18.212 Course Notes

Wanlin Li

Spring 2019

1 Catalan Numbers

1.1 February 6

• Stanley-style (enumerative/algebraic/geometric) vs Erdos-style combinatorics (extremal/probabilistic)

• Enumerative results tend to be of the form A = B, extremal results tend to be asymptotic

• Catalan numbers Cn: number of sequences ε1, ε2, . . . , ε2n with εi = ±1 such that any partial sum ε1 + ··· + εi ≥ 0 and ε1 + ··· + ε2n = 0

• Deﬁnition. Dyck path: path on Cartesian plane starting at (0, 0), ending at (2n, 0), and traveling (1, 1) or (1, −1) at each step always staying in the upper half plane

• Starting at index 0: 1, 1, 2, 5, 14, 42,...

• Drunk man problem: random walk on number line, cliff to one side and infinite in other direction; find probability of survival

• 1 P Cn Probability that man falls is 2 22n n≥0

• Simple solution: probability that person starting at i reaches position N before i falling off cliff at 0 is N ; in original problem i = 1 and N → ∞ so drunk man falls off cliff Pn • Catalan recurrence: Cn = i=1 Ci−1Cn−i with C0 = 1

• 2 Generating√ function for Catalan numbers satisﬁes f(x) = xf(x) + 1 so f(x) = 1− 1−4x 2x

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1.2 February 8 √ 1− 1−4x • Generating function f(x) = 2x for Catalan numbers

• Biased random walk with probability p of moving one step toward safety and 1 − p of moving toward cliﬀ: probability of falling is (1 − p)f(p(1 − p)) which is 1 1 1−p 1 if p ≤ 2 , p if p > 2 n • Proof of generalized binomial expansion (1 + x) by Taylor series expansion

2n • 1 Generating function gives Cn = n+1 n

• Number of total lattice paths from A to B with n up steps and n down steps is 2n n

• Path crosses x-axis at ﬁrst point C, reﬂect path from C to B = (2n, 0) about y = −1 to get path from (0, 0) to (2n, −2) with n−1 up steps and n+1 down steps

2n • Procedure is invertible so number of non-Dyck paths from A to B is n−1 2n 2n 2n • 1 Thus number of Dyck paths is n − n−1 = n+1 n 2n+1 • Necklace and cyclic shift proof: n is number of sequences with n positive 1s and n + 1 −1s

• 2n + 1 cyclic shifts of the sequence

2n+1 • 1 Claim 1: all 2n + 1 shifts are unique, so the number of necklaces is 2n+1 n

• Claim 2: for each necklace there is a unique sequence among all 2n + 1 arrangements such that ε1 + ··· + εi ≥ 0 for i ≤ 2n and ε2n+1 = −1

2n+1 2n • 1 1 Therefore Cn is the number of such necklaces 2n+1 n = n+1 n

• Other interpretations of Cn: number of triangulations of (n + 2)-gon, number of valid parenthesizations of n + 1 letters, number of plane binary trees with n non-leaf vertices and n + 1 leaves

• Bijection between triangulation and binary tree: start with side of (n + 2)-gon, exactly one triangle is adjacent to that side; then children of triangle are the triangles/sides adjacent to it

• Letters correspond to n + 1 leaves of binary tree, binary tree gives order of multiplication

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1.3 February 11

• Definition. Queue: first in first out

• Deﬁnition. Stack: last in ﬁrst out

• Claim: number of queue-sortable permutations of 1, 2, . . . , n is Cn

• Queue-sortable: each number is read and either placed in queue or directly sent to output

• Example. 2, 4, 1, 3 is queue-sortable: 2 goes into queue, 4 goes into queue, 1 goes directly to output, 2 is removed from queue and appended, 3 goes directly to output, 4 comes out of queue and is appended

• Example. 3, 2, 1 is not queue-sortable

• Claim: number of stack-sortable permutations of size n is also Cn

• Example. 4, 1, 3, 2 is stack-sortable: 4 goes into stack, 1 goes directly to output, 3 goes into stack, 2 goes into stack; 2, 3, 4 taken out of stack and appended to output

• Example. 2, 3, 1 is not stack-sortable

• w = (w1, w2, . . . , wn) is a permutation of size n and π = (π1, π2, . . . , πk) is a permutation of size k ≤ n

• Permutation w contains pattern π if there exists some (not necessarily consecutive) subsequence of w whose entries appear in the same relative order as entries of π

• Call w π-avoiding if w does not contain pattern π

• Exercise 1. The queue-sortable permutations are exactly the (321)-avoiding permutations.

• Exercise 2. The stack-sortable permutations are exactly the (231)-avoiding permutations.

• Exercise 3. For all patterns π of size 3, the number of π-avoiding permutations of size n is Cn.

3 18.212 Wanlin Li 2 Young Tableaux

2.1 February 11

• Deﬁnition. Partition: λ = (λ1 ≥ λ2 ≥ · · · ≥ λe > 0) such that n = λ1 + λ2 + ··· + λe

• Young diagram/Ferrers shape: row i contains λi boxes where λi are a non- increasing partition

• Deﬁnition. Standard Young tableau: way of ﬁlling boxes of Young diagram with numbers 1, 2, . . . , n without repetition such that the numbers increase within the rows and columns of the Young diagram

λ • f is the number of standard Young tableaux of shape λ

λ Lemma. Given λ = (n, n), f = Cn.

• Proof: construct a sequence ε1, . . . , ε2n of ±1 such that εi = 1 if i is in the ﬁrst row of the tableaux and −1 if i is in the second row

• Hook-Length Formula due to Frame-Robinson-Thrall 1954

2.2 February 13

λ • f is number of standard Young tableaux of shape λ

Frobenuius-Young Identity. X (f λ)2 = n! λ:|λ|=n

• RHS is number of permutations of Sn, LHS is number of pairs (P,Q) where P and Q are standard Young tableaux of the same shape

• (1961) Schensted correspondence LHS ↔ RHS

• Schensted insertion algorithm: T some intermediate standard Young tableau ﬁlled with numbers in S ⊂ Z≥1, x ∈ Z≥1\S

• T ← x is the SYT with one extra box given by following procedure:

1. x1 := x, i := 1

2. If xi is greater than all the entries in the ith row of T, add a new box at the end of the ith row ﬁlled with xi and stop

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3. Otherwise ﬁnd the smallest entry y in the ith row such that y > xi, replace y by xi and set xi+1 = y, go back to step 1

• P is the insertion tableau, Q is the recording tableau

• w1, w2, . . . , wn ∈ Sn 7→ (P,Q) where P = ((∅ ← w1) ← w2) ← · · · ← wn and Q : i ∈ [n] goes to the box that was added at the ith step in construction of P

• Example. w = 3, 5, 2, 4, 7, 1, 6; P goes in sequence

∅ 3 3 5 2 5 2 4 2 4 7 1 4 7 1 4 6 3 3 5 3 5 2 5 2 5 7 3 3

and Q is 1 2 5 3 4 7 6

Theorem. This map w 7→ (P,Q) gives a bijection between Sn and pairs (P,Q).

• Proof: process is reversible by example

• Deﬁnition. Schensted shape: given w 7→ (P,Q), the shape λ of P and Q is called the Schensted shape of w

Schensted’s Theorem. The length λ1 of the ﬁrst row is size of the longest increasing subsequence in w. The length of the ﬁrst column of λ is the size of the longest decreasing subsequence in w.

• Corollary: the number of (123)-avoiding permutations in Sn is the nth Catalan number.

• Proof: w is (123)-avoiding iﬀ the length of the longest increasing subsequence is at most 2. w matches with a pair (P,Q) of tableaux with the same shape; P and Q can be glued together (Q is ﬂipped by 180◦ and its entry i is replaced by 2n + 1 − i) to form a SYT of shape (n, n), of which there are exactly Cn

Green’s Theorem. λ1 + ··· + λi is the length of the longest subsequence that can be split into i increasing subsequences.

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2.3 February 15

• Schensted correspondence w ∈ Sn 7→ (P,Q)

• Deﬁnition. jth basic subsequence: for j ∈ 1, . . . , λ1,Bj is the set of all entries of permutation w that were originally inserted in the jth position of the ﬁrst row

Lemma. Each Bj is decreasing.

Lemma. For every x ∈ Bj with j ≥ 2, there exists y ∈ Bj−1 such that y < x and y is located to the left of x in w.

• Proof: at the moment of insertion of x, y is the entry located to the left of x.

• Proof of Schensted’s Theorem: λ1 is the number of basic sequences, and if x1 < x2 ··· < xr is an increasing subsequence in w, r < λ1 because each xi must come from a diﬀerent Bj. By the second lemma there does in fact exist an increasing subsequence of size λ1.

• Hook-length formula for number of SYT with shape λ

• Hook-walk proof by induction on n (Greene, Nijenhuis, Wilf 1979)

• H(λ) is product of hook-lengths of λ

λ P λ−v • f = v f where v is a bottom right corner of λ λ (4,4,4,1) (5,4,3,1) (5,4,4) • Example. λ = (5, 4, 4, 1) so f = f + f + f

• Easy to see hook-length holds for n = 1, want to show

n! X (n − 1)! = H(λ) H(λ − v) v

X H(λ) 1 = nH(λ − v) v

• Idea to construct random process with probabilities represented by RHS

• Deﬁnition. Hook walk: randomly pick any box u of λ, jump from u to any other box u0 in the hook of u and repeat procedure until corner v is reached

• Deﬁne P (v) to be the probability that a hook walk ends at corner v, claim P (v) = H(λ) nH(λ−v)

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• Deﬁne P (u, v) to be the probability that a hook walk starting at u ends at v;

X 1 1 P (u, v) = ··· h(u) − 1 h(u0) − 1 u→u0→···→v

• For any rectangle abcd with a in the upper left and d in the lower right corners, h(a) + h(d) = h(b) + h(c)

• When d is a corner, (h(a) − 1) = (h(b) − 1) + (h(c) − 1)

1 • Lattice with A in upper left and B in lower right, row j has weight and column yj i has weight 1 ; every vertex has weight 1 xi xi+yj

• Consider sum of path weights over lattice paths from A to B;

1 • Weight of path is x1...xky1...yl

2.4 February 19

• Consider variables x1, . . . , xk, y1, . . . , yl and the (k +1)×(l +1) grid graph G with vertices weighted 1 + 1 for vertex at position (i, j) with last row and last column xi yj weighted 1

• For lattice path P from A to B, weight of path P is Πvweight(v) for v ∈ P

Lemma. P w(P ) = 1 where P covers all paths from A (position P x1···xky1···yl (1, 1)) to B (position (k + 1, l + 1))

• Proof by induction on k + l

• Hook walks allow jumping around, not the same as lattice paths

• Hook walk in grid graph G is path P starting at any vertex that can jump over rows and columns but still ending at B

Lemma. P w(P ) = 1 + 1 1 + 1 ··· 1 + 1 1 + 1 ··· 1 + 1 over P x1 x2 xk y1 yl hook walks P in G

1 1 • 1 corresponds to skipping row/column, or corresponds to choosing that row xi yj or column

• P (u, v) is probability that hook walk starting at u ends at corner v of λ

7 18.212 Wanlin Li • P 1 1 P (u, v) = h(u)−1 h(u0)−1 ··· P :u→v

• Deﬁne co-hook of v as all boxes in same row or column as v • P 1 By Lemma, ﬁx corner box v and u P (u, v) = Πt 1 + h(t)−1 where t is box in co-hook of v

h(t) H(λ) • Above product is Πt h(t)−1 = H(λ−v) P • For ﬁxed u, v P (u, v) = 1 • P P P (u, v) = n u∈λ v

• Switching order of summation, ! X X X H(λ) P (u, v) = = n H(λ) − v v u∈λ v

as desired

• Deﬁnition. Poset: set P with binary relation “≤” satisfying

1. a ≤ a ∀a ∈ P 2. a ≤ b, b ≤ a ⇒ a = b 3. a ≤ b, b ≤ c ⇒ a ≤ c

• Strict inequality given by a < b : a ≤ b, a 6= b

• Deﬁnition. Covering: a is covered by b, denoted a <· b : a < b and there does not exist c such that a < c < b

• Deﬁnition. Hasse diagram: graph formed by covering relations

• Deﬁnition. Linear extension of poset: bijective map f : P → [n] such that a ≤ b ⇒ f(a) ≤ f(b)

• Young diagram λ maps to linear extension of poset Pλ but not all posets have variant of hook-length formula

• Exercise 1. “Baby hook length formula”: for a rooted tree T or Hasse diagram with n nodes, the number of linear extensions of the poset P is n! where Πa∈T h(a) h(a) is the number of nodes in the subtree rooted at a

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2.5 February 20

• Shifted Young diagram: Young tableaux starting along the diagonal; partition into distinct parts λ1 > λ2 > ···

• Same problem of ﬁlling boxes of shifted Young diagram with indices strictly increasing within rows and columns

Theorem. The number of standard Young tableaux of shifted shape λ of n boxes n! is πh(a) where the “hook length” h(a) resembles the normal hook, except that if the leg of the hook reaches the descending staircase the hook also includes the row beneath the vertical leg

3 q-analogs

3.1 February 20

• Quantum objects depending on parameter q, classical objects with limit q = 1

1−qn • 2 n−1 Quantum analog of integer n given by [n]q = 1 + q + q + ··· + q = 1−q

• Quantum analog of n! is [n]q! = [1]q[2]q ··· [n]q

• When q is clear, subscript can be omitted

n [n]q! • q-binomial coeﬃcients [ ] = k q [k]q![n−k]q!

(1+q+q2+q3)(1+q+q2) • 4 2 2 Example. [ 2 ] = (1+q) = (1 + q )(1 + q + q ) n • Observations: [ k ]q is a palindromic polynomial in q with positive integer coeﬃ- cients ai (Gaussian coeﬃcients)

• Sequence of coeﬃcients is unimodal: a0 ≤ a1 ≤ · · · ≤ am/2 ≥ · · · ≥ am

n n−k n−1 n−1 • Claim [ ] = q + ; proof by expansion k q k−1 q k q

• Let λ ⊆ k × (n − k) be a Young diagram that ﬁts inside a k × (n − k) rectangle

n • Number of such Young tableaux is k

n P |λ| Theorem. [ k ] = λ q

• Palindromic property from theorem by symmetry of k ×(n−k) rectangle, degree of polynomial is k(n − k)

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• Proof by induction on n and checking Pascal’s identity

• Want method of breaking down Young diagram to account for two parts of Pas- cal’s identity

• n−1 Case 1: λ1 < n − k so λ ﬁts inside k × (n − k − 1) rectangle, gives k q

• Case 2: λ1 = n − k so ﬁrst row occupies entirety of rectangle, remainder of diagram ﬁts inside (k − 1) × (n − k) box so this gives qn−k n−1 k−1 q

3.2 February 22

• Generalization of binomial expansion to q-commutative q-binomial expansion

• Variables x, y with yx = qxy and q commuting with each of x, y n n P n k n−k • Claim (x + y) = [ k ]q x y k=0

n n [n]q! 1−q • [ ] = with [n] = k q [k]q![n−k]q! q 1−q

n P |λ| Theorem. [ k ]q = λ⊂k×(n−k) q

• Deﬁnition. Grassmannian: Grk,n is space of k-dimensional linear subspaces in n-dimensional space, with 0 ≤ k ≤ n deﬁned by k × n matrices of rank k modulo row operations (unique row spaces)

r • Fact: the finite fields have size p for some prime p, and for every such power of a prime there exists a unique finite field with pr elements up to isomorphism

• Count number of elements in Grassmannian over Fq; number of k × n matrices over Fq with rank k divided by the number of k × k invertible matrices over Fq n • Matrix A with rows v1, . . . , vk and vi linearly independent in (Fq) n n n 2 • q − 1 possibilities for v1, q − q for v2, q − q for v3, etc. so total number of k × n matrices is (qn − 1)(qn − q)(qn − q2) ··· (qn − qk−1)

k k k k−1 • Number of k × k invertible matrices is (q − 1)(q − q) ··· (q − q )

(qn−1)(qn−q)···(qn−qk−1) n Theorem. The number of Grassmannians over Fq is (qk−1)(qk−q)···(qk−qk−1) = [ k ]q for q a power of a prime

• Second way of computing number of Grassmannians: use Gaussian elimination to put A in reduced row echelon form

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• Rank of matrix is number of pivots; remove all pivot columns to get k × (n − k) matrix which is mirror image of Young diagram

P |λ| • Number of Grassmannians over Fq is λ q because each entry in λ can take on one of q values

n P |λ| • This shows [ k ]q = λ q for any q a power of a prime

• For two rational expressions f(q), g(q) with integer coeﬃcients, if f(q) = g(q) for inﬁnitely many values of q then f and g are identically equal

n P |λ| • Therefore [ k ]q = λ q for any q

• Given permutation w = (w1, . . . , wn) ∈ Sn of (1, 2, . . . , n), pair of indices (i, j) is inversion of w if i < j and wi > wj

• Let inv(w) denote the number of inversions of w

P inv(w) Theorem. [n]q! = q w∈Sn

• Straightforward proof by induction

• Start with permutation u ∈ Sn−1, n places to insert n into u; if n is inserted at the ith position, there are n − i additional inversions

P inv(w) n−1 P inv(u) • Thus q = (1 + q + ··· + q ) q = [n]q! w∈Sn u∈Sn−1

3.3 February 25

• Deﬁnition. q-multinomial coeﬃcients: given nonnegative integers n1, . . . , nr n [n]q! with sum n, [ n ,...,n ] = 1 r q [n1]q!···[nr]q!

• Deﬁnition. Multiset: collection of elements allowing repetition of entries

n n nr • Denote multiset with n1 1s, n2 2s, etc. by S = {1 1 , 2 2 , . . . , r }

• Given w = w1, . . . , wn is a permutation of multiset S, number of inversions inv(w) is number of pairs (i, j) with 1 ≤ i < j ≤ n such that wi > wj

n P inv(w) Theorem. [ n1,...,nr ]q = w q

n • P Corollaries of above: [ n1,...,nr ]q is an integer polynomial in q with degree nanb; 1≤a

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k n−k • Permutations of S = {1 , 2 } are in bijective correspondence with Young diagrams λ ⊆ k × (n − k) by counting lattice paths (2 is movement to the right, 1 is movement up)

• Claim |λ| = inv(w) by matching rows and columns

• Usual permutations w of [n] = {1, . . . , n} can be thought of as bijections from [n] → [n] (symmetric group)

• Statistics on permutations: A : Sn → {0, 1, 2,..., } with generating function P A(w) FA(x) = w x

• Deﬁnition. Equidistribution of statistics: two statistics A, B if FA = FB

• Statistics examples: inv(w), l(w) the minimal number of adjacent transpositions needed to express w, cyc(w) the number of cycles in w counting ﬁxed points

• Exercise 1. l(w) = inv(w).

Theorem.

X inv(w) 2 n−1 q = (1 + q)(1 + q + q ) + ··· (1 + q + ··· + q ) = [n]q!

w∈Sn

Theorem. X xcyc(w) = x(1 + x)(2 + x) ··· (n − 1 + x)

w∈Sn

• Proof by induction: take permutation u of Sn−1 in cycle notation, insert new entry n; n − 1 ways to insert into existing cycle and 1 way to make new cycle so multiplication by (n − 1 + x)

3.4 February 27

• Statistics: inv(w), cyc(w)

P inv(w) P cyc(w) • q = [n]q! and x = x(x + 1) ··· (x + n − 1) for w ∈ Sn

• Deﬁnition. Descent: given w = (w1, . . . , wn) ∈ Sn, descent of w is i ∈ {1, . . . , n − 1} such that wi > wi+1; des(w) is number of descents of w

des(w) • Eulerian polynomial P x w∈Sn

• Many statistics are equidistributed with these three statistics

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• Deﬁnition. Eulerian statistic: equidistributed to descent statistic

• Deﬁnition. Mahonian statistic: equidistributed to inversion statistic P • Deﬁnition. Major index: given permutation w ∈ Sn, maj(w) = i over descents of w

• Example. maj(25731684) = 3 + 4 + 7 = 14

Theorem. The number of inversions and the major index are equidistributed.

• Deﬁnition. Record: given permutation w, a record is an entry greater than all preceeding entries (left-to-right maximum); rec(w) is number of records

Theorem. rec(w) and cyc(w) are equidistributed.

• Proof: construct bijection f : Sn → Sn, w 7→ w˜ such that cyc(w) = rec(w ˜)

• Write w in cycle notation w = (a1 ... )(a2 ... ) ... (ak ... ) with ai the maximum elements of the cycles and a1 < a2 < ··· < ak

• Remove all parentheses to get from w to w˜

• Exercise 1. Find some similar bijection from the number of inversions to the major index.

• Deﬁnition. Exceedance: index i such that wi > i; exc(w) is the number of exceedances in w

• Strict exceedance has wi > i, weak exceedance has wi ≥ i

Theorem. exc(w) and des(w) are equidistributed.

• Deﬁnition. Anti-exceedance: index i such that wi < i

• Proof: exceedance and anti-exceedance are equidistributed because number of exceedances of w is number of anti-exceedances of w−1; same bijection between cyc(w) and rec(w ˜) maps anti-exceedance(w) to des(w ˜)

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3.5 March 4

• Deﬁnition. Eulerian numbers: An,k is number of permutations in Sn with exactly k descents

• Bijection between Sn and increasing binary trees on n nodes

• Example. w = 4 2 8 5 1 3 9 10 6 7; Tw has root 1, left branch with elements to the left of 1, right branch with elements to the right of 1 and continue recursively extracting minimum each time

• Each node is smaller than its children

Theorem. An,k is the number of increasing binary trees on n nodes with k left edges.

• Deﬁnition. Eulerian triangle: triangle with entries An,k for 0 ≤ k < n

• Edges of triangle are all 1

Theorem. An+1,k = (n − k + 1)An,k−1 + (k + 1)An,k

• Number of places to add left edge in increasing binary tree and number of places to add right edge will match up to (k + 1, n − k + 1)

4 Stirling Numbers

4.1 February 27

n−k • Definition. Stirling numbers (first kind): s(n, k) = (−1) c(n, k) where c(n, k) is the number of permutations of size n with exactly k cycles, counting fixed points

• s(n, 0) = 0 for n > 0 and s(0, 0) = 1

Theorem. n X c(n, k)xk = x(x + 1)(x + 2) ··· (x + n − 1) k=0 or equivalently

n X s(n, k)xk = x(x − 1)(x − 2) ··· (x − n + 1) k=0

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• Deﬁnition. Stirling numbers (second kind): number S(n, k) of set partitions of [n] into k non-empty blocks

• Example. π = (125|3478|6) is a possible set partition, looks like cycle notation but cyclic order within set does not matter

Theorem. n X n S(n, k)(x)k = x k=0

where (x)k denotes x(x − 1) ··· (x − k + 1)

4.2 March 1

n−k • Stirling numbers of the ﬁrst kind s(n, k): (−1) times number of permutations in Sn with exactly k cycles

• Stirling numbers of the second kind S(n, k): number of set partitions of n elements into exactly k subsets

4−2 • Example. s(4, 2) = (−1) (4 ∗ 2 + 3) = 11

• Example. S(4, 2) = 4 + 3 = 7

Theorem. n X k s(n, k)x = (x)n k=0

where (x)n denotes the falling power x(x − 1) ··· (x − n + 1). Furthermore

n X n S(n, k)(x)k = x . k=0

n • R[x] has 2 linear bases {x }, {(x)n}

• Corollary: (s(n, k))n,k≥0 is a lower triangular matrix and (S(n, k))n,k≥0 is its in- verse switching between the bases of R[x] P n • Proof of identity for Stirling numbers of the second kind S(n, k)(x)k = x : enough to prove it is true for all x ∈ N

• Let F = {f :[n] → [x]} be the set of functions from an n-element set to an x-element set

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n • |F | = x but can also use function to construct set partition π of [n] such that i, j are in the same block of π iﬀ f(i) = f(j)

• Fix a set partition π with k blocks B1,...,Bk; then the number of functions using set partition π is x(x − 1) ··· (x − k + 1) = (x)k

P n • Counting |F | in both ways gives S(n, k)(x)k = x

• Another interpretation of S(n, k): place non-attacking rooks on triangular chess board (Young diagram) of shape λ = (n − 1, n − 2,..., 1)

• Associate rook placement with set partition: to the right of the row with length l, write number l + 1 and write 1 in bottom box of diagram

• For each rook, hook of rook hits numbers i, j; then i, j will be in the same block of the set partition

Theorem. S(n, k) is the number of non-attacking rook placements on the chess- board of shape λ described above with exactly n − k rooks.

P • Deﬁnition. Bell number: B(n) = k S(n, k) is the total number of set partitions of an n element set or the total number of rook placements on λ

• Example. By counting rook placements, B(4) = 1 + 6 + 7 + 1 = 15

• Arc diagram of a set partition: e.g. π = (146|23|5), draw arc between any adjacent members of same block of partition (increasing order) on number line

• Non-crossing and non-nesting set partitions

• Deﬁnition. Non-crossing set partition: arcs in arc diagram of partition are pairwise non-crossing

• Deﬁnition. Non-nesting set partition: arcs in arc diagram are pairwise non- nesting, i.e. if i < j < k < l, i − l and j − k cannot be simultaneously connected

Theorem. The number of non-crossing set partitions of [n] is equal to the number of non-nesting set partitions of [n], which is also equal to the Catalan number Cn.

• Exercise 1. Prove that the number of non-crossing set partitions of [n] is equal to Cn.

• Claim π is non-nesting iﬀ in the corresponding rook partition, all rooks are in the SW or NE directions of each other

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• Light source in NW of board, every rook casts a shadow bounded by its hook; path between shadow and non-shadow is Dyck path

• Draw triangular array of Stirling numbers (ﬁrst kind or second kind), there exists recurrence relation in array

4.3 March 4

• c(n, k) are signless versions of Stirling numbers of the ﬁrst kind (number of permutations with k cycles)

• S(n, k) the Stirling numbers of the second kind (set partitions of [n] into k blocks)

Theorem. c(n + 1, k) = n · c(n, k) + c(n, k − 1)

Theorem. S(n + 1, k) = k · S(n, k) + S(n, k − 1)

5 Posets and Lattices

5.1 March 4

• Deﬁnition. Lattice: poset L with two binary operations meet ∧ and join ∨ de- ﬁned as follows:

– x ∧ y is the unique maximal element of L which is ≤ x and y – x ∨ y is the unique minimal element which is ≥ x and y

• Axiomatic deﬁnition of lattice:

– (L, ∧, ∨) with ∧, ∨ binary operations that are commutative and asso- ciative – x ∨ x = x ∧ x = x – x ∧ (x ∨ y) = x = x ∨ (x ∧ y) – x ∧ y = x ⇔ x ∨ y = y

• From lattices to posets: x ≤ y ⇔ x ∧ y = x

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5.2 March 11

• Lattice: poset with binary operations ∧, ∨

• Alternatively, poset (L, ≤) admitting ∧, ∨ operations; x ≤ y ↔ x∧y = x ↔ x∨y = y

• Example. Boolean lattice Bn: elements are all subsets of [n] with order S1 ≤ S2 iﬀ S1 ⊆ S2 with S1 ∧ S2 = S1 ∩ S2 and S1 ∨ S2 = S1 ∪ S2

• Any ﬁnite lattice has a unique minimum 0ˆ and unique maximum element 1ˆ

• Bn is skeleton of n-dimensional hypercube

• Example. Partition lattice Πn: elements are set partitions of [n] with order by reﬁnement π ≤ σ if each block of π is contained in a block of σ (π reﬁnes σ or σ coarsens π)

• π ∧σ is common reﬁnement (blocks formed by pairwise intersections of blocks of π, σ), π ∨ σ is common coarsening (if a, b can be connected by some chain of arcs in π and σ, elements are in same block)

• Can view π ∨ σ as connected components of overlayed blocks

• Example. Young’s lattice Y: inﬁnite lattice of all Young diagrams ordered by containment

• Minimal element 0ˆ is the empty Young diagram

• If there are k elements below Y, there are k + 1 elements above Y

• Can ﬁx m, n and let L(m, n) be sublattice of Young lattice of Young diagrams that ﬁt inside m × n box

n P |λ| • Recall [ k ]q = q λ⊂k×(n−k)

• λ ∧ µ = λ ∩ µ, λ ∨ µ = λ ∪ µ

• Deﬁnition. Distributive lattice: lattice L such that x ∨ (y ∧ z) = (x ∨ y) ∧ (x ∨ z) and x ∧ (y ∨ z) = (x ∧ y) ∨ (x ∧ z)

• Prototypical distributive lattice is boolean lattice

• Young’s lattice is also distributive but partition lattice for n ≥ 3 is not

• Deﬁnition. Order ideal: given poset P, subset I of P such that ∀x ∈ I, y ≤ x ⇒ y ∈ I

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• J(P ) is poset of all order ideals in P ordered by containment

Birkhoff’s Fundamental Theorem for Finite Distributive Lattices. P 7→ J(P ) is a bijection between finite posets and finite distributive lattices.

5.3 March 13

• Given poset P,J(P ) is poset of order ideals in P ordered by inclusion

Lemma. J(P ) is a distributive lattice.

Birkhoff’s Fundamental Theorem for Finite Distributive Lattices. Every finite distributive lattice is of the form J(P ) for some finite poset P, and P → J(P ) is a bijection between finite posets and finite distributive lattices.

• Deﬁnition. Join-irreducible element: element x ∈ L a ﬁnite distributive lattice if x 6= 0ˆ (unique minimal element) and x cannot be expressed as y ∨ z for y, z < x

• Given L any ﬁnte distributive lattice, reconstruct P by taking sub-poset of L of all join-irreducible elements

• Exercise 1. L is isomorphic to J(P ) where P is the sub-poset of join-irreducible elements deﬁned above.

• Deﬁnition. Ranked poset: poset P is ranked if there is a function ρ : P → {0, 1, 2,... } such that ρ(x) = 0 for any minimal element x of ρ and ρ(y) = ρ(x)+1 if x <· y

Proposition. Any ﬁnite distributive lattice is ranked with the modularity property ρ(x ∨ y) + ρ(x ∧ y) = ρ(x) + ρ(y).

• Given L = J(P ) and order ideal I ∈ L, ρ(I) = |I|

• Claim if I <· J then |J| = |I| + 1; removing any maximal element of J gives another order ideal

• Deﬁnition. Rank number: for ﬁnite ranked poset P, rank number ri = #{x ∈ P |ρ(x) = i}

• Vector representation (r0, r1, . . . , rN ) of P where N is the maximal rank

• Deﬁnition. Rank-symmetric: ﬁnite ranked poset P with ri = rN−i ∀i

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• Deﬁnition. Unimodular: ﬁnite ranked poset P with r0 ≤ r1 ≤ · · · ≤ rj and rj ≥ rj+1 ≥ · · · ≥ rN

• Deﬁnition. Product of posets: P × Q is poset whose elements are pairs (p, q) with p ∈ P, q ∈ Q and order relation (p, q) ≤ (p0, q0) iﬀ p ≤ p0, q ≤ q0

• Example. [n] is chain 1 < 2 < ··· < n and Hasse diagram of [m] × [n] is m × n grid rotated by 45◦; this poset is rank-symmetric and unimodular

• J([m]×[n]) = L(m, n) is poset of Young diagrams that ﬁt inside a m×n rectangle

• m+n mn Rank numbers ri are Gaussian coeﬃcients [ n ]q = r0 + r1q + ··· + rmnq

Theorem. L(m, n) is rank-symmetric and unimodular.

• Example. J([2] × [n]) has rotated triangular Hasse diagram

• Example. J(J([2] × [n])) is poset of shifted Young diagrams that ﬁt inside the n-triangle

Sperner’s Theorem. Suppose S1,S2,...,SM are diﬀerent subsets of [n] such n that no subset contains another, i.e. for any i 6= j, Si 6⊆ Sj. Then M ≤ n . b 2 c

• Deﬁnition. Chain: set C of elements in poset P such that any two elements of C are compatible, i.e. C is totally ordered

• Deﬁnition. Anti-chain: set A of elements in poset P such that any two elements of A are incompatible

• Deﬁnition. Sperner property: ﬁnite ranked poset with rank numbers (r0, r1, . . . , rN ) such that the maximum size M of an anti-chain in P is the maximal rank number

• Clear that M ≥ max{ri}

• Sperner’s Theorem says the boolean lattice is Sperner

5.4 March 15

• P a ﬁnite ranked poset, ρ : P → Z≥0

• Pi = {x ∈ P |ρ(x) = i} with rank number ri = |Pi|

• P is rank-symmetric if ri = rN−i∀i

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• P is unimodal if sequence has exactly one peak

• P is Sperner if size of the maximal antichain is max(r0, . . . , rN )

Theorem. (Sperner’s Theorem) The boolean lattice Bn is Sperner.

• Deﬁnition. Chain: sequence of totally ordered elements

• Deﬁnition. Saturated chain: sequence of totally ordered elements occupying consecutive ranks, i.e. x0 <· x1 <· x2 <· ··· <· xl

• Deﬁnition. Symmetric chain decomposition: decomposition of elements of P into disjoint union of saturated chains Ci such that ∀Ci = {x0 <· ··· <· xl}, ρ(xl) = N − ρ(x0)

Lemma. If P has a symmetric chain decomposition, then P is rank-symmetric, unimodal, and Sperner.

• Proof: given symmetric chain decomposition C1 ∪C2 ∪· · ·∪Cm, each Ci intersects the middle level and m is the maximal rank number of P

• For each anti-chain A, |A ∩ Ci| ≤ 1 so |A| ≤ m and P is Sperner

• If [n] represents n-element chain, Bn = [2] × [2] × · · · × [2] n times and Bn is skeleton of n-dimensional cube

Theorem. (de Bruijn) Bn has a symmetric chain decomposition.

• Generalization: [a] × [b] × · · · × [c] has a symmetric chain decomposition

Lemma. [a] × [b] has a symmetric chain decomposition.

• Proof by diagram

Lemma. If P,Q have symmetric chain decompositions, then P × Q has a symmetric chain decomposition.

• 0 0 S 0 Proof: P = C1 ∪ · · · ∪ Ck and Q = C1 ∪ · · · ∪ Cl so P × Q = Ci × Cj and there is 0 a symmetric chain decomposition for each Ci × Cj; this forms a symmetric chain decomposition of P × Q

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• Let P be any ﬁnite poset, not necessarily ranked, with M(P ) the maximum number of elements in an anti-chain and m(P ) the minimum number of disjoint chains that cover all elements of P

Dilworth’s Theorem. For any ﬁnite poset P,M(P ) = m(P ).

Minsky’s Dual of Dilworth. For any ﬁnite poset P, the maximum number of elements in a chain of P is equal to the minimum number of disjoint anti-chains covering all elements of P.

• Deﬁnition. k-antichain: union of k anti-chains

• lk is maximum number of elements in {A1 ∪ A2 ∪ · · · ∪ Ak} where Ai are anti- chains, i.e. maximum number of elements covered by a k-antichain

• Example. l1 = M(P )

• mk is maximum number of elements in {C1 ∪ C2 · · · ∪ Ck} where Ci are chains

Greene’s Theorem. Let λ(P ) be a partition (l1, l2 − l1, l3 − l2,... ) and µ(P ) a partition (m1, m2 −m1,... ). Then λ and µ are conjugate, i.e. their Young diagrams are transposes of each other.

5.5 March 18

• Greene’s Theorem relating ﬁnite poset P to Young diagram λ

• Schensted correspondence transforming permutation w ∈ Sn to pair (P,Q) of standard Young tableaux of the same shape

• w ∈ Sn 7→ P = ({1, . . . , n},

• Example. w = (3524716) with Schensted shape (3, 3, 1); poset on w given by 1

• Increasing/decreasing subsequences in w are chains/antichains of poset P

P λ2 • Frobenius-Young identity: λ f = n!

• Young’s lattice Y (isomorphic to poset of order ideals of Z≥0 × Z≥0), Yn the set of all Young diagrams with n boxes, i.e. nth level of Y

• Number of elements of Yn is p(n) (nth partition number)

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• Each SYT corresponds to saturated chain in Hasse diagram of Y from ∅ to λ

λ2 • Can think of f corresponding to path in Y from ∅ to ∅ with n up steps followed by n down steps

• Deﬁne R[Yn] to be the linear space of formal linear combinations of Young dia- p(n) grams with n boxes; R[Yn] isomorphic to R

• R[Y] = ⊕R[Yn]

• Up, down operators U, D : R[Y] → R[Y] deﬁned as follows: P – U : λ 7→ µ:λ<·µ µ P – D : µ 7→ λ:λ<·µ λ

• U adds box to Young diagram and D removes box from Young diagram

n n • Number of (P,Q) of shape λ is coeﬃcient of ∅ in D (U (∅))

• Key identity: commutator [D,U] = DU − UD is the identity operator

• Coeﬃcient of µ in (DU − UD)λ: if λ 6= µ, can add box a and remove box r in either order to get same result

• If λ = µ, diﬀerence between number of “inner” and “outer” corners is 1

• Definition. Differential poset: ranked (not necessarily finite) poset with unique minimum element 0ˆ such that the up and down operators for P satisfy [D,U] is the identity

• Combinatorially, given any two elements x 6= y in the poset, number of elements covering both x and y is the number of elements covered by both x and y

• If x = y, number of upward edges is one more than the number of downward edges in the Hasse diagram

0 • Compare to operators u, d on R[x] where u : f(x) 7→ xf(x) and d : f(x) 7→ f (x)

Stanley’s Theorem. For any diﬀerential poset, DnU n(0)ˆ = n!0ˆ.

Lemma. [D,U] being the identity means DU n = nU n−1 + U nD.

• Proof by induction

n n n−1 n n−1 n−1 n n−1 n−1 • D U (0)ˆ = D (DU )(0)ˆ = D (nU + U D)(0)ˆ = D (nU )(0)ˆ since D(0)ˆ = 0

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5.6 March 20

• Deﬁnition. Diﬀerential poset: ranked poset with unique minimum element 0ˆ with up and down operators satisfying [D,U] being the identity

• Attempt at constructing diﬀerential poset at every level

• To construct level n+1, take mirror image between n and n−1st level, add extra edge for every vertex on nth level

• Level n has Fn nodes

• Construction is forced up to level 5

Theorem. The construction described above gives a lattice, called the Fibonacci lattice.

• Young’s lattice and Fibonacci lattice are both diﬀerential posets

Theorem. For any diﬀerential poset, DnU n0ˆ = n!0ˆ.

• Imagine DD...DDUU...UU0ˆ as collisions/jumping of D’s over U’s, viewing as paths with n up steps followed by n down steps; n! ways to match D’s to U’s

• More generally, pick any word with n ups and n downs; each D must be matched with a U to its right

• These matchings are in bijection with rook placements on board with D representing right step and U representing up step on staircase

• Word of U’s and D’s can be converted to Young diagram ν that ﬁts inside an n×n square

Theorem. Given word of n U’s and n D’s, the coeﬃcient of 0ˆ when the word is applied to 0ˆ is the number of placements of n non-attacking rooks in the shape ν. This is explicitly given by (νn)(νn−1 − 1) ··· (ν1 − n + 1).

• k+l Unimodality of Gaussian coeﬃcients k q

• k+l k q is polynomial in q with coeﬃcients ai

• Let P be the poset that is the product of two chains [k] × [l]; J(P ) is lattice L(k, l) of all Young diagrams that ﬁt inside a k × l rectangle, J(P )n is set of all rank n elements of J(P )

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• kl Need to show #J(P )n+1 ≥ #J(P )n for all n < 2

• For λ ⊂ J(P )n, deﬁne Add(λ) to be the set of all elements of P such that λ∪{x}mλ and Remove(λ) is the set of all elements of P such that λ\{y} <· λ

• Add(λ) is the set of boxes that can be added to λ, Remove(λ) is the set of boxes that can be removed from λ while staying inside the box

Lemma. Fix n and suppose w : P → R≥0 is a weight function such that for any λ ∈ J(P )n, X X w(x) > w(y). x∈Add(λ) y∈Remove(λ)

Then #J(P )n+1 ≥ #J(P )n.

• Modify up and down operators to take into account the weights of the added/removed boxes: X U : λ 7→ pw(x) · µ µ=λ∪{x} x∈Add(λ) X D : λ 7→ pw(y) · µ µ=λ\{y} y∈Remove(λ)

T • D = U

• Claim H = DU − UD is given by diagonal matrix   X X H : λ 7→  w(x) − w(y) λ x∈Add(λ) y∈Remove(λ)

• Easy to see by carrying out operations on same/diﬀerent boxes

T T • DU = UD + H = UU + H with UU positive semideﬁnite and H positive deﬁnite; then DU has positve determinant

5.7 March 22

• Vn is linear space of formal linear combinations of Young diagrams λ ≤ k × l with |λ| = n p • Un is weighted up operator, Dn is weighted down operator (weighted by w(x) where w(x) is weight of box being modiﬁed)

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T • Hn = Dn · Un − Un−1Dn−1 sends Vn to Vn,Dn = Un

• Claim Hn given by diagonal matrix with diagonal entries X X (Hn)λ,λ = w(x) − w(y) x∈Add(λ) y∈Remove(λ)

• If all eigenvalues of H strictly positive, Dn · Un = Un−1Hn−1 + Hn is positive deﬁnite

• Rank of AB ≤ min(m, n, k) where A is m × n and B is n × k matrix

• Dn is an × an+1,Un is an+1 × an and Dn · Un has rank an, forcing an ≤ an+1

• Need to ﬁnd weight function

• Deﬁne w :[k] × [l], w(i, j) = (i − j + l)(j − i + k)

Lemma. For all λ ⊂ k × l, wλ = k · l − 2|λ|.

• k·l Then wλ > 0 if n = |λ| < 2

6 Partitions

6.1 March 22

• p(n) is the number of partitions of n

Theorem. X 1 p(n)qn = (1 − q)(1 − q2)(1 − q3) ··· n≥0

• k+l P |λ| First proof: k q = λ⊆k×l q , take limit as k, l → ∞

• Second proof: λ = (λ1, λ2,... ) and mi = #{j|λj = i} (multiplicities of parts), write λ = (1m1 2m2 ··· )

• Only ﬁnitely many mi are nonzero

n m +2m +··· 1 1 • P p(n)q = P q 1 2 = ··· m1,m2,···≥0 1−q 1−q2

Theorem. The number of partitions of n into odd parts is the same as the number of partitions of n into distinct parts.

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Euler’s Pentagonal Number Theorem.

1 2 3 P n = (1 − q)(1 − q )(1 − q ) ··· n≥0 p(n)q

∞ X = (−1)mqm(3m−1)/2 m=−∞

6.2 April 1

• P n 1 1 p(n)q = (1−q)(1−q2)··· = f(q) n≥0

• Euler’s Pentagonal Number Theorem

∞ X f(q) = (−1)mqm(3m−1)/2 = 1 − q − q2 + q5 + q7 − q12 ··· m=−∞

• Gauss: ∞ X (f(q))3 = (−1)m · m · qm(m+1)/2 m=−∞

2 • f(q) is a total mess

m m • Given partition λ = (λ1, λ2,... ) = (1 1 2 2 ··· ) where mi is number of parts of λ equal to i

2 3 • (1 − q)(1 − q )(1 − q ) ···

X P X = (−1) mi qm1+2m2+··· = (−1)#parts in λ · q|λ|

mi∈{0,1} λ

where the sum is over partitions λ with distinct parts

• even odd Let pdist (n), pdist(n) be the number of partitions of n with distinct parts and an even/odd number of parts

m(3m−1) • even odd m Euler’s Theorem claims pdist (n)−pdist(n) = (−1) if n = 2 and 0 otherwise

• Bijective proof using involution σ on almost all partitions of n with distinct parts such that µ = σ(λ) and λ have diﬀerent parity in number of parts

• Given partition λ, A is the last row of λ and B is the longest diagonal segment starting at the upper right corner

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• a = |A|, b = |B|; if a > b then µ = σ(λ) is obtained by removing the diagonal segment and adding a new row with b boxes; if a ≤ b then µ obtained by removing A and adding new diagonal segment

• Example. A in yellow, B in blue:

λ = σ(µ) = ,

µ = σ(λ) =

• σ is well deﬁned except when A and B overlap and a = b or a = b + 1, these partitions have pentagonal number of boxes

• Deﬁnition. Jacobi’s Triple Product Identity:

∞ Y X 2 (1 − q2n)(1 + q2n−1z)(1 + q2n−1z−1) = qr · zr n≥1 r=−∞

• Special cases: √ √ 1. z = − x, q = x3 gives Euler’s Pentagonal Theorem √ 2. z = −x, q = x gives Gauss formula for f(q)3 Q 1−qm P∞ r r2 3. z = −1 gives m≥1 1+qm = r=−∞(−1) q √ • Substitute q by q and z by qz so Triple Product becomes

∞ P zrqr(r+1)/2 Y r=−∞ (1 + zqn)(1 + z−1qn−1) = Q 1 − qn n≥1 n≥1

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a Q n • Coeﬃcient [z ] in n≥1(1 + zq ) is number of partitions into a distinct parts of q|λ|

b Q −1 n−1 • Coeﬃcient [z ] in n≥1(1 + z q ) is number of partitions into b distinct parts of q|λ|−b

6.3 April 3

• Modiﬁed version of Jacobi Triple Product Identity

n |µ| • Q (1 + zq ) = P q where µ is partition with a parts, all distinct n≥1 µ

−1 n−1 |ν|−b • Q (1 + z q ) = P q where ν is partition with b parts, all distinct n≥1 ν

• Q 1 n≥1 1−qn is generating function for all partitions

• Need bijection (a, b, µ, ν) ↔ (r, λ) with a − b = r (matching powers of z) and r(r+1) |µ| + |ν| − b = 2 + |λ| (matching powers of q)

• Example. a = 5, b = 3, µ = (7, 6, 4, 3, 1) and ν = (6, 5, 3), represent µ and ν by shifted Young diagrams and mark the ﬁrst boxes of each row (along the diagonal)

• Transpose ν and remove the marked boxes to get ν,˜ then glue µ and ν˜ along the marked boxes such that the last marked boxes are aligned

• Gluing results in Young diagram λ with extra triangular pice in the top left with r = a − b rows

7 Trees

7.1 April 3

• Deﬁnition. Labeled tree: connected acyclic graph on n nodes labeled by 1, 2, . . . , n

• Two labeled trees are diﬀerent if they are not isomorphic

Cayley’s Formula. The number of labeled trees on n nodes is nn−2.

• Want to generalize Cayley’s Formula

T dT (1)−1 dT (2)−1 dT (n)−1 • Given tree T, deﬁne weight w(T ) to be monomial x = x1 x2 ··· xn where dT (k) is the degree of the vertex labeled k in T P T • Deﬁne Fn(x1, . . . , xn) = x over all labeled trees on n nodes T

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n−2 Theorem. Fn = (x1 + ··· + xn) .

n−2 • Let Rn = Fn − (x1 + ··· + xn) ; want Rn identically 0

Lemma. Rn is a polynomial in x1, . . . , xn of degree ≤ n − 2.

T • Each monomial x has degree n − 2

Lemma. For each i = 1, . . . , n, Rn|xi=0 = 0.

• Suﬃces to prove lemma for i = n; only surviving monomials correspond to cases where vertex n is a leaf

P T P T 0 0 • Then x = x (x1 + ··· + xn−1) where T are trees with n removed; adding T T 0 edge back increases degree of some vertex by 1

• By induction this is Fn−1 · (x1 + ··· + xn−1) and Rn|xn=0 as desired

Lemma. If f is a polynomial in x1, . . . , xn of degree ≤ n − 1 such that f|xi=0 ∀i, f must be identically 0.

7.2 April 5

Lemma. If f(x1, . . . , xn) is a polynomial of degree at most n − 1 such that

f|xi=0 = 0, then f = 0.

a1 an • Assume for the sake of contradiction that f contains some monomial x1 ··· xn P with nonzero coeﬃcient and ai ≤ n − 1

• One of the ai must be 0, but setting xi = 0 gives contradiction

• Deﬁnition. Prufer encoding: bijection between labeled trees on [n] and Prufer code (c1, . . . , cn−2) with ci ∈ [n] given by the following algorithm: 1. Find leaf v with minimal label 2. Record in the code the label of the parent of v 3. Remove v from T and repeat steps until only two vertices are left

• Vertex of degree d appears d − 1 times in the code, leaves of T are the indices that do not appear in the code

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• Decoding the sequence: need code (c1, . . . , cn−2) and set of labels [n]

• First ﬁnd smallest element l in the set of labels that does not appear in the code, connect l to ﬁrst entry of code

• Remove l from labels and c from code, repeat steps until code is empty and connect remaining two elements in set of labels

• If d1, . . . , dn ≥ 1 the degrees of the tree, the number of labeled trees on [n] with these degrees is the multinomial coeﬃcient n−2 d1−1,d2−1,...,dn−1

• Second combinatorial proof of Cayley’s Formula: assume ﬁrst (minimal) node is root, direct all edges to point to root

• Find path from maximal vertex to root and mark all left-to-right minima of this sequence

• Convert path into permutation cycle using record idea, remove path and replace by edges in the corresponding cycle

• Each vertex has exactly one outgoing edge, can view tree as map from [n] → [n] such that f(i) = j if directed edge (i, j) is in graph

• This is bijection from labeled trees on n nodes to functions f :[n] → [n] with f(1) = 1, f(n) = n

7.3 April 10

• Given graph G = (V,E) without loops (self-edges) but allowing multi-edges

0 • Deﬁnition. Spanning tree: subgraph T = (V,E ) of G which is a tree

• Spanning trees on Kn are equivalent to labeled trees on n vertices

7.4 April 12

• Graph G = (V,E) a graph without self-loops but allowing multi-edges

• Spanning tree of G is subgraph T with same set of vertices V and set of edges E0 ⊂ E such that T is a tree

• Number of spanning trees of G

• Incidence matrix of G: C = (cie) an n × m matrix with cie = 1 if vertex i is incident to edge e and 0 otherwise

• Adjacency matrix of G: A = (aij) an n × n matrix with aij the number of edges connecting i and j

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• Deﬁnition. Laplacian Matrix: L = D−A an n×n matrix where D is a diagonal matrix with diagonal entries (d1, . . . , dn) with di the degree of matrix i

• Row and column sums of Laplacian matrix are 0

Matrix Tree Theorem. Fix i ∈ [n]. Let L˜ = Li be L with the ith row and column removed. Then the number of spanning trees of G is the determinant of L.˜

• L˜ is called the reduced Laplacian matrix

• Deﬁnition. Spectrum of graph: eigenvalues of A(G)

• If G is regular, L = dI − A and eigenvalues µi = d − λi

• Example. G = Kn has L with n − 1 along diagonals and −1 everywhere else; L˜ − nIn−1 is (n − 1) × (n − 1) matrix of all −1s so all eigenvalues except one are 0; only nonzero eigenvalue is −n + 1, then eigenvalues of λL are n, n, . . . , n, 1 and det(L˜) = nn−2 (Cayley)

+ • Let G be a graph on n vertices without multiple edges and G the graph obtained by adding a new vertex 0 connected to all vertices

P dT (0)−1 • Let FG(x) = T x

Fg(0) • Number of spanning trees of G is n

Reciprocity Theorem for Spanning Trees. Let G = (V, E) be the complemen- n−1 tary graph to G. Then FG(x) = (−1) FG(−x − n).

+ • Let On be the empty graph on n vertices, On is a star with exactly one spanning tree

• n−1 n−1 n−1 n−1 Then FOn = x and by Reciprocity, FKn = (−1) (−x − n) = (x + n)

• Let Km ∪Kn be the disjoint union of Km and Kn so FKm∪Kn (x) = FKm (x)·FKn (x)· x = (x + m)m−1 · (x + n)n−1 · x

• Complement of Km ∪ Kn is complete bipartite graph Km,n and FKm,n (x) = (x + n)m−1 · (x + m)n−1 · (x + m + n)

m−1 n−1 • Corollary: number of spanning trees in Km,n is n m

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7.5 April 17

• Matrix Tree Theorem: let G be a graph on n vertices and deﬁne the n × n Lapla- cian matrix L = (Lij) with Lij = degG(i) if i = j and −eij otherwise where eij is the number of edges connecting i, j

i • Reduced Laplacian matrix L˜ = L is the Laplacian matrix with the ith row and column removed

Matrix Tree Theorem. The number of spanning trees of G is det L.˜

• Deﬁnition. Direct product of graphs: given G = (V1,E1) and H = (V2,E2), 0 0 0 0 G × H = (V1 × V2,E3) where E3 contains edges ((i, j), (i , j )) if i = i and (j, j ) ∈ 0 0 E2 or j = j and (i, i ) ∈ E1

Lemma. Let A(G) be the adjacency matrix of G. If A(G) has eigenvalues α1, . . . , αm and A(H) has eigenvalues β1, . . . , βn, then A(G×H) has eigenvalues αi + βj for all i, j.

• Proof: A(G × H) = A(G) ⊗ A(H) (tensor product of matrices)

• If G is a regular graph of degree d1 and H is a regular graph of degree d2,G × H is a regular graph of degree d1 + d2 and the eigenvalues of its Laplacian are d1 + d2 − αi − βj

• Deﬁnition. Hypercube graph: Hd is the one dimensional skeleton of the d-cube; alternatively, the product of d 2-chains C 0 1 • A(C) = with eigenvalues ±1 1 0

d • Hd has 2 eigenvalues of form ε1 + ··· + εd where εi ∈ {1, −1} d • L(Hd) has eigenvalues 2k for k = 0, . . . , d with multiplicity k

Theorem. The number of spanning trees on Hd is

d d 1 Y d 2d−d−1 Y d (2k)(k) = 2 k(k) 2d k=1 k=1

• Proof of matrix tree theorem: direct edges of G in any direction

33 18.212 Wanlin Li

• Deﬁnition. Oriented incidence matrix: graph G has n vertices and m directed edges, B = (bie) for i ∈ V, e ∈ E with bie = 1 if i is the source of e, −1 if i is the target of e, and 0 otherwise

Lemma. The Laplacian matrix of G is given by BBT .

T • (BB )ij is scalar product of ith and jth rows of B

• Corollary: if B˜ is the oriented incidence matrix with the ith row removed, L˜ = B˜B˜T

Cauchy-Binet Formula. Let A be a k × m matrix and B a m × k matrix with k ≤ m. Then X det(AB) = det(AS) det(BS) S⊂[m] |S|=k where AS is the k × k submatrix of A in the column set of S and BS is the k × k submatrix of B in the row set of S.

Ik A A 0 0 AB • Proof: take product of block matrices = 0 Im −Im B −Im B

• Take determinants on both sides, determinant of RHS matrix is ± det AB

• Determinant of second LHS matrix is summation of Cauchy-Binet up to sign (using permutation representation of determinant)

7.6 April 19

• B˜ is oriented (n − 1) × m incidence matrix with one row removed

T T S 2 • L˜ = B˜B˜ and det(L˜) = det(B˜B˜ ) = P (det B˜ ) S⊂E |S|=n−1

Lemma. det(B˜S) is ±1 if S is the set of edges of a spanning tree and 0 otherwise.

• If S contains a cycle, there is some linear combination of columns that is zero (sum over edges of cycle in the appropriate direction) so determinant is 0

S • If S is a spanning tree, after some permutation of rows/columns, B˜ will be lower triangular

34 18.212 Wanlin Li

S • S has leaf so some row of B˜ has only one nonzero entry, expand determinant by this row and induct down

• Weighted version of matrix tree theorem: suppose that each edge e of G has Q weight xe, for a spanning tree T let w(T ) be e∈T xe P • Deﬁne FG = T w(T ) over all spanning trees of G P • Weighted version of Laplacian: Lij = e xe over edges incident to i if i = j, −xe if e = (i, j) and i 6= j, 0 otherwise

Weighted Matrix Tree Theorem. FG = det(L˜).

• If weights are all positive integers, can replace each edge e with xe edges of weight 1 and apply regular matrix tree theorem on new graph Gm

m • Then FG is the number of spanning trees of G and Weighted Matrix Tree The- orem is proved by treating edge weights as variables

• G-digraph (directed graph) has no self-loops but allows multiple edges, has a ﬁxed root r

• Two versions of spanning trees for digraphs: out-trees and in-trees

• Deﬁnition. Out-tree: (also called arborescence) tree rooted at r and growing out of r, i.e. all edges in the shortest path from the root r to a vertex i are directed away from r

• Deﬁnition. In-tree: tree rooted at r and growing into r, i.e. all other edges directed toward r

• Number of in-trees and out-trees might be diﬀerent

in out • Two versions of Laplacian: L and L

out in • Lij is the out-degree of vertex i if i = j, Lij is in-degree in corresponding position

• Both matrices have Lij the negative number of directed edges from i to j if i 6= j

• L = D − A for appropriate corresponding in/out-degree matrix D

ij i+j • Cofactors of L given by L = (−1) det L˜ij where L˜ij is L with the ith row and jth column removed

in ir • The number of out-trees rooted at r is given by (L ) for any i

out ri • The number of in-trees rooted at r is given by (L ) for any i

35 18.212 Wanlin Li

7.7 April 22

• Directed graph G on n vertices without self-loops

• Adjacency matrix (Aij) where aij is the number of directed edges from i to j (any matrix with non-zero diagonal entries)

in out • D is diagonal matrix where diagonal entries are in-degrees of vertices; D is diagonal matrix where diagonal entries are out-degrees vertices

in in out out • L = D − A and L = D − A

ij i+j ij ij • Cofactors of any square matrix L are given by L = (−1) det L˜ where L˜ is L with the ith row and jth column removed

Directed Matrix Tree Theorem. Fix a root r of G. The number of in-trees rooted at r is given by (Lout)ri and the number of out-trees rooted at r is given by (Lin)ir for any i.

• In-tree and out-tree versions are equivalent by ﬂipping the direction of all edges (taking the transpose of the adjacency matrix)

• Directed version of matrix tree theorem can be proved by induction on the number of edges in the graph excluding edges coming out of the root (stronger hy- pothesis)

out rk • Let Inr(G) be the number of in-trees of G rooted at r; need to show this is (L )

out • Base case when all edges come out of the root, number of in-trees is 0 and L has all non-zero entries in the rth row

• Pick an edge e from i to j where i 6= r, let G1 be G\{e} and G2 be the graph with all edges e0 from i to j0 except e removed

• Claim Inr(G) = Inr(G1) + Inr(G2) since every non-rooted vertx in an in-tree has out-degree exactly 1

out • L (G) has rth row a1, . . . , an

out • L (G1) has rth row a1, . . . , ai − 1, . . . , aj + 1, . . . , an because out-degree of ai decreases by 1, number of edges from i to j decreases by 1 and its negative increases by 1

out out out • L (G2) has ith row 0, . . . , ai = 1, . . . , aj = −1,..., 0 L (G) = L (G1) + out L (G2)

out ri • By linearity of the determinant, Inr(G) = (L (G)) as desired

36 18.212 Wanlin Li

• Involution proof: for each edge (i, j) assign variable xij and A = [xij] the adjacency matrix with xii = 0

out out out P • L = D − A where (D )ii = xij j6=i

out rr P • (L ) is T w(T ) where T is in-tree rooted at r and w(T ) is the weight of T by product of edge weights

n • Exercise 1. Use the matrix tree theorem to prove Abel’s Identity: (x + y) = P n k−1 n−k k k y(y + kz) (x − kz) .

7.8 April 24 out P • Variables xij such that xii = 0 and (L )ij = xik if i = j and −xij otherwise k6=i

out rr P Theorem. (L ) = T w(T ) where T is in-tree rooted at r

• WLOG r = n, n−1 out nn X σ(w) Y out (L ) = (−1) (L )iwi w∈Sn−1 i=1 which is equal to the sum over subdivisions of {1, . . . , n − 1} into disjoint cycles C1 ∪ · · · ∪ Cm ∪ {ﬁxed points} of     m  Y  Y X (−1)  xij ·  xfj   i→j f ﬁxed point j6=f edge of cycle

σ(δ) c−1 • For a permutation δ with a single cycle of size c, (−1) = (−1) so each cycle c−1 c out contributes (−1) · (−1) = −1 when accounting for signs in (L )ij

• Sum is X number of red cycles Y (−1) xij H i→j edge of H where H is set of directed graphs on vertices [n] with n − 1 edges colored red or green such that for any vertex i = 1, . . . , n − 1, the out-degree of i is 1 and all red edges form a disjoint union of cycles

• Red edges correspond to cycles of the permutation and green edges correspond to ﬁxed points

37 18.212 Wanlin Li

• In-trees rooted at n are the same as acyclic digraphs with n − 1 edges such that the out-degree of every vertex except the root is exactly 1

• Want to show the contributions of graphs with any cycles will cancel each other out

• Want to construct involution on graphs with at least one cycle Q • Need weight-preserving (preserve xij), sign-reversing involution ω on the e∈H graphs H with at least one cycle of either color, i.e. ﬁnd involution that recolors the graph and changes the parity of the number of red cycles

• Fix a total order of edges i → j (e.g. lexicographically) and totally order all possible cycles (lexicographically), then ﬁnd ﬁrst cycle in H and switch its color

• Contributions of H and ω(H) cancel out, everything except in-trees cancel out

7.9 May 6

• Cn is the number of binary trees on n vertices or full binary trees on 2n + 1 vertices

Theorem.

1. The number of labeled binary trees on n vertices is n!Cn. 2. The number of increasing binary trees on n vertices (labels increase down the tree) is n!.

3. The number of left-increasing binary trees on n vertices (require only that the left child is larger than the parent) is (n + 1)n−1.

4. The number of increasing full binary trees on 2n + 1 vertices is A2n+1, the number of alternating permutations in S2n+1.

• Bijection for 2 from increasing binary trees to permutations in Sn: if i is in left subtree of r, i appears to the left of r in the corresponding permutation; root of subtree is smallest element in the corresponding portion of the permutaiton

• Bijection for 4 from increasing binary tree to alternating permutation: use the same conversion from tree to permutation; resulting permutation is alternating

• Alternating permutation is speciﬁc to up-down or down-up permutations

38 18.212 Wanlin Li 8 Electrical Networks

8.1 April 24

• Given graph G, view edges of graph as resistors and ﬁx orientations of all edges

• For any edge e = (u, v),Ie is current, Re is resistance, Ve = IeRe is voltage drop across edge

• Kirchhoﬀ’s Current Law: sum of currents into a node is equal to the sum of currents out of the node m • P Kirchhoﬀ’s Voltage Law: for all cycles C in G with edges e1, . . . , em, ±Vei = 0 i=1 where “+” is chosen if the orientation of ei agrees with C and “−” otherwise

• Voltage law equivalent to existence of potential function on vertices of the graph

8.2 April 26

• G a connected digraph with two selected vertices A, B

• Suppose resistances of all edges are known and voltage of battery is provided; goal to ﬁnd currents and overall resistance of graph

• By summing currents, for any vertex v and adjacent vertices v1, . . . , vd with Ri d 1 1 P 1 the resistance of the corresponding edge, Φ(v) + ··· + − Φ(Vi) is 0 R1 Rd Ri i=1 if v 6= A, B, −I if v = A, and I if v = B where Φ(v) denotes the potential at vertex v

1 • Define conductance C of edge (i, j) to be if edge (i, j) exists and 0 otherwise ij Rij (infinite resistance) P • Definition. Kirchhoff’s matrix: K an n × n matrix with Kij = cil if i = j and l6=i −cij otherwise  Φ(1)  −I  Φ(2)   0   .   .  •  .   .  Electrical laws correspond to K  .  =  .      Φ(n − 1)  0  Φ(n) I

• Kirchhoﬀ’s matrix is Laplacian for graph with edge weights cij

• Φ not unique, has ambiguity because only potential diﬀerence matters

39 18.212 Wanlin Li

• Rank of K is n minus the number of connected components of the graph

Φ(n)−Φ(1) • RAB(G) = I ; can assume I = 1 and Φ(1) = 0 by grounding ﬁrst vertex

(1) • Matrix K obtained by removing the ﬁrst row and column of K Φ(2) 0 . . • (1)  .  . (1) Electrical laws equation corresponds to K  .  = . with K nondegen- Φ(n) 1 erate

det K˜ (1) • Solve by Cramer’s rule R (G) = Φ(n) = AB det K(1)

(1) 1,n 1,n • det K˜ = det K where K is matrix with ﬁrst and last rows and columns removed

1 1,n P Q Theorem. R (G) = det K = S e∈S Re where S is a spanning tree of G0 AB det K(1) P Q 1 T e∈T Re (graph with A, B glued together) and T is a spanning tree of G.

• Equivalently, numerator is given by sum over “almost spanning trees” (forests with two components such that A, B are in diﬀerent connected components)

• Wheatstone bridge: if all edges have the same resistance, connecting opposite ends by voltage source gives no current along middle edge (potentials are the same)

• Given graph G with two chosen vertices A, B, rescale potentials so that Φ(A) = 0 and Φ(B) = 1; then all other potentials are between 0 and 1

• Consider random walk on G such that the probability p(u, v) of moving from u 1 Ruv to v is P 1 w6=u Ruw

Theorem. Φ(v) is the probability that a random walks starting at vertex v hits vertex B before hitting A.

• Attempting to solve Markov chain gives same system as Kirchhoﬀ’s matrix system

40 18.212 Wanlin Li

8.3 April 29

• Given G a digraph

• Deﬁnition. Eulerian cycle: closed directed walk using every edge of G exactly once

Theorem. G contains an Eulerian cycle iﬀ G is connected and for every vertex v, the in and out-degrees of v are the same. For an undirected graph, the degree of every vertex must be even.

• Proof: continually pick edges to follow out of vertices; can only get stuck at starting vertex when all edges have been used

• Compute number of Eulerian cycles (pick one fixed edge to be the first in the cycle, i.e. fix a root r and outgoing edge e)

BEST Theorem. Let G be a connected digraph with in-degree equal to out- degree at every vertex (G is an Eulerian digraph). The number of Eulerian cycles Q is equal to Inr(G) × v(do(v) − 1)! where Inr(G) is the number of in-trees of G rooted at r and do(v) denotes the out-degree of v.

• BEST = de Bruijn, von Aardenne-Ehrenfest, Smith, Tutte

in out • In this case, Laplacian matrix L = L = L and all row sums and all column sums of L are 0; this implies that all cofactors of L are the same

• Number of in-trees rooted at r is the same as the number of out-trees rooted at r, this number is independent of the choice of r

• Let C be an Eulerian cycle and order the edges of G corresponding to their order in C; denote ordering by

• For any vertex v 6= r, mark the outgoing edge from v that appears last in C

• Let T be the subgraph formed by the marked edges; T is an in-tree rooted at r

• T has n − 1 edges, every vertex has out-degree exactly one in T, no edge comes out of r; just need T to have no directed cycles

• If T has a directed cycle C˜ with r∈ / C,˜ let f be the maximal edge of C˜ in order

• Therefore T is in-tree rooted at r

41 18.212 Wanlin Li

• For any vertex v (including r), let wv be the permutation given by the ordering by

• wv corresponds to permutation of size do(v) − 1

• Claim: the map sending Eulerian cycle C to (T, {wv}) is a bijection

9 Matching

9.1 May 1

• Parking: n cars on a one-way road with n parking spaces, each driver has pre- ferred parking space; preference function f :[n] → [n] such that ith car prefers to park in space f(i)

• If a driver’s favorite space has been taken, the driver parks in the next available space

• Not all cars guaranteed to be able to park

• Deﬁnition. Parking function: preference function f :[n] → [n] if all n cars are able to park along the road

Lemma. f is a parking function iﬀ for any k = 1, . . . , n, the number of drivers with f(i) ≥ n + 1 − k is ≤ k.

• If f is a permutation of 1, 2, . . . , n, then f is a parking function

Lemma. f is a parking function iﬀ there exists a permutation σ ∈ Sn such that σ(i) ≥ f(i) ∀i.

Theorem. There are (n + 1)n−1 parking functions for n cars.

• Proof: use n cars and n + 1 parking spots on a 1-way circular road, preference function now given by f :[n] → [n + 1], all cars always park, one space always remains empty

• Observation: the number of functions f :[n] → [n + 1] such that the ith spot remains empty is the same for all i

• Observation: functions f resulting in a parking such that n + 1 stays empty are exactly parking functions

42 18.212 Wanlin Li

n−1 • Therefore there are exactly (n + 1) parking functions

• Pick random preference function; probability of choosing parking function is (n+1)n−1 1 1 n−1 e nn = n 1 + n ≈ n as n → ∞ P • Statistics on parking function: f 7→ s(f) = i f(i)

n(n+1) n+1 • n ≤ s(f) ≤ 2 ; s(f) = 2 for n! parking functions and s(f) = n for only 1 parking function

• Deﬁnition. Tree inversion: inversion in labeled tree T on n+1 vertices {0, 1, . . . , n} with root 0 is given by pair (i, j) with i < j such that j is on the shortest path from i to the root 0

• inv(T ) is the number of inverisons of T

n+1 Theorem. inv(T ) is equidistributed with 2 − s(f). In other words,

X n+1 −s(f) X inv(T ) x( 2 ) = x . f T

This polynomial is called the inversion polynomial for trees In(x).

n−1 • In(1) = (n + 1) ,In(0) = n!

• Deﬁnition. Increasing trees: trees with no inversions

• Number of increasing trees is n!; there are i places to attach vertex i

• In(−1) = ± number of alternating permutations of size n (w1 < w2 > w3 < w4 ··· )

9.2 May 3

n+1 P P inv(T ) P ( 2 )− i f(i) • Tree inversion polyonmial In(x) = T x = f x where T is over spanning trees of Kn+1 and f :[n] → [n] is a parking function

• Alternating permutation: w1 < w2 > w3 ... ; An also known as Euler numbers

• Plugging in x = −1 gives Euler numbers An

n • P x Deﬁnition. Exponential generating function: n≥0 An n! gives tan(x) + sec(x); odd n gives tan(x) and even n gives sec x

n • P x x Deﬁnition. Bernoulli numbers: n≥0 Bn n! = ex−1

43 18.212 Wanlin Li

• n−1 2n Bn only nonzero if n even or n = 1; B2n = (−1) 42n−22n A2n−1

∞ x−1 (−1)n−1·2(2n)! • 1 u Riemann zeta ζ(x) = Γ(x) 0 eu−1 du has B2n = (2π)2n ζ(2n) at integer values ´

Pn n Proposition. For all n ≥ 1, 2An+1 = k=0 k AkAn−k with A0 = A1 = 1.

• Resembles a “labeled” version of Catalan numbers

• Proof by putting n + 1 in position k + 1

0 2 • 2A (x) = A (x) + 1

• Regular binary tree: each vertex has at most 2 children

• Full binary tre: every vertex has 0 or 2 children (odd number of total vertices)

Proposition. The Catalan number Cn is the number of binary trees on n vertices or the number of full binary trees on 2n + 1 vertices.

• Binary tree on n vertices to full binary tree by adding all missing children; cre- ates full binary tree with n non-leafs and n + 1 leaves

• Bijection between Dyck paths of length 2n and full binary trees with 2n + 1 vertices; given Dyck path, break up into consecutive up-steps with a1, a2,... the lengths of those groups allowing ai = 0 for consecutive down steps

• Attach a1 vertices along left chain with all children, attach a2 vertices starting from bottom-most right child of a1-chain, repeat; if ai = 0, move one up the diagonal and keep going

• Provides way to search through binary tree

10 Noncrossing Path Enumeration

10.1 May 6 a+b • A = (0, 0),B = (a, b); a lattice paths from A to B

• Connect A1,A2 to B1,B2 by noncrossing lattice paths (no common vertices)

• G an acyclic digraph with A1,...,Ak,B1,...,Bk selected vertices in G

• N(A1,...,Ak,B1,...,Bk) is the number of ways to connect sources A1,...,Ak with sinks B1,...,Bk by k pairwise non-crossing paths (no shared vertices)

44 18.212 Wanlin Li

Lindstrom’s Lemma/Gessel-Viennot Method. Consider the special case N(A1,...,Ak; Bw(1),...,Bw(k)) = 0 for all permutations w other than the identity, so that Ai must connect to Bi for all i. Then N(A1,...,Ak; B1,...,Bk) = det(C) where C = (Cij) is a k × k matrix with cij = N(Ai,Bj).

Theorem. Without the assumption that Ai must be connected to Bi for all i, P (−1)σ(w)N(A ,...,A ; B ,...,B ) = det(C). w∈Sn 1 k w(1) w(k)

• Proof by involution principle: let P (A1,...,Ak; B1,...,Bk) be the number of all possible ways to connect Ais with Bis by paths Pi : Ai → Bi without the non- crossing restriction; P (A1,...,Ak; B1,...,Bk) = P (A1,B1) ··· P (Ak,Bk)

σ(w) • Expand det(C) = P (−1) c ··· c which by the above is the same w∈Sn 1w(1) kw(k) as P (−1)σ(w)P (A ,...,A ; B ,...,B ) w∈Sn 1 k w(1) w(k)

• Would like the contribution of all crossing paths to be 0

• Construct sign-reversing involution σ on k-tuples of paths (P1,...,Pk),Pi : Ai → Bw(i) that have at least one crossing

• σ: ﬁnd ﬁrst crossing X of two paths in the family, swap ends of paths after X so paths switch between crossing and touching at X

• Need careful deﬁnition of “ﬁrst” crossing

• Find smallest i such that Pi has an intersection; find first intersection point along Pi, find minimal j such that Pj also passes through this point

10.2 May 8

• G an acyclic directed graph with sources A1,...,Ak and sinks B1,...,Bk

• N(A1,...,Ak; B1,...,Bk) is the number of non-crossing collections of paths {Pi} such that Pi connects Ai to Bj

• If Pi must connect Ai to Bi,N(A1,...,Ak; B1,...,Bk) is the determinant of matrix C with cij = N(Ai,Bj)

• Proof: construct sign-reversing involution σ on “bad” collections of paths

• Weighted version of Lindstrom Lemma: edge e of G has weight xe > 0; deﬁne N(A1,...,Ak; B1,...,Bk) as the sum of weights of non-crossing family of paths with weight of a family given by w(P ) = Q Q x i e∈Pi e

• Lemma still holds with the same proof because σ is weight-preserving

45 18.212 Wanlin Li

• Lindstrom Lemma often used for planar graphs G drawn inside a square such that A1,...,Ak are on the left side of the square ordered top to bottom and B1,...,Bk are on the right side

• det C ≥ 0

• Minors are determinants of square submatrix

• Deﬁnition. Totally positive (nonnegative) matrix: matrix C with real entries such that all its minors are positive (nonnegative)

• All minors of C are also nonnegative

Theorem. A matrix C is totally nonnegative iﬀ it can be represented as by a planar graph of the form G above.

• Consider m × n grid with edges going up and to the right; weights xij only on the horizontal edges to avoid redundancy

• Put A1,...,Am on the bottom and B1,...,Bn on the right (topologically equivalent to left and right)

• C is the m × n matrix with cij the sum of weights of all paths from Ai to Bj

Theorem. Any totally positive matrix can be uniquely presented like C.

mn • Space of totally positive m × n matrices bijective with (R>0)

• Deﬁnition. Plane partition: given m × n rectangle, ﬁll boxes with entries from 1 to k allowing repetition and such that entries are weakly decreasing across rows and columns

• Calculate number of plane partitions

• Example. m = n = k = 2 has 6 plane partitions

m+n • With k = 2, general answer is m because there is a lattice path dividing 1s and 2s

• Paths separating numbers in plane partition have weakly non-crossing paths starting and ending at the same points

• Number of m × n plane partitions with entries ≤ k is the number of weakly non-crossing families of paths (P1,...,Pk−1) in the m × n grid

46 18.212 Wanlin Li

• Translate P2 one unit in the (1, −1) direction and repeat; this gives non-crossing family of paths

m+n • Can explicitly write matrix C and cij = m+i−j

• Matrix has explicit product formula

• Can think of plane partitions as 3D stack of boxes

• Also equivalent to rhombus tilings of hexagons (view as 3D projection)

10.3 May 10

• Let P (m, n, k) be the number of plane partitions of shape m × n with entries in {0, 1, . . . , k}

• By Lindstrom’s Lemma, P (m, n, k) is the determinant of a k × k matrix with m+n cij = m+i−j

• Result of Percy MacMohon:

m n k Y Y Y i + j + l − 1 P (m, n, k) = i + j + l − 2 i=1 j=1 l=1

• Plane partition is 3D analog of Young diagram in m × n × k box; equivalently a rhombus tiling of an equiangular hexagon with opposite sides of m, n, k

• Split hexagon into equilateral triangles, rhombus tiling corresponds to matching triangles

• Planar dual graph of triangular lattice is honeycomb graph

• Rhombus tilings correspond to perfect matchings on the honeycomb graph (all vertices matched)

• Let (m, n) be the number of domino tilings of the m × n rectangle

• At least one of m, n must be even

• Result of Kasteleyn: WLOG n is even, then

m n/2 b 2 c Y Y πk πl π(bm/2c + 1) N(m, n) = 4 cos2 + 4 cos2 + 2 cos if m odd n + 1 m + 1 m + 1 k=1 l=1

47 18.212 Wanlin Li

Theorem. Let m = 2k + 1 and n = 2l + 1. Remove one corner of an m × n grid; then the number of domino tilings is the number of spanning trees of the k × l grid.

• Temperley’s bijection: nodes with both coordinates even form a k × l grid, use dominos to determine direction of outgoing edges

• Number of tilings is the same for all locations of the removed box

10.4 May 15

• Plane partition with weakly decreasing entries from 0 to k

• Also the number of perfect matchings on a honeycomb graph

• Number of plane partitions equal to the number of pseudo-line arrangements (from rhombus tiling)

• Pseudo-line follows surface of rhombus tiling when interpreted as stacks of 3D cubes

• Lines in the same group do not cross, lines in diﬀerent groups cross exactly once

• Many results relate to symmetric polynomials

Fundamental Theorem of Symmetric Polynomials. Any symmetric polynomial f(x1, . . . , xn) can be written uniquely as a polynomial in the elementary symmetric polynomials.

• Deﬁnition. Semi-standard Young tableau: ﬁlling of Young diagram λ with integers 1, 2,... such that the entries weakly increase across rows and strictly increase down columns

• Deﬁnition. Schur polynomial: given λ a Young diagram with at most n rows, P #1 #n sλ(x1, . . . , xn) = x1 ··· xn where sum is over semi-standard Young tableaux SSYT with entries in [n]

• Schur polynomial for column of size k is elementary symmetric polynomial ek

Theorem. Any symmetric polynomial f(x1, . . . , xn) can be written uniquely as a P linear combination f = λ cλsλ(x1, . . . , xn) of Schur polynomials.

48 18.212 Wanlin Li

0 0 0 • Jacobi-Trudi formula: λ = (λ1, . . . , λe) with e ≤ n and λ = (λ1, . . . , λm) the conjugate partition; then sλ(x1, . . . , xn) is the determinant of the matrix M with M = e 0 ij λi+j−i

Theorem. sλ(1,..., 1), the number of semi-standard Young tableaux of shape λ, n Q λi−λj +j−i Q n+c(a) is given by det M where Mij = λ0 +j−i . This is also j−i or h(a) i 1≤i

• Reverse plane partitions: weakly increasing in both rows and columns

• Exact formula for number of reverse plane partitions for rectangular shapes; bijection to SSYTs

Lemma. RPP’s of rectangular shape n×m ﬁlled with 0, 1, . . . , k are in bijection with SSYTs of the same shape n × m ﬁlled with 1, 2, . . . , k + n by adding i to all entries in the ith row.