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?
