PomeDefinition 19.2.3.923345ACA44A Let $\mathcal{S}'(C)$ be the subset of $\mathcal{S}(C)$ consisting of semigroups $S$ such that $S$ generates $\mathbb {Z}^{n+1}$ (as an Abelian group).
19.7.24 Okounkov bodies of semigroupsA6ED691E2495
Given $S\in \hat{\mathcal{S}}(C)$, we will write $C(S)\subseteq C$ for the closed convex cone generated by $S\cup \{ 0\} $. Moreover, for each $k\in \mathbb {Z}_{>0}$, we define
\[ \Delta _ k(S)\coloneqq \mathrm{Conv}\left\{ k^{-1}x\in \mathbb {R}^ n:x\in S_ k \right\} \subseteq \mathbb {R}^ n. \]
Here $\mathrm{Conv}$ denotes the convex hull. Lemma 19.7.25.E41D68AC6F7C Let $S,S'\in \mathcal{S}'(C)$. Assume that $\operatorname{vol}(S\cap S')>0$. Then $S\cap S'\in \mathcal{S}'(C)$, and
\[ C(S\cap S')=C(S)\cap C(S'). \]
Proof. We first observe that the cone $C(S)\cap C(S')$ has full dimension since otherwise $\operatorname{vol}(S\cap S')=0$. Take a full-dimensional subcone $C'$ in $C(S)\cap C(S')$ such that $C'$ intersects the boundary of $C(S)\cap C(S')$ only at $0$. It follows from [Theorem 1, KK12] that there is an integer $N>0$ such that any $x\in \mathbb {Z}^{n+1}\cap C'$ with Euclidean norm at least $N$ lies in $S\cap S'$. Therefore, $S\cap S'\in \mathcal{S}'(C)$. Since $C'$ can be chosen around any ray in the interior of $C(S)\cap C(S')$, the same argument also shows that
\[ C(S\cap S')=C(S)\cap C(S'). \]
This completes the proof. $\square$
Definition 19.7.26.0270FA2C03BD Given $S\in \mathcal{S}'(C)$, its Okounkov body is defined as follows
\[ \Delta (S)\coloneqq \left\{ x\in \mathbb {R}^ n:(x,1)\in C(S) \right\} . \]
Theorem 19.2.5.37D86D761E81 For each $S\in \mathcal{S}'(C)$, we have
4E701FB8EA13
19.2.6
\begin{equation} \operatorname{vol}S=\lim _{k\to \infty } k^{-n}|S_ k|=\operatorname{vol}\Delta (S)>0. \end{equation}
Moreover, as $k\to \infty $,
A57FF2EE0305
19.2.7
\begin{equation} \Delta _{k}(S)\xrightarrow {d_{\mathrm{Haus}}}\Delta (S). \end{equation}
Proof. The equalities (19.2.6) follow from theorem 19.7.20.
It remains to prove (19.2.7). By the argument of [Lemma 4.8, WN14], for any compact set $K\subseteq \operatorname{Int}\Delta (S)$, there is $k_0>0$ such that for any $k\geq k_0$, $\alpha \in K\cap (k^{-1}\mathbb {Z})^ n$ implies that $\alpha \in \Delta _ k(S)$.
To conclude, fix $\epsilon >0$. Choose $\delta >0$ such that $K\coloneqq \Delta (S)_{\delta }$ has positive volume and $d_{\mathrm{Haus}}(K,\Delta (S))<\epsilon /2$. Applying 19.7.7 to $K$, we find that $\Delta _ k(S)$ contains $K_{n^{1/2}k^{-1}}$ for all large $k$. After increasing $k$, we may also assume that $d_{\mathrm{Haus}}(K,K_{n^{1/2}k^{-1}})<\epsilon /2$. Since $\Delta _ k(S)\subseteq \Delta (S)$, this implies (19.2.7). $\square$
Corollary 19.2.8.9FB3E8CE18F7 Let $S,S'\in \mathcal{S}'(C)$. Assume that $\operatorname{vol}(S\cap S')>0$. Then we have
\[ d_{\mathrm{sg}}(S,S')=\operatorname{vol}(S)+\operatorname{vol}(S')-2\operatorname{vol}(S\cap S'). \]
Proof. This is a direct consequence of 19.7.25 and (19.2.6). $\square$
Lemma 19.7.27.B1ED8431B1B9 Given $S\in \mathcal{S}'(C)$, we have $S\sim \operatorname{Reg}(S)$.
Proof. Since $S$ and $\operatorname{Reg}(S)$ have the same Okounkov body, we have $\operatorname{vol}S=\operatorname{vol}\operatorname{Reg}(S)$ by theorem 19.2.5. By corollary 19.2.8 again,
\[ d_{\mathrm{sg}}(\operatorname{Reg}(S),S)=\operatorname{vol}\operatorname{Reg}(S)-\operatorname{vol}S=0. \]
This completes the proof. $\square$
Lemma 19.7.28.323C6B6B61D0 Let $S,S'\in \mathcal{S}'(C)$. Assume that $d_{\mathrm{sg}}(S,S')=0$. Then $\Delta (S)=\Delta (S')$.
Proof. Observe that $\operatorname{vol}(S\cap S')>0$, as otherwise
\[ d_{\mathrm{sg}}(S,S')\geq \operatorname{vol}S+\operatorname{vol}S'>0, \]
which is a contradiction.
It follows from 19.7.25 that $S\cap S'\in \mathcal{S}'(C)$. It suffices to show that $\Delta (S)=\Delta (S\cap S')$. In fact, suppose that this holds, since $\operatorname{vol}\Delta (S')=\operatorname{vol}S'=\operatorname{vol}S=\operatorname{vol}\Delta (S)$, the inclusion $\Delta (S')\supseteq \Delta (S\cap S')=\Delta (S)$ is an equality.
After replacing $S'$ by $S\cap S'$, we still have $d_{\mathrm{sg}}(S,S')=0$ by 19.7.19, and now $S'\subseteq S$. Hence corollary 19.2.8 gives
\[ 0=d_{\mathrm{sg}}(S,S')=\operatorname{vol}S-\operatorname{vol}S', \]
so $\operatorname{vol}S=\operatorname{vol}S'$. Therefore
\[ \operatorname{vol}\Delta (S)=\operatorname{vol}\Delta (S')>0 \]
by (19.2.6). Therefore, they are equal. $\square$
Lemma 19.7.29.B63FC87B8DDF Suppose that $S^ i\in \mathcal{S}'(C)$ is a decreasing sequence such that
\[ \lim _{i\to \infty }\operatorname{vol}S^ i>0. \]
Then there is $S\in \mathcal{S}'(C)$ such that $S^ i\to S$.
Proof. By 19.7.27, we may replace $S^ i$ by its regularization and assume that $S^ i=C(S^ i)\cap \mathbb {Z}^{n+1}$. We define
\[ S=\left( \bigcap _{i=1}^{\infty }C(S^ i) \right) \cap \mathbb {Z}^{n+1}. \]
The convex bodies $\Delta (S^ i)$ form a decreasing sequence and satisfy
\[ \lim _{i\to \infty }\operatorname{vol}\Delta (S^ i)=\lim _{i\to \infty }\operatorname{vol}S^ i>0. \]
By continuity from above for Lebesgue measure,
\[ \operatorname{vol}\bigcap _ i\Delta (S^ i)=\lim _{i\to \infty }\operatorname{vol}\Delta (S^ i)>0. \]
Set
\[ K\coloneqq \bigcap _{i=1}^{\infty }C(S^ i). \]
The cone $K$ is full-dimensional. As noted above, $K\cap \mathbb Z^{n+1}$ generates $\mathbb Z^{n+1}$ as an Abelian group; hence $S=K\cap \mathbb Z^{n+1}$ lies in $\mathcal S'(C)$. Moreover, the closed convex cone generated by $S$ is exactly $K$. Therefore
\[ \Delta (S)=\{ x:(x,1)\in K\} =\bigcap _{i=1}^{\infty }\{ x:(x,1)\in C(S^ i)\} =\bigcap _{i=1}^{\infty }\Delta (S^ i). \]
By corollary 19.2.8 and theorem 19.2.5, we can compute the distance
\[ d_{\mathrm{sg}}(S,S^ i)=\operatorname{vol}S^ i-\operatorname{vol}S=\operatorname{vol}\Delta (S^ i)-\operatorname{vol}\Delta (S), \]
which tends to $0$ by construction. $\square$
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