Algebraic Combinatorics

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Combinatorics

[Math 701, Spring 2021 lecture notes]

Darij Grinberg

January 3, 2022 (unfinished!)

1. What is this?

These are the notes for an introductory course on algebraic combinatorics held in the Spring Quarter 2021 at Drexel University¹. The topics covered are

• formal power series and their use as generating functions (Chapter 3);

• integer partitions andq-binomial coefficients (Chapter 4);

• permutations and their lengths, inversions and cycles (Chapter 5);

• alternating sums, the use of sign-reversing involutions and the combinatorial view on determinants (Chapter 6);

• the basics of symmetric polynomials, particularly Schur polynomials (Chap- ter 7).

Most (but not all) of these chapters are in a finished state (the final few sec- tions of Chapter 3 need details). However, as these notes have been written within a 3-months window, they are improvable both in detail and in coverage.

Some plans for improvement exist, but will likely have to wait for a second iter- ation of the course. Various topics, such as the matrix-tree theorem or the basis theorems for symmetric polynomials, are slated for eventual inclusion. Errors and confusions will be fixed whenever I become aware of them (any assistance is greatly appreciated!²).

Exercises of varying difficulty appear at the end of the text (Chapter A).

1The website of this course ishttps://www.cip.ifi.lmu.de/~grinberg/t/21s/

2Please send comments todarijgrinberg@gmail.com

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Acknowledgments

Thanks to the students in my Math 701 course for what was in essence an alpha-test of these notes. Some exercises have been adapted from collections by Richard Stanley, Martin Aigner, Donald Knuth, Miklós Bóna, Mark Wildon and Igor Pak. A math.stackexchange user named Mindlack has contributed the proof of Proposition 3.11.26.

2. Before we start...

This is a course on algebraic combinatorics. This subject can be viewed either as a continuation of enumerative combinatorics by other means (specifically, algebraic ones), or as the part of algebra where one studies concrete polynomials (more precisely, families of polynomials). For example, the Schur polynomials (which I think are one of the main stars in this course, but it will take us a while to get to them) can be viewed on the one hand as a tool for enumer- ating certain kinds of tableaux (essentially, tabular arrangements of numbers that increase along rows and columns), while on the other hand they form a family of polynomials with a myriad surprising properties, generalizing (e.g.) the Vandermonde determinant. I hope to cover both aspects of the subject to a reasonable amount in this course.

To understand this course, you are assumed to speak the language of rings and fields (we will mostly need the basic properties of polynomials and linear maps; we will define what we need about power series), and to have some basic knowledge of enumerative combinatorics (see below). My Math 533 course from Winter 2021 ( http://www.cip.ifi.lmu.de/~grinberg/t/21w/ ), and the references I gave therein, can help refresh your knowledge of the former. As for the latter, there are dozens of sources available (I made a list at https:

//math.stackexchange.com/a/1454420/ , focussing mostly on texts available online).

We letNdenote the set{0, 1, 2, 3, . . .}of nonnegative integers.

We will need some basics from enumerative combinatorics (see, e.g., [Newste19,

§8.1] for details, and [19fco, Chapters 1 and 2] for more details):

• (addition principle = sum rule) If A and B are two disjoint sets, then

|A∪B| =|A|+|B|.

• (multiplication principle = product rule) If A and B are any two sets, then|A×B| =|A| · |B|.

• (bijection principle)There is a bijection (= bijective map = invertible map

= one-to-one correspondence) between two sets X and Y if and only if

|X| =|Y|.

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• A set withn elements has 2ⁿ subsets, and has n

k

size-k subsets for any k∈ _R.

• A set with nelements hasn! permutations (= bijective maps from this set to itself).

• (dependent product rule)Consider a situation in which you have to make n choices (sequentially). Assume that you have a1 options available in choice 1, and then (after making choice 1) you havea2 options available in choice 2 (no matter which option you chose in choice 1), and then (after both choices 1 and 2) you havea3 options available in choice 3 (no matter which options you chose in choices 1 and 2), and so on. Then, the total # of ways to make allnchoices is a₁a₂· · ·an. (This is formalized in [Newste19, Theorem 8.1.19].)

A few words about binomial coefficients are in order:

Definition 2.0.1. For any numbersn andk, we set n

k

=







n(n−1) (n−2)· · ·(n−k+1)

k! , ifk ∈_N;

0, else.

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Note that “numbers” is to be understood fairly liberally here. In particular, n can be any integer, rational, real or complex number (or, more generally, any element in a Q-algebra), whereas k can be anything (although the only nonzero values of

n k

will be achieved fork ∈N, by the above definition).

Example 2.0.2. For any k∈ _{N, we have} −1

k

= (−1) (−1−1) (−1−2)· · ·(−1−k+1) k!

= (−1) (−2) (−3)· · ·(−k)

k! = (−1)^kk!

k! = (−1)^k. Ifn,k∈ _N_and _n≥_{k, then}

n k

= ^n!

k!(n−k)!. (2)

But this formula only applies to the case whenn,k ∈ _N_and _n ≥k. Our above definition is more general than it.

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Example 2.0.3. Let n∈ _{N. Then,} 2n

n

= ¹·3·5· · · · ·(2n−1) n! ·2ⁿ. Proof of Example 2.0.3. We have

(2n)!=1·2· · · · ·(2n)

= (1·3·5· · · · ·(2n−1))·(2·4·6· · · · ·(2n))

| {z }

=(2·1)·(2·2)···(2·n)

=₂ⁿ(1·2···n)

= (1·3·5· · · · ·(2n−1))·2ⁿ(1·2· · · · ·n)

| {z }

=n!

= (1·3·5· · · · ·(2n−1))·2ⁿn!.

Now, (2) yields 2n

n

= (2n)!

n!(2n−n)! = (2n)!

n!n! = (1·₃·₅· · · · ·(2n−₁))·₂ⁿn!

n!·n!

(since (2n)! = (1·3·5· · · · ·(2n−1))·2ⁿn!)

= (1·3·5· · · · ·(2n−1))·2ⁿ

n! = ¹·3·5· · · · ·(2n−1) n! ·2ⁿ. This proves Example 2.0.3.

Entire books have been written about binomial coefficients and their properties. See [Spivey19] for a recent text (and [GrKnPa94, Chapter 5] and [Grinbe15, Chapter 3] and [Knuth1, §1.2.6] and [Wildon19, Chapter 2] for elementary introductions). Here are two more basic facts that we will need ([19fco, Theorem 1.3.8] and [19fco, Proposition 1.3.6], respectively):

Proposition 2.0.4 (Pascal’s identity, aka recurrence of the binomial coefficients).

We have

m n

=

m−1 n−1

+

m−1 n

(3) for any numbersm and n.

Proposition 2.0.5. Let m,n ∈_Nsatisfy m<n. Then, m

n

=0.

Note that Proposition 2.0.5 really requiresm ∈ N. For example, 1.5 < 2 but 1.5

2

=0.375 6=0.

Yet another useful property of the binomial coefficients is the following ([19fco, Theorem 1.3.11]):

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Theorem 2.0.6 (Symmetry of the binomial coefficients). Let n ∈ _N and k ∈ _R.

Then,

n k

= n

n−k

.

Note the n ∈ _N requirement. Convince yourself that Theorem 2.0.6 would fail forn =−1 andk=0.

3. Generating functions

In this first chapter, we will discuss generating functions: first informally, then on a rigorous footing. You may have seen generating functions already, as their usefulness extends far beyond combinatorics; but they are so important to this course that they are worth covering twice in case of doubt.

Rigorous introductions to generating functions (and formal power series in general) can also be found in [Loehr11, Chapter 7 (in the 1st edition)], in [Henric74, Chapter 1], and (to some extent) in [19s, Chapter 7].³ A quick overview is given in [Niven69], and many applications are found in [Wilf09].

There are furthermore numerous books that explore enumerative combinatorics through the lens of generating functions ([GouJac83], [Wagner08], [Lando03]

and others).

3.1. Examples

Let me first show what generating functions are good for. Then, starting in the next section, I will explain how to rigorously define them. For now, I will work informally; please suspend your disbelief until the next section.

Theideabehind generating functions is easy: Any sequence (a0,a1,a2, . . .)of numbers gives rise to a “power series” a₀+a₁x+a₂x²+· · ·, which is called thegenerating function of this sequence. This “power series” is an infinite sum (an “infinite polynomial” in an indeterminate x), so it is not immediately clear what it means and what we are allowed to do with it; but before we answer such questions, let us first play around with these power series and hope for the best. The following four examples show how they can be useful.

3.1.1. Example 1: The Fibonacci sequence

Example 1. The Fibonacci sequence is the sequence (f0, f1, f2, . . .) of integers defined recursively by

f0 =0, f₁ =1, fn = fn−1+ fn−2 for eachn ≥2.

3Bourbaki’s [Bourba03, §IV.4] contains what might be the most rigorous and honest treatment of formal power series available in the literature; however, it is not the most readable source, as the notation is dense and heavily relies on other volumes by the same author.

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Its entries are known as theFibonacci numbers. Here are the first few of them:

n 0 1 2 3 4 5 6 7 8 9 10 11

fn 0 1 1 2 3 5 8 13 21 34 55 89

Let us see what we can learn about this sequence by considering its generating function

F(x) := f0+f₁x+ f2x²+ f3x³+· · ·

=0+1x+1x²+2x³+3x⁴+5x⁵+· · · . We have

F(x) = f0+ f1x+ f2x²+ f3x³+ f4x⁴+· · ·

=0+1x

| {z }

=x

+ (f₁+f0)x²+ (f2+f₁)x³+ (f3+ f2)x⁴+· · ·

(since f0 =0 and f1=1 and fn = fn−1+ fn−2 for each n≥₂)

=x+ (f1+ f0)x²+ (f2+ f1)x³+ (f3+ f2)x⁴+· · ·

| {z }

=(^f1x²+f₂x³+f₃x⁴+···)⁺(^f0x²+f₁x³+f₂x⁴+···)

(here we are hoping that this manipulation of infinite sums is indeed legitimate)

=x+f₁x²+f₂x³+f₃x⁴+· · ·

| {z }

=x(^f1x+f₂x²+f₃x³+···)

=x(F(x)−f₀)=xF(x) (since f₀=0)

+f₀x²+ f₁x³+ f₂x⁴+· · ·

| {z }

=x²(^f0+f₁x+f₂x²+f₃x³+···)

=x²F(x)

=x+xF(x) +x²F(x) = x+x+x² F(x).

Solving this equation for F(x) (assuming that we are allowed to divide by 1−x−x²), we get

F(x) = ^x

1−_x−_x² = ^x

(1−_φ+x) (1−_φ₋x)^, whereφ+ = ¹+√

5

2 and φ− = ¹−√ 5

2 are the two roots of the quadratic polynomial 1−x−x² (note that φ+ and φ− are sometimes known as the “golden ratios”; we haveφ+ ≈1. 618 and φ− ≈ −0.618). Hence,

F(x) = ^x

(1−φ+x) (1−φ−x)

= √¹

5 · ¹

1−φ+x −√¹

5 · ¹

1−φ−x (4)

(by partial fraction decomposition).

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Now, what are the coefficients of the power series 1

1−αx for anα ∈ _C? Let me first answer this question forα =1. Namely, I claim that

1

1−x =₁+x+x²+x³+· · · _. ₍₅₎ Indeed, this follows by observing that

(1−x)1+x+x²+x³+· · ·

=1+x+x²+x³+· · ·−x

1+x+x²+x³+· · ·

=1+x+x²+x³+· · ·−x+x²+x³+x⁴+· · ·

=1

(again, we are hoping that these manipulations of infinite sums are allowed).

Note that the equality (5) is a version of thegeometric series formulafamiliar from real analysis. Now, for anyα ∈ C, we can substitute αxforx in the equality (5), and thus obtain

1

1−αx =1+αx+ (αx)²+ (αx)³+· · ·

=1+αx+α²x²+α³x³+· · · _. ₍₆₎ Hence, our above formula (4) becomes

F(x) = √¹

5· ¹

1−_φ+x −√¹

5· ¹ 1−_φ₋_x

= √¹

5·1+_φ+x+_φ₊²x²+_φ³₊x³+· · ·−√¹

5·1+_φ−x+_φ²₋x²+_φ₋³x³+· · · (by (6), applied toα =φ+ and again to α =φ−)

= √¹ 5

∑

k≥0

φ₊^k x^k−√¹ 5

∑

k≥0

φ^k₋x^k

=

∑

k≥0

1

√5·_φ^k₊− √¹ 5 ·_φ₋^k

x^k.

Now, for any given n ∈ N, the coefficient of xⁿ in the power series on the left hand side of this equality is fn (since F(x) = f0+ f1x+ f2x²+ f3x³+· · ·), whereas the coefficient ofxⁿ on the right hand side is clearly 1

√5·φⁿ₊−√¹ 5·φⁿ₋.

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Thus, comparing coefficients before xⁿ, we obtain fn = √¹

5 ·_φⁿ₊−√¹ 5 ·_φ₋ⁿ

assuming that “comparing coefficients” is allowed, i.e., that equal power series really have equal coefficients

= √¹

5 · ¹+√ 5 2

!n

−√¹

5· ¹−√ 5 2

!n

for any n ∈ N. This formula is known as Binet’s formula. It has many con- sequences; for example, it implies easily that lim

n→_∞

fn+1

fn

= φ+ = ¹+√ 5

2 ≈

1.618 . . .. Thus, fn ∼φⁿ₊ in the asymptotical sense.

3.1.2. Example 2: Dyck words and Catalan numbers

Before the next example, let us address a warmup question: What is the number of 2n-tuples that containn entries equal to 0 andn entries equal to 1 ?

(For example, for n = 2, these 2n-tuples are (1, 1, 0, 0), (1, 0, 1, 0), (1, 0, 0, 1), (0, 1, 1, 0), (0, 1, 0, 1), (0, 0, 1, 1), so there are 6 of them.)

Answer: The number is 2n

n

, since choosing a 2n-tuple that contains n entries equal to 0 and n entries equal to 1 is tantamount to choosing an n- element subset of {1, 2, . . . , 2n} (and we know that there are

2n n

ways to choose the latter).

Example 2. A Dyck word of length 2n (where n ∈ N) is a 2n-tuple that contains n entries equal to 0 and n entries equal to 1, and has the additional property that for eachk, we have

(# of 0’s among its first k entries)

≤(# of 1’s among its first kentries). (7) (The symbol “#” means “number”.)

Some examples: The tuples

(1, 0, 1, 0), (1, 1, 0, 0), (1, 1, 0, 1, 0, 0), (), (1, 0) are Dyck words. The tuples

(0, 1, 1, 0), (1, 0, 0, 1), (1, 1, 0), (1), (1, 1, 1, 0) are not Dyck words.

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A Dyck path is a path from the point (0, 0) to the point (2n, 0) in the Carte- sian plane that moves only using “NE-steps” (i.e., steps of the form (x,y) → (x+1,y+1)) and “SE-steps” (i.e., steps of the form(x,y)→(x+1,y−₁)) and never falls below the x-axis (i.e., does not contain any point(x,y) with y<0).

Examples: Forn=2, the Dyck paths from(0, 0) to(2n, 0)are

A Dyck path can be viewed as the “skyline” of a “mountain range”. For example:

Dyck path “mountain range”

The names “NE-steps” and “SE-steps” in the definition of a Dyck path refer to compass directions: If we treat the Cartesian plane as a map with the x-axis directed eastwards and the y-axis directed northwards, then an NE-step moves to the northeast, and an SE-step moves to the southeast.

Note that any NE-step and any SE-step increases the x-coordinate by 1 (that is, the step goes from a point with x-coordinate k to a point with x-coordinate k+1). Thus, any Dyck path from(0, 0)to(2n, 0)has precisely 2nsteps. Of these 2n steps, exactly n are NE-steps while the remaining n are SE-steps (because any NE-step increases the y-coordinate by 1, while any SE-step decreases the y-coordinate by 1). Since a Dyck path must never fall below the x-axis, we see that the number of SE-steps up to any given point can never be larger than the number of NE-steps up to this point. But this is exactly the condition (7) from the definition of a Dyck word, except that we are talking about NE-steps and SE-steps instead of 1’s and 0’s. Thus, there is a simple bijection between Dyck words of length 2n and Dyck paths(0, 0)→ (2n, 0):

• send each 1 in the Dyck word to a NE-step in the Dyck path;

• send each 0 in the Dyck word to a SE-step in the Dyck path.

So the # of Dyck words (of length 2n) equals the # of Dyck paths (from (0, 0) to(2n, 0)). But what is this number?

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Example: Forn = 3, this number is 5. Indeed, here are all Dyck paths from (0, 0) to(6, 0), and their corresponding Dyck words:

Dyck path Dyck word

(1, 1, 0, 0, 1, 0)

(1, 1, 1, 0, 0, 0)

(1, 0, 1, 0, 1, 0)

(1, 0, 1, 1, 0, 0)

(1, 1, 0, 1, 0, 0)

(We will soon stop writing the commas and parentheses when writing down words. For example, the word(1, 1, 0, 0, 1, 0) will just become 110010.)

Back to the general question.

For eachn∈ _{N, let}

cn = (# of Dyck paths (0, 0) →(2n, 0))

= (# of Dyck words of length 2n) (as we have seen above). Then, c0 = 1 (since the only Dyck path from (0, 0) to(0, 0) is the trivial path) and c₁ = 1 and c₂ = 2 and c₃ = 5 and c₄ = 14 and so on. These numbers cn

are known as theCatalan numbers. Entire books have been written about them, such as [Stanle15].

Let us first find a recurrence relation for cn. The argument below is best understood by following an example; namely, consider the following Dyck path

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from(0, 0)to(16, 0)(so the corresponding nis 8):

Fix a positive integer n. If D is a Dyck path from (0, 0) to (2n, 0), then the first return of D (this is short for “first return of D to the x-axis”) shall mean the first point onD that lies on the x-axis but is not the origin (i.e., that has the form (i, 0) for some integer i > 0). For instance, in the example that we just gave, the first return is the point(6, 0). IfDis a Dyck path from(0, 0)to(2n, 0), and if(i, 0) is its first return, then i is even⁴, and therefore we have i = 2k for somek ∈ {1, 2, . . . ,n}. Hence, for any Dyck path from(0, 0) to(2n, 0), the first return is a point of the form(2k, 0)for some k∈ {1, 2, . . . ,n}. Thus,

(# of Dyck paths from (0, 0) to (2n, 0))

=

∑

n k=1

(# of Dyck paths from (0, 0) to (2n, 0) whose first return is (2k, 0)). Now, let us fix somek ∈ {1, 2, . . . ,n}. We shall compute the # of Dyck paths from (0, 0) to (2n, 0) whose first return is (2k, 0). Any such Dyck path has a natural “two-part” structure: Its first 2k steps form a path from (0, 0)to(2k, 0), while its last (i.e., remaining) 2(n−k) steps form a path from(2k, 0) to(2n, 0). Thus, in order to construct such a path, we

• first choose its first 2ksteps: They have to form a Dyck path from (0, 0)to (2k, 0)that never returns to the x-axis until(2k, 0). Hence, they begin with a NE-step and end with a SE-step (since any other steps here would cause the path to fall below the x-axis). Between these two steps, the remaining 2k−2 = 2(k−1) steps form a path that not only never falls below the x-axis, but also never touches it (since (2k, 0) is the first return of our Dyck path, so that our Dyck path does not touch the x-axis between(0, 0) and (2k, 0)). In other words, these 2(k−1) steps form a path from (1, 1) to (2k−1, 1) that never falls below the y = 1 line (i.e., below the x-axis shifted by 1 upwards). This means that it is a Dyck path from (0, 0) to (2(k−1), 0) (shifted by (1, 1)). Thus, there are c_k−1 possibilities for this path. Hence, there arec_k−1 choices for the first 2ksteps of our Dyck path.

4Proof. The number of NE-steps before the first return must equal the number of SE-steps before the first return (because these steps have altogether taken us from the origin to a point on the x-axis, and thus must have increased and decreased the y-coordinate an equal number of times). This shows that the total number of steps before the first return is even.

In other words,iis even (because the total number of steps before the first return isi).

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• then choose its last 2(n−k) steps: They have to form a path from (2k, 0) to(2(n−k), 0) that never falls below the x-axis (but is allowed to touch it any number of times). Thus, they form a Dyck path from (0, 0) to (2(n−k), 0) (shifted by (2k, 0)). So there are c_n−k choices for these last 2(n−k) steps.

Thus, there are c_k−1c_n−k many options for such a Dyck path from (0, 0) to (2n, 0)(since choosing the first 2ksteps and choosing the last 2(n−k)steps are independent).

Let me illustrate this reasoning on the Dyck path from(0, 0) to(16, 0)shown above. This Dyck path has first return (6, 0); thus, the corresponding k is 3.

Since this Dyck path does not return to the x-axis before(2k, 0) = (6, 0), its first 2ksteps stay above the yellow trapezoid shown here:

In particular, the first and the last of these 2k steps are uniquely determined, while the steps between them form a diagonally shifted Dyck path that is filled in green here:

Finally, the last 2(n−k) steps form a horizontally shifted Dyck path that is filled in purple here:

Our above argument shows that there arec_k−1 choices for the green Dyck path andc_n−k choices for the purple Dyck path, thereforec_k−1c_n−k options in total.

Forget that we fixedk. Our counting argument above shows that

(# of Dyck paths from (0, 0) to (2n, 0) whose first return is (2k, 0))

=c_k−1c_n−k. (8)

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Now,

cn = (# of Dyck paths from (0, 0) to (2n, 0))

=

∑

n k=1

(# of Dyck paths from (0, 0) to (2n, 0) whose first return is (2k, 0))

| {z }

=ck−1cn−k

(by (8))

=

∑

n k=1

c_k−1c_n−k =c0c_n−1+c₁cn−2+c2cn−3+· · ·+c_n−1c0.

This is a recurrence equation for cn. Combining it with c0 = 1, we can use it to compute any value of cn recursively. Let us, however, try to digest it using generating functions!

Let

C(x):=

∑

n≥0

cnxⁿ =c0+c1x+c2x²+c3x³+· · · . Thus,

C(x) = c0+c1x+c2x²+c3x³+· · ·

=1+ (c0c0)x+ (c0c1+c1c0)x²+ (c0c2+c1c1+c2c0)x³+· · ·

sincec₀=1

and cn =c0cn−1+c₁cn−2+c2cn−3+· · ·+cn−1c0 for eachn>0

=1+x

(c₀c₀) + (c₀c₁+c₁c₀)x+ (c₀c₂+c₁c₁+c₂c₀)x²+· · ·

| {z }

=(^c0+c₁x+c₂x²+···)²

(because if we multiply out(^c0+c₁x+c₂x²+···)²

and collect like powers ofx, we obtain exactly(c₀c₀)+(c₀c₁+c₁c₀)x+(c₀c₂+c₁c₁+c₂c₀)x²+···)

=1+x





c0+c1x+c2x²+· · ·

| {z }

=C(x)







2

=1+x(C(x))².

This is a quadratic equation in C(x). Let us solve it by the quadratic formula (assuming for now that this is allowed – i.e., that the quadratic formula really does apply to our “power series”, whatever they are). Thus, we get

C(x) = ¹±√ 1−4x

2x . (9)

The±sign here cannot be a+sign, because if it was a+, then the power series on top of the fraction would not be divisible by 2x (as its constant term would be 2 and thus nonzero⁵). Thus, (9) becomes

C(x) = ¹−√ 1−4x

2x = ¹

2x

1−(1−4x)^1/2. (10)

5If you find this unconvincing, here is a cleaner way to argue this: Multiplying the equality

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How do we find the coefficients of the power series(1−4x)^1/2? For eachn∈ N, the binomial formula yields

(1+x)ⁿ =

∑

n k=0

n k

x^k =

∑

k≥0

n k

x^k. (11)

(Here, we have replaced the ∑ⁿ

k=0

sign by a ∑

k≥0

sign, thus extending the summa- tion from all k ∈ {0, 1, . . . ,n} to all k ∈ N. This does not change the value of the sum, since all the newly appearing addends are 0, as you can easily check.) Let us pretend that the formula (11) holds not only for n ∈ N, but also for n=1/2. That is, we have

(1+x)^1/2 =

∑

k≥0

1/2 k

x^k. (12)

Now, substitute−4x forxin this equality (here we are making the rather plausible assumption that we can substitute−4x for x in a power series); then, we get

(1−4x)^1/2 =

∑

k≥0

1/2 k

(−4x)^k =

∑

k≥0

1/2 k

(−4)^kx^k

= 1/2

0

| {z }

=1

(−4)⁰

| {z }

=1

x⁰

|{z}

=1

+

∑

k≥1

1/2 k

(−4)^kx^k

=₁+

∑

k≥1

1/2 k

(−₄)^kx^k. Hence,

1−(1−4x)^1/2 =1− 1+

∑

k≥1

1/2 k

(−4)^kx^k

!

=−

∑

k≥1

1/2 k

(−4)^kx^k.

(9) by 2x, we obtain 2xC(x) =1±√

1−4x. The left hand side of this equality has constant term 0, but the right hand side has constant term 1±1 (here, we are making the assumption that√

1−4xis a power series with constant term 1; this is plausible because√

1−4·0=1 and will also be justified further below). Thus, 0=1±1; this shows that the±sign is a− sign.

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Thus, (10) becomes C(x) = ¹

2x

1−(1−4x)^1/2

| {z }

=− _∑

k≥1

1/2 k

(−4)^kx^k

= ¹

2x −

∑

k≥1

1/2 k

(−4)^kx^k

!

=

∑

k≥1

1/2 k

−(−4)^kx^k 2x

| {z }

=2(−4)^k−1x^k−1

=

∑

k≥1

1/2 k

2(−4)^k⁻¹x^k⁻¹

=

∑

k≥0

1/2 k+1

2(−4)^kx^k

(here, we have substitutedk+_{1 for} kin the sum). Comparing coefficients beforexⁿ in this equality gives

cn =

1/2 n+1

2(−₄)ⁿ. (13)

This is an explicit formula forcn (and makes computation ofcn pretty easy!), but it turns out that it can be simplified further. Indeed, the definition of 1/2

n+₁

yields 1/2

n+1

= (1/2) (1/2−₁) (1/2−₂)· · ·(1/2−n) (n+1)!

= 1 2 · −1

2 · −3 2 · −5

2 · · · −(2n−1) 2 (n+1)!

= (1·(−1)·(−3)·(−5)· · · · ·(−(2n−1)))/2ⁿ⁺¹ (n+₁)!

= ((−1)·(−3)·(−5)· · · · ·(−(2n−1)))/2ⁿ⁺¹ (n+1)!

= (−₁)ⁿ(1·₃·₅· · · · ·(2n−₁))/2ⁿ⁺¹ (n+1)! .

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Thus, (13) rewrites as

cn = (−1)ⁿ(1·3·5· · · · ·(2n−1))/2ⁿ⁺¹

(n+₁)! ·2(−4)ⁿ

= ¹·₃·₅· · · · ·(2n−₁) (n+1)!

| {z }

= 1 n+1^·

1·3·5· · · · ·(2n−1)

(since(n+1)!=(n!n+1)·n!)

·(−₁)ⁿ·₂(−₄)ⁿ 2ⁿ⁺¹

| {z }

=2ⁿ

= ¹

n+1· ¹·3·5· · · · ·(2n−1)

n! ·2ⁿ

| {z }

=

2n n

(by Example 2.0.3)

= ¹ n+1

2n n

.

Hence, we have shown that

cn = ¹ n+1

2n n

. Moreover, we can rewrite this further as

cn = 2n

n

− 2n

n−1

(since another binomial coefficient manipulation⁶yields 1 n+1

2n n

= 2n

n

− 2n

n−1

).

Here is the upshot: The # of Dyck words of length 2niscn = ¹ n+1

2n n

. In other words, a 2n-tuple that consists ofn entries equal to 0 andn entries equal to 1 (chosen uniformly at random) is a Dyck word with probability 1

n+1. (There are also combinatorial ways to prove this; see, e.g., [GrKnPa94, §7.5, discussion at the end of Example 4] or [Stanle15, §1.6] or [Martin13] or [Loehr11, Theorem 1.56]⁷ or [Spivey19, §8.5, proofs of Identity 244]⁸.)

6See Exercise A.2.1.2(a)for this.

7Note that the “Dyck paths” in [Loehr11] differ from ours in that they use N-steps (i.e., steps (i,j)7→(i,j+1)) and E-steps (i.e., steps(i,j)7→(i+1,j)) instead of NE-steps and SE-steps, and stay above thex =y line instead of above the x-axis. But this notion of Dyck paths is equivalent to ours, since a clockwise rotation by 45^◦followed by a√

2-homothety transforms it into ours.

8Again, [Spivey19] works not directly with Dyck paths, but rather with paths that use E-steps (i.e., steps (i,j) 7→ (i+_1,j)) and N-steps (i.e., steps (i,j) 7→ (i,j+₁)) instead of NE-steps and SE-steps, and stay below the x = y line instead of above the x-axis. But this kind of Dyck paths is equivalent to our Dyck paths, since a reflection across thex=y, followed by a clockwise rotation by 45^◦followed by a√

2-homothety transforms it into ours.

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Here is a list of the first 12 Catalan numberscn:

n 0 1 2 3 4 5 6 7 8 9 10 11

cn 1 1 2 5 14 42 132 429 1430 4862 16 796 58 786 . 3.1.3. Example 3: The Vandermonde convolution

Example 3: TheVandermonde convolution identity(also known as theChu–Vandermonde identity) says that

a+b n

=

∑

n k=0

a k

b n−k

for any numbers a,b and anyn∈ _N (where “numbers” can mean, e.g., “complex numbers”).

Let us prove this using generating functions. For now, we shall only prove this for a,b ∈ N; later I will explain why it also holds for arbitrary (rational, real or complex) numbersa,b as well.

Indeed, fix a,b∈ N. Recall (from (11)) that (1+x)ⁿ =

∑

k≥0

n k

x^k

for eachn∈ _{N. Hence,}

(1+x)^a =

∑

k≥0

a k

x^k and

(1+x)^b =

∑

k≥0

b k

x^k and

(1+_x)^a⁺^b =

∑

k≥0

a+b k

x^k.

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Thus,

(1+x)^a

| {z }

=_∑

k≥0

a k

x^k

(1+x)^b

| {z }

=_∑

k≥0

b k

x^k

=

∑

k≥0

a k

x^k

!

k

∑

≥0

b k

x^k

!

=

∑

k≥0

a k

x^k

!

`≥

∑

0

b

`

x^`

!

=

∑

k≥0

∑

`≥0

a k

x^k

b

`

x^`

=

∑

k≥0

∑

`≥0

a k

b

`

x^k^+`

=

∑

n≥0

∑

n k=0

a k

b n−k

! xⁿ

(here, we have merged addends in whichxappears in the same power). Hence,

n

∑

≥0

∑

n k=0

a k

b n−k

!

xⁿ = (1+x)^a(1+x)^b = (1+x)^a⁺^b =

∑

k≥0

a+b k

x^k

=

∑

n≥0

a+b n

xⁿ. Comparing coefficients in this equality, we obtain

∑

n k=0

a k

b n−k

=

a+b n

for eachn∈ _N.

This completes the proof of the Vandermonde convolution identity fora,b∈ _N.

3.1.4. Example 4: Solving a recurrence

Example 4. The following example is from [Wilf04, §1.2]. Define a sequence (a0,a1,a2, . . .) of numbers recursively by

a0 =1, an+1 =2an+nfor all n ≥_0.

Thus, the first entries of this sequence are 1, 2, 5, 12, 27, 58, 121, . . .. This sequence appears in the OEIS (= Online Encyclopedia of Integer Sequences) as A000325, with index shifted.

Can we find an explicit formula foran (without looking it up in the OEIS)?

Again, generating functions are helpful. Set

A(x) = a0+a₁x+a2x²+a3x³+· · · .

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Then,

A(x) = a0+a₁x+a2x²+a3x³+· · ·

=1+ (2a0+0)x+ (2a₁+1)x²+ (2a2+2)x³+· · ·

(sincea₀=1 anda_n+1 =2an+n for all n≥0)

=1+2

a₀x+a₁x²+a₂x³+· · ·

| {z }

=xA(x)

+0x+1x²+2x³+· · ·

| {z }

=x(0+1x+2x²+3x³+···)

=1+2xA(x) +x

0+1x+2x²+3x³+· · ·. (14) Thus, it would clearly be helpful to find a simple expression for 0+1x+ 2x²+3x³+· · ·. Here are two ways to do so:

First way: We assume that our power series (whatever they actually are) can be differentiated (as if they were functions). We furthermore assume that these derivatives satisfy the same basic rules (sum rule, product rule, quotient rule, chain rule) as the derivatives in real analysis. (Again, these assumptions shall be justified later on.)

Denoting the derivative of a power series f by f⁰, we then have 1+x+x²+x³+· · ·⁰ =1+2x+3x²+4x³+· · · . Hence,

1+2x+3x²+4x³+· · · =1+x+x²+x³+· · ·⁰ = 1

1−x 0

(since (5) yields 1+x+x²+x³+· · · = ¹

1−x). Using the quotient rule, we can easily find that

1 1−x

0

= ¹

(1−x)²^{, so that} 1+2x+3x²+4x³+· · · =

1 1−x

0

= ¹

(1−x)²^. ⁽¹⁵⁾ The left hand side of this looks very similar to the power series 0+1x+2x²+ 3x³+· · · that we want to simplify. And indeed, we have the following:

0+1x+2x²+3x³+· · · =x

1+2x+3x²+4x³+· · ·

| {z }

= 1

(1−x)²

=x· ¹ (1−x)²

= ^x

(1−x)²^. ⁽¹⁶⁾

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Second way: We rewrite 0+1x+2x²+3x³+· · · as an infinite sum of infinite sums:

0+1x+2x²+3x³+· · ·

= x¹ + x² + x³ + x⁴ + · · · + x² + x³ + x⁴ + · · · + x³ + x⁴ + · · · + x⁴ + · · · . .. ...

=

∑

k≥1

x^k+x^k⁺¹+x^k⁺²+· · ·

| {z }

=x^k·(¹⁺^x⁺^x²⁺^x³^+···)

=_x^k· 1 1−x

(by (5))

=

∑

k≥1

x^k· ¹

1−x = ¹

1−x ·

∑

k≥1

x^k

| {z }

=x¹+x²+x³+···

=x·(¹⁺^x⁺^x²⁺^x³^+···)

=x· 1 1−x

(by (5))

= ¹

1−_x ·x· ¹

1−_x = ^x

(1−x)²^. ⁽¹⁷⁾

(We used some unstated assumptions here about infinite sums – specifically, we assumed that we can rearrange them without worrying about absolute con- vergence or similar issues – but we will later see that these assumptions are well justified. Besides, we obtained the same result as by our first way, which is reassuring.)

Having computed 0+1x+2x²+3x³+· · ·, we can now simplify (14), obtain- ing

A(x) =1+2xA(x) +x

0+1x+2x²+3x³+· · ·

| {z }

= x

(1−x)²

=1+2xA(x) +x· ^x (1−x)²^.

(25)

This is a linear equation in A(x). Solving it yields A(x) = ¹−2x+2x²

(1−x)²(1−2x)

= ²

1−_2x

| {z }

=2 ∑

k≥0

2^kx^k (by (6))

− ¹

(1−x)²

| {z }

=1+2x+3x²+4x³+···

(by (15), or alternatively by dividing the equality (17) byx)

(by partial fraction decomposition)

=2

∑

k≥0

2^kx^k

| {z }

=_∑

k≥0

2^k+1x^k

−1+2x+3x²+4x³+· · ·

| {z }

=_∑

k≥0

(k+1)x^k

=

∑

k≥0

2^k⁺¹x^k−

∑

k≥0

(k+1)x^k =

∑

k≥0

2^k⁺¹−(k+1)x^k. Comparing coefficients, we obtain

an =2ⁿ⁺¹−(n+1) for eachn ∈_N.

This is also easy to prove directly (by induction onn).

3.2. Definitions

The four examples above should have convinced you that generating functions can be useful. Thus, it is worthwhile to put them on a rigorous footing by first defining generating functions and then justifying the manipulations we have been doing to them in the previous section (e.g., dividing them, solving quadratic equations, taking infinite sums, taking derivatives, ...). We are next going to sketch how this can be done (see [Loehr11, Chapter 7 (in the 1st edition)] and [19s, Chapter 7] for some details).

First things first: Generating functions are not actually functions. They are so-called formal power series (short FPSs). Roughly speaking, a formal power series is a “formal” infinite sum of the form a0+a1x+a2x²+· · ·_{, where} _x is an “indeterminate” (we shall soon see what this all means). You cannot substitute x = 2 into such a power series. (For example, substituting x = 2 into 1

1−x = 1+x+x²+x³+· · · would lead to the absurd equality 1

−₁ = 1+₂+₄+₈+₁₆+· · ·.) The word “function” in “generating function” is somewhat of a historical artifact.

3.2.1. Reminder: Commutative rings

In order to obtain a precise understanding of what FPSs are, we go back to abstract algebra. We begin by recalling the concept of a commutative ring. This

(26)

is defined in any textbook on abstract algebra for more details, but we recall the definition for the sake of completeness.

Informally, a commutative ring is a set K equipped with binary operations

⊕, and and elements 0 and 1 that “behave” like addition, subtraction and multiplication (of numbers) and the numbers 0 and 1, respectively. For example, they should satisfy rules like(a⊕_b)_c = (a_c)⊕(b_c).

Formally, commutative rings are defined as follows:

Definition 3.2.1. Acommutative ringmeans a setKequipped with three maps

⊕: K×K →K, : K×K →K, _: _K×_K →_K

and two elements 0∈ K and 1∈ K satisfying the following axioms:

1. Commutativity of addition: We have a⊕b =b⊕afor all a,b ∈ K.

(Here and in the following, we write the three maps⊕, and infix – i.e., we denote the image of a pair(a,b) ∈ K×Kunder the map ⊕by a⊕b rather than by⊕(a,b).)

2. Associativity of addition: We havea⊕(b⊕c) = (a⊕b)⊕cfor alla,b,c∈ K.

3. Neutrality of zero: We have a⊕0=0⊕a= afor all a ∈K.

4. Subtraction undoes addition: Leta,b,c ∈ K. We have a⊕b =cif and only if a=c b.

5. Commutativity of multiplication: We have ab =bafor all a,b ∈ K.

6. Associativity of multiplication: We have a(bc) = (ab)c for all a,b,c∈ K.

7. Distributivity: We have

a(b⊕c) = (ab)⊕(ac) and (a⊕b)c = (ac)⊕(bc) for all a,b,c ∈ K.

8. Neutrality of one: We have a1 =1a =afor all a∈ K.

9. Annihilation: We havea0 =0a =0for all a ∈ K.

[Note: Most authors do not include in the definition of a commutative ring, but instead require the existence of additive inverses for all a∈ K. This is equivalent to our definition, because if additive inverses exist, then we can define a bto be a⊕_b_where_b is the additive inverse of b.]

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The operations ⊕, and are called the addition, the subtraction and the multiplicationof the ringK. This does not imply that they have any connection with the usual addition, subtraction and multiplication of numbers; it merely means that they play similar roles to the latter and behave similarly. When confusion is unlikely, we will denote these three operations ⊕, and by +, −and ·, respectively, and we will abbreviate ab= a·bby ab.

The elements 0 and 1 are called the zero and the unity (or the one) of the ringK. Again, this does not imply that they equal the numbers 0 and 1, but merely that they play analogous roles. We will simply call these elements 0 and 1 when confusion with the corresponding numbers is unlikely.

We will use PEMDAS conventions for the three operations ⊕, and . These imply that the operationhas higher precedence than⊕and , while the operations ⊕ and are left-associative. Thus, for example, “ab+ac”

means (ab) + (ac) (that is, (ab)⊕(ac)). Likewise, “a−b+c” means (a−b) +c = (a b)⊕c.

Here are some examples of commutative rings:

• The setsZ,Q,Rand Care commutative rings. (Of course, the operations

⊕_, _andof these rings are just the usual operations+_, −_and ·_known from high school.)

• The set N is not a commutative ring, since it has no subtraction. (It is, however, something called acommutative semiring.)

• The matrix ringQ^m^×^m (this is the ring of allm×m-matrices with rational entries) is not a commutative ring form >1 (because it fails the “commutativity of multiplication” axiom). However, it satisfies all axioms other than “commutativity of multiplication”. This makes it a noncommutative ring.

• The set

Zh√ 5i

=ⁿa+b√

5 | a,b ∈ _Z^o

is a commutative ring with operations+, −and · inherited from R. This is because anya,b,c,d∈ _Zsatisfy

a+b√ 5

+c+d√ 5

= (a+c) + (b+d)√

5∈_Z^h√ 5i

;

a+_b√ 5

−_c+_d√ 5

= (_a−_c) + (b−_d)√

5∈_Z^h√ 5i

;

a+b√

5 c+d√ 5

= (ac+5bd) + (ad+bc)√

5∈ _Z^h√ 5i

. This is called a subring of R (i.e., a subset of R that is closed under the operations+, − and · and therefore constitutes a commutative ring with these operations inherited fromR).

Algebraic Combinatorics

Combinatorics

Darij Grinberg

January 3, 2022 (unfinished!)

Contents

1. What is this?

Acknowledgments

2. Before we start...

3. Generating functions

3.1. Examples

∑

∑

∑

∑

∑

∑

∑

∑

∑

∑

∑

∑

∑

∑

∑

∑

∑

∑

∑

∑

∑

∑

∑

∑

∑

∑

∑

∑

∑

∑

∑

∑

∑

∑

∑

∑

∑

∑

∑

∑

∑

∑

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∑

∑

∑

∑

3.2. Definitions