A Laurent series proof of the Habsieger-Kadell q-Morris identity

We give a Laurent series proof of the Habsieger-Kadell q-Morris identity, which is a common generalization of the q-Morris identity and the Aomoto constant term identity. The proof allows us to extend the theorem for some additional parameter cases.


Introduction
This paper is closely related to the well-known Dyson's ex-conjecture.The conjecture was made by Freeman Dyson in 1962 when studying statistical theory of energy levels of complex systems [7].
Theorem 1.1.For nonnegative integers a 0 , . . ., a n , where CT x f (x) means to take the constant term in the Laurent expansion of f (x) in the powers of x 0 , x 1 , . . ., x n .
The q-analogous of the Dyson conjecture was made by Andrews [1] in 1975.
Theorem 1.2 (Zeilberger-Bressoud).For nonnegative integers a 0 , a 1 , . . ., a n , CT x 0 i<j n x i x j a i x j x i q a j = (q) a 0 +a 1 +•••+an (q) a 0 (q) a 1 • • • (q) an , where (z) m := (z;q)∞ (zq m ;q)∞ = ( ). Almost all methods for Dyson's ex-conjecture fail to extend for the q version.Up to now, only three different methods succeeded: the combinatorial proof in [23], the short proof in [9] using iterated Laurent series, and the one page proof in [14] using the Combinatorial Nullstellensatz.The methods apply to some constant terms of similar type.
In this paper we study the Habsieger-Kadell q-Morris identity, an important variation of the equal parameter case of the q-Dyson theorem.The original identity studies the constant term of the following Laurent polynomial for m + l n: A q (x 0 , x 1 , . . ., x n ; a, b, k, m, l) q χ(i>m) x i x 0 b+χ(i m)+χ(i n−l+1) where the expression χ(S) is 1 if the statement S is true, and 0 otherwise.In giving a Laurent series proof of the Habsieger-Kadell q-Morris identity, we are able to establish a unified formula that also works for the additional cases m + l > n.The result is stated as follows, where the additional boldfaced part χ(i 2n − m − l) is only effective when m + l > n.
Theorem 1.3.For nonnegative integers a, b, k, m, l satisfying m, l < n, we have where The m = l = 0, q = 1 case of the result is the Morris identity, which is equivalent to the well-known Selberg integral [19].In his thesis [18] Morris established the identity and conjectured the q-analogous identity.The q-Morris identity, or the m = l = 0 case, was proved by Habsieger [12] and later by Zeilberger [24].The m = 0, q = 1 case of the result, called the Aomoto identity, was constructed by Aomoto [3].By extending Aomoto's method Kadell [13] constructed the m + l n case, in the same year of Habsieger's proof.As far as we know, the m + l > n case was not considered before.
Our approach is by extending the proof of the Aomoto identity in [8].The basic idea is to regard both sides of (1.3) as polynomials in q a of degree at most d = nb+m+l.Then to show the equality of the two polynomials, it is sufficient to show that they are equal at d + 1 points.Note that this basic idea was used by Habsieger for q-Selberg integral in [12].The equality at the d vanishing points are not hard to handle by the techniques in [9,16].But in this approach, we have to deal with two problems: i) the multiple roots problem for small k; ii) the d + 1-st suitable point is hard to find.We handle the former problem by a rationality result of Stembridge, and the latter problem by a hard searching process.
We present the major steps of our proof in Section 2. The steps are expanded by the rationality result in Section 3, by the proof of the vanishing lemma in Section 4, and by the proof for the extra point in Section 5.
While we were finishing the presented work, the one page proof of the q-Dyson theorem was published.Moreover, Károlyi and Nagy [15] gave a generalization of Theorem 1.3 in the m = 0 case using the Combinatorial Nullstellensatz.The two approaches are different but have some connections.
2 Proof of the Habsieger-Kadell q-Morris identity Following notations in the introduction, we may assume that 0 m, l < n by the following argument.If m n then which is just A q (x 0 q, . . ., x n ; a − 1, b + 1, k, 0, l).Then by substituting x 0 by x 0 /q, we can see that the constant term is M n (a − 1, b + 1, k, 0, l; q).The case l n is similar: we observe that A q (x 0 , . . ., x n ; a, b, k, m, l) can be rewritten as A q (x 0 , . . ., x n ; a, b + 1, k, m, 0).
Let us rewrite M (q a , q k ) = M n (a, b, k, m, l; q) as We have the following characterization.
Lemma 2.1.For fixed b, n ∈ N and 0 m, l < n, M (q a , q k ) is uniquely determined by the following three properties.
1. M (q a , q k )(q) n k /(q) nk is a polynomial in q a of degree nb + m + l, whose coefficients are rational functions in q k and q.
2. For any k > b + 1, M (q −ξ , q k ) = 0 if ξ belongs to one of the following three sets: (2.2) Proof.Assume M (q a , q k ) also satisfies the above three properties.Then for every ξ ∈ Since both sides are rational functions in q k and they agree at infinitely many points, they are identical as rational functions.Now as polynomials in q a , whose coefficients are rational functions in q k and q, M (q a , q k )(q) n k /(q) nk agrees with M (q a , q k )(q) n k /(q) nk at nb + m + l + 1 distinct ξ's as above, they must be equal to each other.
Note that the condition k > b + 1 can not be dropped, since D 3 has duplicate elements when k b − 1, and Denote by M n (a, b, k, m, l; q) the left-hand-side of (1.3).Then Theorem 1.3 will follow by induction on n if we can show the following three lemmas, whose proofs will be given in later sections.Lemma 2.2.For fixed b, n ∈ N and 0 m, l < n, M n (a, b, k, m, l; q)(q) n k /(q) nk is a polynomial in q a of degree at most nb + m + l, whose coefficients are rational functions in q k and q.
Since M n (a, b, k, m, l; q) is a polynomial in q a , the definition of a can be extended for all integers, in particular for negative integers a. Lemma 2.3 (Vanishing Lemma).For fixed b, n ∈ N, and 0 m, l < n, and k > b + 1, M n (−h, b, k, m, l; q) vanishes when h equals one of the values in (2.2).
Remark 2.4.At this stage we can already claim the truth of Theorem 1.3 for m = 0.In this case D 1 is empty, so that we can choose a = 0 as the extra point.Then M n (0, b, k, 0, l; q) reduces to M n (0, 0, k, 0, 0; q), and the equal parameter case of the q-Dyson theorem applies.
The extra point in the above lemma is found through a hard searching process.It is a surprise for this special h: the constant term M n (−h, b, k, m, l; q) reduces to a single constant term that can be evaluated by Remark 2.4 or the hypothesis.

The polynomial-rational characterization
To prove Lemma 2.2, we need the the following rationality result, which is implicitly due to J.R. Stembridge [20], as can be seen from the proof.The q = 1 case of this result is the equal parameter case of [8,Proposition 2.4].Proposition 3.1.For any n ∈ N and α = (α 1 , . . ., α n ) ∈ Z n with 1 i n α i = 0, we have where R n (q k , q; α) is a rational function in q k and q, and [x α ] refers to take the coefficient of Proof.In [20, Equation 44] Stembridge gave the following equation (set where the summation is taken over some elements whose number is bounded by a function of n and C n [S](q k , q) is a formal power series in q k and q.
By [20, Corollary 3.3] we know that In [20, Page 334, Line 33] Stembridge stated that C n [λ](q k , q) is of the form f λ (q k , q) • C n [∅](q k , q) for some rational function f λ .Therefore, combining with (3.2) and (3.3) we get The desired rational function is then given by Proof of Lemma 2.2.When regarded as Laurent series in x 0 , the equality can be easily shown to hold for all integers a. Rewrite M n (a, b, k, m, l; q) as CT where b * i = b + χ(i n − l + 1).The well-known q-binomial theorem [2, Theorem 2.1] is the identity Setting z = uq n and b = q −n in (3.6), we obtain for all integers n, where n k = (q)n (q) k (q) n−k is the q-binomial coefficient.Using (3.7), we see that for 1 i n, Expanding the first product in (3.5) and taking constant term in x 0 , we see that, by Proposition 3.1, M n (a, b, k, m, l; q) becomes for some rational functions R n (q k , q; k) in q k and q, where is a Laurent polynomial in x 1 , . . ., x n independent of a and the sum ranges over all se- and so is the sum.
The coefficients of M n (a, b, k, m, l; q)(q) n k /(q) nk in q a are clearly rational functions in q k and q.

Proof of the vanishing lemma
We will follow notations in [9,16], where different versions of the vanishing lemma were proposed for dealing with q-Dyson related constant terms.The new vanishing lemma will be handled by the same idea but we have to carry out the details.We will include some basic ingredients for readers' convenience.
In this section, we let K = C(q), and assume that all series are in the field of iterated Laurent series K(( as a working field has been explained in [9]. We emphasize that the field K((x n ))((x n−1 )) • • • ((x 0 )) include the field of rational functions as a subfield, so that every rational function is identified with its unique iterated Laurent series expansion.The series expansions of 1/(1 − q k x i /x j ) will be especially important.
The constant term of the series F (x) in x i , denoted by CT x i F (x), is defined to be the sum of those terms in F (x) that are free of x i .It follows that CT We shall call the monomial M = q k x i /x j small if i < j and large if i > j.Thus the constant term in Constant term operators defined in this way has the important commutativity property: CT The degree of a rational function of x is the degree in x of the numerator minus the degree in x of the denominator.For example, if i = j then the degree of 1 − x j /x i = (x i − x j )/x i is 0 in x i and 1 in x j .A rational function is called proper (resp.almost proper ) in x if its degree in x is negative (resp.zero). Let be a rational function of x k , where p(x k ) is a polynomial in x k , and the α i are distinct monomials, each of the form x t q s .Then the partial fraction decomposition of F with respect to x k has the following form: where p 0 (x k ) is a polynomial in x k , and p 1 (x k ) is a polynomial in x k of degree less than d.
The following lemma has appeared in [16].
Lemma 4.1.Let F be as in (4.2) and (4.3).Then where the sum ranges over all j such that x k /α j is small.In particular, if , where LC x k means to take the leading coefficient with respect to x k .
The following lemma plays an important role in our argument.Lemma 4.2.Let k, b and k 1 , . . ., k s be nonnegative integers.Then for any k 1 , . . ., k s with 0 k i (s − 1)k + b + 1 for all i, either 0 k i b for some i, or 1 − k k j − k i k for some i < j, except only when k i = (s − i)k + b + 1 for i = 1, . . ., s.
Proof.Assume k 1 , . . ., k s to satisfy that for all i, b < k i (s − 1)k + b + 1, and for all i < j, either Let which implies σ(i + 1) > σ(i).This can only occur when σ is the identity permutation. Let By Lemma 2.2, we know that M n (a, b, k, m, l; q) is a polynomial in q a , so the definition of a can be extended to negative integers.Then, since (u The vanishing lemma says that CT x Q(h) = 0 for every h in (2.2).We attack the vanishing lemma by repeated application of Lemma 4.1.This will give a big sum of terms, each will be detected to be 0 by Lemma 4.2.This is better summarized in the following Lemma 4.3.To state the lemma, we need more notations.For any rational function F of x 0 , x 1 , . . ., x n , and for sequences of integers k = (k 1 , k 2 , . . ., k s ) and r = (r 1 , r 2 , . . ., r s ) let E r,k F be the result of replacing x r i in F with x rs q ks−k i for i = 0, 1, . . ., s − 1, where we set r 0 = k 0 = 0. Then for 0 < r 1 < r 2 < . . .< r s n and 0 k i h, we define Note that the product on the right hand side of (4.6) cancels all the factors in the denominator of Q that would be taken to zero by E r,k .If k i = 0 for some i and r i m, then Q(h | r; k) has the factor E r,k [(x r i /x 0 ) b+1+χ(r i n−l+1) ] = 0.If k i = 0 for some i and r i > m, by the definition of Q(h | r; k) in (4.6), the factor 1−x 0 /x r i appears in Q(h | r; k), but it cancels nothing in the denominator of Q(h).Thus it would be taken to zero by As a warm up, it is easy to check that Q(h) is proper in x 0 with degree −nh − m.Thus applying Lemma 4.1 gives This formula is compatible with the following lemma if we treat Lemma 4.3.The rational functions Q(h | r; k) have the following two properties: ) for some i with 1 i s and n > s, then Proof of property (i).
) for all i, the k i 's have to be in one of the following three cases.Case 1: 0 k i b for some the electronic journal of combinatorics 21(3) (2014), #P3.38 In this case we only need the value of k s .Since Case 3 only occurs when s n − l + 1, we have Proof of property (ii).Note that since h k i for all i and h ∈ D 1 D 2 D 3 {(n − l − 1)k + b + 1}, the hypothesis implies that h > sk − χ(s < m).
We only show that Q(h | r; k) is proper in x rs so that Lemma 4.1 applies.The rest is the same as that in the proof of [9,Lemma 5.1].To this end we write the electronic journal of combinatorics 21(3) (2014), #P3.38 is 1 if i ∈ R and j / ∈ R, and is 0 otherwise, as is easily seen by checking the four cases.Thus the part of N contributing to the degree in x rs is E r;k s i=1 j =r 0 ,...,rs x r i x j q χ(r i >j) k , which has degree (n − s)sk, and the part of D contributing to the degree in x rs is E r;k     j =r 0 ,...,rs j m x 0 x j q h h+1 j =r 0 ,...,rs j>m which has degree at least (n − s)h + χ(s < m).
Thus the total degree of Q(h | r; k) in x rs is at most Now we are ready to prove the vanishing lemma.
Proof of the vanishing lemma.Recall that CT x Q(h) = M n (−a, b, k, m, l; q).We prove by induction on n − s that CT the lemma is the case s = 0. (Note that taking the constant term with respect to a variable that does not appear has no effect.)We may assume that s n and 0 < r

Proof for the extra point
We need the following lemma.Proof.We have to split into the following two cases.Case 1: m + l n.Then where For k > b + 1, after cancelations and combinations, we obtain where the last equality is obtained by making the substitution . By Remark 2.4 (or the hypothesis), we obtain which can be routinely checked to be equal to M n (−h, b, k, m, l; q).Case 2: m + l > n.The computation is similar to but more complicated than case 1. Indeed we need the case 1 result in some sense.We omit some details for brevity.We have where and A and B are similar to A and B, with A B simplifies as . (5.9) A similar computation gives (q) (n−l+i+1)k (q) (i+1)k (q) (i+1)k−b−1−χ(i<m−n+l) (q) (n−l+i)k+b+1+χ(i n−m) (q) k = M n (−h, b, k, m, l; q), which is routine.
the electronic journal of combinatorics 21(3) (2014), #P3.38 Note that we can avoid using the induction hypothesis.The truth of Lemma 5.1 in case 1 results in the truth of Theorem 1.3 in case 1, which is needed in the case 2 of Lemma 5.1.Now we are ready to deal with the extra point.
Proof of Lemma 2.5.As we discussed in (4.8), CT x 0 Q(h) can be written as where the sum ranges over all the r's and the k's with 0 < r 1 < • • • < r s n, 0 k 1 , k 2 , . . ., k s h such that Lemma 4.3 does not apply.Note that we may have different s.
Since h = (n−l −1)k +b+1 and 0 k i h, by Lemma 4.2, there leaves only one term for which Lemma 4.3 is not applicable.This term corresponds to k i = (n − l − i)k + b + 1 for i = 1, . . ., n − l and r i = i for i = 1, . . ., n − l.It follows that CT (5.11) The lemma then follows from Lemma 5.1.
The extra point h = (n − l − 1)k + b + 1 in Lemma 2.5 is not easy to find.This h seems to be the only choice of the extra point for which it is not hard to show that CT x Q(h) = M n (−h, b, k, m, l; q).Intuitively a desired extra point must be chosen from boundary values, i.e., values next to the vanishing points listed in (2.x (q/x 1 ) k−b−1 (x 1 ) b+k+1 (1/x 3 ) k−b−1 (x 3 q) b+k+1 1 − 1/(x 1 q b+1 ) Secondly, the boundary values h = mk, (m + 1)k, . . ., (n − 1)k do not work either for a similar reason.

Lemma 2 . 5 .
For fixed b, n ∈ N, and 0 m, l < n, and k > b + 1, if we assume Theorem 1.3 holds for smaller values of n, then M a+b the electronic journal of combinatorics 21(3) (2014), #P3.38