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Apr 13, 2017 at 12:19 history edited CommunityBot
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Jul 31, 2013 at 1:22 comment added Stefan @DavidSpeyer : Thanks for all the time you have obviously put into this problem and your partial results. If I solve this problem one way or the other, with a proof or a counterexample, I will edit my question and leave a comment for you, which you will hopefully be notified of in your e-mail.
Jul 31, 2013 at 1:21 vote accept Stefan
Jul 29, 2013 at 22:36 comment added GH from MO This is all very nice, but in your special case the final sign should be be $(-1)^{n(n-1)/2}$. The problem comes from the fourth display, where you forgot to multiply $\sum t_i\mu_{\sigma(i)}$ and $\mu_j t_i$ by $-1$. The Lemma and its proof should be updated accordingly.
Jul 29, 2013 at 17:03 comment added Stefan if your assertion is correct (I have no reason to doubt it, though I don't understand the proof 100%), I am pretty sure that the determinant of the matrix $M$ has the right sign if $\gamma_{n-1}< \mu_1 < \gamma_{n} < \mu_2$ and and $\gamma_n - \mu_1$ is ``really small''. I'd like to extend this to the case where $\gamma_{n-1}< \mu_1 < \gamma_{n} < \mu_2$ and $\gamma_n - \mu_1$ is not necessarily small.
Jul 27, 2013 at 2:14 history answered David E Speyer CC BY-SA 3.0