Here we suppose that the Delzant polytope $P$ is integral. We consider all the multiples $k\overline{P}$ for integers $k\geq 0$ and let $B_{k}$ be the set of lattice points

$$B_{k}=k\overline{P}\cap{\bf Z}^{n}.$$

Let the number of points in $B_{k}$ be $N_{k}+1$. We can put all these sets together by considering the cone over $P$

$$cone(P)=\{(\underline{x},y)\in{\bf R}^{n+1}:y\geq 0,\underline{x}\in y\overline{P}\}.$$

The disjoint union of the sets $B_{k}$ can be identified with the set $B=cone(P)\cap{\bf Z}^{n+1}$. Now $B$ is an abelian semi-group under addition and we have a corresponding ring $R$ over ${\bf C}$ with one generator $s_{b}$ for each point of $b\in B$ and relations $s_{b}s_{b^{\prime}}=s_{b+b^{\prime}}$. This is a graded ring, $R=\bigoplus R_{k}$, where $R_{k}$ has a basis $s_{\nu}$ corresponding to the points $\nu$ of $B_{k}$. Further, there is an obvious action of the torus $T_{c}^{n}$ on $R$.

All of these definitions make sense for any convex set $P$. The crucial fact is that when the $P$ is an integral polytope the ring is finitely generated. Thus there is a corresponding projective variety $X_{{\rm alg}}={\rm Proj}(R)$, and the group action on $R$ defines an action on $X_{{\rm alg}}$. Second, if $P$ is Delzant, then $X_{{\rm alg}}$ is smooth and of course this recovers the same complex manifold $X_{{\rm cx.}}$. The vector spaces $R_{k}$ are the sections

$$R_{k}=H^{0}(X_{{\rm cx.}},L^{k})$$

and it is not hard to see that for any $k\geq 1$ the sections give an embedding $X_{{\rm cx.}}\rightarrow{\bf P}(R_{k}^{*})$. From this algebro-geometric point of view the integer $\lambda_{r}(\nu)-c_{r}$, for lattice points $\nu\in\overline{P}$, is the order of vanishing of the section $s_{\nu}$ along the corresponding divisor in $X_{{\rm cx.}}$.

Example Let $P$ be the square $(0,1)^{2}\subset{\bf R}^{2}$. The corresponding manifold is the product $S^{2}\times S^{2}$. The points in $B_{1}$ are the four vertices $p_{0}=(0,0),p_{1}=(0,1),p_{2}=(1,0),p_{3}=(1,1)$ so $R_{1}$ has a corresponding basis $s_{0}s_{1},s_{2},s_{3}$ say. The equation $p_{0}+p_{3}=p_{1}+p_{2}$ goes over to the relation $s_{0}s_{3}=s_{1}s_{2}$. The embedding of $X_{{\rm cx.}}$ in ${\bf P}^{3}$ has image the quadric hypersurface cut out by the equation $Z_{0}Z_{1}-Z_{2}Z_{3}=0$.

When the polytope $P$ is integral but not Delzant the variety $X_{{\rm alg}}$ we construct is singular. If each vertex lies on exactly $n$ codimension-1 faces then $X_{{\rm alg}}$ is an orbifold. Much of the theory, including the differential-geometric constructions, extends easily to this case.

To sum up we have three ways—complex, symplectic and algebraic— of constructing a compact manifold associated to an integral Delzant polytope. From now on we will just denote this by $X$.