Math 7409 Lecture Notes 11
Products of Posets
Let (X1,≤1), ..., (Xn,≤n) be posets. The direct product of these posets is the poset (X,≤) where
X = X1 × ... × Xn and (x1,...,xn) ≤ (y1,...,yn) if and only if xi ≤i yi for i = 1, ..., n. That the direct product of posets is a poset is immediate. Note that the direct product of lattices will be a lattice with the corresponding definitions for meet and join.
Proposition: (a) If |X| = n, then the power-set lattice P(X) is the direct product of n copies of the two element lattice {0,1}. = a1 a 2⋯ a r (b) If n p1 p2 pr , where the pi are distinct primes, then the lattice D(n) of divisors a a of n is isomorphic to the direct product of the lattices (chains) D(p1 1), ..., D(pr r).
Proof: (a) Let the elements of X be x1,x2,...,xn. Each subset Y of X will be identified with its characteristic function (e1,...,en) where ei = 1 iff xi in Y and 0 otherwise. This is a bijection between the n power set P(X) and {0,1} . If Y and Z have characteristic functions (e1,...,en) and (f1,...,fn) respectively, then ⊆ ⇔∀ ∈ ⇒ ∈ Y Z i xi Y xi Z ⇔∀ = ⇒ = i ei 1 f i 1 ⇔∀ ≤ i ei f i , so the map is an isomorphism. (b) An exercise.
Möbius Function of a Poset
Let P = (X, ≤) be a finite (locally finite) poset. The incidence algebra I(P) of P is the set of functions f: X × X → ℝ which satisfy f(x,y) = 0 unless x ≤ y. Addition and scalar multiplication are defined pointwise, and two functions are multiplied by the rule: ⋅ = ∑ f g x , y f x , z g z , y . x≤z≤y Example: Consider the divisor lattice P = D(12) where X = {1,2,3,4,6,12}. A function in I(P) can be identified with the 6 × 6 matrix whose rows and columns are indexed by the elements of X and whose (i,j) entry is the function value f(i,j). Two such functions are : 1 1 1 1 1 1 1 2 3 4 5 6 0 1 0 1 1 1 0 7 0 8 9 10 ≡ 0 0 1 0 1 1 ≡ 0 0 11 0 12 13 f 1 and f 2 0 0 0 1 0 1 0 0 0 14 0 15 0 0 0 0 1 1 0 0 0 0 16 17 0 0 0 0 0 1 0 0 0 0 0 18
Computing the values of f1·f2 for (1,6), (4,4) and (2,4): z ∈ [1,6] = {1,2,3,6} f1·f2(1,6) = f1(1,1)f2(1,6) + f1(1,2)f2(2,6) + f1(1,3)f2(3,6) + f1(1,6)f2(6,6) = 1·5+1·9+1·12+1·16 =42
z ∈ [4,4] = {4} f1·f2(4,4) = f1(4,4)f2(4,4) = 1·14 = 14
z ∈ [2,4] = {2,4} f1·f2(2,4) = f1(2,2)f2(2,4) + f1(2,4)f2(4,4) = 1·8 + 1·14 = 22 Proposition: If |X| = n, the incidence algebra I(P) is isomorphic to a subalgebra of the algebra of upper triangular matrices. A function f is invertible if and only if f(x,x) ≠ 0 for all x in X.
Proof: Take a linear extension of P; that is, since X is finite, we can number the elements of X as x1,...,xn where xi ≤ xj only if i ≤ j. Now map f in I(P) to the matrix A = (aij) where aij = f(xi,xj). Clearly A is upper triangular. Also the map is an isomorphism since if matrices A and B correspond to f and g, then the matrix corresponding to f⋅g has (i,j) entry ∑ = ∑ = ∑ f xi , xk g xk , x j aik bkj aik bkj ≤ ≤ ≤ ≤ ≤ ≤ xi xk x j i k j 1 k n where the last equality holds because unless i ≤ k ≤ j, either aik or bkj is 0. In particular, f⋅g(x,y) = 0 unless x ≤ y (since there are no terms in the sum), so f⋅g in I(P). Note that the values f(x,x) are the diagonal elements of the matrix corresponding to f. So, a function satisfying f(x,x) ≠ 0 corresponds to an invertible matrix (since it is upper triangular). We need to know that the function defined by this inverse matrix is in I(P), so we examine an algorithm to obtain the inverse (uniqueness would then imply the result). Suppose that f(x,x) ≠ 0 for all x in X. We calculate the values of g(x,y) of a function g in I(P) by induction on the cardinality of the interval [x,y], as follows: If |[x,y]| = 0 then set g(x,y) = 0. If |[x,y]| = 1 then x = y and we set g(x,x) = f(x,x)-1. If |[x,y]| > 1, we set =− −1 ∑ g x , y f x , x f x , z g z , y . xz≤y The function g is well-defined, because the values of g on the right-hand side of the last equation have the form g(z,y), where x < z ≤ y; so the interval [z,y] is properly contained in [x,y] and the values are defined by induction. Clearly g in I(P). Now we have, f⋅g(x,x) = 1 by definition and f⋅g(x,y) = 0 if x ≠ y since, − − g x , y=− f x , x 1 f⋅g− f x , x g x , y=− f x , x 1 f⋅g g x , y . Thus, g is the inverse of f. ▐
There are three especially important elements of I(P). The first is the characteristic function of equality: = 1 if x= y , e x , y { } 0 otherwise; this is the identity element of I(P) corresponding to the identity matrix. Next is the characteristic function of the partial order (also known as the zeta function): = 1 if x≤ y , i x , y { } 0 otherwise; and finally the inverse of the function i is the Möbius function μ which is given by: ∑ = 1 if x= y , x , z { } x≤z≤y 0 otherwise;
Proposition: The Möbius function is integer-valued. Proof: The inverse function is calculated as in the last theorem with f = i. Now i(x,x) = 1 for all x, so the factor i(x,x)-1 = 1. μ(x,y) is a linear combination of μ(z,y) with integer coefficients (in fact, all -1), where x < z ≤ y; by induction, μ(x,y) is an integer. (The induction starts with μ(x,x).)
Example: The values of the Möbius function μ(1,z) on the divisor lattice D(12) can be calculated recursively or by taking the inverse of the matrix f1 to get:
12 (0)
6 (1) 4 (0)
2 (-1) 3 (-1)
1 (1)
It follows immediately from the definition of the Möbius function that:
Proposition ( Möbius inversion): Let f and g be elements of I(P). TFAE: a f x , y= ∑ g x , z ; ≤ ≤ x z y b g x , y= ∑ f x , z z , y. x≤z≤y
Proof: ∑ f x , z z , y= ∑ ∑ g x ,t z , y x≤z≤y x≤z≤ y x≤t ≤z = ∑ ∑ g x ,tit , z z , y x≤z≤y x≤t =∑ g x ,t ∑ it , z z , y x≤t t ≤z≤y =∑ g x ,tt , y x≤t =g x , y.
Proposition: Let P be a totally ordered set. Then the Möbius function of P is 1 if x=y , ={− } x , y 1 if y covers x, 0 otherwise. Proof: From the algorithm for calculating the inverse function, we have that μ(x,x) = 1 and =− ∑ x , y z , y otherwise. If y covers x, then the only z with x < z ≤ y is z = y, so μ(x,y) = xz≤y - μ(y,y) = -1. Otherwise, we proceed by induction on the length of the interval [x,y], with base case of having a unique z with x < z < y (y covers z). We have μ(x,y) = -(μ(z,y) + μ(y,y)) = -( -1 + 1) = 0. Example: Let P = (P, ≤) be the usual poset of non-negative integers (a total order). Let x = 0 and consider any function f(n) = f(0,n) with the property that there is another function g(i) = g(0,i) such that = ∑ f n g i . 0≤i≤n Then by Möbius inversion we have that g(n) = f(n) - f(n-1). Thus the Möbius function of a totally ordered set can be viewed as the analog of the classical difference operator ∇f(n) = f(n) - f(n-1) and the incidence algebra serves as a calculus of finite differences.
Proposition: Let P1, ..., Pk be posets, and let P = P1 × ... × Pk. Then the Möbius function of P is defined by k ⋯ ⋯ =∏ x1, , x k , y1, , y k xi , yi . i=1 Pf. Since the Möbius function is unique, we only have to show that the RHS has the property that ∑ = 1 if x= y , x , y { } x≤z≤y 0 otherwise;
If z = (z1, ..., zk) then x ≤ z ≤ y if and only if xi ≤ zi ≤yi for i = 1, ..., k; so the sum on the left is over the product of the intervals [xi,yi]. Then the sum factorizes as indicated in the statement. ▐
Theorem: (a) The Möbius function of the Boolean lattice P(X) is given by ∣ ∣−∣ ∣ = −1 Z Y if Y ⊆Z Y , Z { } 0 otherwise. (b) The Möbius function of the lattice D(n) of divisors of n is given by = −1d if z/y is the product of d distinct primes, y , z { } 0 otherwise.
Möbius inversion has many applications in combinatorics and number theory. The following is only a partial listing. From a) above we obtain the Principle of Inclusion and Exclusion. From b) above we obtain the classical number theoretic version of Möbius inversion which leads to the Euler phi-function. We can obtain: 1) a formula for counting connected simple graphs on a fixed number of vertices. 2) the chromatic polynomial for a graph. 3) a formula for counting the number of ways to place non-attacking rooks on a chess board (called the rook polynomial). 4) a formula for the number of nxm matrices of a given rank r over a finite field of order q. 5) the relationship between the Stirling numbers of the first and second kind.