MATH 263A NOTES: ALGEBRAIC COMBINATORICS AND SYMMETRIC ...aaronlan/assets/diaconis... · MATH 263A NOTES: ALGEBRAIC COMBINATORICS AND SYMMETRIC FUNCTIONS7 Deﬁnition 2.12. If is

MATH 263A NOTES: ALGEBRAIC COMBINATORICS ANDSYMMETRIC FUNCTIONS

AARON LANDESMAN

CONTENTS

1. Introduction 42. 10/26/16 52.1. Logistics 52.2. Overview 52.3. Down to Math 52.4. Partitions 62.5. Partial Orders 72.6. Monomial Symmetric Functions 72.7. Elementary symmetric functions 82.8. Course Outline 83. 9/28/16 93.1. Elementary symmetric functions eλ 93.2. Homogeneous symmetric functions, hλ 103.3. Power sums pλ 124. 9/30/16 145. 10/3/16 205.1. Expected Number of Fixed Points 205.2. Random Matrix Groups 225.3. Schur Functions 236. 10/5/16 246.1. Review 246.2. Schur Basis 246.3. Hall Inner product 277. 10/7/16 297.1. Basic properties of the Cauchy product 297.2. Discussion of the Cauchy product and related formulas 308. 10/10/16 328.1. Finishing up last class 328.2. Skew-Schur Functions 338.3. Jacobi-Trudi 369. 10/12/16 37

1

2 AARON LANDESMAN

9.1. Eigenvalues of unitary matrices 379.2. Application 399.3. Strong Szego limit theorem 4010. 10/14/16 4110.1. Background on Tableau 4310.2. KOSKA Numbers 4411. 10/17/16 4511.1. Relations of skew-Schur functions to other fields 4511.2. Characters of the symmetric group 4612. 10/19/16 4913. 10/21/16 5513.1. Review 5513.2. Completing the example from last class 5513.3. Completing the example; back to the Schur functions 5714. 10/24/16 5815. 10/26/16 6116. 10/28/16 6616.1. Plane partitions, RSK, and MacMahon’s generating

function 6617. 10/31/16 7117.1. Announcements and Review 7118. 11/2/16 7418.1. Overview 7418.2. P-partitions 7418.3. The order polynomial 7519. 11/4/16 7819.1. Review 7819.2. A possibly non-politically correct example 7819.3. More on descent 7919.4. Shuffling Cards 7920. 11/7/16 8121. 11/9/16 8321.1. Algebra of the Ai 8321.2. Quasi-Symmetric Functions 8422. 11/11/16 8622.1. Application to symmetric function theory 8722.2. Connection of quasi-symmetric functions to card

shuffling 8822.3. Applications 8923. 11/14/16 9023.1. Combinatorial Hopf Algebras 9023.2. Examples of Hopf Algebras 91

MATH 263A NOTES: ALGEBRAIC COMBINATORICS AND SYMMETRIC FUNCTIONS3

23.3. What did Hopf do? 9224. 11/16/16 9324.1. Definition of combinatorial Hopf algebras 9324.2. Examples 9425. 11/18/16 9525.1. What do Hopf algebras have to do with card shuffling? 9525.2. Lyndon words 9725.3. The standard bracketing of Lyndon words 9726. 11/28/16 9827. 11/30/16 10028. 12/2/16 10228.1. Macdonald Polynomials 10228.2. Proof of Theorem 28.3 10329. 12/5/16 10629.1. Review 10629.2. Defining D 10729.3. Examples of Macdonald polynomials 10729.4. Understanding the operator D in an alternate manner 10830. 12/7/16 11030.1. School 1 11030.2. School 2 11130.3. Persi’s next project 112

4 AARON LANDESMAN

1. INTRODUCTION

Persi Diaconis taught a course (Math 263A) on Algebraic Combi-natorics and Symmetric Function Theory at Stanford in Fall 2016.

These are my “live-TEXed“ notes from the course. Conventionsare as follows: Each lecture gets its own “chapter,” and appears inthe table of contents with the date.

Of course, these notes are not a faithful representation of the course,either in the mathematics itself or in the quotes, jokes, and philo-sophical musings; in particular, the errors are my fault. By the sametoken, any virtues in the notes are to be credited to the lecturer andnot the scribe. Thanks to Lisa Sauermann for taking notes on October17, when I missed class. 1 Please email suggestions to aaronlandesman@

gmail.com.

1This introduction has been adapted from Akhil Matthew’s introduction to hisnotes, with his permission.

[email protected]

[email protected]


2. 10/26/16

2.1. Logistics.(1) Math 263A(2) Algebraic Combinatorics(3) Persi Diaconis(4) office hours Tuesday 2-4, 383 D(5) (No email)

2.2. Overview. This is a course in algebraic combinatorics and sym-metric function theory. We’ll talk about what we want to cover in thecourse. Combinatorics is pretty hard to define. It deals with thingslike sets Xn which are finite, permutations, partitions, graphs, trees.We might try to estimate |Xn|, functions T : Xn → R or

| x ∈ Xn : T(x) = y |.

Here’s a slogan: Symmetric function theory “makes math” out oflots of classical combinatorics.

We’ll try to cover(1) Chapter I of MacDonald’s book symmetric functions and Hall

polynomials(2) More things that weren’t mentioned. . .

Remark 2.1. We’ll have many digressions into “why are we studyingthis” and “what is it good for.”

2.3. Down to Math.

Definition 2.2. For n ∈ Z, a weak composition of n is a partition ofn into (a1,a2, . . .) with

∑∞k=1 ak = nwith ak ≥ 0.

Definition 2.3. Let R be a commutative ring. For example, R =Z, Q, Z[x], . . . Suppose we have infinitely many variables

x1, x2, x3, . . .

then a homogeneous symmetric function of degree n is a formalpower series

f(x1, x2, . . .) = f(x) =∑α

cαxα

where(1) α ranges over all weak compositions of n.(2) cα ∈ R

(3) xα = xα11 xα22 · · ·

(4) f(x1, x2, . . .) = f(xσ(1), xσ(2), . . . for σ ∈ S∞.

6 AARON LANDESMAN

(5) Every term has the same degree.

Example 2.4. (1) f(x) =∑∞n=1 xi is a symmetric function.

(2) f(x) = x21x2 + x21x3 + · · · + x22x2 + x22x3 + · · · is another sym-

metric function.

Definition 2.5. Let ΛnR be all symmetric functions of degree n overR.

Remark 2.6. We often omit the R subscript when it is understood orclear from context.

Remark 2.7. We have Λn ·Λm ⊂ Λn+m.

Definition 2.8. Define ΛR := ⊕∞n=0Λ

nR .

2.4. Partitions.

Definition 2.9. λ is a partition of n, written λ ` n if

λ = (λ1, λ2, . . .)

with λ1 ≤ λ2 ≤ · · · and ∑i

λi = n.

Write |λ| = n, `(λ) := the number of nonzero parts of λ .

Example 2.10. The partitions of 5 are

5, 41, 311, 32, 2111, 11111, 221 .

We will often write

λ = 1n1(λ)2n2(λ)

with ni(λ) equal to the number of parts equal to i. For example,

221 = 122.

These satisfy ∑i

ini(λ) = n.

Example 2.11. We can also draw young diagrams with dots or youngtableaux with boxes. Here is the partition 13223 = 322111 ` 10.


Definition 2.12. If λ is a partition of n, λ ′ (the transpose) is what youobtain when flipping the diagram.

2.5. Partial Orders.

Example 2.13. We can define the partial order by λ ≤ µ if λi ≤ µi forall i.

Example 2.14. One can write down a partial orders. For example,we can define a partial order on these diagrams by saying one canget from one partition to another by moving dots to adjacent rows sothat at every stage one arrives at a partition. Algebraically, the partialorder is majorization order, where λ ≤ µ if

∑ji=1 λi ≥

∑ji=1 µi for

all j.

Fact 2.15. We have λ ≤ µ ⇐⇒ µ ′ ≤ λ ′.

This isn’t too hard, but it’s a little bit finicky, and we’ll come backto proving it later.

Example 2.16. Take λ < µ if |λ| < µ or λ1 = µ1, . . . , λn = µn, λn+1 <µn+1. This is a lexicographic ordering.

2.6. Monomial Symmetric Functions. Suppose λ is some partitionλ = (λ1, λ2, . . .) and mλ =

∑α x

α, the sum over all distinct permuta-tions of λ.

Example 2.17. We have m21 =∑i<j xix

2j + x

2ixj. We have m2 =∑

j x2j . We havem11 =

∑i<j xixj.

Lemma 2.18. Formλ of size |λ| = n, these form a basis for Λn.

Corollary 2.19. We have

dimΛn = p(n).

Proof. Apply the preceding lemma.

8 AARON LANDESMAN

2.7. Elementary symmetric functions.

Definition 2.20. The elementary symmetric functions

ej =∑

i1<···<ij

xi1xi2 · · · xij ,

so e1 =∑i xi, e2 =

∑i<j xixj, . . .. For λ ` n, we define eλ = e

n1(λ)1 e

n2(λ)2 · · · .

Fact 2.21. We’ll see that eλ as λ ` n form a basis of Λ over Z.

Lemma 2.22. We have

eλ =∑µ`n

Mλµmµ

wheremλµ is the number of matrices with row sums λ and column sums µ.

Proof. Say λ = λ1 · · · λr,µ = µ1 · · ·µs.

Example 2.23. [Darwin’s data (see Persi’s paper “sequential MonteCarlo methods for statistical analysis of tables”)] Look at the set ofall tables with the same row sums and the same column sums (thisis the Mµλ) and see where Darwin’s original table fits in. The size ofthis set is sharp-P complete problem, and nobody knows the answer.

Example 2.24. Given a bipartite graph one can represent it as a 01 matrix depending on whether vertex i in the first column is con-nected to vertex j in the second column. The degrees of the verticesare the row and column sums. We show that the generating function∏

i,j

(1+ xiyj) =∑λ,µ

Mλµmλ(x)mµ(y).

You can pick off the λ,µ coefficient of the left by using a Fast Fouriertransform. So, you want to get your hands on the coefficients of xλyµin the left hand product.

2.8. Course Outline.(1) Next time, we’ll talk about the classical bases eλ,hλ,pλ.(2) We’ll discuss the Hall inner product(3) Schur functions(4) Robin Schensted Knuth correspondence(5) Character Theory of the symmetric group. Letting Rm be the

class functions of Sn we get ⊕Rm = Λ. In particular, sλ =∑χµλpµ with χ the characters of the symmetric group.

(6) Random matrix theory (relates to characters of the unitarygroup)


(7) Combinatorial Hopf algebras - can stick things together bycombinatorics and you can also pull them apart. Wheneveryou can pull things apart and stick them together, you canform a Hopf algebra. It turns out that Λ is the terminal objectin Hopf algebra. This ends up explaining all sorts of generat-ing functions.

(8) MacDonald polynomials.There will be two course projects.(1) Everybody from now until November 2, choose 10 problems.

We’ll try and combine them so we have a set of solutions.(2) At the end, there will be a list of things to do a small final pa-

per on. There are lots of interesting applications to algebraicgeometry and so on.

3. 9/28/16

Today we’ll talk about various bases for symmetric functions.

3.1. Elementary symmetric functions eλ. We defined the elemen-tary symmetric functions last time as

ei =∑

1≤i1<i2<···<ir

xi1xi2 · · · xir .

If λ is a partition, we define

eλ := eλ1eλ2 · · · =∞∏i=1

eni(λ)i .

We have

E(t) :=

∞∏i=1

(1+ xit) =

∞∑i=0

eiti,

with e0 = 1. Note thatn∏i=1

(1+ xit) =

n∑i=1

eiti.

Proposition 3.1. For any λ, we have

eλ ′ =∑µ≤λ

aµλmµ

with the ≤ order the dominating order, with aλλ = 1. Remember λ ′ is thetranspose of λ.

10 AARON LANDESMAN

Proof. Say eλ ′ =∏eλ ′i

. Writing

(xi1xi2 · · · )(xi1xi2 · · · ) · · · = xα.

with α = α1α2 · · · , xα =∏xαii and i1 < i2 < · · · < iλ ′1

, j1 < j2 <

· · · < jλ ′2. Draw a tableau of shape λ. λ ′i is the length of the ith

column of λ.

i1 j1 k1

i2 j2 k2

i3 k3

i4

All the values of r appearing in the Tableau have to be in the first rrows. Hence, the size of the first r rows are λ1+ · · ·+λr. This impliesα1 + · · ·+ λr ≤ λ1 + · · ·+ λr. Now, eλ ′ is a symmetric function. Pickany monomial appearing on the right hand side. This shows thatµ ≤ λ. The only nonzero mµ have µ ≤ λ. It’s also easy to see thataλλ = 1 using this interpretation as filling in tableau of shape λ (weneed to have the ith row filled in completely by i’s, assuming thelower rows were filled only by lower elements).

Corollary 3.2. The collection eλ form a basis for λZ.

Proof. Apply the previous proposition. The eλ ′ form a basis, as thechange of basis to mλ is upper triangular, with 1’s on the diagonal.Hence the eλ form a basis since taking the transpose is an involution.

Theorem 3.3 (Fundamental theorem of symmetric functions). Wehave Λ = Z[e1, . . .] and these ei are algebraically independent over Z.

Proof. If the ei were dependent, then there is some polynomial inthem which is 0. Writing out that polynomial would give a finitelinear combination of eλ which evaluates to 0. This contradicts thateλ form a basis for ΛZ.

3.2. Homogeneous symmetric functions, hλ.

Definition 3.4. Define

hn :=∑λ`n

mλ.


Example 3.5. So,

h1 = m1 = e1 =∑i

xi

h2 = m1 +m11 =∑i

x2i +∑i<j

xixj =∑i≤jxixj.

In general

hn =∑

1≤i1≤i2≤···≤in

xi1xi2 · · · xin

and h0 = 1.

Define

H(t) =

∞∑n=1

hiti =

∞∏i=1

(1− xit)−1.

Observe E(t)H(−t) = 1. This allows us to inductively computethe coefficients of H. So,

n∑i=0

(−1)ieihn−i = 0(3.1)

for n ≥ 1.

Example 3.6. Take n = 2. We have

e0h2 − e1h1 + e2h0 = 0.

Expanding this, we have

h2 = e21 − e2

since h0 = e0 = 1, e1 = h1. Verifying this, e21 = (∑i xi)

2 =∑i x2i +

2∑i<j xixj, and indeed h2 =

∑i≤j xixj.

Remark 3.7. The above shows that ei and hi determine each other.

Definition 3.8. Define the map

ω : Λ→ Λ

ei 7→ hi.

This is a well defined homomorphism by Theorem 3.3.

Remark 3.9. By (3.1), we have w2 = id. This implies that ω is aninvolution, so Λ = Z[h1, . . .] which is equivalent to showing thathλ := hλ1hλ2 · · · form a basis.

12 AARON LANDESMAN

We can write

hλ =∑

Nµλmµ

where the Nµλ are always non-negative integer entries.Fix N and let H =

(hi−j

)0≤i,j≤N where ha = 0 if a < 0.

Define

E =((−1)i−jei−j

)0≤i,j≤N

.

Example 3.10. When N = 2, we get 1 0 0h1 1 0h2 h1 1

We have HE = id by (3.1).

Fact 3.11. If A is invertible, then any minor of A equal the comple-mentary cofactor of AT .

Corollary 3.12 (Jacobi-Trudi identity).

Proof. If λ,µ have length ≤ p and λ ′,µ ′ have length ≤ q with p +q = N+ 1, consider the minor of H with row indices λi + p− i for1 ≤ i ≤ p and column indices µi + p − i for 1 ≤ i ≤ p, then byFact 3.11, we have

det(hλi−µj−i+1

)1≤i,j≤p

= det(eλi−λj−i+1

)1≤i,j≤q

Taking µ = 0, we have that

dethλi−i+1 = det eλ ′i−i+1.

3.3. Power sums pλ.

Definition 3.13. The rth power sum is

pr :=∑i

xri

for i ≥ 1.

Warning 3.14. p0 is not defined!


Lemma 3.15. Define

P(t) =

∞∑r=1

prtr−1.

We have P(t) = H ′(t)H(t) . and P(t) = E ′(−t)

E(t) .

Proof. We just verify the first identity.

P(t) =

∞∑r=1

prtr−1

∑r

∑i

xritr−1

=∑i

xi1− xit

=∑i

∂

∂tlog

1

1− xit

=∂

∂tlog∏ 1

1− xit

=∂

∂tlogH(t)

=H ′(t)

H(t).

Corollary 3.16 (Newton’s identities). We have

nhn =

n∑r=1

prhn−r.

and

nen =

n∑r=1

(−1)r−1pren−r

Proof. This follows immediately from expanding the equalities inLemma 3.15.

Remark 3.17. We have hn ∈ Q[p1, . . . ,pn] with pn ∈ Z[h1, . . . ,hn],and h2 = 1

2

(p21 + p2

). Observe that ΛQ = Λ×Z Q = Q[p1,p2, . . .].

Defining pλ =∏i pλi , we obtain that pλ form a basis and pj are alge-

braically independent over Q.

14 AARON LANDESMAN

We can also see that recalling ω as the map sending ei 7→ hi, wehave

ω(pn) = (−1)npn,

and

ω(pλ) = (−1)|λ|−`(λ)pλ

with |λ| the size of λ and `(λ) equal to the length of λ.

4. 9/30/16

Recall we have Λ the ring of symmetric functions, and the variousbases

mλ, eλ,hλ,pλ.

Lemma 4.1. We can write

hn =∑λ`n

1

ξλpλ

and

en =∑λ`n

tλ1

ξλpλ

where

ξλ =∏i

inini!

tλ = (−1)|λ|−`(λ) .

where ni is the number of i’s in λ.

Proof. We first claim

H(t) =

∞∑n=0

hntn = e

∑∞r=1

prtr

r .

To see this, observe it suffices to show

logH(t) =∑

prtr/r

which is equivalent to showing

H ′(t)

H(t)=

∞∑r=1

prtr−1,

which we proved last time.


The right hand side is

e∑∞r=1

prtr

r =∏r

eprt

r

r

=∏r

∞∑mr=0

(prt

r

r

)mr/mr!

=

∞∑n=0

tr∑

m1,...,∑rmr=n

∏i

pmrr1

rmrmr!

=

∞∑n=0

tn∑λ`n

pλ1

ξλ.

Remark 4.2. Here is some motivation for why we are doing thesecalculations. Let Sn be the symmetric group. Consider a typical per-mutation σ ∈ Sn. Say it “looks like?”

(1) For example, how many fixed points does it have?(2) How many cycles does it have?(3) What is the length of the longest cycle(4) What is the order of the cycle? (the order o(σ) is the smallest

k so that σk = id.)Answers:

(1) about 1(2) about logn(3) about .62n

(4) about e(logn)22 .

Definition 4.3. Let u(σ) := 1n! denote the uniform distribution for

the symmetric group.

(1) More precisely,

1

n!# σ ∈ Sn : fp(σ) = j ∼

1

ej!

where fp(σ) is the number of fixed points of σ. So,

P(fp(σ) = 0) ∼1

e.

This models a Poisson distribution.

16 AARON LANDESMAN

(2) We also have

1

n!#

σ ∈ Sn :

c(σ) − logn√logn

∼ Φ(x)

where Φ(x) =∫x−∞ e−t2/2/

√2tdt is the normal distribution,

and c(σ) is the number of cycles.(3) We also have

1

n!#

σ ∈ Sn :logo(σ) − (logn)2

2√(logn)33

∼ Φ(x).

These feature only depend on σ having a given cycle class. Thatis, they only depend on the conjugacy class of σ in Sn. Recall σ isconjugate to τ if and only if they have the same cycle type. Let ai(σ)be the number of cycles in σ of type i.

Remark 4.4. We have the following facts

k∑i=1

iai(σ) = n

fp(σ) = a1(σ)

c(σ) =∑i

ai(σ)

`(σ) = maxiai(σ) > 0

o(σ) = lcmai(σ).

Lemma 4.5 (Cauchy). We have

# σ ∈ Sn : σ has cycle type a1 . . . am =n!∏

i=1 naiai!

=n!

3λ.

Proof. Fix a1, . . . ,an. Observe that Sn acts transitively. The size ofthe cycle class is n!∏

i naiai!

.

Define the cycle indicator

Cn(x1, . . . , xn) =1

n!

∑σ∈Sn

n∏i=1

xai(σ)i ,


and C0(x) = 1. We have another function, also called the cycle indi-cator

C(t) =

∞∑n=0

tnCn(x)

Theorem 4.6 (Polya). We have

C(t) = e∑∞i=1 t

ixi/i

Proof. Observe

e∑∞i=1 t

ixi/i =

∞∏i=1

etixi/i

=

∞∏i=1

∞∑ai=0

(tixi/i

)ai 1ai!

=

∞∑n=0

tn∑

a1,a2,...,:∑i ai=n

∏ xaii

iaiai!

= C(t).

Remark 4.7. There are similar formulas for cycle factorizations ofGLn(Fq), which are q-analogs of the above formulas for Sn.

Definition 4.8. Fix θ ∈ (0,∞) . The Poisson distribution with pa-rameter theta is

Pθ(j) =e−θθj

j!

for 0 ≤ j ≤∞.

Define

Mθ(x) =

∞∑j=0

xjPθ(j)

=

∞∑j=1

xjθje−θ

j!

= e−θ+xθ.

18 AARON LANDESMAN

Definition 4.9. If

P(j), 0 ≤ j <∞is any probability distribution, then we have moments and fallingfactorial moments

Mh =

∞∑j=0

jhP(j),

Vh =

∞∑j=1

j(j− 1) · · · (j− h+ 1)P(j).

These two moments are equivalent, we can find either from theother.

Differentiating k times, and taking P(j) = Pθ(j) we see

Mhθ(x) =

∞∑j=0

j(j− 1) · · · (j− h+ 1)xj−hPθ(j).

For Pθ(j) we get vh = θh. We have

mh(x) = θkexθ−θ

Now, setting x = 1, we see

mk(1) = Vk = θk.

So, the falling factorial moments are these simple numbers.

Example 4.10.

Theorem 4.11. For any k = 1, 2, . . ., for all n ≥ k, the kth moments ofthe number of fixed points of σ ∈ Sn satisfies

1

n!

∑σ∈Sn

fp(σ)k = kth moment of the Poisson P1(j).

Proof. Consider ai(σ) = fp(σ). We want

1

n!

∑σ∈Sn

a1(σ) (a1σ− 1) · · · (a1(σ) − k+ 1)

in

Cn(x1, . . . , ) =1

n!

∑σ∈Sn

∏xai(σ)i .


Set x1 = x1, x2 = · · · = 1. Then,

Cn(x) =1

n!

∑σ∈Sn

xai(σ).

Then, ∑tncn(x) = e

tx+∑∞i=2

ti

i

= etx−t+∑∞i=1

ti

i

=etx−t

1− t.

Differentiate k times in x and set x = 1, we get∑tn1

n!

∑σ∈Sn

a1(σ)(A1(σ)−1 · · · (a1(σ) − k− 1)

=tk

1− t

= tk + tk+1 + tk+2 + · · · .

One can read off the moments from equating powers of t, since theleft hand side is the formula for the moments.

Remark 4.12. If you have two measures, and all moments are equal,then the two measures are close. This is called the method of mo-ments. Since the falling factorial moments in the above theorem areequal, the moments are equal.

The method of moments implies that: Let Pn(j) = PSn fp(σ) = j

then as n→∞, pn(j)→ 1ej!

.

The point of these calculations is the following. Exactly the samecalculations show

(1)

E(ai(σ))k =

∞∑j=0

jkP1/i(x).

So, the number of transpositions has Poisson distribution withparameter 1/2. Here, the subscript k on the expectation meansthe falling factorial moment.

20 AARON LANDESMAN

(2) The joint distribution

E(a1(σ)j1a2(σ)j2 · · ·a`(σ)j`) =∏k=1

E1/k(xjk)

for n ≥∑jii.

Remark 4.13. The fact that the moments are exactly equal is calledstabilization. This will lead to some interpretations in terms of char-acters.

5. 10/3/16

5.1. Expected Number of Fixed Points. Recall from last time wehave

Cn =1

n!

∑σ∈Sn

∏xai(σ)i .

Then we defined

C(t) =

∞∑n=0

tnCn = e∑∞ xi

ti

i .

This has a lot of information in it.

Example 5.1. Suppose c(σ) is the number of cycles of σ. Then,

c(σ) =

n∑i=1

ai(σ).

Setting all xi = x, we have

Cn(x) =1

n!

∑σ

xc(σ).

Then,

C(t) =1

(1− t)x

=

∞∑j=0

(−x

j

)(−t)j

=∑j

tj

j!x(x+ 1) · · · (x+ j− 1).

since∑i xt

i/i is the power series expansion for − log(1− t)


Therefore,

Cn =1

n!x(x+ 1) · · · (x+n− 1)

= x

(x+ 1

2

)(2+ x

3

)· · ·(n− 1+ x

n

)= E(xSn) =

∏Exxi ,

Here Sn denotes the sum, not the symmetric group. Here P(xi =

0) = i−1i and P(xi = 1) = 1

i , and

E(xSn) =

n∑j=0

xjP(Xn = j),

as in general,

E(f(Sn)) =∑

f(j)P(Sn = j)).

Here we took f(j) = xj.So,

AV(Sn) = 1+1

2+ · · ·+ 1

n∼ logn,

VAR(Sn) =

n∑n=1

1

i

(1−

1

i

)∼ logn

P

c(σ) − logn√

logn≤ x → Φ(x).

The coefficient of xj is the number of permutations with j cycles.These happen to be called sterling numbers of the first kind.

For more information, see Shepp, Lloyd, cycles of permutations.

Question 5.2. Who cares about all this stuff with fixed points?

There was a game played where someone took two decks of cardsup to n. People play this game and you get a dollar if the samenumber comes up. The question is a question of the number of fixedpoints. Monmort in 1708 proved the number of fixed points has aPoisson distribution, as we proved last time. Note that we may aswell call the cards on the first deck 1 . . . n, so the number of matchesis just the number of fixed points in a random permutation.

We also have a metric

d(π,σ) = # i : π(i) 6= σ(i) .

22 AARON LANDESMAN

See Diaconis, Goralnick, and Mulman on fixed points of permuta-tions for a classification of possible fixed points of transitive primi-tive actions of the symmetric group.

Definition 5.3. The Caley distance between two permutations

dc(σ,π) = minimum number of transpositions needed to express πσ−1.

I.e., this is the distance in the Caley graph where the vertices arepermutations and the edges join two elements differing by a permu-tations.

Exercise 5.4. We have dc(σ,π) = n− c(σπ−1), where c(σ) is the num-ber of cycles of σ.

Remark 5.5. The above two distance measures are the only two bi-invariant distances that Persi knows of.

5.2. Random Matrix Groups.

Question 5.6. For any sequence of groups, call it Gn, pick g ∈ Gn atrandom. What does it look like?

Example 5.7. Take Gn = GLn(Fq). The conjugacy classes of a ma-trix are indexed by (f, λ(f)) where f is an irreducible polynomial ofdegree d(f) and λ(f) a partition of |λ(f)|, with the restriction that∑

d(f)|λ(f)| = n.

Here λ is a function from irreducible polynomials to partitions ofSn. This is an interesting way to get your hands on the symplecticgroup. This is Jordan form where we’re not assuming the roots arein the field.

There is more information on these for general finite groups of lietype, look at J. Fulman’s thesis: Random matrix theory over finitefields.

Remark 5.8. One can do similar things for(1) On(R)(2) Un(C)(3) Sp2n(R)(4) GLn(Zp).

As long as the group is compact, it make sense to have a descriptionof conjugacy classes. If you wanted to look at this subject, you mightlook at Persi’s paper Patterns and Eigenvalues, in bulletin of the ams.


Example 5.9. In the orthogonal group has conjugacy classes indexedby its eigenvalues. So, this is asking: pick a matrix at random, whatare its eigenvalues?

In fact, G doesn’t have to be compact, but we can stick take an nup to x uniformly, and look at n =

∏p p

ap(n).

Example 5.10. For example, we can take ω(n) =∑p|n 1. This is the

number of prime divisors. So, ω(12) = 2. The Erdos Katz theoremsays

P

ω(n) − log log x√

log log x

→ Φ(t).

This can similarly be done for GLn(R) by chopping it off along com-pact subsets, say with all entries up to n, and let n→∞.

5.3. Schur Functions. Recall we have Λ the ring of symmetric func-tions, with four basesmλ, eλ,hλ,pλ.

We will work with n variables x1, . . . , xn. If α ∈Nn, say

α = (α1, . . . ,αn) ,

define

xα := xα11 · · · xαnn .

Polynomials in n variables have an action of the symmetric groupby permuting variables.

Consider ∑w∈Sn

sgn(w)w(xα) =: aα

These are alternating polynomials under the action of the symmetricgroup. That is,

σaα = sgnσaα.

In particular, aα = 0 if xi = xj for any i 6= j. Therefore,

xi − xj ≡ 0 mod aα.

Let

δ = (n− 1,n− 2, . . . , 1, 0) .

Hence, ∏i<j

xi − xj = det(xji)1≤i≤n,0≤j≤n−1

= aδ

24 AARON LANDESMAN

which is the Vandermonde determinant, divides aα.After reordering, we may as well assume α1 > α2 > · · · > αn ≥ 0.

In this case, we obtain that

α1 − 1 ≥ · · · ≥ αn −n.

and we can write α = λ+ δ.Then,

aα = aλ+δ =∑w∈Sn

sgnww(xλ+δ)

= det(xλj+(n−j)i

)1≤i,j≤n

.

Definition 5.11. The Schur function sλ(x1, . . . , xn) is

sλ(x1, . . . , xn) :=aλ+δaδ

.

Exercise 5.12. If one adds 0’s to λ one gets the same Schur function.

6. 10/5/16

6.1. Review. Last time, we had x1, . . . , xn and took α ∈ Nn. Then,we took

aα = aα(a1, . . . , xn) =∑w∈Sn

ε(w)w(xα),

where ε(w) is the sign of w.We saw that aα is divisible by aδ :=

∏i<j xi − xj, where δ = (n−

1,n− 2, . . . , 1, 0).Now, take α a strictly increasing partition, and write α = λ + δ

with λ a partition of size |λ| and of length at most n. We defined

sλ :=aλ+δaδ

,

which is a symmetric polynomial, being a quotient of two antisym-metric polynomials.

6.2. Schur Basis. The monomials are an obvious basis for the sym-metric functions. Hence is, aα form a basis for the alternating poly-nomials in n variables.

The map

Λn → ALTn

f 7→ aδf


is bijective. Since the aα form a basis in alternating polynomials theSchur functions form a basis for Λn.

Warning 6.1. This map is not degree preserving, but it does showthat the dimension of the symmetric functions in n variables of de-gree d is equal to the dimension of alternating functions of degreed+ degaδ.

Remark 6.2 (Important secret about Schur functions!). The Schurfunctions are the characters of the unitary group. That is,

sλλ`n, at most n parts

are the irreducible polynomial characters of

Un := M ∈Mn×n :MM∗ = I .

That is, given an irreducible representation ρλ of Un, we have

sλ(z1, . . . , zn) = tr(ρλ(M)),

ifM has eigenvalues z1, . . . , zn in S1.

Theorem 6.3 (Jacobi Trudi identity). We have

sλ = det(hλi−i+j

)1≤i,j≤n ,

for any n ≥ `(λ) with the convention that h0 = 1 and hj = 0 for j < 0.Additionally,

sλ = det(eλ ′i−i+j

)1≤i,j≤n

for anym ≥ `(λ ′).Example 6.4. Consider the matrix

hλ1 hλ1+1 hλ1+2hλ2−1 hλ2

. . .hλn

For example, s(j) = hj, since we have a 1 × 1 matrix. We also gets1n = en. For example,

s3,1 = det(h3 h4h0 h1

)= h3h1 − h4.

Proof. Let e(k)j be thenth elementary symmetric function in x1 · · · xk · · · xn.Introduce

M =((−1)n−ie

(k)n−i

)1≤i,k≤n

26 AARON LANDESMAN

and let α = α1, . . . ,αn ∈Nn. Let Aα =(xαij

). Let

Hα =(hαi−n+j

).

Lemma 6.5. We have

Aα = HαM.

Proof. We have

E(k)(t) =

∞∑n=0

e(k)n t

n

=

n∏i=1,i 6=k

(1+ xit)

and recall

H(t) =

n∏i=1

(1− xit)−1 =

∞∑n=1

hntn.

Therefore,

H(t)E(k)(−t) = (1− xit)−1 .

Now, look at the coefficient of tαn on each side. We haven∑j=1

hαi−n+j(t)n−1e

(k)n−j = x

αin

Now, take the determinant of both sides in the lemma. We have

aα = detHα detM.

Taking α = δ = (n− 1, . . . , 1). We get

detHδ = det(hn−i−λ+j) = 1,

since it is upper triangular with diagonal 1. Therefore, detM = aδ.Our formula says aα = detHαaδ, using that aα is the determinant ofAα by definition. Hence,

sλ =aλ+δaδ

= det (Hλ+δ) .

We can prove the other formula directly by this sort of manipulation.


6.3. Hall Inner product. In order to define the hall inner product,we will need three expansions of∏

i,j

(1− xiyj

)−1 ,

where xi,yj are two sets of variables. We have the following identi-ties.

Lemma 6.6. (1)∏i,j

(1− xiyj

)−1=∑λ

z−1λ pλ(x)pλ(y).

(2) ∏i,j

(1− xiyj

)−1=∑λ

hλ(x)mλ(y) =∑λ

hλ(y)mλ(x)

(3) ∏i,j

(1− xiyj

)−1=∑λ

sλ(x)sλ(y).

Proof. We prove these in order(1) Recall

hn =∑λ`n

z−1λ pλ(x).

We have also

H(t) =∑n

tnhn =∏i

(1− xit)−1 .

Set t = 1, and we obtain∏i,j

(1− xiyj)−1 =

∞∑n=0

hn(xy)

=∑λ

zλpλ(x,y)

=∑λ

zλpλ(x)pλ(y).

Since

pr =∑i,j

(xiyj)r =

(∑i

xri

) (yj)r .

28 AARON LANDESMAN

(2) We have ∏i,j

(1− xiyj

)−1=∏j

H(yj)

=∏j

∞∑n=0

hn(x)yrj

=∑α∈Nn

hαyα

=∑λ

hλ(x)mλ(y),

where the last equality uses that hα is symmetric, so it doesn’tdepend on the ordering of α, only on the partition λ associ-ated to α.

(3) The third part is slightly messier, but similar, and we will omitit.

Definition 6.7. We have the Hall inner product 〈, 〉 on Λ so that

〈hµ,mλ〉 = δµλ.

Lemma 6.8. Say uλ, vµ are two bases of Λ. Then,

〈uλ, vµ〉 = δλµif and only if ∏

i,j

(1− xiyj

)−1=∑λ

uλ(x)vλ(y).

We’ll fill in the proof of this lemma next time.

Lemma 6.9. With respect to the Hall inner product, pλ form an orthogonalbasis and the Schur functions sλ form an orthonormal basis.

Proof. This is immediate from Lemma 6.8 and Lemma 6.6.

Remark 6.10 (Important secret fact). We have

〈f,g〉 =∫Un

f(m)g(m)dm,

where by integrating a function we mean integrating the function ofthe eigenvalues of the corresponding matrix.


7. 10/7/16

7.1. Basic properties of the Cauchy product. Last time we saw∏ (1− xiyj

)−1=∑λ

z−1λ pλ(x)pλ(y)(7.1)

=∑λ

mλ(x)hλ(y)(7.2)

=∑λ

sλ(x)sλ(y).(7.3)

Remark 7.1. We had a fairly uninspiring manipulation to prove this,but Dan bump’s textbook has a very helpful group theoretic/representationtheoretic argument for this.

We defined the Cauchy inner product by

〈mλ,hµ〉 = δλ,µ.

Recall from last time, we stated

Proposition 7.2. If uλ , vλ are two bases of Λn. The the following areequivalent:

(1) We have 〈uλ, vµ〉 = δλµ.(2) The Cauchy product

∏ (1− xiyj

)−1=∑λ uλ(x)vλ(y).

Proof. Write

uλ =∑ρ

aλρhρ

vµ =∑σ

bµσmσ.

Observe (1) says ∑ρ

aλρbµρ = δλµ

and (2) says ∑λ

uλ(x)vλ(y) =∑λ

mλ(x)hλ(y).

because ∑λ

aλρbλσ = δρσ.

30 AARON LANDESMAN

So, these two rephrasings are equivalent becauseAB = I ⇐⇒ BA =I.


〈pλ,pµ〉 = zλδλµ

and we have that the sλ are orthonormal.

Proof. Apply Equation 7.1 and Proposition 7.2.

Corollary 7.4. The inner product 〈, 〉 is symmetric and positive definite.

Proof. Write f =∑λ aλ. Then,

〈f, f〉 =∑

a2i .

Recall the map w : Λn → Λn with

ω(hλ) = eλ

ω(pλ) = ±pλ.

This implies

〈pλ,pµ〉 = 〈ωpλ,ωpµ〉〈u, v〉 = 〈ω(u),ω(v)〉.

soω preserves norms (it is an orthogonal transformation). Now, ourthree identities yield∏ (

1+ xiyj)=∑λ

ελz−1λ pλ(x)pλ(y)

=∑λ

mλ(x)eλ(y)

=∑λ

sλ(x)sλ ′(y).

7.2. Discussion of the Cauchy product and related formulas. We’llnow have a discussion on the meaning of the above derived formu-las.

Question 7.5. What do these formulas mean?


(1) Let’s begin by working with x1, . . . , xn. We have

E(x) =

n∏i=1

(1+ xit)

=

n∑n=0

entn.

To see what this means, multiply both sides by 1(1+t)n

. We get

n∏i=1

(1

1+ t+ xi

t

1+ t

)i =

n∑i=1

1(nr

)en(nr

)(t

1+ t

)r1

(1+ t)n−r.

Now, let’s interpret this probabilistically. Flip a coin n times.Say the probability of getting heads is t

1+t . Let x = (x1, . . . , xn)be a binary pattern. Then, the left hand side is

E(∏

xxii ) = E(x

x) =∑ξ

∏xiξP(ξ).

On the other hand, the part of the right hand side(n

r

)(t

1+ t

)r(1

1+ t

)n−ris the probability that you get r successes out of n flips. Also,

1(nr

)er = ErFD (xz)

=∑z∈FD

∏xzip(z).

is the Fermi Dirac statistic (here z is a random binary vector):one drops r balls into n spot with balls in distinct spots. Thefact that this does not depend on t tells us this

∑xi is a suf-

ficient statistic, meaning you only have to keep track of thenumber of 1’s to estimate the probability, not the whole dis-tribution.

(2) Let’s now look at the generating function for the h’s. We have

H(A) =∏

(1− xit)−1 =

∞∑n=0

hn(t)tn.

32 AARON LANDESMAN

Now, multiply both sides by (1− t)n. We have∏ 1− t

1− xit=∑ 1(

n+r−1r

)hr(t)(n+ r− 1

r

)· (1− t)n tr.

To interpret this, recall the geometric distribution with pt(j) =tj(1− t) (it measures the chance you get j heads before gettinga tail). The left hand side generates x1, x2, . . . , xn independentgeometric distributions with parameter t.

The left hand side is

E(xx) =∑z∈Nn

∏xzii P(x = z).

The right hand side is a sort of dot product of the negativebinomial distribution with hr. Here, the negative binomialdistribution is(

n+ r− 1

r

)tn (1− x)n = Pn,t(x).

This measures the chance we get r heads before the nth tail.The coefficient

1(n+r−1r

)hr(t)is the Bose Einstein distribution. That is, this is the generatingfunction for Bose-Einstein statistics.

Here, Bose-Einstein statistics measure dropping r unlabeledballs into n boxes so all configurations are equally likely. Forexample, if you have 10 electrons being put into 2 slots, thereare 11 possibilities of the form i, 10 − i for 0 ≤ i ≤ 10. Inthe Bose-Einstein distribution, all possibilities have chance 1

11 .Chartarjee and Persi are writing a paper on Bose Einstein dis-tributions. The identities above say that by randomizing rwecan make the coefficients independent.

8. 10/10/16

8.1. Finishing up last class. Recall from last time we had

eλ(x1, . . . , xc) = c(λ)E(xc11 · · · x

ctt |Ri = λi)

with λ = λ1 · · · λn. Here the normalizing constant c(λ) is some prod-uct of binomial coefficients. Here, P(Xij = 1) = p and cj =

∑j xij,Ri =∑

j xij.


Similarly,

hλ(x1, . . . , xc) = c ′(λ)E(xc11 · · · x

ccc |Ri = λi

)where P(Xij = h) = ph(1− p) for 0 ≤ n <∞.

Last class, we proved∏i,j

(1+ xiyj

)=∑λ

sλ(x)sλ ′(y).

In Macdonald exercise 1 on orthogonality, he says to set

y1 = y0 . . . yn = t.

Then, he obtains the identity

E(t)n =∑λ

(n

λ

)sλ(x)t

|λ|

where (X

λ

)=∏λ

n− c(x)

n(x)

where

c(x) = column of x - row of xh(x) = hook length of x.

Since this holds for all n, we can view n as a variable, and we obtain

E(t)X =∑λ

(X

λ

)sλ(x)t

|λ|

Similarly

H(t)X =∑λ

(X

λ ′

)sλ(x)t

|λ|

Remark 8.1. These(nλ

)generalize the binomial formula since if λ just

has one shape, this becomes the usual binomial formula.

8.2. Skew-Schur Functions.

Definition 8.2. A Young Tableau is semi-standard is the numbersare weakly increasing from left to right and strictly increasing fromtop to bottom.

34 AARON LANDESMAN

Example 8.3. An example of a semi-standard young tableau of λ =4322 is

1 1 2 3

2 3 4

3 4

5 5

We have sλ =aλ+δaδ

. Combinatorialists think that

sλ(x1, . . . , xn) =∑t

xT

where T ranges over all semi-standard young Tableau of shape λ.

Example 8.4. All semi-standard young tableau of shape 2, 1with n =3 are

1 1

2

1 1

3

1 2

2

1 2

3

1 3

2

1 3

3

2 2

3

2 3

3


It turns out we get

s21 = x21x2 + x

31x3 + x1x

22 + 2x1x2x3 + x1x

23 + x

22x3 + x2x

23

= 2m13 +m21.

Say λ,µ are two partitions we can express any symmetric functionas

f =∑λ

〈f, sλ〉sλ.

Define sλ/µ to satisfy

〈sλ/µ, sν〉 = 〈sλ, sµsν〉.

Definition 8.5. The Littlewood Richardson coefficients are cλµν sothat

sµsν =∑λ

cλµνsλ.

We have

sλ/µ =∑ν

cλµνsν

with zλµν = 0 unless |λ| = |µ|+ |ν| implies sλ/µ = 0 unless |λ| ≥ |µ|.We will soon show sλ/µ = 0 unless µ ≤ λ, meaning λn ≥ µn for alln.

Example 8.6. If

µ =

λ =

Then λ/µ is the shape of λwith µ removed.

36 AARON LANDESMAN

8.3. Jacobi-Trudi.

Theorem 8.7 (Jacobi-Trudi). We have

sλ/µ(x) =∑w∈Sn

ε(w)hλ+δ−w(µ+δ)

= det(hλi−µj−i+j

)1≤i,j≤n

.

Proof. Consider∑λ

sλ/µ(x)sλ(y) =∑λν

cλµνsν(x)sλ(y)

=∑ν

sν(x)sµ(y)sν(y)

= sµ(y)∑ν

sν(x)sν(y)

= sµ(y)∑ν

hν(x)mν(y).

So, ∑λ

sλ/µ(x)sλ(y)aµ+δ(y) =∑n

uhν(x)mν(y)aµ+δ(y)

=∑α

hα(x)∑w∈Sn

yα+w(µ+δ)

Therefore, sλ/µ(y) is the coefficient of yλ+δ on the right hand sideabove. Therefore,

sλ/µ(x) =∑w∈Sn

ε(w)hλ+δ−w(µ+δ)

= det(hλi−µj−i+j

)1≤i,j≤n

.

Remark 8.8. If µ = 0 this reduces to the old Jacobi-Trudi identity.

Corollary 8.9. By duality,

sλ/µ = det(eλ ′i−µ

′j−i+j

).

w(sλ/µ) = sλ ′/µ ′ .

Proof. Follows from Jacobi-Trudi. The point is that the duality mapω switches sλ and sλ ′ and also switches e’s and h’s.


Corollary 8.10. We have sλ/µ = 0 unless µ < λ.

Proof. Follows from Jacobi-Trudi, more details omitted.

Say λr < µr for some r. Then, λi ≤ λr < µr ≤ µj for 1 ≤ j ≤ r ≤ i.This implies λi − µj − i+ j < 0 for these i and j. This implies

hλi−µj−i+j

has a zero block in the lower left hand (n− r+ 1)× r corner.

9. 10/12/16

9.1. Eigenvalues of unitary matrices. Consider the group of unitarymatrices

Un = M ∈Mn×n :MM∗ = I .

Say a given unitary matrix has eigenvalueseiθjnj=1

. The Weyl den-sity of the eigenvalues is

f(θ1, . . . , θn) =1

2π)nn!

∏1≤j<k≤n

|eiθj − eiθk |2.

with respect to dθ1, . . . ,dθn. We could look at

Tn(M) =∑j

eiθj

Tn(M2) =

∑j

e2iθj .

Theorem 9.1 (Diaconis and Dishahshaham). The probability that Tn(Mi) ∈Bi tends to (as n→∞)

n∏j=1

P(zj ∈ Bj

),

where zj are independent standard complex Gaussians,

P(zj ∈ B

)=

∫B

e−|z|2

πdz.

Remark 9.2. We now describe the proof of this to illustrate the utilityof Schur functions.

Proof. The idea for this proof is to use the method of moments.

38 AARON LANDESMAN

Definition 9.3. If µn are probability measures on R, we say µn con-verges to µ if for every bounded probability measure f,∫

f(x)µn(dx)→ ∫ f(x)µ(dx).If µ has a density, this is equivalent to saying µn(ball)→ µ(ball).

If µn have finite moments with

µn(k) =

∫R

xkµn(dx)

and µn(k) goes to µ(k) as n → ∞, this means µn goes to µ, pro-vided µ is determined by its moments. For example, µ is alwaysdetermined by its moments if

∞∑k=1

1

µ(2k)1/2k =∞.

For µn on C, we need

µn(a,b) =∫

C

zaz−bµndz.

Example 9.4. For µ the standard normal moments are∫C

zaz−be−|z|2/πdz = δaba!

ForM ∈ Un, we have

Tn(Mj) =

n∑k=1

eiθk = Pj(eiθ1 , . . . , eiθn) = Pj(M).

We have ∫Un

k∏j=1

(Tn(Mj))aj

k∏j=1

Tn(Mj)bjdM =

∫Un

PλPµdM

= 〈Pµ,Pλ〉

where λ,M are two partitions.Using that the Schur functions are the characters of the unitary

group,

〈sλ, sµ〉 = δλµ.


We know also that

pλ =∑ν

χν(λ)sν

pµ =∑ν ′

χν ′(λ)sν ′

where

χν(λ)

is the νth irreducible character of sn at the λth conjugacy class. (We’llsee this soon in the course). Therefore,∫

Un

k∏j=1

(Tn(Mj))aj

k∏j=1

Tn(Mj)bjdM = 〈Pµ,Pλ〉

=∑ν,ν ′

χν(λ)χν ′(µ)〈sν, sν ′〉

=∑ν

χν(λ)χν ′(µ)

=∏i

naiai!δλµ.

where∏naiai! is the size of the conjugacy class of λ in Sn (ai are the

cycle lengths). But, note that∏naiai! are the moments of the normal

distribution z1√2z2 · · ·

√nzn. We have an exact equality provided

n ≥∑i

iai,∑i

ibi.

This concludes the proof of the theorem on the distribution of theUn. See the paper by Persi and Evans for more details.

9.2. Application.

Remark 9.5. Here is an application: In the talk Persi gave yesterday,take I = (a,b) on S1. Pick a matrix at random and count the numberof eigenvalues in I. This is

∑ni=1 δIe

iθj . So, the number of eigenvaluesin an interval should be proportional to the length of the interval. Wecan ask how they fluctuate.

Theorem 9.6 (Wieand). Let XI =# eigenvalues in I−n2π(b−a)√

logn, then

XI → n(0, 1)

40 AARON LANDESMAN

where n(0, 1) denotes the normal distribution with mean 0 and variance 1.

Question 9.7. What are the correlation between XI,XJ?

It turns out they are 0 unless J and I have an endpoint agreeing, inwhich case it is −1/2 if I∩ J is a point, and 1/2 if I ⊂ J.9.3. Strong Szego limit theorem.

Definition 9.8. A Toeplitz operator is a matrix of the forma b c de a b cf e a bg f e a

.

meaning that the entries are constant on diagonals.

Question 9.9. What are the eigenvalues of Toeplitz matrices?

Consider

|z| = 1, f(z) =∑j∈Z

f(j)zj.

so that f(j) = f(−j). Take the Toeplitz matrixf(0) f(1) · · · f(n− 1)

f(−1). . . . . . ...

... . . . . . . f(1)

f(−n+ 1) · · · f(−1) f(0)

.

Then, taking real eigenvalues λ1, . . . , λn ofM take

µn(m) =1

n

n∑j=1

δλj

Theorem 9.10 (Weak Szego). We have µn(m) → uf−1

with u uniformon S1. That is,

1

n

n∑j=1

ψ(λj)→ 1

2π

∫2π0ψ(f(eiθ)

)dθ

where ψ is any bounded continuous function.

Remark 9.11. This doesn’t look like eigenvalues of random matrices,but this turns out to be the same as saying that the eigenvalues arejointly normal.


10. 10/14/16

Today we’ll talk about skew Schur functions.

Definition 10.1. Define

〈sλ/µ, sν〉 := 〈sλ, sµsν〉.where

sµsν =∑λ

cλµνsλ

sλ/µ =∑λ

cλµνsν.

We have the Jacobi Trudi identities

sλ/µ = det(hλi−µj−i+j

)1≤i,j≤n

with `(λ, `(µ) ≤ n and

sλ/µ = det(hλ ′i−µ

′j−i+j

)here with `(λ ′), `(µ ′) ≤ n.

Lemma 10.2. We have sλ/µ = 0 unless µ ≤ λ, meaning µi ≤ λi for all i.

Proof. We have the following manipulations with three sets of vari-ables ∑

λ,µ

sλ/µ(x)sλ(z)sµ(y) =∑λ,µ,ν

cλµνsν(x)sλ(z)sµ(y)

=∑µ,ν

sν(x)sµ(y)sµ(z)sν(z)

=∑ν

sν(x)sν(z) ·∑µ

sµ(y)sµ(z)

=∑ν

sν(X)sν(z)∏i,j

(1− yizj

)−1=∏i,j

(1− xizj

)−1∏i,j

(1− yizj

)−1=∑λ

sλ(x,y)sλ(z).

This implies

sλ(x,y) =∑µ

sλ/µ(x)sµ(y)(10.1)

42 AARON LANDESMAN

by taking the coefficient of sλ(z) in∑λ,µ

sλ/µ(x)sµ(y)sλ(z) =∑λ

sλ(x,y)sλ(z).

Lemma 10.3. In fact, we have

sλ/µ(x,y) =∑µ≤ν≤λ

sλ/ν(x)sν/µ(y).

Proof. Using (10.1), we have∑µ

sλ/µ(x,y)sµ(z) = sλ(x,y, z)

by taking xy for x, and z for y in (10.1). Similarly,

sλ(x,y, z) =∑ν

sλ/µ(x)sν(y, z).

This implies∑µ

sλ/µ(xy)sµ(z) = sλ(x,y, z)

=∑ν

sλ/ν(x)sν(y, z)

=∑ν,µ

sλ/ν(x)sν/µ(y)sµ(z).

now, taking the coefficient of sµ(z), we get∑µ

sλ/µ(x,y) =∑ν

sλ/ν(x)sν/µ(y).

Remark 10.4. More generally, if x1, . . . , xn are n sets of variables,

sλ/µ(x1, . . . , xn) =

∑ν0≤ν1≤···≤νn

n∏i=1

sνi/νi−1(xi)(10.2)

where the sum is taken over all sequences of partitions µ = ν0 ≤· · · ≤ νn = λ.


10.1. Background on Tableau.

Definition 10.5. A semi-standard Young tableau (SST) in 1, . . . ,nis a placement of 1, . . . ,n into boxes of shape λ which are weaklyincreasing in each row and strictly increasing in each column.

One can make a similar definition for λ/µ.

Example 10.6. Taking λ = 6, 4, 2, 2we have the semi-standard youngtableau

1 1 2 2 4 4

2 2 4 4

3 3

4 4

Definition 10.7. A semi-standard young tableau is standard if it usesthe numbers 1, . . . ,nwithout repetition, and has n boxes.

In (10.2), take xi = xi. The left hand side is sλ/µ(x1, . . . , xn) and theright hand side sνi/νi−1(x) by the Jacobi Trudi restrictions, sλ/µ(x) =0 unless λ/µ is a row hook shape (meaning at most one box in eachcolumn). Then, sλ/µ(x) = x|λ|−|µ|. In general, (10.2) has right handside product xα11 · · · x

αnn where

|νi|− |νi−1| = αi.

and νi are a sequence of row hooks.

Corollary 10.8. This tells us

sλ/µ(x1, . . . , xn) =∑T

xT

where T is a semi-standard young tableau of shape λ/µ in n variables, with

xT =∏i

xTii .

and Ti is the number of i’s in T . In particular,

sλ(x1, . . . , xn) =∑

T SST of shape λ in n variables

xT .

Proof. Follows from the above discussion.

44 AARON LANDESMAN

10.2. KOSKA Numbers. Fix λ ≥ µ,n,ν. we define the Koska num-ber

Kλ/µ,ν

to be the number of SST of shape λ/µwith content ν. We have

sλ/µ(x1, . . . , xn) =∑ν

Kλ/µ,νmν.

Example 10.9. Take λ = 6, 4, 2, 2. Then

1 1 1 2 2 3

2 2 3 3

3 3

4 4

has content 4443333332222111 = 13243543. Take µ = 4, 4, 2 and thefilling of λ/µ

2 3

4 4

has content 4432.

We have

〈sλ/µ,hν〉 = Kλ/µ,ν

which implies

< sλ, sµhν >= Kλ/µ,ν.

So, Pieri’s formula says

sµhν =∑λ

Kλ/µ,νsλ.

Example 10.10. If ν = r is a partition of rwith one part, then

Kλ/µ,r = 0

unless λ/µ is an r row hook shape. In this case, Kλ/µ,r = 1.In particular, we have

sµhν =∑

λ/µ is an r row hook shape

sλ.


Example 10.11. Say µ = 33, r = 2. What λ are possible if λ/µ is a 2row hook shape? We can have λ equal to

equal to

or equal to

11. 10/17/16

11.1. Relations of skew-Schur functions to other fields.(1) Persi didn’t show how to prove Toeplitz matrix stuff from the

Schur functions. If you want to know more about ToeplitzSchur functions, see Persi’s paper with Dan Bump on ToeplitzMinors.

(2) The last thing we did in this class was show how to movebetween skew Schur functions to semi-standard tableau. Thatis, we took

sλ/µ(x1, . . . , xn) =∑T

xT

where T varies over all semi-standard tableau of shape λ/µin the variables 1, . . . ,n.

(3) m × n tableaus correspond to plane partitions, with blockscoming out of the diagram. If you look at a plane partition ata distance, it looks like a tiling of the outline by Rhombus byRhombi. So, these three viewpoints turn out to be equivalent.This has an enormous literature. The probabilistic part endsup having a sort of circle, and each of the corners of the circleare in some way of the same type (the Arctic Circle theorem).Vadim Goren proved the first nice theorems about probabilis-tic viewpoints on distributions of number in shapes λ.

46 AARON LANDESMAN

(4) One of the things you hear a lot about in symplectic geometryis the moment map. The distribution you get is called theDeusman Heckman measure, and has something to do withthese skew tableau and their probabilistic distribution.

Remark 11.1 (Random Remark). The skew Schur shapes index therepresentations of the affine Hecke algebra.

11.2. Characters of the symmetric group. Let G be a finite groupand let A be a commutative ring with unit (often C, R, Z,K). Iff,g : G→ A, define the inner product

〈f,g〉 = 1

|z|

∑x∈G

f(x)g(x−1).

Definition 11.2. Let G denote the irreducible characters of G over C,which are an orthonormal basis with respect to the inner product 〈, 〉above.

Definition 11.3. Given H ⊂ G, we have restriction

ResHG(f) : H→ A

sending f : G→ A to the composition H→ G→ A.

Definition 11.4. Given H ⊂ G, we have induction

IndGH(f)

Sending a function f : H→ A to a function G→ A by sending f to asum of functions over cosets (see Serre’s rep theory of finite groupsbook).

Definition 11.5. Let G = Sn. Define

ψ : Sn → Λn

w 7→ Pρ(w)

where if w has cycle type 1a1(w), . . . ,nan(w) then ρ(w) :=∏iai(w).

Remark 11.6. If u ∈ Sm, v ∈ Sn, acting on different sets of variables,then u× v ∈ Sm × Sn ⊂ Sm+n. (we will think about embedding thisas the firstm blocks and then the last n blocks). u× v has cycle typeρ(u)∪ ρ(w), so the map ψ is multiplicative. That is,

ψ(u× v) = ψ(u)ψ(w)because the power sums are multiplicative. That is,ψ is a homomor-phism.


Definition 11.7. Let Rn denote the class functions on Sn,

Rn :=f : Sn → Z : f(u−1vu) = f(v)

,

the center of the group algebra. We have

Rn = 〈xλ〉λ`m〈δλ〉λ`n.

For each λ there is an character, and we call the character xλ. Here,δλ is, for any partition of cycle type λ, we have

δλ(w) =

1 if ρ(w) = λ0 otherwise.

Definition 11.8. Let, R := ⊕∞n=0R

n. Here, S0 = e is the group with 1element and R0 = Z. Now, R has a product. If f ∈ Rm,g ∈ Rn, we letf · g define

f · g := IndSm+nSm×Snf× g.

on each graded piece, and then extend by linearity.

Remark 11.9. Under the above definition, R is a commutative, as-sociative (by a standard fact about induction) graded algebra withunit. If f =

∑∞n=0 fn,g =

∑∞n=0 gn then

〈f,g〉 :=∑n

〈fn,gn〉.

Goal 11.10. We would like to show in the strongest sense R is iso-morphically isometric to Λ.

Definition 11.11 (Characteristic map). We have a characteristic map

Ch : R→ ΛC = Λ⊗Z C

f 7→ 〈f,ψ〉,where f ∈ Rn. Note that

〈f,ψ〉 = 1

n!

∑w

f(w)ψ(w)

=∑ρ`n

f(ρ)pρz−1ρ .

where n!/zρ is the number of elements with partition size equal toρ. One then extends by linearity.

Lemma 11.12. In fact, Ch(•) is a ring map.

48 AARON LANDESMAN

Proof. That is, we have

Ch(f · g) = 〈IndSn+mSm×Snf× g,ψ〉Sn+m= 〈f× g,ResSn+mSn×Smψ〉Sn×Sm= 〈f,ψ〉Sm〈g,ψ〉Sn .

Lemma 11.13. For f,g ∈ Rn, we have

〈Ch(f),Ch(g)〉 = 〈f,g〉Snand for f,g in different graded pieces, 〈Ch(f),Ch(g)〉 = 0.Proof. We have

〈Ch(f),Ch(g)〉 =∑ρ`n

f(ρ)g(ρ)z−1ρ

=∑w

f(w)g(w)z−1ρ(w)

zρ

n!

= 〈f,g〉Snwhere here we used that pρ were orthogonal and

〈pλ,pµ〉 = zλδλµ.

This starts our goal of showing the characteristic map is a isomor-phism, since being an isometry on each graded piece implies it isinjective.

Theorem 11.14. The characteristic maps is an isometric isomorphism of Ronto ΛC.

Proof. Let ηn be the trivial character of Sn. This is 1 on each permu-tation. So,

Ch(ηn) =∑ρ`n

z−1λ Pρ = hn,

as we proved when we introduced the hλ basis. For λ = λ1, . . . , λr `n, define

ηλ := ηλ1 · · · ηλr= IndSnSλ1×···Sλr

(triv).

Since the map is multiplicative, we have

Ch(ηλ) = hλ.


Finally, define

χλ := det (ηλn−i+1)1≤i,j≤n ∈ Rn.

We’ll finish the proof next time, but Ch(χλ) = sλ.

12. 10/19/16

Today, we’ll back up, and cover character theory more coherently.

Definition 12.1. Let G be a finite group. Let Q be a probability dis-tribution meaning Q(g) ≥ 0, with

∑g∈GQ(g) = 1.

Example 12.2. This is an example which is morally about card shuf-fling by making repeated transpositions. Take G = Sn, let w ∈ Sndenote a permutation. Let

Q(w) =

1n if w = id2/n2 if w = (i, j)0 otherwise

Note, this is a probability distribution because

2

n+

(n

2

)2

n2= 1.

We have

Q ∗Q(g) =∑η

Q(η)Q(gη−1).

Similarly,

Q∗h = Q ∗Q∗(h−1).

We have u(g) = 1|G|

.

Theorem 12.3 (Poincare 1912). We haveQ∗k → u if and only ifQ is notconcentrated in a coset of a subgroup.

(For example if we made the above measure vanish on the identity,convolutions would be concentrated on either even or odd permu-tations, since the sign would alternate at each step.)

Now,

|Q∗h − u| := maxA⊂G

|Q∗h(A) − u(A)|

=1

2

∑g∈G

|Q∗h(g) − u(g)|.

50 AARON LANDESMAN

The math problem is: given g,Q, ε > 0, how large do you need totake k so that |Q∗h − u| < ε.

One way to approach these problems is to use Fourier analysis.Let G be a finite group.

Definition 12.4. A representation ρ of G is a homomorphism

ρ : G→ GL(V)

so that

ρ(st) = ρ(s)ρ(t).

We denote dρ := dimV .

Example 12.5. The trivial representation is the representation withdimV = 1 and ρ(g) = 1.

Example 12.6. If G = Sn,V = R, we have the sign representation isthe 1-dimensional representation satisfying

ρ(w) = sgn(w).

Example 12.7. If G = Sn and V = Rn with basis

〈e1, . . . , en〉,then ρ(w) is a linear map with

ρ(w)(ei) = ew(i).

This is a permutation matrix(δiw(i)

).

Example 12.8. If V = R(n2) with basis ei,j for i, j an unordered pair,we have a representation of the symmetric group defined by

ρ(w)ei,j = ew(i),w(j).

Definition 12.9. For G an arbitrary group, a representation ρ is irre-ducible if there does not exist a strict nontrivial subspace V0 ⊂ V sothat

ρ(g)V0 ⊂ V0for all g ∈ G.

Example 12.10. The n dimensional representation of Sn above is notirreducible for n ≥ 2. We can consider the subspace

V0 :=

v ∈ Rn :

∑i

vi = 0

which is a strict nontrivial subspace fixed by the action of Sn.


Definition 12.11. If Q is a probability distribution on G and ρ a Grepresentation, we define the Fourier transform

Q(ρ) =∑g∈G

Q(g)ρ(g).

Lemma 12.12. We have

Q ∗Q(ρ) = Q(ρ)2.

Proof.

Exercise 12.13. Prove this. Hint: It follows from the definition.


u(ρ) =

1 if ρ = triv

0 if ρ is nontrivial and irreducible,

where u is the uniform distribution.

Proof.

Exercise 12.15. Prove this. Hint: Use Schur’s lemma.

We would like to show Q∗h → u by showing Q(ρ)h → 0. For thiswe will need Fourier inversion and the Plancherel theorem.

Theorem 12.16 (Fourier Inversion). If f : G→ C is a function, we have

f(g) =1

|G|

∑ρ∈G

dρ tr(ρ(g−1)f(ρ)

)where G is the set of all irreducible representations.

Proof. Observe that both sides are linear in f. So, it’s enough to provethis for a basis. We will show it for the basis of δ functions. That is,we only need show for every g0,

δg0(g) =1

|G|

∑ρ∈G

dρ tr(ρ(g−1)δg0(g0)

)=1

|G|

∑ρ∈G

dρ tr(ρ(g−1)ρ(g0)

)=1

|G|

∑ρ∈G

dρ tr(ρ(g−1g0)

)

52 AARON LANDESMAN

and this is the decomposition of the regular representation, definedby V = L2(G) with Tg(f)(x) = f(g−1x). That is, we are using

Theorem 12.17. For V the regular representation, we have

V = ⊕ρ∈GdρVρ.

Taking characters on both sides yields the result, since the charac-ter of the regular representation is 0 if g 6= id and |G| if g = id.

Remark 12.18. In the case G is a cyclic group, this recovers the usualFourier inversion on a circle.

Theorem 12.19 (Plancherel). If f,h : G→ C are two functions, then

∑g

f(g)h(g−1) =1

|G|

∑ρ∈G

dρ tr(f(ρ)h(ρ)

).

Proof. Both sides are linear in f so we only need verify this for δ func-tions. That is, take f(g) = δg0(g). We only need verify this is thiscase. Then,

f(ρ) = ρ(g0).

Then, we have to check

h(g−10 ) =1

|G|

∑ρ∈G

dρ tr(ρ(g0)h(ρ)

).

This holds by Fourier inversion above.

We would like to apply this to bound |Q∗h − u|.


4|Q∗h − u|2 ≤∑

ρ∈G,ρ 6=triv

dρ|Q(ρ)h|

with

|M|2 = tr(MM∗).


Proof. We have

4|Q∗h − u|2 =(∑

|Q∗h(g) − u(g))2

≤ |G|∑g

|Q∗h(g) − u(g)|2

=∑

ρ∈G,ρ 6=triv

dρ tr(Q(ρ)h(Q(ρ)h)∗

)=∑ρ∈G

dρ|Q(ρ)h|.

using Cauchy Schwartz in the second line above, we used Plancherelin the third line and that the Fourier transform of the uniform is 0 forρ nontrivial.

Lemma 12.21 (Schur’s lemma in disguise). Suppose G is a group witha probability measure Q which is a class function (constant on conjugacyclasses).

Example 12.22. For example, take

Q(w) =


Then, if ρ is irreducible,

Q(ρ) = cρid.

Proof. We have

ρ(g−1)Q(ρ)ρ(g) =∑g

Q(x)ρ(g−1xg)

= Q(ρ).

This says this matrix commutes with the action of the group. So, bySchur’s lemma, the only matrices commuting with the action of thegroup are scalars.

Remark 12.23. Further, the constant cρ can be computed by takingthe trace of both sides.

54 AARON LANDESMAN

Example 12.24. Take Q on Sn the distribution

Q(w) =


We have

Q(ρ) =1

nρ(I) +

2

n2

∑ρ(i, j) = cI.

Hence,dρ

n+2

n2

(n

2

)χρ(1, 2) = cdρ.

This implies

c =1

n+n− 1

n

χρ(1, 2)dρ

.

We want to bound

|Q∗k − u|2 ≤∑

ρ∈G,ρ 6=triv

d2ρ

(1

n+n− 1

n

χρ

dρ

)2k.

So, we have one term from the sign representation, we get χρ(i, j) =1, dρ = 1, then (

1

n−n− 1

n

)2k=

(1−

2

n

)2k.

Another term is from the n − 1 dimensional representation withdρ = (n− 1). We get χρ(1, 2) = n− 3. Then, this term becomes

(n− 1)2(1

n+n− 1

n

n− 3

n− 1

)2k= (n− 1)2

(1−

2

n

)2k≤ e2 logn−2k 2n

≤ e−2c

where k = 12n (logn+ c) in the last line above. It turns out that these

terms dominate the whole sum, and this yields the desired answer.

The answer to our card shuffling question ultimately gives

1

2e−c ≤ |Q∗k − u| ≤ 2e−c

with k = 12n (logn+ c).


13. 10/21/16

13.1. Review. Last time, we let G be a finite group. We define Qa probability measure, we defined Q∗k, a convolution, we definedthe uniform measure u. We also defined the distance between theconvolution and the uniform measure,

|Q∗k − u| =1

2

∑g∈G

|Q∗k(g) − u(g)|.

We found

u|Q∗k − u|2 ≤∑ρ

dρ|Q(ρ)|2k2 ,

with |M|2 = tr(MM∗).

13.2. Completing the example from last class. We were looking atthe example of G = Sn and

Q(g) =

1n if g = id2n if g = (i, j)0 otherwise

We had that Q is then a class function, so

Q(ρλ) = cρλI,

c =1

n+n− 1

n

χ(i, j)fλ

with

χλ(1, 2) = tr ρλ(1, 2)

fλ = dim ρλ = χλ(id).

We then had to bound

u|Q∗k − u|2 <∑

λ`n,|λ|=n

(fλ)2 ( 1

n+n− 1

n

χλ(1, 2)fλ

)2k.

To bound this sum say 1n is about 0, n−1n is about 1. We have

∑ (fλ)2

=

n!. If χλ(1,2)fλ

= 12 , then,

∑λ`n,|λ|=n

(fλ)2 ( 1

n+n− 1

n

χλ(1, 2)fλ

)2k.

56 AARON LANDESMAN

is roughly

n!p(n)1

2

2k

= en logn+√n−2k log 2

which is small if k is roughly 10n logn. But, if λ = (n− 1, 1) is thisparticular partition, we have

χn−1,1(1, 2) = n− 3, fn−1,1 = n− 1.

Therefore, in this case,

1

n+n− 1

n

χn−1,1(1, 2)fn−1,1

=

(1−

2

n

).

Then,

(n− 1)2(1−

2

n

)2k≤ e−c

where c is defined so that k = 12n (logn+ c).

In order to prove something, we have to know what fλ are and

what 1n + n−1n

χλ(1,2)fλ

are.To calculate these, we have the following lemma from Macdonald.

Lemma 13.1.

fλ =n!∏

x∈λ h(x)

where hλ is the hook length of x.

Proof. This is some exercise in Macdonald.

Example 13.2. In the case of the standard representation, we have

nn− 2n− 3· · · 11

and so we see fλ = n!n(n−2)! = n− 1.

We haveχλ(1, 2)fλ

=2

n(n− 1)

∑i

(λi2

)−∑j

(λ ′j2

)

=2

n(n− 1)

∑i

(λi2

)− λi(i− 1),

as is shown in Macdonald.


Remark 13.3. See Persi’s work “use of group representations in prob-ability and statistics.”

The final result is that k = 12n (logn+ c) and |Q∗ − u| ≤ 2e−c.

Remark 13.4. Igor Pak has a survey of proofs of the hook lengthformula.

13.3. Completing the example; back to the Schur functions. Weproved

sλ =∑ρ`|λ|

z−1ρ xλρpρ

Here zρ =∏i iaiai!, where ρ has ai parts equal to i. Set xi = 1 for

1 ≤ i ≤ n. Then, pj(1, . . . , 1) =∑1i = n. We have pλ(1) = nρ(λ).

Then, Macdonald shows

sλ(1) =∏x∈λ

n+ c(x)

h(x)

where h(x) is the hook-length of box x, and c(x) is the content of boxx, equal to the column of xminus the row of x.

So, using the hook length formula,

sλ(1) =∏x∈λ

n+ c(x)

h(x)

=fλ

n!

∏x

(n+ c(x)) .

Hence, we obtain∑ρ

z−1ρ xλρpρ = sλ

=fλ

n!

∏x

(n+ c(x)) .

Here 21n−2 denotes a partition for the standard representation of Sn.Now, we can rewrite this as∏

x∈λ(n+ c(x)) =

∑ρ`n

n!z−1ρxλρ

fλn`(ρ).

58 AARON LANDESMAN

Example 13.5. Therefore, taking e = 2(1n−2, `(ρ) = n− 1, we obtainn!

2(n− 2)!=

n!

z(21n−2)xλρfλ

=∑x∈λ

c(x)

=∑x

(λi2

)−∑j

(λj2

)Remark 13.6. (1) Persi says this was the first problem solved us-

ing non-commutative Fourier analysis in probability.(2) This also started the study of cutoff phenomenon, with |Q∗k−

u| sharply cutting off to randomness around1

nn logn.

(3) There is a subject called comparison theory, which says that ifyou know one walk, carefully, then, morally, you know about“any walk” to good approximation.

(4) For more information see Persi’s paper “comparison theo-rems for random walks on finite groups.”

14. 10/24/16

Today, we’ll discuss the Robinson-Schensted-Knuth (RSK) algo-rithm. Persi proceeded to explain the rules of solitaire.

Remark 14.1. Persi says if you can solve solitaire, Persi will get youon the front page of the New York Times.

Solitaire is a hard problem we can’t solve, so in math, we look foran easier problem we can solve.

Remark 14.2. One such problem is called patience. Here are therules.

(1) Start with n cards labeled 1, . . . ,n, shuffled in a random order(2) Turn up one card at a time(3) You can play a low card on a high card (e.g., put a 2 on a 6)(4) If you turn over a card higher than any card showing, you

must start a new pile(5) The goal is to get as few piles as possible

Example 14.3. Start with 423175968 in a 9 card deck. When we playthis game, we get 421, 3, 75, 97, 8. Hence, here, we ended up with 5piles. Note that 23569 is an increasing subsequence of length 5.


8 9 10 11 12 13 14 15 16 17 1854 525 1746 2790 2503 1518 632 188 33 11 1

TABLE 1.

So, Persi says, let’s say we win the game with 9 or fewer piles.

Question 14.4. How should you play, what is the optimal strategy,how many piles do you expect?

Lemma 14.5. Let π ∈ Sn. The number of piles if you play as far to the leftas possible is `(π), which is the length of the longest increasing subsequence.Further, “play to the left” is optimal.

Proof. Say k := `(π). We claim that no mater how we play, cardsin an increasing subsequence must go in separate piles. This is justbecause the only cards that can go on top of a given card are lowercards. Therefore, for any strategy, k is a lower bound on the numberof piles.

To complete the proof, we only need show that this strategy ofplaying to the left achieves (at most) k. When we play a card on thelast pile in the play to the left strategy, there must have been lowernumbers on all previous piles. This means we have constructed anincreasing subsequence of length equal to the number of piles.

Remark 14.6. Provably, this play to the left strategy (patience sort-ing) is the fastest algorithm for computing the longest increasingsubsequence.

Some people say that “patience sorting” is the fastest way for hu-mans to sort n cards. By “patience sorting” we mean that one firstplays the card game, and then look for 1 on top of a pile, then 2 ontop of a pile, and so on.

Example 14.7. Playing the game a few thousand times with 52 cards,Persi found the statistics.

Question 14.8. Say we pick π ∈ Sn at random. What is the distribu-tion of `(π), where `(π) is the length of the longest cycle.

This is related to Ulam’s problem on sorting cards using deletionand insertion operations. This yields a metric, saying the distancebetween two permutations is the minimum number of deletions andinsertions. More mathematically, an insertion and deletion operationis a cycle. So, equivalently, if we let S denote the set of all cycles, this

60 AARON LANDESMAN

distance is the length function with generating set S for Sn. Call thismetric du(σ, τ).

Fact 14.9. It turns out that du(1,π) = n− `(π). (The proof is similarto that of Lemma 14.5.

Theorem 14.10 (Logan-Schepp, Kerov-Virscict, 1965). We have thatEn(`(π)) is about 2

√n, where En denotes the expected value over Sn. Fur-

ther,

1

n

∑π∈Sn

(`(π) − En(π))2 ∼ n1/3.

Finally, the Baik-Deift-Johansson theorem says

P

(`(π) − 2

√n

n1/6 ≤ x)→ A(x)

with

A(x)e−∫∞x (x−t)q2(t)dt,

where q(x) satisfies

q ′′ = xq(x) + 2q2(x)

and q(x) ∼ Ai(x) the airy function.

Pick M = M∗. Form the GUE. This means that M is an n × nmatrix zij = zji with zij iid normal 0, 1/2 and zii normal 0, 1. This isa hermitian matrix, so it has real eigenvalues, λ1, . . . , λn.

Theorem 14.11 (Tracy-Widom). We have

P

(λ1 − 2

√n

n1/6 ≤ x)→ A(x).

There are quite a few other appearances of the Tracy-Widom dis-tribution. For example:

(1) The buses in Cuernevaca (is this spelled correctly?)- there weresmall taxis picking everyone up. The buses got a small amountof regulation. This made the traffic in the city smooth andregular The inter-arrival times looked like Tracy-Widom

(2) the shape of a burning piece of paper(3) the distance between parked cars(4) coffee stains


Definition 14.12. The RSK algorithm assigns to a permutation π apair P(π),Q(π) a pair of standard Young Tableau of the same shapeso that the map

π 7→ (P(π),Q(π))

is bijective.

Example 14.13. Start with the permutation 423175968. We start play-ing 4. Then we play 24. Then, we get 24, 3. The 1 bumps downthe first column. We get 124, 3, 7. We next get 124, 37, 5. Then,124, 37, 5, 9. Then, 124, 37, 59, 6. Then, the 8 starts a new pile 124, 36, 59, 6, 8.In the end, we get,

P(π) =1 3 5 6 8

2 7 9

4

Next, we count where we add a box to get Q(π). This results in

Q(π) =1 3 5 7 9

2 6 8

4

One can run this algorithm backwards to get a bijection. This is acombinatorial proof that n! =

∑λ`n f

2λ, where fλ is the size of the

irrep, and also the number of young tableau of shape λ.

Next time, we’ll examine Staley’s chapter 7 RSK.

15. 10/26/16

Recall that we’re talking about the RSK algorithm, which assignsto a permutation π a pair of standard young tableau (P,Q) of thesame shape λ.

We made P by playing the patience game discussed in the pre-vious lecture. So, λ1 is the length of the longest increasing subse-quence. We get that λ1 + λ2 is the maximum of the length of theunion of the biggest two increasing subsequences (this isn’t preciselythe right statement, since they can have some overlap, for the precisestatement see Curtis Greene’s work).

Remark 15.1. About 30 years ago, all partitions and young tableaucan be said in terms of nilpotent orbits of groups and flags in vectorspaces. There turns out to be a natural metric between flags takingvalues in the permutations. You can say RSK in that formulation.

62 AARON LANDESMAN

But, we prefer to describe it this way, since it came from analyzingsolitaire.

15.0.1. Row Insertion.

Definition 15.2. If P is a semi-standard tableau and k is an integer,we write P ← k (“insert k into P”). We insert k into the first row inthe lowest place possible (or if its bigger, we insert it at the end ofthe row, and end the process). If it is inserted into the row, it bumpsthe current element in the row into the next row. We then continuethe process until a number is the biggest in its row, at which pointit is added to the end of the row, and the process is terminated. Weillustrate this by the following example.

Example 15.3. Let

1 1 2 4 5 5 6

2 3 3 6 8

4 4 6 8

6 7

8 9

Now, let’s say we insert a 4.

1 1 2 4 4 5 6

2 3 3 5 8

4 4 6 6

6 7 8

8 9,

where the numbers with overlines indicate they were bumped in thepath. The insertion path was (15)(24)(34)(43).

Lemma 15.4. Insertion has the following properties:(1) “Stuff moves to the left.” That is, if (rs) ∈ P ← k has (r+ 1t) has

t ≤ s(2) If j ≤ k then I ((P ← j)← h) lies strictly to the left of I (P ← j)

Proof. (1) Start with a semi-standard tableau P. This has weaklyincreasing rows and strictly increasing columns. So, eitherPr+1,s > Pr,s or there is no box under r, s. In the first case,when we insert something, it has to be inserted on somethingstrictly bigger. When you move something from Pr,s to row


r+ 1, it can always go in (r+ 1, s) and maybe to the left. Incases 2, you can’t go to the right of column s.

(2) This argument is similar, see Staley’s book on enumerativecombinatorics, volume 2, chapter 7, RSK.

Corollary 15.5. The insertion P ← k is a semi-standard young tableau.

Proof. We first show that the rows are still weakly increasing afterthe insertion. At each step of the insertion, we replace a number byanother number lower than it. A number a can only bump a largernumber b. By the first part of the above lemma, b does not move tothe right, and so b is inserted below a strictly smaller number.

Definition 15.6 (RSK algorithm). Start with a matrix A =(aij)

withonly finitely many nonzero entries. RSK assigns to A a pair (P,Q)where P,Q are semi-standard young tableau of the same shape.

Further, the column sums of A equal the content of P, and the rowsums equal the content of Q.

Example 15.7. 1 0 20 2 01 1 0

To A assign (

i1 i2 · · · imj1 j2 · · · jm

)with i1 ≤ · · · ≤ im, if ia + ib then ja ≤ jb, and ij appears aij times.Then, form P via solitaire on the bottom row. So, we take the se-quences (

1 1 1 2 2 3 31 3 3 2 2 1 2

)Now, we play a book keeping arrow, showing how we got throughthe elements. We get the following sequence of moves

1

then

1 3

64 AARON LANDESMAN

then

1 3 3

then

1 2 3

3

then

1 2 2

3 3

then

1 1 2

2 3

3

then

1 1 2 2

2 3

3

We next keep track of the sequence of insertions, adding the nextnumber in the first row to the place where the new box was added.

We get the following sequence of moves

1

then

1 1

then

1 1 1

then

1 1 1

2


then

1 1 1

2 2

then

1 1 1

2 2

3

then

1 1 1 3

2 2

3.

At this point, we see the column sums are the content of theQ tableauand the column sums are the content of the P tableau.

Theorem 15.8. We have that RSK is a bijection between finite integer ma-trices with finitely many nonzero entries and pairs (P,Q) of semi-standardyoung tableau of the same shape with the property that the content of P isthe column sums of A and the content of Q is the row sums of A.

Proof. We know that P is semi-standard because we built it by play-ing solitaire. Next, we show Q is semi-standard. Q is gotten fromthe book-keeping array by inserting the first row of(

i1 i2 · · · imj1 j2 · · · jm

)with the ij’s weakly increasing. Therefore, since we’re forming shapesby adding increasing numbers, we obtain that the rows and columnsare both weakly increasing. It remains to check he columns are strictlyincreasing. For the strictness, we use the second part of Lemma 15.4.

We now claim that the content of P is the column sums of A andthe content of Q is the row sums of A. This follows fairly immedi-ately from the construction.

Exercise 15.9. Verify that the statement regarding the row sums andcolumn sums.

Remark 15.10. The correspondence for permutations is given whenwe take the matrices to be permutation matrices, and then we obtainsemi-standard young tableau (with entries 1, . . . ,n) as a result.

66 AARON LANDESMAN

16. 10/28/16

Recall that last time we discussed a fancy version of RSK usingmatrices. Here is an application. We give a bijective proof of theCauchy following equation.

Corollary 16.1. We have∏i,j

(1− xiyj

)−1=∑λ

sλ(x)sλ(y)

Proof. We have ∏i,j

(1− xiyj

)−1=∏i,m

∞∑aij=0

(xiyj

)aijA typical term is xαyβ and arises from some collection of aij with∑j aij = x

αii ,∑i aij = y

βjj . This is equivalent to picking a matrix

with given row and column sums. Using the RSK bijection, this isequivalent to picking P,Q two semi-standard young tableau of thesame shape. The result then follows from the combinatorial defi-nition of sλ(x) as the sum over all semi-standard young tableau ofshape λ of x raised to the content of λ.

16.1. Plane partitions, RSK, and MacMahon’s generating function.For λ ` n, we have λ1 ≥ · · · ≥ λr > 0with

∑i λi = n.

Definition 16.2. A plane partition π = πij is a collection of integersindexed by i and j so that

πij ≥ πi,j+1πi+1,j ≤ πij

That is, the rows are decreasing rightward and the columns are de-creasing downward.

Example 16.3. The following is a plane partition:

7 5 5 3 2 1 1 1

6 5 5 2 1 1

6 3 2 2

We have that this has 18 parts meaning 18 numbers appearing. Wehave the size (the sum of all numbers) with |π| = 59. The number ofrows is 3. The number of columns is 8. The trace of π is

∑i aii = 14.


Definition 16.4. Define P(r, c) to be the set of plane partitions withat most r rows and at most c columns.

In particular, P(1, c) is the number of partitions with at most cparts.Theorem 16.5 (Euler).∑

π∈P(1,c)qtr(π)x|π| =

c∏i=1

(1− qxi

)−1Proof.Exercise 16.6. Prove this by expanding the right hand side.

The following is a higher dimensional case:Theorem 16.7 (MacMahon).∑

π∈P(α,α)

x|π| =

∞∏i=1

(1− xi

)−iThe proof is vintage bijective combinatorics, due to bender and

Knuth, though it was originally proved by MacMahon.We want to prove this. We proceed in several steps.(1) First, we describe a way of sticking together two partitions

with distinct parts λ,µwith the same number of rows to makea new partition

ρ(λ,µ).

with ρ(λ,µ) a partition of |λ|+ |µ|− `(λ).Example 16.8. Consider λ = 532,µ = 631. (It’s not importantthat they happen to be partitions of the same number) bothwith distinct parts and the same number of rows. First, formthe shifted diagram of λ.

• • • • •• • •• •

Similarly, form the shifted diagram of µ,

• • • • • •• • ••

68 AARON LANDESMAN

Next, form the shifted diagram of µ, which is essentially thetranspose of the diagram for µ.

•• •• • •• •• •••

Next, glue λ to the shifted diagram to µ.

• • • • •• • • •• • • •• •• •••

This resulting shape is ρ(λ,µ).

Note that one can recover λ,µ from ρ(λ,µ), and so this mapis a bijection.

(2) We next extend the above ρ(λ,µ) construction between pairs(P,Q) of reverse semi-standard young tableau (meaning strictlydecreasing down columns and weakly decreasing down rows)of the same shape and plane partitions. The bijection is givenby sending P,Q to π whose ith column is ρ(Pi,Qi), for P,Qreverse semi-standard young tableau of the same shape with.

Example 16.9. Take

P =4 4 2 1

3 1 1

2


and

Q =5 3 2 2

4 2 1

1

We then get

π(P,Q) =4 4 2 1

4 2 2 1

4 2

2

2

(3) Now, replace each row of π(P,Q) by its transpose.

Example 16.10. Using the above example

π(P,Q) =4 4 2 1

4 2 2 1

4 2

2

2

we get π ′(P,Q) given by

π ′(P,Q) =4 3 2 2

4 3 1 1

2 2 1 1

1 1

1 1

It’s not hard to see that the resulting map

(P,Q) 7→ π ′ (P,Q)

is a bijection, with |π ′| = |P|+ |Q|+ |Sh(P)|. (Here the shape isthe sum of the lengths of the rows). Furthermore,

diag(π ′) = Sh(P) = Sh(Q)

so tr (π ′) = |Sh(P)|. Further, we have invariants

`1(π′) = max(Q)`2(π

′) = max(P)

70 AARON LANDESMAN

where `1 is the number of rows and `2 is the number of columns.(4) ToA =

(aij)

we assign a pair of reverse semi-standard youngtableau of the same shape.

Example 16.11. Given2 0 10 1 10 3 0

From our RSK algorithm, this yields(

1 1 1 2 2 3 3 31 1 3 2 3 1 2 2

)Reversing this, we get(

3 3 3 2 2 1 12 2 1 3 2 3 1 1

)Playing the solitaire game, we get

3 2 2 2 1

2 1

and the second tableau can be computed similarly as the onerecording where boxes are placed, gotten by RSK where youput a high card on a low card.

From this matrix A, we get P,Q, given by

|P| =∑i,j

jaij|Q| =∑i,j

iaij

with

maxP = maxj : aij 6= 0

maxQ = max

i : aij 6= 0

|Sh(P)| = |Sh(Q)| =

∑aij

Theorem 16.12.∑π∈P(r,c)

qtrπx|π| =

r∏i=1

c∏j=1

(1− qxi+j−1

)−1


Proof. LetMrc be all r by cmatrices. Then,∑π∈P(r,c)

qtr(π)x|π| =∑

A=aij∈Mqaijx

∑(i+j)aij−

∑aij

=

r∏i

c∏j

σ∑aij≥0

qaijx(i+j−1)aij

17. 10/31/16

17.1. Announcements and Review.

Remark 17.1. Happy Halloween!

Last time, we discussed plane partitions. Recall these were collec-tions π = πij so that πij ≥ πi,j+1,πij ≥ πi+1,j. We defined P(r, c) asthe collection of plane partitions fitting in an r× c box.

Question 17.2. Fix a shape or a pair (r, c), and fix n. Consider all πso that n =

∑i,j πi,j. Choose one uniformly at random. What does

this partition “look like”?

Today, we look at the above question for λ ` n (a partition insteadof a plane partition).

Let P(n) be the number of partitions of n. We have∞∑n=0

p(x)xn =

∞∏i=1

(1− xi

)−1Remark 17.3. Hardy and Ramanujan found

P(n) ∼ eπ√2n3 /4n

√3

Remark 17.4. Persi’s test to see how fast something grows is to plugin n = 52. This is about e43. For reference, n! is about e68.

Question 17.5. So, pick a partition at random. How many 1’s does ithave?

Here are some theorems going back to the 1940’s.

Remark 17.6. We have a generating function, and a product form.Look at the singularities of the right hand side. This has singularitiesdense in the unit circle. The circle method was invented to solve thisproblem. Later, Rademacher found an exact formula, in the sensethat given n, there is some formula with finitely many terms, which

72 AARON LANDESMAN

you can compute to find P(n), though the number of terms dependson n. The following theorems are proved using these ideas.

Theorem 17.7 (Erdos-Lehner). Let λ = 1a1 · · ·nan . Then,

P

(πa1√6n≤ x

)∼ 1− e−x.

In particular “a typical partition has around√n 1’s.

Furthermore,

P

(πkak√6n≤ x

)∼ 1− e−x.

Theorem 17.8. Let y1(λ) ≥ y2(λ) ≥ · · · yk(λ) be the sequence of largestparts of λ. Then,

P

(π√6nY1 −

logπ√6n≤ x

)∼ e−e

−x.

for −∞ < x <∞. Further,

P

(π√6nYn −

logπ√6n≤ x

)∼

∫x−∞

(e−e

−x)−kv

(k− 1) !dv

Additionally, we have

P

(π√6n

(a1, 2a2, . . . nan) ≤ (x1, . . . , xn))

∼

n∏i=1

(1− e−xi

).

Theorem 17.9. Fix q ∈ (0, 1). Take

Qq(λ) = q|λ|

∞∏i=1

(1− qi

).

a probability measure on all partitions. Under this measure,

P (a1 = x1, . . . ,an = xn) =

k∏i=1

qixi(1− qi

).

Corollary 17.10. Under Qq, the ai (λ) are independent with

Qq(ai = j) = qij(1− qi).

Corollary 17.11. Pick λ from Qq, let N =∑i iai (λ). Then,

P (λ|N(λ) = n) =1

P(n).

Proof. Use the Borel-Cantelli lemma.


Remark 17.12. The above corollary suggests the following algorithm.Pick λ ` 1000. Pick q close to 1 so that Eq (N(λ)) = 1000. To do this,use

Eq (N (λ)) =

∞∑i=1

iEq (ai)

=

∞∑i=1

iqi

1− qi.

Now, fix this resulting q, and sample repeatedly from Qq. Just waituntil you get λ with N (λ) = 1000. The problem with this is thatN (λ) has a Gaussian distribution, with

√n the standard deviation.

This algorithm will be too slow to get many random partitions ofsize 1000.

The next topic in this line of work is the shape of a random parti-tion.

Question 17.13. What does the shape of a random partition looklike?

Remark 17.14. That is, choose some random partition of n. If n islarge, the partition will have some shape. We know the partition haslength

√n logn, and it also has approximately this height. There is

then some curve describing the locations of the boxes. The curve wedraw has a limit. The limit is

e−πx/√6 + e−πn/

√6 = 1.

A reference for this is Vershik, Statistical mechanics of combinatorialpartitions.

The main other measure used is the following:

Definition 17.15. Let P(n), denote the Plancherel measure, definedby

P(λ) :=f(λ)2

n!

where f(λ) is the dimension of the irreducible representation of Sn ofshape λ, which is

fλ =n!∏

i hook lengths hi.

74 AARON LANDESMAN

The above questions have been asked under this measure, andtheir answers are also known for this measure. For example, weknow

P

(λ1 − 2

√n

n1/6 ≤ x)F(x)

where F is the Tracy Witten distribution.

Remark 17.16. One good reference is Andrei Okunkov is “the usesof random partitions” (from 2003).

18. 11/2/16

18.1. Overview. Today, we’ll discuss three related topics:(1) P-partitions(2) Quasi-symmetric functions(3) Shuffling cards

18.2. P-partitions. To some degree, this has disappeared from mod-ern research, but Stanley and others have done some interesting workon it in the past.

Definition 18.1. A partition of k inton parts is a sequence 1 ≤ f(1) ≤· · · ≤ f(n) with

∑ni=1 f(i) = k.

Definition 18.2. A composition of k inton parts 1 ≤ f(1), f(2), . . . , f(n)with

∑ni=1 f(i) = k. That is, it is a partition without an order.

Question 18.3. What about if we give a partial ordering on the partsof a composition?

Example 18.4. Say we have a composition fwith

f(1) + f(2) + f(3) = n

and 1 ≤ f(2) < f(1), f(2) ≤ f(3). How many such partially orderedcompositions are there?

Definition 18.5. Let P be a partial order on [n] := 1, . . . ,n, withrelation denoted <P. Let <N denote the total ordering given by thenatural numbers. A P-partition is a function

f : [n]→N+.

so that if i <P j then f(i) ≤ f(j) and i <P j but i >N, we havef(i) < f(j).

Example 18.6. If we take P to be a poset which is a set, we get a usualcomposition series.


Example 18.7. Take n = 3 and consider the partial order 2 <P 1, 2 <P3, but 1, 3 are non-comparable. Then, this yields the restrictionsf(2) < f(1) and f(2) ≤ f(3).

Observe there is no constraint on the relative order of 1 and 3

Remark 18.8. Not every set of constraints can be obtained as a P-partition, for some poset P.

Definition 18.9. Let L(P) denote the set of linear extensions of P,meaning chains which extend the partial order of P.

Example 18.10. Taking the partial order P from Example 18.7, theset of linear extensions are 2 < 1 < 3 and 2 < 3 < 1. In the first,we obtain f(2) < f(1) ≤ f(3). In the second case, we obtain f(2) ≤f(3) < f(1).

Lemma 18.11 (Key Lemma). Let P be a partial order on [n]. Let

P(P) := all P-partitions .

Then,

P(P) =∐

π∈L(P)P(π).

Proof. Consider P-partitions with real values, meaning f(i) ∈ R. So,f corresponds to an element of Rn, instead of Nn.

Definition 18.12. Let

Hij :=x ∈ Rn : xi = xj, i < j

denote a given hyperplane. These split Rn into chambers C of pointswhich do not lie on any of the hyperplanes. These split Rn into facesF on some Hij.

A chamber is indexed by n! permutations

Any f ∈ P(P) either lies in a chamber or a face. Generically, it willlie in a chamber. The inequalities are chosen so that if you’re on ahyperplane, you will have a particular equality.

Exercise 18.13. Complete the proof, verifying the outline above.

18.3. The order polynomial. We’d like to enumerate P-partitions.

Remark 18.14. A neat way to enumerate them is to count f : [n] →[m] with the restriction that if i <P j, then f(i) ≤ f(j), and i <P j, i <N

j then f(i) < f(j). We’ll then later ask for which has a given sum k.

76 AARON LANDESMAN

Remark 18.15. Here is an equivalent formulation: Drop n labeledballs into m labeled boxes, and count the number of ways that areconsistent with the partial order.

Definition 18.16. The order polynomial, denotedΩP(m) is the num-ber of such functions as in Remark 18.14.

Remark 18.17. If one drops n balls in at random, the chance that theassignment works out with respect to the partition is

ΩP(m)

mn.

Example 18.18. (1) Say P is poset on 1, . . . ,n which is a set (noadditional relations), thenΩP(m) = mn.

(2) If P is is the chain 1 < 2 < · · · < n, we obtain weakly mono-tone functions. Then,

ΩP(m) =

(n+m− 1

n

)by stars and bars.

(3) Say P = P1 ∪ P2. Then,

ΩP(m) = ΩP1(m)ΩP2(m).

Definition 18.19. Given a permutation π ∈ Sn, the descent set of thepermutation is the set of i so that π(i+ 1) < π(i). d(π) is the numberof descents. We let D(π) denote the descent set of π.

Example 18.20. Taking π = 6431274, this has descent set 1, 2, 3, 6,d(π) = 4, and D(π) = 1, 2, 3, 6.

Example 18.21. What about Ωπ(m), the complete ordering given bythe permutation π? Then, we have

Ωπ(m) =

(m+n− 1− d(π)

n

).

For example, take

π = 14253.

This has d(π) = 2. This is the number of solutions to 1 ≤ f(1) ≤f(4) < f(2) ≤ f(5) < f(3). This is the same as counting

f ≤ f(1) ≤ f(4) ≤ f(2) − 1 ≤ f(5) − 1 ≤ f(3) − 2 ≤ m− 2.

So, this is (m− 2+n− 1

n

)=

(m+n− 1− d(π)

n

).



∞∑m=0

ΩP(m)xm =∑π∈L(P)

xd(π)+1

(1− x)n+1.

Proof. By linearity, it’s enough to prove this for all permutations π,by Lemma 18.11

∞∑m=0

Ωπ(m)xm =

∞∑m=0

(n+m− 1− d(π)

n

)xm

=xd+1

(1− x)n+1.

Example 18.23. If you take the partial order with no constraints, thisbecomes

∞∑m=0

mnxm =∑π∈Sn

xd(π)+1/ (1− x)n+1 .

For example, if n = 0, we obtain∑xn =

1

1− x.

If you differentiate this, you get∑mxm =

x

(1− x)2.

If you further differentiate this, you obtain more special cases, andcan see this satisfies the same recursion as the descent numbers.

Remark 18.24. The polynomial

An(x) =∑π∈Sn

xd(π)+1 =

n−1∑j=0

Anjxj+1

and the latter are the Eulerian number, where the Anj is the numberof permutations of n things with j descents.

78 AARON LANDESMAN

19. 11/4/16

19.1. Review. Last time we were discussing P-partitions. Let P be apartial order on [n]. Let f : [n]→ [m] so that

i <P j =⇒ f(i) ≤ f(j)i <P j and i >N j =⇒ f(i) < f(j).

We introduced

ΩP(m) = # P-partitions

Recall that if P was a poset which is a set of size n, then ΩP(m) =1mn . If P is a linear ordering corresponding to a permutation π, sayπ = 3214, then we require f(3) < f(2) < f(1) ≤ f(4).

Any P has

P(P) =∐

π∈L(P)P(π)

Example 19.1 (Euler). If P is a set of size n, we get∞∑i=0

ΩP(m)tm =

∞∑m=0

mntm = An(t)/ (t− t)n+1 ,

where

An(t) =

n−1∑j=1

An,jtj+1

with An,j the number of permutations in Sn with j descents. Forexample, if n = 2, we get∞∑

m=0

m2tm =t+ t2

(1− t)3.

If n = 4, we get ∑m

m4tm =t+ 11t2 + 11t3 + t4

(1− t)5.

19.2. A possibly non-politically correct example.

Question 19.2. Say we have n1 girls and n2 boys. Let n := n1 + n2.They each have a summer job. There arem salary levels. If a salary in1, . . . ,m is assigned at random, to each person, what is the chancethat all the girls are paid strictly less than all the boys?


Let’s rephrase the above question in our poset language. Set aposet with 2 girls and 3 boys, and consider the complete bipartitegraph K3,2 where the the 3 corresponds to 3 boys, and the 2 corre-sponds to the two boys. We want to count the number of P-partitions.To use the theory developed above, we need to find all linear exten-sions of this poset L(P). There are n2!n1! such linear extensions. So,the answer is

1

mn

∑0≤a≤n1−1,0≤b≤n2−1

(m+n− 1− (a+ b+ 1)

n

)An1,aAn2,b.

To enumerate partitions of r, use

ΩP(q,m) =∑f∈P(P)

n∏i=1

qf(r)

The coefficient of qr in this counts P partitions with parts at mostm,and n parts that sum to r.

Here are some references:(1) Ira Gessel has slides on his webpage of a nice talk he gives,

“P-partitions and permutation enumeration”(2) Kile Petersen “Descent, peaks, and P-partitions”(3) Richard Stanley, Volume 1 and Volume 2 of “enumerative com-

binatorics”

19.3. More on descent. Here are two comments about descents.

Remark 19.3. In Sn,

D(π) = i : π(i+ 1) < π(i) .

For example, D(6412375) = 1, 2, 6 ⊂ [n− 1]. If S ⊂ [n− 1], wedefine As =

∏D(π)=s π.

Fact 19.4 (Solomon).

ASAT =∑

u≤[n−1]cuSTAU.

where c ∈ Z≥0 is a fairly explicit combinatorial constant.

19.4. Shuffling Cards.

Definition 19.5. The Gilberg-Shannon-Reeds model for card shuf-fling is the following: Say we have a deck of n cards. The probability

80 AARON LANDESMAN

of cutting the cards into c and n− c is(nc

)2n

.

If the left has A and the right has B. The change next from the left isA/(A+B).

Remark 19.6. If you shuffle cards 8 times perfectly, they come backto where the start.

The above is provably the highest entropy model for shuffling.

Let Q(π) be the change of the permutation after 1 shuffle. Then,

Q ∗Q(π) =∑η∈Sn

Q(η)Q(πη−1).

Similarly, one can define Q∗k(π) := Q ∗Q∗k−1(π). We have the uni-form distribution u(π) = 1

n! .

Question 19.7. Given ε > 0, how large should we choose k so that

|Q∗k − u| < ε

where we use the distance |Q∗k − u| := maxA⊂Sn |Q∗k(A) − u(A)|.

Here are some other descriptions of Q.(1) Flip a fair coin for each card, and put all 1’s on top.(2) Take the unit interval and place n points down at random.

Label them x1, . . . , xn. Now do the Baker’s transformation,sending x 7→ 2x mod 1. Look at the induced permutation.

Remark 19.8. We also have a-shuffles, by sending x 7→ ax mod 1.So, for example, a 3 shuffle is where we put n points down, andstretch each third out. This corresponds to partitioning cards into3 piles, and then put them together randomly. We have Qa ∗Qb =Qab, using this description. We want to study the k-fold convolution.Hence, we have Q∗k = Q∗k2 = Q2k .

Fact 19.9 (Diaconis-Bayer). We have

Qa(π) =1

an

(a+n− 1− d(π−1)

n

).

So, the mathematics of shuffling is about the same as the mathemat-ics of P partitions, and about the same as the mathematics of quasi-symmetric functions.


20. 11/7/16

Recall that last time we were talking about shuffling.

Definition 20.1. We define

sh (a1, . . . ,an; x1, . . . , xm)

to be the set of all sequences in a1, . . . ,an, x1, . . . , xm with no repeti-tions such that ai comes before aj for i < j and xi comes before xj fori < j.

Remark 20.2. In Coexeter groups, if Sk × Sn−k is the subgroup of Snpermuting the first k and n − k elements separately. The shortestcoset representatives for Sn/Sk × Sn−k are shuffles, where the met-ric is given by the minimum number of transpositions to get to theidentity.

Remark 20.3. If P1,P2 are disjoint total orders, (meaning that a posetis a union of two disjoint chains) then L(P) corresponds to shuffles.

Recall the Gilbert-Shannon-Reeds shuffling scheme, which shuf-fles by cutting the deck randomly, and then randomly dropping fromeach cut pile with probability proportional to the size of the deck sofar.

We defined

Qa ∗Qb =∑

Qa(η)Qb(πη−1).

where Qa is an a shuffle, cutting a pile into a piles. We found Qa ∗Qb = Qab and Q∗h2 = Q2h .

Remark 20.4 (Cool magic trick to impress your friends!). Persi sendsSarah a deck of cards in random order, and he knows the order ofthe deck. When you shuffle a deck of cards once, it’s not very ran-dom. But each half of the cards stay in the same order. So, when youshuffle again you have four sequences. After 3 shuffles, you usuallyhave eight rising sequences. When she puts the card into the mid-dle, she gets a 9th rising sequence. Then Persi plays solitaire. Then,you’ll have eight piles, each about an eighth of the deck, and one pilewhich has a single card.

Theorem 20.5 (Bayer). We have

Qa(π) =

(n+ a− r

n

)/an.

where r = r(π) is equal to the number of rising sequences in π.

82 AARON LANDESMAN

k 1 2 3 4 5 6 7 8 9|Q∗k − u| 1.000 1.000 1.000 1.000 .924 .62 .32 .16 .08

TABLE 2.

Remark 20.6. Note that we also have r(π) = d(π−1) + 1. So, we alsohave

Qa(π) =

(r+ a− 1− d(π−1)

n

)/an.

Using the total variation distance

|Q∗k − u| =1

2

∑|Q∗k(π) −

1

n!|

= maxA⊂Sn

|Q∗k(A) − u(A)|.

Example 20.7. When n = 52, we get

Theorem 20.8. If k = 32 log2 n+ c, then

|Q∗h − u| = 2φ(2−c) − 1+ o

(1√n

)where φ(x) = 1√

2π

∫x∞ e−t2/2dt.

Remark 20.9. If X is a finite set and P(x) is a probability. Then, theentropy of P is −

∑x P(x) logP(x).

Theorem 20.10. If X = Sn and P is any probability supported on singleshuffles, then the entropy of P is at most the entropy of Q2.

Consider

Q2(id) =

(n+2−rn

)2n

=

(n+1n

)2n

=n+ 1

2n.

We have thatQ2 of any non-identity πwill have two rising sequences,yielding 1

2n .

Theorem 20.11. We have

|Q∗k2 − u| =1

2

∑π∈Sn

|Q2k(π) −1

n!|

=

n∑j=1

Bn,j|

(2k+n−jn

)2nk

−1

n!|


where Bn,j are the Eulerian numbers, the number of permutations with jrising sequences, which An,j−1, which is the number of π with j − 1 de-scents.

21. 11/9/16

21.1. Algebra of the Ai. We’ll work in Q [Sn]. Define

Q2 =n+ 1

2nI+

1

2nA2.

We let

A2 =∑σ−1

σ

where σ−1 has one descent. We let

Ai :=∑

σ−1 has i-1 descents

σ.

Theorem 21.1. The set

Aini=1

generates a commutative, semisimple subalgebra of Q [Sn] with idempo-tents

en(`) =

n∑r=1

σ` (n− r, . . . , 1− r)Ar

where σ` is the `th elementary symmetric function and en(`) are the Euler-ian idempotents.

Remark 21.2. Further, the idempotents above en(`) give a hodge de-composition of hochschild homology.

Proof. We know that an A-shuffle corresponds to an element

Ba :=1

an

n∑r=1

(a+n− r

n

)Ar.

Let’s consider B22 (the same as Q2 above). Writing B2 = c1A1 + c2A2with A1 = id We have

B22 = (c1A1 + c2A2)2

= c21A1 + c1c2A2 + c2c1A2 + c22A

22

= B2.

84 AARON LANDESMAN

Then, we see B4 is a linear combination of A1,A2,A3,A4. Next, welook at B2B3, which is a linear combination of A1,A2,A3,A22 andA2A3. Hence, A2A3 =

∑αijAi. Similarly, we see B2B3 = B3B2 = B6.

We also see A2A3 = A3A2, and then we can keep going by inductionto see AiA3 = A3Ai, and so on. This shows the algebra is commuta-tive.

Next, let’s verify semi-simplicity (meaning that it is a direct sumof simple algebras, algebras with no nontrivial ideals). We have

Bm2 = B2m =1

2mn

n∑r=1

(2m +n− r

n

)Ar

=1

n!

n∑`=0

1

2`m

m∑r=1

σ` (n− r, . . . , 1− r)Ar.

Now, B2 acts on this subalgebra by left multiplication as a linear mapwith eigenvalues 1, 12 , . . . , 1

2n−1and eigenvectors of the form 1

2`en(`),

and this turns out to be one equivalent characterization of semisim-ple.

We’ll omit the proof that these are the idempotents, but there’ssome discussion of them in chuck Weibel’s book.

21.2. Quasi-Symmetric Functions. We work in infinitely many vari-ables x1, x2, . . ..

Definition 21.3. A quasi-symmetric function of degree n is a formallinear combination ∑

αn

c(α)xα

where α nmeans α is a composition of n, with

xα = xα1i1 xα2i2· · · xαnin

so that the coefficient[xα1i1· · · xαnin

]f =

[xα1j1· · · xαnjn

]f

for all i1 < · · · < in and j1, . . . , jn. for all n.

Example 21.4. The polynomial∑i<j x

2ixj is a quasi-symmetric func-

tion.

Remark 21.5. We will use a useful correspondence

(21.1)α : α n→ S : S ⊂ [n− 1]


3 12 21 111∅ 1 2 1, 2

TABLE 3.

If S = s1 < s2 < · · · sk, then α1 = s1,α2 = s2 − s1, . . . ,αn = n− sk.From α, we let Sα = α1,α1 +α2, . . . ,α1 + · · ·+αn−1.

Example 21.6. If n = 3, the compositions of 3 are

Definition 21.7. Let Q be the ring of quasi-symmetric functions,

Q = ⊕∞i=1Q

i,

where Qi are quasi-symmetric function of degree i.

Definition 21.8. Let α = (α1, . . . ,αn) with no zero parts. Let

Mα =∑

i1<···<ik

xα1i1· · · xαkik .

These are a basis for quasi-symmetric functions, Qn.

Remark 21.9. We have Λn ⊂ Qn. (There are more quasi-symmetricfunctions than symmetric functions.

Here is another basis for quasi-symmetric functions.

Lα :=∑

i1≤i2≤···≤ik

xi1 · · · xik

with ij < ij+1 if j ∈ Sα.

Example 21.10. If n = 3,

M3 =∑i

x3i

M21 =∑i<j

x2ixj

M12 =∑i<j

xix2j

M111 =∑i<j<k

xixjxk,

86 AARON LANDESMAN

and

L3 =M111 +M12 +M21 +M3

L21 =M111 +M21

L12 =M111 +M12

L111 =M111.

Proposition 21.11. We have

Lα =∑β≥α

Mβ =∑

Sα⊂T⊂[n−1]Mco(T).

Proof. We have

Mα =∑

Sα⊂T⊂[n−1](−1)T/SαLco(T),

so Lααn−1 form a basis.

22. 11/11/16

The purpose of the last few lectures were to get us better acquaintedwith quasi-symmetric functions, card shuffling, and plane partitions.

Now, we’re discussing quasi-symmetric functions. Recall last timewe defined

QSym(x1, . . .)

so that [xα1i1· · · xαrin

]f =

[xα1j1· · · xαrjn

]f,

for i1 < · · · < in, j1 < · · · < jn. We have two bases

Mα =∑

i1<···<in

xαjijLα =

∑ij≤ij+1

xi1 · · · xik

We defined

Sα :=

α1,α1 +α2, . . . ,

n−1∑i=1

αi

.

Recall we defined P partitions as a finite collection X so that xf =∏x∈X x

f(x), for f : X→N, with conditions on the function f depend-ing on the poset P. For π ∈ Sn, we defined π-partitions as functions

f : [n]→N+,


with f(p1) ≤ · · · ≤ f(πn) and f(πi) < f(πi+1) if πi > πi+1. Fix π ∈ Sn.We have ∑

f∈P(π)xf = Lco(π)(x)

where co(π) is the composition of the descent set of π.

Proposition 22.1. Let P be a poset of cardinality n. Then,

KP(x) :=∑f∈P(P)

xf =∑π∈L(P)

Lco(π)(x).

Proof. This follows from the above, and from the fundamental lemmathat we can write a sum over all functions subject to the constraintsof P as a sum over all possible total orderings of all functions subjectto those total orderings.

Definition 22.2 (A twist of notation). Let w ∈ Sn and P is a partialorder on [n]. A (P,w) partition is a function

f : [n]→N+

so that f(π1) ≤ · · · ≤ f(πn), but if s < t and w(s) > w(t) thenf(s) < f(t).

Warning 22.3. This is slightly different from the P partition, since wedon’t just have conditions on adjacent elements.

22.1. Application to symmetric function theory.

Example 22.4. Take a skew symmetric tableau

• • •• • •• • ••

We can now tilt this by 135 degrees, to obtain Pλ/µ. Then, we createwλ/µ by labeling the boxes

8 9 10

3 5 7

2 4 6

1

88 AARON LANDESMAN

which yields a labeling on the tilted Pλ/µ. Observe that(Pλ/µ,wλ/µ

)is a semi-standard young tableau. Further, KPλ/µ,wλ/µ = sλ/µ.

So, we can express

sλ/µ =∑

T SSYT of shape λ/µ

Lco(T)

where co(T) is the composition associated with the descent set of T .We say i is a descent of T is i+ 1 is in a strictly lower row.

Example 22.5. If we have

• • • 2 8

• 1 4 5 10

3 6 9

7.

the corresponding descent set is 2, 8, 5, 6. This is a subset of 10. Thiscorresponds to the composition 23122, gotten by taking the differ-ences in the parts of the subset.

This proves

sλ/µ(1,q,q2, . . . ,qn) =∑T q

maj(T)

(1− q) · · · (1− qn) ,

where the sum is taking over all SSYT of shape λ/µ and

maj(T) =∑i∈D(T)

i.

Remark 22.6. You can then use these definitions to reprove MacMa-hon’s formula that∑

PP(n)xn =

∞∏i=1

(1− xi

)−i.

22.2. Connection of quasi-symmetric functions to card shuffling.Let θ = (θ1, . . . , θn, . . .) where

∑i θi = 1 and θi ≥ 0.

Definition 22.7. One obtains a θ shuffle of n cards by dropping nballs into boxes with probability θi that the ball lands in box i. Then,from the ni cards in the ith barrel, riffle shuffle these by GSR.


Example 22.8. If θ = (1/2, 1/2, 0, . . .) then this is usual 2-shuffles. Ifθ =

(1a , . . . , 1a , 0, . . .

), then this is a-shuffles. We can also have things

like θ = (2/3, 1/3, 0, . . .) which is a weighted shuffle.

Let Pθ(π) be the chance of obtaining π after a θ shuffle.

Fact 22.9. We have

Pθ ∗ Pη(π) = Pθ∗η(π)where

θ ∗ η =(θi · ηj, . . . ,

).

For example, if θ = η = (α,β, 0, . . .) then

θ ∗ η =(α2,αβ,αβ,β2, 0, . . .

).

The key theorem is the following:

Theorem 22.10 (Stanley). We have Pθ(π) = Ldes(π−1)(θ).

22.3. Applications. The following application is due to Assaf, Persi,and sound. Fix α ∈ (0, 1). We have a correspondence sendingDα(π)to (α, 1−α, 0, . . .). We can study convolutions by defining

Sep(k) = maxπ∈Sn

1−n!P∗kα (π),

and define the norm

`∞(k) = maxπ

|1−n!P∗kα (π)|.

Theorem 22.11. We have

Sep(k) = 1−∑w∈Sn

sgn(w)∏

(θi(1− θ)i)kni(π),

and

`∞(k) =∑w∈Sn

∏ (θi + (1− θ)i

)kni(π)− 1,

where

ni(π) = # i cycles in π .

Further, if

k = b 2 logn− log 2+ c

− log(θ2 + (1− θ)2

)cthen sep(k) ∼ 1− c−`

−cand `∞(k) ∼ c−`

−c− 1.

This uses quasi-symmetric function theory.

90 AARON LANDESMAN

23. 11/14/16

23.1. Combinatorial Hopf Algebras. One useful reference is Com-binatorial Hopf Algebras by Aguiar, Bergeron, and Sottile.

Definition 23.1. A Hopf algebraH is an algebra over a field k, mean-ing there is a multiplication map which is k-bilinear, (we will usuallytake k = R,) inducing the map

m : H⊗H→ H∑ij

aijxi ⊗ xj 7→∑ij

aijxixj

with a coproduct

∆ : H→ H⊗H

which is an algebra map, meaning the diagonal is a homomorphism

∆(xy) = ∆(x)∆(y)

and satisfies “coassociativity” meaning that if

∆(x) =∑ij

xi ⊗ xj

then

(1⊗∆)∆(x) = (1⊗∆)∑ij

xi ⊗ xj

=∑ij

xi ⊗(∑

kl

xkj ⊗ xlj

)

=∑ij

(∑kl

xki ⊗ xli

)⊗ xj

= (∆⊗ 1)∑ij

xi ⊗ xj

= (∆⊗ 1)∆(x).

Further, we will assume Hopf algebras are graded and connected(meaning H0 = k · 1).

Remark 23.2. A Hopf algebra is essentially the same as a groupscheme.


Remark 23.3. From coassociativity, Hopf algebras satisfy

∆2 =∑ijk

xi ⊗ xj ⊗ xk

Remark 23.4. We say H is commutative if multiplication is commu-tative, and H is cocommutative if

∆(x) =∑

xi ⊗ xj =∑

xj ⊗ xi.

23.2. Examples of Hopf Algebras.

Example 23.5. One example of a Hopf algebra is the free associativealgebra on x1, . . . , xn given by

k [x1, . . . , xn]

where elements are of the form aww with w = xi1 · · · xij , with mul-tiplication given by w ·w ′ = ww ′. Further,

∆(xi) = 1⊗ xi + xi ⊗ 1.Then,

∆(w) = ∆(xi1) · · ·∆(xij)

=(1⊗ xi1 + xi1 ⊗ 1

)· · ·(1⊗ xij + xij ⊗ 1

)=∑S⊂[j]

∏a∈S

xia ⊗∏b∈[j]\S

xib .

Consider the Hopf square map

m∆ : H→ H

w 7→ ∑S⊂[j]

∏a∈S

xia ⊗∏b∈[j]\S

xib .

This is inverse riffle shuffling!You can use this to find the eigenvectors for riffle shuffling, see

Persi’s paper Hopf algebras and Markov chains, two examples anda theory, with Amy Pang and Arun Ram.

Example 23.6. The ring of symmetric functions is an algebra with thefollowing coproduct. Choose the e basis, for example (or we couldhave chosen h or p with the same formula as follows, with h or preplacing e,) and define ∆ by

∆(en) =

n∑i=0

ei ⊗ en−i.

92 AARON LANDESMAN

Question 23.7. What is the Hopf square mapm∆.

The Hopf square map is Kolmogorov’s model of rock breaking.

Remark 23.8. You can develop all of symmetric function theory fromthe Hopf algebras perspective. Zelevinsky cleaned this up, and thesame arguments give you the representations of GLn(Fq).

Example 23.9. Consider all finite simple unlabeled graphs. This isan algebra by taking linear combinations so that

HG =∑

agg

where product is given by disjoint union. We can grade this by thenumber of vertices. The coproduct is given by

∆(g) =∑s⊂V(g)

gs ⊗ gsc ,

where gs is the induced subgraph. This is both commutative andcocommutative.

Example 23.10. Consider labeled graphs, where each vertex is la-beled with a unique identifier from one up to |V(g)|. Then, ele-ments are finite linear combinations and multiplication is given byg1 · · · g2 = g1

∐g2 with labeling of g2 all higher than labelings of g1.

Comultiplication is given by

∆(g) =∑s⊂V(g)

gs ⊗ gsc

with labelings for the second graph in the tensor product moveddown to 1.

23.3. What did Hopf do? In the 1930s the great mathematicians ofthe world were trying to figure out the topology of the classical groups.That is, they were trying to compute the cohomology of things likeO(n). The homotopy type of the orthogonal group is fairly hopeless,even now, since this is computing homotopy types of spheres. Peo-ple were doing this one group at a time. Cartan would do U(3), andsomeone else would do O(4). Hopf realized that cohomology has acup product. Because one is working with a group, one also has acoproduct on the cohomology algebra. Hatcher discusses this storyis his book, in the section on H-spaces.

Hopf algebras also appear in the development of algebraic groups,and it allows one to work over finite fields and more general fields.

The story of more relevance to us is the following. Next time, wewill define a combinatorial Hopf algebra with a character ρ. This


has to do with the zeta function of the poset, in the poset case. Thetheorem says that there is a terminal object in this category whichis quasi-symmetric functions. When the character map to quasi-symmetric functions, the image is the chromatic polynomial. We’llsee this in more detail next class.

Definition 23.11. For G a graph, the chromatic polynomial PG(r) isthe number of ways to color a graph with r colors.

Definition 23.12. The Stanley chromatic polynomial is PG(x) is de-fined as follows. Suppose P1, . . . ,Pc ≥ 0with

c∑i=1

Pi = 1

and look at the probability that this coloring is proper. This is a sym-metric function of P1, . . . ,Pc.

24. 11/16/16

24.1. Definition of combinatorial Hopf algebras.

Definition 24.1. A Hopf algebra is a connected graded Hopf algebrawith a coproduct

∆ : H→ H⊗H

with ∆(x) ∈∑ni=0Hi ⊗ Hn−i for x ∈ Hn, and ∆(xy) = ∆(x)∆(y)

together with a unit, a counit, and an antipode (though the antipodefollows from the other properties). A combinatorial Hopf algebra isa Hopf algebra together with a character

ζ : H→ k

meaning ζ(ab) = ζ(a)ζ(b).

Remark 24.2. The characters, hom(H,k) form a group by

φ ∗ψ(x) = mkφ⊗ψ∆(x) =∑

φ(x1) ·ψ(x2).

We also have

H∆−→ H⊗H φ⊗ψ−−−→ H0 ⊗H0

ζ−→ k

Remark 24.3. We have X(H) is abelian, if A is cocommutative.

94 AARON LANDESMAN

24.2. Examples.

Definition 24.4. A poset is graded if it has a unique maximal element1 and unique minimal element 0, and all chains from 0 to 1 havethe same length. The rank of a poset is the length of the maximalchain. The rank of an element is the rank of the poset between 0 to x(meaning the sub-poset of all elements less than or equal to x).

Example 24.5. Here are some examples of graded posets:(1) A chain(2) A cube B3, the boolean algebra on 3 elements, graded by the

number of elements in the set.(3) In general Bn, the boolean algebra on n elements which is

graded by the number of elements in the set.(4) The poset of divisors of n. This is a product of chains whose

lengths are the powers of the distinct primes appearing in theexpansion of n.

(5) Set partitions of n. For example, when n = 3, we have 1, 2, 3at the bottom, (12, 3), (13, 2), (1, 23) in the middle and 123.

(6) Given a graded poset and two elements, the poset betweenthose two elements is another graded poset.

Example 24.6 (Rota’s Hopf algebra). Take R to be the set of all formallinear combinations of graded finite posets. We have a product P ·Qis given by the product poset P ×Q, with ordering on the productgiven by (p,q) ≤ (p ′,q ′) if p ≤ p ′ and q ≤ q ′. The coproduct isgiven by sending

∆(p) =∑0≤x≤1

[0, x]⊗ [x, 1]

where [x,y] means the subposet of z with x ≤ z ≤ y. Further, wetake ζ(P) = 1 for all P.

Definition 24.7. This is the zeta function of the poset ζ−1(x) = µ[0, x]with µ[x, x] = 1 and

∑x≤z≤y µ[x, z] = 0. These two relations define µ

uniquely by inducting from the bottom of the poset.

Definition 24.8. Let P be all isomorphism classes of finite posets, notnecessarily graded. We take P ·Q to be the disjoint union P

∐Q. The

coproduct ∆(P) :=∑I⊂P I⊗ P \ I, where I ranges over all downward

closed ideals meaning x ∈ I,y ≤ x =⇒ y ∈ I. Here the character isdefined by ζ(P) = 1.


Remark 24.9. We have a map

J : (P, ζ)→ (,ζ)

by J(P) = I ⊂ P.

Example 24.10. Take the Hopf algebra QSym defined by k[x1, . . .]with xi non-commuting variables.

ForMα,Lα two compositions, the coproduct∆(Mα) =∑α=βγMα⊗

Mγ. We define the product by the usual product in the ring. This isgraded by degree.

Question 24.11. What is the Hopf square mapm∆(x)?

We define

ζQSym(Mα) =

1 if α = (n) or empty0 otherwise

We have Λ ⊂ QSym is a combinatorial Hopf algebra with the samecharacter.

Theorem 24.12. For any combinatorial Hopf algebra (H, ζ), there existsa unique Hopf morphism ψ : (H, ζ) → (QSym, ζQSym). For h ∈ Hn,ψ(h) =

∑α a composition of n ζαmα. If α = (α1, . . . ,αn) a composition of n,

we have

H∆k−1−−−→ H⊗k → Hα1 ⊗ · · ·Hαn

ζ∗k−−→ k

with the middle map being the tensor product of the projections H→ Hαi .

25. 11/18/16

Persi predicts this will be the last lecture about Hopf algebras.

25.1. What do Hopf algebras have to do with card shuffling? Re-call

Qa(w) =

(a+n− 1− d(w−1)

n

)/an.

This has the feature that

Qa ∗Qb = Qab.

This is related to P-partitions, as we saw in previous classes. Thesealso relate to quasi-symmetric functions, as we saw.

96 AARON LANDESMAN

Remark 25.1. We can say this in an alternate way: We have a Markovchain Ka(w,w ′) = Qa(w ′w−1).

It’s a theorem of Persi and Ken Brown that Ka(w,w ′) is diagonal-izable, but first proved by Phil Hanlen. (The proof uses algebraictopology in a funny way.) The eigenvalues are real, and equal to1, 1/a, 1/a2, . . . , 1/an−1 with multiplicity of 1/an−k = c(n,k), with

c(n,k) := # w ∈ Sn : w has k cycles.

Example 25.2. (1) When k = n, we have c(n,n) = 1.(2) When k = n− 1, we have c(n,n− 1) =

(n2

), corresponding to

transpositions.

Question 25.3. What are the eigenvectors of this Ka(w,w ′) matrix?

The Hopf machinery gives us the eigenvectors.

Example 25.4. Here, let’s answer the question: What are the(n2

)eigenvectors with eigenvalue 1/a? We’ll index the eigenvectors byi, j for i < j, and call the wth component fi,j(w). We have

fij(w) =

1 if i and j are adjacent in w in order−1 if i and j are adjacent in w and out of order0 otherwise.

Observe that ∑1≤i<j≤n

fij(w) = 2d(w) −n− 1.

This implies that

Eid(d(xk)) =

n− 1

2

(1−

1

2k

).

where Eid means you start at the identity.We also see P(w) − n−2

3 is an eigenfunction for 1/a2 where P(w)is the number of peaks in w (where the middle of three consecutiveplaces is bigger than the places on both ends. We similarly get

Eid(P(xk))n− 2

3

(1−

1

22k

).


25.2. Lyndon words. Say we have an ordered alphabet of non-commutingvariables, 1, 2, . . ..

Definition 25.5. A word is Lyndon if it is lexicographically strictlysmaller than any cyclic shift.

Example 25.6. We have 1213 is Lyndon, but 1212 is not Lyndon, sinceshifting twice recovers the same word.

Theorem 25.7. Any word w has a unique decomposition w = `1`2 · · · `nwith `1 ≥ `2 ≥ · · · ≥ `k, with lexicographic ordering.

Example 25.8. We have 32114 has a decomposition as 3, 2, 114.

25.3. The standard bracketing of Lyndon words. If L is Lyndon,it has a unique decomposition ` = `1 · · · `2 into nontrivial Lyndonwords with `2 the longest right hand Lyndon factor.

Example 25.9. We can write 13245 = 13, 245 with 245 the longestright hand Lyndon factor.

Definition 25.10. The standard bracketing λ(`) is recursively de-fined as follows:

λ(i) = i

λ(`1, `2) = [λ(`1λ(`2)]

with [w,w ′] = ww ′ −w ′w. We have

λ(13245) = [λ(13), λ(245)][13− 31, 2 (45− 54) − (45− 54) 2]

Definition 25.11. For w = `1 · `2 · · · `k, the symmetrization

Sym(w) =∑σ∈Sk

k∏i=1

λ(`σ(i)).

Fact 25.12. If w has no repetitions, then Sym(w) is a sum of wordswith all distinct letters, with coefficients ±1.

Theorem 25.13. On Sn, Ka(w,w ′) has n! eigenfunctions fw(•) for w inSn, where

fw(w′) = the coefficient of w ′ in Sym(w).

The corresponding eigenvalue is 1an−k

where k is the number of left to rightminima in w.

98 AARON LANDESMAN

Example 25.14. Taking w = 32145 w has three minimal 3, 2, 1. Theseare the record values. The number of permutations with k recordvalues is the sterling number c(n,k).

Remark 25.15. We (Persi, McGrath, and Pitman) showed that anyfunction of w which only depends on the biggest cycles is “close torandom” after one Q2 shuffle.

Remark 25.16. Exactly the same description of the right eigenfunc-tions holds for any combinatorial Hopf algebra Markov chain whereH is cocommutative.

There is a similar description of the left eigenfunctions.

Remark 25.17. To learn more, see what Amy Peng has done.

26. 11/28/16

We are familiar with the classical basesmλ, eλ,pλ,hλ, sλ. But, thereare also

(1) Pλ(x1, . . . , xn; t) the Hall-Littlewood polynomials(2) Jα(x1, . . . , xn) the Jack polynomials(3) Pλ(x1, . . . , xn; t,q)

Today, we’ll talk about Hall-Littlewood polynomials.Hall was studying enumerative group theory. If A is an abelian

group, A is the direct product of its Sylow p subgroups. Let λ =

1n1 · · · rnr . SayMλ = p|λ|. Let

gλµν(p) = # subgroups H ⊂Mλ of type µ and cotype ν .

Now, fix λ,µ,ν and gλµν(Γ) has a formula using the symmetricfunctions uλ with λ a partition and u a variable. We can write Hto denote the algebra given by elements of the form

∑λ aλuλ. Make

H into an algebra by

uµuν =∑λ

gλµν(p)uλ.

He showed this is a commutative associative algebra freely gener-ated by uΓ .

We can now map intoΛ, the space of symmetric functions by uΓ 7→er. Hall observed it is better to map uΓ 7→ p(

n2)er. So, this is not a

map over Z, but it is a map

H→ Λ⊗Z Z [1/p] .

The image of uλ on the right hand side is Pλ(x1, x2, . . . ; t), with t = 1p .


Littlewood defined these pλ(x1, . . . , xn; t) directly, and showed theyfit together compatibly so that one may take n→∞.

Example 26.1. We have

p42(x1, x2; t) = x41x22 + x

21x42 + (1− t)x31x

33.

Fact 26.2. These gλµν(p) and Pλ(x1, . . . ; t) are crucial in writing down

(1) the character theory of GLn(Fq)(2) the spherical functions (i.e., the K bi-invariant functions) on

GLn(k)/Kwhere k is a local field and K is a maximal compactsubgroup.

(3) the distribution of the critical group of a random graph (JasonFulman makes the connection in a nice way between sand-piles and Hall-Littlewood polynomials)

(4) K-P-Z found a nonlinear system of partial differential equa-tions. Alexei Borodin and Ivan Corwin, made sense of theseequations in terms of Hall-Littlewood polynomials.

Here are some more facts of a different kind.

Fact 26.3. (1) The Pλ(x; t) are an orthogonal basis for Λ with co-efficients for Λ⊗Z [t] with 〈pλ,pµ〉 = ζ−1λ (t)δλµ with

ζλ(t) = ζλ

n∏i=1

(1− xλi

)−1and

ζλ =∏i

ididi!

and λ = 1a12a2 · · · . Furthermore, Pλ =∑µ≤λw

λµmµ for some

wλµ with wλλ = 1.(2) When t = 0, Pλ = sλ.(3) When t = 1,Pλ = mλ.

Example 26.4. Take λ = 21 and mλ = c(p2)× c(p), where c(n) :=Z/n.

When µ = 1, we have H = c(p). We can have ν corresponding toc(p)× c(p) = (1, 1), or ν = 2 corresponding to c(p2).

What is g211,11?

100 AARON LANDESMAN

Say p = 3. We have

H = (0, 0) , (0, 1) , (0, 2) ,H = (0, 0) , (3, 1) , (6, 2) ,H = (0, 0) , (6, 1) , (3, 2) .

Here, we have g211,11(p) = p. We also have g2111,2 = 1, with

H = (0, i)0≤i<p .

Fact 26.5. gλµν(p) is a polynomial in p.

Theorem 26.6 (Hall). (1) If cλµν = 0, then gλµν = 0, with cλµν theLittlewood-Richardson coefficients defined by

sλsν =∑λ

cµν6λsλ.

(2) If cλµν 6= 0, then gλµν has degree n(λ) −n(µ) −n(ν), with n(λ) =∑ni=1(i− 1)λi.

(3) We have gλµν = gλνµ.

Remark 26.7. These gλµν arise also naturally in the study of dvr’s R.If M is a finitely generated R-module, then M ∼= ⊕ni=1p/pλi . That is,these are classified by partitions λ. If |R/p| = q, then gλµν(q) is thenumber of modules of shape µ and cotype ν inmλ.

27. 11/30/16

Definition 27.1. A Gelfond pair is a pair (G,H) for G a group andH a subgroup, so that for any two functions on H, the convolutioncommutes.

Proposition 27.2. We have A is equivalent to G,H Gelfond, and B =L(G/H) = V ⊕ V1 ⊕ · · · ⊕ Vk, where the Vi are multiplicity free.

We have Sn,Sk× Sn−k and L(Sn/Sk× Sn−k = V0⊕ · · · ⊕Vk. Here,Vi corresponds to λ = (n− i, i) for 0 ≤ i ≤ k. Then,

L(G/H) = IndGH(1) = V0 ⊕ · · · ⊕ Vk.

Frobenius reciprocity says each Vi has the same 1-dim space of H-fixed vectors, call them s0, s1, . . .. Then,

si(ux) = si(x)

for all u, x, and si(id) = 1. These si from i = 1k are the sphericalfunctions


Example 27.3. If G = O(n),H = O(n − 1), then (G,H) is Gelfondand G/H = Sn−1. Then, L(G/H) = ⊕∞

i=0Vi.

Then, we have Sn/Sk × Sn−k with si(x) and x a k-set of [n]. Thisonly depends on x through all |x ∩ 1, 2, . . . ,n | = d. Then sn(x) =Pni (d) are the Dual-Hahn polynomials. If P(x) is left H invariant,the probably P(hx) = P(x), with P(i) =

∑x si(x)P(x) with si the ith

spherical function.

Remark 27.4. Saxl classified all subgroups H of Sn (and more gener-ally all finite groups of Lie type) where the quotient Sn/H is Gelfond.

Apart from some small particular cases, like the alternating groups,the main cases are H equal to Sk× Sn−k and S2 o Sn/2 (the hyperocta-hedral group) when n is even.

Example 27.5. We have GLn(R) and O(n, R) is a Gelfond pair.Here, in GLn, O(n) acts on Σ with Σ n× n symmetric (with Σq =

qΣqT ). Then, the isotropy subgroup of this action is the orthogonalgroup. Then,

L(GLn/On) = ⊕λVλwhere λ ranges over all partitions with at most n parts.

The spherical functions are called zλ. This is a polynomial in theeigenvalues of Σ.

Say zλ(Σ) are polynomials in the eigenvectors of Σ. Then, forx1, . . . , xn ∈ Rn, we can take

Σ :=1

n!

∑(xi − x)

(xj − x

)T ,

The principle components are the eigenvectors of Σ. So, an impor-tant math problem is: what is the probability distribution of theeigenvectors. If we assume our data comes from a normal distri-bution, which has a true covariance.

It turns out that zλ(x1, . . . , xn) We have

zλ(x1, . . . , xn) =∑µ`n

zµsλ(µ)Pµ,

where Pµ are the zonal polynomials.

Remark 27.6. Zonal polynomials are spherical functions forGLn/O(n).They were invented to do mathematical statistics. They are a basisfor symmetric functions. They are a special case of MacDonald poly-nomials.

102 AARON LANDESMAN

28. 12/2/16

Today we’ll discuss MacDonald polynomials.

28.1. Macdonald Polynomials. For today, we write F := Q(q, t).

Definition 28.1. The inner product

〈pλ,pµ〉 := zλ(q, t)δλµwhere

zλ(q, t) :=`(λ)∑r=1

1− qλi

1− tλi.

Remark 28.2. If you’d like, for intuition, you can pretend q > 0, t <1, which makes these inner products positive.

Theorem 28.3. Let < denote the lexicographic order. (If n = 4 then 14 <212 < 22 < 31 < 4.) For every partition λ, there exist unique uniquepolynomials

Pλ(x;q, t) ∈ ΛFso that

(1)

Pλ = mλ +∑µ<λ

uλµmµ

and(2) If λ 6= µ.

〈Pλ,Pµ〉 = 0Definition 28.4. The polynomials Pλ(x,q, t) are the Macdonald poly-nomials.

Here are some special cases of the theorem.

Example 28.5. (1) If q = t, we have Pλ = sλ.(2) If q = 0, we have Pλ are the Hall-Littlewood polynomials.(3) Let α > 0. Set q = tα, let q→ 1 so that t→ 1. Then,

1− qm

1− tm=1− tαm

1− tm→ α,

so the inner product becomes

〈pλ,pµ〉 = δλµzλa`(λ).These are the Jack symmetric functions. When α = 2, theseare the zonal polynomials from the last lecture.


(4) When t = 1, Pλ(x,q, 1) = mλ.(5) When q = 1, we have P(x, 1, t) = eλ ′ .

We have

zλ(q−1, t−1) = (q−1t)|λ|zλ(q, t).

28.2. Proof of Theorem 28.3. We now prove Theorem 28.3We have inner products

Π(x,y;q, t) =∏i,j

(txiyj;q)∞(xiyj;q)∞

where

(a;q)∞ =

∞∏r=1

(1− aqr) .

Lemma 28.6. We have

Π (x,y;q, t) =∑λ

z−1λ (q, t)Pλ(x)Pλ(y).

Proof. Then,

logπ(x,y;q, t) =∑i,j

∞∑r=0

. log(1− xiyjqr) − log(1− txyjqr)−1

So, ∑i,j

∞∑i=0

∞∑n=1

1

n(xiyjq

r)n (1− tn)

Then,∑i,j

∞∑r=0

∞∑n=1

1

n

(xiyjq

r)n

(1− tn) =∑i,j

∞∑n=1

1

n

1− tn

1− qnxni y

nj

=

∞∑n=1

1

n

1− tn

1− qnPn(x)Pn(y).

Therefore,

Π(x,y;q, t) =∞∏n=1

e1n1−tn

1−qnPn(x)Pn(y)

=∏n

∞∑mn=0

1

m

(1

n

1− tn

1− qnPn(x)Pn(y)

)mn.

104 AARON LANDESMAN

Here’s another fact:

Lemma 28.7. Let uλ and vλ be two bases of ΛnF ¡ working in x1, . . . , xn.Then, the following are equivalent:

(1) We have

〈uλ, vλ〉q,t = δλµ

(2) We have ∑λ

uλ(x)vλ(y) = Π(x,y;q, t)

Proof. Let p∗λ := zλ(q, t)pλ. Then,

〈p∗λ,pµ〉 = δλµ.

Then, expanding,

uλ =∑ρ

aλρp∗ρ.

Also,

vµ =∑σ

bµσpσ.

Then,

〈uλ, vµ〉q,t =∑ρ

aλρbµρ.

Then, vλ,uµ are dual bases if and only if∑ρ

aλρbµρ = δλµ.

Now, note that the second condition of Lemma 28.7 holds if andonly if ∑

λ

uλ(x)vλ(y) =∑ρ

p∗ρ(x)pρ(y),

by Lemma 28.6. This in turn is equivalent to∑λ

aλρbλσ = δρσ.

Exercise 28.8. Verify this claimed equivalence. Hint: Take A =(aλρ)

and B = (bµσ), and use that AB = I if and only if BA = I.


Now, the idea to finish the proof is to work in x1, . . . , xn and con-struct an F linear map D : ΛnF → ΛnF satisfying

(1)

Dmj =∑µ≤λ

cλµmµ

(2)

〈Df,g〉q,t = 〈f,Dg〉q,t

(3) We have λ 6= µ then cλλ 6= cλµ.In other words, D is diagonalizable and self adjoint with distincteigenvalues. The Macdonald polynomials are the eigenvectors ofD.

We’ll construct this D next time, probably, but first let’s checkuniqueness, assuming this D exists satisfying properties.

Proposition 28.9. For every partition λ of length ≤ n, there exists aunique symmetric polynomial Pλ ∈ ΛnF such that

Pλ =∑µ≤λ

uλµmµ

with uλµ ∈ F and uλλ = 1 and DPλ = cλλPλ.

The main content of this is the uniqueness (and existence will bepostponed until next time when we construct D.

Proof. Suppose both conditions in the proposition statement hold.Then,

DPλ =∑µ≤λ

uλµDmµ

=∑ν≤µ≤λ

uλµcµνmν.

Also,

cλλPλ =∑ν

cλλuλνmν.

Therefore, equating coefficients,

cλλuλν =∑ν≤µ≤λ

uλµcµν.

106 AARON LANDESMAN

This implies

(cλλ − cνν)uλν =∑

µ≤µ≤λuλµcµν

if ν ≤ λ. If uλλ = 1, this gives us a recursive way of computing uλµ’s.Then, orthogonality is free because

cλλ〈pλ,pµ〉q,t = 〈pλ,Dpµ〉q,t

= cµµ〈pλ,pµ〉q,t.

29. 12/5/16

Recall that last time we were trying to define Macdonald polyno-mials. We continue with this goal today.

29.1. Review. Let Λn,F denote the ring of symmetric functions in nvariables over the field F = Q(q, t). We have an inner product

〈pλ,pµ〉q,t = δλµzλ

`(λ)∏i=1

1− qλi

1− tλi.

The claim (the existence of Macdonald polynomials) is that for all λthere exist unique polynomials Pλ(x,q, t) ∈ λn,F so that

Pλ(x,q, t) = mλ +∑µ<λ

uµλmµ

with uµλ ∈ F, and

〈Pλ,Pµ〉q,t = 0

for λ 6= µ.To construct these polynomials, Macdonald builds an operator

D : : Λn,F → Λn,F

satisfying(1)

Dmλ =∑µ≤λ

cµλmµ.

(2)

〈Df,g〉q,t = 〈f,Dg〉q,t

for f,g ∈ Λn,F.(3) For λ 6= µwe have cλλ 6= cµµ.


The Macdonald polynomials are then taken to be the eigenvectorsof D.

29.2. Defining D. We now define the operator D explicitly. Let

∆ :=∏

1≤i<j≤n(xi − xj) =

∑w∈Sn

ε(w)xwδ

where δ = (n− 1,n− 2, . . . , 1, 0) . Define

Tt,xif(x1, . . . , xn) = f(x1, . . . , txi, . . . , xn)Tq,xif(x1, . . . , xn) = f(x1, . . . ,qxi, . . . , xn).

Then, define D by

D :=1

∆

n∑i=1

(tt,xi∆) Tq,xi .

Remark 29.1. Observe,

D =1

∆

n∑i=1

(tt,xi∆) Tq,xi

n∑i=1

∏i 6=j

txi − xjxi − xj

Tq,xi .

Then, Macdonald proves the three properties by manipulatorics. There-fore, the Macdonald polynomials exist.

29.3. Examples of Macdonald polynomials.

Example 29.2. We have

P3,1(x,q, t) =(q+ 1)(t− 12)

(3q2t+ q2 + 2qt+ 2qt+ 2q+ t+ 2

)24 (qt− 1)2 (qt+ 1)

p41(x)

+ · · ·−(q+ 1) (q− 1)2 (t+ 1)

(t2 + 1

)4 (qt− 1)2 (qt+ 1)

p4(x).

where pi are the power sums, so p4(x) =∑ni=1 x

ni .

Remark 29.3. Everybody in this field, save Persi, uses the computer.The package to use is John Stembridge’s (at Michigan) Maple pack-age there’s also Mike Zabrocki’s homepage (at York in Canada), andsage programs.

Persi computed this by asking his friend Bergeron.

There’s a particular integral form

Hµ(x;q, t) = Pλ(x

1− t;q, t).

108 AARON LANDESMAN

Example 29.4. In this case, we have

H3,1(x) =1

24(q+ 1)

(q2t+ 3q2 + 2qt+ 3q+ 3t+ 1

)p41

+ · · ·− 14(q+ 1) (q− 1)2 (t− 1)p4.

If we now expand H3,1 in the Schur functions, we obtain

H3,1(x;q, t) = s4 +(q2 + q+ t

)s3,1 +

(q2 + qt

)s22 +

(q3 + q2t+ qt

)2s211 + q

3s14 .

We also have

H3(x;q, t) = s3 + (q2 + q)s21 + q3s13

H21(x;q, t) = s3 + (q+ t)s21 + qs13

H111(x;q, t) = s3 + (t2 + t)s21 + t3s13

In general, we have the Macdonald positivity conjecture:

Theorem 29.5 (Haiman). Defining Kλµ(q, t) by

Hµ =∑λ

Kλµ(q, t)sλ

and Kλµ(q, t) have positive integral coefficients.

Remark 29.6. The proof involves a lot of algebraic geometry andHilbert schemes, surprisingly! However, this means no one has beenable to read it, since no one knows both!

29.4. Understanding the operator D in an alternate manner.

Definition 29.7. Let Pk be the set of partitions of k. Let q, t > 1.Define

πq,t(λ) =Z

zλ(q, t)

Here,

zλ(q, t) = zλ`(λ)∏i=1

1− qλi

1− tλi.

Here,

Z =(q,q)λ(q, t)λ


where (x,y)k is the Pochhammer symbol

(x,y)k :=k−1∏i=0

(1− xyi

).

This determines a probability measure on Pk, meaning that theπq,t(λ) add to 1. Taking q = t, we get

πq,q(λ) =Z

zλ.

Taking

q = tα

and letting q approach 1, we get

πα(λ) =Z

zλα−`(λ).

This is useful in biology, known as Ewen’s sampling formula.

Question 29.8. How do we pick λ at random using πq,t(λ)?

Finding Macdonald’s operatorD translates to a Markov chainM(λ, λ ′)on Pk. That is, from λ, there is a method of picking λ ′. Each row ofthe matrix is a probability distribution. It is easy to run this matrix.Further, πq,t(λ) is a stationary distribution. That is,∑

λ

πq,t(λ)M(λ, λ ′) = πq,t(λ′).

So, πq,t(λ) is a left eigenvector with eigenvalue 1 forM(λ, λ ′).In order to describe the spectral theory ofM(λ, λ ′), we have

P(x;q, t) =1

cλ(q, t)

∑ρ`|λ|

(z−1ρ∏

(1− tρi)Xλρ(q, t))pρ.

with Xλρ defined by

Xλρ(q, t) :=∑µ

χλρKµλ(q, t).

The Kλµ are the Koska numbers as we defined previously. It hap-pens that Kµλ are equal to

Kµλ(q, t) :=∑ρ

z−1ρ χµρX

µρ(q, t).

Finally, χλµ are the character of the symmetric group correspondingto the partition λ.

110 AARON LANDESMAN

Theorem 29.9. M(λ, λ ′) has expected value Rλλ∈Pk . Here,

Rλ =t

qk − 1

`(λ)∑i=1

(qλi − 1

)t−i.

The first eigenvalue is 1, the second eigenvalue is

t

qk − 1

(qk − 1

t+q− 1

t2

).

Here, the eigenfunctions are

fλ(ρ) = Xλρ(q, t)

`(ρ)∏i=1

(1− qρi)

which are orthogonal in L2(πq,t).In our last lecture on Wednesday, we’ll say more about what this

D operator is.

30. 12/7/16

There are two schools applying Macdonald polynomials and othercombinatorial techniques.

There are two schools.

30.1. School 1. The first school is Garsia, Haiman, Haglurd, andothers. These people do combinatorics and manipulatorics. Thereare three main results from this school.

(1) Haglunds formula for Hµ(x;q, t). Hagland’s formula says forµ ` n,

Hµ =∑σ∈Sn

qmaj(π(σ,µ))tinv(π(q,t)Lides(σ).

Here, L is the quasi-symmetric function. ides(σ) is the de-scent set of σ−1 (so ides stands for inverse descent). maj isthe major index, π is the permutation gotten by putting σ inµ from bottom to top. Finally inv.

Example 30.1. Suppose σ = 18362457. We have σ−1 = 15367482.Then, ides(σ) = 2, 5, 7. We associate to ides(σ) = 2, 5, 7 thecomposition 2, 3, 2, 1 (the difference set). Then, for w a com-position, Lw is the corresponding quasi-symmetric function.


Next, we explain what π(σ,µ) is. Say µ = 5, 3, 2, 2 is a par-tition of 12. Then, we construct

Then, take a permutation of 12. Say, 12, 2, 1, 3, 4, 7, 6, 11, 10, 9, 8.9, 5.Then, put this permutation in the shape from bottom to top

11 10 8 9 5

4 7 6

1 3

12 2

To compute maj (for major MacMahon) the major index wehave majπ(σ,µ) above is maj(12, 1, 4, 11) + maj(2, 3, 7, 10) +maj(6, 8) + maj(9) + maj(5) = 2 + 0 + 0 + 0 + 0. In general,to compute maj we sum maj over all columns, and maj(η) =∑i∈des(η) i.Finally, we describe inv For each row, sum 1 for each inver-

sion i < j but si > sj, except if there is an inversion at a < band immediately above a there is a c with a < c < b, thendon’t count that inversion.

(2) Haiman’s proof of Macdonald positivity: He showed you canexpress Hµ(x;q, t) =

∑λ Kq,t(λ,µ)sλ, where Kq,t(λ,µ) are pos-

itive integer polynomials in q and t.(3) The Shuffle conjecture.

Here is a nice quote from Persi about school 1: They’re careful,they write proofs, they write code, they’re great!

Example 30.2. Here is an expression for H21(x;q, t). H12(x;q, t) =tL2 + tL4 + qL1,2 + qL41 + qL2 + L∅.

30.2. School 2. The second school is the DAHA school. Both schoolsthink the other school is crazy. DAHA stands for “double affineHecke algebra.” This is the school of Macdonald, LLT (three au-thors), Arun Ram, Chednick, and so on.

112 AARON LANDESMAN

Example 30.3. IfG is a group andH is a subgroup, the Hecke algebrais the bi-invariant functions on G,

H(G,G) = f : G→ C : f(h1gh2) = f(g) .

These Hecke algebras come from affine root systems. There are a fewcommon types. We have type A with n− 1 dots in a row. These havecoefficients in q and t. Similarly, this works for all types.

Macdonald polynomials turn out to be type A. There are gener-alizations to all types. Roughly, they are the characters of finite di-mensional representations of the corresponding lie groups . The listchapter of Macdonald’s book is amazing. There is a classical analy-sis on lots of families of orthogonal polynomials, such as Hermite-Hahn, Chevychev, Laguare, Jacobi polynomials, and so on. There’sa sort of hierarchy, where higher members specialize to lower mem-bers in the limit. There is a highest member of this hierarchy. TheAskey Wilson polynomials lie at the top of the hierarchy, and prov-ably so.

It’s a fact that the Pλ associated to the root system C1 are exactlythe Askey Wilson polynomials.

30.3. Persi’s next project. Here is the next project Persi’s planningto work on. If you add two numbers, one usually has places whereone carries the one. A natural question is how often one must carrydigits. Say one adds n numbers in base b‘ John Holt looks at frac-tals, and he got interested in the carries process when the numbersare chosen at random. The carries form a Markov chain (mean-ing whether you carry at step m only depends on whether you didat step m − 1). We let M(i, j) denote the matrix whose i, j entryis the chance of carrying at j and following that one carries at i.This was claimed to be an amazing matrix. The eigenvalues are1, 1/b, . . . , 1/bn−1. The eigenfunctions were also quite explicit withEulerian numbers all over the place. The 1/bi are eigenvalues of rif-fle shuffling. Persi showed with Jason Fulman. The descent processof repeated riffle shuffling is exactly the same as the carry process.What does this have to do with this course. Suppose we have a finiteabelian group. Observe Z/p2 ⊃ Z/p. These tell us about the carriesprocess. More generally, suppose we haveMλ a finite abelian groupof type λ. Say Hµ is a subgroup of Mλ/Hµ of type ν. This splits ifand only if the partition splits into two parts.

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