Let \(T\subset\mathbb R^2\) be a closed Euclidean triangle, and let \(G\subset\operatorname{Isom}(\mathbb R^2)\) be the group generated by reflections in the sides of \(T\).
Show that the tiles \[ T_g=g(T), \qquad g\in G, \] cover \(\mathbb R^2\).
Give an example in which every point of \(\mathbb R^2\) belongs to the interior of at most one tile.
Give an example in which every point of \(\mathbb R^2\) belongs to the interiors of infinitely many tiles.
Hint: use the closure of \(G\) as a Lie subgroup of \(\operatorname{Isom}(\mathbb R^2)\).
Solution
The reflected triangles always cover the plane.
An equilateral triangle gives the usual triangular tiling, whose tile interiors are pairwise disjoint.
Take a triangle having one angle \(\alpha\) with \[ \frac{\alpha}{\pi}\notin\mathbb Q. \] Then every point lies in the interiors of infinitely many reflected tiles.
1. Let\[ X=\bigcup_{g\in G}g(T). \]Then \(X\) is open.
Certainly every interior point of a tile is an interior point of \(X\).
Suppose \(x\) lies in the relative interior of a side of a tile \(g(T)\). Reflection in that side is \[ g r g^{-1}\in G, \] where \(r\) is reflection in the corresponding side of \(T\). Thus \(X\) contains the two congruent triangles on both sides of that edge, and hence a neighborhood of \(x\).
Finally suppose \(x\) is a vertex of \(g(T)\), and let the angle there be \(\alpha\in(0,\pi)\). Choose \(\varepsilon>0\) small enough that inside \(B(x,\varepsilon)\) the tile is exactly a sector of angle \(\alpha\). The reflections in the two incident sides belong to \(G\). In polar angle coordinates centered at \(x\), their group sends the original angular interval \[ [0,\alpha] \] to all intervals \[ [k\alpha,(k+1)\alpha], \qquad k\in\mathbb Z. \] Their union is all angular directions, modulo \(2\pi\). Hence the corresponding reflected sectors cover \(B(x,\varepsilon)\). Thus every point of \(X\) is interior, so \(X\) is open.
2. The set \(X\) is also closed.
Let \[ x_n\in X, \qquad x_n\to x. \] Write \[ x_n=g_n(t_n), \qquad g_n\in G, \quad t_n\in T. \] Since \(T\) is compact, after passing to a subsequence we may assume \[ t_n\to t\in T. \]
Write each Euclidean isometry as \[ g_n(y)=R_ny+a_n, \qquad R_n\in O(2). \] The rotations and reflections \(R_n\) lie in the compact group \(O(2)\). Moreover \[ a_n=x_n-R_nt_n \] is bounded. Passing to another subsequence, we therefore have \[ g_n\longrightarrow h \] in \(\operatorname{Isom}(\mathbb R^2)\), for some \(h\) in the closure of \(G\). Necessarily \[ x=h(t). \]
Now put \[ y_n=g_n^{-1}(x). \] Then \[ y_n\to h^{-1}(x)=t. \] By 1, \(X\) is an open neighborhood of \(t\), so \(y_n\in X\) for all sufficiently large \(n\). Since \(X\) is \(G\)-invariant and \(g_n\in G\), \[ x=g_n(y_n)\in X. \] Thus \(X\) is closed.
3. Part (a) follows.
2. Therefore\[ X=\mathbb R^2. \]
4. An equilateral triangle gives the example required in part (b).
Take \[ T=\operatorname{conv}\{0,1,e^{i\pi/3}\}\subset\mathbb C\cong\mathbb R^2. \] Successive reflections in its three sides produce the standard triangular lattice tessellation by congruent equilateral triangles. Equivalently, the three side reflections generate the affine Coxeter group of the equilateral triangular tiling, with \(T\) as a fundamental chamber. Hence distinct tiles have disjoint interiors. Every point therefore belongs to the interior of at most one tile.
5. For part (c), choose a triangle \(T\) with a vertex \(A\) whose angle \(\alpha\) satisfies\[ \alpha/\pi\notin\mathbb Q. \]Then the closure \(H=\overline G\) is the full Euclidean isometry group.
Let \(r_1,r_2\) be reflections in the two sides meeting at \(A\). Their product is a rotation about \(A\) through angle \(2\alpha\). Since \[ \frac{2\alpha}{2\pi}=\frac{\alpha}{\pi} \] is irrational, the powers of this rotation are dense in the full circle group of rotations about \(A\). Hence \(H\) contains every rotation about \(A\).
Let \(s\) be reflection in the side opposite \(A\), and set \[ A'=s(A)\ne A. \] Conjugating by \(s\) shows that \(H\) also contains every rotation about \(A'\).
The composition of a rotation about \(A\) through \(\theta\) with a rotation about \(A'\) through \(-\theta\) is a translation. If \[ d=|A-A'|, \] its translation length is \[ 2d\left|\sin\frac\theta2\right|. \] Conjugating by the rotations about \(A\) gives translations of this length in every direction. To obtain an arbitrary translation vector \(v\), choose \(m\) so large that \[ \frac{|v|}{m}<2d. \] Then choose \(\theta\) with \[ 2d\left|\sin\frac\theta2\right|=\frac{|v|}{m} \] and rotate the resulting translation direction to that of \(v\). Composing this translation with itself \(m\) times gives translation by \(v\). Thus \(H\) contains all translations and all rotations, hence the full orientation-preserving Euclidean group. Since \(G\) itself contains reflections, in fact \[ H=\operatorname{Isom}(\mathbb R^2). \]
6. Every point lies in the interiors of infinitely many tiles for the triangle in
5.
Fix \(x\in\mathbb R^2\) and consider \[ U_x=\{h\in H:x\in\operatorname{int}(h(T))\}. \] This is a nonempty open subset of \(H\): translate any chosen interior point of \(T\) to \(x\), and small perturbations of that isometry still leave \(x\) inside the image triangle.
By 5, \(G\) is dense in the positive-dimensional Lie group \(H\). Hence every nonempty open subset of \(H\) contains infinitely many elements of \(G\); otherwise one could remove its finitely many \(G\)-points and retain a nonempty open set disjoint from the dense subgroup. Thus \(U_x\cap G\) is infinite.
Only finitely many Euclidean isometries can stabilize the nondegenerate triangle \(T\) as a set. Therefore infinitely many elements of \(U_x\cap G\) give distinct tiles. Each of those tiles contains \(x\) in its interior. Since \(x\) was arbitrary, every point of the plane belongs to the interiors of infinitely many tiles.