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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?

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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?

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    $\begingroup$ The first quadrant is unbounded and simple at the origin. It is a polyhedron corresponding to $\mathbb{C}^2$. $\endgroup$ – Hwang Apr 12 '17 at 0:16
  • $\begingroup$ 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? $\endgroup$ – Rob Apr 12 '17 at 17:48
  • $\begingroup$ 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 $\endgroup$ – Rob Apr 12 '17 at 18:04
  • $\begingroup$ Why is being compact related to a free action? It doesn't seem to me. $\endgroup$ – Hwang Apr 13 '17 at 0:17
  • $\begingroup$ 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. $\endgroup$ – Rob Apr 13 '17 at 9:03

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