# Cut and Project Sets Using Hyperbolic Space

One strategy for creating aperiodic sets in $\mathbf{R}$ is to take a line $L$ of irrational slope in $\mathbf{R}^2$ along with a compact window $W \subset \mathbf{R}$ which is thought of as a subset of $L^\perp.$ For simplicity, you can just make $W$ an interval containing 0. We then take any points of $\mathbf{Z}^2$ (say the standard lattice in ${\mathbf{R}^2}$) which are also contained in $L\times W$ and project them orthogonally onto $L.$ If we identify $L$ with $\mathbf{R}$ then we have constructed an aperiodic set in $\mathbf{R}$ which also has a variety of nice dynamical and analytic properties. This construction can be significantly generalized using abelian groups, but I was thinking of a slightly different generalization.

Why not do a similar construction using $\mathbf{H}^2?$ Namely, fix a uniform tiling of $\mathbf{H}^2$ and a geodesic $\gamma$ in $\mathbf{H}^2$ which descends to a geodesic on the (potentially singular) quotient surface. We would probably want to choose a geodesic which is not closed and is dense in some region of the surface. Choose some segment of a geodesic orthogonal to $\gamma$ to act like a window and project any vertices of the uniform tiling orthogonally onto $\gamma$ if they lie within the range of the window. (The region corresponding to $L \times W$ in the hyperbolic setting would look something like a banana in the Poincare disc model of $\mathbf{H}^2$.) After composing with an isometry from $\gamma$ to $\mathbf{R}$ we might get an aperiodic point set.

Is there an obvious reason why this wouldn't yield (at the very least) an aperiodic tiling of $\mathbf{R}$ with finite local complexity?

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## 1 Answer

In general the set of lengths of intervals between consecutive points on $L$ would be an infinite set. The reason is that there can exist such lines $L$ whose image in the unit tangent bundle of the quotient surface is dense in the unit tangent bundle---the geodesic flow has dense leaves. One can therefore find infinitely many pairs of vertices $(x_i,y_i)$, all in the same orbit of vertices under the symmetry group of the tiling, all contained in the banana neighborhood of $L$, with the following properties: the distance between $x_i$ and $L$ accumulates to zero; letting $v_i$ be the tangent vector to the segment $[x_i,y_i]$ at $x_i$, and letting $w_i$ be the parallel translate of $v_i$ along the perpendicular from $x_i$ to $L$, the angle of $w_i$ with $L$ assumes a dense set of values in some open interval of angles. It follows that the length of the projection of $[x_i,y_i]$ to $L$ assumes infinitely many values in a finite interval. And this implies that there exist infinitely many consecutive point distances.

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Thanks you! We were looking at this in the hopes that we could construct tiling spaces which factored over laminations on higher genus surfaces. – mkreisel Nov 19 '12 at 21:01
Are there any particular choices of tiling and geodesic for which the number of lengths of tiles would be finite? – mkreisel Nov 19 '12 at 21:05
My very weak guess is no, unless the geodesic is an axis of a hyperbolic element of the symmetry group of the tiling of $\mathbb{H}^2$, in which case the tiling of the geodesic is periodic. – Lee Mosher Nov 20 '12 at 1:35