3
$\begingroup$

I have seen two different ways of describing a Delzant polytope:

From Canna Da Silva https://people.math.ethz.ch/~acannas/Papers/toric.pdf, a Delzant polytope is a polytope in $\mathbb{R}^{n*}$ satisfying the three following properties:

  • it is simple: there are exactly $n$ edges meeting at each vertex;
  • it is rational: the edges meeting at each vertex are of the form $p + tu_i$, where $t \geq 0$ and $u_i \in \mathbb{Z}^{n*}$;
  • it is smooth: for each vertex, the corresponding $u_1,...,u_n$ form a $\mathbb{Z}$-basis of $\mathbb{Z}^{n*}$.

On the other hand, one can look at a Delzant polytope through its half-space description: a Delzant polytope $\Delta$ with $d$ facets can be written as $$ \Delta := \lbrace x \in \mathbb{R}^{n*} \ | \ \langle x, v_j \rangle + a_j \geq 0, \ j=1,...,d \rbrace, $$ where the $v_j$'s are primitive inward-pointing conormals in $\mathbb{Z}^n$, and $a = (a_1,...,a_d) \in \mathbb{R}_{\geq 0}^{d*} \setminus \{0\}$. In this case, https://arxiv.org/pdf/1308.3224.pdf describe the Delzant's properties by:

  • compactness: each $k$-codimensional face of $\Delta$ is the intersection of exactly $k$ facets;

  • smoothness: the $k$ associated conormals $(v_{l_1},...,v_{l_k})$ can be extended to a $\mathbb{Z}$-basis for the lattice $\mathbb{Z}^n$.

Although both descriptions seem very similar (at least intuitively), I have trouble proving that they are equivalent. Any help will be greatly appreciated.

Thanks in advance!

$\endgroup$
1
  • 1
    $\begingroup$ Related question which you might possibly be interested in: mathoverflow.net/questions/4982/look-into-delzant-polytope. In particular, note the answer of Joel Fine explaining that the Delzant condition is the same as saying that each vertex can be taken to the positive orthant model by some element of $GL(n,\mathbb{Z})$. $\endgroup$ Commented Dec 23, 2019 at 19:36

0

You must log in to answer this question.

Browse other questions tagged .