Delzant theorem for polyhedra

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?

• The first quadrant is unbounded and simple at the origin. It is a polyhedron corresponding to $\mathbb{C}^2$. – Hwang Apr 12 '17 at 0:16
• Thanks, seems like the construction as you said works the same. For the quadrant, the complex space $\mathbb{C}^2$ is found and I've also looked at other examples like the semicylinder. I haven't gone deep into the uniqueness proof of Delzant theorem but when I do so I'll think about your question. Your question is old though ( sorry I didn't notice it), have you gone farther with this? – Rob Apr 12 '17 at 17:48
• My bad, the fact that the polyhedron $\Delta$ is compact is in fact used when showing that the zero level set $Z$ is compact. It is necessary in the proof to show that $G$ acts freely on $Z$ and so apply the Reduction theorem – Rob Apr 12 '17 at 18:04
• Seems right. I though compactness was needed when saying that stabilizers of $z' \in Z$ are always included in stabilizers of points $z$ whose image is a vertex of the polytope $\Delta'$. Not sure, but this seems to work even if $\Delta'$ is not bounded. Interesting topic though, I'm doing my final bachelor project on Delzant theorem, if I have more time after going into relation with integrable systems I'll take a look at all this. Sorry about my bad observations though, should have checked better :D. – Rob Apr 13 '17 at 9:03