# Difference between revisions of "Coloring R 2"

Let $R_1 = {\bf Z}[\omega]$ denote the Eisenstein integers, where $\omega := e^{2\pi i/3}$, thus $\omega+\omega^2+1=0$. The quotient $R_1/2R_1$ is then a finite field of order 4, with elements given by the residue classes of $0, 1, \omega,\omega^2$ mod $2R_1$. In particular $R_1/2R_1$ has characteristic two, so we will not distinguish between addition and subtraction in this field; one also has the complex conjugation automorphism $z \mapsto \overline{z}$ in this field, which is also the Frobenius automorphism $z \mapsto z^2$. We will use this finite field as our choice of colors.

Let $c: {\bf C} \to R_1/2R_1$ be a unit distance 4-coloring. For any $z \in {\bf C}$, let $\rho_z: R_1 \to R_1/2R_1$ be the renormalised coloring function

$\rho_z(a) := c(z+a) + a.\quad (1)$

By construction, we have the translation equivariance

$\rho_z(a+b) = \rho_{z+b}(a) + b \quad (2)$

for all $z \in {\bf C}$ and $a,b \in R_1$; the unit distance coloring property also implies that

$\rho_{z+e^{i\theta}}(a) \neq \rho_z(a) \quad (3)$

for all $z \in {\bf C}, a \in R_1, \theta \in {\bf R}$. By combining (3) with (2) we also see that

$\rho_z(a + \omega^j) \neq \rho_z(a) + \omega^j \quad (4)$

for all $z \in {\bf C}, a \in R_1, j \in {\bf Z}$.

The isometries of the 4-coloring problem give the following symmetries:

• (Translation symmetry) One can replace $c(z)$ by $c(z_0+z)$ and $\rho_z(a)$ by $\rho_{z_0+z}(a)$ for any $z_0 \in {\bf C}$.
• (Rotation symmetry) One can replace $c(z)$ by $\omega^{-j} c(\omega^j z)$ and $\rho_z(a)$ by $\omega^{-j} \rho_{\omega^j z}( \omega^j a )$ for any integer $j$.
• (Conjugation symmetry) One can replace $c(z)$ by $\overline{c(z)}$ and $\rho_z(a)$ by $\overline{\rho_z(\overline{a})}$.
• (Recoloring) One can replace $c(z)$ by $\phi(c(z))$ and $\rho_z(a)$ by $\phi(\rho_z(a)+a)+a$ for any permutation $\phi: R_1/2R_1 \to R_1/2R_1$. In particular, taking $\phi$ to be a translation $\phi(z) := z+h$ for some $h \in R_1/2R_1$, we can replace $\rho_z(a)$ by $\rho_z(a)+h$.
• (Compactness) Given a sequence of colorings $c^{(n)}$ with attendant renormalisations $\rho_z^{(n)}$, one can (by Tychonoff's theorem) find a coloring $c$ which is a limit point of the $c^{(n)}$ in the product topology, hence the renormalisations $\rho_z$ are limit poitns of the $\rho_z^{(n)}$. In particular, for any $z, a$, we have $\rho_z(a) = \rho_z^{(n)}(a)$ for infinitely many $n$.

Lemma 1 (Convexity) If $z \in {\bf C}, a \in R_1, j \in {\bf Z}$ are such that

$\rho_z(a) = \rho_z(a + \omega^j) = \rho_z(a + \omega^{j+2})$

then we also have

$\rho_z(a+\omega^{j+1}) = \rho_z(a).$

Proof From three applications of (3) we have

$\rho_z(a + \omega^{j+1}) \neq \rho_z(a) + \omega^{j+1}$
$\rho_z(a + \omega^{j+1}) \neq \rho_z(a+\omega^j) + \omega^{j+2}$
$\rho_z(a + \omega^{j+1}) \neq \rho_z(a+\omega^{j+2}) + \omega^{j}.$

Since $\omega^j, \omega^{j+1}, \omega^{j+2}$ occupy the three non-zero residue classes in $R_1/2R_1$, the claim follows. $\Box$

Call a subset A of the Eisenstein integers $R_1$ convex if it is connected (in the unit distance graph of $R_1$) and has the property that $a+\omega^{j+1} \in A$ whenever $a, a+\omega_j, a+\omega^{j+2} \in A$ (i.e., whenever the long diagonal and one other vertex of a unit rhombus lie in A, then the remaining vertex also does). It is not difficult to classify all the possible convex subsets of $R_1$:

• The empty set.
• The entire lattice $R_1$.
• A half-plane.
• A strip bounded between two parallel lines.
• A sector of angle 60 or 120 degrees.
• A half-strip bounded by two parallel rays and a line segment making angles of 60 and 120 degrees with these rays.
• A half-infinite region bounded by two parallel rays and two line segments at angles 120 to each other, and with each line segment also making an angle of 120 degrees with one of the rays.
• A truncated sector bounded by two non-intersecting rays and a line segment at angle 120 to each ray.
• An equilateral triangle.
• A parallelogram with angles 60 and 120 degrees.
• A trapezoid with angles 60 and 120 degrees.
• A pentagon with four angles of 120 degrees and one angle of 60 degrees.
• A hexagon with all angles 120 degrees.

In particular, each non-empty convex set is a convex polygon (possibly unbounded) with anywhere from zero to six sides, with the sides having outward normals of the form $i\omega^j$ for various distinct j.

Corollary 2 If $z \in {\bf C}$ and $h \in R_1/2R_1$, then all the connected components of the set $\{ a \in R_1: \rho_z(a) = h \}$ are convex.

Lemma 3 If $z \in {\bf C}, a \in R_1, j \in {\bf Z}$ are such that

$\rho_z(a) = \rho_z(a + \omega^j)$

then

$\rho_z(a + \omega^{j+1}), \rho_z(a+\omega^{j-1}) \in \{ \rho_z(a), \rho_a(z) + \omega^j \}\quad (4).$

If we in fact have

$\rho_z(a) = \rho_z(a + \omega^j) = \rho_z(a + 2 \omega_j)\quad (5)$

then

$\rho_z(a+\omega^{j+1}) = \rho_z(a + \omega^{j+1} + \omega_j) \quad (6)$

and

$\rho_z(a+\omega^{j-1}) = \rho_z(a + \omega^{j-1} + \omega_j). \quad (7)$

Proof As in Lemma 1, we have

$\rho_z(a + \omega^{j+1}) \neq \rho_z(a) + \omega^{j+1}$
$\rho_z(a + \omega^{j+1}) \neq \rho_z(a+\omega^j) + \omega^{j+2}$

and this gives (4) for $\omega^{j+1}$; a similar argument works for $\omega^{j-1}$. Assuming (5), we then have

$\rho_z(a + \omega^{j+1}), \rho_z(a+\omega^{j+1} + \omega^j) \in \{ \rho_z(a), \rho_a(z) + \omega^j \};$

on the other hand, from (3) we have

$\rho_z(a+\omega^{j+1} + \omega^j) \neq \rho_z(a + \omega^{j+1}) + \omega^j$

and this gives (6). A similar argument gives (7). $\Box$

Lemma 4 If $z \in {\bf C}$, and $\rho_z$ takes on exactly two values $c, c+\omega^j$, then $\rho_z$ is $\omega^j$-peroidic, thus $\rho_z(a+\omega^j) = \rho_z(a)$.

Proof By hypothesis, $\rho_z(a+\omega^j)-\rho_z(a)$ can only take on the values $0$ or $c'-c = \omega^j$ (note that $R_1/2R_1$ has characteristic 2). But the latter possibility is excluded by (3), giving the claim. $\Box$

Corollary 5 If $z \in {\bf C}$ and $j \in {\bf Z}$, and the functions $\rho_z$ and $\rho_{z+\omega^j}$ each take on exactly two values, then the images $\rho_z(R_1)$ and $\rho_{z+\omega^j}(R_1)$ are either equal or disjoint.

Proof If this is not the case, then we have $\rho_z(R_1) = \{ c_0, c_0 + \omega^k \}$, $\rho_{z+\omega^j}(R_1) = \{ c_0, c_0 + \omega^l \}$, for some $c_0 \in R_1/2R_1$ and $k,l \in {\bf Z}$ with $k \neq l$. The difference of these two sets is all of $R_1/2R_1$, thus there exist $a_1,a_2 \in R_1$ such that $\rho_z(a_1) - \rho_{z+\omega^j}(a_2) = \omega^j$. But by Lemma 4, $\rho_z$ is $\omega^k$-periodic and $\rho_{z+\omega^j}$ is $\omega^l$-periodic, hence there exists $a \in R_1$ such that $\rho_z(a_1) = \rho_z(a)$ and <maath>\rho_{z+\omega^j}(a_2) = \rho_{z+\omega^j}(a)[/itex], hence $\rho_z(a) - \rho_{z+\omega^j}(a) = \omega^j$. But this contradicts (4). $\Box$

In the case when $\rho_z$ takes on at most three values, e.g. $\{1,\omega,\omega^2\} \hbox{ mod } 2R_1$, we can now get a good classification of the possible behaviours of $\rho_{e^{i\theta}}$:

Proposition 6 Let $z \in {\bf C}$, and suppose that $\rho_z$ takes values in $\{1,\omega,\omega^2\} \hbox{ mod } 2R_1$. For $j \in {\bf Z}$, let $S_j := \{ a \in R_1: \rho_z(a) = \omega^j \}$ denote the color classes, thus $S_{j+3}=S_j$ and $R_1$ is partitioned into $S_0,S_1,S_2$. Then at least one of the following statements hold.

• ($\omega^j$-periodic case) There is an integer $j$ such that $S_j$ is empty, and $S_{j+1}, S_{j+2}$ are $\omega^j$-periodic.
• (Tri-sector case I) There exists a unit triangle $\{a_1,a_{\omega^2},a_{\omega^4}\} \in R_1$ such that $S_j = \{ a_{\omega^j} + k \omega^{j-1} + l \omega^{j+1}: k,l \in {\bf N} \}$ for $j=0,2,4$. (Here ${\bf N} = \{0,1,2,\dots\}$ are the natural numbers.)
• (Tri-sector case II) There exists a unit triangle $\{a_1,a_{\omega^2},a_{\omega^4}\} \in R_1$ such that $S_j = \{ a_{\omega^j} - k \omega^{j-1} - l \omega^{j+1}: k,l \in {\bf N} \}$ for $j=0,2,4$.

Proof Consider a convex component $C$ of one of the $S_j$. As stated above, there are at most six sides to this region, with outward normal in $\{i\omega^k: k \in {\bf Z}\}$. Actually, the outward normal of an edge of $C$ cannot equal $i \omega^j$, since by Lemma 3 $\rho_{z}$ would equal 0 on the other side of this edge, contradicting the hypothesis. Similarly, the outward normal of an edge of $C$ cannot equal $-i \omega^j$. Thus the only possible outward normals are $i\omega^{j+1}, i\omega^{j+2}, -i\omega^{j+1}, -i\omega^{j+2}$.

Now suppose that $C$ two edges with outward normals $i\omega^{j+2},-i\omega^{j+1}$. They must meet at some corner $a$, with and

$\rho_{z}(a-\omega^j), \rho_z(a+\omega^j) \neq \omega^j.$

By Lemma 3, this implies that

$\rho_z(a-\omega^j) = \omega; \quad \rho_z(a+\omega^j) = \omega^{j+2}.$

By (4), we now have

$\rho_z(a+\omega^{j+4}) \neq \omega^{j+1}+\omega^{j+2}, \omega^j+\omega^{j+1}; \quad \rho_z(a+\omega^{j+5}) \neq \omega^{j+2}+\omega^{j+1}, \omega^j+\omega^{j+2}$

and thus (as $\rho_z$ avoids 0)

$\rho_z(a+\omega^{j+4}) = \omega^{j+1}; \quad \rho_z(a+\omega^{j+5}) = \omega^{j+2}$

which contradicts (6). Thus $C$ cannot have two edges with outward normals $i\omega^{j+2}, -i\omega^{j+1}$; a similar argument prohibits $C$ from having two edges with outward normals $-i\omega^{j+2}, i\omega^{j+1}$. This only leaves the following possibilities for $C$:

• The entire lattice $R_1$;
• A half-plane with outward normal $\pm i\omega^{j+1}$ or $\pm i\omega^{j+2}$;
• A strip bounded by two parallel lines with outward normals $+i\omega^{j+1}, -i\omega^{j+1}$;
• A strip bounded by two parallel lines with outward normals $+i\omega^{j+2}, -i\omega^{j+2}$;
• A sector bounded by two rays with outward normals $-i\omega^{j+1}, -i\omega^{j+2}$;
• A sector bounded by two rays with outward normals $i\omega^{j+1}, i\omega^{j+2}$.

The lattice $R_1$ is partitioned into the convex components of $S_0, S_1, S_2$, which are all necessarily disjoint from each other. If one of these components is all of $R_1$ then we are certainly in the first case of the proposition, so we may assume this is not the case. If there is a component of some $S_j$ that is a half-plane or a strip with outward normals in $\{ i, -i \}$, then there can be no components from $S_0$, and the only permissible components from $S_1, S_2$ are also half-planes or strips with normals in $\{ i, -i \}$, since any other possible component would intersect the original component under consideration. In this case the first case of the proposition holds (with $j=0$). Thus we may assume that there are no components that are half-planes or strips with outward normals in $\{-1,+1\}$. Similarly with $\{-1,+1\}$ replaced by $\{-\omega,\omega\}$ or $\{-\omega^2,\omega^2\}$; thus we are left with the case where the only components of any of the $S_j$ are sectors.

By applying a rotation by $\omega^j$ for some $j$ (which does not affect the hypothesis or conclusion of the proposition) we may assume without loss of generality that $S_0$ contains a component which is a sector with outward normals $i\omega, i\omega^2$, then this component takes the form $\{ a_1 + k \omega + l \omega^{-1}: k,l \geq 0\}$ for some $a_1 \in R_1$. By Lemma 3, we then have $\rho_{e^{i\theta}}(a_1 + \omega^2 + k \omega) = \omega^2$ for all $k \geq 0$, and similarly $\rho_{e^{i\theta}}(a_1 + \omega^4 + l \omega^{-1}) = \omega$ for all $l \geq 0$. Since the components of $S_0,S_1,S_2$ are sectors, this forces $S_2$ to consist of the sector $\{ a_{\omega^2} + k \omega - m: k,m \geq 0\}$ and $S_1$ to consist of the sector $\{ a_{\omega} - m + l \omega^{-1}: l,m \geq 0\}$, where $(a_{\omega^2}, a_{\omega})$ is equal to either $(a_1+\omega^2, a_1-1)$ or $(a_1-1,a_1-\omega)$. In either case we are in the second case of the proposition. $\Box$

Now we can analyse the behaviour near a point $z_0$ where $\rho_{z_0}$ takes on just one value. Let $\eta := \exp( \frac{1}{2} \arccos(\frac{5}{6}) )$. We record the identities

$\eta = (\omega^4+\omega^5) \eta^2 + (\omega+\omega^2) \quad (8)$
$\eta^3 = \frac{2}{3} (\omega^4+\omega^5) \eta^2 + (\omega+\omega^2). \quad (9)$

Proposition 7 Let $z_0 \in {\bf C}$ and $c_0 \in R_1/2R_1$ be such that $\rho_{z_0}(a) = c_0$ for all $a \in R_1$, and let $j \in {\bf Z}$.

• If $\rho_{z_0+\omega^j \eta^2}$ is $\omega^j$-periodic, then it is also $2R_1$-periodic.
• $\rho_{z_0+\omega^j \eta^2}$ is $\omega^k$-periodic for some k, and takes on at most two values.
• $\rho_{z_0+\omega^j \eta^2}$ is $2R_1$-periodic.

Proof We begin with the first claim. Using the translation and rotation symmetries we may asssume $z_0=0$, $c_0=0$ and $j=0$, thus $\rho_1$ is $1$-periodic. Suppose for contradiction that $\rho_{\eta^2}$ is not $2R_1$-periodic.

From (4) we see that $\rho_{e^{i\theta}}$ takes values in $\{1,\omega,\omega^2\} \hbox{ mod } 2R_1$ for all $\theta \in {\bf R}$, so in particular Proposition 6 applies. By this proposition (or by Lemma 4) we see that $\rho_{ \eta^2}$ takes values in $\{\omega,\omega^2\} \hbox{ mod } 2R_1$. Note that as $\rho_{ \eta^2}$ is not $2R_1$-periodic, it must attain both of the values $\omega, \omega^2 \hbox{ mod } 2R_1$

Now consider $\rho_{\omega \eta^2}$. It must be one of the configurations in Proposition 6, and also from (4) one has $\rho_{\omega \eta^2}(a) \neq \rho_{\eta^2}(a)$ for all $a$. If $\rho_\omega$ is constant, then this constant value must be $1 \hbox{ mod } 2R_2$ (since $\rho_1$ attains the two values $\omega,\omega^2 \hbox{ mod } 2R_2$). If it is non-constant, it cannot be $\omega$-periodic or $\omega^2$-periodic, thanks to Corollary 5. If it is non-constant and 1-periodic, it must then take the values $\rho_{\omega \eta^2}(a) = \rho_{ \eta^2}(a)+1$ for all $a$. If it is in the tri-sector case I, then by translating $\rho_z(a)$ to $\rho_z(a+n)$ for natural numbers $n$ and using compactness, we may assume without loss of generality that $\rho_{\omega \eta^2}$ is 1-periodic, in which case we must be in one of the previous cases. Similarly in the tri-sector case II. Thus we may assume without loss of generality that the function $\rho_{\omega \eta^2}$ is either equal to the constant function $1$, or to the function $\rho_{ \eta^2}+1$. Similarly (or by conjugation symmetry), $\rho_{-\omega^2 \eta^2}$ is equal to either $1$ or $\rho_{\eta^2}+1$.

Suppose $\rho_{\omega \eta^2},\rho_{-\omega^2 \eta^2}$ are both equal to $\rho_{\eta^2}+1$. From (4), (8) we have

$\rho_{-\omega^2 \eta^2}(a + \omega + \omega^2) \neq \rho_{\omega \eta^2}(a) + \omega + \omega^2$

for all $a \in R_1$, and hence

$\rho_{\eta^2}(a + \omega+\omega^2) \neq \rho_{\eta^2}(a) + 1;$

since $\rho_{\eta^2}$ only takes on the values $\omega, \omega^2 \hbox{ mod } 2R_1$, we conclude that $\rho_{\eta^2}$ is $\omega+\omega^2$-periodic and hence $2R_1$-periodic (since it was already 1-periodic), contradicting the hypothesis. Thus at least one of $\rho_{\omega \eta^2},\rho_{-\omega^2 \eta^2}$ are equal to 1. In fact once one of these is 1 the other will be forced to be also. To show this we will assume $\rho_{\omega \eta^2}=1$, as the other case is similar. Consider the function $\rho_{(1+\omega) \eta^2}$. On the one hand, by (4) we have $\rho_{(1+\omega)\eta^2}(a) \neq \rho_{\omega \eta^2}(a)$ for all $a$, and hence $\rho_{(1+\omega)\eta^2}$ avoids the value $1$. On the other hand, from (4), (8) one has

$\rho_{(1+\omega) \eta^2}(a) \neq \rho_0( a + (1+\omega) ) + \omega^2$

so $\rho_{(1+\omega)\eta^2}$ also avoids the value $\omega^2$. Since $\rho_{\eta^2}$ takes precisely the two values $\omega,\omega^2$, we conclude from Corollary 5 that $\rho_{(1+\omega) \eta^2}$ is constant. It cannot be the constant $\omega$ since this would imply $\rho_{(1+\omega) \eta^2}(a) = \rho_{\eta^2}(a)$ for some $a$, so it must be the constant 0: \

$\rho_{(1+\omega) \eta^2} = 0.$

Now we can apply a similar analysis to $\rho_{2\eta^2}$. It has to avoid 0 entirely, and from (4), (8) we have

$\rho_{2 \eta^2}(a) \neq \rho_{\omega \eta^2}(a + 1 - \omega^2) + \omega$

causing $\rho_{2\eta^2}$ to also avoid $\omega^2$. Using Corollary 5 as before we conclude that

$\rho_{2 \eta^2} = 0.$

Using this and $\rho_0 = 0$ we conclude from (4), (8) that $\rho_{-\omega^2 \eta^2}$ avoids both 0 and $\omega^2$, and thus by Corollary 5 again we have

$\rho_{-\omega^2 \eta^2} = 1$

as claimed. The same arguments then give

$\rho_{(1-\omega^2) \eta^2} = 0.$

Now look at $\rho_{2\omega \eta^2}$. Comparing with $\rho_{\omega \eta^2}=1$ and $\rho_{(1+\omega)\eta^2}=0$ using 4 we see that this function avoids the values $0,1$. On the other hand from (4), (8) one has

$\rho_{2\omega \eta^2}(a) \neq \rho_{\eta^2}(a + \omega + \omega^2) + 1$

and hence

$\rho_{2\omega \eta^2}(a) = \rho_{\eta^2}(a + \omega + \omega^2)$

for all $a$. A similar argument gives

$\rho_{-2\omega^2 \eta^2}(a) = \rho_{\eta^2}(a - \omega - \omega^2).$

Now we study $\rho_{\frac{2}{3}(1+\omega) \eta^2}$. From (4), (9) we have

$\rho_{\frac{2}{3}(1+\omega)\eta^2}(a) \neq \rho_{2\omega\eta^2}(a - \omega - \omega^2) + 1 = \rho_{\eta^2}(a) + 1$
$\rho_{\frac{2}{3}(1+\omega)\eta^2}(a) \neq \rho_0(a + 1 + \omega) + \omega^2$
$\rho_{\frac{2}{3}(1+\omega)\eta^2}(a) \neq \rho_{2\eta^2}(a - 1 + \omega^2) + \omega$

and hence

$\rho_{\frac{2}{3}(1+\omega)\eta^2}(a) = \rho_{\eta^2}(a)$

for all $a \in R_1$. A similar argument gives

$\rho_{\frac{2}{3}(1-\omega^2)\eta^2}(a) = \rho_{\eta^2}(a).$

On the other hand from (4), (8) we have

$\rho_{\frac{2}{3}(1+\omega)\eta^2}(a) \neq \rho_{\frac{2}{3}(1-\omega^2)\eta^2}(a + \omega + \omega^2) + 1$

which forces $\rho_{\eta^2}$ to be $\omega+\omega^2$-periodic, which contradicts the hypothesis as before. This proves the first part of the proposition.

To prove the second part, we again normalise $z_0=0, c_0=0,j=0$. Observe from Proposition 6 that the only way in which this cannot occur is if $\rho_{\eta^2}$ is in one of the tri-sector cases. If it is in the tri-sector case I, we can look at the translates $\rho_z(a+n)$ for natural numbers n and use compactness to reduce to the case where $\rho_{\eta^2}$ is 1-periodic and takes the value $\omega$ in one half-plane and $\omega^2$ in the complementary half-plane; but this contradicts the first part of the proposition. Similarly in tri-sector case II (where we now use the translates $\rho_z(a-n)$.

Now we prove the third part. Again we normalise $z_0=0, c_0=0, j=0$. Assume for contradiction that $\rho_{\eta^2}$ is not $2R_1$-periodic, then by the previous two parts it must be either $\omega$-periodic or $\omega^2$-periodic. Suppose for instance that $\rho_{\eta^2}$ is $\omega$-periodic; by Proposition 6 it attains each of the values $1, \omega \hbox{ mod } 2R_1$. Now look at $\rho_{\omega \eta^2}$. By the previous part of the proposition, it must be either $1$-periodic, $\omega$-periodic, or $\omega^2$-periodic and takes on at most two values. By Corollary 5 (and Proposition 6) it is either constant, or $\omega$-periodic; in the latter case it must take the values $1, \omega^2$ mod $2R_2$ and in fact must equal $\rho_{\omega \eta^2}(a) = \rho_{\eta^2}(a) + \omega$ by (4); but this is ruled out by the first part of the proposition. Thus $\rho_{\omega \eta^2}$ is constant; since $\rho_{\omega \eta^2}(a) \neq \rho_0(a), \rho_{\eta^2}(a)$ for all $a \in R_1$, this forces $\rho_{\omega \eta^2} = \omega$. Now we turn to $\rho_{-\omega^2 \eta^2}$. From (4) we have $\rho_{-\omega^2 \eta^2}(a) \neq \rho_0(a), \rho_{\omega \eta^2}(a - \omega - \omega^2)$ for all $a$, so $\rho_{-\omega^2 \eta^2}$ takes values in $1, \omega \hbox{ mod } 2R_1$. Since $\rho_{\eta^2}$ takes values in $1, \omega^2$, we see from Corollary 5 that $\rho_{-\omega^2 \eta^2}$ must be constant, but this contradicts the first part of this proposition (with $z = -\omega^2 \eta^2$ and $j=1$. Thus $\rho_{\eta^2}$ cannot be $\omega$-periodic; a similar argument rules out the case when $\rho_{\eta^2}$ is $\omega^2$-periodic. $\Box$

Corollary 8 If $\rho_z$ is $2R_1$-periodic for some $z$, then so is $\rho_{z + a \eta^2}$ for any $a \in R_1$.

Proof By induction it suffices to show that if $\rho_z$ is $2R_1$-periodic, then so is $\rho_{z + \omega^j \eta^2}$ for any integer $j$. By recoloring one may assume without loss of generality that $\rho_z=0$, and then the claim follows from the last part of Proposition 7. $\Box$

Corollary 9 If $c$ is $2R_1$-periodic on $R_1$, then it is also $2R_1$-periodic on $R_1 + \eta^2 R_1$.