PomeDefinition 19.7.2.9A491602434D A convex body in $\mathbb {R}^ n$ is a non-empty compact convex set.
19.7 Almost semigroups00B44728CC57
00B44728CC57
This appendix develops the theory of almost semigroups in $\mathbb {Z}^{n+1}$ that underpins the partial Okounkov body construction of section 19.1. Almost semigroups generalize the integral semigroups arising in the classical Lazarsfeld–Mustaţă and Kaveh–Khovanskii constructions to the setting where the underlying graded family is no longer multiplicative. The main result is the construction of a convex body $\Delta (S)\subseteq \mathbb {R}^ n$ out of an almost semigroup $S$. The corresponding statements were stated without proof in section 19.1; the proofs are given here.
19.7.1 Convex bodiesA0D1AD337F41
Fix $n\in \mathbb {N}$.
We write $\mathcal{K}_ n$ for the set of convex bodies in $\mathbb {R}^ n$.
Definition 19.7.3.5FD89093BE0B The Hausdorff metric between $K_1,K_2\in \mathcal{K}_ n$ is given by
\[ d_{\mathrm{Haus}}(K_1,K_2)\coloneqq \max \left\{ \adjustlimits \sup _{x_1\in K_1}\inf _{x_2\in K_2}|x_1-x_2|, \adjustlimits \sup _{x_2\in K_2}\inf _{x_1\in K_1}|x_1-x_2|\right\} . \]
Theorem 19.7.4 (Blaschke selection theorem).BA1F1C1CF20F The metric space $(\mathcal{K}_ n,d_{\mathrm{Haus}})$ is locally compact.
Proof. Fix $R>0$ and let $\mathcal K(R)$ be the nonempty compact convex subsets of the closed ball $\overline B(0,R)$. For $K\in\mathcal K(R)$ its support function $h_K$ on the unit sphere is bounded by $R$ in absolute value and is $R$-Lipschitz. A diagonal selection on a countable dense subset of the sphere, followed by the common Lipschitz bound, gives a uniformly convergent subsequence $h_{K_j}\to h$ from every sequence.
Let $K=\{x:\langle u,x\rangle\leq h(u)\text{ for all unit }u\}$. This is closed, convex and contained in $\overline B(0,R)$. It is nonempty: select $x_j\in K_j$ and pass to a convergent subsequence; the limit satisfies all the inequalities. For every fixed $u$, select $x_j\in K_j$ maximizing $\langle u,x\rangle$ and pass to a subsequence. Its limit lies in $K$ and attains $h(u)$, so $h=h_K$.
For compact convex sets, $d_{\mathrm{Haus}}(K,L)=\sup_{|u|=1}|h_K(u)-h_L(u)|$. One inequality follows by approximating each point of either body within its Hausdorff distance. For the other, an upper bound $\delta$ on the support-function difference gives $K\subset L+\delta\overline B$ and the reverse inclusion by separation of a point from a closed convex set. Hence $K_j\to K$ in Hausdorff distance. We have proved compactness of $\mathcal K(R)$. Every sufficiently small closed Hausdorff ball about a given body lies in some $\mathcal K(R)$ and is closed there, so is compact. In dimension zero the space has just one member.
Theorem 19.7.5.66339C0E3C21 The Lebesgue volume $\operatorname{vol}\colon \mathcal{K}_ n\rightarrow \mathbb {R}_{\geq 0}$ is continuous.
Proof. Let $K_j\to K$ in Hausdorff distance. All bodies lie in a fixed ball. Every point outside $K$ eventually lies outside $K_j$, because its distance from $K$ is positive. If $K$ has interior, every point $x$ of its interior eventually lies in $K_j$: choose $\rho>0$ with $x+\rho\overline B\subset K$; uniform convergence of support functions then gives $\langle u,x\rangle<h_{K_j}(u)$ for every unit $u$, which by separation implies $x\in K_j$.
The boundary of a convex body has Lebesgue measure zero, as proved in the relative-boundary lemma. Thus the indicator functions converge almost everywhere, and dominated convergence on the fixed ball proves volume convergence. If $K$ has empty interior, it lies in a proper affine subspace and has measure zero; the outside-point argument alone again gives indicator convergence almost everywhere. In dimension zero the normalized volume is always $1$.
Theorem 19.7.6.AFBD46E7C0A4 Let $K_ i, K\in \mathcal{K}_ n$ ($i\in \mathbb {N}$). Then $K_ i\xrightarrow {d_{\mathrm{Haus}}}K$ if and only if the following conditions hold:
45F2C2B839F6each point $x\in K$ is the limit of a sequence $x_ i\in K_ i$, and
BA71889FAAC3the limit of any convergent sequence $(x_{i_ j})_{j\in \mathbb {N}}$ with $x_{i_ j}\in K_{i_ j}$ lies in $K$, where $i_ j$ is a strictly increasing sequence in $\mathbb {Z}_{>0}$.
Proof. Hausdorff convergence immediately supplies points $x_i\in K_i$ tending to each $x\in K$, and forces every convergent selection from a subsequence to land in $K$.
Conversely, the two conditions first imply that the $K_i$ are uniformly bounded. Otherwise choose a subsequence containing points $y_i$ tending to infinity, and, by the first condition, points $x_i\in K_i$ tending to a fixed $x\in K$. Choose $R$ with $K\subset B(x,R)$. Convexity gives $z_i\in[x_i,y_i]\subset K_i$ with $|z_i-x|=R$ for large $i$. A subsequence of these sphere points converges outside $K$, contradicting the second condition.
If the Hausdorff distance did not tend to zero, one of two failures would occur along a subsequence. Either there are $y_i\in K_i$ at a fixed positive distance from $K$; uniform boundedness gives a convergent subsequence, contradicting the second condition. Or there are $x_i\in K$ at a fixed positive distance from $K_i$; pass to $x_i\to x\in K$ and use the first condition to find $y_i\in K_i$ with $y_i\to x$, again a contradiction. This proves the converse and explains why convexity is needed in the boundedness step.
For a convex body $K\subseteq \mathbb {R}^ n$ and $\epsilon >0$, we write $K_\epsilon \coloneqq \{ x\in K: B(x,\epsilon )\subseteq K\} $ for the inner $\epsilon $-shrinking of $K$. (Compare with the outer thickening $K^\epsilon $ used in corollary 21.12.35; the two are different operations.)
Lemma 19.7.7.5C4A19C5ED0B Let $K\in \mathcal{K}_ n$ be a convex body with positive volume and $K'\in \mathcal{K}_ n$. Assume that, for some $k\in \mathbb Z_{>0}$, $K'$ contains $K\cap (k^{-1}\mathbb Z)^ n$. Then
\[ K'\supseteq K_{n^{1/2}/k}. \]
Proof. Let $x\in K_{n^{1/2}k^{-1}}$. By assumption, the closed ball $B$ with center $x$ and radius $n^{1/2}k^{-1}$ is contained in $K$. Observe that $x$ can be written as a convex combination of points in $B\cap (k^{-1}\mathbb {Z})^ n$, which are contained in $K'$ by assumption. It follows that $x\in K'$. $\square$
Given a sequence of convex bodies $K_ i$ ($i\in \mathbb {N}$), we set
\[ \varliminf _{i\to \infty }K_ i= \overline{\bigcup _{i=0}^{\infty }\bigcap _{j\geq i}K_ j}. \]
Suppose $K$ is the limit of a subsequence of $K_ i$, we have
This is a simple consequence of theorem 19.7.6.
Lemma 19.7.9.806DFD35CCE5 Assume that $n>0$. Let $K\subseteq \mathbb {R}^ n$ be a convex body. Let
\[ t_{\min }\coloneqq \min \{ t\in \mathbb {R}:\{ x_1=t\} \cap K\neq \varnothing \} ,\quad t_{\max }\coloneqq \max \{ t\in \mathbb {R}:\{ x_1=t\} \cap K\neq \varnothing \} . \]
Then for $t\in [t_{\min },t_{\max }]$, the map
\[ t\mapsto \{ y\in \mathbb {R}^{n-1}:(t,y)\in K\} \]
is continuous with respect to the Hausdorff metric.
Proof. We may assume that $t_{\min }<t_{\max }$ as otherwise there is nothing to prove.
For each $t\in [t_{\min },t_{\max }]$, we write $K_ t=\{ y\in \mathbb {R}^{n-1}:(t,y)\in K\} $. Let $t_ j\to t$ be a convergent sequence in $[t_{\min },t_{\max }]$, we want to show that $K_{t_ j}$ converges to $K_ t$ with respect to the Hausdorff metric. Recall that this amounts to the following two assertions:
4164F65EB6F1For each convergent sequence $y_ j\in K_{t_ j}$ with limit $y$, we have $y\in K_ t$;
5031A2661BB4Given any $y\in K_ t$, up to replacing $t_ j$ by a subsequence, we can find $y_ j\in K_{t_ j}$ converging to $y$.
The first assertion is obvious. Let us prove the second. Take $y\in K_ t$. Up to replacing $t_ j$ by a subsequence and taking the symmetry into account, we may assume that $t_ j> t$ for all $j$. In particular, $t<t_{\max }$.
We can find a point $z=(z^1,z')\in K$ such that $z^1>t$ (for example, there is always such a point with $z^1=t_{\max }$). Replacing $t_ j$ by a subsequence, we may assume that $t_ j\in (t,z^1)$ for all $j$. Then it suffices to take
\[ y_ j=\frac{z^1-t_ j}{z^1-t}y+\frac{t_ j-t}{z^1-t}z'. \]
This completes the proof. $\square$
Lemma 19.7.10.09E448752D43 Assume that $n>0$. The maps $\mathcal{K}_ n\rightarrow \mathbb {R}$ given by
\[ K\mapsto \min _{x\in K}x_1,\qquad K\mapsto \max _{x\in K}x_1 \]
are continuous with respect to the Hausdorff metric.
Proof. We prove the assertion for the maximum; the proof for the minimum is the same. Let $K,L\in \mathcal{K}_ n$ and set $\epsilon =d_{\mathrm{Haus}}(K,L)$. For every $x\in K$, there exists $y\in L$ with $|x-y|\leq \epsilon $, hence
\[ x_1\leq y_1+\epsilon \leq \max _{z\in L}z_1+\epsilon . \]
Taking the maximum over $x\in K$ gives
\[ \max _{x\in K}x_1\leq \max _{z\in L}z_1+\epsilon . \]
Interchanging $K$ and $L$, we get
\[ \left|\max _{x\in K}x_1-\max _{z\in L}z_1\right|\leq d_{\mathrm{Haus}}(K,L). \]
This proves the continuity. $\square$
Lemma 19.7.11.127837115DD3 Let $D_ j\subseteq \mathbb {R}^ n$ $(j\geq 1)$ be a decreasing sequence of bounded convex sets. Assume that $\operatorname{vol}\bigcap _ j D_ j>0$. Then
\[ \overline{\bigcap _{j=1}^{\infty }D_ j}=\bigcap _{j=1}^{\infty }\overline{D_ j}. \]
Proof. The $\subseteq $ direction is clear. By convexity, it suffices to show that both sides have the same positive volume: indeed, a proper inclusion of closed convex sets with positive volume would cut off a non-empty relatively open convex piece by a separating hyperplane, hence would strictly decrease the volume. As the boundary of convex sets has zero Lebesgue measure, it follows that the volumes of both sides are equal to $\lim _{j\to \infty }\operatorname{vol}D_ j$. $\square$
Definition 19.7.12.FBDC5E2EBC8A Let $K,K'\in \mathcal{K}_ n$, their Minkowski sum is given by
\[ K+K'\coloneqq \{ x+x': x\in K,x'\in K'\} . \]
Proposition 19.7.13.F9F053A774FEThe Minkowski sum $\mathcal{K}_ n\times \mathcal{K}_ n\rightarrow \mathcal{K}_ n$ is continuous.
Proof. For compact sets $K,L,K',L'$, \[d_{\mathrm{Haus}}(K+L,K'+L')\leq d_{\mathrm{Haus}}(K,K')+d_{\mathrm{Haus}}(L,L').\] Indeed, given $x+y\in K+L$, select $x'\in K'$ and $y'\in L'$ at their respective distances from $x,y$; compactness ensures minimizers exist. The triangle inequality gives the stated bound on the distance to $K'+L'$. Interchange the pairs for the reverse bound. A Minkowski sum of two nonempty compact convex sets is nonempty, compact and convex. Hence this estimate proves continuity of the asserted map.
Theorem 19.7.14 (Brunn–Minkowski).72547EEEBFE4 Let $K,K'\in \mathcal{K}_ n$. Then for any $t\in (0,1)$, we have
\[ \operatorname{vol}\bigl ((1-t)K'+tK\bigr )\geq \bigl (\operatorname{vol}K'\bigr )^{1-t}\bigl (\operatorname{vol}K\bigr )^ t. \]
Proof. We prove the stated multiplicative form by induction on $n$. If either body has zero volume, its right-hand side is zero and positivity suffices. In dimension zero all volumes are $1$. In dimension one the bodies are intervals: their Minkowski combination has length $(1-t)a+tb\geq a^{1-t}b^t$ by the weighted arithmetic--geometric mean inequality.
Assume the result in dimension $n-1$ and let $A=K'$, $B=K$, $C=(1-t)A+tB$ have positive-volume input bodies. Slice them by the first coordinate and put $f(x)=\operatorname{vol}_{n-1}(A_x)$, $g(y)=\operatorname{vol}_{n-1}(B_y)$ and $h(s)=\operatorname{vol}_{n-1}(C_s)$. Slice inclusion and the induction hypothesis give \[h((1-t)x+ty)\geq f(x)^{1-t}g(y)^t.\] On the interiors of their projection intervals, $f$ and $g$ are positive and continuous. Continuity follows from the proved Hausdorff continuity of slices and volume continuity in dimension $n-1$; endpoint values do not affect integrals.
Normalize $f,g$ to have integral $1$, and divide $h$ by $(\operatorname{vol}A)^{1-t}(\operatorname{vol}B)^t$. Their inequality has the same form. Let $T$ be the increasing map matching the cumulative integrals of $f$ and $g$. On the two open projection intervals it is continuously differentiable and satisfies $T'(x)=f(x)/g(T(x))$. The map $S(x)=(1-t)x+tT(x)$ is increasing, so change of variables and the slice inequality give \[\int h\geq\int h(S(x))S'(x)\,\mathrm dx\geq\int f(x)^{1-t}g(T(x))^t\left((1-t)+t\frac{f(x)}{g(T(x))}\right)\,\mathrm dx\geq\int f=1.\] The last inequality is the weighted arithmetic--geometric mean inequality applied to $1$ and $f(x)/g(T(x))$. The changes of variables can first be made on compact subintervals and then passed to the full intervals by monotone convergence. Undoing normalization and using Fubini proves $\operatorname{vol}C\geq(\operatorname{vol}A)^{1-t}(\operatorname{vol}B)^t$, as required.
19.7.15 The Okounkov bodies of almost semigroupsA31E0D5C79D3
Fix an integer $n\geq 0$. Fix a closed convex cone $C\subseteq \mathbb {R}^{n}\times \mathbb {R}_{\geq 0}$ such that $C\cap \{ x_{n+1}=0\} =\{ 0\} $. Here $x_{n+1}$ is the last coordinate of $\mathbb {R}^{n+1}$.
19.7.16 Generalities on semigroups3A76517E84BF
Write $\hat{\mathcal{S}}(C)$ for the set of subsets of $C\cap \mathbb {Z}^{n+1}$ and $\mathcal{S}(C)$ for the set of subsemigroups $S\subseteq C\cap \mathbb {Z}^{n+1}$. For each $k\in \mathbb {N}$ and $S\in \hat{\mathcal{S}}(C)$, we write
\[ S_ k\coloneqq \left\{ x\in \mathbb {Z}^ n: (x,k)\in S \right\} . \]
Note that $S_ k$ is a finite set by our assumption on $C$.
We introduce a pseudometric on $\hat{\mathcal{S}}(C)$ as follows:
642E2D6850DF
19.7.17
\begin{equation} d_{\mathrm{sg}}(S,S')\coloneqq \varlimsup _{k\to \infty } k^{-n} \left(|S_ k|+|S'_ k|-2|(S\cap S')_ k|\right). \end{equation}
Lemma 19.7.18.7887065F3C7C The above defined $d_{\mathrm{sg}}$ is a pseudometric on $\hat{\mathcal{S}}(C)$.
Proof. Only the triangle inequality needs to be argued. Take $S,S',S''\in \hat{\mathcal{S}}(C)$. We claim that for any $k\in \mathbb {N}$,
\[ |S_ k|+|S'_ k|-2|S_ k\cap S'_ k|+|S''_ k|+|S'_ k|-2|S''_ k\cap S'_ k|\geq |S_ k|+|S''_ k|-2|S_ k\cap S''_ k|. \]
From this the triangle inequality follows. To argue the claim, we rearrange it to the following form:
\[ |S'_ k|-|S_ k\cap S'_ k| \geq |S'_ k\cap S''_ k|-|S_ k\cap S''_ k|, \]
which is obvious. $\square$
Given $S,S'\in \hat{\mathcal{S}}(C)$, we say $S$ is equivalent to $S'$ and write $S\sim S'$ if $d_{\mathrm{sg}}(S,S')=0$. This is an equivalence relation by 19.7.18.
Lemma 19.7.19.680A0890C1EE Given $S,S',S''\in \hat{\mathcal{S}}(C)$, we have
\[ d_{\mathrm{sg}}(S\cap S'',S'\cap S'')\leq d_{\mathrm{sg}}(S,S'). \]
In particular, if $S^ i,S'^ i\in \hat{\mathcal{S}}(C)$ ($i\in \mathbb {N}$) and $S^ i\to S$, $S'^ i\to S'$, then
\[ S^ i\cap S'^ i\to S\cap S'. \]
Proof. Observe that for any $k\in \mathbb {N}$,
\[ |S_ k\cap S_ k''|-|S_ k\cap S_ k'\cap S_ k''|\leq |S_ k|-|S_ k\cap S_ k'|. \]
The same holds if we interchange $S$ with $S'$. It follows that
\[ |S_ k\cap S_ k''|+|S_ k'\cap S_ k''|-2|S_ k\cap S_ k'\cap S_ k''|\leq |S_ k|+|S'_ k|-2|S_ k\cap S_ k'|. \]
The first assertion follows.
Next we compute
\[ \begin{split} d_{\mathrm{sg}}(S^ i\cap S'^ i,S\cap S')\leq & d_{\mathrm{sg}}(S^ i\cap S'^ i,S^ i\cap S')+d_{\mathrm{sg}}(S^ i\cap S',S\cap S')\\ \leq & d_{\mathrm{sg}}(S'^ i,S')+d_{\mathrm{sg}}(S^ i,S) \end{split} \]
and the second assertion follows. $\square$
The volume of $S\in \mathcal{S}(C)$ is defined as
\[ \operatorname{vol}S\coloneqq \lim _{k\to \infty } (ka)^{-n}|S_{ka}|=\varlimsup _{k\to \infty } k^{-n}|S_ k|, \]
where $a$ is a sufficiently divisible positive integer. The existence of the limit and its independence from $a$ both follow from the more precise result [Theorem 2, KK12], a special case of which can be stated as follows: Theorem 19.7.20 (Kaveh–Khovanskii).85641516C8CA Let $S\in \mathcal{S}(C)$. Then, for every sufficiently divisible positive integer $a$, the limit
\[ \lim _{k\to \infty }(ka)^{-n}|S_{ka}| \]
exists, is independent of $a$, and is equal to $\varlimsup _{k\to \infty } k^{-n}|S_ k|$. If moreover $S$ generates $\mathbb {Z}^{n+1}$ as an Abelian group, and if $C(S)$ denotes the closed convex cone generated by $S\cup \{ 0\} $, then
\[ \lim _{k\to \infty } k^{-n}|S_ k| = \operatorname{vol}\{ x\in \mathbb {R}^ n:(x,1)\in C(S)\} . \]
In particular, for every convex body $K\subseteq \mathbb {R}^ n$,
\[ \lim _{k\to \infty } k^{-n}\# \left(kK\cap \mathbb {Z}^ n\right)=\operatorname{vol}K. \]
The same assertion holds with $\mathbb {Z}^ n$ replaced by any lattice, provided the Lebesgue measure is normalized by that lattice.
Lemma 19.7.21.DCB8E8464FB6 Let $S,S'\in \mathcal{S}(C)$. Then
\[ |\operatorname{vol}S-\operatorname{vol}S'|\leq d_{\mathrm{sg}}(S,S'). \]
Proof. Choose a sufficiently divisible positive integer $a$. Taking the limsup in the definition of $d_{\mathrm{sg}}$ along $k=ma$ gives
\[ d_{\mathrm{sg}}(S,S')\geq \operatorname{vol}S+\operatorname{vol}S'-2\operatorname{vol}(S\cap S'). \]
It follows that $\operatorname{vol}S-\operatorname{vol}S'\leq d_{\mathrm{sg}}(S,S')$ and $\operatorname{vol}S'-\operatorname{vol}S\leq d_{\mathrm{sg}}(S,S')$. $\square$
We define $\overline{\mathcal{S}}(C)$ as the closure of $\mathcal{S}(C)$ in $\hat{\mathcal{S}}(C)$ with respect to the topology defined by the pseudometric $d_{\mathrm{sg}}$. By 19.7.21, $\operatorname{vol}\colon \mathcal{S}(C)\rightarrow \mathbb {R}$ admits a unique $1$-Lipschitz extension to
1009A092673C
19.7.22
\begin{equation} \operatorname{vol}\colon \overline{\mathcal{S}}(C)\rightarrow \mathbb {R}. \end{equation}
Lemma 19.7.23.EC9CD63CAB3A Suppose that $S,S'\in \overline{\mathcal{S}}(C)$ and $S\subseteq S'$. Then
\[ \operatorname{vol}S\leq \operatorname{vol}S'. \]
Proof. Take sequences $S^ j,S'^ j$ in $\mathcal{S}(C)$ such that $S^ j\to S$, $S'^ j\to S'$. By 19.7.19, after replacing $S^ j$ by $S^ j\cap S'^ j$, we may assume that $S^ j\subseteq S'^ j$ for each $j$. Then our assertion follows easily. $\square$
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.
Note that for any $S\in \mathcal{S}'(C)$, the cone $C(S)$ has full dimension (i.e., the topological interior is non-empty). Given a full-dimensional subcone $C'\subseteq C$, the semigroup $C'\cap \mathbb {Z}^{n+1}$ lies in $\mathcal{S}'(C)$. This class behaves well under intersections: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
\[ \Delta (S)\coloneqq \left\{ x\in \mathbb {R}^ n:(x,1)\in C(S) \right\} . \]
This is essentially proved in [Lemma 4.8, WN14], which itself follows from a theorem of Khovanskii [Kho92].
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). $\square$
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. $\square$
19.7.30 Okounkov bodies of almost semigroups10CE70BA2854
Definition 19.7.31.B0C65454CC85 We define $\overline{\mathcal{S}'(C)}_{>0}$ to be the set of elements in the closure of $\mathcal{S}'(C)$ in $\hat{\mathcal{S}}(C)$ with positive volume. An element in $\overline{\mathcal{S}'(C)}_{>0}$ is called an almost semigroup in $C$.
Theorem 19.7.32.1551B679B8B7 The Okounkov body map $\Delta \colon \mathcal{S}'(C) \rightarrow \mathcal{K}_ n$ as defined in definition 19.7.26 admits a unique continuous extension
1B239C5F9599
19.7.33
\begin{equation} \Delta \colon \overline{\mathcal{S}'(C)}_{>0}\rightarrow \mathcal{K}_ n. \end{equation}
Moreover, for any $S\in \overline{\mathcal{S}'(C)}_{>0}$, we have
558EBEE0E3F0
19.7.34
\begin{equation} \operatorname{vol}S=\operatorname{vol}\Delta (S). \end{equation}
Proof. The uniqueness of the extension is clear once existence is known. Moreover, (19.7.34) follows easily from theorem 19.2.5 and theorem 19.7.5 by continuity. It remains to argue the existence of the continuous extension. We first construct an extension and prove its continuity.
Step 1. We construct the desired map (19.7.33). Let $S\in \overline{\mathcal{S}'(C)}_{>0}$. We wish to construct a convex body $\Delta (S)\in \mathcal{K}_ n$.
Let $S^ i\in \mathcal{S}'(C)$ be a sequence that converges to $S$ such that
\[ d_{\mathrm{sg}}(S^ i,S^{i+1})\leq 2^{-i}. \]
For each $i,j\geq 0$, we introduce
\[ S^{i,j}=S^ i\cap S^{i+1}\cap \cdots \cap S^{i+j}. \]
Then by 19.7.19,
\[ d_{\mathrm{sg}}(S^{i,j},S^{i,j+1})\leq 2^{-i-j}. \]
Take $i_0>0$ large enough so that, for $i\geq i_0$, we have $\operatorname{vol}S^ i>2^{-1}\operatorname{vol}S$ and $2^{2-i}< \operatorname{vol}S$. Then
\[ \operatorname{vol}S^ i-\operatorname{vol}S^{i,j}\leq d_{\mathrm{sg}}(S^{i,0},S^{i,1})+d_{\mathrm{sg}}(S^{i,1},S^{i,2})+\cdots +d_{\mathrm{sg}}(S^{i,j-1},S^{i,j})\leq 2^{1-i}. \]
It follows that $\operatorname{vol}S^{i,j}> 2^{-1}\operatorname{vol}S-2^{1-i}>0$ whenever $i\geq i_0$. In particular, by 19.7.25, $S^{i,j}\in \mathcal{S}'(C)$ for $i\geq i_0$.
By 19.7.29, for $i\geq i_0$, there exists $T^ i\in \mathcal{S}'(C)$ such that $S^{i,j}\to T^ i$ as $j\to \infty $. Moreover,
\[ d_{\mathrm{sg}}(T^ i,S)=\lim _{j\to \infty } d_{\mathrm{sg}}(S^{i,j},S)\leq \lim _{j\to \infty } d_{\mathrm{sg}}(S^{i,j},S^ i)+d_{\mathrm{sg}}(S^ i,S)\leq 2^{1-i}+d_{\mathrm{sg}}(S^ i,S). \]
Therefore, $T^ i\to S$. We then define
\[ \Delta (S)\coloneqq \overline{\bigcup _{i=i_0}^{\infty } \Delta (T^ i)}. \]
In other words, we have defined
\[ \Delta (S)\coloneqq \varliminf _{i\to \infty }\Delta (S^ i). \]
Indeed, by the construction in the proof of 19.7.29 and the cone equality in 19.7.25,
\[ \Delta (T^ i)=\bigcap _{j\geq 0}\Delta (S^{i,j}) =\bigcap _{j\geq i}\Delta (S^ j). \]
Thus the bodies $\Delta (T^ i)$ are increasing in $i$ and are all contained in the compact slice $\{ x\in \mathbb {R}^ n:(x,1)\in C\} $. In particular, $\Delta (S)$ is a convex body and
\[ \operatorname{vol}\Delta (S)=\lim _{i\to \infty }\operatorname{vol}\Delta (T^ i)=\lim _{i\to \infty }\operatorname{vol}T^ i=\operatorname{vol}S. \]
This is an honest limit: if $\Delta $ is the limit of a subsequence of $\Delta (S^ i)$, then $\Delta (S)\subseteq \Delta $ by (19.7.8). Moreover,
\[ \operatorname{vol}\Delta =\lim _{m\to \infty }\operatorname{vol}\Delta (S^{i_ m})=\lim _{m\to \infty }\operatorname{vol}S^{i_ m}=\operatorname{vol}S. \]
Comparing the volumes, we find that equality holds. So by theorem 19.7.4,
Next we claim that $\Delta (S)$ as defined above does not depend on the choice of the sequence $S^ i$. In fact, suppose that $S'^ i\in \mathcal{S}'(C)$ is another sequence satisfying the same conditions as $S^ i$. By 19.7.19, we have
\[ R^ i=S^{i+1}\cap S'^{i+1}\longrightarrow S\cap S=S. \]
Since $\operatorname{vol}S>0$ and the volume is continuous on $\overline{\mathcal S}(C)$, we have $\operatorname{vol}R^ i>0$ for all large $i$. Thus $R^ i\in \mathcal S'(C)$ by 19.7.25. Moreover,
\[ d_{\mathrm{sg}}(R^ i,R^{i+1}) \leq d_{\mathrm{sg}}(S^{i+1},S^{i+2}) + d_{\mathrm{sg}}(S'^{i+1},S'^{i+2}). \]
Passing simultaneously to subsequences of $S^ i$ and $S'^ i$, we may assume
\[ d_{\mathrm{sg}}(S^{i+1},S^{i+2})+d_{\mathrm{sg}}(S'^{i+1},S'^{i+2})\leq 2^{-i}. \]
Then the construction above applies to $R^ i=S^{i+1}\cap S'^{i+1}$. Since $R^ i\subseteq S^{i+1}$ and $R^ i\subseteq S'^{i+1}$, every Hausdorff limit of $\Delta (R^ i)$ is contained in the corresponding limits of $\Delta (S^ i)$ and $\Delta (S'^ i)$. The volumes of all three limits are $\operatorname{vol}S$, hence the inclusions are equalities. The same is true with $S'^ i$ in place of $S^ i$. So we conclude that $\Delta (S)$ as in (19.7.35) does not depend on the choices we made.
Step 2. It remains to prove the continuity of $\Delta $ defined in Step 1. Suppose that $S^ i\in \overline{\mathcal{S}'(C)}_{>0}$ is a sequence with limit $S\in \overline{\mathcal{S}'(C)}_{>0}$. We want to show that
16236975E9F1
19.7.36
\begin{equation} \Delta (S^ i)\xrightarrow {d_{\mathrm{Haus}}}\Delta (S). \end{equation}
We first reduce to the case where $S^ i\in \mathcal{S}'(C)$. By (19.7.35), for each $i$, we can choose $T^ i\in \mathcal{S}'(C)$ such that $d_{\mathrm{sg}}(S^ i,T^ i)<2^{-i}$ and $d_{\mathrm{Haus}}(\Delta (S^ i),\Delta (T^ i))<2^{-i}$. If we have shown $\Delta (T^ i)\xrightarrow {d_{\mathrm{Haus}}}\Delta (S)$, then (19.7.36) follows immediately.
Next we reduce to the case where $d_{\mathrm{sg}}(S^ i,S^{i+1})\leq 2^{-i}$. Thanks to theorem 19.7.4, in order to prove (19.7.36), it suffices to show that each subsequence of $\Delta (S^ i)$ admits a subsequence that converges to $\Delta (S)$. Hence, we may pass to a sufficiently fast subsequence and reduce to the required case.
After these reductions, (19.7.36) is nothing but (19.7.35). $\square$
Proof. Let $S^ j,S'^ j\in \mathcal{S}'(C)$ be elements such that $S^ j\to S$, $S'^ j\to S'$. Then it follows from 19.7.19 that $S^ j\cap S'^ j \to S$. Since $\operatorname{vol}$ is continuous, for large $j$, $S^ j\cap S'^ j$ has positive volume and hence lies in $\mathcal{S}'(C)$ by 19.7.25. We may therefore replace $S^ j$ by $S^ j\cap S'^ j$ and assume that $S^ j\subseteq S'^ j$. Hence, (19.2.14) follows from the continuity of $\Delta $ proved in theorem 19.7.32. $\square$
Remark 19.7.37.219FB2B3C530 The construction of $\Delta $ is independent of the choice of $C$ in the following sense. If $C'$ is another cone satisfying the same assumptions as $C$ and $C'\supseteq C$, then the Okounkov body map $\Delta \colon \overline{\mathcal{S}'(C')}_{>0}\rightarrow \mathcal{K}_ n$ extends the corresponding map (19.7.33). We will use this fact without further comment.
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