This is about a (seemingly) basic lemma about rational polyhedral cones that is sometimes used when working with toric varieties and is usually "left to the reader". Unfortunately, I could neither prove it myself nor find a complete proof in the literature. So, I am looking for either a proof or a proper reference.

Lemma. Let $V$ be an $\mathbb{R}$-vector space of finite dimension, and let $N$ be a $\mathbb{Z}$-structure on $V$ (i.e., a free abelian group with $N\otimes_{\mathbb{Z}}\mathbb{R}=V$). Let $\sigma$ and $\tau$ be $N$-rational polyhedral cones in $V$, and suppose that $\tau$ is a subset of $\sigma$. Then, the following statements are equivalent:

(i) $\tau$ is a face of $\sigma$;

(ii) If $x,y\in\sigma$ with $x+y\in\tau$, then $x,y\in\tau$;

(iii) If $x,y\in\sigma\cap N$ with $x+y\in\tau$, then $x,y\in\tau$.

Showing that (i) and (ii) are equivalent and imply (iii) is clear. My problem is to show that (iii) implies (i) or (ii), i.e., that it suffices to consider only rational points.

Note 1. Trying to show (iii)$\Rightarrow$(i) analogously to (ii)$\Rightarrow$(i) leads to the question whether $(\sigma-\tau)\cap N=(\sigma\cap N)-(\tau\cap N)$, which has a negative answer in general.

Note 2. Condition (iii) is sometimes expressed by saying that the monoid $\tau\cap N$ is a face of the monoid $\sigma\cap N$, and similarly for (ii).

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    $\begingroup$ I'd think that (iii) implies (ii) as you can approximate $x\in\sigma$ by $x'\in\sigma\cap N$ arbitrarily well. $\endgroup$ – Dima Pasechnik Aug 23 '19 at 8:59
  • $\begingroup$ Dear @Dima: Yes, that's why I think the claim is true. Can you provide details? $\endgroup$ – Fred Rohrer Aug 23 '19 at 9:08

To prove that (iii) implies (i), assume w.l.o.g. that $\tau\neq\sigma$. We first need to show that $\tau\subset \partial \sigma$. If this is not the case then either there exists a hyperplane $H$ containing $\tau$ such that $H$ is not a supporting hyperplane for $\sigma$, or $\tau$ is full-dimensional. If $H$ exists then there are $x,y\in\sigma$ on different sides of $H$, which may be chosen in $N$ as $\sigma$ is rational, so that $x+y\in\tau$, contadicting (iii).

If $H$ does not exist, then $\dim \tau=\dim\sigma$, and there exists $x\in\sigma\setminus\tau$ ($x$ may be chosen to be a generator for an extreme ray of $\sigma$) and $y\in\tau\setminus\partial\tau$ (again, it's possible to choose $x,y\in N$) so that $x+y\in\tau$, again contradicting (iii).

Thus $\tau\subset \partial \sigma$. Let $\sigma'$ be the minimal face of $\sigma$ containing $\tau$. Note that $\sigma'$ is a rational polyhedral cone, and we are basically in the situation as above, with $\sigma$ replaced by $\sigma'$, except that we don't have any more dimensions to spare, and so either $\tau=\sigma'$, as required, or we contradict (iii) as above.

  • $\begingroup$ Dear @Dima, in the case where $H$ does not exist, why can you choose $y$ in $N$? $\endgroup$ – Fred Rohrer Aug 29 '19 at 11:33
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    $\begingroup$ 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 $y_0$). $\endgroup$ – Dima Pasechnik Aug 29 '19 at 18:36
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    $\begingroup$ $H$ 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 $x+y=2u\in\tau$. $\endgroup$ – Dima Pasechnik Sep 2 '19 at 9:34
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    $\begingroup$ ...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 $x+y=2u\in\tau\cap N$. $\endgroup$ – Dima Pasechnik Sep 2 '19 at 9:40
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    $\begingroup$ 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. $\tau^-$. $\endgroup$ – Dima Pasechnik Sep 2 '19 at 9:45

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