# Why are symplectic toric varieties projective?

Let $$X$$ be a symplectic toric manifold meaning a compact symplectic manifold $$(X, \omega)$$ with $$\dim{X} = 2n$$ equipped with a Hamiltonian action of a maximal-dimension torus $$\mathbb{T} = (\mathbb{S}^1)^n$$.

These are classified in terms of (bounded) Delzant polytopes $$\Delta \subset \mathbb{R}^n$$. Given $$X$$, the image of its moment map $$\mu : X \to \mathbb{R}^n$$ is compact and in fact we can show it is a polytope $$\Delta$$. Then $$X$$ is recoverable as the partial symplectic reduction $$X_{\Delta}$$ of a certain torus action on $$\mathbb{C}^d$$ arising from the structure of $$\Delta$$.

Given any polytope $$\Delta \subset \mathbb{R}^n$$ I can also associate to it a toric variety $$X^{\text{alg}}_\Delta$$ which is smooth if $$\Delta$$ is Delzant. If I understand these constructions correctly, $$X_{\Delta}$$ and hence $$X$$ should be the analytification of $$X_{\Delta}^{\text{alg}}$$.

This implies a number of amazing properties of $$X$$. It is actually a projective variety and hence Kahler.

(1) Is there a way to see these properties of $$X$$ without going through the classification?

(1a) Where does the compatible complex structure invariant under the $$\mathbb{T}$$-action arise from? Why is it integrable?

(1b) Why does the $$(\mathbb{S}^1)^n$$-action extend to a holomorphic $$(\mathbb{C}^\times)^n$$-action? If we can choose a vector field perpendicular to the $$S^1$$-orbits we can extend the action but I don't see how to do this in a canonical way.

(2) Why do only projective toric varieties show up? There are non-projective smooth compact toric varieties whose analytification has a top-dimensional torus action compatible with the complex structure. Is there a reason these can never be symplectic? My guess is if the manifold were symplectic compatible with the action this would force the symplectic and complex structures meaning it would be Kahler and hence projective.