Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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

share|improve this question
add comment

1 Answer

up vote 1 down vote accepted

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.

share|improve this answer
    
thanks! Can you share some references about this? –  David Petrecca Feb 1 at 8:23
    
There is a paper of Lerman and Tolman that says that there is one-one correspondence between toric orbifolds and Delzant polytopes with integer weights attached to the faces. arxiv.org/abs/dg-ga/9412005 The conic case generalizes this to the situation where weights are allowed to be non-integer. I dont know if there is a proof of this in literature, but it seems to be well known. At least in many of Donaldson's writings on the subject it seems implicit. –  Ved Datar Feb 4 at 18:28
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.