# The symmetric functions catalog

An overview of symmetric functions and related topics

2023-09-22

## $q$-analogs

We have the following definitions:

$[n]_q \coloneqq \frac{1-q^n}{1-q}, \quad [n]_q! \coloneqq [n]_q [n-1]_q \dotsm [2]_q [1]_q, \quad \qbinom{n}{k}_q \coloneqq \frac{[n]_q!}{[k]_q! [n-k]_q!}.$

Furthermore, for a composition $\alpha \vDash n,$ we let

$\qbinom{n}{\alpha}_q \coloneqq \frac{[n]_q!}{[\alpha_1]_q! [\alpha_2]_q! \dotsm [\alpha_\ell]_q!}$

We have that

$\sum_{\pi \in \symS_n} q^{\maj(\pi)} = \sum_{\pi \in \symS_n} q^{\inv(\pi)} = [n]_q!.$

These identities are true for charge and cocharge as well. Combinatorial statistics on permutations with this distribution are called Mahonian.

Note that

$\qbinom{n}{k}_q = \sum_{w \in BW(n,k)} q^{\maj(w)} = \sum_{w \in BW(n,k)} q^{\inv(w)}$

where $BW(n,k)$ is the set of binary words of length $n$ with exactly $k$ ones. This generalize as follows. Let $W(n,\alpha)$ denote the set of words of length $n,$ with weight $\alpha.$ Then we have the following identity for the $q$-multinomial coefficients:

$\qbinom{n}{\alpha}_q = \sum_{w \in W(n,\alpha)} q^{\maj(w)} = \sum_{w \in W(n,\alpha)} q^{\inv(w)}.$

For proofs, see these lecture notes by D. Foata and G.-N. Han.

We define the Pochhammer symbol as $(a)_n \coloneqq a(a+1)\dotsm (a+n-1)$ and the $q$-Pochhammer symbol as

$(a;q)_n \coloneqq \prod_{j=0}^{n-1}(1-a q^j).$

### $q$-Lucas theorem

The $q$-Lucas theorem first appeared in [Eq. (1.2.4), Oli65] and later in [Prop. 2.2, Dés82]. It was perhaps known by Gauß already. A combinatorial proof was first given by V. Strehl [Str81].

Let $n=n_1 d + n_0$ and $k=k_1 d + k_0,$ where $0\leq n_0, k_0 < d.$ We have that

$\qbinom{n}{k}_q \equiv \binom{n_1}{k_1} \qbinom{n_0}{k_0}_q \mod \Phi_d$

where $\Phi_d$ is the $d^\thsup$ cyclotomic polynomial. In particular

$\qbinom{n}{k}_q = \binom{n_1}{k_1} \qbinom{n_0}{k_0}_q \text{ whenever } q=e^{2\pi i \frac{c}{d}}$

and $\gcd(c,d)=1.$ A nice proof using cyclic sieving is given in [Sag92].

### Various $q$-identities

Cauchy's $q$-binomial theorem states that

$\prod_{j=1}^n (1+yq^j) = \sum_{j=0}^n y^j q^{j(j+1)/2} \qbinom{n}{j}_q$
Problem (See background in I. Pak's survey Combinatorial inequalities, 2019).

Find a combinatorial proof that the coefficient of $q^\ell,$ $1\leq \ell \leq 1 + n k/2$ in $(1-q)\qbinom{n+k}{k}_q$ is non-negative.

The following is a very useful identity.

Theorem ($q$-Vandermonde).

We have that

$\qbinom{a+b}{c}_q = \sum_{j} q^{j(a-c+j)} \qbinom{a}{c-j}_q \qbinom{b}{j}_q.$
Theorem (See [PS19a]).

Let $Cyc(n) \subseteq \symS_n$ be the set of permutations consisting of a long cycle. Then

$\sum_{\sigma \in Cyc(n+1)} q^{\maj(\sigma)} \equiv \mu(n) \qquad \mod \Phi_n(q)$

where $\mu$ is the number-theoretical Möbius function.

## Catalan numbers and Fuß–Catalan numbers

The MacMahon $q$-analog of the Catalan numbers is the following:

\begin{align*} \catalan(n;q) &= \frac{1}{[n+1]_q}\qbinom{2n}{n}_q \\ &= \frac{[2n]_q!}{[n+1]_q![n]_q! } \\ &= q^{-n} \qbinom{2n}{n}_q - \qbinom{2n}{n+1}_q \\ &= \qbinom{2n}{n}_q - q \qbinom{2n}{n-1}_q \end{align*}

Note that

$\catalan(n;q) = \sum_{w \in DP(n)} q^{\maj(w)}$

where $DP(n)$ is the set of binary words of length $2n,$ with $n$ ones, such that the number of ones in any initial segment never exceeds the number of zeros in the same segment.

The $q$-analog of Fuß–Catalan numbers can be expressed as

\begin{align*} \catalan(n,m;q) &\coloneqq \frac{1}{[n]_q} \qbinom{(m+1)n}{n-1}_q \\ &= \frac{1}{[(m+1)n+1]_q}\qbinom{(m+1)n+1}{n}_q \\ &= \qbinom{(m+1)n}{n}_q - q \frac{[mn]_q}{[n]_q} \qbinom{(m+1)n}{n-1}_q. \end{align*}

The rational Catalan numbers $\catalan(a/b;q)$ are defined as

$\catalan(a/b;q)= \frac{1}{[a+b]_q} \qbinom{a+b}{a}_q.$

This is a $q$-analog of the number of paths from $(0,0)$ to $(b,a)$ staying weakly above the line $y=\frac{a}{b}x.$ These objects admit a cyclic sieving phenomena.

### q-Narayana numbers

The $q$-Narayana numbers are defined as

\begin{align*} N(n,k;q) &\coloneqq \frac{q^{k(k-1)}}{[n]_q} \qbinom{n}{k}_q \qbinom{n}{k-1}_q \\ &= \frac{1}{[k]_q} \qbinom{n-1}{k-1}_q \qbinom{n}{k-1}_q \\ &= q^{k(k - 1)- n} \left(\qbinom{n - 1}{k - 1}_q \qbinom{n + 1}{k}_q - \qbinom{n}{k - 1}_q\qbinom{n}{k}_q \right). \end{align*}

Note that $\catalan(n;q) = \sum_{k=1}^n N(n,k;q)$ and that

$\sum_{w \in DP(n)} q^{\maj(w)} t^{\text{peaks}(w)} = \sum_{k=1}^n t^k N(n,k;q).$

where we sum over all Dyck paths.

Example (Table of $N(n,k;q)$).
 $\;$ $\textbf{1}$ $\textbf{2}$ $\textbf{3}$ $\textbf{4}$ $\textbf{1}$ $1$ $\textbf{2}$ $1$ $q^2$ $\textbf{3}$ $1$ $q^2+q^3+q^4$ $q^6$ $\textbf{4}$ $1$ $q^2+q^3+2q^4+q^5+q^6$ $q^6+q^7+2q^8+q^9+q^{10}$ $q^{12}$

In [Thm. 6, Brä04b], it is proved that

$N(n,k+1;q) = \schurS_{2^k}(q,q^2,\dotsc,q^{n-1}).$
Lemma.

Let $\xi$ be a primitive $d^\thsup$ root of unity where $d \mid n$ and $d>1.$ Then

$N(n,k;\xi) = \begin{cases} \binom{n/d}{k/d}\binom{n/d-1}{k/d-1} &\text{ if k \equiv_d 0} \\ \binom{n/d}{(k-1)/d}\binom{n/d-1}{(k-1)/d} &\text{ if k \equiv_d 1} \\ 0 &\text{ otherwise.} \end{cases}$
Proof.

We have that $N(n,k;q)$ is equal to \begin{align*} N(n,k;q) &= q^{k(k - 1)- n} \left(\qbinom{n + 1}{k} \qbinom{n - 1}{k - 1} - \qbinom{n}{k}\qbinom{n}{k - 1} \right). \end{align*}

As $q=\xi,$ $q$-Lucas tells us that at least one of the two factors in the last term vanish. Hence,

$N(n,k;\xi) = \xi^{k(k - 1)} \qbinom{n + 1}{k}_{\xi} \qbinom{n - 1}{k - 1}_{\xi}.$

Using $q$-Lucas on the first factor shows that $d\mid k$ or $d \mid (k-1)$ or the entire thing vanish. Analyzing these two cases immediately gives the formula above.

## $q$-identities and symmetric functions

### $q$-hook formula and Schur polynomials

Theorem (See ).

If $\lambda \vdash n,$ then

\begin{align*} f^\lambda(q) \coloneqq \sum_{T \in \SYT(\lambda)} q^{\maj(T)} &=\sum_{T \in \SYT(\lambda)} q^{\cocharge(T)} \\ &=q^{n(\lambda)} \frac{[n]_q!}{\prod_{\square \in \lambda} [h(\square)]_q} &=q^{n(\lambda)} \frac{\prod_{i=1}^n (1-q^i)}{\prod_{\square \in \lambda} (1-q^{h(\square)})}. \end{align*}

The identity involving cocharge can be found in [p.199, Ber09]. Note that the formula on p. 44 is not correct.

A $q$-analog in the skew case can be found in [Prop. 3.3, MPP18].

We have the following symmetry under conjugation:

$f^{\lambda}(q) = q^{\binom{n}{2}} f^{\lambda'}(1/q).$
Theorem (From [Mac95]).

Let $\lambda$ be a partition of $n.$ Then

$\schurS_\lambda(1,q,\dotsc,q^{m-1})= q^{n(\lambda)} \prod_{(i,j) \in \lambda} \frac{[m+c(i,j)]_q}{ [h(i,j)]_q} = q^{n(\lambda)} \prod_{1 \leq i < j \leq m } \frac{1-q^{\lambda_i-\lambda_j +j-i}}{1-q^{j-i}}$

and

\begin{align*} \schurS_\lambda(1,q,q^2,\dotsc) &= \frac{q^{n(\lambda)}}{ \prod_{\square \in \lambda} [h(\square)]_q } \\ &= \frac{f^\lambda(q)}{(1-q)(1-q^2) \dotsm (1-q^{n})} &= \frac{f^\lambda(q)}{(1-q)^n [n]_q!} \end{align*}

For a short proof, see [BD09].

Theorem (From [p. 363, Sta01]).

Let $\lambda/\mu$ be skew shape with $n$ boxes. Then

\begin{align*} \schurS_{\lambda/\mu}(1,q,q^2,\dotsc) &= \frac{ \sum_{T \in \SYT(\lambda/\mu)} q^{\maj(T)} }{(1-q)(1-q^2) \dotsm (1-q^{n})}. \end{align*}

By using RSK, one can show that

$$\sum_{\sigma \in \symS_n} t^{\maj(\pi)} q^{\maj(\pi^{-1})} = \sum_{\lambda \vdash n} \sum_{P,Q \in \SYT(\lambda)} t^{\maj(P)} q^{\maj(Q)}.$$

The Cauchy identity then gives that

$$\sum_{n \geq 0} \frac{z^n}{(q)_n (t)_n} \sum_{\pi \in \symS_n} t^{\maj(\pi)} q^{\maj(\pi^{-1})} = \prod_{i,j \geq 0} \frac{1}{1-z q^i t^j}.$$

Here, $(q)_n = (1-q)(1-q^2)\dotsb (1-q^n).$ This identity appears in [Eq. (7.117), Sta01]. Note that Stanley uses a different definition for $[n]_q!.$

Theorem (K. Killpatrick, [Kil05]).

Let $W_\lambda \subset \symS_n$ be the set of permutations of type $\lambda.$ Then

$\sum_{\pi \in W_\lambda} q^{\inv(\pi)}= \sum_{\pi \in W_\lambda} q^{\charge(\pi)}.$

### Distribution with fixed number of descents

In [Kei18], some results regarding two-row Young tableaux refined by descents are given. Let

$f^{\lambda/\mu}_d(q) \coloneqq \sum_{\substack{ T \in \SYT(\lambda/\mu) \\ |\DES(T)|=d} } q^{\maj(T)}.$
Theorem (W. Keith, 2018).

We have that for $j\leq m\geq k\geq 0,$ that

\begin{align} f^{(m,k)/(j)}_d(q) = q^{d^2} \left( \qbinom{m-j}{d} \qbinom{k}{d} - \qbinom{m+1}{d} \qbinom{k-j-1}{d} \right). \end{align}

In the same paper, it is shown that for $\lambda \vdash n,$ and $m \geq \lambda_1,$ we have that

\begin{equation*} f^{(m,\lambda)}_{n}(q) = q^{\binom{n+1}{2}}\schurS_{\lambda}(1,q,q^2,\dotsc,q^{m-1}). \end{equation*}

In [Che20], the author studies the major index distribution of so called increasing tableaux, introduced by O. Pechenik [Pec14]. In particular, X. Chen provides nice $q$-binomial formulas in the case of two-row skew shapes.

Theorem (See [Cor. 4.9(ii), Hua13] ).

Let $D \subseteq [n-1],$ and let $\alpha = comp(D)$ be the composition defined by $D$. Furthermore, let $\SYT(\alpha)$ be the set of ribbon standard Young tableaux such that row $i$ from the bottom has length $\alpha_i.$ We let $D= \{s_1,\dotsc,s_\ell\}$ and $s_0 \coloneqq 0$ and note that $s_j = \alpha_1+ \dotsb + \alpha_j.$

Then we have the following equalities.

\begin{align} \sum_{T \in \SYT(\alpha)} q^{\maj(T)} &= \sum_{ \substack{ \sigma \in \symS_n \\ \DES(\sigma) = D }} q^{\cocharge(\sigma)} = \sum_{ \substack{ \sigma \in \symS_n \\ \DES(\sigma) = D }} q^{\inv(\sigma)} = \\ \sum_{\mu \vdash n} \sum_{\substack{ T \in \SYT(\mu) \\ \DES(T) = D}} f^\mu(q) &= [n]_q! \det\left[ \frac{1}{ [s_j-s_{i-1}]_q! } \right]_{1 \leq i,j \leq \ell} \end{align}

We also have a connection with the $q$-multinomials:

$\sum_{ \substack{ \sigma \in \symS_n \\ \DES(\sigma) \subseteq D }} q^{\inv(\sigma)} = \qbinom{n}{\alpha}_q.$

Using properties of the Robinson–Schensted–Knuth correspondence or Jeu-de-taquin, one can show that for any $D\subseteq [n-1],$

\begin{equation*} \sum_{\substack{T \in \SYT(\lambda/\mu) \\ \DES(T) = D}} q^{\maj(T)} = \sum_{\nu \vdash n} \sum_{\substack{T \in \SYT(\nu)\\ \DES(T) = D}} c^{\lambda}_{\mu\nu} q^{\maj(T)}. \end{equation*}

and

\begin{equation*} \sum_{\substack{T \in \SYT(\lambda/\mu) \\ \DES(T) = D}} q^{\cocharge(T)} = \sum_{\nu \vdash n} \sum_{\substack{T \in \SYT(\nu)\\ \DES(T) = D}} c^{\lambda}_{\mu\nu} q^{\cocharge(T)}. \end{equation*}

See also S. Pfannerer–Mittas slides on evaluating similar polynomials at certain roots of unity.

Definition (See [Hua20b]).

A cyclic descent map for $\lambda/\mu$ is a pair $(\cDES, \phi)$ such that $\cDES$ sends elements in $\SYT(\lambda/\mu)$ to subsets of $[n],$ and $\phi : \SYT(\lambda/\mu) \to \SYT(\lambda/\mu)$ is a bijection, and

• $\cDES(T) \cap [n-1] = \DES(T),$
• $\cDES(\phi T) = \cDES(T) + 1,$
• $\emptyset \subsetneq \cDES(\phi T) \subsetneq [n].$

Lemma ().

Let $\lambda/\mu$ be a shape which is not a connected ribbon. Then $\lambda/\mu$ admits a cyclic descent map.

### Evalutations at roots of unity

Theorem (See [Rho10a]).

B. Rhoades observed that [Prop. 4.5 , Spr74] implies the following. Let $\lambda \vdash n$ and define $f^\lambda(q)$ as above. Let $\xi = \exp(2\pi i/n),$ and $d\geq 0.$ Then

$f^\lambda(\xi^d) = \chi^{\lambda}( c_n^d )$

where $\chi^{\lambda}$ is the irreducible $\symS_n$-character associated with $\lambda,$ and $c_n$ is the long cycle $(1,2,\dotsc,n).$ Recall that $\chi^{\lambda}$ can be computed using the Murnaghan–Nakayama rule, and in this particular case, there is a simple hook-formula for border-strip tableaux.

The following proposition generalizes the above identity. It is easy to prove using the identities for the principal specialization of Schur polynomials above.

Proposition (See [SSW11]).

Let $F(\xvec)$ be a homogeneous symmetric function of degree $n = md,$ such that $F(\xvec) = \sum_{\lambda \vdash n} \chi^F_\lambda \frac{\powerSum_\lambda(\xvec)}{z_\lambda} \text{ and define } f(q) \coloneqq \prod_{j=1}^n (1-q^j) F(1,q,q^2,\dotsc).$

Let $\xi$ be a primitive $n^\thsup$ root of unity. Then $f(\xi^d) = \chi^F_{m^d}.$

Theorem (From [Thm. 9.14 , DLT94]).

Let $\lambda \vdash nm.$ Then

$(-1)^{(m-1)n} \schurS_\lambda(1,q,q^2,\dotsc,q^{m-1}) \equiv K_{\lambda,n^m}(q) \text{ mod } \Phi_m(q).$

The paper [DLT94] has several evaluations of the transformed Hall–Littlewood polynomials at roots of unity. In particular, $K_{\lambda/\nu,\mu^k}(\xi)$ is up to some sign equal to $K^{(k)}_{\lambda/\nu,\mu},$ the number $k$-ribbon tableaux of shape $\lambda/\nu$ and weight $\mu.$

See also , where Lie group characters of type $A,$ $B,$ $C$ and $D$ are evaluated at roots of unity.

Theorem (Lee–Oh 2022).

Lee and Oh proves in [Thm. 16, LO22] the following. Let $d \mid n,$ and let $\xi$ be a primitive $d^\thsup$ root of unity. Then $\schurS_{\lambda/\mu}(1,\xi,\xi^2,\dotsc,\xi^{n-1})$ is 0, unless the $d$-quotient of $\lambda/\mu$ exists, in which case

$\schurS_{\lambda/\mu}(1,\xi,\xi^2,\dotsc,\xi^{n-1}) = \sign(\chi^{\lambda/\mu}((d^m))) \prod_{j=0}^{d-1} \schurS_{\lambda^{(j)}/\mu^{(j)}}(\underbrace{1,1,\dotsc,1}_{N/d}),$

where $|\lambda/\mu| = dm,$ and the $d$-quotient of $\lambda/\mu$ is given by $\left( \lambda^{(0)}/\mu^{(0)}, \dotsc, \lambda^{(d-1)}/\mu^{(d-1)} \right).$

The non-skew version of the above statement was first given in [Thm. 16, RSW04].

See also [Kum22] and [Chapter I, §5, Example 24, Mac95]. A generalization of this identity for flagged Schur polynomials is given in [Kum23].

In [Alb22], one notes that the algebra homomorphism $\phi_r$ that sends $\completeH_j$ to $\completeH_{j/r}$ if $r \mid j,$ and $0$ otherwise, corresponds to evaluating the symmetric function at $(x_i^{1/r}, \xi x_i^{1/r}, \xi^2 x_i^{1/r}, \dotsc, \xi^{r-1} x_i^{1/r})$ for each variable $i.$ The above theorem by Lee–Oh can then be lifted to the symmetric functions level, see [Thm. 3.1, Alb22].

## $q$-Pochhammer and basic hypergeometric series

When working with $q$-analogs, it is convenient to introduce some additional notation.

We let the $q$-Pochhammer symbol be defined as

$(a;q)_n \coloneqq (1-a)(1-aq)\dotsm (1-aq^{n-1}), \qquad (a;q)_0 \coloneqq 1.$

This is also known as the $q$-shifted factorial. From the definition, we have the relation

$(q^a;q)_{n+r} = (q^a;q)_{r} (q^{a+r};q)_n.$

By convention, $(a;q)_\infty \coloneqq \prod_{j \geq 0}(1-aq^j),$ so that

$(a;q)_n = \frac{(a;q)_\infty}{(aq^{n};q)_\infty}.$

This allows us to set

$(a;q)_{-n} \coloneqq \frac{1}{ (aq^{-n};q)_n } = \frac{ (-q/a)^n q^{\binom{n}{2}} }{ (q/a;q)_n }.$

One can then prove that

\begin{equation*} (a q^{-n};q)_{n} = q^{-\binom{n}{2}} \left( -\frac{a}{q} \right)^n (q/a;q)_{n}. \end{equation*}

We have the following identities:

\begin{align*} [m]_q &= \frac{ (q;q)_m}{(1-q)(q;q)_{m-1}} = \frac{(q^{m};q)_\infty}{(1-q)(q^{m+1};q)_\infty} \\ [m]_q! &= \frac{ (q;q)_m}{(1-q)^n} \\ \qbinom{m}{r}_q &= \frac{(q^{m-r+1};q)_r}{(q;q)_r} = \frac{(q^{r+1};q)_\infty (q^{m-r+1};q)_\infty }{ (q^{m+1};q)_\infty (q;q)_\infty }. \end{align*}

See the appendix of [GR04] for many more identities. The first edition is available online here.

### $q$-binomial theorem

The $q$-binomial theorem can be reformulated using $q$-hypergeometric series. We have the identity

$\sum_{n=0}^\infty \frac{(a;q)_n}{(q;q)_n}z^n = \frac{(az;q)_\infty}{(z;q)_\infty}.$