Let $P \subset \mathbb R^n$ be a Delzant polytope defined by inequalities $\ell\_i(x) \geq 0, i=1, \ldots, d$.


Of course, from the symplectic point of view, the inequalities $a\_i \ell\_i \geq 0$ still define the same polytope for all positive real $a\_i$.


What I wonder is the following. Does the function $g: P \rightarrow \mathbb R$ (up to affine terms)
$$
g(x) = \sum\_{i=1}^d (a\_i \ell\_i(x) \log \ell\_i(x)) + h(x) 
$$
with $h$ smooth on $P$ define a compatible complex structure on $P$, no matter what the $a\_i>0$ are?


Or are legal potentials only the ones where the $\ell\_i(x)$ are of the form $\langle x, \mu\_i \rangle - \lambda\_i$ where $\mu\_i$ is a primitive element of the integral lattice and points to the interior of the $i$-th face?


thank you
David



Legal potentials on delzant polytopes
Yes, but it might be singular. For instance, if $a\_i = 1/\beta\_i$, then the potential $u$ corresponds to a conical K\"ahler metric with cone angles $2\pi\beta\_i$ along the divisor corresponding to the face $[l\_i=0]$. In particular if you take $a\_i$ to be an integer it will correspond to an orbifold structure.


