Title A $q$-analogue of Catalan Hankel determinants (New Trends in Combinatorial Representation Theory) Author(s) ISHIKAWA, Masao; TAGAWA, Hiroyuki; ZENG, Jiang Citation 数理解析研究所講究録別冊 = RIMS Kokyuroku Bessatsu (2009), B11: 19-41 Issue Date 2009-04 URL http://hdl.handle.net/2433/176780 Right Type Departmental Bulletin Paper Textversion publisher Kyoto University
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Title A $q$-analogue of Catalan Hankel determinants (New Trendsin Combinatorial Representation Theory)
In this article we shall survey the various methods of evaluating Hankel determinants and
as an illustration we evaluate some Hankel determinants of a q-analogue of Catalan numbers.
Here we consider (aq;q)n
(abq2;q)nas a q-analogue of Catalan numbers Cn = 1
n+1
`2nn
´, which is known as
the moments of the little q-Jacobi polynomials. We also give several proofs of this q-analogue, in
which we use lattice paths, the orthogonal polynomials, or the basic hypergeometric series. We
also consider a q-analogue of Schroder Hankel determinants, and give a new proof of Moztkin
Hankel determinants using an addition formula for 2F1.
x 1. Introduction
Given a sequence a0, a1, a2,. . . , we set the Hankel matrix of the sequence to be
(1.1) A(t)n = (ai+j+t)0≤i,j≤n−1 =
0BBB@
at at+1 . . . at+n−1
at+1 at+2 . . . at+n
......
. . ....
at+n−1 at+n . . . at+2n−2
1CCCA .
For n = 0, 1, 2, . . . , let
(1.2) Cn =1
n + 1
2n
n
!,
which are called the Catalan numbers. The generating function for the Catalan numbers is
given byX
n≥0
Cntn =1−√1− 4t
2t.
Received May 8, 2008. Accepted July 6, 2008.2000 Mathematics Subject Classification(s): Primary 05A30; Secondary 05A10, 05E35, 15E15,33D15Key Words: Catalan numbers, determinants, Dyck paths, orthogonal polynomials, continued frac-tions∗Faculty of Education, Tottori University, Koyama, Tottori, Japan.∗∗Faculty of Education, Wakayama University, Sakaedani, Wakayama, Japan.∗∗∗Institut Camille Jordan, Universite Claude Bernard Lyon I, 43, boulevard du 11 novembre 1918,
If we substitute Xi = q−ki , Bl = −aql and Al = −abql+1 into (5.1), then we see that
det (µki+j)0≤i,j≤n−1 =
n−1Yi=0
q(n−1)ki(aq; q)ki
(abq2; q)ki+n−1
Y
0≤i<j≤n−1
“q−ki − q−kj
” Y
1≤i≤j≤n−1
“abqj+1 − aqi
”
One can derive (1.15) easily by a direct computation.
x 5.2. An addition formula for 2F1
In this subsection we give a new proof of (1.7) using an addition formula for 2F1 and
LU-decomposition of Motzkin Hankel matrices. First, we shall prove the following identity.
Lemma 5.2. If i and j are nonnegative integers, then we have
X
k≥0
i
k
! j
k
!2F1
»k−i+1
2, k−i
2
k + 2; 4
–2F1
» k−j+12
, k−j2
k + 2; 4
–= 2F1
» 1−i−j2
, −i−j2
2; 4
–.(5.2)
Proof. Recall the quadratic transformation formula (see [9, (3.1.5)]):
(1− z)a2F1(a, b; 2b; 2z) = 2F1
„a
2,a + 1
2; b +
1
2;
z2
(1− z)2
«.(5.3)
Applying (5.3) with a = k − i, b = k + 3/2 and z = 2 we obtain
2F1
»k−i+1
2, k−i
2
k + 2; 4
–= (−1)k−i
2F1
»k − i, k + 3/2
2k + 3; 4
–.
Substituting i by j yields
2F1
» k−j+12
, k−j2
k + 2; 4
–= (−1)k−j
2F1
»k − j, k + 3/2
2k + 3; 4
–.
Now, applying (5.3) with a = −i− j, b = 3/2 and z = 2 we obtain
2F1
» 1−i−j2
, −i−j2
2; 4
–= (−1)i+j
2F1
»−i− j, 32
3; 4
–.
Therefore we can rewrite (5.2) as follows:
X
k≥0
i
k
! j
k
!2F1
»k − i, k + 3/2
2k + 3; 4
–2F1
»k − j, k + 3/2
2k + 3; 4
–= 2F1
»−i− j, 32
3; 4
–.(5.4)
34 Ishikawa, Tagawa and Zeng
Now we recall a formula of Burchnall and Chaundy [3, (43)]:
2F1
»c− a, c− b
c; x
–=X
k≥0
(c− a)k(a)k(d)k(c− b− d)k
k!(c + k − 1)k(c)2kx2k
× 2F1
»c− a + k, c− b− d + k
c + 2k; x
–2F1
»c− a + k, d + k
c + 2k; x
–.(5.5)
It is then easy to check that the specialization of (5.5) with
a =3
2, b = 3 + i + j, c = 3, d = −j, x = 4
yields (5.4).
Proof of (1.7). Define lij and uij by
lij =
i
j
!2F1
» j−i+12
, j−i2
j + 2; 4
–,(5.6)
uij =
j
i
!2F1
» i−j+12
, i−j2
i + 2; 4
–.(5.7)
Then Ln = (lij)0≤i,j≤n−1 is a lower triangular matrix with all diagonal entries 1, and Un =
(lij)0≤i,j≤n−1 is an upper triangular matrix with all diagonal entries 1. The formula (5.2) gives
the LU-decomposition of Motzkin Hankel matrix:
(Mij) = LnUn.
Hence we conclude that det (Mij) = 1.
x 5.3. A q-analogue of Schroder numbers
We define Sn(q) (n ≥ 0) by the following recurrence:
S0(q) = 1, Sn(q) = q2n−1Sn−1(q) +
n−1X
k=0
q2(k+1)(n−1−k)Sn−1−k(q)Sk(q).
In fact one can show that
Sn(q) =X
P∈S2n,0
ω(P ),
where ω(P ) is the number of triangles below the path P (see Figure 10), and the sum runs over
all Schroder paths from the origin to (2n, 0). As a q-analogue of (1.10) and (1.11) we consider
the matrix
(5.8) S(t)n (q) =
“q(i−j)(i−j−1)Si+j+t(q)
”0≤i,j≤n−1
.
Note that this matrix is not a Hankel matrix, but as a q-analogue of (1.10) and (1.11), the
following theorem holds:
A q-analogue of Catalan Hankel determinants 35
Theorem 5.3. Let n be a positive integer.
(i) If t = 0 or 1, then we have
(5.9) det S(1)n−1(q) = det S(0)
n (q) =
n−1Y
k=1
(q2k−1 + 1)n−k.
(ii) If t = 2, then we have
(5.10) det S(2)n (q) = q−1
nY
k=1
(q2k−1 + 1)n+1−k(
n+1Y
k=1
(q2k−1 + 1)− 1).
To prove this theorem, we define the matrices
bS(t)n (q) =
“q2(n−i)(t+i+j−2)St+i+j−2(q)
”1≤i,j≤n
,
eS(t)n (q) =
“q−(t+i+j)(t+i+j−1)St+i+j−2(q)
”1≤i,j≤n
,
then the following lemma can be easily proven by direct computations:
Lemma 5.4. Let n be a positive integer. Then
det bS(t)n (q) = q
n(n−1)(2n+3t−4)3 det S(t)
n (q),(5.11)
det eSn(q) = q−n(4n2+3(2t+1)n+3t2+3t−1)
3 det S(t)n (q).(5.12)
Lemma 5.5. Let n be a positive integer.
(i) If n ≥ 2, then we have
(5.13) det bS(0)n (q) = q(n−1)(n−2) det bS(1)
n−1(q).
(ii) If n ≥ 2, then we have
(5.14) det bS(1)n (q) = qn2−n+1 det bS(2)
n−1(q) + q2(n−1)2 det bS(1)n−1(q).
(iii) If n ≥ 3, then we have
(5.15) det eS(0)n (q) det eS(2)
n−2(q) = det eS(0)n−1(q) det eS(2)
n−1(q)−n
det eS(1)n−1(q)
o2
.
Proof. We consider the digraph (V, E), in which V is the plane lattice Z2 and E the set of
rise vectors, fall vectors and long level vectors in the above half plane. Let ui = (x0−2(i−1), 0)
and v(t)j = (x0+2(j+t−1), 0) for i, j = 1, 2, . . . , n, t = 0, 1, 2 and a fixed integer x0. It is easy to
see that the n-vertex u = (u1, . . . , un) is D-compatible with the n-vertex v(t) = (v(t)1 , . . . , v
(t)n ).
We assign the weight of each edge as a rise vector, a fall vector and a long level vector whose
origin is (x, y) and ends at (x+1, y+1), (x+1, y−1) and (x+2, y) has weight qx−y−x0+2(n−1),
1 and qx−y−x0+2n−1, respectively, which is visualized in Figure 11. Then, by applying Lemma
2.3, we can obtain
(5.16) GFhP0
“u,v(t)
”i= det bS(t)
n (q).
This is important to prove the following.
36 Ishikawa, Tagawa and Zeng
(i) Assume t = 0 and let u and v be as above. Put eui = (x0−2i+3, 1) and ev(1)j = (x0+2j−3, 1)
for i, j = 2, . . . , n, and let eu = (eu2, . . . , eun) and ev(1) = (ev(1)2 , . . . , ev(1)
n ). Then each n-path
P = (P1, P2, . . . , Pn) from u to v(0) corresponds to an (n− 1)-path eP = ( eP2, . . . , ePn) from euto ev(1) by regarding eP as the subpath of P . In fact, note that P1 is always the path composed
of a single vertex u1 = v1, each Pi always starts from the rise vector ui → eui and ends at the
fall vector ev(1)i → v
(0)i for i = 2, . . . , n. Hence this gives a bijection, and the product of the
weight of the rise vectors ui → eui and the fall vectors ev(1)i → v
(0)i for i = 2, . . . , n is q(n−1)(n−2).
This proves (5.13).
(ii) Assume t = 1 and let u and v be as above, i.e., ui = (x0 − 2(i − 1), 0) and v(1)j =
(x0 + 2j, 0) for 1 ≤ i, j ≤ n (see Figure 12). Put eui = (x0 − 2i + 3, 1) (2 ≤ i ≤ n) and
ev(2)j = (x0 + 2j − 1, 1) (2 ≤ j ≤ n), and let eu = (eu2, . . . , eun) and ev(2) = (ev(2)
2 , . . . , ev(2)n ).
Further, put ui = (x0 − 2i + 4, 2) (2 ≤ i ≤ n) and v(1)j = (x0 + 2j − 2, 2) (2 ≤ j ≤ n), and let
u = (u2, . . . , un) and v(1) = (v(1)2 , . . . , v
(1)n ). Let P = (P1, P2, . . . , Pn) be any non-intersecting
n-paths from u to v(1). Then, it is easy to see that P must satisfy one of the following two
conditions:
(1) P1 is the long level vector whose origin is u1 and ends at v(1)1 , and Pi goes through the
vertices eui and ev(2)i for i = 2, 3, . . . , n.
(2) P1 is a path which goes through only three vertices u1, u1 + (1, 1)(= v(1)1 − (1,−1)) and
v(1)1 , and Pi goes through the vertices eui, ui, v
(1)i and ev(2)
i for i = 2, 3, . . . , n.
By a similar argument as in the proof of (i), we can deduce that
GFhP0(un,v(1)
n )i
= qn2−n+1GFhP0(un−1,v
(2)n−1)
i+ q2(n−1)2GF
hP0(un−1,v
(1)n−1)
i
holds. By the equality (5.16), we obtain the identity (5.14).
(iii) This identity can be proven by applying the Desnanot-Jacobi adjoint matrix theorem (4.8)
to eS(t)n (q).
Proof of Theorem 5.3. (i) The first equality of (5.9) is easily obtained from (5.11) and (5.13).
By applying the equalities (5.11) and (5.12) to (5.14) and (5.15), we have
det S(1)n (q) = q det S
(2)n−1(q) + det S
(1)n−1(q)(5.17)
for n ≥ 2, and we have
det S(0)n (q) det S
(2)n−2(q) = det S
(0)n−1(q) det S
(2)n−1(q)− q2(n−1){det S
(1)n−1(q)}2(5.18)
for n ≥ 3. By the equalities (5.17) and (5.18), for n ≥ 3, the following identity holds:
det S(0)n (q)(det S
(1)n−1(q)− det S
(1)n−2(q))
= det S(0)n−1(q)(det S(1)
n (q)− det S(1)n−1(q))− q2n−1{det S
(1)n−1(q)}2.(5.19)
Moreover, by applying the first equality of (5.9) to (5.19) and replacing n with n−1, we obtain
(5.20) (1 + q2n−3){det S(0)n−1(q)}2 = det S
(0)n−2(q) det S(0)
n (q)
for n ≥ 4. We prove the second equality of (5.9) by induction on n. If n = 1, 2, 3, then it is
easily obtained by direct computations. Assume that (5.9) holds up to n− 1. Then, by (5.20)
and induction hypothesis, we can obtain the second equality (5.9).
(ii) It follows from our result of (i) and the equality (5.17) that (5.10) holds.
A q-analogue of Catalan Hankel determinants 37
By applying the Desnanot-Jacobi adjoint matrix theorem (4.8) to S(1)n+1(1), then we have
(5.21) det S(1)n+1(1) det S
(3)n−1(1) = det S(1)
n (1) det S(3)n (1)− {det S(2)
n (1)}2
for n ≥ 2. Therefore the following identity is easily obtained by induction on n and the formula
(5.21):
Remark. For positive integer n, we have
(5.22) det S(3)n (1) = 2(n+3
2 ) − (2n + 3)2(n+22 ) − 2(n+1
2 ).
Note that there is a relation between domino tilings of the Aztec diamonds and Schroder
paths (see [8, 15]). It might be an interesting problem to consider what this weight means.
x 5.4. Delannoy numbers
The Delannoy numbers D(a, b) are the number of lattice paths from (0, 0) to (b, a) in
which only east (1, 0), north (0, 1), and northeast (1, 1) steps are allowed. They are given by
the recurrence relation
D(a, 0) = D(0, b) = 1,
D(a, b) = D(a− 1, b) + D(a− 1, b− 1) + D(a, b− 1).(5.23)
The first few terms of D(n, n) (n = 0, 1, 2, . . . ) are given by 1, 3, 13, 63, 321, . . . . By a similar
argument we can derive the following result. We may give a proof in another occasion.
Proposition 5.6. Let n be a positive integers. Then the following identities would
hold:
det (D(i + j, i + j))0≤i,j≤n−1 = 2(n+12 )−1,(5.24)
det (D(i + j + 1, i + j + 1))0≤i,j≤n−1 = 2(n+22 )−2 + 2(n+1
2 )−1,(5.25)
det (D(i + j + 2, i + j + 2))0≤i,j≤n−1 = 2(n+32 )−3 + (2n + 1)2(n+2
2 )−2 − 2(n+12 )−1.(5.26)
x 6. Concluding remarks
Since a hyperpfaffian version of (1.3) is obtained in [13], we believe it will be interesting
problem to consider a hyperpfaffian version of Theorem 1.1 and Theorem 1.3. We shall argue on
it in another chance. The authors also would like to express their gratitude to the anonymous
referee for his (her) constructive comments.
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