An alternative evaluation. of the Andrews{Burge determinant. C. Krattenthaler y. Dedicated to Gian-Carlo Rota

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1 An alternative evaluation of the Andrews{Burge determinant C. Krattenthaler y Dedicated to Gian-Carlo Rota Abstract. We give a short, self-contained evaluation of the Andrews{Burge determinant (Pacic J. Math. 58 (994), {4).. Introduction In [9, Theorem ], Andrews and Burge proved a determinant evaluation equivalent to x i j y i j (.) det 0i;j i j i j = () (n3 mod 4) (n ) j= xy j b(j)=c xy bj=c 3n j 3 ; (j) j where the shifted factorial (a) k is given by (a) k := a(a ) (a k ), k, (a) 0 :=, and where (A)= if A is true and (A)=0 otherwise. This determinant identity arose in connection with the enumeration of symmetry classes of plane partitions. The known proofs [9, 0] of (.) require that one knows (.) to hold for x = y. Indeed, the latter was rst established by Mills, Robbins and Rumsey [5, p. 53], in turn using another determinant evaluation, due to Andrews [3], whose proof is rather complicated. In the meantime, simpler proofs of the x = y special case of (.) were found by Andrews [7], Andrews and Stanton [8], and Petkovsek and Wilf [6]. In this note we describe a new, concise, and self-contained proof of (.), see section 3. y Supported in part by EC's Human Capital and Mobility Program, grant CHRX-CT and the Austrian Science Foundation FWF, grant P09-MAT 99 Mathematics Subject Classication. Primary 5A5; Secondary 05A5, 05A7, 33C0. Key words and phrases. determinant evaluations, hypergeometric series, enumeration of symmetry classes of plane partitions. Typeset by AMS-TEX

2 C. Krattenthaler In fact, the main purpose of this note is to popularize the method that I use to prove (.) (see section for a description). This method is simple but powerful. Aside from this note, evidence for this claim can be found e.g. in [,, 3, 4]. Thus, the method enlarges the not at all abundant collection of methods for evaluating determinants. In fact, aside from elementary manipulations by row and column operations, we are just aware of one other method, namely Andrews' \favourite" method of evaluating determinants (cf. [,, 3, 4, 5, 6, 8]), which basically consists of guessing and then proving the LU-factorization of the matrix in question (i.e., the factorization of the matrix into a product of a lower triangular times an upper triangular matrix). We should, however, also point out a limitation of our method. Namely, in order to be able to apply our method, we need a free parameter occuring in the determinant. (In (.) there are even two, x and y.) Andrews' method, on the other hand, might still be applicable if there is no free parameter present. Still, it is safe to speculate that many more applications of our method are going to be found in the future.. The method The method that I use to prove (.) is as follows. Suppose we have a matrix fij (x) with entries 0i;j f ij(x) which are polynomials in x, and we want to prove the explicit factorization of det f ij (x) as a polynomial in x, (.) det fij (x) = C(n) (x a l (n)) ml(n) ; 0i;j l where C(n); a l (n); m l (n) are independent of x, and the a l (n)'s are pairwise dierent for xed n. Then, in the rst step, for each l we nd m l (n) linearly independent linear combinations of the columns, or of the rows, which vanish for x = a l (n). In different, but equivalent terms, we nd m l (n) linearly independent vectors in the kernel of our matrix evaluated at x = a l (n), i.e., in the kernel of f ij (a l (n)), or of its transpose. That this really guarantees that (x a l (n)) m l(n) is a factor of the determinant is a fact that might not be well-known enough. Therefore, for the sake of completeness, we state it as a lemma at the end of this section and provide a proof for it. The nding of a l (n) linearly independent linear combinations of columns or rows which vanish for x = a l (n) (equivalently, linearly independent vectors in the kernel of the matrix, or its transpose, evaluated at x = a l (n)) can be done with some skill (and primarily patience) by setting x = a l (n) in the matrix f ij (x), computing tables for the coecients of the linear combinations for n = ; ; : : : (by solving the respective systems of linear equations on the computer), and nally guessing what the general pattern of the coecients could be. To prove that the guess is correct, in case of binomial determinants (such as the one in (.)) one has to verify certain binomial identities. But this is pure routine today, by means of Zeilberger's algorithm [8, 9].

3 An alternative evaluation of the Andrews{Burge determinant 3 Next, in the second step, one checks the degrees of both sides of (.) as polynomials in x. If it should happen that the degree of det f ij (x) is not larger than P l m l(n), the degree of the right-hand side, then it follows by what we did in the rst step that the determinant det f ij (x) has indeed the form of the right-hand side of (.), where C(n) is some unknown constant. This constant can then be determined in the third step by comparison of coecients of a suitable power of x. Finally, here is the promised lemma, and its proof. Lemma. Let A = A(x) be a matrix whose entries are polynomials in x, and u a number. If dim Ker A(u) k, then u is a root of det A(x) of multiplicity at least k. Proof. Let v ; v ; : : :; v k be k linearly independent (column) vectors in the kernel of A(u). Without loss of generality, we may assume that v ; v ; : : :; v k are such that the matrix [v ; v ; : : :; v k ], formed by gluing the columns v ; v ; : : :; v k to a matrix, is in column-echelon form. In addition, again without loss of generality, we may assume that for any i = ; ; : : :; k the vector v i is of the form v i = (0; : : :; 0; ; : : :) t, i.e., the rst i entries are 0, the i-th entry is, and the remaining entries could be anything. (To justify \without loss of generality" one would possibly have to permute the columns of A.) Now we consider the matrix A ~ = A(x), ~ formed by replacing for i = ; ; : : :; k the i-th column of A by the column Av i. It is an easy observation that A ~ and A are related by elementary column operations. Therefore, their determinants are the same. On the other hand, the i-th column of A(u) ~ is A(u)v i = 0, for any i = ; ; : : :; k. Hence, each entry of Av i, being a polynomial in x, must be divisible by (x u). Therefore, in the determinant det A, ~ we may take (x u) out of the i-th column, i = ; ; : : :; k, with the entries in the remaining determinant still being polynomials in x. This proves that (x u) k divides det A ~ = det A, and, thus, the lemma. 3. Proof of (.) As announced in the previous section, the proof consists of three steps. For convenience, let us denote the determinant in (.) by AB(x; y; n), or sometimes just AB(n) for short. Step. Identication of the factors. We show that the product on the right-hand side of (.), j= xy j b(j)=c xy 3n j 3 bj=c ;

4 4 C. Krattenthaler is indeed a factor of AB(n), in the way that was described in section. Let us rst consider just one part of this product, j= ((x y)= j ) b(j)=c : Let us concentrate on a typical factor (x y j l), j n, l (j )=. We claim that for each such factor there is a linear combination of the columns that vanishes if the factor vanishes. More precisely, we claim that for any j; l with j n, l (j )= there holds (3.) b(jl)=c X s=l (j l ) (j s) (j l s) sl (s l )! (column s of AB(y j l; y; n)) (column j of AB(y j l; y; n)) = 0: To avoid confusion, for j = l it is understood by convention that the sum in the rst line of (3.) vanishes. To establish the claim, we have to check (3.) b(jl)=c X s=l (j l ) (j l s) sl (j s) (s l )! y j l i s y i s i s i s y j l i y i j i j i j = 0: The exceptional case j = l can be treated immediately. By assumption the sum in the rst line of (3.) vanishes for j = l, and, by inspection, also the last line in (3.) vanishes for j = l. So we are left with establishing (3.) for l j=. In terms of the usual hypergeometric notation this means to check (3.3) rf s a ; : : :; a r b ; : : :; b s ; z ( i j l y) il ( i l)! = X k=0 (a ) k (a r ) k k! (b ) k (b s ) k z k ; 4 F j l; j l; i l; i j y 3 j l; i j l y ; i j l y ; ( i 4l y) il ( i l)! l; i l; i l y j l; i l y ; i l y y j l i y i j = 0; i j i j 4F 3 j l; j ;

5 An alternative evaluation of the Andrews{Burge determinant 5 for j n, l j=. This identity can be proved routinely by means of Zeilberger's algorithm [8, 9] and Salvy and Zimmermann's Maple package GFUN [7]. However, it happens that a 4 F 3 -summation is already known that applies to both 4 F 3 -series in (3.3), namely Lemma in [9]. (Here we need the assumption l Q j=.) Little simplication then establishes (3.3) and hence the claim. Thus, j= ((x y)= j ) b(j)=c is a factor of AB(n). Now we prove that j= (x y)= 3n j 3 bj=c is a factor of AB(n). Also here, let us concentrate on a typical factor (x y 6n j l ), j n, l j=. This time we claim that for each such factor there is a linear combination of the columns that vanishes if the factor vanishes. More precisely, we claim that for any j; l with j n, l j= there holds Xnl s= (n j s) jl (n l j) jl (s) nls (3n l j ) nls 4 nls (n l s)! (n j ) nls This means to check Xnl s= (column s of AB(y 6n j l ; y; n)) = 0: (n j s) jl (s) nls (3n l j ) nls (n l j) jl 4 nls (n l s)! (n j ) nls y 6n j l i s y i s i s i s Converting this into hypergeometric notation and cancelling some factors, we see that we have to check (3.4) (4 i j l 6n y) i i; 5 5 F j l 3n; j n; 4 4 j 4n; l n; i j l 3n y ; = (3 i y) i l n; 3 i j l 6n y 5 i j l 3n y ; i; 5 5 F j l 3n; j n; l n; i y 4 4 j 4n; l n; 3 i y ; i y Again, in view of Zeilberger's algorithm, this is pure routine. However, also this identity happens to be already in the literature. It is exactly identity (5.7) = 0 ;

6 6 C. Krattenthaler in [8], with a = Q 3 i j l 3n y, x = 3 j l 3n, z = j n, p = i. Thus, j= (x y)= 3n j 3 is a factor of AB(n). bj=c Step. Bounding the polynomial degrees. The degree of AB(n) as a polynomial in x is obviously at most. But the degree of the product on the n n right-hand side of (.) is exactly. Therefore it follows that x i j y i j (3.5) det 0i;j i j i j = C(n) xy j b(j)=c xy 3n j 3 bj=c ; j= with some C(n) independent of x, and, by symmetry, also independent of y. Step 3. Determining the constant. To compute C(n), on both sides of (3.5) set y = x and then compare coecients of x (n ). On the right-hand side the coecient is () (n3 mod 4) C(n), whereas on the left-hand side the coecient is det 0i;j (i j)! = n i=0 = n i=0 = n i=0 (i)! (i)! (i)! = n(n ) i=0 = (n ) i= det (i j )j 0i;j det (i) j 0i;j 0i<j i! (i)! (i) i ; (j i) where in the step from the rst to the second line we used elementary column operations, and the subsequent step is just the Vandermonde determinant evaluation. This completes the proof of (.). References [] G. E. Andrews, Plane partitions (II): The equivalence of the Bender{Knuth and the MacMahon conjectures, Pacic J. Math. 7 (977), 83{9. [] G. E. Andrews, Plane partitions (I): The MacMahon conjecture, in: Studies in Foundations and Combinatorics, G.-C. Rota (ed.), Adv. in Math. Suppl. Studies, Vol., 978, 3{50.

7 An alternative evaluation of the Andrews{Burge determinant 7 [3] G. E. Andrews, Plane partitions (III): The weak Macdonald conjecture, Inventiones Math. 53 (979), 93{5. [4] G. E. Andrews, Macdonald's conjecture and descending plane partitions, in: Combinatorics, representation theory and statistical methods in groups, oung Day Proceedings, T. V. Narayana, R. M. Mathsen, J. G. Williams (eds.), Lecture Notes in Pure Math., vol. 57, Marcel Dekker, New ork, Basel, 980, 9{06. [5] G. E. Andrews, Plane partitions (IV): A conjecture of Mills{Robbins{ Rumsey, Aequationes Math. 33 (987), 30{50. [6] G. E. Andrews, Plane partitions V: The t.s.s.c.p.p. conjecture, J. Combin. Theory Ser. A 66 (994), 8{39. [7] G. E. Andrews, Pfa's method (I): The Mills{Robbins{Rumsey determinant, preprint. [8] G. E. Andrews and D. W. Stanton, Determinants in plane partition enumeration, preprint. [9] G. E. Andrews and W. H. Burge, Determinant identities, Pacic J. Math. 58 (993), {4. [0] I. P. Goulden and D. M. Jackson, Further determinants with the averaging property of Andrews{Burge, J. Combin. Theory Ser. A 73 (996), 368{375. [] C. Krattenthaler, Determinant identities and a generalization of the number of totally symmetric self-complementary plane partitions, Elect. J. Combin. (to appear). [] C. Krattenthaler, Some q-analogues of determinant identities which arose in plane partition enumeration, Seminaire Lotharingien Combin. (to appear). [3] C. Krattenthaler, A new proof of the MRR-conjecture including a generalization, preprint. [4] G. Kuperberg, Another proof of the alternating sign matrix conjecture, Math. Research Letters (996), 39{50. [5] W. H. Mills, D. H. Robbins and H. Rumsey, Enumeration of a symmetry class of plane partitions, Discrete Math. 67 (987), 43{55. [6] M. Petkovsek and H. Wilf, A high-tech proof of the Mills-Robbins-Rumsey determinant formula, Elect. J. Combin. 3 (no., \The Foata Festschrift") (996), #R9, 3 pp. [7] B. Salvy and P. Zimmermann, GFUN A MAPLE package for the manipulation of generating and holonomic functions in one variable, ACM Trans. Math. Software 0 (994), 63{77. [8] D. Zeilberger, A fast algorithm for proving terminating hypergeometric identities, Discrete Math. 80 (990), 07{. [9] D. Zeilberger, The method of creative telescoping, J. Symbolic Comput. (99), 95{04. Institut fur Mathematik der Universitat Wien, Strudlhofgasse 4, A-090 Wien, Austria. KRATT@Pap.Univie.Ac.At

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