My question concerns an application in physics. By Vandermonde identity I refer to the following statement: take $f_j (z)=z^j$, where $z=x+iy$ is a complex coordinate and $j$ an integer. Make an $N\times N$ determinant where each column $j$ contains $f_j (z_i),j=0,\ldots,N-1,i=1,\ldots,N$. This determinant can be simply evaluated: $\prod_{i\lt j}(z_i-z_j)$, which is the Vandermonde formula.

Now, consider the following generalization: let the function $f$ be

$f_j(x, y) = \sum_{k} e^{i B y - 1/2 (B + x)^2}$

where the sum over $k$ is over all integers, $B=2\pi j/b + k a$, so that $f_j$ is explicitly periodic under $y\mapsto y+b$ and it is periodic up to a phase $e^{i k a}$ under $x\mapsto x+a$. All in all, the function $f$ is quasiperiodic over the rectangle $a \times b$ whose area $ab$ must equal $2\pi M$, $M$ being an integer. This means that $f_{j+M}=f_{j}$, so for fixed $x,y$ there are only $M$ distinct $f_j(x,y)$. This function $f_j$ can be related to one of the Jacobi theta functions in terms of $z=x+iy$.

Take $f_j(x_i,y_i)$ where $j=0,\ldots,N-1; i=1,...,N$ and arrange it into a determinant $N\times N$. Is there a simple formula for the value of this determinant that resembles the Vandermonde expression given above?


1 Answer 1


Yes, there is such a formula. see (A.2.17) of Nijhoff's lecture notes.


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