About a Delzant polytope. (In particular dodecahedron)

Hi. I have a question.

Definition. Delzant polytope $P$ is a rational convex simple polytope with the smooth condition. Here, "smooth" means that for each vertex $v$, the $n$ edges containing $v$ form an element of $SL(n,\mathbb{Z})$, where $n$ is a dimension of $P$.

(If you wonder why this condition is called smooth, See Fulton. Introduction to toric variety chap I)

My question is as follow.

Can dodecahedron be the Delzant polytope? I mean, is there a symplectic toric manifold whose moment map image is combinatorially equivalent to a dodecahedron?

Delzant's classfication theorem of compact symplectic toric manifold is surely very strong. But I think it is very hard to check whether the given polytope (having many faces) is of Delzant type or not. If you know any reference of give me any comment, I really appriciate for your help.

Thank you.

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As far as the classification of toric symplectic manifolds is concerned, the property "having a Delzant polytope of the combinatorial type of the regular dodecahedron" seems unmotivated. Nevertheless, it's a nice question. – André Henriques Mar 15 '11 at 21:46
The obvious first thing to try is a rational pyritohedron, but that fails: at one of the eight vertices with threefold rotational symmetry, the adjacent edges lie along directions which are the even permutations of $(p^2, -pq, q^2)$ for positive coprime integers $p \gt q$, giving $(p^3+q^3)^2$ for a determinant, rather than $1$. – Tracy Hall Mar 15 '11 at 22:50

Elisa Prato, Symplectic Toric Geometry and the Regular Dodecahedron (2015)

Sorry it took so long. Hope you find this helpful. Another article covering all the platonic solids, rational or not, is on its way.

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Dear Elisa Prato, I think your paper is not answer for the question. Your paper discuss about irrational regular dodecahedron. But Cho's question is just about combinatorial type of regular dodecahedron, i.e, Is there a 3-dimensional delzant polytope of same combinatorial type with the regular dodecahedron? – Seonhwa Kim Dec 1 '15 at 13:27

This is just a partial answer giving a solution to a dual problem. I will use the language of toric varieties (http://www3.amherst.edu/~dacox/), but in essence this answer is purely combinatorial. I will construct a smooth 3-dimensional toric variety whose fan has combinatorial structure of icosahedron. Unfortunately, I am not sure that this toric variety is projective (Update David proves in the comment below that this variety is not pojective!*). If it were projective this would of course give a solution to your question but I have doubts that it is not projective (this should not be hard to check). But even if we can not get the dodecahedron this way, we will get a collection of Delzant polytops quite "close" to dodecahedron it terms of thire combinatorial structure.

In terms of combinatorics, fist thing what we will do is the following: we will show how to decompose $\mathbb R^3$ into $20$ rational simplicial cones, where each simplicial cone can be sent to the positive octant by a matrix from $SL(3,\mathbb Z)$. There will be in total $12$ rays and each ray is in the border of $5$ simplicial cones (just like for icosahedron).

Construction. First we need to chose an integral lattice $N$ in $\mathbb R^3$, it will be an index two sublattice in $\mathbb Z^3$; $(a,b,c)\in N$ if $a,b,c\in \mathbb Z$, $a+b+c\in 2\mathbb Z$. Next we specify $12$ points in $N$ that lay on $12$ ray of our fan. These are: $(\pm 1, \pm 1,0)$, $(\pm 1, 0, \pm 1), (0,\pm 1, \pm 1)$. In fact these points are vertices of the cuboctahedron http://en.wikipedia.org/wiki/Cuboctahedron . Cuboctahedron has $8$ triangular faces and $6$ square faces, and to finish the construction we should cut each square face by a diagonal into two triangles. This is done as follows: the faces $z=\pm 1$ are cut into two by the plane $x=0$, faces $y=\pm 1$ by $z=0$, and the faces $x=\pm 1$ by $y=0$. Now it is not hard to see that the obtained triangulation of cuboctahedron gives us a decomposition of $\mathbb R^3$ into $12$ standard simplicial cones (just notice that the triples of vectors ($(1,1,0)$, $(0,1,1)$, $(1,0,1)$) and $((0,\pm 1,1), (1,0,1))$ from integral bases in $N$).

Now we can ask the question. Have we constructed the fan of a projective variety? One can answer this question (but I don't do it here (Update David proved in the comment that this example is not projective)). First of all, if we don't cut square faces of cuboctahedron by diagonals, we have a fan of a singular projective variety with $6$ ordinary double singularities (i.e. singularities given locally by $(x^2+y^2+z^2+t^2=0)$). A moment polytope of this variety is dual to cuboctahedron and is called rombic dodecahedron http://en.wikipedia.org/wiki/Rhombic_dodecahedron . This is not a Delzant polytope because it has $8$ bad vertices. Now, this polytope has $12$ faces and in order to make the polytope Delzant we should just generically perturb the faces (by replacing them by nearby parallel planes). Any such generic perturbation will give us Delzant polytope. One just need to check if among these polytopes there will be the Dodecahedron... In other words, each perturbation corresponds to a symplectic structure on a small resolution of the singular variety, but it is not clear we can get the resolution corresponding to choices of diagonals in squares that we made.

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Sadly, you lose. Recall that a fan is the fan of a projective toric variety if and only if it supports a convex piecewise linear function, which is linear on each cone and strictly convex whenever two cones meet. Suppose that f is such a function. Then the convexity condition gives $f(0,1,1)+f(0,−1,1)<f(1,0,1)+f(-1,0,1)$, and five similar conditions coming from the other squares. Add all six inequalities together, and you get that a sum of twelve terms is less than the same sum of twelve terms, a contradiction. – David Speyer Apr 13 '11 at 20:32
Nice approach though! I definitely hope that some variant of this would work. – David Speyer Apr 13 '11 at 20:32
David, thanks a lot for this comment!! (I was suspecting that something like this will happen, since the picture is too symmetric...) – Dmitri Apr 13 '11 at 20:37