This talk on $q$-analogs was held on Friday February 3, 2017 in MC 4045. The talk was given by Fengyang Wang.

## Abstract

Some of the most interesting results in combinatorics are generalizations of simple problems or theorems to problems or theorems parameterized by a parameter $q$ (often complex-valued). By taking the limit as $q\to1$, the simple problems can be recovered. Surprisingly, many of these $q$-analogs take similar forms, and are seen across a wide variety of problems that may initially seem unrelated.

Concepts from MATH 239 will be used. It is heavily recommended that people who have not taken MATH 239 or an equivalent course read some material on ordinary generating functions.

## Outline

• Review of prerequisite concepts. (15 minutes)

• Partitions of a positive integer.

• The Ferrers diagram.

• Ordinary generating functions.

• Permutations of $\{1\ldots n\}$ with $k$ inversions. (15 minutes)

• The $q$-analog of non-negative integers. (5 minutes)

• The $q$-analog of the gamma function and factorial. (5 minutes)

• Distinct partitions of $k$ that fit in $m\times n$ rectangles. (15 minutes)

• The $q$-analog of binomial coefficients. (15 minutes)

## Partitions of an Integer

Let $n\in\mathbf{N}\cup\{0\}$. Define a partition of $n$ to be a multiset

$p = \{a_1 \dots a_k\}$

with

$\sum_{i=1}^k a_i = n$

and each $a_i\in\mathbf{Z}^+$. We say that the partition $p$ has $k$ parts, the numbers $a_i$ are the parts, and $|p|$ (the size of $p$) is $n$.

As a notational convenience, define

$\operatorname{length}(p) = k$

to be the number of parts, and

$\operatorname{maximum}(p) = \max \{a_i : i \in [1, k]\}$

to be the largest part.

Let $n = 3$. Then the partitions of $n$ are $(3)$, $(2, 1)$, and $(1, 1, 1)$. If we take a partition and arrange it so that the largest part comes first, then we can represent it graphically.

For example, $(7, 5, 3, 1, 1)$ is a partition of $17$. Graphically, we can draw a diagram of this, by making each part its own row:

This diagram is called the Ferrers diagram. If we transpose it, by interchanging rows and columns, we obtain

which is the Ferrers diagram of another partition of $17$: $(5, 3, 3, 2, 2, 1, 1)$. These two partitions of $17$ are called conjugates.

## Permutations

Denote by $S_n$ the set of bijections

$σ: \{1\dots n\}\to\{1\dots n\}$

We can think of permutations in several, equivalent ways. A useful way to think of a permutation is a rearrangement of the numbers $1$ to $n$. We can write a rearrangement out as a list. For example,

$σ = \begin{pmatrix} 1 & 2 & 3 & 4 & 5 \\ 3 & 5 & 1 & 2 & 4 \end{pmatrix} \in S_5$

is a permuatation with $σ(1) = 3$, $σ(2) = 5$, $σ(3) = 1$, $σ(4) = 2$, and $σ(5) = 4$.

A permutation may also be thought of as defining a new ordering on the items. For instance, we may think of the permutation $σ$ above as defining a new ordering $\prec$ with $3 \prec 5 \prec 1 \prec 2 \prec 4$. We may sometimes be interested in how different the order $\prec$ is compared to the standard order $<$. We can measure this degree of difference by looking at how many pairs of numbers are out of order.

More formally, define an inversion of permutation $σ$ to be an unordered pair $\{i < j\}$ such that $σ(i) > σ(j)$. The inversions are precisely the pairs of indices where $i < j$ but $j \prec i$ according to the order induced by $σ$.

For instance, the inversions in the permutation above are $\{1, 3\}, \{1, 4\}, \{2, 3\}, \{2, 4\}, \{2, 5\}$. The total number of inversions of permutation $σ$ is $5$, so we may write $\operatorname{inv}(σ) = 5$. In general, we define $\operatorname{inv}(σ)$ to be the number of inversions of permutation $σ$.

As an exercise, show that

$\max \{ \operatorname{inv}(σ) : σ \in S_n \} = \binom{n}{2}$

## Generating Functions

(During the talk, a brief overview of generating functions was given. This content is subsumed by the linked summary.)

## Counting Inversions

Fix $n$. We shift our focus to $S_n$, and we want to consider the weight function $\operatorname{inv}: S_n \to \mathbf{N}\cup\{0\}$.

Write

$F_n(q) = \sum_{σ\in S_n} q^{\operatorname{inv}(σ)}$

so that $F_n(q)$ is the ordinary generating function for $S_n$ with weight function $\operatorname{inv}$. Note that this generating function is actually a polynomial, because $S_n$ is a finite set.

To accomplish this task, we will first consider a recursive decomposition of permutations. Let $σ ∈ S_n$. We split $σ$ up into two pieces of information, which together are enough to recover $σ$:

• $σ(1)$; i.e., where $1$ gets sent to by $σ$. Note that $σ(1) ∈ \{1, \dots, n\}$.

• $σ' ∈ S_{n-1}$, a permutation that describes where $2$ to $n$ are sent by $σ$.

A sensible definition of $σ'$ would be:

$σ'(k) = \begin{cases} σ(k+1) & \text{if }σ(k-1) < m \\ σ(k+1) - 1 & \text{otherwise} \end{cases}$

Importantly, we can see that this defines a bijection

$φ: S_n → \{1, \dots, n\} \times S_{n-1}$

We will use this bijection to find a functional equation that allows us to determine $F_n(q)$. By applying the bijection, we find

\begin{align*} F_n(q) &= \sum_{σ\in S_n} q^{\operatorname{inv}(σ)} \\ &= \sum_{m=1}^n \sum_{σ'\in S_{n-1}} q^{\operatorname{inv}(φ^{-1}(m, σ'))} \\ \end{align*}

At this point, we ask: what is $\operatorname{inv}(φ^{-1}(m, σ'))$? Can we find a formula for this in terms of $\operatorname{inv}(σ')$? Indeed, we can. Note that, if $σ = φ^{-1}(m, σ')$, then $σ(1) = m$. But this is inverted with exactly $m-1$ other positions. The remaining inversions of $σ$ do not involve $σ(1)$, and hence must be inversions of $σ'$.

We therefore obtain the formula

$\operatorname{inv}(φ^{-1}(m, σ')) = m - 1 + \operatorname{inv}(σ')$

Substituting into our earlier work, we obtain

\begin{align*} F_n(q) &= \sum_{m=1}^n \sum_{σ'\in S_{n-1}} q^{\operatorname{inv}(φ^{-1}(m, σ'))} \\ &= \sum_{m=1}^n \sum_{σ'\in S_{n-1}} q^{m-1+\operatorname{inv}(σ')} \\ &= \sum_{m=1}^n \sum_{σ'\in S_{n-1}} q^{m-1} q^{\operatorname{inv}(σ')} \\ &= \sum_{m=1}^n q^{m-1} \sum_{σ'\in S_{n-1}} q^{\operatorname{inv}(σ')} \end{align*}

But now we recognize that the inner sum is in fact simply $F_{n-1}(q)$! Hence,

\begin{align*} F_n(q) &= \sum_{m=1}^n q^{m-1} F_{n-1}(q) \\ &= F_{n-1}(q) \sum_{m=1}^n q^{m-1} \\ &= F_{n-1}(q) \frac{1-q^n}{1-q} \end{align*}

so the functional equation we obtain is

$F_n(q) = \frac{1-q^n}{1-q} F_{n-1}(q)$

which is a recurrence relation. We know the empty permutation has no inversions. So in the degenerate (n = 0) case, we simply have

$F_0(q) = \sum_{σ\in S_0} q^{\operatorname{inv}(σ)} = q^0 = 1$

We now have a recurrence relation and a base case; therefore we can see that in general,

$F_n(q) = \prod_{k=1}^n \frac{1-q^k}{1-q}$

(NB: this summary is incomplete and under construction.)