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\begin{document}

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\title[]{Gaussian inequality}
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\author{Tewodros~Amdeberhan  and David~Callan}

\address{Department of Mathematics,
Tulane University, New Orleans, LA 70118, USA}
\email{tamdeber@tulane.edu}

\address{Department of Statistics, 
University of Wisconsin-Madison, Madison, WI 53715, USA}
\email{callan@stat.wisc.edu}

\subjclass[2010]{11B65, 11A07.}

\date{\today}





\begin{abstract} We prove some special cases of Bergeron's inequality involving two Gaussian polynomials (or $q$-binomials).
\end{abstract}

\maketitle

\section{Introduction}

\bigskip
\noindent
We begin by recalling the $q$-analogues $[n]!_q=\prod_{j=1}^n\frac{1-q^j}{1-q}$ of factorials and the $q$-analogue
$\binom{n}k_q=\frac{[n]!_q}{[k]!_q[n-k]!_q}$ of binomial coefficients. Adopt the convention $[0]!_q=1$. It is well-known that these rational functions $\binom{n}k_q$ are polynomials, in $q$, also called \emph{Gaussian polynomials}, having non-negative coefficients which are also \emph{unimodal} and symmetric. Furthermore, there are several combinatorial interpretations of which we state two of them.

\bigskip
\noindent
A \emph{word} of length $n$ over the \emph{alphabet} set $\{0,1\}$ is a finite sequence $w=a_1\cdots a_n$. Construct 
$$\mathcal{W}_{n,k}=\{w=a_1\cdots a_n: \text{$w$ has $k$ zeros and $n-k$ ones}\}$$
and the \emph{inversion set} of $w$ as $\text{Inv}(w)=\{(i,j): \text{$i<j$ and $a_i>a_j$}\}$.
The corresponding \emph{inversion number} of $w$ will be denoted $\text{inv}(w)=\#\text{Inv}(w)$. Then, we have
$$\binom{n}k_q=\sum_{w\in\mathcal{W}_{n,k}}q^{\text{inv}(w)}.$$


\noindent
Yet, another formulation which would come to appeal to many combinatorialists is
$$\binom{a+d}a_q=\sum_{T}q^{area(T)}$$
where $T$ is a lattice path inside an $a\times d$ box and $area(T)$ is area above the curve $T$. 

\bigskip
\noindent
Given two polynomials $f(q)$ and $g(q)$, we write $f(q)\geq g(q)$ provided that $f(q)-g(q)$ has non-negative coefficients in the powers of $q$. 

\smallskip
\noindent
The well-known \emph{Foulkes conjecture} (see, for instance \cite{B2}) was generalized by Vessenes \cite{V}. She conjectured that 
\begin{align}
(h_b\circ h_c) - (h_a \circ h_d)
\end{align}
is \emph{Schur positive} (expands with positive integer coefficients in the Schur basis $\{s_{\mu}\}_{\mu\vdash n}$ of symmetric polynomials) whenever $a\leq b\leq c \leq d$, with $n=ad=bc$, and one writes $(h_n\circ h_k)$ for the \emph{plethysm} of complete homogeneous symmetric functions. A well-known fact is that $(h_n\circ h_k)(1,q)=\binom{n+k}{k}_q$. Moreover, any non-zero evaluation of a Schur function at $1$ and $q$ is of the form $q^i+q^{i+1}+\dots + q^j$ for some $i<j$. Exploiting these facts on the occasion of \cite{B}, and assuming that (1) holds, F. Bergeron (see also [4]) underlined that the evaluation of the difference in (1), at $1$ and $q$, would imply the following: 

\noindent
\begin{conjecture} Assume $0<a\leq b\leq c\leq d$ are positive integers with $ad=bc$. Then, the following difference of two Gaussian polynomials is symmetric and satisfies
\begin{align} \binom{b+c}b_q-\binom{a+d}a_q\geq0. \end{align}
\end{conjecture}

\smallskip
\noindent
One can associate a direct combinatorial meaning to Vessenes' conjecture in the context of representation theory of $GL(V)$. Indeed, if it holds true, it would signify that there is an embedding of the composite of symmetric powers $S^a(S^d(V))$ inside $S^b(S^c(V)) $, as $GL(V)$-modules. It may however be more natural to state that there is a surjective $GL(V)$-module morphism the other way around (which is also equivalent). Therefore each $GL(V)$-irreducible occurs with smaller multiplicity in $S^a(S^d(V))$ than it does in $S^b(S^c(V))$, and the conjecture reflects this at the level of the corresponding characters (with Schur polynomials appearing as characters of irreducible representations). 

\bigskip
\noindent
The sole attempt \cite{Z} toward resolving Conjecture 1 was made by F. Zanello, who attends to the special case $a\leq 3$, including the property of symmetry and unimodality.  A sequence of numbers is \emph{unimodal} if it does not increase strictly after a strict decrease. The author in \cite{Z} offers a strengthening of Conjecture 1 to the effect that

\begin{conjecture} Preserve the hypothesis in Conjecture 1. Then, the coefficients of the symmetric polynomial
\begin{align*} \binom{b+c}b_q-\binom{a+d}a_q \end{align*}
are non-negative and unimodal.
\end{conjecture}

\noindent
Notice that symmetry is clear, since both $\binom{b+c}b_q$ and $\binom{a+d}a_q$ are symmetric polynomials of the same degree, $ad=bc$.
We started out this project with the goal of  proving the below $3$-parameter special case of Conjecture 1 which we dubbed the \emph{$\beta$-Conjecture}. Namely,

\begin{conjecture} For integers $0<a< b$ and $\beta\geq1$, we have
\begin{align}\binom{b+\beta a}b_q\geq \binom{a+\beta b}a_q. \end{align} \end{conjecture}

\smallskip
\noindent
The case $\beta=1$ is trivial. However, our journey in this effort failed short of capturing the $\beta$-Conjecture in its fullest. In the sequel, we supply the details of our success in settling the particular instance $\beta=2$. Let's commence by stating one useful identity.

\begin{theorem} ($q$-analogue Vandermonde-Chu). The following holds true
\begin{align}\sum_{j\geq0}\binom{X}{Z-j}_q\binom{Y}j_qq^{j(X-Z+j)}=\binom{X+Y}Z_q.\end{align} 
\end{theorem}
\begin{remark} In view of (4), Conjecture 1 tantamount
\begin{align*}
\binom{b+c}b_q=\sum_{k=0}^{\pmb{b}}\binom{b}k_q\binom{c}k_qq^{k^2} \geq \sum_{k=0}^{\pmb{a}}\binom{a}k_q\binom{d}k_qq^{k^2}=\binom{a+d}a_q. \end{align*}
\end{remark}

\section{THE CASE $\beta=2$ and $q=1$}

\noindent
In this section, we wish to explain the resolution of the $\beta$-Conjecture for the ordinary binomial coefficients ($q=1$) while $\beta=2$, which elaborates a natural development. 

\smallskip
\noindent
For $a<b<c<d$ with $ad=bc$ and special case $c=2a,d=2b$, say $b=a+i,\,i\ge 1$, it would be desirable to find a bijective proof for
$$\binom{3a+i}{a+i}\geq \binom{3a+2i}{a}\, .$$
An injection from a set counted by the smaller number to one counted by the larger number would be nice but a better proof would be an expression for the difference as a sum of obviously positive terms. For $i=1$, we have 
$$\binom{3a+1}{a+1}-\binom{3a+2}{a} = \binom{3a+1}{a-1}\, ,$$
and the right-hand side is clearly positive. It seems for general $i=1,2,\dots,$ 
$$\binom{3a+i}{a+i} - \binom{3a+2i}{a} = \sum_{k=1}^i c_k(i) \binom{3a+i}{a-k}$$
for integers $c_k(i)$ and, furthermore, the $c_k(i)$ \emph{are all positive}. Here is a table for $c_k(i)$ when $1\leq k\leq i\leq8$:
\[
\begin{array}{cccccccc}
 1 & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} \\
 3 & 1 & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} \\
 6 & 6 & 1 & \text{} & \text{} & \text{} & \text{} & \text{} \\
 10 & 19 & 9 & 1 & \text{} & \text{} & \text{} & \text{} \\
 15 & 45 & 39 & 12 & 1 & \text{} & \text{} & \text{} \\
 21 & 90 & 120 & 66 & 15 & 1 & \text{} & \text{} \\
 28 & 161 & 301 & 250 & 100 & 18 & 1 & \text{} \\
 36 & 266 & 658 & 755 & 450 & 141 & 21 & 1 \\
\end{array}
\]
but appeared hard to get a handle on them. The evolution of our next progress begins with the discovery of
$$c_k(i)=\binom{i+k-1}{2k}+2\binom{i+k-1}{2k-1}-\binom{i}k.$$
Let's contract these coefficients as $c_k(i)=\frac{i+3k}{i+k}\binom{i+k}{2k}-\binom{i}k$, for $i, k\geq1$. Notice $c_0(i)=0$. We need some preliminary results.

\begin{lemma} We have
$$\binom{3a+2i}a=\sum_{k\geq0}\binom{i}k\binom{3a+i}{a-k}.$$ \end{lemma}
\begin{proof} This follows from the Vandermonde-Chu identity (Theorem 1 for $q=1$)
$$\binom{X+Y}Z=\sum_{k\geq0}\binom{X}k\binom{Y}{Z-k}$$
applied to $\binom{3a+2i}a=\binom{i+3a+i}a$ with $X=i, Y=3a+i$ and $Z=a$.
\end{proof}

\begin{lemma} We have 
$$\binom{3a+i}{a+i}=\sum_{k\geq0} \frac{i+3k}{i+k}\binom{i+k}{2k}\binom{3a+i}{a-k}
=\sum_{k\geq0} \left[\binom{i+k}{2k}+\binom{i+k-1}{2k-1}\right]\binom{3a+i}{a-k}.$$ \end{lemma}
\begin{proof} We implement Zeilberger's algorithm (from the Wilf-Zeilberger theory). Define 
$$F(i,k)=\frac{i+3k}{i+k}\cdot \frac{\binom{i+k}{2k}\binom{3a+i}{a-k}}{\binom{3a+i}{a+i}} \qquad \text{and} \qquad
G(i,k)=-\frac{\binom{i+k-1}{2k-2}\binom{3a+i}{a-k}}{\binom{3a+i}{a+i}}.$$
Check that $F(i+1,k)-F(i,k)=G(i,k+1)-G(i,k)$ and sum both sides over all integer values $k$. Then, notice the right-hand side vanishes and hence we obtain a sum $\sum_k F(i,k)$ that is \it constant \rm in the variable $i$. Determine this constant by substituting, say $i=1$,
$$\sum_{k=0}^1F(1,k)=\frac{\binom{3a+1}a}{\binom{3a+1}{a+1}}+\frac{2\binom{3a+1}{a-1}}{\binom{3a+1}{a+1}}=\frac{a+1}{2a+1}+\frac{a}{2a+1}=1.$$ 
Therefore, $\sum_kF(i,k)=1$, identically, for all $i\geq1$. The proof follows.
\end{proof}

\noindent
We now state the main result of this section.

\begin{theorem} We have
$$\binom{3a+i}{a+i}-\binom{3a+2i}a
=\sum_{k\geq1}\left\{\frac{i+3k}{i+k}\binom{i+k}{2k}-\binom{i}k \right\}\binom{3a+i}{a-k}.$$ \end{theorem}
\begin{proof} Immediate from Lemma 1 and Lemma 2.
\end{proof}

\begin{lemma} For $k\geq1$, the coefficients $c_k(i)$ are non-negative. \end{lemma}
\begin{proof} We may look at it in two different ways: 

\noindent
(1) $c_k(i)=\frac{2k}{i+k}\binom{i+k}{2k}+\binom{i+k}{2k}-\binom{i}k=\frac{2k}{i+k}\binom{i+k}{2k}+\binom{i+k}{i-k}-\binom{i}{i-k}$. Obviously  $\binom{i+k}{i-k}\geq\binom{i}{i-k}$,
 therefore $c_k(i)\geq0$.

\noindent
(2) $c_k(i)=\binom{i+k}{2k}+\binom{i+k-1}{2k-1}-\binom{i}k=\binom{i+k}{2k}+\sum_{r=0}^{k-2}\binom{i+r}{k+1+r}$ shows clearly that $c_k(i)\geq0$. The identity $\binom{i+k-1}{2k-1}=\binom{i}k+\sum_{r=1}^{k-1}\binom{i+r-1}{k+r}$ results from a cascading effect of the familiar binomial recurrence $\binom{u}v+\binom{u}{v-1}=\binom{u+1}{v}$.
\end{proof}

\section{ $q$-ANALOGUES when $\beta=2$}

\noindent
In the present section, we aim to generalize our proofs given in the preceding section by lifting the argument from the ordinary binomials to Gaussian polynomials.

\begin{lemma} We have
$$\binom{3a+2i}a_q=\sum_{k\geq0} q^{(a-k)(i-k)}\binom{i}k_q\binom{3a+i}{a-k}_q.$$ \end{lemma}
\begin{proof} This follows from the \it Vandermonde-Chu \rm identity (Theorem 1)
$$\binom{X+Y}Z_q=\sum_{k\geq0}q^{(Z-k)(X-k)}\binom{X}k_q\binom{Y}{Z-k}_q$$
on $\binom{3a+2i}a_q=\binom{i+3a+i}a_q$ with $X=i, Y=3a+i$ and $Z=a$.
\end{proof}

\begin{lemma} We have
\begin{align*}
\binom{3a+i}{a+i}_q
&=\sum_{k\geq0}q^{(a-k)(i-k)}\left[\binom{i+k}{2k}_q+q^{a+i}\binom{i+k-1}{2k-1}_q\right]\binom{3a+i}{a-k}_q.
\end{align*} \end{lemma}
\begin{proof} Let's rewrite $\binom{i+k}{2k}_q+q^{a+i}\binom{i+k-1}{2k-1}_q=\left[1+\frac{q^{a+i}(1-q^{2k})}{1-q^{i+k}}\right]\binom{i+k}{2k}_q$ and define the functions
\begin{align*} 
F(i,k)&=q^{(a-k)(i-k)}\left[1+q^{a+i}\cdot\frac{1-q^{2k}}{1-q^{i+k}}\right]\frac{\binom{i+k}{2k}_q\binom{3a+i}{a-k}_q}{\binom{3a+i}{a+i}_q}, \qquad \text{and} \\
G(i,k)&=-q^{(a-k+1)(i-k+1)}\cdot\frac{\binom{i+k-1}{2k-2}_q\binom{3a+i}{a-k}_q}{\binom{3a+i}{a+i}_q}.
\end{align*}
Divide both sides of the intended identity by $\binom{3a+i}{a+i}_q$. Our goal is to prove $\sum_kF(i,k)=1$ by adopting the Wilf-Zeilberger technique. To this end, calculate the following two ratios 
$$A(i,k):=\frac{F(i+1,k)}{F(i,k)}-1 \qquad \text{and} \qquad B(i,k):=\frac{G(i,k+1)}{F(i,k)}-\frac{G(i,k)}{F(i,k)}$$ 
resulting in
\begin{align*}
A(i,k)&=\frac{q^{a-k}(1-q^{i+k})(1-q^{a+i+1})(1-q^{i+k+1}+q^{a+i+1}-q^{a+i+2k+1})}{(1-q^{i-k+1})(1-q^{2a+i+k+1})(1-q^{i+k}+q^{a+i}-q^{a+i+2k})}-1 \qquad \text{and} \\
B(i,k)&=\left[-\frac{1-q^{a-k}}{1-q^{2a+ i+ k+1}}+\frac{q^{a+i-2k+1}(1-q^{2k})(1-q^{2k-1})}{(1-q^{i+k})(1-q^{i-k+1})}\right]\cdot
\frac{1-q^{i+k}}{1-q^{i+k}+q^{a+i}-q^{a+i+2k}}.
\end{align*}
Verify routinely $A(i,k)=B(i,k)$. Thus $F(i+1,k)-F(i,k)=G(i,k+1)-G(i,k)$. Now, sum both sides over all integer values $k$. Then, notice that the right-hand side vanishes and hence we obtain a sum $\sum_k F(i,k)$ that is \it constant \rm in the variable $i$. Determine this constant by substituting, say $i=1$ and proceed with some simplifications leading to 
$$\sum_{k=0}^1F(1,k)=q^a\cdot \frac{1-q^{a+1}}{1-q^{2a+1}}+\frac{1-q^a}{1-q^{2a+1}}=1.$$ 
Therefore, $\sum_kF(i,k)=1$, identically, for all $i\geq1$. The assertion follows.
\end{proof}

\begin{lemma} We have the identity
\begin{align*}
\binom{i+k}{2k}_q=\binom{i}k_q+\sum_{r=1}^kq^{k+r}\binom{i+r-1}{k+r}_q.
\end{align*}
\end{lemma}
\begin{proof} Use the recursive relations $\binom{a}b_q=\binom{a-1}b_q+q^{a-b}\binom{a-1}{b-1}_q=q^b\binom{a-1}b_q+\binom{a-1}{b-1}_q$.
\end{proof}

\begin{lemma} We have the inequality $\binom{i+k}{i-k}_q\geq\binom{i}{i-k}_q$.
\end{lemma}
\begin{proof} We use the interpretation of the Gaussian polynomials as the \emph{inversion number} generating function for all bit
strings of length $n$ with $k$ zeroes and $n-k$ ones, that is 
$$\binom{n}k_q=\sum_{w\in\, 0^k1^{n-k}}q^{\text{inv}(w)}.$$
Let $w'\in\, 0^{i-k}1^k\sqcup 1^k$ denote a bit where the last $k$ digits are all ones. In this sense, we get
\begin{align*}\binom{i+k}{i-k}_q=\sum_{w\in\,0^{i-k}1^{2k}}q^{\text{inv}(w)}
&=\sum_{w'\in\,0^{i-k}1^k\sqcup 1^k}q^{\text{inv}(w')}+\sum_{w'\not\in\,0^{i-k}1^k\sqcup 1^k}q^{\text{inv}(w')} \\
&=\sum_{w\in\,0^{i-k}1^k}q^{\text{inv}(w)}+\sum_{w'\not\in\,0^{i-k}1^k\sqcup 1^k}q^{\text{inv}(w')} \\
&=\binom{i}{i-k}_q+\sum_{w'\not\in\,0^{i-k}1^k\sqcup 1^k}q^{\text{inv}(w')}
\end{align*}
where we note that $\text{inv}(w')=\text{inv}(w)$ if the word $w'\in\,0^{i-k}1^k\sqcup 1^k$ is associated with $w\in\,0^{i-k}1^k$ found by dropping the last $k$ ones.
The assertion is now immediate.
\end{proof}

\noindent
We prove the main result of this section and our paper, the $\beta$-Conjecture for $\beta=2$.

\begin{theorem} The polynomial $P(q):=\binom{3a+i}{a+i}_q-\binom{3a+2i}a_q$ has non-negative coefficients. \end{theorem}
\begin{proof} From Lemma 4 and Lemma 5, we infer
\begin{align*}
P(q)=\sum_{k\geq1}q^{(a-k)(i-k)}\left[\binom{i+k}{2k}_q+q^{a+i}\binom{i+k-1}{2k-1}-\binom{i}k_q\right]\binom{3a+i}{a-k}_q.
\end{align*}
It suffices to verify positivity of the terms inside the inner parenthesis on the right-hand side. We may pair up these terms and compliment the lower index to the effect that
\begin{align*}
\binom{i+k}{2k}_q-\binom{i}k_q+q^{a+i}\binom{i+k-1}{2k-1}_q
=\binom{i+k}{i-k}_q-\binom{i}{i-k}_q+q^{a+i}\binom{i+k-1}{2k-1}_q.
\end{align*}
To reach the conclusion, simply apply Lemma 6 or Lemma 7.
\end{proof}

\section{Final remarks}

\noindent
In the present section, we close our discussion with one conjecture as a codicil of certain calculations we encountered while digging up ways to prove the $\beta$-Conjecture.

\begin{conjecture} For each $0\leq k\leq a<b$, we have
$$\binom{a}k_q\binom{a+b}{b-k}_q\geq \binom{b}k_q\binom{a+b}{a-k}_q \qquad \text{or} \qquad 
\binom{a}k_q\binom{b}k_q\binom{b+a}b_q\left[\frac1{\binom{a+k}k_q}-\frac1{\binom{b+k}k_q}\right]
\geq0.$$
\end{conjecture}

\noindent
The next elementary result might be helpful if one decides to engage this conjecture.

\begin{lemma} For $0\leq k\leq a<b$, we have
\begin{align*}
\frac1{\binom{a+k}k_q}-\frac1{\binom{b+k}k_q}
&=\sum_{i=1}^kq^{a+i}\frac{1-q^{b-a}}{1-q^{b+i}}\prod_{j=i}^k\frac{1-q^j}{1-q^{a+j}}\prod_{j=1}^{i-1}\frac{1-q^j}{1-q^{b+j}}.
\end{align*}
\end{lemma}
\begin{proof} This results from partial fractions. \end{proof}

\begin{example}
\begin{align*}
\frac1{\binom{a+1}1_q}-\frac1{\binom{b+1}1_q}
&=\frac{q^{a+1}(1-q)(1-q^{b-a})}{(1-q^{a+1})(1-q^{b+1})} \\
\frac1{\binom{a+2}2_q}-\frac1{\binom{b+2}2_q}
&=\frac{q^{a+1}(1-q)(1-q^2)(1-q^{b-a})}{(1-q^{a+1})(1-q^{a+2})(1-q^{b+1})}+\frac{q^{a+2}(1-q)(1-q^2)(1-q^{b-a})}{(1-q^{a+2})(1-q^{b+1})(1-q^{b+2})}.
\end{align*}
\end{example}

\begin{example}
\begin{align*}
\frac1{\binom{a+3}3_q}-\frac1{\binom{b+3}3_q}
&=\frac{q^{a+1}(1-q)(1-q^2)(1-q^3)(1-q^{b-a})}{(1-q^{a+1})(1-q^{a+2})(1-q^{a+3})(1-q^{b+1})} \\
&+\frac{q^{a+2}(1-q)(1-q^2)(1-q^3)(1-q^{b-a})}{(1-q^{a+2})(1-q^{a+3})(1-q^{b+1})(1-q^{b+2})} \\
&+\frac{q^{a+3}(1-q)(1-q^2)(1-q^3)(1-q^{b-a})}{(1-q^{a+3})(1-q^{b+1})(1-q^{b+2})(1-q^{b+3})}. \\
\end{align*}
\end{example}


\noindent
\begin{remark}
As a side note, we recall that G. E. Andrews \cite{GA} expresses $\binom{n}k_q-\binom{n}{k-1}_q$ as the generating function for partitions with particular \emph{Frobenius symbols}, while L. M. Butler \cite{LB} does this with the help of the \emph{Kostka-Foulkes polynomials} to show non-negativity of the coefficients. We shall provide an alternative algebraic approach.
\end{remark}

\begin{lemma} For $0\leq 2k\leq n$, we have $\binom{n}k_q-\binom{n}{k-1}_q\geq0$.
\end{lemma}
\begin{proof} Let $n=\alpha k+d$ where $0\leq d<k$. Rewrite
\begin{align*}
\binom{n}k_q-\binom{n}{k-1}_q
&=q^k\binom{n}{k-1}_q\frac{1-q^{(\alpha-2)k}}{1-q^k}
+q^{(\alpha-1)k}\binom{n}{k-1}_q\frac{1-q^{d+1}}{1-q^k}.
\end{align*}
Observe $\frac{1-q^{(\alpha-2)k}}{1-q^k}$ is already a polynomial with non-negative coefficients. Furthermore, since $U(q):=\binom{n}{k-1}_q$ is unimodal \cite{A}, \cite{HJ}, \cite{S}, 
the coefficient of $q^j$ in $U(q)\cdot(1-q^{d+1})$ is non-negative as long as $2j\leq\deg(U)$. The same is true for $U(q)\frac{1-q^{d+1}}{1-q^k}$ as a formal power series. 
Since the \it polynomial \rm $U(q)\frac{1-q^{d+1}}{1-q^k}$ is \it symmetric, \rm having degree no greater than $U(q)$,
all remaining coefficients of $U(q)\frac{1-q^{d+1}}{1-q^k}$ are non-negative. \end{proof}

\section*{Acknowledgement}

\noindent
The first author thanks R. P. Stanley for bringing Bergeron's conjecture to his attention, also F. Bergeron for his generous explanation of the problem studied in the present work.


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