# Inequality for Gaussian polynomials III

Recall the constructions $$[n]_q=\frac{1-q^n}{1-q}, [n]!_q=[1]_q[2]_q\cdots[n]_q$$ with $$[0]!_q:=1$$ and the $$q$$-binomials (Gaussian polynomials) $$\binom{n}k_q=\frac{[n]!_q}{[k]!_q[n-k]!_q}.$$ Given two polynomials $$f(q)$$ and $$g(q)$$, we write $$f(q)\geq g(q)$$ provided that $$f(q)-g(q)$$ is a polynomial having non-negative coefficients.

QUESTION. Suppose $$0\leq k\leq a are integers. Is it true that $$\binom{b+a}{b-k}_q\geq\binom{a+b}{a-k}_q$$?

• There is a closely related question on unimodality discussed here: arxiv.org/pdf/1410.7087.pdf The cool thing is that a similar difference gets an interpretation from representation theory, in terms of Kronecker coefficients. Oct 11, 2022 at 6:09

We know that there is a $$q$$-unimodality of the $$q$$-binomial coefficients. That is, $$\binom{n}{k}_q - \binom{n}{k-1}_q$$ has nonnegative coefficients for $$k \leq n/2$$. This was shown by Lynne M. Butler in A unimodality result in the enumeration of subgroups of a finite abelian group (in more generality than just $$q$$-binomials).
Now we just observe that $$b-k$$ is closer to the center peak at $$(a+b)/2$$ then $$a-k$$ is. So, $$\binom{b+a}{b-k}_q - \binom{a+b}{a-k}$$ has nonnegative coefficients as desired.
• A paper by G.E.Andrews, On the difference of successive Gaussian polynomials, J. Statistical Planning and Inference, 1993 gives a combinatorial proof regrading $\binom{n}k_q-\binom{n}{k-1}_q\geq0$. The last line in your argument seems to require more justification (although it is plausible, in principle). Oct 10, 2022 at 19:29
Just to add an alternating approach to that of Butler and Andrews, let's show that $$\binom{n}k_q-\binom{n}{k-1}_q\geq0$$, provided $$2k\leq n$$.
Let $$n=\alpha k+d$$ where $$0\leq d. Rewrite \begin{align*} \binom{n}k_q-\binom{n}{k-1}_q &=q^k\binom{n}{k-1}_q\frac{1-q^{(\alpha-2)k}}{1-q^k} +q^{(\alpha-1)k}\binom{n}{k-1}_q\frac{1-q^{d+1}}{1-q^k}. \end{align*} Observe $$\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$$ is unimodal, the coefficient of $$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}$$ as a formal power series. Since the polynomial $$U(q)\frac{1-q^{d+1}}{1-q^k}$$ is symmetric, having degree no greater than $$U(q)$$, all remaining coefficients of $$U(q)\frac{1-q^{d+1}}{1-q^k}$$ are non-negative.