About a Delzant polytope. (In particular dodecahedron) - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T22:48:16Z http://mathoverflow.net/feeds/question/58573 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/58573/about-a-delzant-polytope-in-particular-dodecahedron About a Delzant polytope. (In particular dodecahedron) YCho 2011-03-15T20:53:38Z 2011-04-13T21:37:32Z <p>Hi. I have a question. </p> <p>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 $P$.</p> <p>(If you wonder why this condition is called smooth, See Fulton. Introduction to toric variety chap I)</p> <p>My question is as follow. </p> <p>Can dodecahedron be the Delzant polytope? I mean, is there a symplectic toric manifold whose moment map image is combinatorially equivalent to a dodecahedron?</p> <p>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. </p> <p>Thank you.</p> http://mathoverflow.net/questions/58573/about-a-delzant-polytope-in-particular-dodecahedron/61587#61587 Answer by Dmitri for About a Delzant polytope. (In particular dodecahedron) Dmitri 2011-04-13T20:12:18Z 2011-04-13T21:37:32Z <p>This is just a partial answer giving a <em>solution</em> to a <em>dual</em> 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 (<strong>Update</strong> 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. </p> <p><strong>In terms of combinatorics,</strong> 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). </p> <p><strong>Construction.</strong> 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 <a href="http://en.wikipedia.org/wiki/Cuboctahedron" rel="nofollow">http://en.wikipedia.org/wiki/Cuboctahedron</a> . 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 $N$). </p> <p>Now we can ask the question. <em>Have we constructed the fan of a projective variety?</em> One can answer this question (but I don't do it here (<strong>Update</strong> 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 <em>rombic dodecahedron</em> <a href="http://en.wikipedia.org/wiki/Rhombic_dodecahedron" rel="nofollow">http://en.wikipedia.org/wiki/Rhombic_dodecahedron</a> . 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.</p>