Delzant theorem says that there is a 1-1 correspondence between compact toric symplectic manifolds (modulo equivariant symplectomorphism) and the Delzant polytopes (modulo lattice isomorphism). The polytope is given by the image of a moment map of the torus action.
My question is
Given a Delzant polyhedron(may not be bounded), is there a unique toric symplectic manifold corresponding to the polyhedron?
Following the book "Lectures on Symplectic Geometry" by Ana Cannas da Silva (with different sign convention), the construction of a toric symplectic manifold from a Delzant polytope $P$ is as follows. Let $v_1, \dots, v_m$ denote inward primitive vectors of facets of $P$. The polypote $P$ is written as $\{x \in (\mathbb{R}^n)^\vee\mid \langle x, v_i \rangle \geq -\lambda_i\}$. Define $A\colon \mathbb{R}^m \rightarrow \mathbb{R}^n$ by $A(e_i) = v_i$. We have an exact sequence
$$0 \rightarrow K \rightarrow^{j} \mathbb{R}^m \rightarrow^{A} \mathbb{R}^n \rightarrow 0.$$ Dually, $$ 0 \rightarrow (\mathbb{R}^n)^\vee \rightarrow^{A^t} (\mathbb{R}^m)^\vee \rightarrow^{j^*} K^\vee \rightarrow 0.$$
Let $\Phi\colon \mathbb{C}^m \rightarrow (\mathbb{R}^m)^\vee$ be a moment map given by $\Phi(z) = (\pi|z_1|^2 - \lambda_1, \dots, \pi|z_m|^2 - \lambda_m)$. Now $Q:=Ker(\mathbb{R}^m/\mathbb{Z}^m \rightarrow^A \mathbb{R}^n/\mathbb{Z}^n)$ acts on $\mathbb{C}^m$ with moment map $j^* \circ \Phi$. The symplectic reduction $M$ at $0$ has residual $T^m/Q$ action whose moment map image is $P$.
It seems to me that we didn't use the fact that $P$ is bounded, so a toric manifold $M$ can be constructed from a polyhedron $P$. Do we need to assume $P$ to be bounded to say such $M$ is unique?
Maybe the condition of being Delzant (I'd say its enough with the "simple" condition) implies that the polytope is bounded. Do you have an example of Delzant polytope which is not bounded?