# Coincidence indices of sublattices and coincidences of colorings

• Manuel Joseph C. Loquias and Peter Zeiner

## Abstract

Even though a lattice and its sublattices have the same group of coincidence isometries, the coincidence index of a coincidence isometry with respect to a lattice Λ1 and to a sublattice Λ2 may differ. Here, we examine the coloring of Λ1 induced by Λ2 to identify how the coincidence indices with respect to Λ1 and to Λ2 are related. This leads to a generalization of the notion of color symmetries of lattices to what we call color coincidences of lattices. Examples involving the cubic and hypercubic lattices are given to illustrate these ideas.

Corresponding author: Peter Zeiner, Fakultät für Mathematik, Universität Bielefeld, Universitätsstraße 25, 33501, Germany, Tel.: +49-521-1064791, Fax: +49-521-1066481, E-mail:

## Acknowledgments

M.J.C. Loquias would like to thank the German Academic Exchange Service (DAAD) for financial support during his stay in Bielefeld. This work was supported by the German Research Foundation (DFG), within the CRC 701.

## 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=Rq ∈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 c11(R)i, c21(R)j, and c31(R)k. Indeed, for j∈{1, 2, 3}, cj ∉Λ2 since Σ1(R) is odd, and cj ∈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 with ∈{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 2j 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: 2r||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 thenqq̅∈Λ2.

Proof. Statement (i) is clear. If q∈𝕃 then qq̅∈Λ2 by (i). On the other hand, if q∈𝕁\𝕃 then qq̅∈(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 xj =i, j, k if j=1, 2, 3, respectively. Facts 4, 5, and 6(i) imply that rxjr̅∈Λ2, and rxjs¯+sxjr¯=rxjs¯rxjs¯¯Λ2 by Fact 6(ii). Hence, R(cj )∈sxjs̅+Λ2=xj2=cj2 for j∈{1, 2, 3}. Similarly, for the other three possibilities, R(cj )∈cj2 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 xj =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 qj are of the same parity, where j∈{1, 2, 3}, then R fixes both colors c0 and cj 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 xj =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 Tq ∈OC(Λ1), the color permutation that it effects is the same as that of the coincidence rotation Rq .

Example 3

Recall that Λ1=𝕁, Λ2=𝕃, and R=Rq,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 RHΛ12 .

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 RHΛ12 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¯2L.

By Fact 5, r1wr̅2+r1ws̅2+s1wr̅2∈2𝕁⊆𝕃, which implies that s1ws̅2∈𝕃. Thus, v=RwRΛ2 and Λ2∩Λ1(R)⊆RΛ2∩Λ1(R). It follows then that RΛ2∩Λ1(R)=Λ2∩Λ1(R), and so RHΛ12 .

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 RHΛ12 .

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¯RΛ2Λ1(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), RHΛ12 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¯22JL and (1/2)r1ws¯1,(1/2)s1wr¯2(1,1,0,0)JL by Fact 5. Hence, (1/2)s1ws¯1=s1ws11L meaning ws1–1𝕃s1=𝕃 for all three possible values of s1. It follows then that vRΛ2 and RΛ2∩Λ1(R)=Λ2∩Λ1(R). Thus, RHΛ12 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, RxRΛ2∩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 RHΛ12 .

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)JL. Therefore, w∈𝕃 for both possible values of s1. Hence, vRΛ2 which implies that RΛ2∩Λ1(R)=Λ2∩Λ1(R). Therefore, RHΛ12 whenever 〈q, p〉 is divisible by 4.

The results above also hold for coincidence reflections Tq,p =Rq,p ·T1,1∈OC(Λ1).

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