Prove that $$\det\bigg(\frac{1-s^{i+j-1}}{1-t^{i+j-1}}\bigg)^n_{i,j=1}=t^{n^3/3-n^2/2+n/6}\prod_{1\leq i<j\leq n}(1-t^{j-i})^2\prod_{i,j=1}^n\frac{1-st^{j-i}}{1-t^{i+j-1}}$$ In his book, D. Bressoud also gives some hints on how to prove this. Below are the hints (and my progress so far in brackets):
- Show that the determinant is a polynomial of degree at most $n^2$ in $s$ (if I form the product of all elements on the main diagonal, I find a polynomial of order at most $n^2$ in $s$, but I can't see how to extend this result to the whole determinant)
- Next identify the roots. The polynomial has a root of order at least $n$ at $s=1$ (this I can prove; every entry in the determinant has a factor $s-1$, which I can extract from each row to find $(s-1)^n$ in front of the determinant)
- For each $1\leq k<n$, show that each row vector of the matrix $$(1-t^{k(i+j-1)})/({1-t^{i+j-1}})$$ is a linear combination of the vectors $$(1,1,\dots,1)\\(1,t,t^2,\dots,t^{n-1})\\(1,t^2,t^4,\dots,t^{2(n-1)})\\ \dots\\(1,t^{k-1},t^{2(k-1)},\dots,t^{(n-1)(k-1)})$$ and therefore this matrix has rank at most $k$ (indeed, I can prove this, and since this is a list of $k$ row vectors, the rank of the matrix above is at most $k$)
- It follows that $s=t^k$ is a root of order at least $n-k$ (not a clue here)
- Similarly, there is a root of order at least $n-k$ at $s=t^{-k}$ (again, no idea why this is true; but if I sum over $k$, it turns out that the roots found in steps 2,4 and 5 add up to $n^2$, which together with step 1 shows that all roots of the polynomial have been found)
- Finally, use the fact that $$\det\bigg(\frac{1}{1-x_iy_j}\bigg)_{i,j=1}^{n}=\prod_{1<i\leq j<n}(x_i-x_j)(y_i-y_j)\prod_{i,j=1}^{n}(1-x_iy_j)^{-1}$$ to prove that the constant term in this polynomial is $$t^{n^3/3-n^2/2+n/6}\prod_{1\leq i<j\leq n}(1-t^{j-i})^2\prod_{i,j=1}^{n}\frac{1}{1-t^{i+j-1}}$$
Since $s=1$, $s=t^k$ and $s=t^{-k}$ are the sought roots, the polynomial has the form $A(s-1)^n\prod_{k=1}^{n-1}(s-t^k)^{n-k}(s-t^{-k})^{n-k}$, where $A$ is a constant. The constant term in this polynomial, according to my computations, is $A(-1)^n$. But I wouldn't know how to use the formula in step 6 to prove the equality. Any help is appreciated, thanks in advance!