Coincidence indices of sublattices and coincidences of colorings

Appendix

Here, we give the promised justifications and proofs in Examples 2 and 3.

Example 2

Recall that Λ1 = ∪j = 03(cj + Λ2), where Λ₁=Im(𝕃), Λ₂=2Im(𝕁), and c₀=0. Let R=R_q ∈SOC(Λ₁), where q=(q₀, q₁, q₂, q₃) is a primitive quaternion. For sure, R is a color coincidence that fixes color c₀ because of Theorem 4. We shall now determine how R acts on the other colors c₁, c₂, c₃, and thus, the color permutation that R generates.

Since R is a color coincidence, it is enough to consider a representative from each coset of Λ₂ in Λ₁ that is in Λ₁(R^–1), and afterwards identify to which coset of Λ₂ the representative is sent by R. Take c₁=Σ₁(R)i, c₂=Σ₁(R)j, and c₃=Σ₁(R)k. Indeed, for j∈{1, 2, 3}, c_j ∉Λ₂ since Σ₁(R) is odd, and c_j ∈Λ₁(R^–1) because Σ₁(R)=den(R)=gcd{k∈ℕ: kR is an integral matrix} (see [30]). One obtains

R(c1) = Σ1(R)|q|2qiq¯,R(c2) = Σ1(R)|q|2qjq¯,R(c3) = Σ1(R)|q|2qkq¯,

where Σ₁(R)/|q|²=1/2^ℓ with ℓ∈{0, 1, 2}. This gives rise to three different cases.

Before we proceed, we take note of the following facts that will be used in the computations thereafter.

Fact 4.For every q∈ℍ and x∈Im(ℍ), we have qxq̅∈ Im(ℍ).

If q∈𝕁 and j is the highest power of 2 such that 2^j divides |q|², then q=(1+i)^jp for some p∈𝕁 of odd norm [[36], page 37]. The next statement follows from this result.

Fact 5.The set (1+i)^r𝕁={q∈𝕁: 2^r||q|²} is a two-sided ideal of 𝕁 for all r∈ℕ.

Fact 6.Let Λ₂=2Im(𝕁)=2Im(𝕃)∪[(0, 1, 1, 1)]+2Im(𝕃)]. Then the following hold:

1. 2𝕁 ∩Im(ℍ)=2Im(𝕃)⊆Λ₂

2. Ifq∈𝕁 thenq–q̅∈Λ₂.

Proof. Statement (i) is clear. If q∈𝕃 then q–q̅∈Λ₂ by (i). On the other hand, if q∈𝕁\𝕃 then q–q̅∈(0, 1, 1, 1)+2Im(𝕃)⊆Λ₂. This proves (ii).  □

The three possible ratios of Σ₁(R)/|q|² are investigated below.

Case I: Σ₁(R)/|q|²=1, that is, |q|²≡1 (mod 4) and either one or three among the components of q is/are odd.

For instance, suppose q₀ is odd while q₁, q₂, q₃ are even, or q₀ is even while q₁, q₂, q₃ are odd. In both instances, one can write q=r+s, where r∈2𝕁 and s=e. One obtains

R(cj) = qxjq¯ = rxjr¯ + rxjs¯ + sxjr¯ + sxjs¯

where x_j =i, j, k if j=1, 2, 3, respectively. Facts 4, 5, and 6(i) imply that rx_jr̅∈Λ₂, and rxjs¯ + sxjr¯ = rxjs¯ − rxjs¯¯ ∈ Λ2 by Fact 6(ii). Hence, R(c_j )∈sx_js̅+Λ₂=x_j +Λ₂=c_j +Λ₂ for j∈{1, 2, 3}. Similarly, for the other three possibilities, R(c_j )∈c_j +Λ₂ for j∈{1, 2, 3}. Therefore, in all instances, R fixes all colors.

Case II: Σ1(R)/|q|2 = 1/2, that is, |q|²≡2(mod 4) and exactly two components of q are odd.

Consider the instance when both q₀ and q₁ are odd, or when both q₂ and q₃ are odd. Either way, one can express q as q=r+s where r∈2𝕁 and s=(1, 1, 0, 0). One has in this case

R(cj) = 12(rxjr¯ + rxjs¯ + sxjr¯ + sxjs¯)

where x_j =i, j, k if j=1, 2, 3, respectively. Now, (1/2)rxjr¯ ∈ 2J because 4 divides |(1/2)rxjr¯|2. This, with Facts 4 and 6(i), implies that (1/2)rxjr¯ ∈ Λ2. Since (1/2)rxjs¯ ∈ J, one obtains that (1/2)rxjs¯ + (1/2)sxjr¯ ∈ Λ2 by Fact 6(ii). Therefore,

R(cj) ∈ (1/2)sxjs¯ + Λ2 = {c1 + Λ2,if j = 1c3 + Λ2,if j = 2c2 + Λ2,if j = 3.

Thus, R induces the permutation (c₂c₃) of colors. Similar computations for the other two possibilities yield that if q₀ and q_j are of the same parity, where j∈{1, 2, 3}, then R fixes both colors c₀ and c_j and swaps the other two colors.

Case III: Σ1(R)/|q|2 = 1/4, that is, |q|²≡0 (mod 4) and all components of q are odd.

Suppose an even number of components of q are congruent to 1 modulo 4. Then, one can write q=r+s, where r∈(1, 1, 0, 0)2𝕁 and s=(1, 1, 1, 1). Thus,

R(cj) = 14(rxjr¯ + rxjs¯ + sxjr¯ + sxjs¯)

where x_j =i, j, k if j=1, 2, 3, respectively. Since 4 divides |(1/4)rxjr¯|2, one has (1/4)rxjr¯ ∈ 2J and together with Facts 4 and 6(i), (1/4)rxjr¯ ∈ Λ2. Also, by Fact 6(ii) one concludes that (1/4)rxjs¯ + (1/4)sxjr¯ ∈ Λ2 because (1/4)rxjs¯ ∈ J. Finally,

R(cj) ∈ 14sxjs¯ + Λ2 = {c2 + Λ2,if j = 1c3 + Λ2,if j = 2c1 + Λ2,if j = 3.

Hence, R generates the permutation (c₁c₂c₃) of colors. On the other hand, if an odd number of components of q are congruent to 1 modulo 4, then similar arguments show that R induces the permutation (c₁c₃c₂) of colors.

Given a coincidence reflection T_q ∈OC(Λ₁), the color permutation that it effects is the same as that of the coincidence rotation R_q .

Example 3

Recall that Λ₁=𝕁, Λ₂=𝕃, and R=R_q,p ∈SOC(Λ₁) is parametrized by the admissible pair (q, p) of primitive quaternions. The following looks at the conditions that the quaternion pair (q, p) should satisfy so that R∈H_Λ1,Λ2 .

Going through the different possible admissible quaternion pairs (q, p) results in the following cases. In each case, the sets RΛ₂∩Λ₁(R) and Λ₂∩Λ₁(R) are compared in order to ascertain whether R∈H_Λ1,Λ2 or not (see Theorem 4).

Case I: |q|² and |p|² are odd

Suppose v∈Λ₂∩Λ₁(R) and write v=qwp̅/|qp̅| for some w∈𝕁. This means that |qp̅|v=qwp̅∈𝕃. Since |q|² and |p|² are odd, one can express q=r₁+s₁ and p=r₂+s₂, where r₁, r₂∈2𝕁, and s₁, s₂ ∈{e, i, j, k}. With this, one obtains

qwp¯ = r1wr¯2 + r1ws¯2 + s1wr¯2 + s1ws¯2 ∈ L.

By Fact 5, r₁wr̅₂+r₁ws̅₂+s₁wr̅₂∈2𝕁⊆𝕃, which implies that s₁ws̅₂∈𝕃. Thus, v=Rw∈RΛ₂ and Λ₂∩Λ₁(R)⊆RΛ₂∩Λ₁(R). It follows then that RΛ₂∩Λ₁(R)=Λ₂∩Λ₁(R), and so R∈H_Λ1,Λ2 .

Case II: |q|² is odd and |p|²≡0 (mod 4), or |q|²≡0 (mod 4) and |p|² is odd

Consider x = (1/2)|qp¯| ∈ L. One has R(x) = qxp¯/|qp¯| = (1/2)qp¯ ∈ J. Thus, R(x)∈RΛ₂∩Λ₁(R). However, the first component of qp̅, which is equal to the inner product 〈q, p〉 of q and p (or the standard scalar product of q and p as vectors in ℝ⁴), is odd. This implies that (1/2)qp¯ ∉ L and R(x)∉Λ₂∩Λ₁(R). Hence, RΛ₂∩Λ₁(R)≠Λ₂∩Λ₁(R) and R∉H_Λ1,Λ2 .

Case III: |q|²≡|p|²≡2 (mod 4)

Write q=r₁+s₁ and p=r₂+s₂ where r₁, r₂∈2𝕁, and

s1, s2 ∈ {(1, 1, 0, 0), (1, 0, 1, 0), (1, 0, 0, 1)}.

Note that 〈q, p〉 is even if and only if s₁=s₂.

Consider again x = (1/2)|qp¯| ∈ L so that R(x) = (1/2)qp¯ ∈ RΛ2 ∩ Λ1(R). Now, (1/2)qp¯ ∉ L if and only if 〈q, p〉 is odd. Since (1/2)qp¯ ∉ L implies that RΛ₂∩Λ₁(R)≠Λ₂∩Λ₁(R), R∉H_Λ1,Λ2 whenever 〈q, p〉 is odd.

It remains to check the case s₁=s₂. Take v∈Λ₂∩Λ₁(R). Write v=qwp̅/|qp̅| for some w∈𝕁. One has

12|qp¯|v = 12(r1wr¯2 + r1ws¯1 + s1wr¯2 + s1ws¯1) ∈ L.

Now, (1/2)r1wr¯2 ∈ 2J ⊆ L and (1/2)r1ws¯1, (1/2)s1wr¯2 ∈ (1, 1, 0, 0)J ⊆ L by Fact 5. Hence, (1/2)s1ws¯1 = s1ws1−1 ∈ L meaning w∈s₁^–1𝕃s₁=𝕃 for all three possible values of s₁. It follows then that v∈RΛ₂ and RΛ₂∩Λ₁(R)=Λ₂∩Λ₁(R). Thus, R∈H_Λ1,Λ2 if 〈q, p〉 is even.

Case IV: |q|²≡|p|²≡0(mod 4)

Write q=r₁+s₁ and p=r₂+s₂ where r₁, r₂∈(1, 1, 0, 0) 2𝕁 and

s1, s2 ∈ {(1, 1, 1, 1), (1, 1, 1, −1)}.

Note that 4|〈q, p〉 if and only if s₁=s₂.

Take x = (1/4)|qp¯| ∈ L. One obtains R(x) = ((1/2)q)((1/2)p¯) ∈ J. Hence, Rx∈RΛ₂∩Λ₁(R). Observe however, that (1/4)qp¯ ∉ L if and only if 4∤〈q, p〉. Since (1/4)qp¯ ∉ L means that RΛ₂∩Λ₁(R)≠Λ₂∩Λ₁(R), 4∤〈q, p〉 implies R∉H_Λ1,Λ2 .

Again, it remains to check the instance when s₁=s₂. Let v∈Λ₂∩Λ₁(R). If one writes v=qwp̅/|qp̅| for some w∈𝕁, one gets

14|qp¯|v = 14(r1wr¯2 + r1ws¯1 + s1wr¯2 + s1ws¯1) ∈ L.

By Fact 5, (1/4)r1wr¯2, (1/4)r1ws¯1, and (1/4)s1wr¯2 ∈ (1, 1, 0, 0)J ⊆ L. Therefore, w∈𝕃 for both possible values of s₁. Hence, v∈RΛ₂ which implies that RΛ₂∩Λ₁(R)=Λ₂∩Λ₁(R). Therefore, R∈H_Λ1,Λ2 whenever 〈q, p〉 is divisible by 4.

The results above also hold for coincidence reflections T_q,p =R_q,p ·T_1,1∈OC(Λ₁).