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A quasi-lattice polytope is a polytope obtained by reflections, translations, and rotations of lattice polytopes. In a tiling of a lattice polytope by quasi-lattice polytopes, are all quasi-lattice polytopes necessarily lattice polytopes?

If we allow dilations in the general sense to enter the "quasi club", then any figure may just be obtained from dilating a unit hypercube, so the result is obviously false. What if we restrict dilations to transformations induced by matrices with integer entries? This stronger question is true in 1 dimension since $[a,b]$ having integral endpoints implies $[ac,bc]$ has integral endpoints for all $c \in \mathbb{Z}.$

Let's be even more general. Fix a dimension $n,$ let $S$ be a set of affine transformations, and define $S$-lattice polytopes as those in the image of $s: P \to P$ for some $s \in S$ where $P$ is the set of $n$ dimensional lattice polytopes. How large can $S$ be such that the statement "in a tiling of lattice polytopes by $S$-lattice polytopes, all $S$-lattice polytopes are lattice polytopes" is true?

For $n=1, S = \{x \to ax+b | a \in T, b \in \mathbb{R}\}$ where $T = \mathbb{Z}$ works. In fact, $T$ can be replaced by any extension $R \supseteq T$ such that $R$ is linearly independent over $\mathbb{Z}$ (defining an infinite set to be linearly independent iff every finite subset is), and this characterizes all maximal $S$ completely.

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  • $\begingroup$ You can make $S$ pretty large, e.g. you can let it contain all transformations $T$ so that $\smash{T(\Bbb Z^n)\cap \Bbb Z^n=\varnothing}$. Then no lattice polytope can be decomposed into $S$-lattice polytopes and the statement is trivially satisfied. $\endgroup$
    – M. Winter
    Commented May 29, 2020 at 15:14

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The answer to your first question is No.

Consider the 5x5 square as a lattice polytope in two different ways as seen above. Each tiling of one of these sqares gives a tiling of the other. But you can choose the tiles in such a way, so that they are lattice polytope in only one of the squares.

So the right square is tiled with quasi-lattice polytopes that are not lattice polytopes.

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  • $\begingroup$ I considered a 3-4-5 triangle before, but didn't find this counterexample because I only thought about rotating subtiles instead of the entire tile. Your approach should generalize to exclude all rotations with angles $\theta$ such that $\tan \theta = b/a$ where $a, b, \sqrt{a^2+b^2} \in \mathbb{Z}.$ Furthermore, angles whose tangents are irrational will never map a lattice polygon to another lattice polygon, so any counterexamples involving angles not of the aforementioned form, if they exist, will be much harder to find. $\endgroup$
    – Display name
    Commented May 29, 2020 at 15:27
  • $\begingroup$ @Displayname Irrational angles might be safe. Consider my comment under your post: if you compose $S$ of rotations by irrational angles, then no $S$-lattice polytope can have two vertices on $\Bbb Z^n$ and so no decomposition of a lattice polytope into such is possible. $\endgroup$
    – M. Winter
    Commented May 29, 2020 at 15:30
  • $\begingroup$ It's true that a polytope created from a rotation by an angle with irrational tangent cannot have two lattice points. But it may have 1 lattice point. What prevents us from using a different $S$-polytope for every corner and then miraculously filling in the interior? $\endgroup$
    – Display name
    Commented May 29, 2020 at 15:33
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    $\begingroup$ @Displayname The edges of a lattice polytope must have rational directions (that is, the tangent of the angle with the axes is rational), and so an $S$-lattice polytope (that is, after an irrational rotation) must have irrational edge directions only. But you cannot decompose a lattice polytope only into polytope with irrational edge directions, because some tile must share (a part of) an edge with the lattice polytope, whose edges have rational directions. $\endgroup$
    – M. Winter
    Commented May 29, 2020 at 15:41

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