Some curious Hankel determinants

Question

Let $f(n,q)=\prod_{j=1}^na(q^j)$ for a polynomial $a(q)$ and let $d(n)=\det(f(i+j,q))_{i,j=0}^n$ be its Hankel determinant.

Computer experiments suggest that $$\lim_{q\to1}\frac{d(n)}{(q-1)^\binom{n+1}{2}}=(a(1)a'(1))^\binom{n+1}{2}\prod_{j=0}^nj!.$$ Has anyone an idea how to prove this?

Remark: For $a(q)=a+qb$ it is easy to verify that $$d(n)=((q-1)b)^\binom{n+1}{2}q^\frac{n(n+1)(2n+1)}{6}{\prod_{j=1}^n[j]_{q}!(a+q^jb)^{n+1-j}},$$ if $[n]_{q}=\frac{1-q^n}{1-q}$ and $[n]_q!=[1]_q[2]_q\dots[n]_q.$

Therefore the conjecture is true for linear polynomials and also for $a(q)=q^m.$

Answer 1

The identity is clearly invariant to scaling $a$ by a constant. (One has to check the constant appears in the determinant to power $n(n+1)$). So wlog we may assume $a(1)=1$.

We can expand $\log a$ as a formal power series in $\log q$, $$ a(q)= e^{ \sum_{i} a_i (\log q)^i}$$ $$a (q^j) = e ^{ \sum_i a_i j^i (\log q)^j}$$ $$\prod_{j=1}^n a(q^j) = e^{ \sum_i a_i (\sum_{j=0}^n j^i) (\log q)^j }$$

the key thing being that $(\sum_{j=0}^n j^i)$ is a polynomial in $n$ of degree $\leq n+1$. Hence if we expand this product $f(n,q)$ out as a power series in $\log q$, the coefficient of the $i$th power of $\log q$ will be a polynomial in $n$ of degree $\leq 2i$, whose leading term is a power of $a_1$.

Hence the coefficient of $\log q^i$ in the power series of $f(x+y,q)$ is a sum of monomials in $x$ times monomials in $y$ of total degree $\leq i$.

When we take a determinant of a sum of rank one matrices, in this case a sum indexed by powers of $\log q$ and then pairs of monomials, it's the same as the sum of the determinants of all sums of $n+1$ matrices from the set. These determinants are nonzero only if the degreees of all the monomials in $x$ and monomials in $y$ appearing are distinct, which means the total degree in each variable must be at least ${n+1 \choose 2}$, for a total degree of at least $2 {n+1 \choose 2}$, and thus total degree in $\log q$ of at least ${n+1 \choose 2}$. This is only sharp if the monomials we study are the highest degree terms, which we saw come only from $a_1$.

So the answer depends only on $a_1$, and the calculations you include in your answer finish the job.

Answer 2

Let $$ D_{n}(q)=\left[\prod_{k=1}^{i+j}a(q^k)\right]_{i,j=0}^{n}, $$ $$c_n(q)=a(q)...a(q^{2n})$$ and $$ {\bf b}_{n}(q)=\left[a(q)...a(q^{n}), a(q)...a(q^{n+1}), \cdots , a(q)...a(q^{2n-1})\right]$$ for $n=1,2,...$. Then, we have \begin{align*} D_{n+1}(q)&=\left( \begin{array}{ccccccc} D_n(q) & {\bf b}_{n+1}(q)^t\\ {\bf b}_{n+1}(q)& c_{n+1}(q). \end{array} \right). \end{align*} By the well known result \begin{align*} \frac{\det \left(D_{n+1}(q)\right)}{(1-q)^{\frac{(n+1)(n+2)}{2}}}&=\frac{1}{(1-q)^{n/2+1}}\det \left( \begin{array}{ccccccc} D_n(q)/(1-q)^{n/2} & {\bf b}_{n+1}(q)^t/(1-q)^{n/2}\\ {\bf b}_{n+1}(q)/(1-q)^{n/2}& c_{n+1}(q)/(1-q)^{n/2} \end{array} \right)\\ &=\frac{\det\left(D_n(q)\right)}{(1-q)^{n(n+1)/2}}\frac{1}{(1-q)^{n/2+1}}\\ &\times\left(\frac{c_{n+1}(q)}{(1-q)^{n/2}}-\frac{{\bf b}_{n+1}(q)}{(1-q)^{n/2}}\left(\frac{D_n(q)}{(1-q)^{n/2}}\right)^{-1}\left(\frac{{\bf b}_{n+1}(q)}{(1-q)^{n/2}}\right)^{t}\right) \end{align*} if matix $\left({D_n(q)}/{(1-q)^{n/2}}\right)$ is invertible. Hence combine with mathematical induction we just need to show that $$F_{n}(q):=\frac{1}{(1-q)^{n}}\left(c_{n}(q)-{\bf b}_{n}(q)\left(D_{n-1}(q)\right)^{-1}{\bf b}_{n}(q)^{t}\right)$$ tends to $$[a(1)a'(1)]^{n}n!$$ as $q\rightarrow 1$. Notice that the blockwise inversion formula here, say \begin{align*} \left(D_{n+1}(q)\right)^{-1}&=\left( \begin{array}{ccccccc} D_n(q) & {\bf b}_{n+1}(q)^t\\ {\bf b}_{n+1}(q)& c_{n+1}(q). \end{array} \right)^{-1}\\ &=\begin{bmatrix} D_n^{-1}+G_{n+1}D_n^{-1}{\bf b}_{n+1}^t{\bf b}_{n+1}D_n^{-1} & -D_n^{-1}\mathbf{b}_{n+1}^tG_{n+1} \\ -G_{n+1}\mathbf{b}_{n+1}D_n^{-1} & G_{n+1} \end{bmatrix}, \end{align*} where $$G_{n}^{-1}=(1-q)^nF_n(q)=c_{n}(q)-{\bf b}_{n}(q)\left(D_{n-1}(q)\right)^{-1}{\bf b}_{n}(q)^{t}.$$ Thus one can use mathematical induction again.