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Oct 11, 2018 at 20:33 comment added Mostafa - Free Palestine Thanks Noam, I learnt so much from your answer!
Oct 11, 2018 at 20:28 history bounty ended Mostafa - Free Palestine
Oct 11, 2018 at 20:28 vote accept Mostafa - Free Palestine
Oct 10, 2018 at 17:34 comment added Noam D. Elkies Thanks / you're welcome. I've now incorporated some of the explanation into my answer.
Oct 10, 2018 at 17:33 history edited Noam D. Elkies CC BY-SA 4.0
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Oct 10, 2018 at 16:16 comment added Iosif Pinelis Very nice, thank you for this detail!
Oct 10, 2018 at 16:10 comment added Noam D. Elkies Sorry, I meant the image of those vectors $(1,\ldots,1,1-n)$ etc. under the appropriate permutation. If $x_1 \geq x_2 \geq \ldots \geq x_n$ then the map taking $x$ to $(x_1-x_2,x_2-x_3,\ldots,x_{n-1}-x_n)$ identifies the this cone with the closed positive orthant, and those extreme rays with the coordinate axes.
Oct 10, 2018 at 15:53 comment added Iosif Pinelis I like the homogenization idea, as well as the idea of fixing an order of the coordinates. However, it is unclear to me how and why "[s]uch a choice limits each of $x$ and $y$ to a cone whose extreme rays are generated by $(1,\ldots,1,1-n)$, $(2,\ldots,2,2-n,2-n)$, etc."; for any choice of "one of the $n!^2$ possible orders of the coordinates of $x$ and $y$"? In particular, the vectors $(1,\ldots,1,1-n)$, $(2,\ldots,2,2-n,2-n)$, etc. are compatible only with some of the $n!^2$ possible orders. Can you provide details on this?
Oct 10, 2018 at 15:02 history answered Noam D. Elkies CC BY-SA 4.0