The poset of rational cones

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Problem: For $d\ge1$, let $\operatorname{Cones}(d)$ be the set of pointed rational polyhedral cones $C\subset\mathbb R^d$, ordered by the transitive closure of elementary extensions

$$C\subsetneq D,\qquad D\cap\mathbb Z^d=(C\cap\mathbb Z^d)+\mathbb Z_+x$$

for some $x\in(D\setminus C)\cap\mathbb Z^d$.

Question 2.3 asks whether the same order is generated by either height $1$ extensions or Hilbert basis extensions. There is a mild ambiguity: this may ask whether each class separately suffices, or whether at least one does. I resolve the Hilbert-basis part, which already answers the literal disjunctive question.

Result: Hilbert basis extensions generate exactly $\operatorname{Cones}(d)$. In fact, they are precisely the elementary extensions.

Let $M(C)=C\cap\mathbb Z^d$.

First suppose $C\subsetneq D$ is elementary, so

$$M(D)=M(C)+\mathbb Z_+x$$

with $x\in M(D)\setminus M(C)$. Since a rational cone is generated by its lattice points,

$$D=\mathbb R_+M(D)=C+\mathbb R_+x.$$

Thus $\mathbb R_+x$ is an extremal ray of $D$; otherwise all extremal rays of $D$ would already lie in $C$, forcing $D=C$.

We show $x\in\operatorname{Hilb}(D)$. If $x=p+q$ with nonzero $p,q\in M(D)$, write

$$p=a+mx,\qquad q=b+nx$$

with $a,b\in M(C)$, $m,n\in\mathbb Z_+$. Then

$$a+b=(1-m-n)x.$$

If $m+n=0$, then $x\in C$, impossible. If $m+n=1$, then $a+b=0$, so $a=b=0$ because $C$ is pointed, and then one of $p,q$ is zero. If $m+n\ge2$, then $-x\in D$, contradicting pointedness of $D$. Hence $x$ is indecomposable, so $x\in\operatorname{Hilb}(D)$.

Now let $y\in\operatorname{Hilb}(D)$, $y\ne x$. Write $y=a+nx$, $a\in M(C)$, $n\in\mathbb Z_+$. If $n>0$, then $y$ decomposes nontrivially unless $a=0,n=1$, which would give $y=x$. Hence $n=0$, so $y\in C$. Therefore

$$\mathbb R_+\bigl(\operatorname{Hilb}(D)\setminus\{x\}\bigr)\subset C,$$

so $C\subset D$ is a Hilbert basis extension.

Conversely, suppose $C\subsetneq D$ is a Hilbert basis extension: for some extremal generator $v$ of $D$,

$$\mathbb R_+\bigl(\operatorname{Hilb}(D)\setminus\{v\}\bigr)\subset C.$$

Every element of $M(D)$ is a nonnegative integer combination of $\operatorname{Hilb}(D)$, so it has the form $c+nv$ with $c\in M(C)$, $n\in\mathbb Z_+$. Hence

$$M(D)=M(C)+\mathbb Z_+v,$$

so the extension is elementary.

Thus elementary extensions and Hilbert basis extensions coincide, and their transitive closures give exactly the same poset $\operatorname{Cones}(d)$.

Citation: Definitions are from Gubeladze–Michałek, “The poset of rational cones,” Pacific J. Math. 292 (2018), 103–115. No external prior resolution is used here.