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MATH 630–600. Enumerative Combinatorics
Solution of Assignment 5.
- Let m 1 ≥ m 2 ≥ · · · ≥ mn be a sequence of positive integers, and M = (mi,j ) be an n × n matrix whose ijth entry is mij =
(mi − i + j)!
(Assume that all mi − i + j are nonnegative.) Prove that the determinant of M is
det(M ) =
1 ≤i<j≤n(mi^ −^ mj^ +^ j^ −^ i) (m 1 + n − 1)!(m 2 + n − 2)! · · · mn!
Solution. Take the factor (^) (mi+^1 n−i)! from the i-th row, we have a matrix N whose i, j-entry is (yi)n−j , where yi = mi + n − i. For the matrix N , proceed like the Vandermonde matrix: If yi = yj , then the determinant is 0. Hence the determinant has a factor of yi − yj , for each pair i neqj. Comparing the degree, one has
det(N ) = C
1 ≤i<j≤n
(yi − yj ),
for some constant C. Now comparing the coefficient of y 1 n −^1 yn 2 −^2 · · · yn, one gets C = 1.
- Find the rank-generating function F (P, q) for the following posets.
(a) DN , the set of all positive integral divisors of n, where i ≤ j if i is a divisor of j. Assume that the prime factorization of n is pα 1 1 pα 2 2 · · · pα k k, where p 1 ,... , pk are distinct primes, and αi are positive integers.
Solution. Elements of rank n are those that can be factored as pa 1 1 pa 22 · · · pa kk where each ai is an integer in [0, αi, and a 1 + a 2 + · · · + ak = n. Hence F (DN , q) is the generating functions of the number of solutions to the above equation, which is
F (DN , q) =
∏^ k
i=
(1 + q + q^2 + · · · + qαi^ ).
(b) The set Πn of all partitions of [n], ordered by refinement.
Solution. A partition is of rank k iff it has n − k blocks. There are S(n, n − k) many such partitions. Hence
F (Πn, q) =
n∑− 1
k=
S(n, n − k)qk.
(c) The set Ln(q) that consists of all subspaces of an n-dimensional vector space Vn(q) over the q-element field Fq, ordered by inclusion.
Solution. F (Ln(q), q) =
∑^ n
k=
n k
q
qk.
Note that it is not a q-analog of binomial theory, since one can not simplify the right-hand side. The q-binomial theorem is
(1 + xq)(1 + xq^2 ) · · · (1 + xqn) =
∑^ n
k=
n k
q
q(
k+ 2 )xk.
- If posets P and Q are graded with rank generating functions F (P, q) and F (Q, q), then prove
F (P × Q, q) = F (P, q)F (Q, q), and F (P ⊗ Q, q) = F (P, qr+1)F (Q, q). Solution. Assume P is of rank n, and there are Pi many elements of rank i. Assume Q is of rank m, and there are Qj many elements of rank j. First it is needed to show that both posets are graded. It can be done by analyzing the structure of cover relation and maximal chains, (as we did in class). You should get that the rank of P × Q is n + m, and the rank of P ⊗ Q is nm + m + n. (1) In the poset∑ P × Q, an element (x, y) is of rank t iff ρP (x) + ρQ(y) = t. There are i+j=t PiQj^ many such elements. Hence^ F^ (P^ ×^ Q, q) =^ F^ (P, q)F^ (Q, q). (2) In the poset P ⊗ Q is graded. An element (x, y) is of rank ρP (x)(m + 1) + ρQ(y). Hence
F (P ⊗ Q, q) =
x∈P,y∈Q
qρP^ (x)(m+1)+ρQ(y)
x∈P
(qm+1)ρP^ (x)^
y∈Q
qρQ(y)
= F (P, qr+1)F (Q, q).
- Check the following rules of cardinal arithmetic:
RP^ +Q^ = RP^ × RQ, (RQ)P^ = RQ×P^. Proof. Prove by giving order-preserving bijections. For the first one, for any order-preserving map f from P + Q to R, let θ(f ) = (f |P , f |Q). For the second one, given any f ∈ (RQ)P^ , i.e., for any p ∈ P , f (p) is an order-preserving map from Q to R, let θ(f ) be the map from Q × P to R, whose action on Q × P is: θ(f ) : (q, p) → f (p)(q). I leave it to you to check that the above defined maps are order-preserving bijections.
- Construct an infinite meet-semilattice P with ˆ1, such that P is not a lattice.
Solution. Here is one example. Let P = { 0 , − 1 , − 2 , · · · } ∪ {α, β, ˆ 0 } where the order is: (1) −n < −m if n > m for any two natural integers n, m. (2) α < −n, β < −n for all natural integers n. α and β are incomparable. (3) ˆ0 is the minimal element, which is less than anything else. This is a meet-semilattice, where ˆ1 = 0. But α ∨ β does not exist.
