Timeline for A characterisation of faces of rational polyhedral cones
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
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| Sep 3, 2019 at 14:47 | comment | added | Fred Rohrer | Dear @Dima, thanks a lot! Now everything is clear. | |
| Sep 2, 2019 at 9:45 | comment | added | Dima Pasechnik | if you prefer you may think about the cones $\tau^+$, resp. $\tau^-$, rational cones spanned by the generators of $\tau$, and the generators of $\sigma$ on the "+" side of $H$, resp. "-" side of $H$. Then $x$ and $y$ may be chosen in $\tau^+$, resp. $\tau^-$. | |
| Sep 2, 2019 at 9:40 | comment | added | Dima Pasechnik | ...then there exists smalll positive rational $q$ so that $x=u+qh\in\sigma\cap N$. Similarly, $y=u-qh\in\sigma\cap N$. And so $x+y=2u\in\tau\cap N$. | |
| Sep 2, 2019 at 9:34 | comment | added | Dima Pasechnik | $H$ is affine span of rational vectors: generators of $\tau$, and, maybe, some generators of $\sigma$, if needed. Thus the normal vector $h$ to $H$ is rational. Take a non-zero $u\in\tau\cap N$, then there exists smalll positive rational $q$ so that $x=u+qh\in\sigma\$. Similarly, $y=u-qh\in\sigma\cap N$. And so $x+y=2u\in\tau$. | |
| Sep 2, 2019 at 8:41 | comment | added | Fred Rohrer | @Dima: Sorry for all the questions, but there is a last point that is unclear to me. In the case where $H$ exists, I know how to choose rational $x$ and $y$ on both sides of $H$ such that $x+y$ lies in $H$. How do you get them such that $x+y$ lies in $\tau$? | |
| Sep 2, 2019 at 8:32 | comment | added | Fred Rohrer | @Dave: I don't see the relevance of Prop. 10 to my question. Prop. 8 claims that (i)-(iii) are equivalent, but the point in question here is not explained. May I ask you to elaborate? | |
| Aug 30, 2019 at 19:57 | comment | added | Dave Anderson | For a reference, see Mustata's lectures on toric varieties, Lecture 2, Proposition 10. (The definition of "face of a semigroup" is in Lecture 1.) Here is the link: www-personal.umich.edu/~mmustata/toric_var.html | |
| Aug 30, 2019 at 19:20 | history | edited | Dima Pasechnik | CC BY-SA 4.0 |
added more details for the case where there is no $H$
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| Aug 29, 2019 at 18:36 | comment | added | Dima Pasechnik | If there is no $H$ then $\tau$ is full-dimensional; pick any rational $x$ in the interior of $\tau$. Let $y_0\not\in\tau$ be a rational generator of a ray of $\sigma$. Then there exists a positive integer $m$ so that $y:=\frac{1}{m}y_0$ satisfies $x+y\in\tau$. (You can check this in the plane generated by $x$ and $y_0$). | |
| Aug 29, 2019 at 13:56 | comment | added | Fred Rohrer | Yes, this yields an element in $\sigma\cap N\setminus\tau$, called $x$ in your answer. But we also need an element of $\tau\cap N$, called $y$ in your answer, with $x+y\in\tau$. | |
| Aug 29, 2019 at 13:44 | comment | added | Dima Pasechnik | if $H$ does not exist, then there is a ray of $\sigma$ which is not in $\tau$, and this ray is rational, so its generator $y$ is in $N$. | |
| Aug 29, 2019 at 11:33 | comment | added | Fred Rohrer | Dear @Dima, in the case where $H$ does not exist, why can you choose $y$ in $N$? | |
| Aug 28, 2019 at 9:00 | vote | accept | Fred Rohrer | ||
| Aug 23, 2019 at 20:18 | history | edited | Dima Pasechnik | CC BY-SA 4.0 |
added a case I missed
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| Aug 23, 2019 at 19:58 | history | answered | Dima Pasechnik | CC BY-SA 4.0 |