Timeline for Computing a determinant involving roots of unity

6 events

when toggle format	what		by	license	comment
Jan 3, 2019 at 20:26	history	edited	T. Amdeberhan	CC BY-SA 4.0	added 1008 characters in body
Jun 24, 2017 at 22:24	comment	added	efs		Of course. I should read more carefully the other answers before posting. Thanks for the response. (I was entertaining myself with this determinant in Mathematica while finishing some pending bureaucratic work in the computer.)
Jun 24, 2017 at 21:23	history	edited	T. Amdeberhan	CC BY-SA 3.0	added 226 characters in body
Jun 24, 2017 at 19:08	comment	added	GH from MO		Your first line should be corrected as $\det\Phi=\prod_{0\leq i<j < d} (\xi^j-\xi^i)$. Also, Zach Teitler is right: distinguishing between $\pm i$ is subtle here, and it is pretty much the same problem as evaluating the Gauss sum precisely (not only up to $\pm$ sign which is an easier task). The point is that $(\det\Phi)^2$ has two square-roots, and you need to pin down which one is $\det\Phi$.
Jun 24, 2017 at 17:55	comment	added	Zach Teitler		You say $\det \Phi = (-1)^{\binom{d}{2}} \left( d^d (-1)^{\binom{d-1}{2}} \right)^{1/2} = (-1)^{\binom{d}{2}} d^{\frac{d}{2}} i^{\binom{d-1}{2}}$. Is it obvious that we get $i$ instead of $-i$ at the end? Or do we need something like the accepted answer's argument (pun intended)?
Jun 24, 2017 at 16:27	history	answered	T. Amdeberhan	CC BY-SA 3.0