Fundamental domains for rhombic lattices with dihedral symmetry of order 8

We show by construction that every rhombic lattice $\Gamma$ in $\mathbb{R}^{2}$ has a fundamental domain whose symmetry group contains the point group of $\Gamma$ as a subgroup of index $2$. This solves the last open case of a question raised in [4] on fundamental domains for planar lattices whose symmetry groups properly contain the point groups of the lattices.


Introduction
A lattice Γ in R d is the Z-span of d linearly independent vectors in R d . The lattice Γ forms a group that is isomorphic to the free abelian group of rank d. has the property that its symmetry group Sym(Q) = P (Γ). By the symmetry group of X ⊆ R d , we mean the group Sym(X) of Euclidean motions in R d that leaves X invariant.
In order to discuss symmetries or geometric properties of fundamental domains, it is convenient to allow repetitions of representatives on sets of measure zero. In particular, we restrict ourselves to consider compact fundamental domains. Thus, in this paper, a fundamental domain for a lattice Γ is a compact set F such that (i) the Minkowski sum F + Γ = {p + v p ∈ F, v ∈ Γ} equals R d and (ii) for any nonzero v ∈ Γ, int(F ) ∩ int(F + v) = ∅.
The following fact generalizes the earlier example: For any lattice Γ, the Voronoi region V of the origin is a fundamental domain for Γ such that Sym(V ) equals P (Γ) [4]. Recall that the Voronoi region of a point v of a lattice Γ in R d is the collection of points in R d closest to v than to any other point in Γ. The following question then naturally arises: Is there a fundamental domain F for Γ that has a symmetry group larger than P (Γ), in the sense that S(F ) properly contains P (Γ)?
About the year 2000 Veit Elser constructed a fundamental domain for Z 2 exhibiting eightfold dihedral symmetry rather than the expected fourfold dihedral symmetry [2] (see also his homepage [3]). For the hexagonal lattice in R 2 , a version with twelvefold dihedral symmetry was studied in [1]. This idea was generalized in [4] to several types of lattices in R 2 and R 3 .
For planar lattices, the only case missing in [4] is that of rhombic lattices and this paper is dedicated to fill this gap.
A rhombic lattice is a planar lattice for which there exists a basis consisting of two vectors of equal length such that the angle between the vectors is not π 2 (square lattice), or π 3 or 2π subregions and adding different ones at each step, the construction we employ here simply adds regions at each step. We prove our main result (Theorem 1) as follows: First, we prove that it is enough to consider rhombic lattices whose diagonals have ratio at most 3 (Lemma 1). We then show that in order to construct F it suffices to select a suitable subregion of a rectangle that satisfies certain conditions (see Rh1-Rh3 below). It turns out from Lemma 3 that it is enough to consider rectangles with edge ratio at most 3. We then establish in Lemma 4 that after taking care of a particular subregion of the rectangle, the remaining portion of the rectangle may be dealt with by considering a smaller rectangle and finding a corresponding subregion satisfying analogues of conditions Rh1-Rh3, thereby creating an iterative process. Finally, we argue that the limit obtained by this iteration gives rise to a fundamental domain for Γ satisfying the desired properties.

Main result
Proof. Let Γ be a rhombic lattice. Because the properties we will be dealing with are invariant under similarities, we assume without loss of generality that Γ has basis {( 2m 0 ) , ( m n )}, where 0 < m < n. Here, a basis for Γ consisting of vectors of equal length is {( m −n ) , ( m n )}. Let α be the reflection in the y-axis and β be the half-turn about the origin. Then, P (Γ) = ⟨α, β⟩ ≅ D 2 .
Suppose first that n ≤ 3m. Let Q be the triangular region with vertices ( 0 0 ), X = ( m 0 ), and W = ( 0 m ). The union S of the images of Q under P (Γ) forms a region with symmetry group D 4 . Translates of S by distinct vectors in Γ are either disjoint or intersect on sets of measure zero. See Figure 1. Now, consider the rectangular region R with vertices W , X, Y , and Z, where Y Z lies on the line through ( 0 n ) and V = ( m n−m ). We claim that there is a subset of R such that if E is the union of the images of this subset under P (Γ), then the closure F = cl(S ∪ E) is a fundamental domain for Γ. If in addition the subset of R is preserved by the mirror reflection σ in the perpendicular bisector of W X, then F has symmetry group ⟨α, σ⟩ ≅ D 4 .
Since n ≤ 3m, the ratio W X XY = 2m (n−m) is at least 1. Partition R into the two sets: A, the square XY V U , and B = cl(R ∖ A), the rectangle W U V Z (see Figure 1). Note that if n = 3m, then V = Z and B is empty.
Consider the images of A and B under P (Γ) and their translates by Γ. Equivalently, these are the elements of Sym(Γ)(A ∪ B). The only element of Sym(Γ)A that overlaps with A is α(A)+( 2m 0 ). In fact, the two are equal. Similarly, B = β(B)+( m n ), and B is disjoint with its other copies in Sym(Γ)B. Thus, it suffices to find K ⊆ A and L ⊆ B such that K ⊍ (α(K) + ( 2m 0 )) = A, L ⊍ (β(L) + ( m n )) = B, and K ⊍ L is invariant under σ, where ⊍ denotes a union of sets whose intersection has measure zero. We take the aforementioned E to be the union of the images of K ⊍ L under P (Γ).
Let τ be the mirror reflection in XV and let ρ be the half-turn about the center of B. The previous discussion tells us that our main problem reduces to finding K ⊆ A and L ⊆ B satisfying the following properties.
In the degenerate case n = 3m, since B = ∅, the problem reduces to finding K ⊆ A that satisfies Rh1 and that is invariant under σ.
We now argue that the case where the ratio n m of the diagonals is greater than 3 may be dealt with similarly. Lemma 1. It suffices to consider the case when n ≤ 3m.
Proof of Lemma 1. If n > 3m, let t be the largest integer such that n − 2mt > m. For each i from 1 to t, construct the square [m(i − 1), mi] × [m(i − 1), mi]. Obtain the union C of the images of these squares under P (Γ). We note that Sym(C) ≅ D 4 . By the choice of t, C does not intersect with any of its translates by non-horizontal vectors in Γ. See for example the squares in Figure 2(a), where t = 3 and C is the union of the red squares. Note that the encircled dots represent lattice points. The points not encircled mark the furthest corners of the Γ-translates of C. Figure 2. The case when the ratio of the diagonals is greater than 3. Here, the smaller encircled dots represent lattice points while larger dots represent the furthest corners of copies of C.
Furthermore, C does not overlap with its horizontal Γ-translates because the squares of the Thus, C does not overlap with its translates by nontrivial vectors in Γ. Moreover, the portion of the plane left uncovered by C + Γ is a union of horizontal strips of width n − 2mt. Consider the points marked by dots on the boundary of one such horizontal strip, say the white strip in Figure 2(a). Each point on the lower boundary is both an upper-right corner and an upper-left corner of two horizontally-separated copies of C, while each vertex on the upper boundary is both a lower-right and lower-left corner of two horizontally-separated copies of C. The upper layer of points is the translation of the lower layer by ( m n−2mt ), and so the points on these two layers are spaced in the same way that the points of a rhombic lattice with basis ( 2m 0 ) and ( m n−2mt ) are spaced. Note that by the choice of t, m < n − 2mt ≤ 3m. In Figure 2 . This time, let Q be the triangular region with vertices D = ( mt mt ), W = mt m(t+1) , and X = m(t+1) mt . Since m < n − 2mt, Q lies entirely in the strip. Let S be the union of the images of Q under P (Γ). Extending the argument used for C, S does not overlap with its horizontal translates. Furthermore, because m < n − 2mt, S does not overlap with its non-horizontal translates. Let R be the rectangle W XY Z with Y Z on the line through ( mt n−mt ) and V = m(t+1) n−m(t+1) . Again, we can partition R into A, the square XY V U , and B = cl(R ∖ A), the (possibly empty) rectangle W U V Z.
Consider the images of A and B under P (Γ) and the translates of these by Γ. Similar to the case when n ≤ 3m, . Thus, if there exist K ⊆ A and L ⊆ B satisfying conditions Rh1, Rh2, and Rh3, and E is formed accordingly, then This confirms that it suffices to consider the case when the ratio n m of the diagonals of the rhombic lattice is at most 3. In fact, we use the same Q, K, and L (up to similarity) for two rhombic lattices if the difference between their diagonal ratios is an even integer. ∎ We now construct the sets K and L, where we assume 1 < n m ≤ 3. Let a and b, with a ≥ b, be the edge lengths of R so that 2m (n−m) = a b . For convenience, we reorient R such that U is at the Recall that B is empty when a = b, which happens when n = 3m. Recall also that σ is the reflection in the perpendicular bisector of W X, τ is the reflection in the line through XV , and ρ is the half-turn about the center of B. In the present orientation of R the axis of σ is vertical and the axis of τ has slope −1.
Lemma 2. There exist K and L satisfying Rh1, Rh2, and Rh3 whenever the ratio a b is an integer.
Proof of Lemma 2. Consider the case when a b is an integer. In each of the rectangles in Figure 3, the rightmost square is A. For a b = 1, B is empty and it suffices to find K ⊆ A = R such that K ⊍ τ (K) = A and σ(K) = K. One way to do this is to draw the two diagonals of A to divide it into four congruent triangles, and choose K to be the union of any two non-adjacent triangles, say the triangles marked red in Figure 3 If a b = 2, B is also a square, and B equals σ(A). This together with Rh3 implies that σ(K) must be L. Note that the chosen K for a b = 1 cannot be used, because its image under σ violates Rh2. Rather, one can take K to be one of the triangles into which the axis of τ divides A, as illustrated in Figure 3 We now consider the case when a b ≥ 3. Because a b is an integer, we can divide R into a row of a b congruent squares each of edge length b. If a b is odd, color triangles in each square as in the case when a b = 1, such that no two red triangles share a common edge (orientations of red triangles alternate). If a b is even, group the squares into adjacent pairs and color triangles in each pair as in the case when a b = 2. This is illustrated in Figures 3(c)   We shall use the same idea for more general rectangles R, that is, we subdivide R into squares and take suitable subregions. If the rectangle has irrational edge ratio, the number of squares is necessarily infinite [5]. The method we develop below works for any real value of the ratio, terminating after a finite number of steps precisely when the edge ratio is rational. First, we reduce the cases we have to consider. Lemma 3. It suffices to construct K and L for rectangles R with edge ratio a b not exceeding 3.
Proof of Lemma 3. Suppose R = A ⊍ B has edge ratio a b ≥ 1, and K and L satisfy Rh1, Rh2, and Rh3. To prove the lemma, we show that for R ′ = A ⊍ B ′ with edge ratio a b + 2 and corresponding ρ ′ and σ ′ , one can construct L ′ ⊆ B ′ from K and L such that K and L ′ satisfy the corresponding analogues of Rh1, Rh2, and Rh3.
Note Let σ ′ = ρσρ −1 = ρσρ, that is, σ ′ is the reflection in the line formed by applying ρ to the axis of σ. Then, Set B ′ = B ∪ ρ(A) ∪ σ ′ (A) and R ′ = A ⊍ B ′ . Observe that the rectangle R ′ has edges with ratio a b + 2. Let ρ ′ = σ ′ ρ(σ ′ ) −1 = σ ′ ρσ ′ , that is, ρ ′ is the half-turn about the image of the center of ρ under σ ′ . It is easy to verify that Figure 5. Constructing rectangle R ′ from rectangle R with edge ratio increased by 2.

It then follows that
Finally, we deduce that Thus, for R ′ = A ⊍ B ′ , τ , ρ ′ , and σ ′ , and K ⊆ A and L ′ ⊆ B ′ as above, the corresponding analogues of conditions Rh1, Rh2, and Rh3 are satisfied. This shows how one can obtain from a given construction for a given rectangle a corresponding construction for a rectangle with edge ratio increased by 2. Hence, it does indeed suffice to consider those with edge ratio at most 3. ∎ Remark 1. In Lemma 3, the edge ratio a b may be an integer. The construction for the integer edge ratio case discussed in Lemma 2 is precisely what is obtained when the procedure in the proof of Lemma 3 is applied inductively starting with the constructions for a b = 2 and a b = 3. The reasoning in the proof of Lemma 3 applies as well to subsets of A, B, K, and L that satisfy analogous conditions. We note them in the following remark as they will be useful later.
We have seen how one can directly construct K and L when a b is an integer. For non-integer values of a b , we will construct K and L in a countable number of steps. In particular, we will show that there are regions A * i , B * i , K * i , L * i with the following properties. D1. For every i,  1 = a, b 1 where i 1 and j 1 are nonnegative integers such that i 1 + j 1 ≤ q 1 − 1. See for example Figure 6(a), where q 1 = 3. The squares all have edge length v 1 , and they form a triangular array of squares propped against the left and upper edges of A. We say that the array is anchored at the upper-left vertex of A, and is of order q 1 , with edge length v 1 . If in addition, q 1 > 1, construct a second array of squares anchored at the lower-right vertex of A of order q 1 − 1 with edge length v 1 . This is the largest order of a second array that can be constructed such that the two arrays do not overlap. The union A * 1 of the two arrays is invariant under τ . Next, form two arrays within B, both of order q 1 − 1 and edge length v 1 , one anchored at the upper-right vertex, and another at the lower-left vertex of B. Let B * 1 be the union of these two arrays. Then ρ(B * 1 ) = B * 1 and σ(A * Figure 6. First steps in constructing K and L for (a) 1 < a b < 2 and (b) 2 < a b < 3.
Suppose 2 < a 1 b 1 < 3. This time, B is larger than A and we set v 1 = (a 1 − b 1 ) − b 1 = a 1 − 2b 1 and Figure 6(b). Within A, construct two arrays of edge length v 1 , one with order q 1 anchored at the lower-right vertex of A, and the other with order q 1 − 1 (if positive) anchored at the upper-left vertex of A. We denote the union of these arrays by A * 1 . Then form two arrays within B, both of order q 1 and edge length v 1 , one anchored at the upper-right vertex of B, and another at the lower-left vertex of B. Let B * 1 be the union of these two arrays. Again, D1a is satisfied.
To construct K * 1 and L * 1 , draw the diagonals of each square in each array, thereby dividing each square into four congruent triangles (as done in the odd-integer-ratio case). Consider the arrays in A 1 and B 1 anchored at a shared vertex of A and B. For each square in these arrays, color the upper and lower triangles red. For the remaining two arrays, color the left and right triangles of each square red as in Figure 7. Let K * 1 and L * 1 be the union of the red regions contained in A * 1 and B * 1 , respectively. By construction, K * 1 and L * 1 satisfy D1b. Let b 2 = b 1 − v 1 q 1 . This is the uniform width of the "path"    Proof of Lemma 4. Suppose 1 < a 1 b 1 < 2. Then,R ⊆ A. Consider the sequences {R j } and {F j } defined by R 1 =R =Â ⊍B, F 1 =K ⊍L , and for 1 ≤ j ≤ 4q 1 , The rectangles R j generated are labeled in Figure 9 by their indices. Note that the orientations of the last rectangles in the sequence are not necessarily the same as those in the figure. The longer edges of R 4q 1 −3 and R 4q 1 −2 may be horizontal and vertical, respectively. Regardless of the orientations,the following properties hold.
(e) The union R j ⊆ B, and We see that these, with τ , ρ, and σ, satisfy analogues of Rh1, Rh2, and Rh3. Moreover, A * i , B * i , K * i , and L * i for i = 1, 2 satisfy conditions D1 and D2.  Figure 9. Filling up the uncovered path S 1 in R by applying a sequence of σ, τ , and ρ onR. The two vertices of rectangle R i marked by big dots are the corresponding images of the two vertices marked in R i−1 .
Step 1: • Without loss of generality, assume that the rectangle R 1 ∶= R has a non-integer edge ratio between 1 and 3. Otherwise we can use Lemma 2 or Lemma 3. • Fill R 1 with A * 1 , B * 1 , K * 1 , and L * 1 as described above, leaving the unfilled path S 1 , with central rectangle R 2 .
• Fill portions of S 1 using the sequence defined in the proof of Lemma 4 to create A * 2 , B * 2 , K * 2 , and L * 2 , leaving copies ofŜ 2 . The big dots in Figure 9 are at the ends of the aforementioned copies ofŜ 2 , thus the union S 2 of these copies is a connected unfilled path.
Step n, n ≥ 3: This is treated analogously as Step 2.
• Without loss of generality, assume that the central rectangle R n inŜ n−1 has a noninteger edge ratio between 1 and 3. Otherwise we can use Lemma 2 or Lemma 3. • Fill R n withÂ * n ,B * n ,K * n , andL * n as described above, leaving the unfilled pathŜ n . • Fill portions of S n−1 using the corresponding sequences of those in the proof of Lemma 4, filling out portions of, in order,Ŝ n−1 ,Ŝ n−2 , . . .,Ŝ 2 , andŜ 1 ∶= S 1 , to create A * n , B * n , K * n , and L * n . Arguing as before, this leaves an empty path S n which is composed of copies ofŜ n .
Note that the process terminates after a finite number of steps if and only if a b is rational. At each step, the squares used to pack the rectangle R do not overlap with any of the squares used in previous steps. As R continues to be packed by squares (the sequences {A * i } and {B * i }), the width of the strips S i left unfilled at each step approaches zero. This is because at step i, this width is bounded above by the remainder obtained when the division algorithm is applied to numbers smaller than the edge lengths of R i .
It follows that the set of all A * i 's and B * i 's provide a tiling of R by a countable number of polygons. This also holds true for the K * i 's and L * i 's together with their complements in the A * i 's and B * i 's. Also, these regions satisfy the required conditions in D1 and D2 at each step. Therefore, as constructed, and regardless of rationality of a b , the sequences {A * i }, {B * i }, {K * i }, and {L * i } satisfy conditions D1 and D2. Performing the procedure on A ⊍ B of Figure 1, we see that K = ⋃ i K * i ⊆ A and L = ⋃ i L * i ⊆ B satisfy Rh1, Rh2, and Rh3. Forming S as in Figure 1 and E = P (Γ)(K ⊍ L), we have that F = cl(S ∪ E) is a compact fundamental domain for Γ with [S(F ) ∶ P (Γ)] = 2. This completes the proof of Theorem 1.
In Figure 10, we show a portion of the tiling by the fundamental domain for the rhombic lattice Γ with basis {( 50 0 ) , ( 25 11 )} generated by the procedure described.