q xk y n k. , provided k < n. (This does not hold for k n.) Give a combinatorial proof of this recurrence by means of a bijective transformation.

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1 Math 880 Alternative Challenge Problems Fall 2016 A1. Given, n 1, show that: m1 m 2 m = ( ) n , where the sum ranges over all positive integer solutions (m 1,..., m ) of m m = n. Give both an algebraic proof by generating functions, and a combinatorial proof by bijective transformations. A2. Given, n 1, count the number of sets S = {s 1 < s 2 <... < s } [n] such that s 1, s 3,... are odd numbers, and s 2, s 4,... are even numbers. A3. Consider a non-commutative ring R containing the integers Z, and generated by elements x, y with the relation yx = qxy, where q is a given integer. Show that: (x + y) n = n [ n =0 ] q x y n. A4. Let B (n) = #{w S n inv(w) = }, the permutations of n having inversions. From the nown formula 0 B(n) x = [n]! q, deduce the recurrence: B (n) = B (n 1) + B (n) 1, provided < n. (This does not hold for n.) Give a combinatorial proof of this recurrence by means of a bijective transformation. A5. Let F (x) C[x] with F (0) = 1. Show that we can factor F (x) = A(x)B(x) for unique series A(x), B(x) C[x] with A(0) = B(0) = 1 and A( x) = A(x), B( x) = 1 B(x). Give simple formulas for A(x), B(x) in terms of F (x). A6. Let V, W F n q be subspaces of dimension dim(v ) =, dim(w ) = l, which are transverse, meaning V W = {0}. For fixed, l, n, find a formula for the number of such pairs (V, W ). Hint: The answer is a power of q times a q-multinomial coefficient. A7. For fixed n, let S = {S [2n] #S = }, and define the weight ΣS = i S i, and the sign sgn(s) = ( 1)ΣS. Show that: { #S + #S = 0 ) if is odd, if is even. ( 1) /2( n /2 A8. Recall the Stirling partition numbers { } n from Notes 9/21 and the Stirling cycle numbers [ n ] from Notes 10/5. (Both are on the Twelvefold Way chart in Notes 10/7.) Use the Involution Principle to show that, for n > m 1, we have: n [ ] { } n ( 1) n = 0, m and similarly =m n { ( 1) m n } [ ] = 0. m =m (These are proved algebraically in Notes 10/5.)

2 A9. Use the Involution Principle to prove the multiplication formula for determinants of n n matrices: det(ab) = det(a) det(b). When expanded into monomials, the left side has n!n n terms, the right side (n!) 2 terms, so probably the left side should correspond to a large signed set, and the right side to its fixed points. A10. Use the Lindstrom-Gessel-Viennot Theorem to prove the version of the Jacobi-Trudi Formula with elementary symmetric functions. That is, for a partition λ = (λ 1,..., λ n ) with λ 1 n, we have the transposed (or conjugate) partition λ = (λ 1,..., λ n), and the formula: s λ = det(e λ i +j i) n i,j=1. Hint: Modify the lattice graph so that no horizontal step can be repeated at the same level. A11. We now that n i=1 (1 + tx i) = n =0 t e i (x 1,..., x n ), where e i is an elementary symmetric polynomial. Now write the product n i=1 (1 + x i + x 2 i ) in terms of e i (x 1,..., x n ). A12. Recall the Gaussian binomial coefficient [ n ]q = N =0 a q for N = (n ), where a i is the number of -Grassmannian permutations of n having i inversions. Show that the sequence a 0, a 1,..., a N is symmetric (a = a N ) and unimodal (a 0 a 1 a M a N 1 a N for some M). Note: This problem is probably too hard. See American Math. Monthly: Proctor (Vol 89 #10, p. 721) and Zeilberger (Vol 96 #7 p. 590). Here is a similar, and also quite difficult problem: Show that the coefficients of (1+x) n are unimodal, meaning ( ) ( n n ) +1 for < 1 2n. Explicitly find an injection from -element subsets to (+1)-element subsets. (It may help to wor with n-element bit strings instead.) A13. Kerber, Finite Group Actions, Lemma 2.3.2, p. 70 uses Polya Theory to discuss the enumeration of unlabeled graphs with n vertices, whose counting sequence begins: a 0 = 1, a 1 = 1, a 2 = 2, a 3 = 4, a 4 = 11, a 5 = 33 ( We also have the connected graphs with n vertices, in which each pair of vertices is connected by a path along edges, whose counting sequence begins: c 0 = 0, c 1 = 1, c 2 = 1, c 3 = 2, c 4 = 6, c 5 = 21 ( Either referring to Kerber, or on your own, compute a n and c n as explicitly as possible. (Probably the best you can do is an algorithm for a n and a recursive formula for c n assuming nown a n.) A14. In HW 10 #1, we discussed n-colorings of the = 6 faces of the cube, the mappings F = {f : [6] [n]}, and their rotation symmetry classes F. Now consider the problem of restricted colorings in which n = (the number of colors equals the number of faces) and we allow only those colorings F opp in which opposite sides have colors adding to n+1, as with ordinary dice: f(n+1 i) = n+1 f(i). Modify our treatment of Polya s Theorem to compute the restricted pattern inventory F opp (x 1,..., x n ) in general. Wor out the example of the cube, and determine how many physically different restricted dice are possible.

3 A15. For a finite group G, a permutation representation is a homomorphism φ : G S n, that is φ Hom(G, S n ). Let j d (G) be the number of subgroups H G with index [G : H] = #G #H = d. Show that: #Hom(G, S n ) xn = exp j d (G) xd. n! d n 0 d 1 Hint: First do the cyclic group G = C = g with g = 1. Thus, a homomorphism φ : G S n is determined by any permutation φ(g) = w with w = id. A16. Define combinatorial Bruhat order on the permutations S n as follows. For -subsets A = {a 1 < < a } and B = {b 1 < < b } of [n], define A B to mean a i b i for all i; and for permutations u = u 1... u n and w = w 1... w n, define u w to mean {u 1, u 2,..., u } {w 1, w 2,..., w } for all. Show that the Mobius function of the Bruhat order is: µ(u, w) = ( 1) inv(u)+inv(w). Hints: The statement is equivalent to saying that for any u < w, we have x [u,w] ( 1)inv(x) = 0; that is, in any non-trivial interval, the number of elements with an even number of inversions is equal to the number of elements with an odd number of inversions. Find a sign-reversing involution adapted to each interval. D.N. Verma s original proof (of the generalization to all Coxeter groups) uses Deodhar s Zigzag Lemma: Let u, w S n and s i = (i, i+1) a transposition of adjacent letters in [n]. If u us i, w ws i, and u ws i, then u w and us i ws i. A17. Continuing the previous problem, show that the covering relations u w are precisely when w = u r ij for r ij = (ij) any 2-cycle, where inv(w) = inv(u)+1, or equivalently, where u i < u j and u i+1, u i+2,..., u j 1 / [u i, u i +1,..., u j ]. A18. First: For a finite poset P, suppose φ : P P is an order-preserving bijection. Then φ is an automorphism, meaning the inverse φ 1 is also orderpreserving. Second: Give a simple infinite poset P for which this is false. A19. Let P be a finite poset with no isolated points (i.e. no a P which are incomparable with all other b P). Suppose P has a longest chain of length l, and every covering a b is contained in a chain of length l. Show that every maximal chain of P has length l. A20. In a poset P, suppose every chain (set of elements which are all comparable) and every anti-chain (set of incomparable elements) are finite. Then is P necessarily finite? A21. Let P = (N) be the power set consisting of all S N, ordered by inclusion. Show that P has both countable and uncountable maximal chains.

4 A22. Recall that Young s lattice Y is the set of all partitions λ = (λ 1 λ 2 0) ordered by inclusion of their Young diagrams (or by elementwise comparison of the parts λ i ). Consider the sublattice O consisting of partitions with only odd parts (all λ i odd). By the Fundamental Theorem of Distributive Lattices ([St] Thm & 3.4.3), O must be the poset of finite order ideals J fin (P) for some poset P. Explicitly describe P. A23. Poset constructions versus algebra constructions. Each part should be enough to count for 1 problem (but do more if it seems too easy). Define the incidence algebra of a (finite) poset as the vector space I(P) = a b C [a, b], where the interval [a, b] P is considered as a basis element of I(P), with multiplication: [a, b] [c, d] = [a, d] if b = c, and zero otherwise. Alternatively, I(P) = {f : Int(P) C}, where the basis element [a, b] corresponds to the function δ ab with δ ab ([x, y]) = 1 if [a, b] = [x, y] and zero otherwise, and the multiplication is the convolution: (f g)[a, b] = c [a,b] f(a, c)g(c, b). a. Show that the ideals of I(P) are all intersections of the sets: I [a,b] = [x,y] [a,b] C [x, y], where [a, b] is any interval of P, and the sum runs over all intervals [x, y] which are not contained in [a, b]. Hint: First show that all ideals are of the form C [x, y] where S is a set of intervals. [x,y] S b. For any posets P, Q, [St] Ch 3.2 gives several operations to combine them into a poset R. Express each incidence algebra I(R) by performing algebra and module constructions involving I(P), I(Q), C P, C Q. Give explicit isomorphisms. (i) R = P + Q (ii) R = P Q (iii) R = P Q (iv) R = P Q Note: The first three are easy, the last might be hard. c. Do part (b) for R = Q P, the poset of order-preserving mappings f : P Q, with f g defined by f(a) g(a) for all a P. Hint: It seems the algebra homomorphisms Hom(I(P ), I(Q)) are relevant. A24. HW 13 #1 discusses a q-analog for Items 1 3 in the Twelvefold Way, counting linear mappings L : F q F n q. Do the same for all (or most) of Items What about Pascal-lie recurrences, and generating functions? Hint: Try to find the best q-analog of maing the elements of [] or of [n] indistinguishable from each other. (What was the formal definition for this in the original Twelvefold Way?) Can you mae natural definitions which specialize to the original numbers when we tae q = 1 or q 1? A25. Show that if a geometric lattice L has µ(ˆ0, ˆ1) = ±1, then L is isomorphic to a Boolean lattice B n. Hints: Show that every a ˆ1 is a meet of co-atoms. Improve [St] Prop by showing that ( 1) ρ(b) ρ(a) µ(a, b) > 0 for all a < b. A26. Let P be a finite poset with an automorphism of prime order: an orderpreserving σ : P P with σ p = id for prime p. Also assume σ(a) a for all a P. Then the completion ˆP = P {ˆ0, ˆ1} has Mobius function: Apply this to the case ˆP = Π p. µ ˆP(ˆ0, ˆ1) 1 (mod p).

5 A27. Say as much as you can about the q-exponential function: x n exp q (x) = [n]! n 0 q = 1 + x + x2 1(1+q) + x 3 1(1+q)(1+q+q 2 ) +. Note that it is the zeta-function of the binomial poset of subspaces of F q. Try to find q-analogs of various formulas for exp(x). Also, if we substitute x x 1 q, it is closely related to the bivariate generating function of partitions: λ q λ x l(λ) = 1 i 1 1 xq, where λ is the number of i boxes in the Young diagram λ, and l(λ) is the number of rows. Try to deduce partition idenities (or bijections) involving formulas for exp q (x). A28. The Principle of Inclusion-Exclusion can be formulated as follows (Wilf Ch 2). Given subsets B 1,..., B n A, we suppose that for any I [n], the value α(i) = # ( i I B i) depends only on = #I, and we may write α(i) = α. Also, define β = β(i) = {a A a B i i I}. Then we have: α = ( n i) βi, β = ( 1) i( n i) αi. i i In terms of the generating functions A(x) = a 0 + a 1 x + + a n x n and B(x) = b 0 + b 1 x + + b n x n, the above equations are equivalent to: A(x) = B(x+1) and B(x) = A(x 1). problem: Can you explain or interpret the equation A(x) = B(x+1) in terms of a reduced incidence algebra for the Boolean poset P = [n]? A29. [St] Ex 3.45 Let L (n) denote the number of -element order ideals of the boolean algebra B n. Show that for fixed, L (n) is a polynomial function of n of degree 1 and leading coefficient 1/( 1)!. Moreover, the differences i L (0) are all nonnegative integers. (Stanley gives a list, including L 3 (n) = ( n 2), L4 (n) = ( n 2) + ( n 3).)

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