℓ_{j} = −1 for some j), let N(k_{1}, ℓ_{1}, …, k_{n}, ℓ_{n}) = 0.

The following is immediate from [1, Proposition 8.4].

Proposition 2.2. (i) N(k_{1}, ℓ_{1}, …, k_{n}, ℓ_{n}) = N(ℓ_{1}, k_{2}, ℓ_{2}, …, k_{n}, ℓ_{n}, k_{1}). (ii) N(k_{1}, ℓ_{1}, …, k_{n}, ℓ_{n}) = N(ℓ_{n}, k_{n}, ℓ_{n−1}, k_{n−1}, …, ℓ_{1}, k_{1}).

(iii) If n ≥ 2, if k_{1}, …, k_{n}, ℓ_{1}, …, ℓ_{n} ≥ 0 and if k_{1} = 0, then

N(k_{1}, ℓ_{1}, …, k_{n}, ℓ_{n}) = N(k_{2}, ℓ_{2}, …, k_{n−1}, ℓ_{n−1}, k_{n}, ℓ_{n} + ℓ_{1}).

Theorem 2.3. If n ∈ N and k_{1}, …, k_{n}, ℓ_{1}, …, ℓ_{n} ∈ N ∪ {0}, then

N(k_{1}, ℓ_{1}, …, k_{n}, ℓ_{n}) = ∑_{r=1}^{n} ∑_{1≤j(1)<···<j(r)≤n} Nom(ℓ_{1}, …, ℓ_{n}; j(1), …, j(r))  (10)

(N(k_{1}, ℓ_{1}, …, k_{j(1)−1}, ℓ_{j(1)−1}, k_{j(1)} − 1, ℓ_{j(r)} − 1, k_{j(r)+1}ℓ_{j(r)+1}, …, k_{n}, ℓ_{n})

∏_{i=1}^{r−1} N(ℓ_{j(i)} − 1, k_{j(i)+1}, ℓ_{j(i)+1}, …, k_{j(i+1)−1}, ℓ_{j(i+1)−1}, k_{j(i+1)} − 1)),

where Nom(ℓ_{1}, …, ℓ_{n}; j(1), …, j(r)) is the multinomial coefficient

\binom{ℓ_{1} + ⋯ + ℓ_{n}}{ℓ_{1} + ⋯ + ℓ_{j(1)−1} + ℓ_{j(r)} + ⋯ + ℓ_{n}, ℓ_{j(1)} + ⋯ + ℓ_{j(2)−1}, …, ℓ_{j(r−1)} + ⋯ + ℓ_{j(r)−1}}.

Proof. If k_{1}+⋯+k_{n} ≠ ℓ_{1}+⋯+ℓ_{n} then both sides of the equality are zero. So suppose k_{1} + ⋯ + k_{n} = ℓ_{1} + ⋯ + ℓ_{n}. If k_{1}, …, k_{n}, ℓ_{1}, …, ℓ_{n} ≥ 1 then the equality follows directly from the recursion formula [1, Theorem 8.5]. To verify the equality (10) in general, we will proceed by induction on n. Let N′(k_{1}, ℓ_{1}, …, k_{n}, ℓ_{n}) be the RHS of (10). In the case n = 1, we need only check that N′(0, 0) = 1, which is clear. Assume n ≥ 2. Writing k_{1}, ℓ_{1}, …, k_{n}, ℓ_{n} on a circle, it is clear that both sides of equation (10) are invariant under the following permutation of order n:

k_{1} → k_{2} → ⋯ → k_{n} → k_{1}, ℓ_{1} → ℓ_{2} → ⋯ → ℓ_{n} → ℓ_{1}.

Therefore, it will be enough to consider three cases: (i) k_{1} = 0, ℓ_{n} ≠ 0, (ii) k_{1} ≠ 0, ℓ_{n} = 0, (iii) k_{1} = ℓ_{n} = 0.

In case (i), all terms in the sum (10) having j(1) = 1 are zero. If j(1) ≥ 2, then using Proposition 2.2 we have

N(k_{1}, ℓ_{1}, …, k_{j(1)−1}, ℓ_{j(1)−1}, k_{j(1)} − 1, ℓ_{j(r)} − 1, k_{j(r)+1}ℓ_{j(r)+1}, …, k_{n}, ℓ_{n}) =

= \begin{cases}\begin{matrix}N( & k_{2}, ℓ_{2}, …, k_{j(1)−1}, ℓ_{j(1)−1}, k_{j(1)} − 1, \\ ℓ_{j(r)} − 1, k_{j(r)+1}ℓ_{j(r)+1}, …, k_{n−1}, ℓ_{n−1}, k_{n}, ℓ_{n} + ℓ_{1}) \end{matrix} & if j(r) < n \\ N(k_{2}, ℓ_{2}, …, k_{j(1)−1}, ℓ_{j(1)−1}, k_{j(1)} − 1, ℓ_{j(r)} + ℓ_{1} − 1) & if j(r) = n. \end{cases}

Therefore,

N′(k_{1}, ℓ_{1}, …, k_{n}, ℓ_{n}) = N′(k_{2}, ℓ_{2}, …, k_{n−1}, ℓ_{n−1}, k_{n}, ℓ_{n} + ℓ_{1})
= N(k_{2}, ℓ_{2}, …, k_{n−1}, ℓ_{n−1}, k_{n}, ℓ_{n} + ℓ_{1})
= N(0, ℓ_{1}, k_{2}, ℓ_{2}, …, k_{n}, ℓ_{n}),

where the second equality is by the induction hypothesis and the third is from Proposition 2.2.

In case (ii), (k_{1} ≠ 0, ℓ_{n} = 0), one similarly shows

N′(k_{1}, ℓ_{1}, …, k_{n}, ℓ_{n}) = N′(k_{1} + k_{n}, ℓ_{1}, k_{2}, ℓ_{2}, …, k_{n−1}, ℓ_{n−1})
= N(k_{1} + k_{n}, ℓ_{1}, k_{2}, ℓ_{2}, …, k_{n−1}, ℓ_{n−1})
= N(k_{1}, ℓ_{1}, , … k_{n−1}, …, k_{n−1}, ℓ_{n−1}, k_{n}, 0).

In case (iii), (k_{1} = ℓ_{n} = 0), given 1 ≤ r ≤ n and 1 < j(1) < ⋯ < j(r) ≤ n, let t(j(1), …, j(r)) be the corresponding term in the sum (10) for N′(k_{1}, ℓ_{1}, …, k_{n}, ℓ_{n}), and, if j(r) < n, let t̃(j(1), …, j(r)) be the term corresponding to 1 ≤ j(1) < ⋯ < j(r) ≤ n − 1 in the analogous sum for N′(k_{2}, ℓ_{2}, …, k_{n−1}, ℓ_{n−1}, k_{n}, ℓ_{1}). If j(1) > 1 and j(r) < n, then using Proposition 2.2 we have

N(k_{1}, ℓ_{1}, …, k_{j(1)−1}, ℓ_{j(1)−1}, k_{j(1)} − 1, ℓ_{j(r)} − 1, k_{j(r)+1}, ℓ_{j(r)+1}, …, k_{n}, ℓ_{n}) =

N(k_{2}, ℓ_{2}, …, k_{j(1)−1}, ℓ_{j(1)−1}, k_{j(r)} − 1, ℓ_{j(r)} − 1, k_{j(r)+1}, ℓ_{j(r)+1}, …, k_{n−1}, ℓ_{n−1}, k_{n}, ℓ_{1}),

which, taken in the sum (10), shows

t(j(1), …, j(r)) = t̃(j(1) − 1, …, j(r) − 1).

If j(1) = 1 and j(r) < n or if j(1) > 1 and j(r) = n, then t(j(1), …, j(r)) = 0. If j(1) = 1 and j(r) = n then since

N(−1, −1)N(ℓ_{1} − 1,k_{2}, ℓ_{2}, …, k_{j(2)−1}, ℓ_{j(2)−1}, k_{j(2)} − 1)
= N(k_{2}, ℓ_{2}, …, k_{j(2)−1}, ℓ_{j(2)−1}, k_{j(2)} − 1, ℓ_{1} − 1),

we have

t(j(1), …, j(r) = t̃(j(2) − 1, …, j(r − 1) − 1, n − 1).

Therefore,

N′(k_{1}, ℓ_{1}, …, k_{n}, ℓ_{n}) = N′(k_{2}, ℓ_{2}, …, k_{n−1}, ℓ_{n−1}, k_{n}, ℓ_{1})
= N(k_{2}, ℓ_{2}, …, k_{n−1}, ℓ_{n−1}, k_{n}, ℓ_{1})
= N(0, ℓ_{1}, k_{2}, ℓ_{2}, …, k_{n−1}, ℓ_{n−1}, k_{n}, 0).
□

Recall the generating functions F_{n}, defined in the introduction. The following theorem is a recursion formula for these, using the operation ⊙ = ⊙_{2} introduced in §1. Here, as promised, we are using objects defined in §1 with N = 2.

Theorem 2.4. If n ∈ N and n ≥ 2, then

(1 − z_{1}w_{1} − ⋯ − z_{n}w_{n})F_{n}(z_{1}, w_{1}, …, z_{n}, w_{n}) = ∑_{r=2}^{n} ∑_{1≤j(1)<···<j(r)≤n}  (11)

(z_{j(1)}w_{j(r)}F̃_{j(1)+n−j(r)}(z_{1}, w_{1}, …, z_{j(1)−1}, w_{j(1)−1}, z_{j(1)}, w_{j(r)}, z_{j(r)+1}, w_{j(r)+1}, …, z_{n}, w_{n}))

⊙ ( ⊙_{i=1}^{r−1} (w_{j(i)}z_{j(i+1)}F̃_{j(i+1)−j(i)}(w_{j(i)}, z_{j(i)+1}, w_{j(i)+1}, …, z_{j(i+1)−1}, w_{j(i+1)−1}, z_{j(i+1)}))),

where

F̃_{j}(x_{1}, y_{1}, …, x_{j}, y_{j}) := \begin{cases}F_{j}(x_{1}, y_{1}, …, x_{j}, y_{j}) & if j ≥ 2 \\ F_{1}(x_{1}, y_{1})/(x_{1}y_{1}) & if j = 1. \end{cases}

Proof. We will use the recursion formula (10) for the coefficients in the definition (2) of F_{n}. Since n ≥ 2, if j ∈ {1, …, n}, then

∑_{\begin{matrix}k_{1},...,k_{n}≥0 \\ ℓ_{1},...,ℓ_{n}≥0 \end{matrix}} N(k_{1}, ℓ_{1}, …, k_{j−1}, ℓ_{j−1}, k_{j} − 1, ℓ_{j} − 1, k_{j+1}, ℓ_{j+1}, …, k_{n}, ℓ_{n})z_{1}^{k_{1}}w_{1}^{ℓ_{1}} … z_{n}^{k_{n}}w_{n}^{ℓ_{n}}

= z_{j}w_{j}F_{n}(z_{1}, w_{1}, …, z_{n}, w_{n}).

This takes care of all terms in the sum (10) with r = 1, and these together with the LHS of equation (10) give the LHS of formula (11). Let 2 ≤ r ≤ n and 1 ≤ j(1) < ⋯ < j(r) ≤ n. Then

∑_{\begin{matrix}k_{1},...,k_{n}≥0 \\ ℓ_{1},...,ℓ_{n}≥0 \end{matrix}} Nom(ℓ_{1}, …, ℓ_{n}; j(1), …, j(r))·

· (N(k_{1}, ℓ_{1}, …, k_{j(1)−1}, ℓ_{j(1)−1}, k_{j(1)} − 1, ℓ_{j(r)} − 1, k_{j(r)+1}ℓ_{j(r)+1}, …, k_{n}, ℓ_{n})

∏_{i=1}^{r−1} N(ℓ_{j(i)} − 1, k_{j(i)+1}, ℓ_{j(i)+1}, …, k_{j(i+1)−1}, ℓ_{j(i+1)−1}, k_{j(i+1)} − 1))·

· z_{1}^{k_{1}}w_{1}^{ℓ_{1}} … z_{n}^{k_{n}}w_{n}^{ℓ_{n}}

= ∑_{p_{1},...,p_{r}≥0} \binom{p_{1} + ⋯ + p_{r}}{p_{1}, …, p_{r}}

( ( ∑_{\begin{matrix}k_{1}+ℓ_{1}+···+k_{j(1)−1}+ℓ_{j(1)−1}+k_{j(1)}+ \\ +ℓ_{j(r)}+k_{j(r)+1}+ℓ_{j(r)+1}+···+k_{n}+ℓ_{n}=p_{1} \end{matrix}}  (12)

N(k_{1}, ℓ_{1}, …, k_{j(1)−1}, ℓ_{j(1)−1}, k_{j(1)} − 1, ℓ_{j(r)} − 1, k_{j(r)+1}, ℓ_{j(r)+1}, …, k_{n}, ℓ_{n})·

· z_{1}^{k_{1}}w_{1}^{ℓ_{1}} … z_{j(1)−1}^{k_{j(1)−1}}w_{j(1)−1}^{ℓ_{j(1)−1}}z_{j(1)}^{k_{j(1)}}w_{j(r)}^{ℓ_{j(r)}}z_{j(r)+1}^{k_{j(r)+1}}w_{j(r)+1}^{ℓ_{j(r)+1}} … z_{n}^{k_{n}}w_{n}^{ℓ_{n}})·

· ( ∏_{i=1}^{r−1} ∑_{\begin{matrix}ℓ_{j(i)}+k_{j(i)+1}+ℓ_{k(i)+1}+···+ \\ +k_{j(i+1)−1}+ℓ_{j(i+1)−1}+k_{j(i+1)}=p_{i+1} \end{matrix}}  (13)

N(ℓ_{j(i)} − 1, k_{j(i)+1}, ℓ_{j(i)+1}, …, k_{j(i+1)−1}, ℓ_{j(i+1)−1}, k_{j(i+1)} − 1)·

· w_{j(i)}^{ℓ_{j(i)}}z_{j(i)+1}^{k_{j(i)+1}}w_{j(i)+1}^{ℓ_{j(i)+1}} … z_{j(i+1)−1}^{k_{j(i+1)−1}}w_{j(i+1)−1}^{ℓ_{j(i+1)−1}}z_{j(i+1)}^{k_{j(i+1)}}) )

Consider a summation beginning at (13) for some fixed i. If j(i + 1) − j(i) ≥ 2 then all terms with either ℓ_{j(i)} = 0 or k_{j(i+1)} = 0 vanish and the summation is equal to the 2p_{i+1}–homogeneous part of

w_{j(i)}z_{j(i+1)}F_{j(i+1)−j(i)}(w_{j(i)}, z_{j(i)+1}, w_{j(i)+1}, …, z_{j(i+1)−1}, w_{j(i+1)−1}, z_{j(i+1)}).

On the other hand, if j(i + 1) − j(i) = 1, then, since N(k − 1, ℓ − 1) = N(k, ℓ) for all k, ℓ ∈ N ∪ {0}, this summation is equal to the 2p_{i+1}–homogenous part of F_{1}(w_{j(i)}, z_{j(i+1)}). In either case, the summation is equal to the 2p_{i+1}–homogeneous part of

w_{j(i)}z_{j(i+1)}F̃_{j(i+1)−j(i)}(w_{j(i)}, z_{j(i)+1}, w_{j(i)+1}, …, z_{j(i+1)−1}, w_{j(i+1)−1}, z_{j(i+1)}).

Similarly, we see that the summation beginning at (12) is equal to the 2p_{1}– homogeneous part of

z_{j(1)}w_{j(r)}F̃_{j(1)+n−j(r)}(z_{1}, w_{1}, …, z_{j(1)−1}, w_{j(1)−1}, z_{j(1)}, w_{j(r)}, z_{j(r)+1}, w_{j(r)+1}, …, z_{n}, w_{n}).

Now using the definition of the operation ⊙ and summing over 2 ≤ r ≤ n and all 1 ≤ j(1) < ⋯ < j(r) ≤ n, one proves the recursion formula (10). □

It is straightforward from the recursion formula that

F_{1}(z_{1}, w_{1}) = \frac{1}{1 − z_{1}w_{1}} .

Starting from this, one can (in principle) compute F_{n} for arbitrary given n, using the recursion formula (10). Below are the results for n = 2, 3, 4. We were motivated to find and write down F_{4}, although it is rather long, because, while we were able to find F_{2} and F_{3} using ad hoc methods to work out, in the notation of Proposition 1.5,

\frac{1}{1 − u_{1}} ⊙ \frac{1}{1 − v}  and  \frac{a}{(1 − u_{1})(1 − u_{2})} ⊙ \frac{1}{1 − v} ,

we were unable to find, in the notation of Proposition 1.6,

\frac{a}{(1 − u_{1})(1 − u_{2})} ⊙ \frac{b}{(1 − v_{1})(1 − v_{2})},

which is needed in computing F_{4}, until we discovered the ℰ–transform.

Examples 2.5. (n = 2):

F_{2}(z_{1}, w_{1}, z_{2}, w_{2}) = \frac{1}{1 − u_{1}^{(2)}} (F_{1}(z_{1}, w_{2}) ⊙ F_{1}(w_{1}, z_{2}))

= \frac{1}{1 − u_{1}^{(2)}} (\frac{1}{1 − z_{1}w_{2}} ⊙ \frac{1}{1 − w_{1}z_{2}} )

= \frac{1}{(1 − u_{1}^{(2)})(1 − u_{2}^{(2)})}

where u_{1}^{(2)} = z_{1}w_{1} + z_{2}w_{2} and u_{2}^{(2)} = z_{1}w_{2} + z_{2}w_{1}.

12 T-moments

(n = 3):

F_{3}(z_{1}, w_{1}, z_{2}, w_{2}, z_{3}, w_{3}) =

= \frac{1}{1 − u_{1}^{(3)}} \begin{matrix}( & (z_{1}w_{2}F_{2}(z_{1}, w_{2}, z_{3}, w_{3})) ⊙ F_{1}(w_{1}, z_{2}) \\ + F_{1}(z_{1}, w_{3}) ⊙ (w_{1}z_{3}F_{2}(w_{1}, z_{2}, w_{2}, z_{3})) \\ + (z_{2}w_{3}F_{2}(z_{1}, w_{1}, z_{2}, w_{3})) ⊙ F_{1}(w_{2}, z_{3}) \\ + F_{1}(z_{1}, w_{3}) ⊙ F_{1}(w_{1}, z_{2}) ⊙ F_{1}(w_{2}, z_{3})) \end{matrix}

= \frac{1}{(1 − u_{1}^{(3)})(1 − u_{2}^{(3)})}(1 + \frac{z_{1}w_{2}}{1 − u_{3}^{(3)}} + \frac{z_{2}w_{3}}{1 − u_{4}^{(3)}} + \frac{z_{3}w_{1}}{1 − u_{5}^{(3)}} )

where

u_{1}^{(3)} = z_{1}w_{1} + z_{2}w_{2} + z_{3}w_{3} u_{2}^{(3)} = z_{1}w_{3} + z_{2}w_{1} + z_{3}w_{2}
u_{3}^{(3)} = z_{1}w_{2} + z_{2}w_{1} + z_{3}w_{3} u_{4}^{(3)} = z_{1}w_{1} + z_{2}w_{3} + z_{3}w_{2}
u_{5}^{(3)} = z_{1}w_{3} + z_{2}w_{2} + z_{3}w_{1}.

(n = 4):

F_{4}(z_{1},w_{1}, z_{2}, w_{2}, z_{3}, w_{3}, z_{4}, w_{4}) =

= \frac{1}{1 − u_{1}^{(4)}} ((z_{1}w_{2}F_{3}(z_{1}, w_{2}, z_{3}, w_{3}, z_{4}, w_{4})) ⊙ F_{1}(w_{1}, z_{2})

+ (z_{1}w_{3}F_{2}(z_{1}, w_{3}, z_{4}, w_{4})) ⊙ (w_{1}z_{3}F_{2}(w_{1}, z_{2}, w_{2}, z_{3})) + F_{1}(z_{1}, w_{4}) ⊙ (w_{1}z_{4}F_{3}(w_{1}, z_{2}, w_{2}, z_{3}, w_{3}, z_{4})) + (z_{2}w_{3}F_{3}(z_{1}, w_{1}, z_{2}, w_{3}, z_{4}, w_{4})) ⊙ F_{1}(w_{2}, z_{3}) + (z_{2}w_{4}F_{2}(z_{1}, w_{1}, z_{2}, w_{4})) ⊙ (w_{2}z_{4}F_{2}(w_{2}, z_{3}, w_{3}, z_{4})) + (z_{3}w_{4}F_{3}(z_{1}, w_{1}, z_{2}, w_{2}, z_{3}, w_{4})) ⊙ F_{1}(w_{3}, z_{4}) + (z_{1}w_{3}F_{2}(z_{1}, w_{3}, z_{4}, w_{4})) ⊙ F_{1}(w_{1}, z_{2}) ⊙ F_{1}(w_{2}, z_{3}) + F_{1}(z_{1}, w_{4}) ⊙ F_{1}(w_{1}, z_{2}) ⊙ (w_{2}z_{4}F_{2}(w_{2}, z_{3}, w_{3}, z_{4})) + F_{1}(z_{1}, w_{4}) ⊙ (w_{1}z_{3}F_{2}(w_{1}, z_{2}, w_{2}, z_{3})) ⊙ F_{1}(w_{3}, z_{4}) + (z_{2}w_{4}F_{2}(z_{1}, w_{1}, z_{2}, w_{4})) ⊙ F_{1}(w_{2}, z_{3}) ⊙ F_{1}(w_{3}, z_{4}) + F_{1}(z_{1}, w_{4})) ⊙ F_{1}(w_{1}, z_{2}) ⊙ F_{1}(w_{2}, z_{3}) ⊙ F_{1}(w_{3}, z_{4}))

= \frac{1}{(1 − u_{1}^{(4)})(1 − u_{2}^{(4)})}(1 + \frac{z_{3}w_{4}}{1 − u_{3}^{(4)}} + \frac{z_{2}w_{3}}{1 − u_{5}^{(4)}} + \frac{z_{4}w_{1}}{1 − u_{6}^{(4)}} + \frac{z_{1}w_{2}}{1 − u_{8}^{(4)}} +

+ \frac{z_{2}w_{4}}{1 − u_{9}^{(4)}} + \frac{z_{3}w_{1}}{1 − u_{10}^{(4)}} + \frac{z_{4}w_{2}}{1 − u_{11}^{(4)}} + \frac{z_{1}w_{3}}{1 − u_{12}^{(4)}} + \frac{z_{1}z_{3}w_{2}w_{4}}{(1 − u_{3}^{(4)})(1 − u_{14}^{(4)})}+

+ \frac{z_{2}z_{3}w_{4}^{2}}{(1 − u_{3}^{(4)})(1 − u_{9}^{(4)})} + \frac{z_{3}^{2}w_{1}w_{4}}{(1 − u_{3}^{(4)})(1 − u_{10}^{(4)})} + \frac{z_{2}z_{4}w_{2}w_{4}}{(1 − u_{4}^{(4)})(1 − u_{9}^{(4)})}+

+ \frac{z_{2}z_{4}w_{2}w_{4}}{(1 − u_{4}^{(4)})(1 − u_{11}^{(4)})} + \frac{z_{2}^{2}w_{3}w_{4}}{(1 − u_{5}^{(4)})(1 − u_{9}^{(4)})} + \frac{z_{1}z_{2}w_{3}^{2}}{(1 − u_{5}^{(4)})(1 − u_{12}^{(4)})}+

+ \frac{z_{2}z_{4}w_{1}w_{3}}{(1 − u_{5}^{(4)})(1 − u_{13}^{(4)})} + \frac{z_{3}z_{4}w_{1}^{2}}{(1 − u_{6}^{(4)})(1 − u_{10}^{(4)})} + \frac{z_{4}^{2}w_{1}w_{2}}{(1 − u_{6}^{(4)})(1 − u_{11}^{(4)})}+

+ \frac{z_{2}z_{4}w_{1}w_{3}}{(1 − u_{6}^{(4)})(1 − u_{13}^{(4)})} + \frac{z_{1}z_{3}w_{1}w_{3}}{(1 − u_{7}^{(4)})(1 − u_{10}^{(4)})} + \frac{z_{1}z_{3}w_{1}w_{3}}{(1 − u_{7}^{(4)})(1 − u_{12}^{(4)})}+

+ \frac{z_{1}z_{4}w_{2}^{2}}{(1 − u_{8}^{(4)})(1 − u_{11}^{(4)})} + \frac{z_{1}^{2}w_{2}w_{3}}{(1 − u_{8}^{(4)})(1 − u_{12}^{(4)})} + \frac{z_{1}z_{3}w_{2}w_{4}}{(1 − u_{8}^{(4)})(1 − u_{14}^{(4)})}),

where

u_{1}^{(4)} = z_{1}w_{1} + z_{2}w_{2} + z_{3}w_{3} + z_{4}w_{4} u_{2}^{(4)} = z_{1}w_{4} + z_{2}w_{1} + z_{3}w_{2} + z_{4}w_{3}
u_{3}^{(4)} = z_{1}w_{1} + z_{2}w_{2} + z_{3}w_{4} + z_{4}w_{3} u_{4}^{(4)} = z_{1}w_{1} + z_{2}w_{4} + z_{3}w_{3} + z_{4}w_{2}
u_{5}^{(4)} = z_{1}w_{1} + z_{2}w_{3} + z_{3}w_{2} + z_{4}w_{4} u_{6}^{(4)} = z_{1}w_{4} + z_{2}w_{2} + z_{3}w_{3} + z_{4}w_{1}
u_{7}^{(4)} = z_{1}w_{3} + z_{2}w_{2} + z_{3}w_{1} + z_{4}w_{4} u_{8}^{(4)} = z_{1}w_{2} + z_{2}w_{1} + z_{3}w_{3} + z_{4}w_{4}
u_{9}^{(4)} = z_{1}w_{1} + z_{2}w_{4} + z_{3}w_{2} + z_{4}w_{3} u_{10}^{(4)} = z_{1}w_{4} + z_{2}w_{2} + z_{3}w_{1} + z_{4}w_{3}
u_{11}^{(4)} = z_{1}w_{4} + z_{2}w_{1} + z_{3}w_{3} + z_{4}w_{2} u_{12}^{(4)} = z_{1}w_{3} + z_{2}w_{1} + z_{3}w_{2} + z_{4}w_{4}
u_{13}^{(4)} = z_{1}w_{4} + z_{2}w_{3} + z_{3}w_{2} + z_{4}w_{1} u_{14}^{(4)} = z_{1}w_{2} + z_{2}w_{1} + z_{3}w_{4} + z_{4}w_{3}.

Theorem 2.6. For every n ∈ N with n ≥ 2, F_{n}(z_{1}, w_{1}, …, z_{n}, w_{n}) is a rational function in variables z_{1}, …, z_{n}, w_{1}, …, w_{n}. It is, moreover, a sum of terms of the form

\frac{z_{i_{1}} … z_{i_{k}} w_{j_{1}} … w_{j_{k}}}{(1 − u_{1}) … (1 − u_{ℓ})}

for some ℓ ∈ {2, …, n}, k ∈ {0, 1, …, ℓ − 1}, i_{1}, …, i_{k}, j_{1}, …, j_{k} ∈ {1, …, n} and for u_{1}, …, u_{ℓ} of the form

u_{p} = ∑_{i=1}^{n} z_{i}w_{σ_{p}(i)}

for distinct permutations σ_{1}, …, σ_{ℓ} of {1, …, n}.

Proof. This follows by induction on n from the recursion formula (10) and Proposition 1.6. □

Examining the power series of F_{n}, for each n the conjecture (1) yields a conjectured identity involving multinomial coefficients. For example, when n = 3 we get the conjecture

3^{3p} = ∑_{\begin{matrix}j,k≥0 \\ j+k=p \end{matrix}} \binom{3j}{j, j, j}\binom{3k}{k, k, k}+

+ 3 ∑_{\begin{matrix}j,k,ℓ≥0 \\ j+k+ℓ=p−1 \end{matrix}} ∑_{\begin{matrix}k′,ℓ′≥0 \\ k′+ℓ′=k+ℓ+1 \end{matrix}} ∑_{\begin{matrix}j′,ℓ′′≥0 \\ j′+ℓ′′=j+ℓ+1 \end{matrix}} \binom{2j + j′}{j, j, j′}\binom{2k + k′}{k, k, k′}\binom{ℓ + ℓ′ + ℓ′′}{ℓ, ℓ′, ℓ′′}

for all p ∈ N.

3. Contour integration

It follows from Theorem 2.6 that F_{n}(z_{1}, w_{1}, …, z_{n}, w_{n}) is a holomorphic function of complex variables z_{1}, w_{1}, …, z_{n}, w_{n} in a suitably small ball around the origin. It is possible, at least in principle, to verify the conjecture mentioned at equation (1) in the introduction for a given value of n from the generating function F_{n} by performing 2n − 1 contour integrations. For example, letting

G_{n}(x_{1}, …, x_{n}) = ∑_{k_{1},...,k_{n}≥0} N(k_{1}, k_{1}, k_{2}, k_{2}, …, k_{n}, k_{n})x_{1}^{k_{1}} … x_{n}^{k_{n}},

for suitably small ϵ > 0 we have

G_{n}(y_{1}^{2}, …, y_{n}^{2}) =

= \frac{1}{(2πi)^{n}} ∫_{|ζ_{1}|=ϵ} … ∫_{|ζ_{n}|=ϵ} ζ_{1}^{−1} … ζ_{n}^{−1}F_{n}(ζ_{1}y_{1}, ζ_{1}^{−1}y_{1}, …, ζ_{n}y_{n}, ζ_{n}^{−1}y_{n})dζ_{n} … dζ_{1}.

Letting

H_{n}(a) = ∑_{k=0}^{∞} N(\underbrace{k, k, …, k, k}_{2n times})a^{k},

we have

H_{n}(b^{n}) =

= \frac{1}{(2πi)^{n−1}} ∫_{|η_{2}|=ϵ} … ∫_{|η_{n}|=ϵ} η_{2}^{−1} … η_{n}^{−1}G_{n}(η_{2} … η_{n}b, η_{2}^{−1}b, η_{3}^{−1}b, …, η_{n}^{−1}b)dη_{n} … dη_{2}.

The conjecture (1) is that

H_{n}(a) = \frac{1}{1 − n^{n}a}

for every n ∈ N.

Although the counjecture can be verified without difficulty in the case n = 2 using combinatorial methods, we will illustrate the method described above in checking it. However, even for n = 3, we have been unable to prove the conjecture using the method of contour integration.

We have

G_{2}^{(1)}(y_{1}, z_{2}, w_{2}) :=\frac{1}{2πi} ∫_{|ζ|=ϵ} F_{2}(ζy_{1}, ζ^{−1}y_{1}, z_{2}, w_{2})ζ^{−1}dζ

=\frac{1}{2πi(1 − y_{1}^{2} − z_{2}w_{2})} ∫_{|ζ|=ϵ} \frac{1}{ζ − ζ^{2}y_{1}w_{2} − y_{y}z_{2}} dζ

and ζ − ζ^{2}y_{1}w_{2} − y_{1}z_{2} = y_{1}w_{1}(r_{1} − ζ)(ζ − r_{2}) with

r_{1}, r_{2} = \frac{1 ± \sqrt{1 − 4y_{1}^{2}z_{2}w_{2}}}{2y_{1}w_{2}} .

If |y_{1}|, |z_{2}|, |w_{2}| < δ for δ > 0 small enough, r_{1} can be forced arbitrarily close to zero and r_{2} can be forced close to ∞. By the residue theorem,

G_{2}^{(1)}(y_{1}, z_{2}, w_{2}) = \frac{1}{(1 − y_{1}^{2} − z_{2}w_{2})y_{1}w_{2}(r_{1} − r_{2})}

= \frac{1}{(1 − y_{1}^{2} − z_{2}w_{2})\sqrt{1 − 4y_{1}^{2}z_{2}w_{2}}} .

Therefore,

G_{2}(x_{1}, x_{2}) = \frac{1}{(1 − x_{1} − x_{2})\sqrt{1 − 4x_{1}x_{2}}} .

Finally,

H_{2}(b^{2}) = \frac{1}{2πi} ∫_{|η|=1} η^{−1}G_{2}(ηb, η^{−1}b)dη

= \frac{1}{2πi\sqrt{1 − 4b^{2}}} ∫_{|η|=1} \frac{1}{η − b − η^{2}b}dη

= \frac{1}{1 − 4b^{2}} .

References

in preparation.

demic Press, 1983. 3368, USA

[1] K. Dykema, U. Haagerup, DT–operators and decomposability of Voiculescu’s circular operator, [2] R.V. Kadison, J.R. Ringrose, Fundamentals of the theory of operator algebras, Volume I, Aca-Department of Mathematics, Texas A&M University, College Station TX 77843–

E-mail address: kdykema@math.tamu.edu

E-mail address: cyan@math.tamu.edu