MATH 580, exam II SOLUTIONS

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1 MATH 580, exam II SOLUTIONS Solve the problems in the spaces provided on the question sheets. If you run out of room for an answer, continue on the back of the page or use the Extra space empty page at the end. Answers without proofs or incomplete proofs will not receive full credit. NO notes, books, or any electronic devices are allowed. Name: Penn ID number: Question Points Score Total: 40

2 1. (10 points) For any set of positive integers S, let f S (n) be the number of integer partitions λ n with at most p nonzero parts, such that (*) λ i > λ i+1 for i S (and otherwise λ i λ ı+1 ). (a) Let S be the set of all positive even integers. Find a product formula for the generating function F S (q) = n f S (n)q n. Solution: Let λ = (λ 1, λ 2,...) and let λ = (1 m1 2 m2... p mp ) be the conjugate (transpose Young diagram) of λ, i.e. the Young diagram λ has m i columns of height i. We have that m i = λ i λ i+1 (assume λ p+1 ) = 0). The condition (*) is then equivalent to m i > 0 for all i S. We thus have F S (q) = q λ = q i imi λ (m 1,m 2,...,m p) m i>0,i S = ( ( ) q mii) q mii i S m i S m i=0 = q p i S i 1 p 1 q i = 1 q p/2 ( p/2 +1) 1 q i (b) Let S be as in (a). Find a product formula for the generating function of g S (n) = R S f R(n), i.e. G S (q) = n g S (n)q n. Solution: By the same method as above, we have = g S (n)q n = F R (q) = q p i R i 1 1 q i n R S R S p 1 1 q i p/2 p (1 + q a 1 ) = 1 q i (1 + q 2i ) a S Page 2

3 2. (10 points) (a) Let P be a finite ranked (graded) poset with rank function ρ, for which µ(s, t) = ( 1) l(s,t), where l(s, t) is the length of the interval [s, t]. For each interval [s, t] set g(s, t) = #{u [s, t]}. For any u, t P find the value of g(u, s)( 1) ρ(s) =? Solution: Let f(t) = 1 for all t, then g(t) := g(u, t) = 1 = f(t) s t for all t. By Möbius inversion for the interval [u, t] we have 1 = f(t) = g(s)µ [u,t] (s, t). Since l(s, t) = ρ(t) ρ(s) we have g(u, s)( 1) ρ(s) = g(u, s)µ(s, t)( 1) l (u, t) = f(t)( 1) l(u,t) = ( 1) l(u,t) (b) Let P be a finite lattice. Find the value of µ(s, t) =?. s t Solution: This is HW problem 88. If we want to use the linear algebra approach let M be the p p matrix corresponding to the Möbius function µ, let V = [1, 1,..., 1] the 1 p vector of all 1s. then the sum is equal to V MV T. Let now Z be the matrix corresponding to the ζ function. Since the poset has a ˆ0, the first row of Z must be V, and because of the ˆ1 the last column of Z is V T, thus V MV T = (ZMZ) 1,p = (ZZ 1 Z) 1,p = Z 1,p = ζ(ˆ0, ˆ1) = 1. Page 3

4 3. (10 points) Fix n and let Q be the subposet of Z 2 consisting of the elements {(i, i), (i+1, i), (i, i+1), i = 0... n 1} {(n, n)} (so its Hasse diagram looks like a chain of diamonds). (a) Find a poset P, such that Q = J(P ). Solution: The recipe for constructing P out of Q is to take all elements of Q which cover exactly one element, i.e. t, s.t. there is a unique s t and the subposet of such elements is P. (this is from the proof that every distributive lattice is J(P ) for some P ) These elements are exactly the ones which correspond to the principal order ideals Λ t = {s : s t} of P. The subposet P we get this way is (1 + 1) (1 + 1) ordinal sum of n 2-element an- }{{} tichains. It can also be described as the poset on (s 1, t 1, s 2, t 2,..., s n, t n ) with s i, t i < s j, t j for all j > i. (b) For the poset P defined in (a) (found or not), determine the number of linear extensions of P e(p ) =? Solution: e(p ) = 2 n. If you found P, then it is clear that every linear extension would have {s i, t i } {2i 1, 2i} and for each i there are 2 possible assignments of values. Otherwise, we know that e(p ) is the number of maximal chains in J(P ). Every maximal chain has to contain (i, i) for all i and then contain either i + 1, i) or (i, i + 1), we have 2 choices for each i independent of the other is. n Page 4

5 (c) Determine the Möbius function of Q, µ(s, t), for any two elements s t Q. Solution: This follows either from facts about Möbius functions of lattices, or can be computed via the recurrence directly by induction. First we do the covering relations for every i: µ((i, i), (i + 1, i)) = µ((i, i), (i, i + 1)) = 1, µ((i, i), (i + 1, i + 1)) = 1 and µ((i + 1, i), (i + 1, i + 1)) = µ((i, i + 1), (i + 1, i + 1)) = 1. Then by induction µ(s, t) = 0 for all other pairs of elements. Page 5

6 4. (10 points) Recall that the zeta polynomial Z(P, n) counts the number of n 1 element multichains in P. (a) Let P and Q be 2 posets and f(n) = Z(P, n), g(n) = Z(Q, n). Express (with proof!) the polynomials Z(P + Q, n), Z(P Q, n) and Z(P Q, n) in terms of f(i), g(j) for any suitable values of i, j. Solution: We have that Z(P + Q, n) = Z(P, n) + Z(Q, n), since every chain of P + Q is either a chain in P or in Q. Z(P Q, n) = n f(i)g(n + 1 i), where we set f(0) = g(0) = 1. C is a chain in P Q iff C = C 1 C 2, where C 1 is a chain in P and C 2 is a chain in Q. Let i = #C 1 + 1, then #C 2 = n 1 (i 1) = n i and the pairs (C 1, C 2 ) are counted by f(i + 1)g(n + 1 i). Z(P Q, n) = f(n)g(n) Every chain C in P Q is of the form (t 1, s 1 ) (t 2, s 2 ), so t 1 t 2 is a chain in P and s 1, s 2,... a chain in Q, and vice versa. (b) Given a poset P, let Int(P ) = {[s, t], s P t} be the poset of intervals of P ordered by inclusion, i.e. [s, t] [a, b] is [s, t] [a, b]. Show that Z(Int(P ), n) = Z(P, 2n 1). Solution: We have that [s, t] [a, b] a s t b. So [s 1, t 1 ] [s 2, t 2 ] [s n 1, t n 1 ] is an n 1 element chain in Int(P ) iff s n 1 s n 2... s 1 t 1 t 2... t n 1 a 2n 2element chain in P. As this is a bijection between 2n 2 element chains of P and n 1 element chains in Int(P ), we get the equality. Page 6

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