I am looking for examples of smooth projective toric threefolds $\mathbb P_\Delta$ such that the rational polytope $\Delta$ has only pentagonal faces and hexagonal faces. I quickly searched for references but only found cases where some triangular faces or square faces show up.
Does any one know an example or any references?
This is rather a comment/question than an answer, but I am not allowed to comment, yet:
I just know the following for a lattice polytope $\Delta$: $\mathbb P_{\Delta}$ is smooth if and only if $\Delta$ is smooth. If I am not mistaken, this means that every vertex of $\Delta$ has exactly three edges meeting in it. When you dualize $\Delta$ (in 3 dim), the vertices of $\Delta^*$ correspond to facets of $\Delta$, edges correspond to edges, and facets of $\Delta^*$ correspond to vertices of $\Delta$. Thus, $\Delta$ being smooth implies that the facets of $\Delta^*$ are triangles, and a 5gon or 6gon is impossible. Since you have an example with square facets: What did I get wrong? Is there something different starting with a rational $\Delta$?
If you are willing to work with smooth stacks, rather than smooth toric varieties (or alternatively consider toric varieties with quotient singularities) then it is definitely possible.
Combinatorially, this broader setup allows you to look for $\Delta$ with rational vertices, 5gon and 6gon faces only and three edges out of each vertex. If you take a regular dodecahedron, you will have the desired combinatorial structure, but without rationality. However, if you approximate the defining equations of the facets of the dodecahedron by rational hyperplanes well enough, the combinatorial structure will not change and you will get the desired (complicated) singular toric variety or smooth toric DM stack.
Is it possible to stay within smooth toric varieties? I find this unlikely but I don't have an argument. It is easy to see that one needs to have exactly 12 pentagon faces, but I don't know how to proceed further.
