ASYMPTOTICS FOR RANDOM WALKS IN ALCOVES OF AFFINE WEYL GROUPS

C. Krattenthaler^{†}

Institut Girard Desargues, Université Claude Bernard Lyon-I 21, avenue Claude Bernard, F-69622 Villeurbanne Cedex, France

                WWW: http://igd.univ-lyon1.fr/~kratt

Dédié à Alain Lascoux

Abstract. Asymptotic results are derived for the number of random walks in alcoves of affine Weyl groups (which are certain regions in n-dimensional Euclidean space bounded by hyperplanes), thus solving problems posed by Grabiner [J. Combin. Theory Ser. A 97 (2002), 285–306]. These results include asymptotic expressions for the number of vicious walkers on a circle, as well as for the number of vicious walkers in an interval. The proofs depart from the exact results of Grabiner [loc. cit.], and require as diverse means as results from symmetric function theory and the saddle point method, among others.

1. Introduction

The enumeration of random walks in lattice regions bounded by hyperplanes is a classical and frequently studied subject in combinatorics and related fields. Its attractiveness stems from the fact that this problem has implications to many other, often seemingly unrelated problems, and, thus, to several different fields. To mention some examples, random walk interpretations exist for ballot problems (see e.g. [7, 34, 45]), standard Young tableaux (see e.g. [45]), semistandard tableaux and plane partitions (see e.g. [22, 25, 28] and [38, Sec. 8]), symplectic tableaux (see e.g. [12, 28]), oscillating tableaux (see e.g. [2, 26]), cylindric partitions (see [15]), non-intersecting lattice paths and vicious walkers (see e.g. [8, 19, 22, 28, 29]), and are therefore used for the solution of problems in these areas (see e.g. [17, 21, 38] for applications in representation theory, and e.g. [8] for applications in statistical physics), as well as in the analysis

2000 Mathematics Subject Classification. Primary 05A16; Secondary 05E05 05E15 60G50 82B23 82B41.

Key words and phrases. random walks, reflection principle, alcoves of affine Weyl groups, vicious walkers, non-intersecting lattice paths, cylindric partitions, Schur functions, symplectic characters, orthogonal characters.

† Research partially supported by the Austrian Science Foundation FWF, grant P13190-MAT, as well as grant S9607-N13, in the framework of the National Research Network “Analytic Combinatorics and Probabilistic Number Theory,” and by EC’s IHRP Programme, grant HPRN-CT-2001-00272, “Algebraic Combinatorics in Europe.”

Current address: Fakultät für Mathematik, Universität Wien, Nordbergstraße 15, A-1090 Vienna, Austria.

of non-parametric statistics in probability theory (see [34] for an introduction to that area).

Clearly, at the very beginning stands the problem of enumerating all lattice paths in the plane integer lattice between two given points, which consist of positive unit steps and do not cross a given diagonal line. (In fact, the original formulation is in terms of a two-candidate ballot problem.) As is well-known, a solution to this problem is provided by the famous reflection principle, which is usually attributed to André [1] (see e.g. [6, p. 22]). It is more than a hundred years later, when Gessel and Zeilberger showed in [16] how far one can go by using the reflection principle. Their main result gives (under certain conditions) the number of random walks in regions of n-dimensional Euclidean space which are bounded by hyperplanes. Their formula involves the elements of the reflection group which is generated by the reflections with respect to the hyperplanes which bound these regions. (The same formula for the case of finite Weyl groups has been independently discovered by Biane [4]. We refer the reader to [24] for an introduction to reflection groups and Weyl groups.) It covers numerous formulae that occurred in the literature earlier (and even afterwards …).^{1}

Recently, Grabiner [19] has revisited the problem of enumerating random walks in alcoves of affine (i.e., infinite) Weyl groups. (See the next section for precise definitions.) To be precise, he considered three types of random walks in these regions: (1) lattice walks consisting of positive unit steps e_{j} (with e_{j} denoting the j-th standard unit vector),

(2) lattice walks consisting of positive and negative unit steps ±e_{j}, and (3) lattice walks consisting of steps of the form ±\frac{1}{2}e_{1}± \frac{1}{2}e_{2}±⋯± \frac{1}{2}e_{n} (where any sign pattern is allowed). We will refer to these three types of walks as walks with standard steps in the positive direction (or walks with positive standard steps, for short), walks with standard steps, and walks with diagonal steps, respectively. Starting from the result of Gessel and Zeilberger, Grabiner derived interesting determinantal formulae for the enumeration of these three types of walks in alcoves of types Ã_{n−1}, B̃_{n}, C̃_{n}, and D̃_{n}. From the Ã_{n−1} results he was also able to derive determinantal formulae for the enumeration of walks on the circle (see the next section for the precise definition), which includes the enumeration of n non-colliding particles on a circle.

All of Grabiner’s formulae are exact results. Hence, as an afterthought, he posed the problem of determining the asymptotic behaviour of the number of these walks if the number of steps becomes large. It happens that this had already been done independently in [29] for walks with diagonal steps in the alcove of type C̃_{n}, albeit in a completely different language, the language of vicious walkers.

The purpose of this paper is to carry out the asymptotic analysis of the number of random walks in alcoves of affine Weyl groups in all the other cases, and also for the number of random walks on the circle. To be precise, we determine the asymptotic behaviour of the number of random walks in an alcove as the number of steps tends to infinity for the case that starting and end point are held fixed, as well as for the case where the end point can be arbitrary. Frequently, the results depend heavily on the parities of the involved parameters, a phenomenon which distinguishes the discrete case from the continuous case. (This phenomenon does also not occur for the corresponding problem for the walks in chambers of finite Weyl groups; see the last paragraph of the Introduction.) While, from an analytic point of view, the order of magnitude is always rather straight-forward to determine, for which very basic tools (if at all), such as Stirling’s formula, or, in one case, a rather standard application of the saddle point method, suffice, the determination of the multiplicative constant poses quite frequently a substantial challenge. Interestingly, carrying out the latter task requires quite often some advanced facts from symmetric function theory (see the proofs in Sections 4– 7). In particular, identities for classical group characters from [27] come in handy at many places. It should be observed that the proofs show that the errors are always exponentially small, with the exception of Theorem 15, where the error is dictated by the Stirling approximation of the binomial coefficient in (3.10), and of Theorem 16, where the error is dominated by those coming from the saddle point approximation given in Lemma A in Appendix A.

1There are in fact only very few known results on the enumeration of walks in regions bounded by hyperplanes that are not covered by this result. For earlier results see [34, Ch. 1 and 2]. More recent results include for example [3, 23, 30, 35, 36, 37, 42, 43, 44], where, from a conceptual point of view, the papers [3] and [35, 36, 37] have to be emphasized most: in [3] the so-called kernel method is exploited (which seems to be especially well-suited for this type of problem), whereas in [35, 36, 37] it is the umbral calculus which is systematically applied to solve lattice path enumeration problems.

In the next section we provide the basic definitions, in particular, the definitions of the alcoves to which our walks are confined, and we summarize all the exact results that exist for the enumeration of the three types of walks in these alcoves. These will be the starting points for our asymptotic calculations, which we carry out in the subsequent sections. The results for the alcove of type Ã_{n−1} are given in Section 3, the results for the enumeration of walks on the circle are the subject of Section 4, we give the results for the alcove of type C̃_{n} in Section 5, in Section 6 there follow the results for the alcove of type B̃_{n}, and, finally, we present the results for the alcove of type D̃_{n} in Section 7. Auxiliary results that are needed in the proofs of the theorems are collected in three appendices.

In concluding the introduction, it is probably useful to review the state of affairs for finite Weyl groups, i.e., the known results on the asymptotic behaviour of walks in chambers of finite Weyl groups as the number of steps of the walks becomes large. (We refer the reader again to the book [24] for definitions and more information on finite Weyl groups.) In fact, since, as we already indicated, the random walk problems considered in this paper can be seen from various different angles, numerous results can be found scattered in the combinatorics, probability, physics, and even representation theory literature. If starting and end point are fixed, the asymptotics of walks with diagonal steps in Weyl chambers of types A_{n−1} and C_{n} were determined (in the language of vicious walkers) by Rubey [40, Ch. 2, Sections 3 and 4], [41, Sections 3 and 4], with previous results in special cases given in [28, Sections 2, 4, 7]. For the case that the starting point is the origin and the end point is fixed, a result of Biane [5] on the asymptotics of multiplicities of irreducible representations in tensor powers of irreducible representations of semisimple Lie groups, combined with an observation due to Grabiner and Magyar [21, Sec. 3.3] that, under mild restrictions, the number of random walks in Weyl chambers is equal to such multiplicities, implies a uniform asymptotic formula for random walks in Weyl chambers of any type, with the exception of walks with standard steps in a Weyl chamber of type A_{n−1}. For the case that the starting point is arbitrary but fixed and the end point is not fixed, Grabiner [20] has recently shown that, by combining a result of Kuperberg [31, Theorem 1.2.1] on the approximation of sums of random variables defined on lattices by the corresponding Brownian motion, and of himself [18] on Brownian motion in Weyl chambers, one obtains the dominating term of the asymptotic behaviour for all types and for all possible step sets at once. We want to remark that walks with standard steps in the positive direction in a chamber of the Weyl group of type A_{n−1} are equivalent to skew standard Young tableaux with at most n rows, the shape depending on starting and end point of the walk. The asymptotic behaviour of the number of non-skew standard Young tableaux with at most n rows is covered by the celebrated earlier (and more general) result of Regev [39]. More precise results than that of Grabiner’s (i.e., with bounds on the errors also) in the case of walks with diagonal steps in a Weyl chamber of type A_{n−1} were found (again in the language of vicious walkers) by Rubey [40, Ch. 2, Sec. 3], [41, Sec. 3], with previous results in special cases given in [28, Sections 2, 3, 6]. In addition, Rubey [40, Ch. 2, Sec. 4], [41, Sec. 4] also provides more precise results in the case of walks with diagonal steps in a Weyl chamber of type C_{n}. (Again, results in special cases can already be found in [28, Sec. 4, 5, 7].)

2. A summary of exact results of random walks in alcoves of affine Weyl groups

In this section we summarize the exact results for random walks in alcoves of affine Weyl groups, also including two results for random walks on a circle, which are the starting points for our asymptotic calculations which are to follow in the later sections. We also use the opportunity to point out, in each case, equivalent formulations of the walk problems (in case they exist).

Before we state the results, let us recall the definitions of these alcoves. Let m be some given positive integer or half-integer. (By definition, a half-integer is an odd number divided by 2.) We define the alcove AÃ_{n−1} ; m of type Ã_{n−1} to be the region

AÃ_{n−1} ; m := {(x_{1}, x_{2}, …, x_{n}) : x_{1} > x_{2} > ⋯ > x_{n} > x_{1} − m}.  (2.1)

(Strictly speaking, this is a scaled alcove.) The (scaled) alcove of type C̃_{n} is defined by

AC̃_{n} ; m := {(x_{1}, x_{2}, …, x_{n}) : m > x_{1} > x_{2} > ⋯ > x_{n} > 0}.  (2.2)

The (scaled) alcove of type B̃_{n} is defined by

AB̃_{n} ; m := {(x_{1}, x_{2}, …, x_{n}) : x_{1} > x_{2} > ⋯ > x_{n} > 0 and x_{1} + x_{2} < 2m}.  (2.3)

Finally, the (scaled) alcove of type D̃_{n} is the region

AD̃_{n} ; m := {(x_{1}, x_{2}, …, x_{n}) : x_{1} > x_{2} > ⋯ > x_{n−1} > |x_{n}|, and x_{1} + x_{2} < 2m}. (2.4)

We begin with results for the enumeration of walks in AÃ_{n−1} ; m. The first result is originally due to Filaseta [7]. It is however covered by the general result [16]. In the statement of the theorem, and also subsequently, given a vector η = (η_{1}, η_{2}, …, η_{n}) we use the symbol |η| to denote the sum of its components, i.e., |η| := η_{1} + η_{2} + ⋯ + η_{n}.

Theorem 1 ([7]). Let m be a positive integer. Furthermore, let η = (η_{1}, η_{2}, …, η_{n}) and λ = (λ_{1}, λ_{2}, …, λ_{n}) be vectors of integers in the alcove AÃ_{n−1} ; m of type Ã_{n−1} (defined in (2.1)). Then the number of random walks from η to λ, which consist entirely of standard steps in the positive direction, and which stay in the alcove AÃ_{n−1} ; m, is given by

(|λ| − |η|)! ∑_{k_{1}+···+k_{n}=0} det_{1≤h,t≤n} (\frac{1}{(λ_{t} − η_{h} + mk_{h})!}) .  (2.5)

The corresponding result for positive and negative standard steps is also a direct consequence of the general result [16], and is stated explicitly in [19].

Theorem 2 ([19, Eq. (34)]). Let m be a positive integer. Furthermore, let η = (η_{1}, η_{2}, …, η_{n}) and λ = (λ_{1}, λ_{2}, …, λ_{n}) be vectors of integers in the alcove AÃ_{n−1} ; m of type Ã_{n−1} (defined in (2.1)). Then the number of random walks from η to λ with exactly k standard steps, which stay in the alcove AÃ_{n−1} ; m, is given by the coefficient of x^{k}/k! in

∑_{k_{1}+···+k_{n}=0} det_{1≤h,t≤n} (I_{λ_{t}−η_{h}+mk_{h}}(2x)) ,  (2.6)

where I_{α}(x) is the modified Bessel function of the first kind

I_{α}(x) = ∑_{j=0}^{∞} \frac{(x/2)^{2j+α}}{j! (j + α)!}.

This result has also a different interpretation: by considering each of the n coordinates as a separate walk, the walks with standard steps in AÃ_{n−1} ; m can be seen as n separate particles on the integer line, where at each tick of the clock exactly one particle moves to the right or to the left by one unit (a move to the right, respectively to the left, of the j-th particle corresponding to a step +e_{j}, respectively −e_{j}), under the constraint that at no time two particles occupy the same lattice site, and such that in addition a shift by m of any of the particles never collides with any of the other particles. Thus we obtain a sub-model of Fisher’s [8] random turns vicious walker model.^{2}

Similarly, random walks with diagonal steps in AÃ_{n−1} ; m with given starting and end point can be seen in several ways: by considering each of the n coordinates as a separate walk, such random walks can be seen as n separate particles on the integer line, where at each tick of the clock each particle moves one unit step to the right (corresponding to a change of +\frac{1}{2} in the corresponding coordinate) or to the left (corresponding to a change of −\frac{1}{2} in the corresponding coordinate), such that they never collide, and such that in addition a shift by 2m of any of the particles never collides with any of the other particles. Thus we obtain a sub-model of Fisher’s [8] lock step vicious walker model.^{3} An alternative, two-dimensional picture arises if we convert the movements in each coordinate of the random walk to a separate path in the plane integer lattice, identifying a change by +\frac{1}{2} in a coordinate with an up-step (1, 1) and a change by −\frac{1}{2} with a down-step (1, −1) of the corresponding path. Thus, such random walks can be seen to be equivalent to families of non-intersecting^{4} lattice paths in the plane integer lattice with steps (1, 1) and (1, −1) (the starting points of which being aligned along a vertical line, as well as the end points) where a shift of the bottom-most path dominates the top-most path. The latter objects are in turn in bijection with (special) cylindric partitions (as defined in [15]) of rectangular shape (see [15, Sec. 3] for that translation; to obtain the presentation of the lattice paths in [15], the above described picture has to be rotated by 45^{∘}).

2In statistical physics, a model of n walkers on the integer line where at each tick of the clock exactly one walker moves to the right or to the left, under the constraint that at no time two walkers occupy the same lattice site, is called the random turns vicious walker model.

3In statistical physics, a model of n walkers on the integer line where at each tick of the clock each walker moves to the right or to the left, under the constraint that at no time two walkers occupy the same lattice site, is called the lock step vicious walker model.

The following result is at the same time a direct consequence of the general result in [16] and of Theorem 3 in [15]. It is stated explicitly in [19]. It is however important to note that it is only true for integral m (as well as the “m-circle result” Theorem 5 for diagonal steps which it implies, as opposed to the companion results Theorems 7, 9, and 11 for the types C̃_{n}, B̃_{n}, and D̃_{n}). This is because the reflection argument from [16] (repeated in [19], and in an equivalent form in [15]) only guarantees that (using the vicious walker picture) particles never occupy the same site, respectively a particle shifted by 2m never occupies the same site as another particle. If m is a half-integer, this does not exclude that a shifted particle changes sides with another particle, and thus the formula (2.7) below would also include walks which violate the condition x_{n} > x_{1} − m which is contained in the definition (2.1) of the alcove AÃ_{n−1} ; m.

Theorem 3 ([19, Eq. (35)]). Let m be a positive integer. Furthermore, let η = (η_{1}, η_{2}, …, η_{n}) and λ = (λ_{1}, λ_{2}, …, λ_{n}) be vectors of integers or of half-integers in the alcove AÃ_{n−1} ; m of type Ã_{n−1} (defined in (2.1)). Then the number of random walks from η to λ with exactly k diagonal steps, which stay in the alcove AÃ_{n−1} ; m, is given by

∑_{k_{1}+···+k_{n}=0} det_{1≤h,t≤n} (\binom{k}{\frac{k}{2} + λ_{t} − η_{h} + mk_{h}}) .  (2.7)

As observed by Grabiner in [19], the above results can be used to derive results on the enumeration of random walks on the m-circle, where by “random walks on the m-circle” we mean random walks in n-dimensional Euclidean space, where each coordinate is reduced modulo m (i.e., a point (x_{1}, x_{2}, …, x_{n}) is identified with (x_{1} + k_{1}m, x_{2} + k_{2}m, …, x_{n} + k_{n}m) for any integers k_{1}, k_{2}, …, k_{n}).

Whereas in the case of standard steps in the positive direction this does not define a different model, it does for standard steps in the positive and negative direction, and also for diagonal steps. The result from [19] for standard steps reads as follows.

Theorem 4 ([19, Eq. (32)]). Let m be a positive integer. Furthermore, let η = (η_{1}, η_{2}, …, η_{n}) be a vector of integers with m > η_{1} > η_{2} > ⋯ > η_{n} ≥ 0, and let λ = (λ_{1}, λ_{2}, …, λ_{n}) be a vector of integers with m > λ_{s+1} > ⋯ > λ_{n} > λ_{1} > ⋯ > λ_{s} ≥ 0, for some s. Then the number of random walks on the m-circle from η to λ with exactly k standard steps, such that at no time two coordinates of a point on the random walk are equal, is given by

\frac{1}{n} ∑_{u=0}^{n−1} e^{−2πius/n} det_{1≤h,t≤n} (\frac{1}{m} ∑_{r=0}^{m−1} e^{−2πi(u+nr)(λ_{t}−η_{h})/mn} exp(2x cos(2π(u + nr)/mn))).

(2.8)

4A family of paths is called non-intersecting if no two paths from the family have any common points.

In the same way as explained above for random walks in AÃ_{n−1} ; m, this result can also seen as counting n non-colliding particles moving on the integer circle of length m (the interval [0, m] with 0 and m identified), where at each tick of the clock exactly one particle moves to the right or to the left by one unit.

Similarly, random walks with diagonal steps on the circle, with the property that at no time two coordinates of a point on the walk are equal can be equivalently seen as the movements of n non-colliding particles on a circle, where at each tick of the clock each particle moves one unit step to the right or to the left. This version of the lock step vicious walker model had been first considered by Forrester [10]. He solved the problem of counting the number of ways n such particles in this model may move from given starting points to given end points in the case that n is odd, however, where the particles may reach the end points in any (cyclic) order (see [10, Sec. 2.2]). An analogous formula for the case that n is even has been recently found by Fulmek [11]. Thus, the result from Grabiner’s paper [19], which we state below, constitutes a refinement of Forrester’s and Fulmek’s formulae, as in Grabiner’s formula the order in which the particles arrive at the end points is fixed. In the statement below, a small typo from [19] has been corrected (in the determinant in Eq. (33) in [19] the term ζ^{−(u+nr)(λ_{j}−η_{i})} has to be replaced by ζ^{−2(u+nr)(λ_{j}−η_{i})}).

Theorem 5 ([19, Eq. (33)]). Let m be a positive integer or half-integer. Furthermore, let η = (η_{1}, η_{2}, …, η_{n}) be a vector of integers or of half-integers with m > η_{1} > η_{2} > ⋯ > η_{n} ≥ 0, and let λ = (λ_{1}, λ_{2}, …, λ_{n}) be a vector of integers or of half-integers with m > λ_{s+1} > ⋯ > λ_{n} > λ_{1} > ⋯ > λ_{s} ≥ 0, for some s. Then the number of random walks on the m-circle from η to λ with exactly k diagonal steps, such that at no time two coordinates of a point on the random walk are equal, is given by

\frac{1}{n} ∑_{u=0}^{n−1} e^{−2πius/n} det_{1≤h,t≤n} (\frac{2^{k−1}}{m} ∑_{r=0}^{2m−1} e^{−2πi(u+nr)(λ_{t}−η_{h})/mn} cos^{k}(π(u + nr)/mn)). (2.9)

Next we quote the two results from [19] on the enumeration of random walks in alcoves of type C̃_{n}. In this case, there is no separate result for positive standard steps, since for such walks the condition m > x_{1}, which appears in the definition (2.2) of the alcove AC̃_{n} ; m, is without meaning, as well as the condition x_{n} > 0, so that the problem of enumerating random walks with positive standard steps between two given points which stay in AC̃_{n} ; m is equivalent to counting random walks with positive standard steps which stay in the Weyl chamber of type A_{n−1}, the latter being defined by

{(x_{1}, x_{2}, …, x_{n}) : x_{1} > x_{2} > ⋯ > x_{n}}.  (2.10)

As we mentioned in the Introduction, this problem has been dealt with in [20, 28, 40, 41]. On the other hand, random walks in AC̃_{n} ; m from η to λ with standard steps in the positive and negative direction are equivalent to oscillating tableaux from (η_{1} − n, η_{2} − (n − 1), …, η_{n} − 1) to (λ_{1} − n, λ_{2} − (n − 1), …, λ_{n} − 1) with at most n rows and at most m − n columns, as is easily seen by identifying a step +e_{j} with the augmentation of the i-th row of the Ferrers diagram by a box, respectively identifying a step −e_{j} with the deletion of a box from the i-th row of the Ferrers diagram. (The reader should recall that an oscillating tableau is a sequence of Ferrers diagrams where successive diagrams in the sequence differ by exactly one box.) The corresponding result from [19] reads as follows.

Theorem 6 ([19, Eq. (23)]). Let m be a positive integer. Furthermore, let η = (η_{1}, η_{2}, …, η_{n}) and λ = (λ_{1}, λ_{2}, …, λ_{n}) be vectors of integers in the alcove AC̃_{n} ; m of type C̃_{n} (defined in (2.2)). Then the number of random walks from η to λ with exactly k standard steps, which stay in the alcove AC̃_{n} ; m, is given by the coefficient of x^{k}/k! in

det_{1≤h,t≤n} (\frac{1}{m} ∑_{r=0}^{2m−1} sin \frac{πrλ_{t}}{m} · sin \frac{πrη_{h}}{m} · exp (2x cos \frac{πr}{m})) .  (2.11)

On the other hand, random walks with diagonal steps in AC̃_{n} ; m are equivalent (by means of the translation explained earlier for random walks in AÃ_{n−1} ; m) to the movements of n non-colliding particles in an interval, where at each tick of the clock each particle moves one unit step to the right or to the left (see also [19, Sec. 5]). Equivalently, these may seen as families of non-intersecting lattice paths consisting of up- and down-steps between two horizontal boundaries (see [19, Sec. 5] and [29]). The corresponding result from [19] is the following.

Theorem 7 ([19, Eq. (18)]). Let m be a positive integer or half-integer. Furthermore, let η = (η_{1}, η_{2}, …, η_{n}) and λ = (λ_{1}, λ_{2}, …, λ_{n}) be vectors of integers or of half-integers in the alcove AC̃_{n} ; m of type C̃_{n} (defined in (2.2)). Then the number of random walks from η to λ with exactly k diagonal steps, which stay in the alcove AC̃_{n} ; m, is given by

det_{1≤h,t≤n} (\frac{2^{k−1}}{m} ∑_{r=0}^{4m−1} sin \frac{πrλ_{t}}{m} · sin \frac{πrη_{h}}{m} · cos^{k} \frac{πr}{2m}) .  (2.12)

The next two results concern the enumeration of random walks in alcoves of type B̃_{n}. While the first result, the result for standard steps, is stated explicitly in [19], the second, the result for diagonal steps, is not made explicit there, although it is made clear how to derive it. We state it here for the sake of completeness. Again, there is no separate result for positive standard steps, since for such walks the condition 2m > x_{1} + x_{2}, which appears in the definition (2.3) of the alcove AB̃_{n} ; m, is without meaning, as well as the condition x_{n} > 0, so that the problem of enumerating random walks with positive standard steps between two given points which stay in AB̃_{n} ; m is again equivalent to counting random walks with positive standard steps which stay in the Weyl chamber of type A_{n−1}, the latter being defined by (2.10).

Theorem 8 ([19, Eq. (43)]). Let m be a positive integer or half-integer. Furthermore, let η = (η_{1}, η_{2}, …, η_{n}) and λ = (λ_{1}, λ_{2}, …, λ_{n}) be vectors of integers in the alcove AB̃_{n} ; m of type B̃_{n} (defined in (2.3)). Then the number of random walks from η to λ with exactly k standard steps, which stay in the alcove AB̃_{n} ; m, is given by the coefficient of x^{k}/k! in

\frac{1}{2} det_{1≤h,t≤n} (\frac{1}{m} ∑_{r=0}^{2m−1} sin \frac{πrλ_{t}}{m} · sin \frac{πrη_{h}}{m} · exp (2x cos \frac{πr}{m}))

+ \frac{1}{2} det_{1≤h,t≤n} (\frac{1}{m} ∑_{r=0}^{2m−1} sin \frac{π(2r + 1)λ_{t}}{2m} · sin \frac{π(2r + 1)η_{h}}{2m} · exp (2x cos \frac{π(2r + 1)}{2m})) . (2.13)

Theorem 9 ([19]). Let m be a positive integer or half-integer. Furthermore, let η = (η_{1}, η_{2}, …, η_{n}) and λ = (λ_{1}, λ_{2}, …, λ_{n}) be vectors of integers or of half-integers in the alcove AB̃_{n} ; m of type B̃_{n} (defined in (2.3)). Then the number of random walks from η to λ with exactly k diagonal steps, which stay in the alcove AB̃_{n} ; m, is given by

\frac{1}{2} det_{1≤h,t≤n} (\frac{2^{k−1}}{m} ∑_{r=0}^{4m−1} sin \frac{πrλ_{t}}{m} · sin \frac{πrη_{h}}{m} · cos^{k} \frac{πr}{2m})

+ \frac{1}{2} det_{1≤h,t≤n} (\frac{2^{k−1}}{m} ∑_{r=0}^{4m−1} sin \frac{π(2r + 1)λ_{t}}{2m} · sin \frac{π(2r + 1)η_{h}}{2m} · cos^{k} \frac{π(2r + 1)}{4m}) . (2.14)

The final two results concern the enumeration of random walks in alcoves of type D̃_{n}. Again, the first result, the result for standard steps, is stated explicitly in [19], while the second, the result for diagonal steps, is not made explicit there, although, again, it is clearly described how to derive it. We state it here for the sake of completeness. Also here, there is no separate result for positive standard steps, since for such walks the condition 2m > x_{1} + x_{2}, which appears in the definition (2.4) of the alcove AD̃_{n} ; m, is without meaning, as well as the condition x_{n−1} > |x_{n}|, so that the problem of enumerating random walks with positive standard steps between two given points which stay in AD̃_{n} ; m is again equivalent to counting random walks with positive standard steps which stay in the Weyl chamber of type A_{n−1}, the latter being defined by (2.10).

Theorem 10 ([19, Eq. (46)]). Let m be a positive integer or half-integer. Furthermore, let η = (η_{1}, η_{2}, …, η_{n}) and λ = (λ_{1}, λ_{2}, …, λ_{n}) be vectors of integers in the alcove AD̃_{n} ; m of type D̃_{n} (defined in (2.4)). Then the number of random walks from η to λ with exactly k standard steps, which stay in the alcove AD̃_{n} ; m, is given by the coefficient of x^{k}/k! in

\frac{1}{4} det_{1≤h,t≤n} (\frac{1}{m} ∑_{r=0}^{2m−1} sin \frac{πrλ_{t}}{m} · sin \frac{πrη_{h}}{m} · exp (2x cos \frac{πr}{m}))

+ \frac{1}{4} det_{1≤h,t≤n} (\frac{1}{m} ∑_{r=0}^{2m−1} sin \frac{π(2r + 1)λ_{t}}{2m} · sin \frac{π(2r + 1)η_{h}}{2m} · exp (2x cos \frac{π(2r + 1)}{2m}))

+ \frac{1}{4} det_{1≤h,t≤n} (\frac{1}{m} ∑_{r=0}^{2m−1} cos \frac{πrλ_{t}}{m} · cos \frac{πrη_{h}}{m} · exp (2x cos \frac{πr}{m}))

+ \frac{1}{4} det_{1≤h,t≤n} (\frac{1}{m} ∑_{r=0}^{2m−1} cos \frac{π(2r + 1)λ_{t}}{2m} · cos \frac{π(2r + 1)η_{h}}{2m} · exp (2x cos \frac{π(2r + 1)}{2m})) . (2.15)

Theorem 11 ([19]). Let m be a positive integer or half-integer. Furthermore, let η = (η_{1}, η_{2}, …, η_{n}) and λ = (λ_{1}, λ_{2}, …, λ_{n}) be vectors of integers or of half-integers in the alcove AD̃_{n} ; m of type D̃_{n} (defined in (2.4)). Then the number of random walks from η to λ with exactly k diagonal steps, which stay in the alcove AD̃_{n} ; m, is given by

\frac{1}{4} det_{1≤h,t≤n} (\frac{2^{k−1}}{m} ∑_{r=0}^{4m−1} sin \frac{πrλ_{t}}{m} · sin \frac{πrη_{h}}{m} · cos^{k} \frac{πr}{2m})

+ \frac{1}{4} det_{1≤h,t≤n} (\frac{2^{k−1}}{m} ∑_{r=0}^{4m−1} sin \frac{π(2r + 1)λ_{t}}{2m} · sin \frac{π(2r + 1)η_{h}}{2m} · cos^{k} \frac{π(2r + 1)}{4m})

+ \frac{1}{4} det_{1≤h,t≤n} (\frac{2^{k−1}}{m} ∑_{r=0}^{4m−1} cos \frac{πrλ_{t}}{m} · cos \frac{πrη_{h}}{m} · cos^{k} \frac{πr}{2m})

+ \frac{1}{4} det_{1≤h,t≤n} (\frac{2^{k−1}}{m} ∑_{r=0}^{4m−1} cos \frac{π(2r + 1)λ_{t}}{2m} · cos \frac{π(2r + 1)η_{h}}{2m} · cos^{k} \frac{π(2r + 1)}{4m}) . (2.16)

3. Asymptotics for random walks in alcoves of type Ã

This section is devoted to finding the asymptotic behaviour of the number of walks from a given starting point to a given end point which stay in the alcove AÃ_{n−1} ; m of type Ã_{n−1} as the number of steps becomes large, as well as the asymptotic behaviour of the number of those walks which start at a given point but may terminate anywhere. In technical terms, we determine the asymptotic behaviour of the expressions given by Theorems 1–3 as k becomes large (in the case of Theorem 1 the role of k is played by |λ| − |η|), and as well if these expressions are summed over all possible end points of the walks.

We begin with the walks with standard steps in the positive direction.

Theorem 12. Let m be a positive integer. Furthermore, let η = (η_{1}, η_{2}, …, η_{n}) and λ = (λ_{1}, λ_{2}, …, λ_{n}) be vectors of integers in the alcove AÃ_{n−1} ; m of type Ã_{n−1} (defined in (2.1)). Then, for large |λ|, the number of random walks from η to λ, which consist entirely of standard steps in the positive direction, and which stay in the alcove AÃ_{n−1} ; m, is asymptotically

\frac{2^{n^{2}−n}}{m^{n−1}} (\begin{matrix}sin & \frac{nπ}{m} \\ sin & \frac{π}{m} \end{matrix} )^{|λ|−|η|} ∏_{1≤h<t≤n} (sin \frac{π(η_{h} − η_{t})}{m} · sin \frac{π(λ_{h} − λ_{t})}{m}) .  (3.1)

Proof. We want to estimate the expression (2.5) for |λ| large. In order to accomplish that, we write it in the form

⟨z^{0}⟩ (|λ| − |η|)! ∑_{k_{1},...,k_{n}=−∞}^{∞} z^{m ∑_{j=1}^{n} k_{j}} det_{1≤h,t≤n} (\frac{1}{(λ_{t} − η_{h} + mk_{h})!})

= ⟨z^{0}⟩ (|λ| − |η|)! det_{1≤h,t≤n} ( ∑_{k_{h}=−∞}^{∞} \frac{z^{mk_{h}}}{(λ_{t} − η_{h} + mk_{h})!}) ,

where here, and in the sequel, the notation ⟨f⟩F stands for the coefficient of f in (an appropriate expansion of) F. (I.e., here, ⟨z^{0}⟩ g(z) denotes the constant coefficient in the Laurent series g(z).) With ω = e^{2πi/m}, we can rewrite this expression as

⟨z^{0}⟩ (|λ| − |η|)! det_{1≤h,t≤n} (\frac{1}{m} ∑_{r_{h}=0}^{m−1} ∑_{k_{h}=−∞}^{∞} \frac{(zω^{r_{h}})^{k_{h}}}{(λ_{t} − η_{h} + k_{h})!}) ,

because ∑_{r=0}^{m−1} ω^{rk} is equal to m if k is divisible by m, and it vanishes otherwise. Evaluating the sum over k_{h}, we obtain the expression

⟨z^{0}⟩(|λ| − |η|)! det_{1≤h,t≤n} (\frac{1}{m} ∑_{r_{h}=0}^{m−1} (zω^{r_{h}})^{η_{h}−λ_{t}} exp (zω^{r_{h}}))

= ⟨z^{|λ|−|η|}⟩ \frac{(|λ| − |η|)!}{m^{n}} ∑_{r_{1},...,r_{n}=0}^{m−1} exp (z ∑_{j=1}^{n} ω^{r_{j}}) det_{1≤h,t≤n} (ω^{r_{h}(η_{h}−λ_{t})})

= \frac{1}{m^{n}} ∑_{r_{1},...,r_{n}=0}^{m−1} W (r)^{|λ|−|η|} det_{1≤h,t≤n} (ω^{r_{h}(η_{h}−λ_{t})}) ,  (3.2)

where in the last line W (r) is an abbreviation for ∑_{j=1}^{n} ω^{r_{j}}.

The sum in (3.2) is a finite sum of the form ∑_{ℓ} c_{ℓ}b_{ℓ}^{k}, with k = |λ| − |η|, where the b_{ℓ}’s are of the form W (r), and the c_{ℓ}’s are bounded. Thus, the asymptotic behaviour of this sum as k → ∞ is dominated by the terms c_{ℓ}b_{ℓ}^{k} for which |b_{ℓ}| is maximal (and c_{ℓ} ≠ 0 of course).

Now, if two summation indices r_{h} and r_{t}, for h ≠ t, should be equal, then the determinant in the summand in (3.2) vanishes. Therefore we may restrict the sum in (3.2) to indices r_{1}, r_{2}, …, r_{n} which are pairwise distinct. Among the latter, the sets of indices {r_{1}, r_{2}, …, r_{n}} for which W (r) has largest modulus are those for which the r_{j}’s are as “close” together as possible, i.e., the sets

{r_{1}, r_{2}, …, r_{n}} = {ℓ, ℓ + 1, …, ℓ + n − 1},  (3.3)

for some ℓ between 0 and m−1. (On the right-hand side, the elements must be reduced modulo m.) Hence, let r_{j} = ℓ + σ(j) − 1, j = 1, 2, …, n, for some permutation σ ∈ S_{n}, with S_{n} denoting the symmetric group of order n. For this choice of indices, we have

W (r)^{|λ|−|η|} det_{1≤h,t≤n} (ω^{r_{h}(η_{h}−λ_{t})})

= |∑_{j=1}^{n} ω^{j−1}|^{|λ|−|η|} ω^{\frac{n−1}{2} (|λ|−|η|)} det_{1≤h,t≤n} (ω^{(σ(h)−1)(η_{h}−λ_{t})})

= (\begin{matrix}sin & \frac{nπ}{m} \\ sin & \frac{π}{m} \end{matrix} )^{|λ|−|η|} ω^{\frac{n−1}{2} (|λ|−|η|)}ω^{∑_{j=1}^{n}(σ(j)−1)η_{j}} det_{1≤h,t≤n} (ω^{−(σ(h)−1)λ_{t}}) .

Thus, if we combine all our findings, we obtain that, as (|λ| − |η|) → ∞, the expression (3.2) is asymptotically

\frac{1}{m^{n}} (\begin{matrix}sin & \frac{nπ}{m} \\ sin & \frac{π}{m} \end{matrix} )^{|λ|−|η|} ∑_{ℓ=0}^{m−1} ∑_{σ∈S_{n}} ω^{\frac{n−1}{2} (|λ|−|η|)}ω^{∑_{j=1}^{n}(σ(j)−1)η_{j}}(sgn σ) det_{1≤h,t≤n} (ω^{−(h−1)λ_{t}})

= \frac{1}{m^{n−1}} (\begin{matrix}sin & \frac{nπ}{m} \\ sin & \frac{π}{m} \end{matrix} )^{|λ|−|η|} ω^{\frac{n−1}{2} (|λ|−|η|)} det_{1≤h,t≤n} (ω^{(h−1)η_{t}}) det_{1≤h,t≤n} (ω^{−(h−1)λ_{t}}) .

Both determinants are Vandermonde determinants, and are therefore easily evaluated. The resulting expression is exactly (3.1). □

Next we address the question of determining the asymptotic behaviour of the number of all walks which start in a given point and proceed for k standard steps in the positive direction, always staying in the alcove AÃ_{n−1} ; m of type Ã_{n−1}. Clearly, this amounts to summing the expression (3.1) over all λ with |λ| = |η|+k. A moment’s reflection shows that this is in fact a finite sum, with the number of terms bounded by (2m)^{m} (to give a very crude bound), a quantity which is independent of k. Thus, it is obvious that the order of magnitude of the number of all these walks is (sin \frac{nπ}{m}/sin \frac{π}{m})^{k}. However, determining the multiplicative constant poses a formidable challenge, in particular if one attempts to do it directly from summing up the expression (3.1). An elegant way to bypass (some of this) difficulty is to set up a relationship between the enumeration of walks with positive standard steps and the enumeration of walks with arbitrary standard steps, which we do in Lemma 14.

Theorem 13. Let m be a positive integer. Furthermore, let η = (η_{1}, η_{2}, …, η_{n}) be a vector of integers in the alcove AÃ_{n−1} ; m of type Ã_{n−1} (defined in (2.1)). Then, as k tends to infinity, the number of random walks which start at η and proceed for exactly k standard steps in the positive direction, which stay in the alcove AÃ_{n−1} ; m, is asymptotically

\frac{2^{\binom{n}{2}}}{m^{n/2}} (\begin{matrix}sin & \frac{nπ}{m} \\ sin & \frac{π}{m} \end{matrix} )^{k} ∏_{1≤h<t≤n} (sin \frac{π(η_{h} − η_{t})}{m})

× (∏_{h=1}^{n/2} cot \frac{(2h − 1)π}{2m} + (−1)^{|η|+k+ \frac{n}{2}} ∏_{h=1}^{n/2} tan \frac{(2h − 1)π}{2m} ) (3.4)

if both n and m are even, it is asymptotically

\frac{2^{\binom{n}{2}}}{m^{n/2}} (\begin{matrix}sin & \frac{nπ}{m} \\ sin & \frac{π}{m} \end{matrix} )^{k} ∏_{1≤h<t≤n} (sin \frac{π(η_{h} − η_{t})}{m}) ∏_{h=1}^{n/2} cot \frac{(2h − 1)π}{2m}  (3.5)

if n is even and m is odd, and it is asymptotically

\frac{2^{\binom{n}{2}}}{m^{(n−1)/2}} (\begin{matrix}sin & \frac{nπ}{m} \\ sin & \frac{π}{m} \end{matrix} )^{k} ∏_{1≤h<t≤n} (sin \frac{π(η_{h} − η_{t})}{m}) ∏_{h=1}^{(n−1)/2} cot \frac{hπ}{m}  (3.6)

if n is odd (regardless of m).

Proof. Our original proof proceeded as indicated in the paragraph before the statement of the theorem, namely by summing the expression (3.1) over all possible λ with |λ| = |η| + k. Although feasible, this path turned out to be a thorny one. After the result had been obtained, the surprising observation was that the asymptotic behaviour of walks in AÃ_{n−1} ; m with positive standard steps is, up to a factor of 2^{k}, identical with the asymptotic behaviour of walks on the m-circle with arbitrary (i.e., positive and negative) standard steps (cf. Theorem 18). As we show in Lemma 14 below, this is even true non-asymptotically. (This is indeed the assertion of Lemma 14 since it does not matter whether we are in the alcove AÃ_{n−1} ; m or on the m-circle if we let the end point of the walks be arbitrary.) The theorem now follows by the (independent) proof of Theorem 18 given in Section 4. □

Lemma 14. The number of random walks which start at η, proceed for exactly k standard steps in the positive direction, and stay in the alcove AÃ_{n−1} ; m, is equal to 2^{−k} times the number of random walks which start at η, proceed for exactly k standard steps, and stay in the alcove AÃ_{n−1} ; m.

Proof. We are going to show this inductively. Let η = (η_{1}, η_{2}, …, η_{n}) be given. We decompose η into its maximal circular subsequences of consecutive elements. The meaning of “circular” is that η_{n} and η_{1} are considered to be “consecutive” if η_{n} = η_{1} − m + 1. In the sequel, we shall use the short hand maximal circular subsequences for these subsequences. For example, if m = 11 there are 3 maximal circular subsequences in (9, 8, 5, 4, 3, 1), namely

(9, 8), (5, 4, 3), (1).  (3.7)

On the other hand, if m = 9, then the maximal circular subsequences are

(1, 9, 8), (5, 4, 3).  (3.8)

Let the maximal circular subsequences in the decomposition of η have respectively the lengths a_{1}, a_{2}, …, a_{ℓ}, with gaps b_{1}, b_{2}, …, b_{ℓ}. More precisely, the gap b_{j} is the difference between the smallest element in the j-th sub-sequence and the largest element in the (j + 1)-st sub-sequence, reduced modulo m. Here, (j + 1) has to be interpreted as 1 if j = ℓ, Thus, in the example (3.7) we have a_{1} = 2, a_{2} = 3, a_{3} = 1, and b_{1} = 3, b_{2} = 2, b_{3} = 3, while in the example (3.8) we have a_{1} = 3, a_{2} = 3, and b_{1} = 3, b_{2} = 2.

It is obvious that, starting from such an η, there are exactly ℓ ways to move by a positive standard step. To be precise, one would increase the largest element of a maximal circular sub-sequence by 1. Similarly, there are exactly 2ℓ ways to move by an arbitrary standard step, namely the ℓ possibilities of positive standard steps described above, together with the ℓ possibilities of decreasing a smallest element of a maximal circular sub-sequence by 1. Thus, for one step, i.e., for k = 1, our claim is true.

The question is whether this persists. As we have seen, the number of possibilities to walk is (for positive standard steps, as well as for arbitrary standard steps) a multiple of the number of maximal circular subsequences. Thus, if we are able to show that the total number of maximal circular sub-sequences in the set of all possible points that we reached from η by walking one positive standard step is exactly one half of the corresponding number of maximal circular sub-sequences in the set of all possible points that we reached from η by walking one arbitrary standard step, then we are sure that the ratio of 1 : 2 will continue to hold for each step.

The latter claim is easy to establish: if we increase the largest element of the j-th maximal circular sub-sequence of η by 1, then we obtain a point whose decomposition has

ℓ + 1 − χ(a_{j} = 1) − χ(b_{j−1} = 2)

maximal circular subsequences, where we used the notation χ(A)=1 if A is true and χ(A)=0 otherwise, and where b_{0} has to be interpreted as b_{ℓ}. On the other hand, if we decrease the smallest element of the j-th maximal circular sub-sequence of η by 1, then we obtain a point whose decomposition has

ℓ + 1 − χ(a_{j} = 1) − χ(b_{j} = 2)

maximal circular subsequences. Thus, the total number of maximal circular subsequences in the set of points reached from η by walking one positive standard step is

ℓ(ℓ + 1) − ∑_{j=1}^{ℓ} (χ(a_{j} = 1) + χ(b_{j} = 1)),

while the total number of maximal circular subsequences in the set of points reached from η by walking one arbitrary standard step is exactly twice of that. This finishes the proof of the lemma. □

Now we turn our attention to walks with positive and negative standard steps. The theorem below gives the asymptotic behaviour of the walks in AÃ_{n−1} ; m with fixed starting and end point. If the end point is allowed to be arbitrary, then the enumeration of the corresponding walks is equivalent to the enumeration of walks with standard steps on the m-circle, and, thus, its asymptotic behaviour is given by Theorem 18 in Section 4.

Theorem 15. Let m be a positive integer. Furthermore, let η = (η_{1}, η_{2}, …, η_{n}) and λ = (λ_{1}, λ_{2}, …, λ_{n}) be vectors of integers in the alcove AÃ_{n−1} ; m of type Ã_{n−1} (defined in (2.1)). Then, as k tends to infinity such that k ≡ |η| + |λ| mod 2, the number of random walks from η to λ with exactly k standard steps, which stay in the alcove AÃ_{n−1} ; m, is asymptotically

\frac{2^{n^{2}−n}}{m^{n−1}} \sqrt{\frac{2}{πk}} (2\begin{matrix}sin & \frac{nπ}{m} \\ sin & \frac{π}{m} \end{matrix} )^{k} ∏_{1≤h<t≤n} (sin \frac{π(η_{h} − η_{t})}{m} · sin \frac{π(λ_{h} − λ_{t})}{m}) .  (3.9)

Proof. We have to determine the asymptotics of the coefficient of x^{k}/k! in (2.6) as k → ∞. To begin with, we rewrite this coefficient as

⟨z^{0} \frac{x^{k}}{k!} ⟩ ∑_{k_{1},...,k_{n}=−∞}^{∞} z^{m(k_{1}+···+k_{n})} det_{1≤h,t≤n} (I_{λ_{t}−η_{h}+mk_{h}}(2x))

= ⟨z^{0} \frac{x^{k}}{k!} ⟩ det_{1≤h,t≤n} ( ∑_{k_{h}=−∞}^{∞} I_{λ_{t}−η_{h}+mk_{h}}(2x) z^{mk_{h}}) ,

Here, ⟨z^{0} \frac{x^{k}}{k!} ⟩g(x, z) denotes the coefficient of z^{0}x^{k}/k! in g(x, z), which is in accordance with our earlier general definition of the coefficient notation. As in the proof of Theorem 12, with ω = e^{2πi/m} we may rewrite this expression as

⟨z^{0} \frac{x^{k}}{k!} ⟩ det_{1≤h,t≤n} (\frac{1}{m} ∑_{r_{h}=0}^{m−1} ∑_{k_{h}=−∞}^{∞} I_{λ_{t}−η_{h}+k_{h}}(2x) (ω^{r_{h}}z)^{k_{h}}) .

Now using the easily verified fact that

∑_{j=−∞}^{∞} I_{j}(2x) z^{j} = exp (x (z + z^{−1}) ),

we obtain

⟨z^{0} \frac{x^{k}}{k!} ⟩ det_{1≤h,t≤n} (\frac{1}{m} (ω^{r_{h}}z)^{η_{h}−λ_{t}} ∑_{r_{h}=0}^{m−1} exp (x (ω^{r_{h}}z + (ω^{r_{h}}z)^{−1}) ))

= ⟨z^{|λ|−|η|} \frac{x^{k}}{k!} ⟩ \frac{1}{m^{n}} ∑_{r_{1},...,r_{n}=0}^{m−1} det_{1≤h,t≤n} (ω^{r_{h}(η_{h}−λ_{t})})

· exp (x(z ∑_{j=1}^{n} ω^{r_{j}} + z^{−1} ∑_{j=1}^{n} ω^{−r_{j}}))

= ⟨z^{|λ|−|η|}⟩ \frac{1}{m^{n}} ∑_{r_{1},...,r_{n}=0}^{m−1} det_{1≤h,t≤n} (ω^{r_{h}(η_{h}−λ_{t})}) · (z ∑_{j=1}^{n} ω^{r_{j}} + z^{−1} ∑_{j=1}^{n} ω^{−r_{j}})^{k}

= \frac{1}{m^{n}} \binom{k}{\frac{1}{2}(k + |λ| − |η|)} ∑_{r_{1},...,r_{n}=0}^{m−1} det_{1≤h,t≤n} (ω^{r_{h}(η_{h}−λ_{t})})

· (∑_{j=1}^{n} ω^{r_{j}})^{^{\frac{1}{2}} (k+|λ|−|η|)} (∑_{j=1}^{n} ω^{−r_{j}})^{^{\frac{1}{2}} (k−|λ|+|η|)}

for the coefficient of x^{k}/k! in (2.6). Writing again W (r) for ∑_{j=1}^{n} ω^{r_{j}}, we have obtained that the coefficient of x^{k}/k! in (2.6) is equal to

\frac{1}{m^{n}} \binom{k}{\frac{1}{2}(k + |λ| − |η|)}

× ∑_{r_{1},...,r_{n}=0}^{m−1} |W (r)|^{k} (W (r)/\overline{W (r)})^{\frac{1}{2} (|λ|−|η|)} det_{1≤h,t≤n} (ω^{r_{h}(η_{h}−λ_{t})}) . (3.10)

Stirling’s formula implies that the binomial coefficient in this expression is asymptotically 2^{k}\sqrt{\frac{2}{kπ}} as k → ∞.

The sum in (3.10), on the other hand, is again a finite sum of the form ∑_{ℓ} c_{ℓ}b_{ℓ}^{k}, with the b_{ℓ}’s of the form |W (r)|, and both the c_{ℓ}’s and b_{ℓ}’s independent of k. Thus, the asymptotic behaviour of this sum as k → ∞ is dominated by the terms c_{ℓ}b_{ℓ}^{k} for which |b_{ℓ}| is maximal (and c_{ℓ} ≠ 0 of course).

If we compare the expression (3.10) that we have obtained so far with the expression (3.2), then we see that we are in a very similar situation here as at the analogous place in the proof of Theorem 12. Therefore, if we apply the arguments given there to our situation, we obtain that, as k → ∞, the expression (3.10) is asymptotically

\frac{1}{m^{n}} \sqrt{\frac{2}{πk}} (2\begin{matrix}sin & \frac{nπ}{m} \\ sin & \frac{π}{m} \end{matrix} )^{k} ∑_{ℓ=0}^{m−1} ∑_{σ∈S_{n}} ω^{\frac{n−1}{2} (|λ|−|η|)}ω^{∑_{j=1}^{n}(σ(j)−1)η_{j}}(sgn σ) det_{1≤h,t≤n} (ω^{−(σ(h)−1)λ_{t})})

= \frac{1}{m^{n−1}} \sqrt{\frac{2}{πk}} (2\begin{matrix}sin & \frac{nπ}{m} \\ sin & \frac{π}{m} \end{matrix} )^{k} ω^{\frac{n−1}{2} (|λ|−|η|)} det_{1≤h,t≤n} (ω^{(h−1)η_{t}}) det_{1≤h,t≤n} (ω^{−(h−1)λ_{t}}) .

Both determinants are Vandermonde determinants, and are therefore easily evaluated. The resulting expression is exactly (3.9). □

As the final issue in this section, we consider the asymptotic behaviour of walks in the alcove AÃ_{n−1} ; m consisting of diagonal steps. The theorem below provides the solution of the problem if the starting and end point are fixed. Exceptionally, to resolve this problem, we need to apply a more advanced asymptotic method, the saddle point method (although a rather basic instance of it). Should the end point be allowed to be arbitrary, then the corresponding enumeration problem is equivalent to the enumeration of walks with diagonal steps on the m-circle, and, thus, its asymptotic behaviour is given by Theorem 20 in Section 4.

Theorem 16. Let m be a positive integer or half-integer. Furthermore, let η = (η_{1}, η_{2}, …, η_{n}) and λ = (λ_{1}, λ_{2}, …, λ_{n}) be vectors of integers or of half-integers in the alcove AÃ_{n−1} ; m of type Ã_{n−1} (defined in (2.1)). Then, as k tends to infinity such that k ≡ 2η_{j} + 2λ_{j} mod 2, the number of random walks from η to λ with exactly k diagonal steps, which stay in the alcove AÃ_{n−1} ; m, is asymptotically

\frac{1}{m^{n−1}} \frac{2^{n^{2}−n}}{\sqrt{2πc_{0}k}} (2^{n} ∏_{j=1}^{n} cos \frac{π (j − \frac{n+1}{2})}{m})^{k} ∏_{1≤h<t≤n} (sin \frac{π(η_{h} − η_{t})}{m} · sin \frac{π(λ_{h} − λ_{t})}{m}) ,

(3.11)

where

c_{0} = ∑_{j=1}^{n} (2 cos \frac{π (j − \frac{n+1}{2})}{m})^{−2} .

Proof. We have to determine the asymptotics of the expression (2.7) as k → ∞. To begin with, we write (2.7) in the form

⟨z^{0}⟩ ∑_{k_{1},...,k_{n}=−∞}^{∞} z^{m(k_{1}+···+k_{n})} det_{1≤h,t≤n} (\binom{k}{\frac{k}{2} + λ_{t} − η_{h} + mk_{h}})

= ⟨z^{0}⟩ det_{1≤h,t≤n} ( ∑_{k_{h}=−∞}^{∞} \binom{k}{\frac{k}{2} + λ_{t} − η_{h} + mk_{h}}z^{mk_{h}}) ,

where, again, the notation ⟨z^{0}⟩g(z) denotes the coefficient of z^{0} in g(z). Arguing in the same way as before in the proofs of Theorems 12 and 15, with ω = e^{2πi/m} we can rewrite this expression as

⟨z^{0}⟩ det_{1≤h,t≤n} (\frac{1}{m} ∑_{r_{h}=0}^{m−1} ∑_{k_{h}=−∞}^{∞} \binom{k}{\frac{k}{2} + λ_{t} − η_{h} + k_{h}} (ω^{r_{h}}z)^{k_{h}})

= ⟨z^{0}⟩ \frac{1}{m^{n}} det_{1≤h,t≤n} (∑_{r_{h}=0}^{m−1} (ω^{r_{h}}z)^{− \frac{k}{2} −λ_{t}+η_{h}} (1 + ω^{r_{h}}z)^{k})

= \frac{1}{m^{n}} ∑_{r_{1},...,r_{n}=0}^{m−1} ω^{− \frac{k}{2} |r|} det_{1≤h,t≤n} (ω^{r_{h}(η_{h}−λ_{t})}) ⟨z^{\frac{nk}{2} +|λ|−|η|}⟩ ∏_{j=1}^{n}(1 + ω^{r_{j}}z)^{k}, (3.12)

where in the next-to-last line we used the binomial theorem.

We have again obtained a finite sum. Therefore the task now is to isolate the summands which are asymptotically largest as k → ∞. First of all, if two summation indices r_{h} and r_{t}, for h ≠ t, should be equal, then the determinant in the summand in (3.12) vanishes. Therefore we may restrict the sum in (3.12) to the summands corresponding to indices r_{1}, r_{2}, …, r_{n} which are pairwise distinct.

According to Lemma A in Appendix A, among the latter, those will be asymptotically largest for which

∏_{j=1}^{n} |cos \frac{π(θ_{0} + r_{j})}{m}|  (3.13)

is largest (where θ_{0} is a solution of (A.4)). As is not difficult to see, these are those sets of indices {r_{1}, r_{2}, …, r_{n}} for which the r_{j}’s are as “close” together as possible, i.e., again the sets as given in (3.3), for some ℓ between 0 and m − 1.

Let ℓ be fixed. Let {r_{1}, r_{2}, …, r_{n}} be a set of indices as in (3.3), i.e., r_{j} = ℓ+σ(j)−1, j = 1, 2, …, n, for some permutation σ ∈ S_{n}. For this set, there is a unique θ_{0} such that (3.13) is maximal, namely θ_{0} = −ℓ − \frac{n−1}{2}. Thus, using Lemma A in (3.12), we obtain that the expression (3.12) is asymptotically

\frac{1}{m^{n}} ∑_{ℓ=0}^{m−1} ∑_{σ∈S_{n}} det_{1≤h,t≤n} (ω^{(σ(h)−1)(η_{h}−λ_{t})}) \frac{ω^{\frac{n−1}{2} (|λ|−|η|)}}{\sqrt{2πc_{0}k}} ∏_{j=1}^{n} (2 cos \frac{π (j − \frac{n+1}{2})}{m})^{k}

= \frac{1}{m^{n−1}} \frac{ω^{\frac{n−1}{2} (|λ|−|η|)}}{\sqrt{2πc_{0}k}} ∏_{j=1}^{n} (2 cos \frac{π (j − \frac{n+1}{2})}{m})^{k}

× det_{1≤h,t≤n} (ω^{(h−1)η_{t}}) det_{1≤h,t≤n} (ω^{−(h−1)λ_{t}}) ,

where c_{0} is given as in the statement of the theorem. Again, both determinants are Vandermonde determinants, and are therefore easily evaluated. The resulting expression is exactly (3.11). □

4. Asymptotics for random walks on the circle

In this section we find the asymptotic behaviour of the number of walks from a given starting point to a given end point on the m-circle as the number of steps becomes large, as well as the asymptotic behaviour of the number of those walks which start at a given point but may terminate anywhere. In technical terms, we determine the asymptotic behaviour of the expressions given by Theorems 4 and 5 as k becomes large, and as well if these expressions are summed over all possible end points of the walks.

Before we state the next theorem, which gives the asymptotic behaviour of walks with standard steps between two fixed points on the m-circle, we need to discuss under which conditions such walks can exist. In the theorem below we consider walks from η = (η_{1}, η_{2}, …, η_{n}) to λ = (λ_{1}, λ_{2}, …, λ_{n}) where λ_{s+1} > ⋯ > λ_{n} > λ_{1} > ⋯ > λ_{s}, which means that, when we interpret such walks as the movements of n separate particles, the first s particles (the particles which start at η_{1}, η_{2}, …, η_{s}) wind themselves once more around the circle than the other particles. Thus, if we want to get from η to λ in k steps, we must have

k ≡ |λ| − |η| + Nnm + sm (mod 2),  (4.1)

where the integer N is the number of times the latter particles wind around the circle. If m is even, then this condition reduces to the familiar k ≡ |λ| + |η| mod 2. However, if m is odd, then there are two possibilities. If in addition n is even, then, depending on whether k is even or odd, there are walks only for every second s. If both m and n are odd, then there is no restriction for k.

We are now ready to state the theorem. As we discussed in Section 2, this theorem gives at the same time the asymptotic behaviour of n non-colliding particles on the circle in the random turns vicious walker model.

Theorem 17. Let m be a positive integer. Furthermore, let η = (η_{1}, η_{2}, …, η_{n}) be a vector of integers with m > η_{1} > η_{2} > ⋯ > η_{n} ≥ 0, and let λ = (λ_{1}, λ_{2}, …, λ_{n}) be a vector of integers with m > λ_{s+1} > ⋯ > λ_{n} > λ_{1} > ⋯ > λ_{s} ≥ 0, for some s. Then, as k tends to infinity such that (4.1) holds for some integer N, the number of random walks on the m-circle from η to λ with exactly k standard steps, such that at no time two coordinates of a point on the random walk are equal, is asymptotically

\frac{2^{n^{2}−n+1}}{nm^{n}} (2\begin{matrix}sin & \frac{nπ}{m} \\ sin & \frac{π}{m} \end{matrix} )^{k} ∏_{1≤h<t≤n} (sin \frac{π(η_{h} − η_{t})}{m} · |sin \frac{π(λ_{h} − λ_{t})}{m}|)  (4.2)

if n is even, and as well if n is odd and m is even, and it is asymptotically

\frac{2^{n^{2}−n}}{nm^{n}} (2\begin{matrix}sin & \frac{nπ}{m} \\ sin & \frac{π}{m} \end{matrix} )^{k} ∏_{1≤h<t≤n} (sin \frac{π(η_{h} − η_{t})}{m} · |sin \frac{π(λ_{h} − λ_{t})}{m}|)  (4.3)

if both n and m are odd.

Proof. We have to determine the asymptotic behaviour of the coefficient of x^{k}/k! in (2.8). We expand the determinant by linearity in the rows and obtain

\frac{1}{nm^{n}} ∑_{u=0}^{n−1} ∑_{r_{1},...,r_{n}=0}^{m−1} exp (2x ∑_{j=1}^{n} cos(2π(u + nr_{j})/mn))

· e^{−2πius/n} det_{1≤h,t≤n} (e^{−2πi(u+nr_{h})(λ_{t}−η_{h})/mn}). (4.4)

In this expression, we have to extract the coefficient of x^{k}/k! to obtain the number of walks with exactly k steps. The expression that we obtain is a finite sum of the form ∑_{ℓ} c_{ℓ}b_{ℓ}^{k}, with the b_{ℓ}’s of the form 2 ∑_{j=1}^{n} cos(2π(u + nr_{j})/mn), and both the c_{ℓ}’s and b_{ℓ}’s independent of k. Also in (4.4) there is the constraint that we must have r_{h} ≠ r_{t} for h ≠ t in order to obtain a non-vanishing summand, because otherwise the determinant is zero. It is not difficult to see that, because of that, the maximal modulus of such a b_{ℓ} is equal to

2 ∑_{j=−n/2}^{n/2−1} cos \frac{(2j + 1)π}{m} = 4 ∑_{j=0}^{n/2−1} cos \frac{(2j + 1)π}{m}

if n is even, and is equal to

2 ∑_{j=−(n−1)/2}^{(n−1)/2} cos \frac{2jπ}{m} = 2 + 4 ∑_{j=1}^{(n−1)/2} cos \frac{2jπ}{m}

if n is odd. Since we have

2 ∑_{j=0}^{n/2−1} cos \frac{(2j + 1)π}{m} = \begin{matrix}sin & \frac{nπ}{m} \\ sin & \frac{π}{m} \end{matrix}

and also

1 + 2 ∑_{j=1}^{(n−1)/2} cos \frac{2jπ}{m} = \begin{matrix}sin & \frac{nπ}{m} \\ sin & \frac{π}{m} \end{matrix} ,

this maximal modulus is 2 sin(nπ/m)/ sin(π/m) in both cases.

If n is even, then the maximal modulus is attained by choosing

(e1) u = n/2 and \{r_{1}, r_{2}, …, r_{n}\} = \{0, 1, …, \frac{n}{2}−1, m−\frac{n}{2}, …, m−2, m−1\}, regardless

of m; (e2) u = n/2 and \{r_{1}, r_{2}, …, r_{n}\} = \{\frac{m}{2} − \frac{n}{2}, \frac{m}{2} − \frac{n}{2} + 1, …, \frac{m}{2} + \frac{n}{2} − 1\} if m is even; (e3) u = 0 and \{r_{1}, r_{2}, …, r_{n}\} = \{\frac{m+1}{2} − \frac{n}{2}, \frac{m+1}{2} − \frac{n}{2} + 1, …, \frac{m+1}{2} + \frac{n}{2} − 1\} if m is

odd. If n is odd, then the maximal modulus is attained by choosing (o1) u = 0 and \{r_{1}, r_{2}, …, r_{n}\} = \{0, 1, …, \frac{n−1}{2}, m− \frac{n−1}{2}, …, m−2, m−1\}, regardless

of m; (o2) u = 0 and \{r_{1}, r_{2}, …, r_{n}\} = \{\frac{m}{2} − \frac{n−1}{2}, \frac{m}{2} − \frac{n−1}{2} + 1, …, \frac{m}{2} + \frac{n−1}{2}\} if m is even.

(If both n and m are odd, then there is no additional choice beyond (o1).) Case (e1) yields the contribution

\frac{(−1)^{s}}{nm^{n}} (2\begin{matrix}sin & \frac{nπ}{m} \\ sin & \frac{π}{m} \end{matrix} )^{k} det_{1≤h,t≤n} (e^{−2πi((n+1)/2−h)λ_{t}/m}) det_{1≤h,t≤n} (e^{2πi((n+1)/2−t)η_{h}/m})

to the asymptotics of (2.8). The two determinants are again Vandermonde-type determinants. They are therefore easily evaluated. Thus we obtain

\frac{2^{n^{2}−n}}{nm^{n}} (2\begin{matrix}sin & \frac{nπ}{m} \\ sin & \frac{π}{m} \end{matrix} )^{k} ∏_{1≤h<t≤n} (sin \frac{π(η_{t} − η_{h})}{m} · |sin \frac{π(λ_{t} − λ_{h})}{m}|)  (4.5)

as the contribution of Case (e1) to the asymptotics. As similar computations show, the contributions of Cases (e2), (e3), (o1) and (o2) are also equal to (4.5). The claims (4.2) and (4.3) follow now upon adding up the corresponding terms in each case. □

Now we are in the position to address the question of determining the asymptotic behaviour of the number of all walks on the m-circle which start in a given point and proceed for k standard steps (and, thus, at the same time for the number of n noncolliding particles on the circle in the random turns vicious walker model). Clearly, this amounts to summing the expressions given in Theorem 17 over all possible λ. Again, this is just a finite sum, with the number of terms bounded by nm^{m}, a quantity which is independent of k. Thus, it is obvious that the order of magnitude of the number of all these walks is (2sin \frac{nπ}{m}/sin \frac{π}{m})^{k}. In order to determine the multiplicative constant, we have to make use of identities featuring Schur functions and odd orthogonal characters.

Theorem 18. Let m be a positive integer. Furthermore, let η = (η_{1}, η_{2}, …, η_{n}) be a vector of integers with m > η_{1} > η_{2} > ⋯ > η_{n} ≥ 0. Then, as k tends to infinity, the number of random walks on the m-circle which start at η and proceed for exactly k standard steps, such that at no time two coordinates of a point on the random walk are equal, is asymptotically

\frac{2^{\binom{n}{2}}}{m^{n/2}} (2\begin{matrix}sin & \frac{nπ}{m} \\ sin & \frac{π}{m} \end{matrix} )^{k} ∏_{1≤h<t≤n} (sin \frac{π(η_{h} − η_{t})}{m})

× (∏_{h=1}^{n/2} cot \frac{(2h − 1)π}{2m} + (−1)^{|η|+k+ \frac{n}{2}} ∏_{h=1}^{n/2} tan \frac{(2h − 1)π}{2m} ) (4.6)

if both n and m are even, it is asymptotically

\frac{2^{\binom{n}{2}}}{m^{n/2}} (2\begin{matrix}sin & \frac{nπ}{m} \\ sin & \frac{π}{m} \end{matrix} )^{k} ∏_{1≤h<t≤n} (sin \frac{π(η_{h} − η_{t})}{m}) ∏_{h=1}^{n/2} cot \frac{(2h − 1)π}{2m}  (4.7)

if n is even and m is odd, and it is asymptotically

\frac{2^{\binom{n}{2}}}{m^{(n−1)/2}} (2\begin{matrix}sin & \frac{nπ}{m} \\ sin & \frac{π}{m} \end{matrix} )^{k} ∏_{1≤h<t≤n} (sin \frac{π(η_{h} − η_{t})}{m}) ∏_{h=1}^{(n−1)/2} cot \frac{hπ}{m}  (4.8)

if n is odd (regardless of m).

Proof. As we already explained, in view of Theorem 17, we have to compute the sum of (4.2), respectively of (4.3), over all possible choices (λ_{1}, λ_{2}, …, λ_{n}). That is, we have to sum these expressions over all integers λ_{1}, λ_{2}, …, λ_{n} with m > λ_{1} > λ_{2} > ⋯ > λ_{n} ≥ 0, and its cyclic permutations, such that walks are possible from η to λ = (λ_{1}, λ_{2}, …, λ_{n}), respectively to its cyclic permutations, in k steps. What the latter means, was discussed in the paragraph containing (4.1). Thus, we have to distinguish between several cases.

Before we list these cases, the reader should observe that the expressions (4.2) and (4.3) are invariant under cyclic permutations of λ_{1}, λ_{2}, …, λ_{n}. Thus, as the first step, we will multiply them by n, respectively by n/2, depending on whether any cyclic permutation of λ can be reached from η in k steps, or only every second.

Now, if both m and n are odd, then any λ, and any cyclic permutation of it can be reached from η if k is large enough. Thus, in this case, we have to multiply the expression (4.3) by n, and subsequently sum it over all integers λ_{1}, λ_{2}, …, λ_{n} with m > λ_{1} > λ_{2} > ⋯ > λ_{n} ≥ 0.

If m is odd but n is even, then only every second cyclic permutation of a given λ can be reached from η for a given k which is large enough. Thus, in this case, we have to multiply the expression (4.2) by n/2, and subsequently sum it over all integers λ_{1}, λ_{2}, …, λ_{n} with m > λ_{1} > λ_{2} > ⋯ > λ_{n} ≥ 0.

On the other hand, if m is even, then (4.1) implies that |λ| ≡ |η| + k mod 2. In particular, given a λ satisfying this condition, every cyclic permutation of λ can be reached from η for a given k which is large enough. Thus, in this case, we have to multiply the expression (4.2) by n, and subsequently sum it over all integers λ_{1}, λ_{2}, …, λ_{n} with m > λ_{1} > λ_{2} > ⋯ > λ_{n} ≥ 0 and |λ| ≡ |η| + k mod 2.

In summary, what we need is, on the one hand, the sum

S_{1} = ∑_{m>λ_{1}>···>λ_{n}≥0} ∏_{1≤h<t≤n} sin \frac{π(λ_{h} − λ_{t})}{m},  (4.9)

and, on the other hand, the same sum (4.9), but restricted to those λ for which |λ| ≡ |η| + k mod 2. The latter sum is equal to \frac{1}{2}(S_{1} + (−1)^{|η|+k}S_{2}), with

S_{2} = ∑_{m>λ_{1}>···>λ_{n}≥0}(−1)^{|λ|} ∏_{1≤h<t≤n} sin \frac{π(λ_{h} − λ_{t})}{m}.  (4.10)

We may again write the summand in the sums S_{1} and S_{2} as a Vandermonde-type determinant, so that (4.9) becomes

(2i)^{−\binom{n}{2}} ∑_{m>λ_{1}>···>λ_{n}≥0} det_{1≤h,t≤n} (e^{2πi((n+1)/2−h)λ_{t}/m}),

and the expression (4.10) becomes

(2i)^{−\binom{n}{2}} ∑_{m>λ_{1}>···>λ_{n}≥0} det_{1≤h,t≤n} (( − e^{2πi((n+1)/2−h)/m})^{λ_{t}}).

Upon replacing λ_{j} by λ_{j} + n − j, these expressions are transformed to

(2i)^{−\binom{n}{2}} ∑_{m−n≥λ_{1}≥···≥λ_{n}≥0} det_{1≤h,t≤n} ((e^{2πi((n+1)/2−h)/m})^{λ_{t}+n−t}),  (4.11)

respectively

(2i)^{−\binom{n}{2}} ∑_{m−n≥λ_{1}≥···≥λ_{n}≥0} det_{1≤h,t≤n} (( − e^{2πi((n+1)/2−h)/m})^{λ_{t}+n−t}).  (4.12)

We may rewrite the latter two determinants using Schur functions (see (C.31) for the definition). Using this notation, we may write (4.11) and (4.12) in the form

(2i)^{−\binom{n}{2}} det_{1≤h,t≤n} ((εq^{n+1−2h})^{n−t}) ∑_{m−n≥λ_{1}≥···≥λ_{n}≥0} s_{λ}(εq^{n−1}, εq^{n−3}, …, εq^{−n+3}, εq^{−n+1}),

(4.13)

where q = e^{πi/m}, with ε = 1 to yield equality with (4.11), and ε = −1 to yield equality with (4.12). The determinant in this expression is a Vandermonde determinant, and is therefore easily evaluated. The sum over Schur functions, on the other hand, can be evaluated by means of (see [28, proof of Theorem 2] for a discussion of this identity, with references to various proofs)

∑_{p≥λ_{1}≥···≥λ_{n}≥0} s_{λ}(x_{1}, x_{2}, …, x_{n}) = (x_{1}x_{2} ⋯ x_{n})^{p/2} so_{((p/2)^{n})}^{odd}(x_{1}, x_{2}, …, x_{n}), (4.14)

where so_{λ}^{odd}(x_{1}, x_{2}, …, x_{n}) is an odd orthogonal character. (See (C.1) for the definition. The notation ((p/2)^{n}) in (4.14) means a vector of n components, all of them equal to p/2.) Thus, the expression (4.13) becomes

(ε^{\binom{n}{2}} ∏_{1≤h<t≤n} sin \frac{π(t − h)}{m})ε^{n(m−n)/2} so_{(((m−n)/2)^{n})}^{odd}(εq^{n−1}, εq^{n−3}, …, εq^{−n+3}, εq^{−n+1}).

(4.15)

If ε = 1, the odd orthogonal character, specialized in this manner, is evaluated in Lemma C1, while for ε = −1 this is done in Lemma C2. If the results are substituted in (4.15), which, as we argued above, is in fact equal to the sum (4.9), respectively to (4.10), the sums that we wanted to evaluate, then the claims (4.6)–(4.8) follow after some further straight-forward (but tedious) calculations. □

The next two results address the asymptotic behaviour of the number of walks on the m-circle consisting of diagonal steps. As we discussed in Section 2, these theorems give at the same time the asymptotic behaviour of n non-colliding particles on the circle in the lock-step vicious walker model. As before, we begin with the result which described the asymptotic behaviour of the number of walks with fixed starting and end point.

Theorem 19. Let m be a positive integer. Furthermore, let η = (η_{1}, η_{2}, …, η_{n}) be a vector of integers or of half-integers with m > η_{1} > η_{2} > ⋯ > η_{n} ≥ 0, and let λ = (λ_{1}, λ_{2}, …, λ_{n}) be a vector of integers or of half-integers with m > λ_{s+1} > ⋯ > λ_{n} > λ_{1} > ⋯ > λ_{s} ≥ 0, for some s. Then, as k tends to infinity such that k ≡ 2η_{j} +2λ_{j} mod 2, the number of random walks on the m-circle from η to λ with exactly k diagonal steps, such that at no time two coordinates of a point on the random walk are equal, is asymptotically

\frac{2^{n^{2}−n}}{nm^{n}} (2^{n} ∏_{j=1}^{n} cos \frac{π (j − \frac{n+1}{2})}{m})^{k} ∏_{1≤h<t≤n} (sin \frac{π(η_{h} − η_{t})}{m} · |sin \frac{π(λ_{h} − λ_{t})}{m}|) . (4.16)

Proof. Clearly, this time we want to determine the asymptotic behaviour of the expression (2.9). First of all, if k ≡ 2η_{j} + λ_{j} mod 2, i.e., in the case where k, the η_{j}’s and the λ_{j}’s are chosen so that walks exist, then we may replace the sum over r in (2.9) by twice the sum of the same summand, but where r runs from 0 to m − 1 (instead of 2m − 1; here it is important that m is an integer). Then we expand again the determinant by linearity, and obtain

\frac{1}{nm^{n}} ∑_{u=0}^{n−1} ∑_{r_{1},...,r_{n}=0}^{m−1} ∏_{j=1}^{n} (2 cos(π(u + nr_{j})/mn))^{k}

· e^{−2πius/n} det_{1≤h,t≤n} (e^{−2πi(u+nr_{h})(λ_{t}−η_{h})/mn}). (4.17)

This is again a finite sum of the form ∑_{ℓ} c_{ℓ}b_{ℓ}^{k}, with the b_{ℓ}’s of the form 2^{n} ∏_{j=1}^{n} cos(π(u+ nr_{j})/mn), and both the c_{ℓ}’s and b_{ℓ}’s independent of k. Also in (4.17) there is the constraint that we must have r_{h} ≠ r_{t} for h ≠ t in order to obtain a non-vanishing summand, because otherwise the determinant is zero. Because of that, one discovers that in order to have b_{ℓ} with maximal modulus we must have u = n/2 and

\{r_{1}, r_{2}, …, r_{n}\} = \{0, 1, …, \frac{n}{2} − 1, m − \frac{n}{2}, …, m − 2, m − 1\}

if n is even, and we must have u = 0 and

\{r_{1}, r_{2}, …, r_{n}\} = \{0, 1, …, \frac{n−1}{2}, m − \frac{n−1}{2}, …, m − 2, m − 1\}

if n is odd.

Let first n be even. Then we obtain for the asymptotics of (4.17) the expression

\frac{(−1)^{s}}{nm^{n}} (2^{n} ∏_{j=1}^{n} cos \frac{π (j − \frac{n+1}{2})}{m})^{k}

× det_{1≤h,t≤n} (e^{−2πi((n+1)/2−h)λ_{t}/m}) det_{1≤h,t≤n} (e^{2πi((n+1)/2−t)η_{h}/m}).

Both determinants are essentially Vandermonde determinants and are therefore easily evaluated. The result is exactly (4.16).

If n is odd then we obtain for the asymptotics of (4.17) the expression

\frac{1}{nm^{n}} (2^{n} ∏_{j=1}^{n} cos \frac{π (j − \frac{n+1}{2})}{m})^{k} det_{1≤h,t≤n} (e^{−2πi((n+1)/2−h)λ_{t}/m}) det_{1≤h,t≤n} (e^{2πi((n+1)/2−t)η_{h}/m}),

which becomes again (4.16) if the Vandermonde-type determinants are evaluated. □

By summation of the corresponding expressions, the previous result allows us now to derive the asymptotic behaviour of walks with a fixed starting point but with arbitrary end point. The sums that need to be carried out are equivalent to some of those that we already evaluated in the proof of Theorem 18.

Theorem 20. Let m be a positive integer. Furthermore, let η = (η_{1}, η_{2}, …, η_{n}) be a vector of integers or of half-integers with m > η_{1} > η_{2} > ⋯ > η_{n} ≥ 0. Then, as k tends to infinity, the number of random walks on the m-circle which start at η and proceed for exactly k diagonal steps, such that at no time two coordinates of a point on the random walk are equal, is asymptotically

\frac{2^{\binom{n}{2}}}{m^{n/2}} (2^{n} ∏_{j=1}^{n} cos \frac{π (j − \frac{n+1}{2})}{m})^{k} ∏_{1≤h<t≤n} (sin \frac{π(η_{h} − η_{t})}{m}) ∏_{h=1}^{n/2} cot \frac{(2h − 1)π}{2m} (4.18)

if n is even, and it is asymptotically

\frac{2^{\binom{n}{2}}}{m^{(n−1)/2}} (2^{n} ∏_{j=1}^{n} cos \frac{π (j − \frac{n+1}{2})}{m})^{k} ∏_{1≤h<t≤n} (sin \frac{π(η_{h} − η_{t})}{m}) ∏_{h=1}^{(n−1)/2} cot \frac{hπ}{m} (4.19)

if n is odd.

Proof. Clearly, we have to sum (4.16) over all possible choices of λ. Depending on the parity of k + η_{j}, this means to take the sum over all integers λ_{1}, λ_{2}, …, λ_{n} with m > λ_{1} > λ_{2} > ⋯ > λ_{n} ≥ 0, or over all half-integers with the same property, and over all their cyclic permutations. Since the expression (4.16) is independent of s, every cyclic permutation yields the same value. Therefore we have to multiply this expression by n, and subsequently sum it over all integers, respectively half-integers, λ_{1}, λ_{2}, …, λ_{n} with m > λ_{1} > λ_{2} > ⋯ > λ_{n} ≥ 0.

So, what we need is the sum

∑_{m>λ_{1}>···>λ_{n}≥0} ′  ∏_{1≤h<t≤n} sin \frac{π(λ_{h} − λ_{t})}{m}

= (2i)^{−\binom{n}{2}}  ∑_{m>λ_{1}>···>λ_{n}≥0} ′  det_{1≤h,t≤n} (e^{2πi((n+1)/2−h)λ_{t}/m}), (4.20)

where the sum ∑ ′ is over all integral λ_{1}, λ_{2}, …, λ_{n}, but also the sum (4.20) where ∑ ′ is restricted to half-integral λ_{1}, λ_{2}, …, λ_{n}. (The equality of the two expressions in (4.20) follows again from the Vandermonde determinant evaluation.)

The sum over all integers λ_{1}, λ_{2}, …, λ_{n} has already been evaluated in the proof of Theorem 18, when we evaluated S_{1}. If we want to form the sum (4.20) over all halfintegers λ_{1}, λ_{2}, …, λ_{n}, then we may replace λ_{j} by λ_{j} + n − j + \frac{1}{2}, and rewrite it as

(2i)^{−\binom{n}{2}} ∑_{m−n≥λ_{1}≥···≥λ_{n}≥0} det_{1≤h,t≤n} (e^{2πi((n+1)/2−h)(λ_{t}+n−t+1/2)/m})

= (2i)^{−\binom{n}{2}} ∑_{m−n≥λ_{1}≥···≥λ_{n}≥0} det_{1≤h,t≤n} (e^{2πi((n+1)/2−h)(λ_{t}+n−t)/m}),

where the sums are now over integral λ_{1}, λ_{2}, …, λ_{n}. The sum in the last line is exactly the same sum as (4.11), which is in turn equal to S_{1} (and, thus, also to the sum (4.20) when ∑ ′ is taken over all integral λ_{1}, λ_{2}, …, λ_{n}). Therefore, regardless of the parity of k, the result is the same. Since the evaluation of S_{1} in the proof of Theorem 18 yielded two different expressions depending on whether n is even or odd, we obtain the two cases in the statement of the theorem. □

5. Asymptotics for random walks in alcoves of type C̃

The subject of this section is the determination of the asymptotic behaviour of the number of walks from a given starting point to a given end point which stay in the alcove AC̃_{n} ; m of type C̃_{n} as the number of steps becomes large, as well as the asymptotic behaviour of the number of those walks which start at a given point but may terminate anywhere. In technical terms, we determine the asymptotic behaviour of the expressions given by Theorems 6 and 7 as k becomes large, and as well if these expressions are summed over all possible end points of the walks. In fact, for Theorem 7, i.e., for the case of diagonal steps, this had already been carried out in [29], so that we only copy the corresponding results for the sake of completeness; see Theorems 23 and 24 below.

Before, however, we address the case of standard steps. As we discussed in Section 2, this case is also equivalent to the movements of n non-colliding particles in an interval according to the random turns vicious walker model. We begin, as usual, with the corresponding results when starting and end point are fixed.

Theorem 21. Let m be a positive integer. Furthermore, let η = (η_{1}, η_{2}, …, η_{n}) and λ = (λ_{1}, λ_{2}, …, λ_{n}) be vectors of integers in the alcove AC̃_{n} ; m of type C̃_{n} (defined in (2.2)). Then, as k tends to infinity such that k ≡ |η| + |λ| mod 2, the number of random walks from η to λ with exactly k standard steps, which stay in the alcove AC̃_{n} ; m, is asymptotically

\frac{2^{2n^{2}−n+1}}{m^{n}} (\frac{2 sin \frac{nπ}{2m} cos \frac{(n+1)π}{2m}}{sin \frac{π}{2m}} )^{k} ∏_{1≤h<t≤n} (sin \frac{π(η_{h} − η_{t})}{2m} · sin \frac{π(λ_{h} − λ_{t})}{2m})

× ∏_{1≤h≤t≤n} (sin \frac{π(η_{h} + η_{t})}{2m} · sin \frac{π(λ_{h} + λ_{t})}{2m}) . (5.1)

Proof. We have to determine the asymptotic behaviour of the coefficient of x^{k}/k! in (2.11). If we expand expression (2.11), i.e., if we use linearity of the determinant in the rows, then we obtain the expression

\frac{1}{m^{n}} ∑_{r_{1},...,r_{n}=1}^{2m−1} exp (2x ∑_{j=1}^{n} cos(πr_{j}/m))( ∏_{j=1}^{n} sin(πr_{j}η_{j}/m))

· det_{1≤h,t≤n} ( sin(πr_{h}λ_{t}/m)). (5.2)

As we said, in this expression we have to extract the coefficient of x^{k}/k! to obtain the number of walks with exactly k steps. The expression that we obtain is a finite sum of the form ∑_{ℓ} c_{ℓ}b_{ℓ}^{k}, with the b_{ℓ}’s of the form 2 ∑_{j=1}^{n} cos(πr_{j}/m), and both the c_{ℓ}’s and b_{ℓ}’s independent of k.

Finding the asymptotics of (5.2) means to find the b_{ℓ}’s with largest modulus. This, in turn, means to choose the parameters r_{h} either close to the lower limit of the summation, 1, respectively close to the upper limit, 2m − 1, or close to m − 1, respectively close to m + 1. (In the first case, all the cosines cos(πr_{h}/m) will be close to 1, whereas in the second case all of them will be close to −1. If some r_{j} is equal to m, then the corresponding term vanishes because of the expression sin(πr_{j}η_{j}/m) occurring in the summand.) There are again restrictions however: if r_{h} = r_{t} for h ≠ t then the determinant in (5.2) vanishes, as well as if r_{h} = 2m − r_{t} for some h and t. Therefore we may restrict ourselves to the cases where r_{h} ≠ r_{t} and r_{h} ≠ 2m − r_{t} for all h and t.

Hence, we will choose the set {r_{1}, r_{2}, …, r_{n}} either from

{1, 2, …, n, 2m − n, …, 2m − 2, 2m − 1}

or from

{m − n, …, m − 2, m − 1, m + 1, m + 2, …, m + n},

in such a way that the r_{j}’s are distinct and r_{h} ≠ 2m − r_{t} for all h and t. Clearly, there are 2^{n} sets of the first type, and 2^{n} sets as well of the second type. As is not difficult to see, for each fixed set, the sum of the corresponding terms c_{ℓ}b_{ℓ}^{k} is equal to

\frac{1}{m^{n}} (2 ∑_{j=1}^{n} cos(πj/m))^{k} · det_{1≤h,t≤n} ( sin(πtη_{h}/m)) · det_{1≤h,t≤n} ( sin(πhλ_{t}/m))  (5.3)

in both cases.

Now we have

∑_{j=1}^{n} cos(πj/m) = \frac{sin(nπ/2m) cos((n + 1)π/2m)}{sin(π/2m)}

and

det_{1≤h,t≤n} ( sin(πhλ_{t}/m)) = 2^{n^{2}−n} ∏_{1≤h<t≤n} sin \frac{π(λ_{h} − λ_{t})}{2m} ∏_{1≤h≤t≤n} sin \frac{π(λ_{h} + λ_{t})}{2m}. (5.4)

The latter identity follows from writing the determinant as

det_{1≤h,t≤n} ( sin(πhλ_{t}/m)) = (2i)^{−n} det_{1≤h,t≤n} (e^{πihλ_{t}/m} − e^{−πihλ_{t}/m}),

and evaluating it by means of (B.4). Substituting this in (5.3), and multiplying the resulting expression by 2 · 2^{n} = 2^{n+1} (the number of these sets {r_{1}, r_{2}, …, r_{n}}), we obtain (5.1). □

Having accomplished the asymptotic analysis of the walks with fixed starting and end point, we can now turn to the analysis of the walks with fixed starting point but arbitrary end point. Again, this amounts to a summation problem, namely summing expression (5.1) over all possible λ. To carry out this task, we make use of identities featuring Schur functions and symplectic characters.

Theorem 22. Let m be a positive integer. Furthermore, let η = (η_{1}, η_{2}, …, η_{n}) be a vector of integers in the alcove AC̃_{n} ; m of type C̃_{n} (defined in (2.2)). Then, as k tends to infinity, the number of random walks which start at η and proceed for exactly k standard steps, which stay in the alcove AC̃_{n} ; m, is asymptotically

\frac{2^{2n^{2}−n}}{m^{n}} (\frac{2 sin \frac{nπ}{2m} cos \frac{(n+1)π}{2m}}{sin \frac{π}{2m}} )^{k} ∏_{1≤h<t≤n} (sin \frac{π(η_{h} − η_{t})}{2m} · sin \frac{π(t − h)}{2m})

× ∏_{1≤h≤t≤n} (sin \frac{π(η_{t} + η_{h})}{2m} · sin \frac{π(t + h)}{2m})

× (∏_{h=0}^{n} ∏_{t=1}^{n} \begin{matrix}sin & \frac{π(t−h+m−1)}{2m} \\ sin & \frac{π(t−h+n)}{2m} \end{matrix} + (−1)^{|η|+k+ \frac{n}{2}} 2^{−n^{2}} \begin{matrix}∏_{h=1}^{n/2} & tan & 2 & \frac{(2h−1)π}{2m} \\ ∏_{h=1}^{n+1} & ∏_{t=1}^{n} & |sin \frac{(2t−2h+1)π}{2m} | \end{matrix}) (5.5)

if both m and n are even, it is asymptotically

\frac{2^{2n^{2}−n}}{m^{n}} (\frac{2 sin \frac{nπ}{2m} cos \frac{(n+1)π}{2m}}{sin \frac{π}{2m}} )^{k} ∏_{1≤h<t≤n} (sin \frac{π(η_{h} − η_{t})}{2m} · sin \frac{π(t − h)}{2m})

× ∏_{1≤h≤t≤n} (sin \frac{π(η_{t} + η_{h})}{2m} · sin \frac{π(t + h)}{2m})

× (∏_{h=0}^{n} ∏_{t=1}^{n} \begin{matrix}sin & \frac{π(t−h+m−1)}{2m} \\ sin & \frac{π(t−h+n)}{2m} \end{matrix} + (−1)^{|η|+k+ \frac{n+1}{2}} 2^{−n^{2}} \begin{matrix}∏_{h=1}^{(n+1)/2} & tan & 2 & \frac{(2h−1)π}{2m} \\ ∏_{h=1}^{n+1} & ∏_{t=1}^{n} & |sin \frac{(2t−2h+1)π}{2m} | \end{matrix}) (5.6)

if m is even and n is odd, it is asymptotically

\frac{2^{2n^{2}−n}}{m^{n}} (\frac{2 sin \frac{nπ}{2m} cos \frac{(n+1)π}{2m}}{sin \frac{π}{2m}} )^{k} ∏_{1≤h<t≤n} (sin \frac{π(η_{h} − η_{t})}{2m} · sin \frac{π(t − h)}{2m})

× ∏_{1≤h≤t≤n} (sin \frac{π(η_{t} + η_{h})}{2m} · sin \frac{π(t + h)}{2m})

× (∏_{h=0}^{n} ∏_{t=1}^{n} \begin{matrix}sin & \frac{π(t−h+m−1)}{2m} \\ sin & \frac{π(t−h+n)}{2m} \end{matrix} + (−1)^{|η|+k+ \frac{n}{2}} 2^{−n^{2}} ∏_{h=1}^{n/2} \begin{matrix}sin & 2 & \frac{(2h−1)π}{2m} \\ cos & 2 & \frac{hπ}{m} \end{matrix} \frac{1}{∏_{h=1}^{n+1} ∏_{t=1}^{n} |sin \frac{(2t−2h+1)π}{2m} |}) (5.7)

if m is odd and n is even, and it is asymptotically

\frac{2^{2n^{2}−n}}{m^{n}} (\frac{2 sin \frac{nπ}{2m} cos \frac{(n+1)π}{2m}}{sin \frac{π}{2m}} )^{k} ∏_{1≤h<t≤n} (sin \frac{π(η_{h} − η_{t})}{2m} · sin \frac{π(t − h)}{2m})

× ∏_{1≤h≤t≤n} (sin \frac{π(η_{t} + η_{h})}{2m} · sin \frac{π(t + h)}{2m}) (∏_{h=0}^{n} ∏_{t=1}^{n} \begin{matrix}sin & \frac{π(t−h+m−1)}{2m} \\ sin & \frac{π(t−h+n)}{2m} \end{matrix} ) (5.8)

if both m and n are odd.

Proof. As we already observed, in view of Theorem 21, we have to carry out the sum of (5.1) over all possible λ, i.e., over all m > λ_{1} > λ_{2} > ⋯ > λ_{n} > 0, where |λ| ≡ k + |η| mod 2. Leaving away factors which are independent of λ, the problem is to compute the sum

∑_{m>λ_{1}>···>λ_{n}>0} ′  ∏_{1≤h<t≤n} (sin \frac{π(λ_{h} − λ_{t})}{2m}) ∏_{1≤h≤t≤n} (sin \frac{π(λ_{h} + λ_{t})}{2m}) ,  (5.9)

where the sum ∑ ′ is either restricted to those λ for which |λ| is even, or to those for which |λ| is odd, depending on the parity of k + |η|. This task will be accomplished if we are able to evaluate the (complete) sum

T_{1} = ∑_{m>λ_{1}>···>λ_{n}>0} ∏_{1≤h<t≤n} (sin \frac{π(λ_{h} − λ_{t})}{2m}) ∏_{1≤h≤t≤n} (sin \frac{π(λ_{h} + λ_{t})}{2m}) ,

and its “signed variant”

T_{2} = ∑_{m>λ_{1}>···>λ_{n}>0}(−1)^{|λ|} ∏_{1≤h<t≤n} (sin \frac{π(λ_{h} − λ_{t})}{2m}) ∏_{1≤h≤t≤n} (sin \frac{π(λ_{h} + λ_{t})}{2m}) .

The sum (5.9) is then equal to \frac{1}{2}(T_{1} + T_{2}) if the sum ∑ ′ is restricted to the λ’s for which |λ| is even, and it is equal to \frac{1}{2}(T_{1} − T_{2}) if the sum ∑ ′ is restricted to the λ’s for which |λ| is odd.

A sum equivalent to T_{1} had already been evaluated in [29, first part of the proof of Theorem 6]. The result is

T_{1} = ∏_{1≤h<t≤n} sin \frac{π(t − h)}{2m} ∏_{1≤h≤t≤n} sin \frac{π(t + h)}{2m} ∏_{h=0}^{n} ∏_{t=1}^{n} \begin{matrix}sin & \frac{π(t−h+m−1)}{2m} \\ sin & \frac{π(t−h+n)}{2m} \end{matrix} .

In order to evaluate T_{2}, we proceed in a manner similar to the evaluation in [29]. By means of (5.4), we may rewrite T_{2} as

∑_{m>λ_{1}>···>λ_{n}>0}(−1)^{|λ|}2^{−n^{2}+n} det_{1≤h,t≤n} ( sin(πhλ_{t}/m))

= ∑_{m>λ_{1}>···>λ_{n}>0} \frac{1}{2^{n^{2}}i^{n}} det_{1≤h,t≤n} ((−e^{πih/m})^{λ_{t}} − (−e^{πih/m})^{−λ_{t}}).

Replacing λ_{j} by λ_{j} + n − j + 1, j = 1, 2, …, n, we obtain the expression

∑_{m−n−1≥λ_{1}≥···≥λ_{n}≥0} \frac{1}{2^{n^{2}}i^{n}} det_{1≤h,t≤n} ((−e^{πih/m})^{λ_{t}+n−t+1} − (−e^{πih/m})^{−(λ_{t}+n−t+1)}). (5.10)

This determinant can be expressed in terms of a symplectic character. Given a partition λ = (λ_{1}, λ_{2}, …, λ_{n}) (i.e., a non-increasing sequence of non-negative integers), the symplectic character sp_{λ}(x_{1}, x_{2}, …, x_{n}) is defined by (see [13, (24.18)])

sp_{λ}(x_{1}, x_{2}, …, x_{n}) = \frac{det_{1≤h,t≤n}(x_{h}^{λ_{t}+n−t+1} − x_{h}^{−(λ_{t}+n−t+1)})}{det_{1≤h,t≤n}(x_{h}^{n−t+1} − x_{h}^{−(n−t+1)})}.  (5.11)

Therefore, writing again q for e^{πi/m}, the sum in (5.10) equals

\frac{1}{2^{n^{2}}i^{n}} det_{1≤h,t≤n} ((−q^{h})^{n−t+1} − (−q^{h})^{−(n−t+1)}) ∑_{m−n−1≥λ_{1}≥···≥λ_{n}≥0} sp_{λ}(−q, −q^{2}, …, −q^{n}).

(5.12)

Now we appeal to the formula (see [27, (3.4)]),

s_{(c^{r})}(x_{1}, x_{1}^{−1}, …, x_{n}, x_{n}^{−1}, 1) = ∑_{c≥ν_{1}≥ν_{2}≥···≥ν_{r}≥0} sp_{ν}(x_{1}, …, x_{n}),  (5.13)

which is valid for r ≤ n, where on the left-hand side we have again a Schur function (cf. (C.31) for the definition). The notation (c^{r}) is short for the vector in which the first r components are equal to c, followed by 2n + 1 − r components all of which are 0. Use of this formula in (5.12) yields the equivalent expression

\frac{1}{2^{n^{2}}i^{n}} det_{1≤h,t≤n} ((−q^{h})^{n−t+1} − (−q^{h})^{−(n−t+1)})
× s_{((m−n−1)^{n})}(−q^{n}, −q^{n−1}, …, −q, 1, −q^{−1}, …, −q^{−n+1}, −q^{n})

= (−1)^{(m−n−1)n} \frac{1}{2^{n^{2}}i^{n}} det_{1≤h,t≤n} ((−q^{h})^{n−t+1} − (−q^{h})^{−(n−t+1)})

× s_{((m−n−1)^{n})}(q^{n}, q^{n−1}, …, q, −1, q^{−1}, …, q^{−n+1}, q^{n}).

Clearly, the determinant is easily evaluated by means of (B.4). The specialized Schur function is evaluated in Lemma C3. If everything is combined and simplified, the claimed formulae (5.5)–(5.8) are eventually obtained. □

We conclude this section by reporting the results from [29] on the asymptotic behaviour of walks in the alcove AC̃_{n} ; m which consist entirely of diagonal steps. These results have been stated there in an equivalent form, namely in the language of walkers in the lock-step vicious walkers model, which are bounded by two walls.

Theorem 23 ([29, Theorem 4]). Let m be a positive integer or half-integer. Furthermore, let η = (η_{1}, η_{2}, …, η_{n}) and λ = (λ_{1}, λ_{2}, …, λ_{n}) be vectors of integers or of half-integers in the alcove AC̃_{n} ; m of type C̃_{n} (defined in (2.2)). Then, as k tends to infinity such that k ≡ 2η_{j} + 2λ_{j} mod 2, the number of random walks from η to λ with exactly k diagonal steps, which stay in the alcove AC̃_{n} ; m, is asymptotically

\frac{4^{n^{2}}}{(2m)^{n}} (2^{n} ∏_{j=1}^{n} cos \frac{jπ}{2m})^{k} ∏_{1≤h<t≤n} (sin \frac{π(η_{h} − η_{t})}{2m} · sin \frac{π(λ_{h} − λ_{t})}{2m})

× ∏_{1≤h≤t≤n} (sin \frac{π(η_{h} + η_{t})}{2m} · sin \frac{π(λ_{h} + λ_{t})}{2m}) . (5.14)

Theorem 24 ([29, Theorem 6]). Let m be a positive integer or half-integer. Furthermore, let η = (η_{1}, η_{2}, …, η_{n}) be a vector of integers or of half-integers in the alcove AC̃_{n} ; m of type C̃_{n} (defined in (2.2)). Then, as k tends to infinity, the number of random walks which start at η and proceed for exactly k diagonal steps, which stay in the alcove AC̃_{n} ; m, is asymptotically

\frac{4^{n^{2}}}{(2m)^{n}} (2^{n} ∏_{j=1}^{n} cos \frac{jπ}{2m})^{k} ∏_{1≤h<t≤n} (sin \frac{π(η_{h} − η_{t})}{2m} · sin \frac{π(t − h)}{2m})

× ∏_{1≤h≤t≤n} (sin \frac{π(η_{h} + η_{t})}{2m} · sin \frac{π(t + h)}{2m}) ∏_{h=0}^{n} ∏_{t=1}^{n} \begin{matrix}sin & \frac{π(t−h+⌊m⌋)}{2m} \\ sin & \frac{π(t−h+n)}{2m} \end{matrix}

× ∏_{h=1}^{n} \begin{matrix}sin & \frac{π(h+⌊m⌋−n)}{2m} \\ sin & \frac{π(2h+⌊m⌋−n)}{2m} \end{matrix} , (5.15)

if k + 2η_{j} is odd, and

\frac{4^{n^{2}}}{(2m)^{n}} (2^{n} ∏_{j=1}^{n} cos \frac{jπ}{2m})^{k} ∏_{1≤h<t≤n} (sin \frac{π(η_{h} − η_{t})}{2m} · sin \frac{π(t − h)}{2m})

× ∏_{1≤h≤t≤n} (sin \frac{π(η_{h} + η_{t})}{2m} · sin \frac{π(t + h)}{2m}) ∏_{h=0}^{n} ∏_{t=1}^{n} \begin{matrix}sin & \frac{π(t−h+⌈m−1⌉)}{2m} \\ sin & \frac{π(t−h+n)}{2m} \end{matrix} , (5.16)

if k + 2η_{j} is even.

6. Asymptotics for random walks in alcoves of type B̃

This section is devoted to finding the asymptotic behaviour of the number of walks from a given starting point to a given end point which stay in the alcove AB̃_{n} ; m of type B̃_{n} as the number of steps becomes large, as well as the asymptotic behaviour of the number of those walks which start at a given point but may terminate anywhere. In technical terms, we determine the asymptotic behaviour of the expressions given by Theorems 8 and 9 as k becomes large, and as well if these expressions are summed over all possible end points of the walks.

The following two theorems address the case of walks with standard steps. The result for fixed starting and end point is the subject of the first of the two.

Theorem 25. Let m be a positive integer or half-integer. Furthermore, let η = (η_{1}, η_{2}, …, η_{n}) and λ = (λ_{1}, λ_{2}, …, λ_{n}) be vectors of integers in the alcove AB̃_{n} ; m of type B̃_{n} (defined in (2.3)). Then, as k tends to infinity such that k ≡ |η| + |λ| mod 2, the number of random walks from η to λ with exactly k standard steps, which stay in the alcove AB̃_{n} ; m, is asymptotically

\frac{4^{n^{2}}}{(2m)^{n}} (\begin{matrix}sin & \frac{nπ}{m} \\ sin & \frac{π}{2m} \end{matrix} )^{k} ∏_{1≤h<t≤n} ( sin \frac{π(η_{h} − η_{t})}{2m} · sin \frac{π(λ_{h} − λ_{t})}{2m}

· sin \frac{π(η_{h} + η_{t})}{2m} · sin \frac{π(λ_{h} + λ_{t})}{2m}) ∏_{h=1}^{n} (sin \frac{πη_{h}}{2m} · sin \frac{πλ_{h}}{2m} ) . (6.1)

Proof. The analysis is analogous to the one in the proof of Theorem 21. Here, we have to estimate the coefficient of x^{k}/k! in (2.13). Expanding the two determinants in (2.13) by linearity in the rows, we obtain

\frac{1}{2m^{n}} ∑_{r_{1},...,r_{n}=1}^{2m−1} (2 ∑_{j=1}^{n} cos \frac{πr_{j}}{m} )^{k} (∑_{j=1}^{n} sin \frac{πr_{j}η_{j}}{m} ) det_{1≤h,t≤n} (sin \frac{πr_{h}λ_{t}}{m} )

+ \frac{1}{2m^{n}} ∑_{r_{1},...,r_{n}=0}^{2m−1} (2 ∑_{j=1}^{n} cos \frac{π(2r_{j} + 1)}{2m})^{k}

· (∏_{j=1}^{n} sin \frac{π(2r_{j} + 1)η_{j}}{2m} ) det_{1≤h,t≤n} (sin \frac{π(2r_{h} + 1)λ_{t}}{2m} ) .

Again, this is a finite sum of the form ∑_{ℓ} c_{ℓ}b_{ℓ}^{k}, with both the c_{ℓ}’s and b_{ℓ}’s independent of k.

We have to locate the b_{ℓ}’s with largest modulus. As can be seen in a manner similar to the considerations in the proof of Theorem 21, the b_{ℓ}’s with largest modulus come from the expansion of the second determinant, when we choose distinct r_{1}, r_{2}, …, r_{n} from

{0, 1, …, n − 1, 2m − n, 2m − n + 1, …, 2m − 1},

or from

{m − n, m − n + 1, …, m − 1, m, …, m + n − 1},

such that r_{h} ≠ 2m−1−r_{t} for all h and t. Clearly, there are 2·2^{n} such sets {r_{1}, r_{2}, …, r_{n}}. For each fixed set, the sum of the corresponding terms c_{ℓ}b_{ℓ}^{k} is equal to

\frac{1}{2m^{n}} (2 ∑_{j=1}^{n} cos \frac{π(2j − 1)}{2m})^{k} det_{1≤h,t≤n} (sin \frac{π(2h − 1)η_{t}}{2m} ) det_{1≤h,t≤n} (sin \frac{π(2h − 1)λ_{t}}{2m} ) .

(6.2)

We have on the one hand

∑_{j=1}^{n} cos \frac{π(2j − 1)}{2m} = \frac{sin \frac{nπ}{m}}{2 sin \frac{π}{2m}} .

On the other hand, the first determinant in (6.2) can be rewritten in the form

(2i)^{−n} det_{1≤h,t≤n} (e^{\frac{πi(2h−1)η_{t}}{2m}} − e^{− \frac{πi(2h−1)η_{t}}{2m}} ) ,

and can thus be evaluated by means of (B.2). After some simplification, the result is

det_{1≤h,t≤n} (sin \frac{π(2h − 1)η_{t}}{2m} )

= 2^{n^{2}−n} ∏_{1≤h<t≤n} ( sin \frac{π(η_{h} − η_{t})}{2m} · sin \frac{π(η_{h} + η_{t})}{2m}) ∏_{h=1}^{n} (sin \frac{πη_{h}}{2m} ) . (6.3)

Clearly, the second determinant in (6.2) is equal to the same expression with η_{j} replaced by λ_{j}, j = 1, 2, …, n. If all this is substituted in (6.2), and if the result is multiplied by 2 · 2^{n}, then we obtain exactly (6.1). □

If the starting point is fixed but the end point is not, we have the following result.

Theorem 26. Let m be a positive integer or half-integer. Furthermore, let η = (η_{1}, η_{2}, …, η_{n}) be a vector of integers in the alcove AB̃_{n} ; m of type B̃_{n} (defined in (2.3)). Then, as k tends to infinity, the number of random walks which start at η and proceed for exactly k standard steps, which stay in the alcove AB̃_{n} ; m, is asymptotically

\frac{4^{n^{2}}}{2(2m)^{n}} (\begin{matrix}sin & \frac{nπ}{m} \\ sin & \frac{π}{2m} \end{matrix} )^{k} ∏_{1≤h<t≤n} ( sin \frac{π(η_{h} − η_{t})}{2m} · sin \frac{π(t − h)}{2m}

· sin \frac{π(η_{h} + η_{t})}{2m} · sin \frac{π(t + h)}{2m}) ∏_{h=1}^{n} (sin \frac{πη_{h}}{2m} · sin \frac{πh}{2m})

× ( ∏_{h=1}^{2n} \begin{matrix}sin & \frac{π(m−n+h)}{4m} \\ sin & \frac{πh}{4m} \end{matrix} ∏_{h=1}^{n−1} ∏_{t=1}^{n} \begin{matrix}sin & \frac{π(m+t−h)}{2m} \\ sin & \frac{π(n+t−h)}{2m} \end{matrix} + ∏_{h=1}^{2n} \begin{matrix}sin & \frac{π(m−n+h−1)}{4m} \\ sin & \frac{πh}{4m} \end{matrix} ∏_{h=1}^{n−1} ∏_{t=1}^{n} \begin{matrix}sin & \frac{π(m+t−h−1)}{2m} \\ sin & \frac{π(n+t−h)}{2m} \end{matrix}

+ (−1)^{|η|+k+\binom{n+1}{2}} ∏_{h=1}^{n} \frac{cos \frac{π(m−n+2h−1)}{4m} · sin \frac{π(m−n+2h)}{4m}}{cos \frac{π(2h−1)}{4m} · sin \frac{πh}{2m}} ∏_{h=1}^{n−1} ∏_{t=1}^{n} \begin{matrix}sin & \frac{π(m+t−h)}{2m} \\ sin & \frac{π(n+t−h)}{2m} \end{matrix}

+ (−1)^{|η|+k+\binom{n}{2}} ∏_{h=1}^{n} \frac{sin \frac{π(m−n+2h−2)}{4m} · cos \frac{π(m−n+2h−1)}{4m}}{cos \frac{π(2h−1)}{4m} · sin \frac{πh}{2m}} ∏_{h=1}^{n−1} ∏_{t=1}^{n} \begin{matrix}sin & \frac{π(m+t−h−1)}{2m} \\ sin & \frac{π(n+t−h)}{2m} \end{matrix} ), (6.4)

if m is an integer with parity equal to that of n, it is asymptotically

\frac{4^{n^{2}}}{2(2m)^{n}} (\begin{matrix}sin & \frac{nπ}{m} \\ sin & \frac{π}{2m} \end{matrix} )^{k} ∏_{1≤h<t≤n} ( sin \frac{π(η_{h} − η_{t})}{2m} · sin \frac{π(t − h)}{2m}

· sin \frac{π(η_{h} + η_{t})}{2m} · sin \frac{π(t + h)}{2m}) ∏_{h=1}^{n} (sin \frac{πη_{h}}{2m} · sin \frac{πh}{2m})

× ( ∏_{h=1}^{2n} \begin{matrix}sin & \frac{π(m−n+h)}{4m} \\ sin & \frac{πh}{4m} \end{matrix} ∏_{h=1}^{n−1} ∏_{t=1}^{n} \begin{matrix}sin & \frac{π(m+t−h)}{2m} \\ sin & \frac{π(n+t−h)}{2m} \end{matrix} + ∏_{h=1}^{2n} \begin{matrix}sin & \frac{π(m−n+h−1)}{4m} \\ sin & \frac{πh}{4m} \end{matrix} ∏_{h=1}^{n−1} ∏_{t=1}^{n} \begin{matrix}sin & \frac{π(m+t−h−1)}{2m} \\ sin & \frac{π(n+t−h)}{2m} \end{matrix}

+ (−1)^{|η|+k+\binom{n}{2}} ∏_{h=1}^{n} \frac{sin \frac{π(m−n+2h−1)}{4m} · cos \frac{π(m−n+2h)}{4m}}{cos \frac{π(2h−1)}{4m} · sin \frac{πh}{2m}} ∏_{h=1}^{n−1} ∏_{t=1}^{n} \begin{matrix}sin & \frac{π(m+t−h)}{2m} \\ sin & \frac{π(n+t−h)}{2m} \end{matrix}

+ (−1)^{|η|+k+\binom{n+1}{2}} ∏_{h=1}^{n} \frac{cos \frac{π(m−n+2h−2)}{4m} · sin \frac{π(m−n+2h−1)}{4m}}{cos \frac{π(2h−1)}{4m} · sin \frac{πh}{2m}} ∏_{h=1}^{n−1} ∏_{t=1}^{n} \begin{matrix}sin & \frac{π(m+t−h−1)}{2m} \\ sin & \frac{π(n+t−h)}{2m} \end{matrix} ), (6.5)

if m is an integer with parity different from that of n, and it is asymptotically

\frac{4^{n^{2}}}{(2m)^{n}} (\begin{matrix}sin & \frac{nπ}{m} \\ sin & \frac{π}{2m} \end{matrix} )^{k} ∏_{1≤h<t≤n} ( sin \frac{π(η_{h} − η_{t})}{2m} · sin \frac{π(t − h)}{2m}

· sin \frac{π(η_{h} + η_{t})}{2m} · sin \frac{π(t + h)}{2m}) ∏_{h=1}^{n} (sin \frac{πη_{h}}{2m} · sin \frac{πh}{2m})

× ∏_{h=1}^{2n} \begin{matrix}sin & \frac{π(⌊m⌋−n+h)}{4m} \\ sin & \frac{πh}{4m} \end{matrix} ∏_{h=1}^{n−1} ∏_{t=1}^{n} \begin{matrix}sin & \frac{π(⌊m⌋+t−h)}{2m} \\ sin & \frac{π(n+t−h)}{2m} \end{matrix} , (6.6)

if m is a half-integer.

Proof. Here, in view of Theorem 25, we have to carry out the sum of (6.1) over all possible λ, i.e., over all λ_{1} > λ_{2} > ⋯ > λ_{n} > 0 with λ_{1} + λ_{2} < 2m, where |λ| ≡ k + |η| mod 2. Leaving away factors which are independent of λ, the problem is to compute the sum

∑_{λ_{1}>···>λ_{n}>0} ; λ_{1}+λ_{2}<2m ′  ∏_{1≤h<t≤n} ( sin \frac{π(λ_{h} − λ_{t})}{2m} · sin \frac{π(λ_{h} + λ_{t})}{2m}) ∏_{h=1}^{n} sin \frac{πλ_{h}}{2m} ,  (6.7)

where the sum ∑ ′ is either restricted to those λ for which |λ| is even, or to those for which |λ| is odd, depending on the parity of k + |η|.

Let first m be an integer. In order to get rid of the restriction λ_{1} + λ_{2} < 2m in the sum in (6.7), we observe that the summand remains unchanged if we replace λ_{1} by 2m − λ_{1} and, moreover, that (λ_{1}, λ_{2}, …, λ_{n}) satisfies the conditions on the summation indices if and only if (2m−λ_{1}, λ_{2}, …, λ_{n}) does. Hence, we may rewrite the sum in (6.7) as

∑_{m≥λ_{1}>···>λ_{n}>0} ′  ∏_{1≤h<t≤n} ( sin \frac{π(λ_{h} − λ_{t})}{2m} · sin \frac{π(λ_{h} + λ_{t})}{2m}) ∏_{h=1}^{n} sin \frac{πλ_{h}}{2m}

+ ∑_{m−1≥λ_{1}>···>λ_{n}>0} ′  ∏_{1≤h<t≤n} ( sin \frac{π(λ_{h} − λ_{t})}{2m} · sin \frac{π(λ_{h} + λ_{t})}{2m}) ∏_{h=1}^{n} sin \frac{πλ_{h}}{2m} , (6.8)

where, again, the sums ∑ ′ are either restricted to those λ for which |λ| is even, or to those for which |λ| is odd, depending on the parity of k + |η|.

The task of evaluating the sums in (6.8) will be accomplished if we are able to evaluate the (complete) sum

U_{1}(c) = ∑_{c≥λ_{1}>···>λ_{n}>0} ∏_{1≤h<t≤n} ( sin \frac{π(λ_{h} − λ_{t})}{2m} · sin \frac{π(λ_{h} + λ_{t})}{2m}) ∏_{h=1}^{n} sin \frac{πλ_{h}}{2m} ,

and its “signed variant”

U_{2}(c) = ∑_{c≥λ_{1}>···>λ_{n}>0}(−1)^{|λ|} ∏_{1≤h<t≤n} ( sin \frac{π(λ_{h} − λ_{t})}{2m} · sin \frac{π(λ_{h} + λ_{t})}{2m}) ∏_{h=1}^{n} sin \frac{πλ_{h}}{2m} ,

for c = m and c = m − 1. The expression (6.8) is then equal to

\frac{1}{2}(U_{1}(m) + U_{1}(m − 1) + U_{2}(m) + U_{2}(m − 1))

if the sum ∑ ′ is restricted to the λ’s for which |λ| is even, and it is equal to

\frac{1}{2}(U_{1}(m) + U_{1}(m − 1) − U_{2}(m) − U_{2}(m − 1))

if the sum ∑ ′ is restricted to the λ’s for which |λ| is odd.

In order to evaluate U_{1}(c) and U_{2}(c), we use (6.3) to rewrite them uniformly as

∑_{c≥λ_{1}>···>λ_{n}>0}ε^{|λ|}2^{−n^{2}+n} det_{1≤h,t≤n} ( sin(π(2h − 1)λ_{t}/2m))

= ∑_{c≥λ_{1}>···>λ_{n}>0} \frac{1}{2^{n^{2}}i^{n}} det_{1≤h,t≤n} ((εe^{πi(2h−1)/2m})^{λ_{t}} − (εe^{πi(2h−1)/2m})^{−λ_{t}}),

with ε = 1 or ε = −1 depending on whether we want to express U_{1}(c) or U_{2}(c). Replacing λ_{t} by λ_{t} + n − t + 1, t = 1, 2, …, n, we obtain the expression

∑_{c−n≥λ_{1}≥···≥λ_{n}≥0} \frac{1}{2^{n^{2}}i^{n}} det_{1≤h,t≤n} ((εe^{πi(2h−1)/2m})^{λ_{t}+n−t+1} − (εe^{πi(2h−1)/2m})^{−(λ_{t}+n−t+1)}). (6.9)

As in the proof of Theorem 22, this determinant can be expressed in terms of a symplectic character as defined in (5.11). Specifically, writing q for e^{πi/2m}, the sum in (6.9) equals

\frac{1}{2^{n^{2}}i^{n}} det_{1≤h,t≤n} ((εq^{2h−1})^{n−t+1} − (εq^{2h−1})^{−(n−t+1)}) ∑_{c−n≥λ_{1}≥···≥λ_{n}≥0} sp_{λ}(εq, εq^{3}, …, εq^{2n−1}).

(6.10)

Use of Formula (5.13) in (6.10) yields the equivalent expression

\frac{1}{2^{n^{2}}i^{n}} det_{1≤h,t≤n} ((εq^{2h−1})^{n−t+1} − (εq^{2h−1})^{−(n−t+1)})

× s_{((c−n)^{n})}(εq^{2n−1}, εq^{2n−3}, …, εq^{3}, εq, 1, εq^{−1}, εq^{−3}, …, εq^{−2n+3}, εq^{−2n+1})

= ε^{(c−n)n} \frac{1}{2^{n^{2}}i^{n}} det_{1≤h,t≤n} ((εq^{2h−1})^{n−t+1} − (εq^{2h−1})^{−(n−t+1)})

× s_{((c−n)^{n})}(q^{2n−1}, q^{2n−3}, …, q^{3}, q, ε, q^{−1}, q^{−3}, …, q^{−2n+3}, q^{−2n+1}). (6.11)

Clearly, the determinant is easily evaluated by means of (B.2). The specialized Schur function is evaluated in Lemma C5 in the case that ε = 1, and in Lemma C6 in the case that ε = −1. If everything is combined and simplified, the claimed formulae (6.4) and (6.5) are eventually obtained.

In the case that m is a half-integer, an adaption of the above argument of replacement of λ_{1} by 2m − λ_{1} in (6.7) shows that (6.7) equals

∑_{⌊m⌋≥λ_{1}>···>λ_{n}>0} ′  ∏_{1≤h<t≤n} ( sin \frac{π(λ_{h} − λ_{t})}{2m} · sin \frac{π(λ_{h} + λ_{t})}{2m}) ∏_{h=1}^{n} sin \frac{πλ_{h}}{2m}

+ ∑_{⌊m⌋≥λ_{1}>···>λ_{n}>0} ′′  ∏_{1≤h<t≤n} ( sin \frac{π(λ_{h} − λ_{t})}{2m} · sin \frac{π(λ_{h} + λ_{t})}{2m}) ∏_{h=1}^{n} sin \frac{πλ_{h}}{2m} ,

where one of the sums ∑ ′ and ∑ ′′ is restricted to those λ for which |λ| is even while the other to those for which |λ| is odd, depending on the parity of k + |η|. Hence, it equals U_{1}(⌊m⌋), regardless of the parity of k + |η|. On the other hand, we have seen above that U_{1}(⌊m⌋) can be rewritten as (6.11) with ε = 1 and c = ⌊m⌋. Application of Lemma C5 with c = ⌊m⌋ − n and some simplification then lead to (6.6). □

We proceed with the corresponding theorems for the case of walks with diagonal steps.

Theorem 27. Let m be a positive integer or half-integer. Furthermore, let η = (η_{1}, η_{2}, …, η_{n}) and λ = (λ_{1}, λ_{2}, …, λ_{n}) be vectors of integers or of half-integers in the alcove AB̃_{n} ; m of type B̃_{n} (defined in (2.3)). Then, as k tends to infinity such that k ≡ 2η_{j} + 2λ_{j} mod 2, the number of random walks from η to λ with exactly k diagonal steps, which stay in the alcove AB̃_{n} ; m, is asymptotically

\frac{4^{n^{2}}}{2(2m)^{n}} (2^{n} ∏_{j=1}^{n} cos \frac{(2j − 1)π}{4m} )^{k} ∏_{1≤h<t≤n} ( sin \frac{π(η_{h} − η_{t})}{2m} · sin \frac{π(λ_{h} − λ_{t})}{2m}

· sin \frac{π(η_{h} + η_{t})}{2m} · sin \frac{π(λ_{h} + λ_{t})}{2m}) ∏_{h=1}^{n} (sin \frac{πη_{h}}{2m} · sin \frac{πλ_{h}}{2m} ) . (6.12)

Proof. We proceed in the same way as in the proof of Theorem 25. This time we have to estimate the two determinants in (2.14). As a preparatory step, we replace the two sums over r in the determinants by 2 times the same sums, but restricted to r from 0 to 2m − 1, which can be safely done when η, λ and k are chosen so that paths from η to λ in k steps exist. (We already did an analogous transformation in the proof of Theorem 19). Then, by expanding the two determinants, we obtain

\frac{1}{2m^{n}} ∑_{r_{1},...,r_{n}=1}^{2m−1} (2^{n} ∏_{j=1}^{n} cos \frac{πr_{j}}{2m})^{k} (∏_{j=1}^{n} sin \frac{πr_{j}η_{j}}{m} ) det_{1≤h,t≤n} (sin \frac{πr_{h}λ_{t}}{m} )

+ \frac{1}{2m^{n}} ∑_{r_{1},...,r_{n}=0}^{2m−1} (2^{n} ∏_{j=1}^{n} cos \frac{π(2r_{j} + 1)}{4m})^{k}

· (∏_{j=1}^{n} sin \frac{π(2r_{j} + 1)η_{j}}{2m} ) det_{1≤h,t≤n} (sin \frac{π(2r_{h} + 1)λ_{t}}{2m} ) ,

which is again a finite sum of the form ∑_{ℓ} c_{ℓ}b_{ℓ}^{k}, with both the c_{ℓ}’s and b_{ℓ}’s independent of k.

As it turns out, also here the dominating terms (the terms for which b_{ℓ} has largest modulus) come from the second determinant. More precisely, these are the terms corresponding to the subsets {r_{1}, r_{2}, …, r_{n}} of

{0, 1, …, n, 2m − n, 2m − n + 1, …, 2m − 1},

with the property that all r_{h}’s are distinct and r_{h} ≠ 2m − 1 − r_{t} for all h and t. Clearly, there are 2^{n} such sets {r_{1}, r_{2}, …, r_{n}}. For each fixed set, the sum of the corresponding terms c_{ℓ}b_{ℓ}^{k} is equal to

\frac{1}{2m^{n}} (2^{n} ∏_{j=1}^{n} cos \frac{π(2j − 1)}{4m})^{k} det_{1≤h,t≤n} (sin \frac{π(2h − 1)λ_{t}}{2m} ) det_{1≤h,t≤n} (sin \frac{π(2h − 1)η_{t}}{2m} ) .

(6.13)

The two determinants in this expression have already been evaluated in (6.3). If this is substituted in (6.13), and if the result is multiplied by 2^{n}, then we obtain exactly (6.12). □

Theorem 28. Let m be a positive integer or half-integer. Furthermore, let η = (η_{1}, η_{2}, …, η_{n}) be a vector of integers or of half-integers in the alcove AB̃_{n} ; m of type B̃_{n} (defined in (2.3)). Then, as k tends to infinity the number of random walks which start at η and proceed for exactly k diagonal steps, which stay in the alcove AB̃_{n} ; m, is asymptotically

\frac{4^{n^{2}}}{2(2m)^{n}} (2^{n} ∏_{j=1}^{n} cos \frac{(2j − 1)π}{4m} )^{k} ∏_{1≤h<t≤n} ( sin \frac{π(η_{h} − η_{t})}{2m} · sin \frac{π(t − h)}{2m}

· sin \frac{π(η_{h} + η_{t})}{2m} · sin \frac{π(t + h)}{2m}) ∏_{h=1}^{n} (sin \frac{πη_{h}}{2m} · sin \frac{πh}{2m})

× ( ∏_{h=1}^{2n} \begin{matrix}sin & \frac{π(m−n+h)}{4m} \\ sin & \frac{πh}{4m} \end{matrix} ∏_{h=1}^{n−1} ∏_{t=1}^{n} \begin{matrix}sin & \frac{π(m+t−h)}{2m} \\ sin & \frac{π(n+t−h)}{2m} \end{matrix} + ∏_{h=1}^{2n} \begin{matrix}sin & \frac{π(m−n+h−1)}{4m} \\ sin & \frac{πh}{4m} \end{matrix} ∏_{h=1}^{n−1} ∏_{t=1}^{n} \begin{matrix}sin & \frac{π(m+t−h−1)}{2m} \\ sin & \frac{π(n+t−h)}{2m} \end{matrix} ), (6.14)

if m is an integer and k + 2η_{j} is even, it is asymptotically

\frac{4^{n^{2}}}{(2m)^{n}} (2^{n} ∏_{j=1}^{n} cos \frac{(2j − 1)π}{4m} )^{k} ∏_{1≤h<t≤n} ( sin \frac{π(η_{h} − η_{t})}{2m} · sin \frac{π(t − h)}{2m}

· sin \frac{π(η_{h} + η_{t})}{2m} · sin \frac{π(t + h)}{2m}) ∏_{h=1}^{n} (sin \frac{πη_{h}}{2m} · sin \frac{πh}{2m})

× ∏_{h=1}^{2n} \begin{matrix}sin & \frac{π(⌊m⌋−n+h)}{4m} \\ sin & \frac{πh}{4m} \end{matrix} ∏_{h=1}^{n−1} ∏_{t=1}^{n} \begin{matrix}sin & \frac{π(⌊m⌋+t−h)}{2m} \\ sin & \frac{π(n+t−h)}{2m} \end{matrix} , (6.15)

if m is a half-integer and k + 2η_{j} is even, it is asymptotically

\frac{4^{n^{2}}}{(2m)^{n}} (2^{n} ∏_{j=1}^{n} cos \frac{(2j − 1)π}{4m} )^{k} ∏_{1≤h<t≤n} ( sin \frac{π(η_{h} − η_{t})}{2m} · sin \frac{π(t − h)}{2m}

· sin \frac{π(η_{h} + η_{t})}{2m} · sin \frac{π(t + h − 1)}{2m}) ∏_{h=1}^{n} (sin \frac{πη_{h}}{2m} · sin \frac{π(2h − 1)}{4m})

× ∏_{h=1}^{2n} \begin{matrix}sin & \frac{π(m−n+h)}{4m} \\ sin & \frac{πh}{4m} \end{matrix} ∏_{h=1}^{n} ∏_{t=1}^{n} \begin{matrix}sin & \frac{π(m+t−h)}{2m} \\ sin & \frac{π(n+t−h)}{2m} \end{matrix} ∏_{h=1}^{n} \begin{matrix}sin & \frac{πh}{2m} \\ sin & \frac{π(m−n+2h)}{4m} \end{matrix} \begin{matrix}cos & \frac{π(2h−1)}{4m} \\ cos & \frac{π(m−n+2h−1)}{4m} \end{matrix} , (6.16)

if m is an integer and k + 2η_{j} is odd, and it is asymptotically

\frac{4^{n^{2}}}{2(2m)^{n}} (2^{n} ∏_{j=1}^{n} cos \frac{(2j − 1)π}{4m} )^{k} ∏_{1≤h<t≤n} ( sin \frac{π(η_{h} − η_{t})}{2m} · sin \frac{π(t − h)}{2m}

· sin \frac{π(η_{h} + η_{t})}{2m} · sin \frac{π(t + h − 1)}{2m}) ∏_{h=1}^{n} (sin \frac{πη_{h}}{2m} · sin \frac{π(2h − 1)}{4m})

× ( ∏_{h=1}^{2n} \begin{matrix}sin & \frac{π(⌈m⌉−n+h)}{4m} \\ sin & \frac{πh}{4m} \end{matrix} ∏_{h=1}^{n} ∏_{t=1}^{n} \begin{matrix}sin & \frac{π(⌈m⌉+t−h)}{2m} \\ sin & \frac{π(n+t−h)}{2m} \end{matrix} ∏_{h=1}^{n} \begin{matrix}sin & \frac{πh}{2m} \\ sin & \frac{π(⌈m⌉−n+2h)}{4m} \end{matrix} \begin{matrix}cos & \frac{π(2h−1)}{4m} \\ cos & \frac{π(⌈m⌉−n+2h−1)}{4m} \end{matrix}

+ ∏_{h=1}^{2n} \begin{matrix}sin & \frac{π(⌊m⌋−n+h)}{4m} \\ sin & \frac{πh}{4m} \end{matrix} ∏_{h=1}^{n} ∏_{t=1}^{n} \begin{matrix}sin & \frac{π(⌊m⌋+t−h)}{2m} \\ sin & \frac{π(n+t−h)}{2m} \end{matrix} ∏_{h=1}^{n} \begin{matrix}sin & \frac{πh}{2m} \\ sin & \frac{π(⌊m⌋−n+2h)}{4m} \end{matrix} \begin{matrix}cos & \frac{π(2h−1)}{4m} \\ cos & \frac{π(⌊m⌋−n+2h−1)}{4m} \end{matrix} ), (6.17)

if m is a half-integer and k + 2η_{j} is odd.

Proof. Here, in view of Theorem 27, we have to carry out the sum of (6.12) over all possible λ, i.e., over all λ_{1} > λ_{2} > ⋯ > λ_{n} > 0 with λ_{1} + λ_{2} < 2m, where k ≡ 2η_{j} + 2λ_{j} mod 2. Leaving away factors which are independent of λ, the problem is to compute the sum

∑_{λ_{1}>···>λ_{n}>0} ; λ_{1}+λ_{2}<2m ′  ∏_{1≤h<t≤n} ( sin \frac{π(λ_{h} − λ_{t})}{2m} · sin \frac{π(λ_{h} + λ_{t})}{2m}) ∏_{h=1}^{n} sin \frac{πλ_{h}}{2m} ,  (6.18)

where the sum ∑ ′ is restricted to integral, respectively to half-integral λ_{1}, λ_{2}, …, λ_{n}, depending on whether k + 2η_{j} is even or odd.

As in previous proofs, we have to distinguish between several cases. Let us first restrict the sum (6.18) to integral λ_{1}, λ_{2}, …, λ_{n}. Then, by repeating the argument from the proof of Theorem 26 of replacement of λ_{1} by 2m−λ_{1}, and using notation from that proof, we obtain that the sum (6.18) equals U_{1}(m) + U_{1}(m − 1) if m is an integer, and it equals 2U_{1}(⌊m⌋) if m is a half-integer. In the proof of Theorem 26 it was shown how to evaluate the sum U_{1}(c) by means of Lemma C5. If the result is substituted, we arrive at (6.14) and (6.15).

If we restrict the sum (6.18) to half-integral λ_{1}, λ_{2}, …, λ_{n}, then we are faced with the problem of evaluating the sum

V (c) = ∑_{c≥λ_{1}>···>λ_{n}>0} ′  ∏_{1≤h<t≤n} ( sin \frac{π(λ_{h} − λ_{t})}{2m} · sin \frac{π(λ_{h} + λ_{t})}{2m}) ∏_{h=1}^{n} sin \frac{πλ_{h}}{2m} ,

where the sum ∑ ′ is over all half-integral λ_{1}, λ_{2}, …, λ_{n}. More specifically, using the replacement of λ_{1} by 2m − λ_{1} another time, one sees readily that the sum (6.18) equals 2V (m − \frac{1}{2}) if m is an integer, and it equals V (m) + V (m − 1) if m is a half-integer.

In order to evaluate V (c), where c is a half-integer, we use (6.3) to rewrite it as

∑_{c≥λ_{1}>···>λ_{n}>0} ; ′ 2^{−n^{2}+n} det_{1≤h,t≤n} ( sin(π(2h − 1)λ_{t}/2m))

= ∑_{c≥λ_{1}>···>λ_{n}>0} ′  \frac{1}{2^{n^{2}}i^{n}} det_{1≤h,t≤n} ((e^{πi(2h−1)/2m})^{λ_{t}} − (e^{πi(2h−1)/2m})^{−λ_{t}}).

Replacing λ_{t} by λ_{t} + n − t + \frac{1}{2}, t = 1, 2, …, n, we obtain the expression

∑_{⌈c−n⌉≥λ_{1}≥···≥λ_{n}≥0} \frac{1}{2^{n^{2}}i^{n}} det_{1≤h,t≤n} ((e^{πi(2h−1)/2m})^{λ_{t}+n−t+1/2} − (e^{πi(2h−1)/2m})^{−(λ_{t}+n−t+1/2)}).

(6.19)

(Note that the last sum is over integral λ_{1}, λ_{2}, …, λ_{n}.) This determinant can be expressed in terms of an odd orthogonal character. (See (C.1) for the definition.) Writing again q for e^{πi/2m}, the sum in (6.19) equals

\frac{1}{2^{n^{2}}i^{n}} det_{1≤h,t≤n} ((q^{2h−1})^{n−t+ \frac{1}{2}} − (q^{2h−1})^{−(n−t+ \frac{1}{2} )}) ∑_{⌈c−n⌉≥λ_{1}≥···≥λ_{n}≥0} so_{λ}^{odd}(q^{2n−1}, q^{2n−3}, …, q).

(6.20)

Again there is a formula which allows us to evaluate the sum in the last line (see [27, (3.2)]),

s_{((a^{n−p},(a−1)^{p})}(x_{1}, x_{1}^{−1}, …, x_{p}, x_{p}^{−1}, 1) = _{∑_{a≥ν_{1}≥···≥ν_{n}≥0} ; oddrows((a^{r})/ν)=p} so_{ν}^{odd}(x_{1}, …, x_{p}),

where the notation (a^{n−p}, (a − 1)^{p}) is a short notation for the vector in which the first n−p components are a, the next p components are a−1, followed by n+1 components all of which are 0, and where oddrows((a^{n})/ν) = p means that the number of rows of odd length in the skew shape (a^{n})/ν equals exactly p. Use of this formula in (6.20) gives

\frac{1}{2^{n^{2}}i^{n}} det_{1≤h,t≤n} ((q^{2h−1})^{n−t+ \frac{1}{2}} − (q^{2h−1})^{−(n−t+ \frac{1}{2} )})

× ∑_{p=0}^{n} s_{(⌈c−n⌉^{n−p},⌈c−n−1⌉^{p})}(q^{2n−1}, q^{2n−3}, …, q^{3}, q, 1, q^{−1}, q^{−3}, …, q^{−2n+3}, q^{−2n+1}).

We now use Lemma C5 to evaluate the Schur function in the last line to obtain the expression

\frac{1}{2^{n^{2}}i^{n}} det_{1≤h,t≤n} ((q^{2h−1})^{n−t+ \frac{1}{2}} − (q^{2h−1})^{−(n−t+ \frac{1}{2} )})

× ∏_{h=1}^{2n} \frac{(q^{\frac{⌈c⌉−n+h}{2}} − q^{− \frac{⌈c⌉−n+h}{2}} )}{(q^{\frac{h}{2}} − q^{− \frac{h}{2}} )} ∏_{h=1}^{n} ∏_{t=1}^{n} \frac{(q^{⌈c⌉+t−h} − q^{−⌈c⌉−t+h})}{(q^{n+t−h} − q^{−n−t+h})} ∏_{h=1}^{n} (q^{h} − q^{−h})^{2}

× ∑_{p=0}^{n} ∏_{h=1}^{n} \frac{1}{(q^{⌈c⌉−n+p+h} − q^{−⌈c⌉+n−p−h})} \frac{1}{∏_{h=1}^{p} (q^{h} − q^{−h}) ∏_{h=1}^{n−p} (q^{h} − q^{−h})}

· \frac{(q^{\frac{⌈c⌉−n}{2}} − q^{− \frac{⌈c⌉−n}{2}} )(q^{\frac{⌈c⌉−n}{2} +p} + q^{− \frac{⌈c⌉−n}{2} −p})}{(q^{⌈c⌉−n+p} − q^{−⌈c⌉+n−p})}.

In terms of the standard basic hypergeometric notation

rϕ_{s}[\begin{matrix}a_{1}, …, a_{r} \\ b_{1}, …, b_{s} \end{matrix}; q, z] = ∑_{ℓ=0}^{∞} \frac{(a_{1}; q)_{ℓ} ⋯ (a_{r}; q)_{ℓ}}{(q; q)_{ℓ}(b_{1}; q)_{ℓ} ⋯ (b_{s}; q)_{ℓ}} ((−1)^{ℓ}q^{\binom{ℓ}{2}})^{s−r+1} z^{ℓ} ,  (6.21)

where the shifted q-factorials (a; q)_{ℓ} are defined by (a; q)_{ℓ} := (1−a)(1−aq) ⋯ (1−aq^{ℓ−1}), ℓ ≥ 1, (a; q)_{0} := 1, this can be written in the form

\frac{1}{2^{n^{2}}i^{n}} det_{1≤h,t≤n} ((q^{2h−1})^{n−t+ \frac{1}{2}} − (q^{2h−1})^{−(n−t+ \frac{1}{2} )})

× ∏_{h=1}^{2n} \frac{(q^{\frac{⌈c⌉−n+h}{2}} − q^{− \frac{⌈c⌉−n+h}{2}} )}{(q^{\frac{h}{2}} − q^{− \frac{h}{2}} )} ∏_{h=1}^{n} ∏_{t=1}^{n} \frac{(q^{⌈c⌉+t−h} − q^{−⌈c⌉−t+h})}{(q^{n+t−h} − q^{−n−t+h})} ∏_{h=1}^{n} (q^{h} − q^{−h})^{2}

× \frac{q^{⌈c⌉n+n}}{(q^{2}; q^{2})_{n} (q^{2⌈c⌉−2n+2}; q^{2})_{n}}_{3}ϕ_{2}[\begin{matrix}q^{2⌈c⌉−2n}, −q^{⌈c⌉−n+2}, q^{−2n} \\ −q^{⌈c⌉−n}, q^{2⌈c⌉+2} \end{matrix}; q^{2}, −q^{2n+1}] . (6.22)

The determinant is again easily evaluated by means of (B.2). On the other hand, the 3ϕ_{2}-series in the last line can be evaluated by a limit case of Jackson’s very-well-poised 8ϕ_{7}-summation (see [14, (2.6.2); Appendix (II.22)])

8ϕ_{7}[\begin{matrix}A, \sqrt{A}q, −\sqrt{A}q, B, C, D, A^{2}q^{1+N}/BCD, q^{−N} \\ \sqrt{A}, −\sqrt{A}, Aq/B, Aq/C, Aq/D, BCD/Aq^{N}, Aq^{1+N} \end{matrix}; q, q]

= \begin{matrix}(Aq; q) & N & (Aq/BC; q) & N & (Aq/BD; q) & N & (Aq/CD; q) & N \\ (Aq/B; q) & N & (Aq/C; q) & N & (Aq/D; q) & N & (Aq/BCD; q) & N \end{matrix} , (6.23)

where N is a nonnegative integer. Namely, if in (6.23) we let N → ∞, put D = \sqrt{A} and C = −\sqrt{A}, and finally replace q by q^{2}, we are left with

3ϕ_{2}[\begin{matrix}A, −\sqrt{A}q^{2}, B \\ −\sqrt{A}, Aq^{2}/B \end{matrix}; q^{2}, −\frac{q}{B} ] = \frac{(−q; q^{2})_{∞} (−\sqrt{A}q/B; q^{2})_{∞} (Aq^{2}; q^{2})_{∞} (\sqrt{A}q^{2}/B; q^{2})_{∞}}{(−\sqrt{A}q; q^{2})_{∞} (−q/B; q^{2})_{∞} (\sqrt{A}q^{2}; q^{2})_{∞} (Aq^{2}/B; q^{2})_{∞}} .

(6.24)

The _{3}ϕ_{2}-series in (6.22) is a special case of the above _{3}ϕ_{2}-series in which A = q^{2⌈c⌉−2n} and B = q^{−2n}. If we substitute the corresponding right-hand side of (6.24) in (6.22), we have evaluated the sum V (c). This, in its turn, leads to the expressions (6.16) and (6.17). □

7. Asymptotics for random walks in alcoves of type D̃

In this final section we turn our attention to the walks in the alcove AD̃_{n} ; m of type D̃_{n}. We determine the asymptotic behaviour of the number of walks from a given starting point to a given end point which stay in AD̃_{n} ; m as the number of steps becomes large, as well as the asymptotic behaviour of the number of those walks which start at a given point but may terminate anywhere. In technical terms, we determine the asymptotic behaviour of the expressions given by Theorems 10 and 11 as k becomes large, and as well if these expressions are summed over all possible end points of the walks.

The first two theorems in this section give our results for the case of walks with standard steps.

Theorem 29. Let m be a positive integer. Furthermore, let η = (η_{1}, η_{2}, …, η_{n}) and λ = (λ_{1}, λ_{2}, …, λ_{n}) be vectors of integers in the alcove AD̃_{n} ; m of type D̃_{n} (defined in (2.4)). Then, as k tends to infinity such that k ≡ |η| + |λ| mod 2, the number of random walks from η to λ with exactly k standard steps, which stay in the alcove AD̃_{n} ; m, is asymptotically

\frac{4^{n^{2}}}{(8m)^{n}} (2\begin{matrix}sin & \frac{nπ}{2m} & cos & \frac{(n−1)π}{2m} \\ sin & \frac{π}{2m} \end{matrix} )^{k}

× ∏_{1≤h<t≤n} ( sin \frac{π(η_{h} − η_{t})}{2m} · sin \frac{π(λ_{h} − λ_{t})}{2m} · sin \frac{π(η_{h} + η_{t})}{2m} · sin \frac{π(λ_{h} + λ_{t})}{2m}). (7.1)

Proof. We proceed as in the proof of Theorem 25. Since we have done similar calculations already several times, we shall be brief here.

We have to find the asymptotic behaviour of (2.15) (instead of the rather similar (2.13), with which we dealt in the proof of Theorem 25). By applying arguments very similar to those in the proof of Theorem 25, we infer that the dominating terms in the expansion of (2.15) come from the expansion of the third determinant. To be precise, these dominating terms add up to 2 · 2^{n−1} times

\frac{1}{4m^{n}} (2 ∑_{j=0}^{n−1} cos \frac{πj}{m} )^{k} det_{1≤h,t≤n} (cos \frac{π(h − 1)η_{t}}{m} ) det_{1≤h,t≤n} (cos \frac{π(h − 1)λ_{t}}{m} ) . (7.2)

We have

∑_{j=0}^{n−1} cos \frac{πj}{m} = \frac{sin \frac{nπ}{2m} · cos \frac{(n−1)π}{2m}}{sin \frac{π}{2m}} .

On the other hand, the first determinant in (7.2) can be rewritten in the form

2^{−n} det_{1≤h,t≤n} (e^{\frac{πi(h−1)η_{t}}{m}} + e^{− \frac{πi(h−1)η_{t}}{m}} ) ,

and can thus be evaluated by means of (B.3). The result is that

det_{1≤h,t≤n} (cos \frac{π(h − 1)η_{t}}{m} ) = 2^{n^{2}−2n+1} ∏_{1≤h<t≤n} ( sin \frac{π(η_{h} − η_{t})}{2m} · sin \frac{π(η_{h} + η_{t})}{2m}). (7.3)

If this is substituted in (7.2) (and as well the analogous evaluation of the second determinant in (7.2)), and if the result is multiplied by 2 · 2^{n−1} = 2^{n}, then we obtain exactly (7.1). □

If the starting point is fixed but the end point is not, we have the following result.

Theorem 30. Let m be a positive integer or half-integer. Furthermore, let η = (η_{1}, η_{2}, …, η_{n}) be a vector of integers in the alcove AD̃_{n} ; m of type D̃_{n} (defined in (2.4)). Then, as k tends to infinity, the number of random walks which start at η and proceed for exactly k standard steps, which stay in the alcove AD̃_{n} ; m, is asymptotically

\frac{4^{n^{2}}}{2^{n}(8m)^{n}} (2\begin{matrix}sin & \frac{nπ}{2m} & cos & \frac{(n−1)π}{2m} \\ sin & \frac{π}{2m} \end{matrix} )^{k}

× ∏_{1≤h<t≤n} ( sin \frac{π(η_{h} − η_{t})}{2m} · sin \frac{π(t − h)}{2m} · sin \frac{π(η_{h} + η_{t})}{2m} · sin \frac{π(t + h − 2)}{2m})

× (\frac{1}{2^{n−1}} \begin{matrix}∏ & {n \atop h=1} & sin & \frac{π(m−n+2h)}{2m} \\ ∏ & {n−1 \atop h=1} & sin & \frac{πh}{2m} \end{matrix} ∏_{1≤h<t≤n} \frac{sin \frac{π(m−n+h+t−1)}{2m} · sin \frac{π(m−n+t+h)}{2m}}{sin^{2} \frac{π(t+h−2)}{2m}}

· ∑_{k=1}^{n} \frac{(−1)^{n−k}(m − n + 2k)}{∏_{h=1}^{n} sin \frac{π(m−n+k+h)}{2m} ∏_{h=1}^{k−1} sin \frac{πh}{2m} ∏_{h=1}^{n−k} sin \frac{πh}{2m}}

40 C. KRATTENTHALER

+ \frac{1}{2^{n−1}} \begin{matrix}∏ & {n \atop h=1} & sin & \frac{π(m−n+2h−1)}{2m} \\ ∏ & {n−1 \atop h=1} & sin & \frac{πh}{2m} \end{matrix} ∏_{1≤h<t≤n} \frac{sin \frac{π(m−n+h+t−2)}{2m} · sin \frac{π(m−n+t+h−1)}{2m}}{sin^{2} \frac{π(t+h−2)}{2m}}

· ∑_{k=1}^{n} \frac{(−1)^{n−k}(m − n + 2k − 2)}{∏_{h=1}^{n} sin \frac{π(m−n+k+h−1)}{2m} ∏_{h=1}^{k−1} sin \frac{πh}{2m} ∏_{h=1}^{n−k} sin \frac{πh}{2m}}

+ \frac{1}{2^{n−1}} \begin{matrix}∏ & {n \atop h=1} & sin & \frac{π(m−n+2h−1)}{2m} \\ ∏ & {n−1 \atop h=1} & sin & \frac{πh}{2m} \end{matrix} ∏_{1≤h<t≤n} \frac{sin \frac{π(m−n+h+t)}{2m} · sin \frac{π(m−n+t+h−1)}{2m}}{sin^{2} \frac{π(t+h−2)}{2m}}

· ∑_{k=1}^{n} \frac{(−1)^{n−k}(m − n + 2k − 1)}{∏_{h=1}^{n} sin \frac{π(m−n+k+h−1)}{2m} ∏_{h=1}^{k−1} sin \frac{πh}{2m} ∏_{h=1}^{n−k} sin \frac{πh}{2m}}

+ \frac{1}{2^{n−1}} \begin{matrix}∏ & {n \atop h=1} & sin & \frac{π(m−n+2h−2)}{2m} \\ ∏ & {n−1 \atop h=1} & sin & \frac{πh}{2m} \end{matrix} ∏_{1≤h<t≤n} \frac{sin \frac{π(m−n+h+t−1)}{2m} · sin \frac{π(m−n+t+h−2)}{2m}}{sin^{2} \frac{π(t+h−2)}{2m}}

· ∑_{k=1}^{n} \frac{(−1)^{n−k}(m − n + 2k − 3)}{∏_{h=1}^{n} sin \frac{π(m−n+k+h−2)}{2m} ∏_{h=1}^{k−1} sin \frac{πh}{2m} ∏_{h=1}^{n−k} sin \frac{πh}{2m}}

+ (−1)^{mn+k+|η|} ∏_{h=1}^{n−1} \frac{1}{cos \frac{π(h−1)}{2m}} ∏_{1≤h<t≤n} \frac{sin \frac{π(m−n+h+t−1)}{2m} · sin \frac{π(m−n+t+h)}{2m}}{sin^{2} \frac{π(t+h−2)}{2m}}

+ (−1)^{mn+k+|η|} ∏_{h=1}^{n−1} \frac{1}{cos \frac{π(h−1)}{2m}} ∏_{1≤h<t≤n} \frac{sin \frac{π(m−n+h+t−2)}{2m} · sin \frac{π(m−n+t+h−1)}{2m}}{sin^{2} \frac{π(t+h−2)}{2m}}

+ (−1)^{mn+k+|η|} ∏_{h=1}^{n−1} \frac{1}{cos \frac{π(h−1)}{2m}} ∏_{1≤h<t≤n} \frac{sin \frac{π(m−n+h+t)}{2m} · sin \frac{π(m−n+t+h−1)}{2m}}{sin^{2} \frac{π(t+h−2)}{2m}}

+(−1)^{mn+k+|η|} ∏_{h=1}^{n−1} \frac{1}{cos \frac{π(h−1)}{2m}} ∏_{1≤h<t≤n} \frac{sin \frac{π(m−n+h+t−1)}{2m} · sin \frac{π(m−n+t+h−2)}{2m}}{sin^{2} \frac{π(t+h−2)}{2m}} ) (7.4)

if m is an integer, and it is asymptotically

\frac{4^{n^{2}−n+1}}{(8m)^{n}} (2\begin{matrix}sin & \frac{nπ}{2m} & cos & \frac{(n−1)π}{2m} \\ sin & \frac{π}{2m} \end{matrix} )^{k}

× ∏_{1≤h<t≤n} ( sin \frac{π(η_{h} − η_{t})}{2m} · sin \frac{π(t − h)}{2m} · sin \frac{π(η_{h} + η_{t})}{2m} · sin \frac{π(t + h − 2)}{2m})

(\begin{matrix}∏ & {n \atop h=1} & sin & \frac{π(⌊m⌋−n+2h)}{2m} \\ ∏ & {n−1 \atop h=1} & sin & \frac{πh}{2m} \end{matrix} ∏_{1≤h<t≤n} \frac{sin \frac{π(⌊m⌋−n+h+t−1)}{2m} · sin \frac{π(⌊m⌋−n+t+h)}{2m}}{sin^{2} \frac{π(t+h−2)}{2m}}

· ∑_{k=1}^{n} \frac{(−1)^{n−k}(⌊m⌋ − n + 2k)}{∏_{h=1}^{n} sin \frac{π(⌊m⌋−n+k+h)}{2m} ∏_{h=1}^{k−1} sin \frac{πh}{2m} ∏_{h=1}^{n−k} sin \frac{πh}{2m}}

+ \begin{matrix}∏ & {n \atop h=1} & sin & \frac{π(⌊m⌋−n+2h−1)}{2m} \\ ∏ & {n−1 \atop h=1} & sin & \frac{πh}{2m} \end{matrix} ∏_{1≤h<t≤n} \frac{sin \frac{π(⌊m⌋−n+h+t)}{2m} · sin \frac{π(⌊m⌋−n+t+h−1)}{2m}}{sin^{2} \frac{π(t+h−2)}{2m}}

· ∑_{k=1}^{n} \frac{(−1)^{n−k}(⌊m⌋ − n + 2k − 1)}{∏_{h=1}^{n} sin \frac{π(⌊m⌋−n+k+h−1)}{2m} ∏_{h=1}^{k−1} sin \frac{πh}{2m} ∏_{h=1}^{n−k} sin \frac{πh}{2m}} ) (7.5)

if m is a half-integer.

Proof. Here, in view of Theorem 29, we have to carry out the sum of (7.1) over all possible λ, i.e., over all λ_{1} > λ_{2} > ⋯ > λ_{n−1} > |λ_{n}| with λ_{1} + λ_{2} < 2m, where |λ| ≡ k + |η| mod 2. Leaving away factors which are independent of λ, the problem is to compute the sum

∑_{λ_{1}>λ_{2}>···>λ_{n−1}>|λ_{n}|} ; λ_{1}+λ_{2}<2m ′  ∏_{1≤h<t≤n} ( sin \frac{π(λ_{h} − λ_{t})}{2m} · sin \frac{π(λ_{h} + λ_{t})}{2m}),  (7.6)

where the sum ∑ ′ is either restricted to those λ for which |λ| is even, or to those for which |λ| is odd, depending on the parity of k + |η|.

Let first m be an integer. As in the proof of Theorem 26, instead of a sum ∑ ′ where we have to deal with the unwieldy constraints λ_{1} + λ_{2} < 2m and λ_{n−1} > |λ_{n}|, we would rather prefer sums where the only constraint is of the form λ_{1} > λ_{2} > ⋯ > λ_{n}. In order to get rid of the constraint λ_{1} + λ_{2} < 2m, we apply again the argument of replacement of λ_{1} by 2m − λ_{1} that we already used in the proofs of Theorems 26 and 28. Similarly, in order to get rid of the constraint λ_{n−1} > |λ_{n}|, we observe that the summand in (7.6) remains invariant under the replacement of λ_{n} by −λ_{n}, and that (λ_{1}, λ_{2}, …, λ_{n}) satisfies the conditions on the summation indices if and only if (λ_{1}, λ_{2}, …, −λ_{n}) does. Hence, we may rewrite the sum in (7.6) as

∑_{m≥λ_{1}>···>λ_{n}≥0} ′  ∏_{1≤h<t≤n} ( sin \frac{π(λ_{h} − λ_{t})}{2m} · sin \frac{π(λ_{h} + λ_{t})}{2m})

+ ∑_{m−1≥λ_{1}>···>λ_{n}≥0} ′  ∏_{1≤h<t≤n} ( sin \frac{π(λ_{h} − λ_{t})}{2m} · sin \frac{π(λ_{h} + λ_{t})}{2m})

+ ∑_{m≥λ_{1}>···>λ_{n}>0} ′  ∏_{1≤h<t≤n} ( sin \frac{π(λ_{h} − λ_{t})}{2m} · sin \frac{π(λ_{h} + λ_{t})}{2m})

+ ∑_{m−1≥λ_{1}>···>λ_{n}>0} ′  ∏_{1≤h<t≤n} ( sin \frac{π(λ_{h} − λ_{t})}{2m} · sin \frac{π(λ_{h} + λ_{t})}{2m}), (7.7)

where, again, the sums ∑ ′ are either restricted to those λ for which |λ| is even, or to those for which |λ| is odd, depending on the parity of k + |η|.

The task of evaluating the sums in (7.7) will be accomplished if we are able to evaluate the (complete) sums

W_{1}(c) = ∑_{c≥λ_{1}>···>λ_{n}≥0} ∏_{1≤h<t≤n} ( sin \frac{π(λ_{h} − λ_{t})}{2m} · sin \frac{π(λ_{h} + λ_{t})}{2m})

and

W_{2}(c) = ∑_{c≥λ_{1}>···>λ_{n}>0} ∏_{1≤h<t≤n} ( sin \frac{π(λ_{h} − λ_{t})}{2m} · sin \frac{π(λ_{h} + λ_{t})}{2m}),

and their “signed variants”

W_{3}(c) = ∑_{c≥λ_{1}>···>λ_{n}≥0}(−1)^{|λ|} ∏_{1≤h<t≤n} ( sin \frac{π(λ_{h} − λ_{t})}{2m} · sin \frac{π(λ_{h} + λ_{t})}{2m})

and

W_{4}(c) = ∑_{c≥λ_{1}>···>λ_{n}>0}(−1)^{|λ|} ∏_{1≤h<t≤n} ( sin \frac{π(λ_{h} − λ_{t})}{2m} · sin \frac{π(λ_{h} + λ_{t})}{2m})

for c = m and c = m − 1. The expression (7.7) is then equal to

\frac{1}{2}(W_{1}(m)+W_{1}(m−1)+W_{2}(m)+W_{2}(m−1)+W_{3}(m)+W_{3}(m−1)+W_{4}(m)+W_{4}(m−1))

if the sum ∑ ′ is restricted to the λ’s for which |λ| is even, and it is equal to

\frac{1}{2}(W_{1}(m)+W_{1}(m−1)+W_{2}(m)+W_{2}(m−1)−W_{3}(m)−W_{3}(m−1)−W_{4}(m)−W_{4}(m−1))

if the sum ∑ ′ is restricted to the λ’s for which |λ| is odd.

We first show how to evaluate W_{1}(c) and W_{3}(c). Using (7.3), we rewrite them uniformly as

∑_{c≥λ_{1}>···>λ_{n}≥0}ε^{|λ|}2^{−n^{2}+2n−1} det_{1≤h,t≤n} ( cos(π(h − 1)λ_{t}/m))

= ∑_{c≥λ_{1}>···>λ_{n}≥0} \frac{1}{2^{n^{2}−n+1}} det_{1≤h,t≤n} ((εe^{πi(h−1)/m})^{λ_{t}} + (εe^{πi(h−1)/m})^{−λ_{t}}),

with ε = 1 or ε = −1 depending on whether we want to express W_{1}(c) or W_{3}(c). Replacing λ_{t} by λ_{t} + n − t, t = 1, 2, …, n, we obtain the expression

∑_{c−n+1≥λ_{1}≥···≥λ_{n}≥0} \frac{1}{2^{n^{2}−n+1}} det_{1≤h,t≤n} ((εe^{πi(h−1)/m})^{λ_{t}+n−t} + (εe^{πi(h−1)/m})^{−(λ_{t}+n−t)}). (7.8)

The determinant in the summand appears as a part in the definition of an even orthogonal character given in (C.49). Specifically, if we write q for e^{πi/m}, the sum in (7.8) equals

\frac{1}{2^{n^{2}−n+1}} det_{1≤h,t≤n} ((εq^{h−1})^{n−t} + (εq^{h−1})^{−(n−t)})

× ∑_{c−n+1≥λ_{1}≥···≥λ_{n}≥0} so_{λ}^{even}(εq^{n−1}, εq^{n−2}, …, εq, ε). (7.9)

(The reader should note that the second determinant in (C.49) vanishes if one of the variables x_{t}, t = 1, 2, …, n, is 1 or −1, which is the case in our situation.) The sum in (7.9) can be simplified by means of another character identity from [27]: given nonnegative integers or half-integers a, b with a ≥ b, the identity [27, (3.15)] implies

so_{(a^{n})}^{even}(x_{1}, x_{2}, …, x_{n}) · ∑_{p=−b}^{b} so_{(b^{n−1},p)}^{even}(x_{1}, x_{2}, …, x_{n})

= ∑_{a+b≥ν_{1}≥ν_{2}≥···≥ν_{n}≥a−b} so_{ν}^{even}(x_{1}, x_{2}, …, x_{n}), (7.10)

with the understanding that the sum on the left-hand side ranges over half-integral p if b is a half-integer, and that the sum on the right-hand side ranges over integers ν_{t} if a + b is an integer, and over half-integers ν_{t} if a + b is a half-integer. The notation (a^{n}) is short for the vector consisting of n components all of which equal to a, while the notation (b^{n−1}, p) means the vector in which the first n − 1 components are b, followed by a component p. If we use (7.10), then (7.9) becomes

\frac{1}{2^{n^{2}−n+1}} det_{1≤h,t≤n} ((εq^{h−1})^{n−t} + (εq^{h−1})^{−(n−t)}) so_{(((c−n+1)/2)^{n})}^{even}(εq^{n−1}, εq^{n−2}, …, εq, ε)

× ∑_{p=−(c−n+1)/2}^{(c−n+1)/2} so_{(((c−n+1)/2)^{n−1},p)}^{even}(εq^{n−1}, εq^{n−2}, …, εq, ε)

= ε^{(c−n+1)n} \frac{1}{2^{n^{2}−n+1}} det_{1≤h,t≤n} ((εq^{h−1})^{n−t} + (εq^{h−1})^{−(n−t)})

× so_{(((c−n+1)/2)^{n})}^{even}(q^{n−1}, q^{n−2}, …, q, 1)

× ∑_{p=−(c−n+1)/2}^{(c−n+1)/2} ε^{(c−n+1−2p)/2} so_{(((c−n+1)/2)^{n−1},p)}^{even}(q^{n−1}, q^{n−2}, …, q, 1).

Clearly, the first determinant can be evaluated by means of (B.3). The even orthogonal character of shape (((c − n + 1)/2)^{n}) can be evaluated by using the definition (C.49) and (B.3). (It should be observed that the second determinant in the numerator on the right-hand side of (C.49) vanishes if x_{n} = ±1.) Finally, the sum over p of even orthogonal characters is evaluated in Lemma C7 if ε = 1, respectively in Lemma C8 if ε = −1.

In order to evaluate W_{2}(c) and W_{4}(c), we proceed in a completely analogous fashion. In particular, the two sums W_{2}(c) and W_{4}(c) are equal to

∑_{c−n+1≥λ_{1}≥···≥λ_{n}>0} \frac{1}{2^{n^{2}−n+1}} det_{1≤h,t≤n} ((εe^{πi(h−1)/m})^{λ_{t}+n−t} + (εe^{πi(h−1)/m})^{−(λ_{t}+n−t)}), (7.11)

with ε = 1 or ε = −1 depending on whether we want to express W_{2}(c) or W_{4}(c). By the use of (5.13), this sum becomes

\frac{1}{2^{n^{2}−n+1}} det_{1≤h,t≤n} ((εq^{h−1})^{n−t} + (εq^{h−1})^{−(n−t)}) ∑_{c−n+1≥λ_{1}≥···≥λ_{n}≥1} so_{λ}^{even}(ε, εq, …, εq^{n−1}). (7.12)

The only difference to (7.9) is that in the sum λ_{n} must be positive, instead of just non-negative. Hence, the remaining steps are the same, with the small modification that we use (7.10) with a = (c − n + 2)/2 and b = (c − n)/2 to see that (7.12) equals

ε^{(c−n+1)n} \frac{1}{2^{n^{2}−n+1}} det_{1≤h,t≤n} ((εq^{h−1})^{n−t} + (εq^{h−1})^{−(n−t)})

× so_{(((c−n+2)/2)^{n})}^{even}(q^{n−1}, q^{n−2}, …, q, 1)

× ∑_{p=−(c−n)/2}^{(c−n)/2} ε^{(c−n−2p)/2} so_{(((c−n)/2)^{n−1},p)}^{even}(q^{n−1}, q^{n−2}, …, q, 1).

Again, the first determinant can be evaluated by means of (B.3), the even orthogonal character of shape (((c−n+2)/2)^{n}) can be evaluated by using the definition (C.49) and (B.3), while the sum over p of even orthogonal characters is evaluated in Lemma C7, respectively in Lemma C8.

If everything is combined and simplified, the claimed formula (7.4) is eventually obtained.

In the case that m is a half-integer, an adaption of the above arguments shows that (7.6) equals

∑_{⌊m⌋≥λ_{1}>···>λ_{n}≥0} ∏_{1≤h<t≤n} ( sin \frac{π(λ_{h} − λ_{t})}{2m} · sin \frac{π(λ_{h} + λ_{t})}{2m})

+ ∑_{⌊m⌋≥λ_{1}>···>λ_{n}>0} ∏_{1≤h<t≤n} ( sin \frac{π(λ_{h} − λ_{t})}{2m} · sin \frac{π(λ_{h} + λ_{t})}{2m}),

regardless of the parity of k + |η|. Hence, it equals W_{1}(⌊m⌋) + W_{2}(⌊m⌋). On the other hand, we have seen above how to evaluate W_{1}(⌊m⌋) and W_{2}(⌊m⌋). If this is substituted and the resulting expressions are simplified, we eventually obtain (7.5). □

We conclude the section with the analogous results for the case of walks with diagonal steps.

Theorem 31. Let m be a positive integer or half-integer. Furthermore, let η = (η_{1}, η_{2}, …, η_{n}) and λ = (λ_{1}, λ_{2}, …, λ_{n}) be vectors of integers or of half-integers in the alcove AD̃_{n} ; m of type D̃_{n} (defined in (2.4)). Then, as k tends to infinity such that k ≡ 2η_{j} + 2λ_{j} mod 2, the number of random walks from η to λ with exactly k diagonal steps, which stay in the alcove AD̃_{n} ; m, is asymptotically

\frac{4^{n^{2}}}{2(8m)^{n}} (2^{n} ∏_{j=0}^{n−1} cos \frac{jπ}{2m})^{k}

× ∏_{1≤h<t≤n} ( sin \frac{π(η_{h} − η_{t})}{2m} · sin \frac{π(λ_{h} − λ_{t})}{2m} sin \frac{π(η_{h} + η_{t})}{2m} · sin \frac{π(λ_{h} + λ_{t})}{2m}). (7.13)

Proof. We proceed as in the proof of Theorem 27. We shall again be brief here.

We have to find the asymptotic behaviour of (2.16) (instead of the rather similar (2.14), with which we dealt in the proof of Theorem 27). By applying arguments very similar to those in the proof of Theorem 27, we infer that the dominating terms in the expansion of (2.16) come from the expansion of the third determinant. To be precise, these dominating terms add up to 2^{n−1} times

\frac{1}{4m^{n}} (2^{n} ∏_{j=0}^{n−1} cos \frac{πj}{2m})^{k} det_{1≤h,t≤n} (cos \frac{π(h − 1)λ_{t}}{m} ) det_{1≤h,t≤n} (cos \frac{π(h − 1)η_{t}}{m} ) . (7.14)

We have already evaluated these determinants in (7.3). If we use these evaluations in (7.14), and if the result is multiplied by 2^{n−1}, then we obtain exactly (7.13). □

Theorem 32. Let m be a positive integer or half-integer. Furthermore, let η = (η_{1}, η_{2}, …, η_{n}) be a vector of integers or of half-integers in the alcove AD̃_{n} ; m of type D̃_{n} (defined in (2.4)). Then, as k tends to infinity, the number of random walks which start at η and proceed for exactly k diagonal steps, which stay in the alcove AD̃_{n} ; m, is asymptotically

\frac{2 · 4^{n^{2}−n}}{(8m)^{n}} (2^{n} ∏_{j=0}^{n−1} cos \frac{jπ}{2m})^{k}

× ∏_{1≤h<t≤n} ( sin \frac{π(η_{h} − η_{t})}{2m} · sin \frac{π(t − h)}{2m} sin \frac{π(η_{h} + η_{t})}{2m} · sin \frac{π(t + h − 2)}{2m})

× (\begin{matrix}∏ & {n \atop h=1} & sin & \frac{π(m−n+2h)}{2m} \\ ∏ & {n−1 \atop h=1} & sin & \frac{πh}{2m} \end{matrix} ∏_{1≤h<t≤n} \frac{sin \frac{π(m−n+h+t−1)}{2m} · sin \frac{π(m−n+t+h)}{2m}}{sin^{2} \frac{π(t+h−2)}{2m}}

· ∑_{k=1}^{n} \frac{(−1)^{n−k}(m − n + 2k)}{∏_{h=1}^{n} sin \frac{π(m−n+k+h)}{2m} ∏_{h=1}^{k−1} sin \frac{πh}{2m} ∏_{h=1}^{n−k} sin \frac{πh}{2m}}

+ \begin{matrix}∏ & {n \atop h=1} & sin & \frac{π(m−n+2h−1)}{2m} \\ ∏ & {n−1 \atop h=1} & sin & \frac{πh}{2m} \end{matrix} ∏_{1≤h<t≤n} \frac{sin \frac{π(m−n+h+t−2)}{2m} · sin \frac{π(m−n+t+h−1)}{2m}}{sin^{2} \frac{π(t+h−2)}{2m}}

· ∑_{k=1}^{n} \frac{(−1)^{n−k}(m − n + 2k − 2)}{∏_{h=1}^{n} sin \frac{π(m−n+k+h−1)}{2m} ∏_{h=1}^{k−1} sin \frac{πh}{2m} ∏_{h=1}^{n−k} sin \frac{πh}{2m}}

+ \begin{matrix}∏ & {n \atop h=1} & sin & \frac{π(m−n+2h−1)}{2m} \\ ∏ & {n−1 \atop h=1} & sin & \frac{πh}{2m} \end{matrix} ∏_{1≤h<t≤n} \frac{sin \frac{π(m−n+h+t)}{2m} · sin \frac{π(m−n+t+h−1)}{2m}}{sin^{2} \frac{π(t+h−2)}{2m}}

· ∑_{k=1}^{n} \frac{(−1)^{n−k}(m − n + 2k − 1)}{∏_{h=1}^{n} sin \frac{π(m−n+k+h−1)}{2m} ∏_{h=1}^{k−1} sin \frac{πh}{2m} ∏_{h=1}^{n−k} sin \frac{πh}{2m}}

+ \begin{matrix}∏ & {n \atop h=1} & sin & \frac{π(m−n+2h−2)}{2m} \\ ∏ & {n−1 \atop h=1} & sin & \frac{πh}{2m} \end{matrix} ∏_{1≤h<t≤n} \frac{sin \frac{π(m−n+h+t−1)}{2m} · sin \frac{π(m−n+t+h−2)}{2m}}{sin^{2} \frac{π(t+h−2)}{2m}}

· ∑_{k=1}^{n} \frac{(−1)^{n−k}(m − n + 2k − 3)}{∏_{h=1}^{n} sin \frac{π(m−n+k+h−2)}{2m} ∏_{h=1}^{k−1} sin \frac{πh}{2m} ∏_{h=1}^{n−k} sin \frac{πh}{2m}} ) (7.15)

if m is an integer and k + 2η_{j} is even, it is asymptotically

\frac{4^{n^{2}−n+1}}{(8m)^{n}} (2^{n} ∏_{j=0}^{n−1} cos \frac{jπ}{2m})^{k}

× ∏_{1≤h<t≤n} ( sin \frac{π(η_{h} − η_{t})}{2m} · sin \frac{π(t − h)}{2m} sin \frac{π(η_{h} + η_{t})}{2m} · sin \frac{π(t + h − 2)}{2m})

46 C. KRATTENTHALER

× (\begin{matrix}∏ & {n \atop h=1} & sin & \frac{π(⌊m⌋−n+2h)}{2m} \\ ∏ & {n−1 \atop h=1} & sin & \frac{πh}{2m} \end{matrix} ∏_{1≤h<t≤n} \frac{sin \frac{π(⌊m⌋−n+h+t−1)}{2m} · sin \frac{π(⌊m⌋−n+t+h)}{2m}}{sin^{2} \frac{π(t+h−2)}{2m}}

· ∑_{k=1}^{n} \frac{(−1)^{n−k}(⌊m⌋ − n + 2k)}{∏_{h=1}^{n} sin \frac{π(⌊m⌋−n+k+h)}{2m} ∏_{h=1}^{k−1} sin \frac{πh}{2m} ∏_{h=1}^{n−k} sin \frac{πh}{2m}}

+ \begin{matrix}∏ & {n \atop h=1} & sin & \frac{π(⌊m⌋−n+2h−1)}{2m} \\ ∏ & {n−1 \atop h=1} & sin & \frac{πh}{2m} \end{matrix} ∏_{1≤h<t≤n} \frac{sin \frac{π(⌊m⌋−n+h+t)}{2m} · sin \frac{π(⌊m⌋−n+t+h−1)}{2m}}{sin^{2} \frac{π(t+h−2)}{2m}}

· ∑_{k=1}^{n} \frac{(−1)^{n−k}(⌊m⌋ − n + 2k − 1)}{∏_{h=1}^{n} sin \frac{π(⌊m⌋−n+k+h−1)}{2m} ∏_{h=1}^{k−1} sin \frac{πh}{2m} ∏_{h=1}^{n−k} sin \frac{πh}{2m}} ) (7.16)

if m is a half-integer and k + 2η_{j} is even, it is asymptotically

\frac{2 · 4^{n^{2}−n+1}}{(8m)^{n}} (2^{n} ∏_{j=0}^{n−1} cos \frac{jπ}{2m})^{k}

× ∏_{1≤h<t≤n} ( sin \frac{π(η_{h} − η_{t})}{2m} · sin \frac{π(t − h)}{2m} sin \frac{π(η_{h} + η_{t})}{2m} · sin \frac{π(t + h − 2)}{2m})

× \begin{matrix}∏ & {n \atop h=1} & sin & \frac{π(m−n+2h−1)}{2m} \\ ∏ & {n−1 \atop h=1} & sin & \frac{πh}{2m} \end{matrix} ∏_{1≤h<t≤n} \begin{matrix}sin & 2 & \frac{π(m−n+h+t−1)}{2m} \\ sin & 2 & \frac{π(t+h−2)}{2m} \end{matrix}

× ∑_{k=1}^{n} \frac{(−1)^{n−k}(m − n + 2k − 1)}{∏_{h=1}^{n} sin \frac{π(m−n+k+h−1)}{2m} ∏_{h=1}^{k−1} sin \frac{πh}{2m} ∏_{h=1}^{n−k} sin \frac{πh}{2m}} (7.17)

if m is an integer and k + 2η_{j} is odd, and it is asymptotically

\frac{4^{n^{2}−n+1}}{(8m)^{n}} (2^{n} ∏_{j=0}^{n−1} cos \frac{jπ}{2m})^{k}

× ∏_{1≤h<t≤n} ( sin \frac{π(η_{h} − η_{t})}{2m} · sin \frac{π(t − h)}{2m} sin \frac{π(η_{h} + η_{t})}{2m} · sin \frac{π(t + h − 2)}{2m})

× (\begin{matrix}∏ & {n \atop h=1} & sin & \frac{π(m−n+2h−1/2)}{2m} \\ ∏ & {n−1 \atop h=1} & sin & \frac{πh}{2m} \end{matrix} ∏_{1≤h<t≤n} \begin{matrix}sin & 2 & \frac{π(m−n+h+t−1/2)}{2m} \\ sin & 2 & \frac{π(t+h−2)}{2m} \end{matrix}

× ∑_{k=1}^{n} \frac{(−1)^{n−k}(m − n + 2k − 1/2)}{∏_{h=1}^{n} sin \frac{π(m−n+k+h−1/2)}{2m} ∏_{h=1}^{k−1} sin \frac{πh}{2m} ∏_{h=1}^{n−k} sin \frac{πh}{2m}}

+ \begin{matrix}∏ & {n \atop h=1} & sin & \frac{π(m−n+2h−3/2)}{2m} \\ ∏ & {n−1 \atop h=1} & sin & \frac{πh}{2m} \end{matrix} ∏_{1≤h<t≤n} \begin{matrix}sin & 2 & \frac{π(m−n+h+t−3/2)}{2m} \\ sin & 2 & \frac{π(t+h−2)}{2m} \end{matrix}

× ∑_{k=1}^{n} \frac{(−1)^{n−k}(m − n + 2k − 3/2)}{∏_{h=1}^{n} sin \frac{π(m−n+k+h−3/2)}{2m} ∏_{h=1}^{k−1} sin \frac{πh}{2m} ∏_{h=1}^{n−k} sin \frac{πh}{2m}} ) (7.18)

if m is a half-integer and k + 2η_{j} is odd.

Proof. Here, in view of Theorem 31, we have to carry out the sum of (7.13) over all possible λ, i.e., over all λ_{1} > λ_{2} > ⋯ > λ_{n−1} > |λ_{n}| with λ_{1} + λ_{2} < 2m, where k ≡ 2η_{j} + 2λ_{j} mod 2. Leaving away factors which are independent of λ, the problem is to compute the sum

∑_{λ_{1}>λ_{2}>···>λ_{n−1}>|λ_{n}|} ; λ_{1}+λ_{2}<2m ′  ∏_{1≤h<t≤n} ( sin \frac{π(λ_{h} − λ_{t})}{2m} · sin \frac{π(λ_{h} + λ_{t})}{2m}),  (7.19)

where the sum ∑ ′ is restricted to integral, respectively to half-integral λ_{1}, λ_{2}, …, λ_{n}, depending on whether k + 2η_{j} is even or odd.

As in previous proofs, we have to distinguish between several cases. Let us first restrict the sum (7.19) to integral λ_{1}, λ_{2}, …, λ_{n}. Then, by repeating the argument from the proof of Theorem 30 of replacement of λ_{1} by 2m − λ_{1} and of replacement of λ_{n} by −λ_{n}, and using notation from that proof, we obtain that the sum (7.19) equals W_{1}(m) + W_{1}(m − 1) + W_{2}(m) + W_{2}(m − 1) if m is an integer, and it equals 2W_{1}(⌊m⌋) + 2W_{2}(⌊m⌋) if m is a half-integer. In the proof of Theorem 30 it was shown how to evaluate the sum W_{1}(c) and W_{2}(c) by means of Lemmas C7 and C8. If the results are substituted, we arrive at (7.15) and (7.16).

If we restrict the sum (7.19) to half-integral λ_{1}, λ_{2}, …, λ_{n}, then we are faced with the problem of evaluating the sum

X(c) = ∑_{c≥λ_{1}>···>λ_{n}>0} ′  ∏_{1≤h<t≤n} ( sin \frac{π(λ_{h} − λ_{t})}{2m} · sin \frac{π(λ_{h} + λ_{t})}{2m})

where the sum ∑ ′ is over all half-integral λ_{1}, λ_{2}, …, λ_{n}. More specifically, using the replacement of λ_{1} by 2m − λ_{1} and the replacement of λ_{n} by −λ_{n} another time, one sees readily that the sum (7.19) equals 4X(m − \frac{1}{2}) if m is an integer, and it equals 2X(m) + 2X(m − 1) if m is a half-integer.

In order to evaluate X(c), where c is a half-integer, we use (7.3) to rewrite it as

∑_{c≥λ_{1}>···>λ_{n}>0} ; ′ 2^{−n^{2}+2n−1} det_{1≤h,t≤n} ( cos(π(h − 1)λ_{t}/m))

= ∑_{c≥λ_{1}>···>λ_{n}>0} ′  \frac{1}{2^{n^{2}−n+1}} det_{1≤h,t≤n} ((e^{πi(h−1)/m})^{λ_{t}} + (e^{πi(h−1)/m})^{−λ_{t}}).

Replacing λ_{t} by λ_{t} + n − t, t = 1, 2, …, n, we obtain the expression

∑_{c−n+1≥λ_{1}≥···≥λ_{n}≥ \frac{1}{2}} ′  \frac{1}{2^{n^{2}−n+1}} det_{1≤h,t≤n} ((e^{πi(h−1)/m})^{λ_{t}+n−t} + (e^{πi(h−1)/m})^{−(λ_{t}+n−t)}). (7.20)

Again, this determinant can be expressed in terms of an even orthogonal character. (See (C.49) for the definition.) Writing, as before, q for e^{πi/2m}, the sum in (7.20) equals

\frac{1}{2^{n^{2}−n+1}} det_{1≤h,t≤n} ((q^{h−1})^{n−t} + (q^{h−1})^{−(n−t)}) ∑_{c−n+1≥λ_{1}≥···≥λ_{n}≥ \frac{1}{2}} ′  so_{λ}^{even}(q^{n−1}, q^{n−2}, …, 1),

(7.21)

where, still, the sum ∑ ′ is over all half-integral λ_{1}, λ_{2}, …, λ_{n}. (The reader should note that the second determinant in (C.49) vanishes if one of the variables x_{t}, t = 1, 2, …, n, is 1, which is the case in our situation.) The sum in (7.21) can also be simplified by means of (7.10). Use of this formula in (7.21) gives

\frac{1}{2^{n^{2}−n+1}} det_{1≤h,t≤n} ((q^{h−1})^{n−t} + (q^{h−1})^{−(n−t)})

× so_{(((2c−2n+3)/4)^{n})}^{even}(q^{n−1}, q^{n−2}, …, 1)

× ∑_{p=−(2c−2n+1)/4}^{(2c−2n+1)/4} so_{(((2c−2n+3)/4)^{n−1},p)}^{even}(q^{n−1}, q^{n−2}, …, 1).

Again, the first determinant can be evaluated by means of (B.3), the even orthogonal character of shape (((2c − 2n + 3)/4)^{n}) can be evaluated by using the definition (C.49) and (B.3), while the sum over p of even orthogonal characters is evaluated in Lemma C7. Substitution of the result and simplification eventually leads to the expressions (7.17) and (7.18). □

Acknowledgements

I am indebted to the referee for an extremely careful reading of the manuscript, for the many suggestions which improved the presentation, and for pointing out a flaw in the original statement and proof of Lemma A.

Appendix A. A saddle point approximation

The subject of this appendix is the saddle point approximation which is needed in the proof of Theorem 16.

Lemma A. Let m be a positive integer and d be an integer or half-integer. Let ω = e^{2πi/m}, and let {r_{1}, r_{2}, …, r_{n}} be a subset of {0, 1, …, m−1}, where the r_{j}’s are pairwise distinct. Furthermore, let

C = C(r_{1}, …, r_{n}; m) := max_{θ} ∏_{j=1}^{n} 2 |cos (\frac{π(θ + r_{j})}{m})| ,  (A.1)

and let θ_{1}, θ_{2}, …, θ_{s} be the values of θ where the maximum in (A.1) is attained. Then, as k tends to infinity such that d + \frac{nk}{2} is an integer, the coefficient of z^{d+ \frac{nk}{2}} in

∏_{j=1}^{n}(1 + ω^{r_{j}}z)^{k}  (A.2)

is asymptotically

\frac{1}{\sqrt{2πk}} C^{k} ∑_{ℓ=1}^{s} ε(r_{1}, …, r_{n}; m)\frac{ω^{\frac{k}{2} ∑_{j=1}^{n} r_{j}−dθ_{ℓ}}}{\sqrt{c_{0}(θ_{ℓ})}} ,  (A.3)

where

ε(r_{1}, …, r_{n}; m) = sgn ∏_{j=1}^{n} cos (\frac{π(θ + r_{j})}{m})

and

c_{0}(θ) = ∑_{j=1}^{n} (2 cos \frac{π(θ + r_{j})}{m})^{−2} ,

except if the sum in (A.3) vanishes, in which case the asymptotic order of the coefficient of z^{d+ \frac{nk}{2}} in (A.2) is strictly less than C^{k}.

Remark. The proof of the lemma below shows that, in fact, each θ_{ℓ} is a solution of the equation

∑_{j=1}^{n} tan \frac{π(θ + r_{j})}{m} = 0.  (A.4)

Proof of Lemma A. We apply the saddle point method (see [9]). (We would in fact like to directly apply a general “law of large powers,” such as for example Theorem 6.5 in [9]. However, I was not able to find an applicable theorem in the literature. In particular, Theorem 6.5 from [9] does not apply in our situation since it requires that the coefficients of the series of which the powers are formed are positive.)

We begin by writing the coefficient of z^{d+ \frac{nk}{2}} in the product (A.2) as an integral,

\frac{1}{2πi} ∫_{C} \frac{∏_{j=1}^{n}(1 + ω^{r_{j}}z)^{k}}{z^{d+ \frac{nk}{2}}} \frac{dz}{z} ,  (A.5)

where C is some contour encircling the origin counter-clockwise. The task is to choose a contour C such that the contribution of the integrand is concentrated very close to a finite number of points on the contour (the saddle points), whereas otherwise the contributions are negligible. The equation for the saddle points is (see [9, Sec. 6.3])

\frac{d}{dz} (\frac{∏_{j=1}^{n}(1 + ω^{r_{j}}z)^{k}}{z^{d+ \frac{nk}{2}}} ) = 0.

Equivalently, this is

\frac{k}{2} ∑_{j=1}^{n} \frac{ω^{r_{j}}z − 1}{ω^{r_{j}}z + 1} − d = 0.  (A.6)

For the saddle point method to work, it is often not necessary to determine the solution(s) of the saddle point equation exactly. It is often sufficient to find a suitable approximation for large k. This is also the case here. Clearly, if k is large, then the coefficient of k/2 in (A.6) must be very small. Therefore, as a first approximation, we consider solutions of the equation

∑_{j=1}^{n} \frac{ω^{r_{j}}z − 1}{ω^{r_{j}}z + 1} = 0.  (A.7)

This is (after denominators have been cleared) a polynomial equation of degree n. Therefore we must expect (up to) n different solutions to (A.7).

We claim that all n solutions to (A.7) are distinct and moreover have modulus 1. To see this, we substitute e^{2πiθ/m} for z in (A.7). After a little manipulation, we obtain the equivalent equation

∑_{j=1}^{n} tan \frac{π(θ + r_{j})}{m} = 0.  (A.8)

The left-hand side of (A.8) is defined for 0 ≤ θ ≤ m except at the values (which have to be taken modulo m)

θ = −r_{1} + \frac{m}{2}, −r_{2} + \frac{m}{2}, …, −r_{n} + \frac{m}{2},

all of which are distinct by assumption. These values are simple poles of the function on the left-hand side of (A.8). Between two successive poles the function is monotone increasing and continuous (here, we regard the interval [0, m] as a circular interval, identifying 0 and m), ranging there from −∞ to +∞. Hence, in each open interval bounded by two successive poles there lies exactly one solution to (A.8). Since the number of poles, and, thus, of such intervals, is n, there are n values of θ in the range [0, m) satisfying equation (A.8). In turn, each solution to (A.8) produces a solution to (A.7) with modulus 1. This establishes that, indeed, all n solutions to (A.7) have modulus 1 and are pairwise distinct.

Let θ_{1}, θ_{2}, …, θ_{n} be the above solutions to (A.8), and, for a given θ_{ℓ}, let z_{ℓ} = e^{2πiθ_{ℓ}/m} be the corresponding solution to (A.7). Obviously, we have 1 + ω^{r_{j}}z_{ℓ} ≠ 0 for all ℓ and j.

We shall need a slightly better approximation of the saddle points, which will approximate (A.6) up to second order. Let y_{ℓ} := z_{ℓ}(1 + \frac{c_{ℓ}}{k}). Substitution of y_{ℓ} in (A.6) leads to the equation

\frac{k}{2} ∑_{j=1}^{n} \frac{ω^{r_{j}}y_{ℓ} − 1}{ω^{r_{j}}y_{ℓ} + 1} − d

= \frac{k}{2} ∑_{j=1}^{n} \frac{ω^{r_{j}}z_{ℓ} − 1}{ω^{r_{j}}z_{ℓ} + 1} (1 + \frac{c_{ℓ}ω^{r_{j}}z_{ℓ}}{k(ω^{r_{j}}z_{ℓ} − 1)} − \frac{c_{ℓ}ω^{r_{j}}z_{ℓ}}{k(ω^{r_{j}}z_{ℓ} + 1)} + O (k^{−2})) − d

= c_{ℓ} ∑_{j=1}^{n} \frac{ω^{r_{j}}z_{ℓ}}{(ω^{r_{j}}z_{ℓ} + 1)^{2}} − d + O (k^{−1})

= c_{ℓ} ∑_{j=1}^{n} \frac{1}{(2 cos \frac{π(r_{j}+θ_{ℓ})}{m})^{2}} − d + O (k^{−1})

= 0,

where we used (A.7) in the third line. Thus, y_{ℓ} will approximate a solution to (A.6) up to second order, if we choose

c_{ℓ} = \frac{d}{∑_{j=1}^{n} (2 cos \frac{π(r_{j} + θ_{ℓ})}{m})^{−2}} .

Returning to the integral (A.5) that we want to approximate, we choose as the contour C the circle

\{(1 + \frac{c_{ℓ}}{k} ) e^{it} : 0 ≤ t ≤ 2π\} ,  (A.9)

which contains the (approximated) saddle points y_{ℓ}. Thus, writing θ̄ for 2πθ_{ℓ}/m for short, the portion of the integral (A.5) contributed by a neighbourhood [θ̄− ε, θ̄+ ε] of θ̄ becomes

\frac{1}{2π} ∫_{−ε}^{ε} \begin{matrix}∏ & {n \atop j=1} & (1 + ω^{r_{j}}(1 + \frac{c_{ℓ}}{k})e^{i(θ̄+t)}) & k \\ ((1 + \frac{c_{ℓ}}{k})e^{i(θ̄+t)}) & d+ \frac{nk}{2} \end{matrix} dt.  (A.10)

In order to estimate this integral, we compute a Taylor expansion of

log (1 + ω^{r_{j}} (1 + \frac{c_{ℓ}}{k} ) e^{i(θ̄+t)}) ,

thereby also neglecting terms which are of order O(k^{−2}):

log (1 + ω^{r_{j}} (1 + \frac{c_{ℓ}}{k} ) e^{i(θ̄+t)})

= log(1 + e^{iθ̄}ω^{r_{j}}) + \frac{c_{ℓ}}{k} \frac{e^{iθ̄}ω^{r_{j}}}{(1 + e^{iθ̄}ω^{r_{j}})} + it (\frac{e^{iθ̄}ω^{r_{j}}}{1 + e^{iθ̄}ω^{r_{j}}} + \frac{c_{ℓ}}{k} \frac{e^{iθ̄}ω^{r_{j}}}{(1 + e^{iθ̄}ω^{r_{j}})^{2}} )

− \frac{t^{2}}{2} (\frac{e^{iθ̄}ω^{r_{j}}}{(1 + e^{iθ̄}ω^{r_{j}})^{2}} + \frac{c_{ℓ}}{k} \frac{e^{iθ̄}ω^{r_{j}} (1 − e^{iθ̄}ω^{r_{j}})}{(1 + e^{iθ̄}ω^{r_{j}})^{3}} ) + O(t^{3}) + O (k^{−2}) .

This expansion is only valid for

|\frac{(1 + \frac{c_{ℓ}}{k} ) e^{i(θ̄+t)} − e^{iθ̄}ω^{r_{j}}}{1 + e^{iθ̄}ω^{r_{j}}} | < 1,  (A.11)

whence, in general, it will only be valid for t in a neighbourhood of 0 and k large enough. More precisely, choose k_{0} so that the left-hand side of (A.11) with t = 0 is less than 1, i.e.,

|\frac{c_{ℓ}}{k_{0}(1 + e^{iθ̄}ω^{r_{j}})}| < 1.

Then the above expansion of the logarithm will be valid for −ε ≤ t ≤ ε and k ≥ k_{0} for some ε which is independent of k.

If this is substituted in (A.10), then, after some simplification based on (A.8) and the definition of c_{ℓ}, we obtain

\frac{1}{2πe^{iθ̄(d+ \frac{nk}{2} )}} ∏_{j=1}^{n}(1 + e^{iθ̄}ω^{r_{j}})^{k}

× ∫_{−ε}^{ε} exp (−\frac{kt^{2}}{2} ( ∑_{j=1}^{n} \frac{e^{iθ̄}ω^{r_{j}}}{(1+e^{iθ̄}ω^{r_{j}})^{2}} + \frac{c_{ℓ}}{k} ∑_{j=1}^{n} \frac{e^{iθ̄}ω^{r_{j}}(1−e^{iθ̄}ω^{r_{j}})}{(1+e^{iθ̄}ω^{r_{j}})^{3}}) + O(kt^{3}) + O (k^{−1})). (A.12)

Next we estimate the integral in (A.12). One argues as is usual in such a situation: the essential contribution to the integral comes from the range |t| < (log k)/\sqrt{k}, the rest being of order O (k^{−1}). In this range, the error terms O(kt^{3}) + O (k^{−1}) are of the order O((log^{3} k)/\sqrt{k}). Finally, one extends the integral to the complete range −∞ < t < ∞, upon making another error of the order O (k^{−1}). Leaving these details to the reader, up to an error of O((log^{3} k)/\sqrt{k}) the integral in (A.12) is asymptotically

∫_{−∞}^{∞} exp (−S \frac{kt^{2}}{2} ) = \sqrt{\frac{2π}{kS}} ,  (A.13)

where

S = ∑_{j=1}^{n} \frac{e^{iθ̄}ω^{r_{j}}}{(1 + e^{iθ̄}ω^{r_{j}})^{2}} + \frac{c_{ℓ}}{k} ∑_{j=1}^{n} \frac{e^{iθ̄}ω^{r_{j}} (1 − e^{iθ̄}ω^{r_{j}})}{(1 + e^{iθ̄}ω^{r_{j}})^{3}} = c_{0}(θ_{ℓ}) + O (\frac{1}{k}) .

with c_{0}(θ_{ℓ}) as given in the statement of the theorem. The evaluation (A.13) is valid since c_{0} > 0, and hence ℜ𝔢 S > 0 for k large enough.

Clearly, the second term in the definition of S makes only a negligible contribution to the asymptotic behaviour of (A.13). If the result is substituted in (A.12), we obtain that the portion of the integral (A.5) near the saddle point z_{ℓ}, given explicitly in (A.10), contributes

\frac{∏_{j=1}^{n}(1 + ω^{r_{j}+θ_{ℓ}})^{k}}{ω^{θ_{ℓ}(d+ \frac{nk}{2} )}} \frac{1}{\sqrt{2πc_{0}(θ_{ℓ})k}} ,

as k tends to infinity, up to an error of O((log^{3} k)/\sqrt{k}). Summation of these contributions, neglecting those which are asymptotically smaller, yields indeed the expression (A.3) after some manipulation.

It remains to show that the contribution of the remaining portions of the integral (A.5) is negligible. This is a routine matter, the details of which we leave to the reader. We content ourselves to indicate that one may start with the observation that the modulus of the integrand of (A.5) along the contour (A.9) is bounded by

|\begin{matrix}∏ & {n \atop j=1} & (1 + ω^{r_{j}}(1 + \frac{c_{ℓ}}{k})e^{it}) & k \\ ((1 + \frac{c_{ℓ}}{k})e^{it}) & d+ \frac{nk}{2} \end{matrix} | < constant^{\sqrt{k}} |∏_{j=1}^{n} (1 + ω^{r_{j}}e^{it})^{k}| ,  (A.14)

the constant depending on r_{1}, r_{2}, …, r_{n} but not on k, as long as t stays away from tiny neighbourhoods of π − \frac{2πr_{j}}{m}, j = 1, 2, …, n, determined by the condition that |1 + ω^{r_{j}}e^{it}| ≤ 1/\sqrt{k}, and that the right-hand side of (A.14) is exponentially smaller than C^{k} for any t in these remaining portions, while the contributions in these tiny neighbourhoods is even super-exponentially small.. □

Appendix B. Some determinants

In our computations we need frequently the following determinant evaluations. All of them are readily proved by the standard argument that proves Vandermonde-type determinant evaluations.

Lemma B. Let n by a non-negative integer. Then

det_{1≤h,t≤n} (x_{h}^{t−1/2} + x_{h}^{1/2−t}) = (x_{1}x_{2} ⋯ x_{n})^{−n+1/2} ∏_{1≤h<t≤n}(x_{h} − x_{t})(1 − x_{h}x_{t}) ∏_{h=1}^{n}(x_{h} + 1), (B.1)

det_{1≤h,t≤n} (x_{h}^{t−1/2} − x_{h}^{−t+1/2}) = (x_{1}x_{2} ⋯ x_{n})^{−n+1/2} ∏_{1≤h<t≤n}(x_{h} − x_{t})(1 − x_{h}x_{t}) ∏_{h=1}^{n}(x_{h} − 1), (B.2)

det_{1≤h,t≤n}(x_{h}^{t−1} + x_{h}^{−(t−1)}) = 2 · (x_{1} ⋯ x_{n})^{−n+1} ∏_{1≤h<t≤n}(x_{h} − x_{t})(1 − x_{h}x_{t}),  (B.3)

det_{1≤h,t≤n}(x_{h}^{t} − x_{h}^{−t}) = (x_{1} ⋯ x_{n})^{−n} ∏_{1≤h<t≤n}(x_{h} − x_{t})(1 − x_{h}x_{t}) ∏_{h=1}^{n}(x_{h}^{2} − 1). (B.4)

Appendix C. Odd and even orthogonal characters, and Schur functions at special values of the arguments

In this appendix we provide the evaluations of odd orthogonal characters, Schur functions of rectangular shape and nearly rectangular shape, and of certain sums of even orthogonal characters, where the variables are specialized in peculiar ways. The evaluations of the odd orthogonal characters are needed for the proof of Theorem 18 (on which, in turn, hinges also the proof of Theorem 13) and the proof of Theorem 20, the evaluations of the special Schur functions are needed for the proofs of Theorems 22, 26, and 28, while the evaluations of the sums of specialized even orthogonal characters are needed for the proofs of Theorems 30 and 32.

Recall that, given a partition λ = (λ_{1}, λ_{2}, …, λ_{n}) (i.e., a non-increasing sequence of non-negative integers) or half-partition (which, by definition, is a non-increasing sequence of positive odd integers divided by 2), the odd orthogonal character so_{λ}^{odd}(x_{1}, x_{2}, …, x_{n}) is defined by (see [13, (24.28)])

so_{λ}^{odd}(x_{1}, x_{2}, …, x_{n}) = \frac{det_{1≤h,t≤n}(x_{h}^{λ_{t}+n−t+1/2} − x_{h}^{−(λ_{t}+n−t+1/2)})}{det_{1≤h,t≤n}(x_{h}^{n−t+1/2} − x_{h}^{−(n−t+1/2)})}.  (C.1)

It is not difficult to see that the denominator in (C.1) does in fact cancel out, so that any odd orthogonal character so_{λ}^{odd}(x_{1}, x_{2}, …, x_{n}) is in fact a Laurent polynomial in x_{1}, x_{2}, …, x_{n} (i.e., a polynomial in x_{1}, x_{1}^{−1}, x_{2}, x_{2}^{−1}, …, x_{n}, x_{n}^{−1}), and is thus well-defined for any choice of the variables x_{1}, x_{2}, …, x_{n} such that all of them are non-zero.

As earlier in Section 4, we shall use the notation ((p/2)^{n}) for the vector of n components, all of them equal to p/2.

Lemma C1. Let m and n be positive integers with m ≥ n, and let q = e^{πi/m}. Then we have

so_{(((m−n)/2)^{n})}^{odd}(q^{n−1}, q^{n−3}, …, q^{−n+3}, q^{−n+1}) = \frac{m^{n/2} ∏_{h=1}^{n/2} cot \frac{(2h − 1)π}{2m}}{2^{\binom{n}{2}} ∏_{1≤h<t≤n} sin \frac{(t − h)π}{m}}  (C.2)

if n is even, and

so_{(((m−n)/2)^{n})}^{odd}(q^{n−1}, q^{n−3}, …, q^{−n+3}, q^{−n+1}) = \frac{m^{(n+1)/2} ∏_{h=1}^{(n−1)/2} cot \frac{hπ}{m}}{2^{\binom{n}{2}} ∏_{1≤h<t≤n} sin \frac{(t − h)π}{m}}  (C.3)

if n is odd.

Proof. In principle, we would like to specialize in the definition (C.1) of the odd orthogonal character. However, we face the difficulty that, because of the peculiar choice of the variables x_{1}, x_{2}, …, x_{n} in (C.2) and (C.3), direct substitution gives an indeterminate expression 0/0. More precisely, the problem is the pairs of reciprocal variables (such as q^{n−1} and q^{−n+1}, q^{n−3} and q^{−n+3}, etc.). In the case that n is odd, we have the additional problem that one of the variables is q^{0} = 1.

To overcome this problem, we have recourse to de l’Hôpital’s rule. We have to distinguish between two cases, depending on whether n is even or odd.

Let first n be even. In that case we must compute the limit as

x_{1} → q^{n−1}, x_{2} → q^{n−3}, …, x_{n/2} → q  (C.4)

of the right-hand side of (C.1) with x_{n/2+1} = q^{−1}, …, x_{n−1} = q^{−n+3}, x_{n} = q^{−n+1}. Upon a simultaneous rearrangement and change of signs of the rows of the two determinants in (C.1), one sees that this is equivalent to computing the limit (C.4) of (det A_{1})/(det B_{1}), where

A_{1} = (\begin{matrix}x_{h}^{\frac{m+n+1}{2} −t} − x_{h}^{−( \frac{m+n+1}{2} −t)} & 1 ≤ h ≤ n/2 \\ q^{(2n−2h+1)( \frac{m+n+1}{2} −t)} − q^{−(2n−2h+1)( \frac{m+n+1}{2} −t)} & n/2 < h ≤ n \end{matrix})_{1≤h,t≤n}

and

B_{1} = (\begin{matrix}x_{h}^{n−t+ \frac{1}{2}} − x_{h}^{−(n−t+ \frac{1}{2} )} & 1 ≤ h ≤ n/2 \\ q^{(2n−2h+1)(n−t+ \frac{1}{2} )} − q^{−(2n−2h+1)(n−t+ \frac{1}{2} )} & n/2 < h ≤ n \end{matrix})_{1≤h,t≤n} .

Now applying de l’Hôpital’s rule, this limit is equal to the limit (C.4) of

\frac{det A_{2}}{\frac{∂}{∂x_{1}} \frac{∂}{∂x_{2}} ⋯ \frac{∂}{∂x_{n/2}} det B_{1}} ,  (C.5)

where

A_{2} = (\begin{matrix}(\frac{m+n+1}{2} − t)(q^{(n−2h+1)( \frac{m+n+1}{2} −t−1)} + q^{−(n−2h+1)( \frac{m+n+1}{2} −t+1)}) & 1 ≤ h ≤ n/2 \\ q^{(2n−2h+1)( \frac{m+n+1}{2} −t)} − q^{−(2n−2h+1)( \frac{m+n+1}{2} −t)} & n/2 < h ≤ n \end{matrix})

= (\begin{matrix}(−1)^{\frac{n}{2} −h}i q^{−(n−2h+1)}(\frac{m+n+1}{2} − t) \\ ×(q^{(n−2h+1)( \frac{n+1}{2} −t)} − q^{−(n−2h+1)( \frac{n+1}{2} −t)}) & 1 ≤ h ≤ n/2 \\ (−1)^{n−h}i (q^{(2n−2h+1)( \frac{n+1}{2} −t)} + q^{−(2n−2h+1)( \frac{n+1}{2} −t)}) & n/2 < h ≤ n \end{matrix}) .  (C.6)

The limit of the denominator of (C.5) is readily obtained, because the determinant of B_{1} can actually be evaluated by means of (B.2). To be precise, with

x_{n/2+1} = q^{n−1}, …, x_{n−1} = q^{3}, x_{n} = q,  (C.7)

we have

det B_{1} = (x_{1}x_{2} ⋯ x_{n})^{−n+1/2} ∏_{1≤h<t≤n}(x_{t} − x_{h})(1 − x_{h}x_{t}) ∏_{h=1}^{n}(x_{h} − 1).  (C.8)

Thus, if we differentiate the product on the right-hand side of (C.8) with respect to x_{1}, say, (using the product rule, of course), and subsequently set x_{1} = q^{n−1} (which is exactly what we want to do; see (C.4)), then it is just one term in the derivative which contributes, namely

\frac{1}{x_{1} − x_{n/2+1}} (x_{1}x_{2} ⋯ x_{n})^{−n+1/2} ∏_{1≤h<t≤n}(x_{t} − x_{h})(1 − x_{h}x_{t}) ∏_{h=1}^{n}(x_{h} − 1),

in which the factor (x_{1}−x_{n/2+1}) cancels; all other terms vanish because of the occurrence of the factor (x_{1} − x_{n/2+1}). (Recall that x_{n/2+1} = q^{n−1}; see (C.7).) An analogous argument applies to the other pairs of variables. Hence, the denominator of (C.5) is equal to

(−1)^{\binom{n/2}{2}}(q^{n^{2}/2})^{−n+1/2} ∏_{1≤h<t≤n/2}(q^{n−2t+1} − q^{n−2h+1})^{4}(1 − q^{2n−2h−2t+2})^{4}

× ∏_{h=1}^{n/2}(q^{n−2h+1} − 1)^{2}(1 − q^{2n−4h+2}). (C.9)

Next we devote ourselves to the evaluation of the determinant of A_{2}, with A_{2} given by (C.6). For t = 1, 2, …, \frac{n}{2} we subtract column n − t + 1 from column t. As a result we obtain that det A_{2} is equal to det A_{3}, where A_{3} is the n × n block matrix

A_{3} = (\begin{matrix}A_{3}^{(1)} & ∗ \\ 0 & A_{3}^{(2)} \end{matrix}) ,  (C.10)

with A_{3}^{(1)} the \frac{n}{2} × \frac{n}{2} matrix

A_{3}^{(1)} = ((−1)^{\frac{n}{2} −h}i q^{−(n−2h+1)}m(q^{(n−2h+1)( \frac{n+1}{2} −t)} − q^{−(n−2h+1)( \frac{n+1}{2} −t)}))_{1≤h,t≤n/2},

and A_{3}^{(2)} the \frac{n}{2} × \frac{n}{2} matrix

A_{3}^{(2)} = ((−1)^{\frac{n}{2} −h}i (q^{(n−2h+1)( \frac{1}{2} −t)} + q^{−(n−2h+1)( \frac{1}{2} −t)}))_{1≤h,t≤n/2}.

Clearly, the determinant det A_{3} is equal to the product (det A_{3}^{(1)}) · (det A_{3}^{(2)}). Using again (B.2), we have

det A_{3}^{(1)} = i^{n/2}q^{−n^{2}/4}m^{n/2}(q^{n^{2}/4})^{−n/2+1/2}

× ∏_{1≤h<t≤n/2}(q^{n−2h+1} − q^{n−2t+1})(1 − q^{2n−2h−2t+2}) ∏_{h=1}^{n/2}(q^{n−2h+1} − 1), (C.11)

whereas by means of (B.1) we have

det A_{3}^{(2)} = (−1)^{\binom{n/2}{2}}i^{n/2}(q^{n^{2}/4})^{(1−n)/2}

× ∏_{1≤h<t≤n/2}(q^{n−2h+1} − q^{n−2t+1})(1 − q^{2n−2h−2t+2}) ∏_{h=1}^{n/2}(1 + q^{n−2h+1}). (C.12)

If we now combine (C.10), (C.11), (C.12), and (C.9), use the fact that det A_{2} = det A_{3}, and substitute all this in (C.5), then, after some simplification, we obtain

so_{(((m−n)/2)^{n})}^{odd}(q^{n−1}, q^{n−3}, …, q^{−n+3}, q^{−n+1})

= \frac{(−1)^{n/2}m^{n/2} ∏_{h=1}^{n/2}(1 + q^{n−2h+1})}{∏_{1≤h<t≤n/2}(q^{h−t} − q^{t−h})^{2} ∏_{1≤h,t≤n/2}(q^{−n+h+t−1} − q^{n−h−t+1}) ∏_{h=1}^{n/2}(1 − q^{n−2h+1})},

which can be rewritten as (C.2).

Now let n be odd. We proceed in a completely analogous manner. Again, the task is to compute the specialized odd orthogonal character, by means of a limit of its definition (C.1). Here, the difficulty is not only the pairs of reciprocal variables, but also that one of the variables is q^{0} = 1.

What we must compute is the limit as

x_{1} → q^{n−1}, x_{2} → q^{n−3}, …, x_{(n−1)/2} → q^{2}, x_{(n+1)/2} → 1,  (C.13)

of the right-hand side of (C.1) with x_{(n+3)/2} = q^{−2}, …, x_{n−1} = q^{−n+3}, x_{n} = q^{−n+1}. Upon a simultaneous rearrangement and change of signs of the rows of the two determinants in (C.1), one sees that this is equivalent to computing the limit (C.13) of (det A_{4})/(det B_{4}), where

A_{4} = (\begin{matrix}x_{h}^{\frac{m+n+1}{2} −t} − x_{h}^{−( \frac{m+n+1}{2} −t)} & 1 ≤ h ≤ (n + 1)/2 \\ q^{(2n−2h+2)( \frac{m+n+1}{2} −t)} − q^{−(2n−2h+2)( \frac{m+n+1}{2} −t)} & (n + 1)/2 < h ≤ n \end{matrix})_{1≤h,t≤n}

and

B_{4} = (\begin{matrix}x_{h}^{n−t+ \frac{1}{2}} − x_{h}^{−(n−t+ \frac{1}{2} )} & 1 ≤ h ≤ (n + 1)/2 \\ q^{(2n−2h+2)(n−t+ \frac{1}{2} )} − q^{−(2n−2h+2)(n−t+ \frac{1}{2} )} & (n + 1)/2 < h ≤ n \end{matrix})_{1≤h,t≤n} .

Applying de l’Hôpital’s rule, this limit is equal to the limit (C.13) of

\frac{det A_{5}}{\frac{∂}{∂x_{1}} \frac{∂}{∂x_{2}} ⋯ \frac{∂}{∂x_{(n+1)/2}} det B_{4}} ,  (C.14)

where

A_{5} = (\begin{matrix}(\frac{m+n+1}{2} − t)(q^{(n−2h+1)( \frac{m+n+1}{2} −t−1)} + q^{−(n−2h+1)( \frac{m+n+1}{2} −t+1)}) & 1 ≤ h ≤ (n + 1)/2 \\ q^{(2n−2h+2)( \frac{m+n+1}{2} −t)} − q^{−(2n−2h+2)( \frac{m+n+1}{2} −t)} & (n + 1)/2 < h ≤ n \end{matrix})

= (\begin{matrix}(−1)^{\frac{n+1}{2} −h}q^{−(n−2h+1)}(\frac{m+n+1}{2} − t) \\ ×(q^{(n−2h+1)( \frac{n+1}{2} −t)} + q^{−(n−2h+1)( \frac{n+1}{2} −t)}) & 1 ≤ h ≤ (n + 1)/2 \\ (−1)^{n+1−h}(q^{(2n−2h+2)( \frac{n+1}{2} −t)} − q^{−(2n−2h+2)( \frac{n+1}{2} −t)}) & (n + 1)/2 < h ≤ n \end{matrix}) . (C.15)

The limit of the denominator of (C.14) is obtained in the same way as before, by using the complete factorization of det B_{4} by means of (B.2). (It should be noted that the difference between the denominator of (C.14) and that of (C.5) is the number of differentiations, and the number of variables with respect to which the limit is taken.) After a small calculation, it turns out that the denominator of (C.14) is equal to

(−1)^{\binom{(n+1)/2}{2}}(q^{(n^{2}−1)/2})^{−n+1/2} ∏_{1≤h<t≤(n−1)/2}(q^{n−2t+1} − q^{n−2h+1})^{4}(1 − q^{2n−2h−2t+2})^{4}

× ∏_{h=1}^{(n−1)/2}(q^{n−2h+1} − 1)^{6}(1 − q^{2n−4h+2}). (C.16)

Next we devote ourselves to the evaluation of the determinant of A_{5}, with A_{5} given by (C.15). For t = 1, 2, …, \frac{n−1}{2} we add column n − t + 1 to column t. As a result we obtain that det A_{5} is equal to det A_{6}, where A_{6} is the n × n block matrix

A_{6} = (\begin{matrix}A_{6}^{(1)} & ∗ \\ 0 & A_{6}^{(2)} \end{matrix}) ,  (C.17)

with A_{6}^{(1)} the \frac{n+1}{2} × \frac{n+1}{2} matrix

A_{6}^{(1)} = ((−1)^{\frac{n+1}{2} −h}q^{−(n−2h+1)}m

× \begin{cases}q^{(n−2h+1)( \frac{n+1}{2} −t)} + q^{−(n−2h+1)( \frac{n+1}{2} −t)} & 1 ≤ t ≤ \frac{n−1}{2} \\ 1 & t = \frac{n+1}{2} \end{cases} )_{1≤h,t≤(n+1)/2},

and A_{6}^{(2)} the \frac{n−1}{2} × \frac{n−1}{2} matrix

A_{6}^{(2)} = ((−1)^{\frac{n+1}{2} −h}(q^{(n−2h+1)(−t)} − q^{−(n−2h+1)(−t)}))_{1≤h,t≤(n−1)/2}.

(It should be noted that the entries in column (n+1)/2 of A_{6}^{(1)} are exactly a half of what would result from substituting t = (n + 1)/2 in the definition of the entries of the other columns.) Clearly, the determinant det A_{6} is equal to the product (det A_{6}^{(1)}) · (det A_{6}^{(2)}). Using (B.3), we have

det A_{6}^{(1)} = (−1)^{(n^{2}−1)/8}q^{−(n^{2}−1)/4}m^{(n+1)/2}(q^{(n^{2}−1)/4})^{−n/2+1/2}

× ∏_{1≤h<t≤(n+1)/2}(q^{n−2h+1} − q^{n−2t+1})(1 − q^{2n−2h−2t+2}), (C.18)

whereas by means of (B.4) we have

det A_{6}^{(2)} = q^{−(n^{2}−1)/4}(q^{(n^{2}−1)/4})^{−(n−1)/2}

× ∏_{1≤h<t≤(n−1)/2}(q^{n−2h+1} − q^{n−2t+1})(1 − q^{2n−2h−2t+2}) ∏_{h=1}^{(n−1)/2}(q^{2n−4h+2} − 1). (C.19)

If we now combine (C.17), (C.18), (C.19), and (C.16), use the fact that det A_{5} = det A_{6}, and substitute all this in (C.14), then, after some simplification, we obtain

so_{(((m−n)/2)^{n})}^{odd}(q^{n−1}, q^{n−3}, …, q^{−n+3}, q^{−n+1})

= \frac{m^{(n+1)/2}}{∏_{1≤h<t≤(n−1)/2}(q^{h−t} − q^{t−h})^{2} ∏_{1≤h,t≤(n−1)/2}(q^{n−h−t+1} − q^{−n+h+t−1})}

× ∏_{h=1}^{(n−1)/2} \frac{(q^{\frac{n+1}{2} −h} + q^{− \frac{n+1}{2} +h})}{(q^{\frac{n+1}{2} −h} − q^{− \frac{n+1}{2} +h})^{3}} ,

which can be rewritten as (C.3). □

The next lemma provides a similar evaluation of a rectangularly shaped odd orthogonal character, the special values of the arguments at which the character is evaluated being exactly the negative values of those in Lemma C1. This evaluation is needed in the proof of Theorem 18, however only in the case that m is even. For the sake of completeness, we state also the corresponding result for odd m without proof.

Lemma C2. Let m and n be positive integers with m ≥ n, and let q = e^{πi/m}. Then we have

so_{(((m−n)/2)^{n})}^{odd}(−q^{n−1}, −q^{n−3}, …, −q^{−n+3}, −q^{−n+1}) = \frac{m^{n/2} ∏_{h=1}^{n/2} tan \frac{(2h − 1)π}{2m}}{2^{\binom{n}{2}} ∏_{1≤h<t≤n} sin \frac{(t − h)π}{m}} (C.20)

if n is even (regardless of m),

so_{(((m−n)/2)^{n})}^{odd}(−q^{n−1}, −q^{n−3}, …, −q^{−n+3}, −q^{−n+1}) = \frac{(−1)^{(m−n)/2}m^{(n−1)/2} ∏_{h=1}^{(n−1)/2} tan \frac{hπ}{m}}{2^{\binom{n}{2}} ∏_{1≤h<t≤n} sin \frac{(t − h)π}{m}}

(C.21)

if both n and m are odd, and

so_{(((m−n)/2)^{n})}^{odd}(−q^{n−1}, −q^{n−3}, …, −q^{−n+3}, −q^{−n+1}) = 0  (C.22)

if n is odd and m is even.

Proof. Again, when we would directly specialize in the definition (C.1) of the odd orthogonal character, then we face the difficulty that we obtain an indeterminate expression 0/0. As we mentioned before the statement of the theorem, we are only going to discuss the case that m is even. The arguments are however completely analogous if m is odd.

Let first n be even. In that case we must compute the limit as

x_{1} → −q^{n−1}, x_{2} → −q^{n−3}, …, x_{n/2} → −q  (C.23)

of the right-hand side of (C.1) with x_{n/2+1} = −q^{−1}, …, x_{n−1} = −q^{−n+3}, x_{n} = −q^{−n+1}. Upon a simultaneous rearrangement and change of signs of the rows of the two determinants in (C.1), and using the equality −1 = q^{m}, one sees that this is equivalent to computing the limit (C.23) of (det A_{7})/(det B_{7}), where

A_{7} = (\begin{matrix}x_{h}^{\frac{m+n+1}{2} −t} − x_{h}^{−( \frac{m+n+1}{2} −t)} & 1 ≤ h ≤ n/2 \\ q^{(2n+m−2h+1)( \frac{m+n+1}{2} −t)} − q^{−(2n+m−2h+1)( \frac{m+n+1}{2} −t)} & n/2 < h ≤ n \end{matrix})_{1≤h,t≤n}

and

B_{7} = (\begin{matrix}x_{h}^{n−t+ \frac{1}{2}} − x_{h}^{−(n−t+ \frac{1}{2} )} & 1 ≤ h ≤ n/2 \\ q^{(2n+m−2h+1)(n−t+ \frac{1}{2} )} − q^{−(2n+m−2h+1)(n−t+ \frac{1}{2} )} & n/2 < h ≤ n \end{matrix})_{1≤h,t≤n} .

Now applying de l’Hôpital’s rule, this limit is equal to the limit (C.23) of

\frac{det A_{8}}{\frac{∂}{∂x_{1}} \frac{∂}{∂x_{2}} ⋯ \frac{∂}{∂x_{n/2}} det B_{7}} ,  (C.24)

where

A_{8} = (\begin{matrix}(\frac{m+n+1}{2} − t)(q^{(n+m−2h+1)( \frac{m+n+1}{2} −t−1)} + q^{−(n+m−2h+1)( \frac{m+n+1}{2} −t+1)}) & 1 ≤ h ≤ n/2 \\ q^{(2n+m−2h+1)( \frac{m+n+1}{2} −t)} − q^{−(2n+m−2h+1)( \frac{m+n+1}{2} −t)} & n/2 < h ≤ n \end{matrix})

= (\begin{matrix}(−1)^{\frac{m}{2} −h−t}q^{−(n−2h+1)}(\frac{m+n+1}{2} − t) \\ ×(q^{(n−2h+1)( \frac{n+1}{2} −t)} + q^{−(n−2h+1)( \frac{n+1}{2} −t)}) & 1 ≤ h ≤ n/2 \\ (−1)^{\frac{n+m}{2} −h−t+1}(q^{(2n−2h+1)( \frac{n+1}{2} −t)} − q^{−(2n−2h+1)( \frac{n+1}{2} −t)}) & n/2 < h ≤ n \end{matrix}) . (C.25)

The limit of the denominator of (C.24) is again readily obtained, because the determinant of B_{7} can actually be evaluated by means of (B.2). The result is that the denominator of (C.24) is equal to

(q^{n^{2}/2})^{−n+1/2} ∏_{1≤h<t≤n/2}(q^{n−2t+1} − q^{n−2h+1})^{4}(1 − q^{2n−2h−2t+2})^{4}

× ∏_{h=1}^{n/2}(q^{n−2h+1} + 1)^{2}(1 − q^{2n−4h+2}). (C.26)

Next we devote ourselves to the evaluation of the determinant of A_{8}, with A_{8} given by (C.25). For t = 1, 2, …, \frac{n}{2} we subtract column n − t + 1 from column t. As a result we obtain that det A_{8} is equal to det A_{9}, where A_{9} is the n × n block matrix

A_{9} = (\begin{matrix}A_{9}^{(1)} & ∗ \\ 0 & A_{9}^{(2)} \end{matrix}) ,  (C.27)

with A_{9}^{(1)} the \frac{n}{2} × \frac{n}{2} matrix

A_{9}^{(1)} = ((−1)^{\frac{m}{2} −h−t}q^{−(n−2h+1)}m(q^{(n−2h+1)( \frac{n+1}{2} −t)} + q^{−(n−2h+1)( \frac{n+1}{2} −t)}))_{1≤h,t≤n/2},

and A_{9}^{(2)} the \frac{n}{2} × \frac{n}{2} matrix

A_{9}^{(2)} = ((−1)^{\frac{m}{2} −h−t+1}(q^{(n−2h+1)( \frac{1}{2} −t)} − q^{−(n−2h+1)( \frac{1}{2} −t)}))_{1≤h,t≤n/2}.

Clearly, the determinant det A_{9} is equal to the product (det A_{9}^{(1)}) · (det A_{9}^{(2)}). Using (B.1), we have

det A_{9}^{(1)} = (−1)^{\binom{n/2+1}{2}}q^{−n^{2}/4}m^{n/2}(q^{n^{2}/4})^{−n/2+1/2}

× ∏_{1≤h<t≤n/2}(q^{n−2h+1} − q^{n−2t+1})(1 − q^{2n−2h−2t+2}) ∏_{h=1}^{n/2}(q^{n−2h+1} + 1), (C.28)

whereas by means of (B.2) we have

det A_{9}^{(2)} = (q^{n^{2}/4})^{1/2−n/2}

× ∏_{1≤h<t≤n/2}(q^{n−2h+1} − q^{n−2t+1})(1 − q^{2n−2h−2t+2}) ∏_{h=1}^{n/2}(q^{n−2h+1} − 1). (C.29)

If we now combine (C.27), (C.28), (C.29), and (C.26), use the fact that det A_{8} = det A_{9}, and substitute all this in (C.24), then, after some simplification, we obtain

so_{(((m−n)/2)^{n})}^{odd}(q^{n−1}, q^{n−3}, …, q^{−n+3}, q^{−n+1})

= \frac{m^{n/2} ∏_{h=1}^{n/2}(1 − q^{n−2h+1})}{∏_{1≤h<t≤n/2}(q^{h−t} − q^{t−h})^{2} ∏_{1≤h,t≤n/2}(q^{−n+h+t−1} − q^{n−h−t+1}) ∏_{h=1}^{n/2}(1 + q^{n−2h+1})},

which can be rewritten as (C.20).

Now let n be odd. We proceed in a completely analogous manner. Again, the task is to compute the specialized odd orthogonal character in (4.15), by means of a limit of its definition (C.1).

What we must compute is the limit as

x_{1} → −q^{n−1}, x_{2} → −q^{n−3}, …, x_{(n−1)/2} → −q^{2},  (C.30)

of the right-hand side of (C.1) with x_{(n+1)/2} = −1, x_{(n+3)/2} = −q^{−2}, …, x_{n−1} = −q^{−n+3}, x_{n} = −q^{−n+1}. Upon a simultaneous rearrangement and change of signs of the rows of the two determinants in (C.1), one sees that this is equivalent to computing the limit (C.30) of (det A_{10})/(det B_{10}), where

A_{10} = (\begin{matrix}x_{h}^{\frac{m+n+1}{2} −t} − x_{h}^{−( \frac{m+n+1}{2} −t)} & 1 ≤ h ≤ (n − 1)/2 \\ q^{(2n+m−2h)( \frac{m+n+1}{2} −t)} − q^{−(2n+m−2h)( \frac{m+n+1}{2} −t)} & (n − 1)/2 < h ≤ n \end{matrix})_{1≤h,t≤n}

and

B_{10} = (\begin{matrix}x_{h}^{n−t+ \frac{1}{2}} − x_{h}^{−(n−t+ \frac{1}{2} )} & 1 ≤ h ≤ (n − 1)/2 \\ q^{(2n+m−2h)(n−t+ \frac{1}{2} )} − q^{−(2n+m−2h)(n−t+ \frac{1}{2} )} & (n − 1)/2 < h ≤ n \end{matrix})_{1≤h,t≤n} .

The determinant det B_{10} can again be evaluated explicitly by means of (B.2), and it turns out to be non-zero. However, the determinant det A_{10} vanishes because all the entries in its n-th row are 0 (because of q^{m} = −1). Therefore the quotient (det A_{10})/(det B_{10}) is zero, whence also its limit (C.30). Thus, the claim (C.22) is established. □

Our next lemma provides special evaluations of so-called Schur functions, which are needed in the proof of Theorem 22. Recall that for any partition λ = (λ_{1}, λ_{2}, …, λ_{N}) (i.e., a non-increasing sequence of non-negative integers) the Schur function s_{λ}(x_{1}, x_{2}, …, x_{N}) is defined by (see [13, p. 403, (A.4)], [33, I, (3.1)], or [32, Prop. 1.4.4])

s_{λ}(x_{1}, x_{2}, …, x_{N}) = \frac{det_{1≤h,t≤N}(x_{h}^{λ_{t}+N−t})}{det_{1≤h,t≤N}(x_{h}^{N−t})}.  (C.31)

Again, it is not difficult to see that the denominator in (C.31) does in fact cancel out, so that any Schur function s_{λ}(x_{1}, x_{2}, …, x_{N}) is in fact a polynomial in x_{1}, x_{2}, …, x_{N}, and is thus well-defined for any choice of the variables x_{1}, x_{2}, …, x_{N}.

In the lemma below, the notation ((m − n − 1)^{n}) is a short notation for the partition in which the first n parts are m − n − 1, followed by n + 1 parts all of which are 0.

Lemma C3. Let q = e^{πi/m}. Then we have

s_{((m−n−1)^{n})}(q^{n}, q^{n−1}, …, q, −1, q^{−1}, …, q^{−n+1}, q^{−n}) = 2^{−n^{2}} \begin{matrix}∏_{h=1}^{n/2} & tan & 2 & \frac{(2h−1)π}{2m} \\ ∏_{h=1}^{n+1} & ∏_{t=1}^{n} & |sin \frac{(2t−2h+1)π}{2m} | \end{matrix}

(C.32)

if both m and n are even,

s_{((m−n−1)^{n})}(q^{n}, q^{n−1}, …, q, −1, q^{−1}, …, q^{−n+1}, q^{−n}) = 2^{−n^{2}} \begin{matrix}∏_{h=1}^{(n+1)/2} & tan & 2 & \frac{(2h−1)π}{2m} \\ ∏_{h=1}^{n+1} & ∏_{t=1}^{n} & |sin \frac{(2t−2h+1)π}{2m} | \end{matrix}

(C.33)

if m is even and n is odd,

s_{((m−n−1)^{n})}(q^{n}, q^{n−1}, …, q, −1, q^{−1}, …, q^{−n+1}, q^{−n})

= 2^{−n^{2}} ∏_{h=1}^{n/2} \begin{matrix}sin & 2 & \frac{(2h−1)π}{2m} \\ cos & 2 & \frac{hπ}{m} \end{matrix} \frac{1}{∏_{h=1}^{n+1} ∏_{t=1}^{n} |sin \frac{(2t−2h+1)π}{2m} |} (C.34)

if m is odd and n is even, and

s_{((m−n−1)^{n})}(q^{n}, q^{n−1}, …, q, −1, q^{−1}, …, q^{−n+1}, q^{−n}) = 0  (C.35)

if both m and n are odd.

Proof. We have to evaluate (C.31) with N = 2n+1, λ = (m−n−1, m−n−1, …, m− n − 1, 0, …, 0) (where m − n − 1 is repeated n times), x_{1} = q^{n}, x_{2} = q^{n−1}, …, x_{n} = q, x_{n+1} = −1, x_{n+2} = q^{−1}, …, x_{2n} = q^{−n+1}, x_{2n+1} = q^{−n}. The denominator of (C.31) can be easily evaluated as it is just a Vandermonde determinant. If we use that q^{m} = −1, under these specializations the numerator becomes

det (\begin{matrix}D_{11} & D_{12} \\ l_{1} & l_{2} \\ D_{21} & D_{22} \end{matrix}) ,  (C.36)

where D_{11} is the n × n matrix

D_{11} = ((−1)^{n+1−h}(q^{n+1−h})^{n−t})_{1≤h,t≤n} ,

D_{12} is the n × (n + 1) matrix

D_{12} = ((q^{n+1−h})^{n+1−t})_{1≤h≤n, 1≤t≤n+1} ,

l_{1} is the (row) vector

l_{1} = ((−1)^{m+n−t})_{1≤t≤n} ,

l_{2} is the (row) vector

l_{2} = ((−1)^{n+1−t})_{1≤t≤n+1} ,

D_{21} is the n × n matrix

D_{21} = ((−1)^{h}(q^{−h})^{n−t})_{1≤h,t≤n} ,

and D_{22} is the n × (n + 1) matrix

D_{22} = ((q^{−h})^{n+1−t})_{1≤h≤n, 1≤t≤n+1} .

For the evaluation of the determinant (C.36), we have to distinguish between four cases, depending on the parities of m and n.

If both m and n are even, we subtract column n+1+t from column t, t = 1, 2, …, n. The result is that the first n entries in the odd numbered rows become zero. If we rearrange the rows so that these zeroes are moved to the bottom (i.e., row 2 is moved up to first position, row 4 is moved up to second position, etc.), then we obtain that the determinant (C.36) is equal to

(−1)^{\binom{n}{2}} det (\begin{matrix}E_{1} & ∗ \\ 0 & E_{2} \end{matrix}) ,  (C.37)

where E_{1} is the n × n matrix

E_{1} = (2(−1)^{n+1−2h}(q^{n+1−2h})^{n−t})_{1≤h,t≤n} ,

and E_{2} is the (n + 1) × (n + 1) matrix

E_{2} = (\begin{matrix}(q^{n+2−2h})^{n+1−t} & h ≠ \frac{n}{2} + 1 \\ (−1)^{n+1−t} & h = \frac{n}{2} + 1 \end{matrix})_{1≤h,t≤n+1} .

Clearly, the determinant in (C.37) (and thus the determinant in (C.36) that we want to evaluate) is equal to the product (det E_{1}) · (det E_{2}). Both of the latter determinants are Vandermonde determinants, and are therefore easily evaluated. If the result is substituted in (C.31), together with the denominator evaluation, the claimed expression (C.32) is obtained after some simplification.

In the other three cases we proceed in a similar manner. There, in the determinant (C.36) we add column n + 1 + t to column t, t = 1, 2, …, n. If the parities of m and n are different, then, again, the rows can be rearranged so that a block form is obtained (this rearrangement is different in the two cases), in which both the upper-left and the lower-right blocks are Vandermonde matrices. Finally, if both m and n are odd, then, after addition of the columns as described above, the first n entries in n + 2 (!) rows become zero (to be precise, these are the odd numbered rows and the (n + 1)-st row). Thus, the determinant of this matrix vanishes. □

Next, we turn to two further special evaluations of Schur functions, which are needed in the proofs of Theorems 26 and 28. These special evaluations are given in Lemmas C5 and C6 below). In the proofs, we make use of the standard basic hypergeometric notation introduced earlier in (6.21), and in particular of the following summation formula.

Lemma C4. For any non-negative integer N and any indeterminate b, we have

2ϕ_{1}[\begin{matrix}q^{−2N}, b \\ q^{4−2N}/b \end{matrix}; q^{2}, \frac{q^{3}}{b} ] = \frac{(b/q; q)_{N} (q^{2}; q^{2})_{N}}{q^{N} (b/q^{2}; q^{2})_{N} (q; q)_{N}} .  (C.38)

Proof. We write the hypergeometric series on the left-hand side of (C.38) as a sum over k, say. The reader should observe that, because of the upper parameter q^{−2N}, the sum is in fact a finite sum, with k running from 0 up to N. Now we reverse the order of summation in the sum, i.e., we replace k by N − k. If we rewrite the resulting sum in basic hypergeometric notation, then we obtain

\frac{(1 − bq^{2N−2})}{q^{N} (1 − bq^{−2})}_{2}ϕ_{1}[\begin{matrix}q^{−2N}, b/q^{2} \\ q^{2−2N}/b \end{matrix}; q^{2}, \frac{q^{3}}{b} ] .

This series can be summed by means of the summation formula (see ([14, Ex. 1.8])

2ϕ_{1}[\begin{matrix}a^{2}, a^{2}/b \\ b \end{matrix}; q^{2}, \frac{bq}{a^{2}} ] = \frac{1}{2} (\frac{(−b/a; q)_{∞} (a; q)_{∞} (q; q^{2})_{∞}}{(b; q^{2})_{∞} (bq/a^{2}; q^{2})_{∞}} + \frac{(b/a; q)_{∞} (−a; q)_{∞} (q; q^{2})_{∞}}{(b; q^{2})_{∞} (bq/a^{2}; q^{2})_{∞}} ) ,

upon letting a tend to q^{−N} and replacing b by q^{2−2N}/b. After some simplification, one arrives at the right-hand side of (C.38). □

Similar to earlier notational conventions, in the lemma below, the notation (c^{n−p}, (c− 1)^{p}) is a short notation for the partition in which the first n − p parts are c, the next p parts are c − 1, followed by n + 1 parts all of which are 0.

Lemma C5. Let n and c be positive integers, let p be a non-negative integer with 0 ≤ p ≤ n, and let q be an indeterminate. Then we have

s_{(c^{n−p},(c−1)^{p})}(q^{2n−1}, q^{2n−3}, …, q^{3}, q, 1, q^{−1}, q^{−3}, …, q^{−2n+3}, q^{−2n+1})

= ∏_{h=1}^{2n} \frac{(q^{\frac{c+h}{2}} − q^{− \frac{c+h}{2}} )}{(q^{\frac{h}{2}} − q^{− \frac{h}{2}} )} ∏_{h=1}^{n} ∏_{t=1}^{n} \frac{(q^{c+n+t−h} − q^{−c−n−t+h})}{(q^{n+t−h} − q^{−n−t+h})} ∏_{h=1}^{n} \frac{(q^{h} − q^{−h})^{2}}{(q^{c+p+h} − q^{−c−p−h})}

× \frac{1}{∏_{h=1}^{p} (q^{h} − q^{−h}) ∏_{h=1}^{n−p} (q^{h} − q^{−h})} \frac{(q^{\frac{c}{2}} − q^{− \frac{c}{2}} )(q^{\frac{c}{2} +p} + q^{− \frac{c}{2} −p})}{(q^{c+p} − q^{−c−p})}. (C.39)

Proof. Using the symmetry of the Schur function, we have to evaluate (C.31) with N = 2n + 1, λ = (c, …, c, c − 1, …, c − 1, 0, …, 0) (where c is repeated n − p times and c − 1 is repeated p times), x_{1} = q^{2n−1}, x_{2} = q^{2n−3}, …,x_{n−1} = q^{3}, x_{n} = q, x_{n+1} = q^{−1}, x_{n+2} = q^{−3}, …, x_{2n−1} = q^{−2n+3}, x_{2n} = q^{−2n+1}, x_{2n+1} = 1. The denominator of (C.31) can be easily evaluated as it is just a Vandermonde determinant. On the other hand, under these specializations the numerator becomes

det (\begin{matrix}F_{1} & F_{2} & F_{3} \\ l_{3} & l_{4} & l_{5} \end{matrix}) ,

where F_{1} is the (2n) × (n − p) matrix

F_{1} = ((q^{2n+1−2h})^{c+2n+1−t})_{1≤h≤2n, 1≤t≤n−p} ,

F_{2} is the (2n) × p matrix

F_{2} = ((q^{2n+1−2h})^{c+n+p−t})_{1≤h≤2n, 1≤t≤p} ,

F_{3} is the (2n) × (n + 1) matrix

F_{3} = ((q^{2n+1−2h})^{n+1−t})_{1≤h≤2n, 1≤t≤n+1} ,

l_{3} is the (row) vector of length n − p consisting entirely of 1’s, l_{4} is the (row) vector of length p consisting entirely of 1’s, and l_{5} is the (row) vector of length n + 1 consisting entirely of 1’s.

We consider first the case where p is strictly between 0 and n, that is, where 0 < p < n. We subtract the (t + 1)-st column from the t-th column, t = 1, 2, …, 2n. Clearly, this makes the last row become (0, 0, …, 0, 1). So, if we subsequently expand the determinant with respect to the last row and factor q^{2n+1−2h} − 1 out of the h-th row, h = 1, 2, …, 2n, we obtain the expression

( ∏_{h=1}^{2n}(q^{2n+1−2h} − 1)) det (G_{1} G_{2} G_{3} G_{4} G_{5}) ,  (C.40)

where G_{1} is the (2n) × (n − p − 1) matrix

G_{1} = ((q^{2n+1−2h})^{c+2n−t})_{1≤h≤2n, 1≤t≤n−p−1} ,

G_{2} is the (2n) × 1 matrix (i.e., column of length 2n)

G_{2} = ((q^{2n+1−2h})^{c+n+p−1} (1 + q^{2n+1−2h}))_{1≤h≤2n} ,

G_{3} is the (2n) × (p − 1) matrix

G_{3} = ((q^{2n+1−2h})^{c+n+p−t−1})_{1≤h≤2n, 1≤t≤p−1} ,

G_{4} is the (2n) × 1 matrix (i.e., column of length 2n)

G_{4} = ((q^{2n+1−2h})^{n} ∑_{k=0}^{c−1} q^{(2n+1−2h)k})_{1≤h≤2n} ,

and G_{5} is the (2n) × n matrix

G_{5} = ((q^{2n+1−2h})^{n−t})_{1≤h≤2n, 1≤t≤n} .

We now use linearity in the (n − p)-th and n-th columns to convert (C.40) into

( ∏_{h=1}^{2n}(q^{2n+1−2h} − 1))( ∑_{k=0}^{c−1} det (G_{1} G_{2}^{(1)} G_{3} G_{4}^{(k)} G_{5})

+ ∑_{k=0}^{c−1} det (G_{1} G_{2}^{(2)} G_{3} G_{4}^{(k)} G_{5}) ), (C.41)

where G_{1}, G_{3} and G_{5} are as above, where G_{2}^{(1)} is the (2n) × 1 matrix

G_{2}^{(1)} = ((q^{2n+1−2h})^{c+n+p})_{1≤h≤2n} ,

where G_{2}^{(2)} is the (2n) × 1 matrix

G_{2}^{(2)} = ((q^{2n+1−2h})^{c+n+p−1})_{1≤h≤2n} ,

and where G_{4}^{(k)} is the (2n) × 1 matrix

G_{4}^{(k)} = ((q^{2n+1−2h})^{n+k})_{1≤h≤2n} .

The determinants in (C.41) are determinants of the form

det (X_{t}^{2n+1−2h})_{1≤h,t≤2n} = ( ∏_{t=1}^{2n} x_{t}^{2n−1}) det ((X_{t}^{−2})^{h−1})_{1≤h,t≤2n} .  (C.42)

Since the last determinant is a Vandermonde determinant, we can evaluate the determinants in (C.41). Thus, under our specializations, the numerator in (C.31) becomes

( ∏_{h=1}^{2n}(q^{2n+1−2h} − 1))( ∏_{1≤h<t≤n}(q^{t−h} − q^{−t+h})^{2})( ∏_{h=1}^{n} ∏_{t=1}^{n}(q^{c+n+t−h} − q^{−c−n−t+h}))

× (∑_{k=0}^{c−1} \frac{∏_{h=1}^{n}(q^{c+n−h−k} − q^{−c−n+h+k}) ∏_{h=1}^{n}(q^{h+k} − q^{−h−k})}{(q^{c+p−k−1} − q^{−c−p+k+1}) ∏_{h=1}^{p−1}(q^{h} − q^{−h}) ∏_{h=1}^{n−p}(q^{h} − q^{−h}) ∏_{h=1}^{n}(q^{c+p+h−1} − q^{−c−p−h+1})}

+ ∑_{k=0}^{c−1} \frac{∏_{h=1}^{n}(q^{c+n−h−k} − q^{−c−n+h+k}) ∏_{h=1}^{n}(q^{h+k} − q^{−h−k})}{(q^{c+p−k} − q^{−c−p+k}) ∏_{h=1}^{p}(q^{h} − q^{−h}) ∏_{h=1}^{n−p−1}(q^{h} − q^{−h}) ∏_{h=1}^{n}(q^{c+p+h} − q^{−c−p−h})}) . (C.43)

The task now is to simplify the two sums. In the first sum in (C.43) we replace k by k − 1. After this replacement, we put the two sums together, that is, for any fixed k we add the k-th summands of the two sums. Thus, the expression (C.43) simplifies to

\frac{( ∏_{h=1}^{2n}(q^{2n+1−2h} − 1))( ∏_{1≤h<t≤n}(q^{t−h} − q^{−t+h})^{2})( ∏_{h=1}^{n} ∏_{t=1}^{n}(q^{c+n+t−h} − q^{−c−n−t+h}))}{( ∏_{h=1}^{p}(q^{h} − q^{−h}))( ∏_{h=1}^{n−p}(q^{h} − q^{−h}))( ∏_{h=0}^{n}(q^{c+p+h} − q^{−c−p−h}))}

× ( ∑_{k=0}^{c} ( ∏_{h=1}^{n−1}(q^{c+n−h−k} − q^{−c−n+h+k})(q^{h+k} − q^{−h−k}))(q^{n} − q^{−n})

·(q^{−k}(q^{−c−n−p}+q^{c+n−p}−q^{c+p−n}−q^{c+n+p})−q^{k}(q^{−c−n−p}+q^{−c+n−p}−q^{−c+p−n}−q^{c+n+p}))). (C.44)

In what follows, we concentrate on the sum in (C.44). We split the sum in two sums according to the additive decomposition of the last factor of the summand and rewrite the two sums in basic hypergeometric notation,

(−1)^{n}q^{cn−c−n}(q^{−2n−2c+2}; q^{2})_{n−1} (q^{2}; q^{2})_{n}

× ((q^{−c−n−p} + q^{c+n−p} − q^{c+p−n} − q^{c+n+p}) _{2}ϕ_{1}[\begin{matrix}q^{−2c}, q^{2n} \\ q^{2−2n−2c} \end{matrix}; q^{2}, q^{1−2n}]

− (q^{−c−n−p} + q^{−c+n−p} − q^{−c+p−n} − q^{c+n+p}) _{2}ϕ_{1}[\begin{matrix}q^{−2c}, q^{2n} \\ q^{2−2n−2c} \end{matrix}; q^{2}, q^{3−2n}] ). (C.45)

We write the second _{2}ϕ_{1}-series as a sum over k, say, reverse the order of summation, that is, we replace k by c − k, and then we write the resulting sum again in basic hypergeometric notation. We obtain a _{2}ϕ_{1}-series which turns out to be identical with the first _{2}ϕ_{1}-series in (C.45). If we put everything together and simplify, then (C.45) becomes

(−1)^{n}q^{cn−2c−2n−p}(q^{−2n−2c+2}; q^{2})_{n−1} (q^{2}; q^{2})_{n}(1 − q^{c})(1 − q^{c+2n})(1 + q^{c+2p})

×_{2}ϕ_{1}[\begin{matrix}q^{−2c}, q^{2n} \\ q^{2−2n−2c} \end{matrix}; q^{2}, q^{1−2n}] . (C.46)

Next we apply the contiguous relation

2ϕ_{1}[\begin{matrix}a, b \\ c \end{matrix}; q, z] = _{2}ϕ_{1}[\begin{matrix}a, bq \\ c \end{matrix}; q, z] − bz \frac{(1 − a)}{(1 − c)}_{2}ϕ_{1}[\begin{matrix}aq, bq \\ cq \end{matrix}; q, z]

to the _{2}ϕ_{1}-series in (C.46). This transforms the expression (C.46) into

(−1)^{n}q^{cn−2c−2n−p}(q^{−2n−2c+2}; q^{2})_{n−1} (q^{2}; q^{2})_{n}(1 − q^{c})(1 − q^{c+2n})(1 + q^{c+2p})

× (_{2}ϕ_{1}[\begin{matrix}q^{−2c}, q^{2n+2} \\ q^{2−2n−2c} \end{matrix}; q^{2}, q^{1−2n}] − q \frac{1 − q^{−2c}}{1 − q^{2−2n−2c}}_{2}ϕ_{1}[\begin{matrix}q^{−2c+2}, q^{2n+2} \\ q^{4−2n−2c} \end{matrix}; q^{2}, q^{1−2n}]) . (C.47)

Both _{2}ϕ_{1}-series in the last line can be evaluated by means of Lemma C4. If we substitute the result in (C.47), put this back in (C.44), and divide the result by the numerator in (C.31) subject to our specializations, which we evaluated by means of the Vandermonde determinant evaluation, we arrive at the right-hand side of (C.39) after some simplification.

If p = 0, then we can proceed in a completely analogous manner. In fact, the computations are somewhat simpler in this case, so that we leave the details to the reader. That the formula works also for p = n can then be checked by verifying that (C.39) for p = n agrees with (C.39) for p = 0 and c replaced by c − 1.

This completes the proof of the lemma. □

Lemma C6. Let n and c be positive integers, and let q be an indeterminate. Then we have

s_{(c^{n})}(q^{2n−1}, q^{2n−3}, …, q^{3}, q, −1, q^{−1}, q^{−3}, …, q^{−2n+3}, q^{−2n+1})

= ∏_{h=1}^{2n} \frac{(q^{\frac{c+h}{2}} − (−1)^{c+h}q^{− \frac{c+h}{2}} )}{(q^{\frac{h}{2}} − (−1)^{h}q^{− \frac{h}{2}} )} ∏_{h=1}^{n−1} ∏_{t=1}^{n} \frac{(q^{c+n+t−h} − q^{−c−n−t+h})}{(q^{n+t−h} − q^{−n−t+h})}. (C.48)

Proof. We replace q by −q in the p = 0 case of Lemma C5. Then, the left-hand side of (C.39) agrees exactly with the left-hand side of (C.48). The products on the right-hand sides are not in the same form, but they are equivalent because the extra terms in Lemma C5 cancel out if p = 0.

□

The final two lemmas concern sums of even orthogonal characters, which are needed in the proofs of Theorems 30 and 32. Given a non-increasing sequence λ = (λ_{1}, λ_{2}, …, λ_{n}) of integers or half-integers with λ_{n−1} ≥ |λ_{n}|, the even orthogonal character so_{λ}^{even}(x_{1}, x_{2}, …, x_{n}) is defined by (see [13, (24.40)])

so_{λ}^{even}(x_{1}, x_{2}, …, x_{n}) = \frac{det_{1≤h,t≤n}(x_{h}^{λ_{t}+n−t} + x_{h}^{−(λ_{t}+n−t)}) + det_{1≤h,t≤n}(x_{h}^{λ_{t}+n−t} − x_{h}^{−(λ_{t}+n−t)})}{det_{1≤h,t≤n}(x_{h}^{n−t} + x_{h}^{−(n−t)})}.

(C.49)

Lemma C7. Let n be a positive integer and c be a non-negative half-integer. Then we have

∑_{p=−c}^{c} so_{((c^{n−1},p))}^{even}(q^{n−1}, q^{n−2}, …, 1)

= ∏_{1≤h<t≤n} \frac{q^{(2c+t+h−1)/2} − q^{−(2c+t+h−1)/2}}{q^{(t+h−2)/2} − q^{−(t+h−2)/2}} \begin{matrix}∏_{h=1}^{n} & (q^{c+h−1/2} − q^{−(c+h−1/2)}) \\ ∏_{h=1}^{n−1} & (q^{h/2} − q^{−h/2}) \end{matrix}

× ∑_{k=1}^{n} \frac{(−1)^{k−1}(2c + 2k − 1)}{∏_{h=1}^{n} (q^{(2c+k+h−1)/2} − q^{−(2c+k+h−1)/2}) ∏_{h=1}^{k−1} (q^{h/2} − q^{−h/2}) ∏_{h=1}^{n−k} (q^{h/2} − q^{−h/2})}. (C.50)

Proof. For convenience, let us write s̃o_{λ}(x_{1}, x_{2}, …, x_{n}) for just “one half” in the definition (C.49) of even orthogonal characters, that is,

s̃o_{λ}(x_{1}, x_{2}, …, x_{n}) = \frac{det_{1≤h,t≤n}(x_{h}^{λ_{t}+n−t} + x_{h}^{−(λ_{t}+n−t)})}{det_{1≤h,t≤n}(x_{h}^{n−t} + x_{h}^{−(n−t)})}.  (C.51)

It should be observed that we have

s̃o_{λ}(x_{1}, …, x_{n−1}, 1) = so_{λ}^{even}(x_{1}, …, x_{n−1}, 1),  (C.52)

since the second determinant in (C.49) vanishes if x_{n} = 1. Now, by (C.51) and the determinant evaluation (B.3), we have

∑_{p=−c}^{c} s̃o_{((c^{n−1},p))}(x_{1}, x_{2}, …, x_{n})

= \frac{1}{2}( ∏_{1≤h<t≤n} \frac{1}{(xh^{\frac{1}{2}} x_{t}^{− \frac{1}{2}} − x_{h}^{− \frac{1}{2}} xt^{\frac{1}{2}} )(xh^{\frac{1}{2}} xt^{\frac{1}{2}} − x_{h}^{− \frac{1}{2}} x_{t}^{− \frac{1}{2}} )})

× ∑_{p=−c}^{c} det_{1≤h,t≤n} (\begin{matrix}x_{h}^{c+n−t} + x_{h}^{−(c+n−t)} & 1 ≤ t < n \\ x_{h}^{p} + x_{h}^{−p} & t = n \end{matrix})

= \frac{1}{2}( ∏_{1≤h<t≤n} \frac{1}{(xh^{\frac{1}{2}} x_{t}^{− \frac{1}{2}} − x_{h}^{− \frac{1}{2}} xt^{\frac{1}{2}} )(xh^{\frac{1}{2}} xt^{\frac{1}{2}} − x_{h}^{− \frac{1}{2}} x_{t}^{− \frac{1}{2}} )})

× det_{1≤h,t≤n} (\begin{matrix}x_{h}^{c+n−t} + x_{h}^{−(c+n−t)} & 1 ≤ t < n \\ 2\frac{x_{h}^{c+ \frac{1}{2}} − x_{h}^{−(c+ \frac{1}{2} )}}{xh^{\frac{1}{2}} − x_{h}^{− \frac{1}{2}}} & t = n \end{matrix})

= ( ∏_{1≤h<t≤n} \frac{1}{(xh^{\frac{1}{2}} x_{t}^{− \frac{1}{2}} − x_{h}^{− \frac{1}{2}} xt^{\frac{1}{2}} )(xh^{\frac{1}{2}} xt^{\frac{1}{2}} − x_{h}^{− \frac{1}{2}} x_{t}^{− \frac{1}{2}} )})( ∏_{h=1}^{n} \frac{1}{xh^{\frac{1}{2}} − x_{h}^{− \frac{1}{2}}} )

× det_{1≤h,t≤n} (\begin{matrix}x_{h}^{c+n−t+ \frac{1}{2}} − x_{h}^{−(c+n−t+ \frac{1}{2} )} + x_{h}^{c+n−t− \frac{1}{2}} − x_{h}^{−(c+n−t− \frac{1}{2} )} & 1 ≤ t < n \\ x_{h}^{c+ \frac{1}{2}} − x_{h}^{−(c+ \frac{1}{2} )} & t = n \end{matrix})

= ( ∏_{1≤h<t≤n} \frac{1}{(xh^{\frac{1}{2}} x_{t}^{− \frac{1}{2}} − x_{h}^{− \frac{1}{2}} xt^{\frac{1}{2}} )(xh^{\frac{1}{2}} xt^{\frac{1}{2}} − x_{h}^{− \frac{1}{2}} x_{t}^{− \frac{1}{2}} )})( ∏_{h=1}^{n} \frac{1}{xh^{\frac{1}{2}} − x_{h}^{− \frac{1}{2}}} )

× det_{1≤h,t≤n} (x_{h}^{c+n−t+ \frac{1}{2}} − x_{h}^{−(c+n−t+ \frac{1}{2} )}) ,

where the last line arises by obvious elementary column operations from the next-to-last line. In this identity, we want to put x_{1} = q^{n−1}, x_{2} = q^{n−2}, …, x_{n} = 1. However, we cannot directly put x_{n} = 1, because this would lead to an expression 0/0. Therefore instead, we have to take the limit x_{n} → 1. Thereby, using (C.52), we get

∑_{p=−c}^{c} so_{((c^{n−1},p))}^{even}(q^{n−1}, q^{n−2}, …, 1)

= ( ∏_{1≤h<t≤n} \frac{1}{(q^{(t−h)/2} − q^{−(t−h)/2})(q^{(h+t−2)/2} − q^{−(h+t−2)/2}} )( ∏_{h=1}^{n−1} \frac{1}{q^{h/2} − q^{−h/2}} )

× det_{1≤h,t≤n} (\begin{matrix}q^{(n−h)(c+n−t+ \frac{1}{2} )} − q^{−(n−h)(c+n−t+ \frac{1}{2} )} & 1 ≤ h < n \\ 2c + 2n − 2t + 1 & h = n \end{matrix}) .  (C.53)

Now we expand the determinant along the last row. Then each of the appearing minors can be evaluated by means of (B.4). If we substitute the result in (C.53), we obtain

∑_{p=−c}^{c} so_{((c^{n−1},p))}^{even}(q^{n−1}, q^{n−2}, …, 1)

= ( ∏_{1≤h<t≤n} \frac{1}{(q^{(t−h)/2} − q^{−(t−h)/2})(q^{(h+t−2)/2} − q^{−(h+t−2)/2})})( ∏_{h=1}^{n−1} \frac{1}{q^{h/2} − q^{−h/2}} )

× ∑_{k=1}^{n}(−1)^{n+k}(2c + 2n − 2k + 1)_{∏_{t=1}^{n} ; t≠k}(q^{c+n−t+ \frac{1}{2}} − q^{−(c+n−t+ \frac{1}{2} )})

·_{∏_{1≤h<t≤n} ; h,t≠k}(q^{(t−h)/2} − q^{−(t−h)/2})(q^{(2c+2n−h−t+1)/2} − q^{−(2c+2n−h−t+1)/2}).

After replacing k by n + 1 − k and performing some simplification, we arrive at the claimed expression. □

Lemma C8. Let n be a positive integer and c be a non-negative half-integer. Then we have

∑_{p=−c}^{c}(−1)^{c−p} so_{((c^{n−1},p))}^{even}(q^{n−1}, q^{n−2}, …, 1)

= ∏_{1≤h<t≤n} \frac{q^{(2c+t+h−1)/2} − q^{−(2c+t+h−1)/2}}{q^{(t+h−2)/2} − q^{−(t+h−2)/2}} ∏_{h=1}^{n−1} \frac{1}{(q^{h/2} + q^{−h/2})}. (C.54)

Proof. According to the definition (C.49) of even orthogonal characters and the determinant evaluation (B.3), we have

∑_{p=−c}^{c}(−1)^{c−p} s̃o_{((c^{n−1},p))}(x_{1}, x_{2}, …, x_{n})

= \frac{1}{2}( ∏_{1≤h<t≤n} \frac{1}{(xh^{\frac{1}{2}} x_{t}^{− \frac{1}{2}} − x_{h}^{− \frac{1}{2}} xt^{\frac{1}{2}} )(xh^{\frac{1}{2}} xt^{\frac{1}{2}} − x_{h}^{− \frac{1}{2}} x_{t}^{− \frac{1}{2}} )})

× ∑_{p=−c}^{c}(−1)^{c−p} det_{1≤h,t≤n} (\begin{matrix}x_{h}^{c+n−t} + x_{h}^{−(c+n−t)} & 1 ≤ t < n \\ x_{h}^{p} + x_{h}^{−p} & t = n \end{matrix})

= \frac{1}{2}( ∏_{1≤h<t≤n} \frac{1}{(xh^{\frac{1}{2}} x_{t}^{− \frac{1}{2}} − x_{h}^{− \frac{1}{2}} xt^{\frac{1}{2}} )(xh^{\frac{1}{2}} xt^{\frac{1}{2}} − x_{h}^{− \frac{1}{2}} x_{t}^{− \frac{1}{2}} )})

× det_{1≤h,t≤n} (\begin{matrix}x_{h}^{c+n−t} + x_{h}^{−(c+n−t)} & 1 ≤ t < n \\ 2\frac{x_{h}^{c+ \frac{1}{2}} + x_{h}^{−(c+ \frac{1}{2} )}}{xh^{\frac{1}{2}} + x_{h}^{− \frac{1}{2}}} & t = n \end{matrix})

= ( ∏_{1≤h<t≤n} \frac{1}{(xh^{\frac{1}{2}} x_{t}^{− \frac{1}{2}} − x_{h}^{− \frac{1}{2}} xt^{\frac{1}{2}} )(xh^{\frac{1}{2}} xt^{\frac{1}{2}} − x_{h}^{− \frac{1}{2}} x_{t}^{− \frac{1}{2}} )})( ∏_{h=1}^{n} \frac{1}{xh^{\frac{1}{2}} + x_{h}^{− \frac{1}{2}}} )

× det_{1≤h,t≤n} (\begin{matrix}x_{h}^{c+n−t+ \frac{1}{2}} + x_{h}^{−(c+n−t+ \frac{1}{2} )} + x_{h}^{c+n−t− \frac{1}{2}} + x_{h}^{−(c+n−t− \frac{1}{2} )} & 1 ≤ t < n \\ x_{h}^{c+ \frac{1}{2}} + x_{h}^{−(c+ \frac{1}{2} )} & t = n \end{matrix})

= ( ∏_{1≤h<t≤n} \frac{1}{(xh^{\frac{1}{2}} x_{t}^{− \frac{1}{2}} − x_{h}^{− \frac{1}{2}} xt^{\frac{1}{2}} )(xh^{\frac{1}{2}} xt^{\frac{1}{2}} − x_{h}^{− \frac{1}{2}} x_{t}^{− \frac{1}{2}} )})( ∏_{h=1}^{n} \frac{1}{xh^{\frac{1}{2}} + x_{h}^{− \frac{1}{2}}} )

× det_{1≤h,t≤n} (x_{h}^{c+n−t+ \frac{1}{2}} + x_{h}^{−(c+n−t+ \frac{1}{2} )}) ,

where s̃o_{λ}(x_{1}, x_{2}, …, x_{n}) is defined in (C.51), and where the last line arises by obvious elementary column operations from the next-to-last line. Now we put x_{1} = q^{n−1}, x_{2} = q^{n−2}, …, x_{n} = 1 in this identity, in which case, due to (C.52), the “incomplete” even orthogonal character s̃o_{((c^{n−1},p))}(x_{1}, x_{2}, …, x_{n}) becomes the specialized even orthogonal

character so_{((c^{n−1},p))}^{even}(q^{n−1}, q^{n−2}, …, 1), and we use (B.3) to evaluate the determinant

in the last line. After some simplification we arrive at (C.54). □

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