5
$\begingroup$

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?

$\endgroup$

1 Answer 1

9
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.