Proof By hypothesis, G is nonempty. Given $\sigma ,\tau \in G$ and $x_1,x_2\in X$ , we have

$$\begin{align*}P(\sigma\tau, x_1) = P(\sigma) P(\tau,x_1) = P(\sigma) P(\tau,x_2) = P(\sigma\tau, x_2),\end{align*}$$

and this common value lies in $H_0$ , since $P(\sigma ), P(\tau ,x_1) \in H_0$ . Thus, $\sigma \tau \in G$ . In addition,

$$\begin{align*}P(\sigma) P(\sigma^{-1}, x_1) = P(\sigma\sigma^{-1}, x_1) = P(e, x_1) = P(e, x_2){=} P(\sigma\sigma^{-1}, x_2) {=} P(\sigma) P(\sigma^{-1}, x_2) .\end{align*}$$

Multiplying both sides on the left by $P(\sigma )^{-1}\in H_0$ , we have

$$\begin{align*}P(\sigma^{-1}, x_1) = P(\sigma^{-1}, x_2) = P(\sigma)^{-1} P(e, x_1) \in H_0, \end{align*}$$

proving that $\sigma ^{-1}\in G$ . Thus, G is a subgroup of $\Gamma $ ; and by equation (6), the map $\sigma \mapsto P(\sigma )$ is a homomorphism.

Proof We proceed by induction on n. For $n=1$ , we have $f_x(0)=x$ as desired. Assuming it holds for n, then modulo 2, we have

$$\begin{align*}f_x^{n+1}(0) = \big(f_x^n(0))^2 + x \equiv x+\sum_{i=0}^{n-1} \Big(x^{2^i}\Big)^2 = \sum_{i=0}^{n} x^{2^i}. \end{align*}$$

Proof Since the $r+1$ listed iterates are distinct, c must be a root of the relevant Misiurewicz polynomial (see Section 1 of [Reference GokselGok20] or equation (1) of [Reference Benedetto and GokselBG23]). Statement (1) is therefore the content of Theorem 1.3 of [Reference GokselGok20] with $i=n=r-s$ .

Define $\displaystyle g(x):=x^{2^{n-1}} + x^{2^{n-2}} + \cdots + x$ . Then Lemma 2.3 gives us

(7) $$ \begin{align} F_{r,s}(x) \equiv \sum_{i=s}^{r-1} x^{2^i} \equiv \big( g(x) \big)^{2^s} \pmod{2}. \end{align} $$

Since $g'(x)\equiv 1 \pmod {2}$ , we have that $g(x)$ is separable modulo 2. Therefore, it follows from equation (7) that the desired ramification index satisfies $e(\mathbb {Q}_2(c)/\mathbb {Q}_2)\leq 2^s$ .

If $n\nmid s$ , then statement (1) of this lemma says that $v_2(f_c^{n}(0))=2^{-s}$ , and hence $e(\mathbb {Q}_2(c)/\mathbb {Q}_2)$ is divisible by $2^s$ . Since we just showed that $e(\mathbb {Q}_2(c)/\mathbb {Q}_2)\leq 2^s$ , we must have $e(\mathbb {Q}_2(c)/\mathbb {Q}_2)=2^s$ , proving statement (2).

If $n|s$ and $s\geq 2$ , then statement (1) of this lemma shows that $e(\mathbb {Q}_2(c)/\mathbb {Q}_2)$ is divisible by $2^s - 1$ . Combined with the above fact $e(\mathbb {Q}_2(c)/\mathbb {Q}_2)\leq 2^s$ , we have $e(\mathbb {Q}_2(c)/\mathbb {Q}_2)=2^s-1$ , proving statement (3) when $s\geq 2$ .

Finally, if $n|s$ and $s=1$ , then we must have $r=2$ , so that $F_{r,s}(x)=x^2+2x =x(x+2)$ , and hence $c\in \{0,-2\}$ . (Actually, the case $c=0$ is excluded by the assumption that the first $r+1$ iterates $f_c^i(0)$ are distinct.) Thus, $\mathbb {Q}_2(c)=\mathbb {Q}_2$ , so that $e(\mathbb {Q}_2(c)/\mathbb {Q}_2)=1 = 2^s-1$ , completing the proof of statement (3).

Proof We will prove the stronger result that $\sqrt {2}\not \in \mathbb {Q}_2(c)$ . Let $n:=r-1\geq 2$ , and let $a:=f_c^{n-1}(0)$ and $y:= f_c(a) = f_c^n(0)$ , so that $a,y \in \mathbb {Q}_2(c)$ . By Lemma 2.4(1), we have $v_2(a)=0$ and $v_2(y)=1/2$ .

Because $y^2=f_c(y)-c=f_c^r(0)-c$ and $a^2=f_c(a)-c=y-c$ , we have

$$\begin{align*}\big(y-a\sqrt{2}\big)\big(y+a\sqrt{2}\big) = y^2 - 2a^2 = f_c^r(0) - c - 2( y- c ) = F_{r,1}(c) - 2y = -2y ,\end{align*}$$

where the final equality is because $F_{r,1}(c)=0$ . Thus, if $\sqrt {2}\in \mathbb {Q}_2(c)$ , we would have

(8) $$ \begin{align} v_2\big( y - a\sqrt{2} \big) + v_2\big( y + a\sqrt{2} \big) = 1 + v_2(y) = \frac{3}{2}. \end{align} $$

If $v_2( y - a\sqrt {2})> v_2( y + a\sqrt {2} )$ , then by the triangle equality for valuations, we would have

$$\begin{align*}v_2\big(y-a\sqrt{2}\big)> v_2\big(y+a\sqrt{2}\big) = v_2\big( 2a\sqrt{2} \big) = \frac{3}{2},\end{align*}$$

contradicting equation (8). Similarly, we cannot have $v_2( y - a\sqrt {2}) < v_2( y + a\sqrt {2} )$ , either.

Thus, $v_2( y - a\sqrt {2}) = v_2( y + a\sqrt {2} )$ , and hence equation (8) yields $v_2( y - a\sqrt {2})=3/4$ . Since $y-a\sqrt {2}\in \mathbb {Q}_2(c)$ , it would follow that the ramification index of $\mathbb {Q}_2(c)/\mathbb {Q}_2$ is at least $4$ , contradicting Lemma 2.4(2), which says that $e(\mathbb {Q}_2(c)/\mathbb {Q}_2)=2$ . Therefore, we must have $\sqrt {2}\not \in \mathbb {Q}_2(c)$ .

Proof Let $n:=r-s$ . If $n|s$ , then by Lemma 2.4(3), the extension $\mathbb {Q}_2(c)/\mathbb {Q}_2$ has odd ramification degree. Therefore, $\mathbb {Q}_2(c)$ cannot contain any of $\sqrt {2},\sqrt {-1},\sqrt {-2}$ , each of which has ramification index 2 over $\mathbb {Q}_2$ . For the remainder of the proof, then, we assume that $n\nmid s$ .

Pick $\ell \geq 1$ such that $n|(s+\ell )$ , let $\alpha :=f_c^\ell (0)$ , and let $a:=\alpha ^{2^{s-1}}$ . Since $n\nmid s$ , we also have $n\nmid \ell $ , and therefore $v_2(a)=v_2(\alpha )=0$ by Lemma 2.4(1).

Let $\omega :=f_c^n(0)$ and $y:=\omega ^{2^{s-1}}$ , so that by Lemma 2.4(1), we have $v_2(\omega )=2^{-s}$ and hence $v_2(y)=1/2$ .

Also define $R:=f_c^{r-1}(0)$ and $S:=f_c^{s-1}(0)$ . By Lemma 2.3, we have

$$\begin{align*}y + S \equiv \bigg( \sum_{i=0}^{n-1} c^{2^i} \bigg)^{2^{s-1}} + \sum_{i=0}^{s-2} c^{2^i} \equiv \sum_{i=s-1}^{r-2} c^{2^i} + \sum_{i=0}^{s-2} c^{2^i} = \sum_{i=0}^{r-2} c^{2^i} \equiv R\ \pmod{2\mathbb{Z}[c]} , \end{align*}$$

and hence

$$\begin{align*}y^2 + 2yS + S^2 = (y+S)^2 \equiv R^2\ \pmod{4\mathbb{Z}[c]} .\end{align*}$$

On the other hand, we also have $R^2+S^2 + 2c = F_{r,s}(c)=0$ , and therefore

(9) $$ \begin{align} y^2 = y^2 - F_{r,s}(c) \equiv \big(\! R^2 - S^2 - 2yS \!\big) - \big(\!R^2 + S^2 + 2c \!\big) \equiv 2(c+ S^2 + yS)\ \pmod{4\mathbb{Z}[c]}. \end{align} $$

Let $\gamma $ be any of $\sqrt {2},\sqrt {-2}$ , or $1+\sqrt {-1}$ . Let $\tilde {\gamma }:=-\gamma $ if $\gamma =\sqrt {\pm 2}$ , or $\tilde {\gamma }:=2-\gamma $ if $\gamma =1+\sqrt {-1}$ . Define $Q(t)\in \mathbb {Z}[c][t]$ to be

$$\begin{align*}Q(t) := (t-a\gamma)(t-a\tilde{\gamma}) = \begin{cases} t^2 \mp 2a^2 & \text{ if } \gamma=\sqrt{\pm 2}, \\ t^2 - 2at+2a^2 & \text{ if } \gamma=1+\sqrt{-1}. \end{cases} \end{align*}$$

Noting that $0 < 2v_2(\omega )\leq v_2(y) < 1$ , then, and working modulo $2\omega ^2\mathbb {Z}[c]$ , we have $2y\equiv 0$ , and hence

$$\begin{align*}Q(y) &\equiv y^2 + 2a^2 \equiv 2(c + S^2 + a^2) \equiv 2\bigg( c + \sum_{i=1}^{s-1} c^{2^i} + \sum_{i=s}^{s+\ell-1} c^{2^i} \bigg)\\ & \equiv 2 f_c^{s+\ell}(0) \ \pmod{2\omega^2\mathbb{Z}[c]} ,\end{align*}$$

where the second congruence is by equation (9), and the third and fourth are by Lemma 2.3 and the definition of a. Since $n | (s+\ell )$ , we have $v_2(f_c^{s+\ell }(0))=2^{-s} = v_2(\omega )$ by Lemma 2.4(1), and therefore $v_2(Q(y)) = 1 + 2^{-s}$ .

If $\gamma \in \mathbb {Q}_2(c)$ , then we would have

(10) $$ \begin{align} v_2\big( y - a\gamma \big) + v_2\big( y - a\tilde{\gamma} \big) = v_2(Q(y)) = 1+2^{-s}. \end{align} $$

If $v_2( y - a\gamma )> v_2( y + a\tilde {\gamma } )$ , then as in the proof of Proposition 2.5, we would have

$$\begin{align*}v_2\big(y-a\gamma\big)> v_2\big(y-a\tilde{\gamma}\big) = v_2\big( a (\gamma-\tilde{\gamma}) \big) \geq 1, \end{align*}$$

contradicting equation (10). Similarly, we cannot have $v_2( y - a\gamma ) < v_2( y + a\tilde {\gamma } )$ , either.

Thus, $v_2( y - a\gamma ) = v_2( y + a\tilde {\gamma } )$ , so that equation (10) yields $v_2( y - a\gamma )=1/2 + 2^{-s-1}$ . Since $y-a\gamma \in \mathbb {Q}_2(c)$ , it would follow that the ramification index of $\mathbb {Q}_2(c)/\mathbb {Q}_2$ is at least $2^{s+1}$ , contradicting Lemma 2.4(2), which says that $e(\mathbb {Q}_2(c)/\mathbb {Q}_2)=2^s$ . Therefore, we must have $\gamma \not \in \mathbb {Q}_2(c)$ .

Proof By Proposition 2.1 and the fact that $f^r(0)=-f^s(0)$ , we have

$$\begin{align*}\gamma^2 = \frac{f^r(0) - y}{f^s(0)-(-y)} = \frac{-f^s(0) - y}{f^s(0)+y} = -1. \\[-42pt] \end{align*}$$

Proof We will label the tree inductively, starting from the root point $x_0$ . Label the tree (using the symbols a and b) arbitrarily up to level r.

For each successive $n\geq r+1$ , suppose that we have labeled $T_{n-1}$ in such a way that for every node x of $T_{n-1}$ up to level $n-r-2$ , the identity of equation (11) holds. (Note that this condition is vacuously true for $n= r+1$ .) For each node y at level $n-1$ , label the two points of $f^{-1}(y)$ arbitrarily as $ya$ and $yb$ . We will now adjust these labels that we have just applied at the nth level of the tree.

For each node x at level $n-r-1$ , consider the first ratio $\gamma _1$ in equation (11), which is a product of $2^{r-1}$ nodes sitting r levels above $xa$ , divided by a product of $2^{s-1}$ nodes sitting s levels above $xb$ . By Lemma 3.1, we have $\gamma _1^2=-1$ , so $\gamma _1=\pm \zeta _4$ . If $\gamma _1=-\zeta _4$ , then exchange the labels of the two level-n nodes $xaa\ldots aa$ and $xaa\ldots ab$ ; otherwise, make no change to the labels of nodes above $xa$ .

Similarly, let $\gamma _2$ denote the second ratio in equation (11), which is a product of $2^{r-1}$ nodes sitting r levels above $xb$ , divided by a product of $2^{s-1}$ nodes sitting s levels above $xa$ . Again by Lemma 3.1, we have $\gamma _2^2=-1$ , so $\gamma _2=\pm \zeta _4$ . If $\gamma _2=-\zeta _4$ , then exchange the labels of the two level-n nodes $xba\ldots aa$ and $xba\ldots ab$ ; otherwise, make no change to the labels of nodes above $xb$ .

Having made these (possible) label changes to various nodes at the nth level of the tree, we have labeled $T_n$ so that for every node x at every level $0\leq \ell \leq n-r-1$ of $T_n$ , the identity of equation (11) holds. Thus, our inductive construction of the desired labeling is complete.

Proof Step 1. We claim that for all $\sigma \in M_{r,s,\infty }$ , all $\tau \in \operatorname {\mathrm {Aut}}(T_\infty )$ , and all nodes x of the tree, we have

(12) $$ \begin{align} P_{r,s}(\sigma) + P^a_{r,s}(\tau,x) = P^a_{r,s}(\sigma \tau,x) \quad\text{and}\quad P_{r,s}(\sigma) + P^b_{r,s}(\tau,x) = P^b_{r,s}(\sigma \tau,x). \end{align} $$

Indeed, given such $\sigma $ , $\tau $ , and x, suppose first that $\operatorname {\mathrm {Par}}(\tau ,x)=0$ . Then $\tau (xa)=\tau (x) a$ and $\tau (xb)=\tau (x) b$ , and hence, writing $P_{r,s}(\sigma )$ as $P^a_{r,s}(\sigma ,\tau (x))$ , we have

$$ \begin{align*} P_{r,s}(\sigma) &= \sum_{w\in\{a,b\}^{r-1}} \operatorname{\mathrm{Par}}(\sigma,\tau(x)aw) + \sum_{w'\in\{a,b\}^{s-1}} \operatorname{\mathrm{Par}}(\sigma,\tau(x)bw') \\ &= \sum_{w\in\{a,b\}^{r-1}} \operatorname{\mathrm{Par}}(\sigma,\tau(xa)w) + \sum_{w'\in\{a,b\}^{s-1}} \operatorname{\mathrm{Par}}(\sigma,\tau(xb)w') \\ &= \sum_{w\in\{a,b\}^{r-1}} \operatorname{\mathrm{Par}}(\sigma,\tau(xaw)) + \sum_{w'\in\{a,b\}^{s-1}} \operatorname{\mathrm{Par}}(\sigma,\tau(xbw')). \end{align*} $$

Therefore, since we are working in $\mathbb {Z}/2\mathbb {Z}$ , equation (3) yields

(13) $$ \begin{align} P_{r,s}(\sigma) + P^a_{r,s}(\tau,x) &= \sum_{w\in\{a,b\}^{r-1}} \big[ \operatorname{\mathrm{Par}}(\sigma,\tau(xaw)) + \operatorname{\mathrm{Par}}(\tau, xaw) \big] \notag \\ & \quad + \sum_{w'\in\{a,b\}^{s-1}} \big[ \operatorname{\mathrm{Par}}(\sigma,\tau(xbw')) + \operatorname{\mathrm{Par}}(\tau, xbw') \big] \notag \\ &= \sum_{w\in\{a,b\}^{r-1}} \operatorname{\mathrm{Par}}(\sigma\tau, xaw) + \sum_{w'\in\{a,b\}^{s-1}} \operatorname{\mathrm{Par}}(\sigma\tau, xbw')\nonumber\\ & = P^a_{r,s}(\sigma \tau,x), \end{align} $$

proving the first identity of equation (12). The second follows by a similar calculation, writing $P_{r,s}(\sigma )$ as $P^b_{r,s}(\sigma ,\tau (x))$ .

The only other possibility is that $\operatorname {\mathrm {Par}}(\tau ,x)=1$ , and hence that $\tau (xa)=\tau (x) b$ and $\tau (xb)=\tau (x) a$ . Therefore, writing $P_{r,s}(\sigma )$ as $P^b_{r,s}(\sigma ,\tau (x))$ , we have

$$ \begin{align*} P_{r,s}(\sigma) &= \sum_{w\in\{a,b\}^{r-1}} \operatorname{\mathrm{Par}}(\sigma,\tau(x)bw) + \sum_{w'\in\{a,b\}^{s-1}} \operatorname{\mathrm{Par}}(\sigma,\tau(x)aw') \\ &= \sum_{w\in\{a,b\}^{r-1}} \operatorname{\mathrm{Par}}(\sigma,\tau(xa)w) + \sum_{w'\in\{a,b\}^{s-1}} \operatorname{\mathrm{Par}}(\sigma,\tau(xb)w') \\ &= \sum_{w\in\{a,b\}^{r-1}} \operatorname{\mathrm{Par}}(\sigma,\tau(xaw)) + \sum_{w'\in\{a,b\}^{s-1}} \operatorname{\mathrm{Par}}(\sigma,\tau(xbw')). \end{align*} $$

The first identity of equation (12) then follows by exactly the calculation of equation (13). The second follows by a similar calculation, writing $P_{r,s}(\sigma )$ as $P^a_{r,s}(\sigma ,\tau (x))$ , and proving our claim.

Step 2. In the notation of Proposition 2.2, let $\Gamma = \operatorname {\mathrm {Aut}}(T_{\infty })$ , let X be the set of nodes of $T_{\infty }$ , let $H=\mathbb {Z}/2\mathbb {Z} \times \mathbb {Z}/2\mathbb {Z}$ , let $H_0\subseteq H$ be the diagonal subgroup, and define $P:\Gamma \times X\to H$ by $P(\sigma , x):=( P_{r,s}^a (\sigma ,x), P_{r,s}^b (\sigma ,x) )$ .

Then $M_{r,s,\infty }$ is precisely the subset G of Proposition 2.2, and equation (12) gives precisely hypothesis (6). Moreover, the identity $e\in \operatorname {\mathrm {Aut}}(T_{\infty })$ satisfies $\operatorname {\mathrm {Par}}(e,y)=0$ for all nodes y, so that $P_{r,s}^a(e,x) = P_{r,s}^b(e,x)=0$ for all nodes x, and hence $e\in M_{r,s,\infty }$ .

Therefore, by Proposition 2.2, $M_{r,s,\infty }$ is a subgroup of $\operatorname {\mathrm {Aut}}(T_{\infty })$ , and P is a homomorphism. Moreover, $P_{r,s}$ is the composition of P with the unique isomorphism $H_0\to \mathbb {Z}/2\mathbb {Z}$ , so that $P_{r,s}$ is a homomorphism. By definition, its kernel is $B_{r,s,\infty }$ , which is therefore a subgroup of $M_{r,s,\infty }$ .

Proof It suffices to prove equation (14) for all $x\in \operatorname {\mathrm {Orb}}_f^-(x_0)$ and all $\sigma \in G_{\infty }$ . Indeed, in that case, equation (4) would hold for all such x viewed as nodes of the tree, since $\sigma (\zeta _4)/\zeta _4 \in \{\pm 1\}$ must be independent of x. That is, every $\sigma \in G_{\infty }$ would be in $M_{r,s,\infty }$ ; and if $P_{r,s}(\sigma )=0$ for all $\sigma \in G_{\infty }$ , then by definition every $\sigma $ is in $B_{r,s,\infty }$ .

Because of the specified labeling of the tree, for any such $\sigma $ and x, we have

(15) $$ \begin{align} \prod_{w\in\{a,b\}^{r-1}} \big[ \sigma(xawa) \big] = \sigma(\zeta_4) \prod_{w' \in\{a,b\}^{s-1}} \big[ \sigma(xbw'a) \big]. \end{align} $$

Note that for each node y of the tree, we have

$$\begin{align*}\big[ \sigma(ya) \big] = (-1)^{\operatorname{\mathrm{Par}}(\sigma,y)} \big[ \sigma(y) a \big] ,\end{align*}$$

and hence equation (15) becomes

$$\begin{align*}(-1)^{P^a_{r,s}(\sigma,x)}\prod_{w\in\{a,b\}^{r-1}} \big[ \sigma(xaw)a \big] = \sigma(\zeta_4) \prod_{w' \in\{a,b\}^{s-1}} \big[ \sigma(xbw')a \big]. \end{align*}$$

Since each product above is over all $w\in \{a,b\}^{r-1}$ or all $w' \in \{a,b\}^{s-1}$ , this equation becomes either

$$\begin{align*}(-1)^{P^a_{r,s}(\sigma,x)}\prod_{w\in\{a,b\}^{r-1}} \big[ \sigma(x)awa \big] = \sigma(\zeta_4) \prod_{w' \in\{a,b\}^{s-1}} \big[ \sigma(x)bw'a \big] \end{align*}$$

if $\operatorname {\mathrm {Par}}(\sigma ,x)=0$ , or

$$\begin{align*}(-1)^{P^a_{r,s}(\sigma,x)}\prod_{w\in\{a,b\}^{r-1}} \big[ \sigma(x)bwa \big] = \sigma(\zeta_4) \prod_{w' \in\{a,b\}^{s-1}} \big[ \sigma(x)aw'a \big] \end{align*}$$

if $\operatorname {\mathrm {Par}}(\sigma ,x)=1$ . Either way, by Lemma 3.2 applied to $\sigma (x)$ in place of x, we have

$$\begin{align*}\sigma(\zeta_4) = (-1)^{P^a_{r,s}(\sigma,x)} \zeta_4. \end{align*}$$

By similar reasoning, we also have $\sigma (\zeta _4) = (-1)^{P^b_{r,s}(\sigma ,x)} \zeta _4$ .

Proof Choose a labeling of the tree according to Lemma 3.2. We will now adjust this labeling, working inductively up from the root point $x_0$ .

For each $n\geq 0$ , suppose that under our current labeling of the tree, the identities (11) and (26) hold for every node x of the tree up to level $n-1$ . This condition is vacuously true for $n=0$ .

For each node x at level n, denote by $\delta (x)$ the expression on the left side of equation (26). We first claim that $\delta (x)$ makes sense, in that its denominator is nonzero. Equivalently, by equation (25), we are claiming that neither $U_a(x)$ nor $U_b(x)$ is $-2$ . However, if $U_a(x)=-2$ , then

$$\begin{align*}u_1 u_2 = -\frac{2}{c} = c^2 + 2c+2 \end{align*}$$

by equation (16), and hence

$$\begin{align*}f^2(0)-[xa] = (u_1 u_2)^2 = c^4 + 4c^3 + 8c^2 + 8c + 4 = 2c^2 + 2c = 2f^2(0),\end{align*}$$

again by Proposition 2.1 and equation (16). But then we would have

$$\begin{align*}[xa]=-f^2(0) = f^3(0), \end{align*}$$

and hence $x=f^4(0)$ would be postcritical, which we assumed does not happen. By this contradiction, we have $U_a(x)\neq -2$ . By a similar argument, we also have $U_b(x)\neq -2$ , proving our claim.

Combining equations (23) and (25) yields

(27) $$ \begin{align} \delta(x)^2 = \frac{ 64 (c^2+c+1)^2 (2+U_a(x)) (2+U_b(x))}{32(c^2+c+1)^2 (2+U_a(x))(2+U_b(x)) } = 2, \end{align} $$

so that $\delta (x)=\pm \sqrt {2}$ . If $\delta (x)=\sqrt {2}$ , then make no change to the labeling above x. On the other hand, if $\delta (x)=-\sqrt {2}$ , then swap the labels of the nodes $xaaaa$ and $xaaab$ , and also swap the labels of the nodes $xaaba$ and $xaabb$ . (That is, in the notation of Figure 2, make two parity changes four levels above x: specifically, by switching the labels of $\pm t_{11}$ , and also switching the labels of $\pm t_{12}$ . At five levels above x, this also forces swapping $s_{11}=[xaaaaa]$ with $s_{12}=[xaaaba]$ , as well as $-s_{11}=[xaaaab]$ with $-s_{12}=[xaaabb]$ , and at the same time swapping $s_{13}=[xaabaa]$ with $s_{14}=[xaabba]$ , and $-s_{13}=[xaabab]$ with $-s_{14}=[xaabbb]$ .)

This change reverses the sign of $\gamma _{aa}(x)$ but no other factors in the expression for $\delta (x)$ , which therefore becomes $+\sqrt {2}$ as desired. In addition, the change does not affect the value of expression (26) at any other node (in place of x) at or below level n of the tree.

Moreover, the only nodes for which equation (11) might be affected by these swaps are x (for the numerator of the first expression in that equation) and $xa$ (for the denominator of the second). In both cases, however, the signs of exactly two terms of the relevant product change, and hence the value of the full expression is unaffected. Thus, the change of labels above preserves property that equation (11) holds at every node of the whole tree.

Having made this adjustment above every node x at level n, then, our labeling satisfies equation (26) at every node up to and including level n, and also still satisfies equation (11) at every node of the whole tree. Our inductive construction of the desired labeling is therefore complete.

Proof We will prove half of equation (31), that $V_{3,2}(\sigma ,xaa) = \operatorname {\mathrm {Par}}(\sigma ,xba) + \operatorname {\mathrm {Par}}(\sigma ,xbb)$ . The proof of the other half, and the proofs of both halves of equation (32), are similar.

By definition of $P_{3,2}$ (from Definition 1.1) written as $P^a_{3,2}(\sigma ,xa)$ , bearing in mind that we are working in $\mathbb {Z}/2\mathbb {Z}$ , we have

$$\begin{align*}\sum_{w\in\{a,b\}^2} \operatorname{\mathrm{Par}}(\sigma,xaaw) = P_{3,2}(\sigma) + \sum_{w'\in\{a,b\}} \operatorname{\mathrm{Par}}(\sigma,xabw'),\end{align*}$$

and hence

$$ \begin{align*} V_{3,2}(\sigma,xaa) & = \sum_{w\in\{a,b\}^2} \operatorname{\mathrm{Par}}(\sigma,xaaw) + \sum_{w'\in \{a,b\}} \operatorname{\mathrm{Par}}(\sigma,xaaw') \\ &= P_{3,2}(\sigma) + \sum_{w'\in\{a,b\}} \operatorname{\mathrm{Par}}(\sigma,xabw') + \sum_{w'\in \{a,b\}} \operatorname{\mathrm{Par}}(\sigma,xaaw') \\ & = P_{3,2}(\sigma) + \sum_{w\in \{a,b\}^2} \operatorname{\mathrm{Par}}(\sigma,xaw) = 2P_{3,2}(\sigma) + \sum_{w'\in \{a,b\}} \operatorname{\mathrm{Par}}(\sigma,xbw') \\ & = \operatorname{\mathrm{Par}}(\sigma,xba) + \operatorname{\mathrm{Par}}(\sigma,xbb), \end{align*} $$

where the fourth equality is by writing $P_{3,2}(\sigma ) = P^a_{3,2}(\sigma ,x)$ , and the fifth is because we are working in $\mathbb {Z}/2\mathbb {Z}$ .

Proof of Theorem 4.3

Step 1. We claim that for all $\sigma \in \tilde {M}_{3,2,\infty }$ , all $\tau \in \operatorname {\mathrm {Aut}}(T_{\infty })$ , and all nodes x of the tree, we have

(33) $$ \begin{align} R_{3,2}(\sigma,\tau(x)) + R_{3,2}(\tau,x) = R_{3,2}(\sigma\tau,x). \end{align} $$

Indeed, for such $\sigma ,\tau , x$ , the left side of equation (33) is

(34) $$ \begin{align} \sum_{w\in\{a,b\}^2} & \big[ \operatorname{\mathrm{Par}}(\sigma,\tau(x) w) +\operatorname{\mathrm{Par}}(\tau, x w) \big] + \sum_{w\in\{a,b\}^2} \big[ \operatorname{\mathrm{Par}}(\sigma,\tau(x)wb) +\operatorname{\mathrm{Par}}(\tau, x wb) \big] \notag \\ & + \sum_{w\in\{a,b\}^2} \bigg[ \sum_{t\in\{a,b\} } \operatorname{\mathrm{Par}}(\sigma,\tau(x)wat) +\operatorname{\mathrm{Par}}(\tau, x wat) \bigg] \notag \\ & + \big[ \operatorname{\mathrm{Par}}(\sigma,\tau(x)aa) + \operatorname{\mathrm{Par}}(\sigma,\tau(x)ab) \big] \big[ \operatorname{\mathrm{Par}}(\sigma,\tau(x)ba) + \operatorname{\mathrm{Par}}(\sigma,\tau(x)bb) \big] \notag \\ & + \big[ \operatorname{\mathrm{Par}}(\tau,xaa) + \operatorname{\mathrm{Par}}(\tau,xab) \big] \big[ \operatorname{\mathrm{Par}}(\tau,xba) + \operatorname{\mathrm{Par}}(\tau,xbb) \big] \notag \\ = \sum_{w\in\{a,b\}^2} & \big[ \operatorname{\mathrm{Par}}(\sigma,\tau(xw)) +\operatorname{\mathrm{Par}}(\tau, x w) \big] + \sum_{w\in\{a,b\}^2} \big[ \operatorname{\mathrm{Par}}(\sigma,\tau(xw)b) +\operatorname{\mathrm{Par}}(\tau, x wb) \big] \notag \\ & + \sum_{w\in\{a,b\}^2} \bigg[ \sum_{t\in\{a,b\} } \operatorname{\mathrm{Par}}(\sigma,\tau(xw)at) +\operatorname{\mathrm{Par}}(\tau, x wat) \bigg] \\ & + \big[ \operatorname{\mathrm{Par}}(\sigma,\tau(xaa)) + \operatorname{\mathrm{Par}}(\sigma,\tau(xab)) \big] \big[ \operatorname{\mathrm{Par}}(\sigma,\tau(xba)) + \operatorname{\mathrm{Par}}(\sigma,\tau(xbb)) \big] \notag \\ & + \big[ \operatorname{\mathrm{Par}}(\tau,xaa) + \operatorname{\mathrm{Par}}(\tau,xab) \big] \big[ \operatorname{\mathrm{Par}}(\tau,xba) + \operatorname{\mathrm{Par}}(\tau,xbb) \big]. \notag \end{align} $$

By equation (3), the first sum on the right side of equation (34) is $\sum _{w\in \{a,b\}^2} \operatorname {\mathrm {Par}}(\sigma \tau ,xw)$ .

We consider the second and third sums together. The contribution of each $w\in \{a,b\}^2$ to these sums is

(35) $$ \begin{align} \operatorname{\mathrm{Par}}(\sigma,\tau(xw)b) + \operatorname{\mathrm{Par}}(\tau, xwb) + \sum_{t\in\{a,b\} } \big[ \operatorname{\mathrm{Par}}(\sigma,\tau(xw)at) +\operatorname{\mathrm{Par}}(\tau, x wat) \big]. \end{align} $$

If $\operatorname {\mathrm {Par}}(\tau ,xw)=0$ , so that $\tau (xw)a=\tau (xwa)$ and $\tau (xw)b=\tau (xwb)$ , then expression (35) is

$$ \begin{align*} \operatorname{\mathrm{Par}}(\sigma,\tau(xwb)) &+ \operatorname{\mathrm{Par}}(\tau, xwb) + \sum_{t\in\{a,b\} } \big[ \operatorname{\mathrm{Par}}(\sigma,\tau(xwat)) +\operatorname{\mathrm{Par}}(\tau, x wat) \big] \\ & = \operatorname{\mathrm{Par}}(\sigma\tau,xwb) + \sum_{t\in\{a,b\} } \operatorname{\mathrm{Par}}(\sigma\tau,xwat). \end{align*} $$

On the other hand, if $\operatorname {\mathrm {Par}}(\tau ,xw)=1$ , so that $\tau (xw)a=\tau (xwb)$ and $\tau (xw)b=\tau (xwa)$ , then a similar computation shows that expression (35) is

$$\begin{align*}V_{3,2}(\sigma,\tau(xw)) + \operatorname{\mathrm{Par}}(\sigma\tau,xwb) + \sum_{t\in\{a,b\} } \operatorname{\mathrm{Par}}(\sigma\tau,xwat), \end{align*}$$

where $V_{3,2}$ is as in equation (30). In either case, then, expression (35) is

$$\begin{align*}V_{3,2}(\sigma,\tau(xw)) \operatorname{\mathrm{Par}}(\tau,xw) + \operatorname{\mathrm{Par}}(\sigma\tau,xwb) + \sum_{t\in\{a,b\} } \operatorname{\mathrm{Par}}(\sigma\tau,xwat). \end{align*}$$

Hence, the second and third sums of equation (34) together are

$$ \begin{align*} \sum_{w\in\{a,b\}^2} & \bigg[ V_{3,2}(\sigma,\tau(xw)) \operatorname{\mathrm{Par}}(\tau,xw)+ \operatorname{\mathrm{Par}}(\sigma\tau,xwb) + \sum_{t\in\{a,b\} } \operatorname{\mathrm{Par}}(\sigma\tau,xwat) \bigg] \\ &= \sum_{w\in\{a,b\}^2} \bigg[ \operatorname{\mathrm{Par}}(\sigma\tau,xwb) + \sum_{t\in\{a,b\} } \operatorname{\mathrm{Par}}(\sigma\tau,xwat) \bigg] \\ &\quad + \big[ \operatorname{\mathrm{Par}}(\sigma,\tau(xba)) + \operatorname{\mathrm{Par}}(\sigma,\tau(xbb)) \big] \big[ \operatorname{\mathrm{Par}}(\tau,xaa) + \operatorname{\mathrm{Par}}(\tau,xab) \big] \\ &\quad + \big[ \operatorname{\mathrm{Par}}(\sigma,\tau(xaa)) + \operatorname{\mathrm{Par}}(\sigma,\tau(xab)) \big] \big[ \operatorname{\mathrm{Par}}(\tau,xba) + \operatorname{\mathrm{Par}}(\tau,xbb) \big], \end{align*} $$

where the equality is because by Lemma 4.4, we have

$$\begin{align*}V_{3,2}(\sigma,\tau(xa)a) = V_{3,2}(\sigma,\tau(xa)b) = \operatorname{\mathrm{Par}}(\sigma,\tau(xb)a) +\operatorname{\mathrm{Par}}(\sigma,\tau(xb)b) \end{align*}$$

and

$$\begin{align*}V_{3,2}(\sigma,\tau(xb)a) = V_{3,2}(\sigma,\tau(xb)b) = \operatorname{\mathrm{Par}}(\sigma,\tau(xa)a) +\operatorname{\mathrm{Par}}(\sigma,\tau(xa)b). \end{align*}$$

Thus, combining all three sums and the last two expressions from equation (34), we have

$$\begin{align*}R_{3,2}(\sigma,\tau(x)) + R_{3,2}(\tau,x) = R_{3,2}(\sigma\tau, x) + Z_{3,2}(\sigma,\tau,x), \end{align*}$$

where $Z_{3,2}(\sigma ,\tau ,x)$ is

$$ \begin{align*} \big[ \operatorname{\mathrm{Par}}( & \sigma,\tau(xba)) + \operatorname{\mathrm{Par}}(\sigma,\tau(xbb)) \big] \big[ \operatorname{\mathrm{Par}}(\tau,xaa) + \operatorname{\mathrm{Par}}(\tau,xab) \big] \\ & + \big[ \operatorname{\mathrm{Par}}(\sigma,\tau(xaa)) + \operatorname{\mathrm{Par}}(\sigma,\tau(xab)) \big] \big[ \operatorname{\mathrm{Par}}(\tau,xba) + \operatorname{\mathrm{Par}}(\tau,xbb) \big] \\ & + \big[ \operatorname{\mathrm{Par}}(\sigma,\tau(xaa)) + \operatorname{\mathrm{Par}}(\sigma,\tau(xab)) \big] \big[ \operatorname{\mathrm{Par}}(\sigma,\tau(xba)) + \operatorname{\mathrm{Par}}(\sigma,\tau(xbb)) \big] \\ & + \big[ \operatorname{\mathrm{Par}}(\tau,xaa) + \operatorname{\mathrm{Par}}(\tau,xab) \big] \big[ \operatorname{\mathrm{Par}}(\tau,xba) + \operatorname{\mathrm{Par}}(\tau,xbb) \big] \\ & - \big[ \operatorname{\mathrm{Par}}(\sigma\tau,xaa) + \operatorname{\mathrm{Par}}(\sigma\tau,xab) \big] \big[ \operatorname{\mathrm{Par}}(\sigma\tau,xba) + \operatorname{\mathrm{Par}}(\sigma\tau,xbb) \big]. \end{align*} $$

Applying (3) to expand each instance of $\operatorname {\mathrm {Par}}(\sigma \tau ,y)$ in the last line above as $\operatorname {\mathrm {Par}}(\sigma ,\tau (y)) + \operatorname {\mathrm {Par}}(\tau ,y)$ , and expanding each of the resulting products, all of the terms cancel. That is, $Z_{3,2}(\sigma ,\tau ,x)=0$ , proving the claimed identity (33).

Step 2. In the notation of Proposition 2.2, let $\Gamma = M_{3,2,\infty }$ , let X be the set of nodes of $T_{\infty }$ , let $H=H_0=\mathbb {Z}/2\mathbb {Z}$ , and let $P= R_{3,2}$ . Then $\tilde {M}_{r,s,\infty }$ is precisely the subset G of Proposition 2.2, the identity element $e\in M_{3,2,\infty }$ clearly satisfies $R_{3,2}(e,x)=0$ for all nodes x, and equation (33) gives hypothesis (6). Therefore, by Proposition 2.2, $\tilde {M}_{r,s,\infty }$ is a subgroup of $M_{3,2,\infty }$ , and $R_{3,2}$ is a homomorphism. By definition, the kernel of $R_{3,2}$ is $\tilde {B}_{r,s,\infty }$ , which is therefore a subgroup of $\tilde {M}_{r,s,\infty }$ .

Proof Step 1. We begin with several technical identities needed to prove equation (36). For every node $x\in \operatorname {\mathrm {Orb}}_f^-(x_0)$ , word $w\in \{a,b\}^2$ , and symbol $t\in \{a,b\}$ , define the quantities $\gamma _w:=\gamma _w(x)$ and $U_t:=U_t(x)$ as in equations (19) and (24). Further define

$$\begin{align*}\gamma^{\prime}_w = \gamma^{\prime}_{w}(x) := [xw aaa] [xw aba] [xw ba] - [xw baa] [xw bba] [xw aa] .\end{align*}$$

Then, borrowing the notation of Figure 2, we have

$$ \begin{align*} \frac{\gamma^{\prime}_{aa} \gamma^{\prime}_{ab}}{\gamma_{aa} \gamma_{ab}} &= \frac{\gamma^{\prime}_{aa} \gamma_{aa} \gamma^{\prime}_{ab}\gamma_{ab}}{\gamma_{aa}^2\gamma_{ab}^2} = \frac{[ (s_{11} s_{12} t_{12})^2 - (s_{13} s_{14} t_{11})^2 ] [ (s_{21} s_{22} t_{22})^2 - (s_{23} s_{24} t_{21})^2 ]} { 8(c^2+c+1) (2 + U_b) } \\ &= \frac{\prod_{i=1}^2[(f^2(0) - u_i)(-c-u_i) - (f^2(0) + u_i)(-c+u_i)]} { 8(c^2+c+1) (2 + U_b) } \\ &= \frac{ [2 (c-f^2(0))u_1] [2 (c-f^2(0))u_2]}{ 8(c^2+c+1) (2 + U_b) } = \frac{4 c^4 u_1 u_2 }{ 8(c^2+c+1) (2 + U_b) } = - \frac{U_a }{2 + U_b}, \end{align*} $$

by equations (16) and (22), along with Proposition 2.1. Similarly, we also have

$$\begin{align*}\frac{\gamma^{\prime}_{ba} \gamma^{\prime}_{bb}}{\gamma_{ba} \gamma_{bb}} = - \frac{ U_b }{2 + U_a} .\end{align*}$$

Proposition 2.1 and equation (16) also give us

$$ \begin{align*} (2-U_a-U_b)(2+U_a+U_b) &= 4 - c^2 (u_1^2 u_2^2 + u_3^2 u_4^2) - 2 U_a U_b \\ &= 4 - c^2\big( (f^2(0)- [xa]) + (f^2(0)- [xb]) \big) - 2 U_a U_b \\ &= 4 - 2c^2 f^2(0) - 2 U_a U_b = - 2 U_a U_b, \end{align*} $$

using the fact that $[xb]=-[xa]$ . Thus, we have

$$\begin{align*}\frac{2 + U_a + U_b}{2 -U_a - U_b} &= \frac{(2 + U_a + U_b)^2}{(2 -U_a - U_b)(2+U_a+U_b)} = \frac{2(2+U_a)(2+U_b)}{-2U_a U_b}\\ & = - \frac{(2+U_a)(2+U_b)}{U_a U_b} , \end{align*}$$

by equation (25). We also have

$$ \begin{align*} (2+U_a-U_b)& (2+U_a+U_b) = (2+U_a)^2 - c^2 u_3^2 u_4^2 = (2+U_a)^2 - c^2 (f^2(0)-[xb]) \\ &= (2+U_a)^2 - c^2 \big( f^2(0) + [xa] \big) = (2+U_a)^2 {-} c^2 \big( f^2(0) {+} f^2(0) {-} u_1^2 u_2^2 \big) \\ &= 4 -2c^2f^2(0) + 4U_a + 2U_a^2 = 4U_a + 2U_a^2 = 2U_a (2+U_a), \end{align*} $$

and hence

$$\begin{align*}\frac{2 + U_a + U_b}{2 + U_a - U_b} = \frac{(2 + U_a + U_b)^2}{(2 +U_a - U_b)(2+U_a+U_b)} = \frac{2(2+U_a)(2+U_b)}{2U_a (2+U_a)} = \frac{2+U_b}{U_a} .\end{align*}$$

Reversing the roles of a and b, a similar argument yields

$$\begin{align*}\frac{2+U_a+U_b}{2-U_a+U_b} = \frac{2+U_a}{U_b}.\end{align*}$$

Step 2. Given any $\sigma \in G_{\infty }$ , by Theorem 3.4, we have $\sigma \in M_{3,2,\infty }$ . For each $t\in \{a,b\}$ , let $\tilde {t}\in \{a,b\}$ satisfy $\sigma (xt)=\sigma (x)\tilde {t}$ . Let $t'\in \{a,b\}$ be the opposite symbol of t (meaning that $\{t,t'\} = \{a,b\}$ ), and let $\tilde {t}'\in \{a,b\}$ be the opposite symbol of $\tilde {t}$ . Then

(37) $$ \begin{align} \sigma\big(U_t(x)\big) = (-1)^{V(t)} c [\sigma(xt)aa][\sigma(xt)ba] = (-1)^{V(t)} U_{\tilde{t}}( \sigma(x) ), \end{align} $$

where the parity $V(t)$ of the sign in equation (37) is

$$\begin{align*}V(t) := \operatorname{\mathrm{Par}}(\sigma,xta) + \operatorname{\mathrm{Par}}(\sigma,xtb) =V_{3,2}(\sigma,xt'a) = V_{3,2}(\sigma,xt'b), \end{align*}$$

where the second and third equalities are by Lemma 4.4.

Thus, if the sign in equation (37) is $+1$ (i.e., if the exponent $V(t)$ is even), then

$$\begin{align*}\sigma\big( \gamma_{t'a}(x) \gamma_{t'b}(x) \big) = \pm \gamma_{\tilde{t}'a}(\sigma(x)) \gamma_{\tilde{t}'b}(\sigma(x)), \end{align*}$$

since $V_{3,2}(\sigma ,xt'a)=V_{3,2}(\sigma ,xt'b)=0$ , and hence an even number of $-$ signs will appear across the two summands that comprise each $\gamma _w(x)$ . On the other hand, if the sign in equation (37) is $-1$ (i.e., if the exponent $V(t)$ is odd), then

$$\begin{align*}\sigma\big( \gamma_{t'a}(x) \gamma_{t'b}(x) \big) = \pm \gamma^{\prime}_{\tilde{t}'a}(\sigma(x)) \gamma^{\prime}_{\tilde{t}'b}(\sigma(x)), \end{align*}$$

since an odd number of $-$ signs will appear in each $\gamma _w(x)$ . That is, the sign that $\sigma $ applies to $U_t(x)$ determines whether the $\gamma $ terms above $xt'$ are mapped to plus-or-minus $\gamma $ terms or to plus-or-minus $\gamma '$ terms.

Step 3. We now compute $\sigma (\sqrt {2})$ according to formula (26) for $\sqrt {2}$ . In light of Step 2, we have four cases to consider.

Case 1: The parities $V(a)$ and $V(b)$ in equation (37) are both even, that is,

(38) $$ \begin{align} \operatorname{\mathrm{Par}}(\sigma,xaa) + \operatorname{\mathrm{Par}}(\sigma,xab) = \operatorname{\mathrm{Par}}(\sigma,xba) + \operatorname{\mathrm{Par}}(\sigma,xbb) = 0 \in\mathbb{Z}/2\mathbb{Z}. \end{align} $$

In formula (28) defining $R_{3,2}(\sigma ,x)$ , then, the four terms $\operatorname {\mathrm {Par}}(\sigma ,xw)$ sum to $0$ , while the product on the second line of that formula is also $0$ . It remains to consider the 12 terms $\operatorname {\mathrm {Par}}(\sigma ,xwb)$ and $\operatorname {\mathrm {Par}}(\sigma ,xwat)$ .

Moreover, as discussed in Step 2, $\sigma $ maps each $\gamma _w(x)$ to $\pm \gamma _{\tilde {w}}(\sigma (x))$ (where $\sigma (xw)=\sigma (x)\tilde {w}$ ). In fact, the sign with which $\sigma $ maps $\gamma _w(x)$ is the sign with which it maps either one of the terms in the sum defining $\gamma _w$ , namely,

(39) $$ \begin{align} \operatorname{\mathrm{Par}}(\sigma,xwaa) + \operatorname{\mathrm{Par}}(\sigma,xwab) + \operatorname{\mathrm{Par}}(\sigma,xwb). \end{align} $$

Thus, $\sigma (\prod _w \gamma _w) = \pm \prod _w \gamma _w$ , where the product is over all $w\in \{a,b\}^2$ , and the parity of the $\pm $ sign is the sum of expression (39) across all four such words w. But that sum is precisely the sum of the remaining 12 terms of $R_{3,2}(\sigma ,x)$ . Combining these observations with the $+$ signs in equation (37), it follows that the image of expression (26) under $\sigma $ is $(-1)^{R_{3,2}(\sigma ,x)}\sqrt {2}$ , as desired.

Case 2: The parities in equation (37) are $V(a)=0$ and $V(b)=1$ , that is,

(40) $$ \begin{align} \operatorname{\mathrm{Par}}(\sigma,xaa) + \operatorname{\mathrm{Par}}(\sigma,xab) = 0, \quad \operatorname{\mathrm{Par}}(\sigma,xba) + \operatorname{\mathrm{Par}}(\sigma,xbb) = 1 \in\mathbb{Z}/2\mathbb{Z}. \end{align} $$

In formula (28) defining $R_{3,2}(\sigma ,x)$ , then, the four terms $\operatorname {\mathrm {Par}}(\sigma ,xw)$ sum to $1$ , while the product on the second line of that formula is $0$ .

Because $V(a)=0$ , Step 2 shows that $\sigma $ maps $\gamma _{ba}(x)\gamma _{bb}(x)$ to $\pm \gamma _{\tilde {b}a}(\sigma (x))\gamma _{\tilde {b}b}(\sigma (x))$ , with the parity of the $\pm $ sign being the sum of expression (39) for $w=ba$ and $w=bb$ . On the other hand, because $V(b)=1$ , Step 2 shows that $\sigma $ maps $\gamma _{aa}(x)\gamma _{ab}(x)$ to $\pm \gamma ^{\prime }_{\tilde {a}a}(\sigma (x))\gamma ^{\prime }_{\tilde {a}b}(\sigma (x))$ . This time, the parity of the $\pm $ sign on each individual $\gamma _w\mapsto \gamma ^{\prime }_{\tilde {w}}$ is

(41) $$ \begin{align} \operatorname{\mathrm{Par}}(\sigma,xw) + \operatorname{\mathrm{Par}}(\sigma,xwaa) + \operatorname{\mathrm{Par}}(\sigma,xwab) + \operatorname{\mathrm{Par}}(\sigma,xwb), \end{align} $$

where the extra $\operatorname {\mathrm {Par}}(\sigma ,xw)$ term (as compared with expression (39)) is because swapping the two terms comprising $\gamma '$ introduces a factor of $-1$ .

Thus, $\sigma (\gamma _{aa}\gamma _{ab}\gamma _{ba}\gamma _{bb}) = \pm \gamma ^{\prime }_{\tilde {a}a}\gamma ^{\prime }_{\tilde {a}b}\gamma _{\tilde {b}a}\gamma _{\tilde {b}b}$ , where the parity of the sign is the sum of expression (39) for $w=ba,bb$ plus the sum of expression (41) for $w=aa,ab$ . By equation (40), this sum is off by $1$ from $R_{3,2}(\sigma ,x)$ . On the other hand, equations (37) and (40) dictate that $\sigma (U_a)=U_{\tilde {a}}$ but $\sigma (U_b)=-U_{\tilde {b}}$ . We observe that

$$ \begin{align*} \frac{ \gamma^{\prime}_{\tilde{a}a} \gamma^{\prime}_{\tilde{a}b} \gamma_{\tilde{b}a} \gamma_{\tilde{b}b} } { 4(c^2+c+1) (2+U_{\tilde{a}} - U_{\tilde{b}} )} \cdot & \frac{4(c^2+c+1) (2+U_{a} + U_{b})} {\gamma_{aa}\gamma_{ab}\gamma_{ba}\gamma_{bb}} \\ & = \frac{\gamma^{\prime}_{\tilde{a}a} \gamma^{\prime}_{\tilde{a}b} }{\gamma_{\tilde{a}a} \gamma_{\tilde{a}b}} \cdot \frac{2 + U_{\tilde{a}} + U_{\tilde{b}}}{2+U_{\tilde{a}} - U_{\tilde{b}}} = \bigg(\! - \frac{U_{\tilde{a}}}{2 + U_{\tilde{b}} }\! \bigg) \cdot \bigg(\! \frac{2 + U_{\tilde{b}} }{U_{\tilde{a}}} \!\bigg) =-1, \end{align*} $$

by the identities from Step 1. Once again, then, the image of expression (26) under $\sigma $ is $(-1)^{R_{3,2}(\sigma ,x)}\sqrt {2}$ .

Case 3: The parities in equation (37) are $V(a)=1$ and $V(b)=0$ . This case is the same as Case 2 with the roles of a and b reversed.

Case 4: The parities $V(a)$ and $V(b)$ in equation (37) are both odd, that is,

(42) $$ \begin{align} \operatorname{\mathrm{Par}}(\sigma,xaa) + \operatorname{\mathrm{Par}}(\sigma,xab) = \operatorname{\mathrm{Par}}(\sigma,xba) + \operatorname{\mathrm{Par}}(\sigma,xbb) = 1 \in\mathbb{Z}/2\mathbb{Z}. \end{align} $$

In formula (28) defining $R_{3,2}(\sigma ,x)$ , then, the four terms $\operatorname {\mathrm {Par}}(\sigma ,xw)$ sum to $0$ , while the product on the second line of that formula is $1$ .

Because $V(a)=V(b)=1$ , $\sigma $ maps each $\gamma _w(x)$ to $\pm \gamma ^{\prime }_{\tilde {w}}(\sigma (x))$ , where the parity of the $\pm $ sign is given by expression (41), for the reasons discussed in Case 2. Summing these expressions together is off from $R_{3,2}(\sigma ,x)$ by $1$ , because of the aforementioned expression on the second line of formula (28).

At the same time, equations (37) and (42) dictate that $\sigma (U_a)=-U_{\tilde {a}}$ and $\sigma (U_b)=-U_{\tilde {b}}$ . The relevant quotient is therefore

$$ \begin{align*} \frac{ \gamma^{\prime}_{\tilde{a}a} \gamma^{\prime}_{\tilde{a}b} \gamma^{\prime}_{\tilde{b}a} \gamma^{\prime}_{\tilde{b}b} } { 4(c^2+c+1) (2-U_{\tilde{a}} - U_{\tilde{b}} )} & \cdot \frac{4(c^2+c+1) (2+U_{a} + U_{b})} {\gamma_{aa}\gamma_{ab}\gamma_{ba}\gamma_{bb}}\\& = \frac{\gamma^{\prime}_{aa} \gamma^{\prime}_{ab} }{\gamma_{aa} \gamma_{ab}} \cdot \frac{\gamma^{\prime}_{ba} \gamma^{\prime}_{bb} }{\gamma_{ba} \gamma_{bb}} \cdot \frac{2 + U_a + U_b}{2 - U_a - U_b} \\ & = \bigg( - \frac{U_a}{2 + U_b} \bigg) \cdot \bigg( - \frac{U_b}{2 + U_a} \bigg)\\ & \cdot \bigg( - \frac{(2 + U_a)(2+U_b)}{U_a U_b} \bigg) =-1, \end{align*} $$

by the identities from Step 1. Yet again, then, the image of expression (26) under $\sigma $ is $(-1)^{R_{3,2}(\sigma ,x)}\sqrt {2}$ , as desired.

Step 4. Having verified equation (36) for every $\sigma \in G_{\infty }$ , it follows that $R_{3,2}(\sigma ,x_1)=R_{3,2}(\sigma ,x_2)$ for all nodes $x_1$ , $x_2$ of the tree, and hence that $\sigma \in \tilde {M}_{3,2,\infty }$ . (Recall that we already observed that $\sigma \in M_{3,2,\infty }$ because of its action on $\zeta _4$ , by Theorem 3.4.)

Finally, if $\zeta _4,\sqrt {2}\in K$ , then $\sigma $ must fix both of these elements, and hence $\sigma \in B_{3,2,\infty }$ with $R_{3,2}(\sigma )=0$ . That is, $\sigma \in \tilde {B}_{3,2,\infty }$ .

Proof By Proposition 2.1 and the fact that $f^r(0)=-f(0)$ , we have

$$\begin{align*}\big( \alpha_1 \dots \alpha_{m} \alpha^{\prime}_1 \dots \alpha^{\prime}_{m} \big)^2 = f^{r}(0) - y = -y - f(0) = w_2^2. \end{align*}$$

Equation (43) follows immediately.

Assume for the remainder of the proof that the product in equation (43) is $w_2$ . Let $\gamma \in K$ be the expression in equation (44). Then by Proposition 2.1 again, we have

$$\begin{align*}\gamma^2 = \frac{(f^{r-1}(0) - w_1) + 2w_2 + (f^{r-1}(0) + w_1)} {f^{r-1}(0) + w_2} = \frac{2(f^{r-1}(0)+w_2)}{f^{r-1}(0) + w_2} = 2. \\[-42pt] \end{align*}$$

Proof We will label the tree inductively, starting from the root point $x_0$ . Label the tree (using the symbols a and b) arbitrarily up to level r.

For each successive $n\geq r+1$ , suppose that we have labeled $T_{n-1}$ in such a way that for every node x of $T_{n-1}$ up to level $n-r-2$ , the desired identities hold. (This condition is vacuously true for $n = r+1$ .)

For each node y at level $n-1$ , label the two points of $f^{-1}(y)$ arbitrarily as $ya$ and $yb$ . We will now adjust these labels, if necessary.

For each node x at level $n-r-1$ , we have

$$\begin{align*}E_{xaa}E_{xab} = \pm [xba] \quad\text{and}\quad E_{xba}E_{xbb} = \pm [xaa], \end{align*}$$

by equation (43) of Lemma 5.1. If the first is $-[xba]=[xbb]$ , then exchange the labels $ya$ and $yb$ above a single node $y\in f^{-(r-1)}([xa])$ , and the adjusted labeling then satisfies $E_{xaa}E_{xab} = [xba]$ . If the second is $-[xaa]=[xab]$ , make a similar swap above a single node $y\in f^{-(r-1)}([xb])$ . Equation (45) now holds for x.

Still for the same node x at level $n-r-1$ , having made these adjustments, call the first four quantities in equation (46) $A_1,A_2,B_1,B_2$ , respectively. By equation (44) of Lemma 5.1, each is $\pm \sqrt {2}$ . (The negative sign in the numerator of $A_2$ is there because $E_{xaa} E_{xab}=[xba]=-[xbb]$ , so to get the desired value $[xbb]$ on the right side of equation (43), we replace $\alpha ^{\prime }_1 = [xabaa\dots a]$ by its negative. Similar reasoning yields the negative sign in the numerator of $B_2$ .) By Proposition 2.1, we have

$$\begin{align*}A_1 A_2 = \frac{ E_{xab}^2 {-} E_{xaa}^2}{E_{xba} E_{xbb}} = \frac{(f^{r-1}(0) - [xab]) - (f^{r-1}(0) {-} [xaa])}{[xaa]} = \frac{[xaa] {+} [xaa]}{[xaa]} = 2,\end{align*}$$

since $[xab]=-[xaa]\neq 0$ . Since both are square roots of $2$ , it follows that $A_1=A_2$ . Similarly, $B_1=B_2$ . We also have

$$ \begin{align*} \frac{A_1}{B_1} &= \frac{E_{xaa}E_{xab} + E_{xab}^2}{E_{xba}E_{xbb} + E_{xbb}^2} = \frac{[xba] + (f^{r-1}(0) - [xab])}{[xaa] + (f^{r-1}(0) - [xbb])} \\ &= \frac{f^{r-1}(0) + [xaa] + [xba]}{f^{r-1}(0) + [xba] + [xaa]} = 1, \end{align*} $$

where the second equality is by Proposition 2.1 and (the now verified) equation (45), and the third equality is because $[xab]=-[xaa]$ and $[xbb]=-[xba]$ . Thus, we have $A_1=A_2=B_1=B_2=\pm \sqrt {2}$ .

If this common value is $-\sqrt {2}$ , then pick $y_1\in f^{-(r-2)}([xaa])$ , and exchange the labels of $y_1 a$ and $y_1 b$ . Also pick $y_2\in f^{-(r-2)}([xab])$ , and exchange the labels of $y_2 a$ and $y_2 b$ . These two switches have the effect of sending each of $E_{xaa}$ and $E_{xab}$ to its negative. Thus, equation (45) remains true, but each of $A_1,A_2,B_1,B_2$ is replaced by its negative, that is, equation (46) holds.

Having made these (possible) label changes to various nodes at the nth level of the tree, we have labeled $T_n$ so that for every node x at every level $0\leq \ell \leq n-r-1$ of $T_n$ , equations (45) and (46) hold. Thus, our inductive construction of the desired labeling is complete.

Proof Step 1. We claim that for all $\sigma \in \tilde {M}_{r,1,\infty }$ , all $\tau \in \operatorname {\mathrm {Aut}}(T_\infty )$ , and all nodes x of the tree, we have

(48) $$ \begin{align} \operatorname{\mathrm{Par}}(\tau,xa) \operatorname{\mathrm{Par}}\big(\sigma,\tau(xb) \big) + \!\sum_{w\in \{a,b\}^{r-2}} \operatorname{\mathrm{Par}}\big( \sigma,\tau(xab)w \big) = \!\sum_{w\in \{a,b\}^{r-2}} \operatorname{\mathrm{Par}}\big( \sigma,\tau(xa)bw \big), \end{align} $$

where we continue to work modulo 2.

If $\operatorname {\mathrm {Par}}(\tau ,xa)=0$ , then $\tau (xab)=\tau (xa)b$ , and the first term in equation (48) is zero, so our claim holds trivially.

Otherwise, we have $\operatorname {\mathrm {Par}}(\tau ,xa)=1$ , and hence $\tau (xab)=\tau (xa)a$ . Then because we are working modulo 2, the difference between the two sides of equation (48) is

$$ \begin{align*} \operatorname{\mathrm{Par}}\big( \sigma,\tau(xb) \big) & + \sum_{w\in \{a,b\}^{r-2}} \Big( \operatorname{\mathrm{Par}}\big( \sigma,\tau(xa)aw \big) + \operatorname{\mathrm{Par}}\big( \sigma,\tau(xa)bw \big) \Big) \\ &= \operatorname{\mathrm{Par}}\big( \sigma,\tau(xb) \big) + \sum_{w'\in \{a,b\}^{r-1}} \operatorname{\mathrm{Par}}\big( \sigma,\tau(xa)w' \big) \\ &= \begin{cases} P_{r,1}^a (\sigma,\tau(x)) \text{ if } \operatorname{\mathrm{Par}}(\tau,x)=0, \\ P_{r,1}^b (\sigma,\tau(x)) \text{ if } \operatorname{\mathrm{Par}}(\tau,x)=1. \end{cases} \end{align*} $$

Since $\sigma \in B_{r,1,\infty }$ , this value is $0$ , and hence equation (48) holds, proving our claim.

By similar reasoning, we also have

(49) $$ \begin{align} \operatorname{\mathrm{Par}}(\tau,xb) \operatorname{\mathrm{Par}}\big(\sigma,\tau(xa) \big) + \!\sum_{w\in \{a,b\}^{r-2}} \operatorname{\mathrm{Par}}\big( \sigma,\tau(xbb)w \big) =\! \sum_{w\in \{a,b\}^{r-2}} \operatorname{\mathrm{Par}}\big( \sigma,\tau(xb)bw \big). \end{align} $$

Step 2. Again with $\sigma ,\tau ,x$ as in Step 1, we now claim that

(50) $$ \begin{align} R_{r,1}(\sigma) + R_{r,1}(\tau,x) = R_{r,1}(\sigma \tau,x). \end{align} $$

Indeed, by equation (3) and the definition of $R_{r,1}$ , we have

$$ \begin{align*} R_{r,1}(\sigma\tau,x) &= \Big( \operatorname{\mathrm{Par}}\big(\sigma,\tau(xa) \big) + \operatorname{\mathrm{Par}}(\tau,xa) \Big) \Big( \operatorname{\mathrm{Par}}\big(\sigma,\tau(xb) \big) + \operatorname{\mathrm{Par}}(\tau,xb) \Big) \\ & \phantom{\operatorname{\mathrm{Par}}} + \sum_{w\in\{a,b\}^{r-2}} \big( \operatorname{\mathrm{Par}}(\tau,xabw) + \operatorname{\mathrm{Par}}(\tau,xbbw) \big) \\ & \phantom{\operatorname{\mathrm{Par}}} + \sum_{w\in\{a,b\}^{r-2}} \Big( \operatorname{\mathrm{Par}}\big(\sigma,\tau(xabw)\big) + \operatorname{\mathrm{Par}}\big(\sigma,\tau(xbbw)\big) \Big) \\ &= R_{r,1}(\tau,x) + \operatorname{\mathrm{Par}}\big(\sigma,\tau(xa) \big) \operatorname{\mathrm{Par}}\big(\sigma,\tau(xb) \big) \\ & \phantom{\operatorname{\mathrm{Par}}} + \operatorname{\mathrm{Par}}(\tau,xa) \operatorname{\mathrm{Par}}\big(\sigma,\tau(xb) \big) + \sum_{w\in \{a,b\}^{r-2}} \operatorname{\mathrm{Par}}\big( \sigma,\tau(xab)w \big) \\ & \phantom{\operatorname{\mathrm{Par}}} + \operatorname{\mathrm{Par}}(\tau,xb) \operatorname{\mathrm{Par}}\big(\sigma,\tau(xa) \big) + \sum_{w\in \{a,b\}^{r-2}} \operatorname{\mathrm{Par}}\big( \sigma,\tau(xbb)w \big) \\ &= R_{r,1}(\tau,x) + \operatorname{\mathrm{Par}}\big(\sigma,\tau(xa) \big) \operatorname{\mathrm{Par}}\big(\sigma,\tau(xb) \big) \\ & \phantom{\operatorname{\mathrm{Par}}} + \sum_{w\in \{a,b\}^{r-2}} \Big( \operatorname{\mathrm{Par}}\big( \sigma,\tau(xa)bw \big) + \operatorname{\mathrm{Par}}\big( \sigma,\tau(xb)bw \big) \Big) \\ &= R_{r,1}(\tau,x) + \operatorname{\mathrm{Par}}\big(\sigma,\tau(x)a \big) \operatorname{\mathrm{Par}}\big(\sigma,\tau(x)b \big) \\ & \phantom{\operatorname{\mathrm{Par}}} + \sum_{w\in \{a,b\}^{r-2}} \Big( \operatorname{\mathrm{Par}}\big( \sigma,\tau(x)abw \big) + \operatorname{\mathrm{Par}}\big( \sigma,\tau(x)bbw \big) \Big) \\ &=R_{r,1}(\tau,x) +R_{r,1}\big(\sigma,\tau(x)\big), \end{align*} $$

where the second equality is by grouping the terms of $R_{r,1}(\tau ,x)$ and rearranging the sums; the third is by equations (48) and (49); and the fourth is by observing that the two-element sets $\{\tau (xa),\tau (xb)\}$ and $\{\tau (x)a,\tau (x)b\}$ coincide. Thus, because $R_{r,1}(\sigma ,\tau (x))=R_{r,1}(\sigma )$ , we have proven equation (50).

Step 3. Once again, we will finish our proof via Proposition 2.2. Let $\Gamma :=B_{r,1,\infty }$ , which we know to be a group by Theorem 3.3. Let $H=H_0:=\mathbb {Z}/2\mathbb {Z}$ , and let $P(\sigma ,x):=R_{r,1}(\sigma ,x)$ .

Clearly, the identity $e\in \operatorname {\mathrm {Aut}}(T_{\infty })$ satisfies $R_{r,1}(e,x)=0$ for all nodes x, so that $e\in \tilde {M}_{r,1,\infty }$ , which is the group G of Proposition 2.2. In addition, equation (50) provides precisely hypothesis (6). By Proposition 2.2, then, $\tilde {M}_{r,1,\infty }$ is a subgroup of $B_{r,1,\infty }$ , and $R_{r,1}$ is a homomorphism. By definition, its kernel is $\tilde {B}_{r,1,\infty }$ , which is therefore a subgroup of $\tilde {M}_{r,1,\infty }$ .

Proof It suffices to prove equation (51) for all $x\in \operatorname {\mathrm {Orb}}_f^-(x_0)$ and $\sigma \in G_{\infty }$ . Indeed, in that case, we have that $\sigma \in B_{r,1,\infty }$ from Definition 1.1, and that equation (47) holds for all nodes $x_1,x_2$ of the tree, since $\sigma (\sqrt {2})/\sqrt {2} \in \{\pm 1\}$ must be independent of x. That is, equation (51) implies that every $\sigma \in G_{\infty }$ is in $\tilde {M}_{r,1,\infty }$ . Furthermore, if $\sqrt {2}\in K$ , then it would follow that $R_{r,1}(\sigma )=0$ for all $\sigma \in G_{\infty }$ , and hence that every such $\sigma $ lies in $\tilde {B}_{r,1,\infty }$ .

Note that for any $y\in \operatorname {\mathrm {Orb}}_f^-(x_0)$ , we have

(52) $$ \begin{align} \sigma(E_y) = (-1)^{R(\sigma,y)} E_{\sigma(y)}, \quad\text{where } R(\sigma,y) := \sum_{w\in\{a,b\}^{r-2}} \operatorname{\mathrm{Par}}(\sigma,yw). \end{align} $$

Indeed, equation (52) is immediate from the definition of $E_y$ in Lemma 5.2, combined with the fact that for each node u of the tree, we have

$$\begin{align*}\sigma\big([ua]\big) = (-1)^{\operatorname{\mathrm{Par}}(\sigma,u)} \big[ \sigma(u) a \big] .\end{align*}$$

Step 1. We begin by proving the first part of equation (51). Given $x\in \operatorname {\mathrm {Orb}}_f^-(x_0)$ and $\sigma \in G_{\infty }$ , if $\operatorname {\mathrm {Par}}(\sigma ,x)=0$ , so that $\sigma (xa)=\sigma (x)a$ and $\sigma (xb)=\sigma (x)b$ , then by the first equation of (45) and the fact that the two-element sets $\{\sigma (xaa),\sigma (xab)\}$ and $\{\sigma (x)aa,\sigma (x)ab\}$ coincide, we have

$$ \begin{align*} 1 &= \sigma(1) = \frac{\sigma(E_{xaa}) \sigma(E_{xab})}{\sigma( [xba] ) } = (-1)^{R(\sigma,xaa) + R(\sigma,xab)} \frac{E_{\sigma(x)aa} E_{\sigma(x)ab} } {(-1)^{\operatorname{\mathrm{Par}}(\sigma,xb)} [\sigma(x)ba]} \\ &= (-1)^{P_{r,1}^a(\sigma,x)} \frac{E_{\sigma(x)aa} E_{\sigma(x)ab} }{[\sigma(x)ba]} = (-1)^{P_{r,1}^a(\sigma,x)}. \end{align*} $$

(Here, the second and fifth equalities are by the choice of labeling of the tree, the third is by equation (52), and the fourth is by the definitions of $P_{r,1}^a$ and R.) On the other hand, if $\operatorname {\mathrm {Par}}(\sigma ,x)=1$ , then $\sigma (xa)=\sigma (x)b$ and $\sigma (xb)=\sigma (x)a$ , and hence

$$ \begin{align*} 1 &= \sigma(1) = \frac{\sigma(E_{xaa}) \sigma(E_{xab})}{\sigma( [xba] ) } = (-1)^{R(\sigma,xaa) + R(\sigma,xab)} \frac{E_{\sigma(x)ba} E_{\sigma(x)bb} } {(-1)^{\operatorname{\mathrm{Par}}(\sigma,xb)} [\sigma(x)aa]} \\ &= (-1)^{P_{r,1}^a(\sigma,x)} \frac{E_{\sigma(x)ba} E_{\sigma(x)bb} }{[\sigma(x)aa]} = (-1)^{P_{r,1}^a(\sigma,x)}. \end{align*} $$

Either way, then, we have $P_{r,1}^a(\sigma ,x)=0$ in $\mathbb {Z}/2\mathbb {Z}$ . By similar reasoning applied to the second equation of (45), we also have $P_{r,1}^b(\sigma ,x)=0$ , verifying the first portion of equation (51).

Step 2. Given $x\in \operatorname {\mathrm {Orb}}_f^-(x_0)$ and $\sigma \in G_{\infty }$ , then by equation (52) and property (46) of the specified labeling of the tree, we have

$$\begin{align*}(-1)^{R(\sigma,xaa)} E_{\sigma(xaa)} + (-1)^{R(\sigma,xab)} E_{\sigma(xab)} =\sigma(\sqrt{2}) \cdot (-1)^{R(\sigma,xbb)} E_{\sigma(xbb)} .\end{align*}$$

Multiplying both sides of this equation by $(-1)^{R(\sigma ,xab)} / E_{\sigma (xbb)}$ , and applying the identity $P^a_{r,1}=0$ proven in Step 1, that is, $R(\sigma ,xaa)+R(\sigma ,xab)=\operatorname {\mathrm {Par}}(\sigma ,xb)$ , we obtain

(53) $$ \begin{align} (-1)^{R(\sigma,xab)+R(\sigma,xbb)} \sigma(\sqrt{2}) = \frac{ E_{\sigma(xab)} + (-1)^{\operatorname{\mathrm{Par}}(\sigma,xb)} E_{\sigma(xaa)} } {E_{\sigma(xbb)} }. \end{align} $$

We consider three cases. First, if $\operatorname {\mathrm {Par}}(\sigma ,xb)=0$ , then equation (53) becomes

$$\begin{align*}(-1)^{R(\sigma,xab)+R(\sigma,xbb)} \sigma(\sqrt{2}) = \frac{ E_{\sigma(xa)a} + E_{\sigma(xa)b} } {E_{\sigma(xb)b} } = \sqrt{2}, \end{align*}$$

by considering the first expression in equation (46) if $\operatorname {\mathrm {Par}}(\sigma ,x)=0$ , or the third if $\operatorname {\mathrm {Par}}(\sigma ,x)=1$ . Here, we have used the fact that the two-element sets $\{\sigma (xaa),\sigma (xab)\}$ and $\{\sigma (xa)a,\sigma (xa)b\}$ coincide.

Second, if $\operatorname {\mathrm {Par}}(\sigma ,xa)=0$ and $\operatorname {\mathrm {Par}}(\sigma ,xb)=1$ , then equation (53) becomes

$$\begin{align*}(-1)^{R(\sigma,xab)+R(\sigma,xbb)} \sigma(\sqrt{2}) = \frac{ E_{\sigma(xa)b} - E_{\sigma(xa)a} } {E_{\sigma(xb)a} } = \sqrt{2}, \end{align*}$$

by considering the second expression in equation (46) if $\operatorname {\mathrm {Par}}(\sigma ,x)=0$ , or the fourth if $\operatorname {\mathrm {Par}}(\sigma ,x)=1$ .

Third, if $\operatorname {\mathrm {Par}}(\sigma ,xa)=\operatorname {\mathrm {Par}}(\sigma ,xb)=1$ , then equation (53) becomes

$$\begin{align*}(-1)^{R(\sigma,xab)+R(\sigma,xbb)} \sigma(\sqrt{2}) = \frac{ E_{\sigma(xa)a} - E_{\sigma(xa)b} } {E_{\sigma(xb)a} } = - \sqrt{2}, \end{align*}$$

again by considering the second or fourth expression in equation (46).

Recalling that

$$\begin{align*}R_{r,1}(\sigma,x)=\operatorname{\mathrm{Par}}(\sigma,xa)\operatorname{\mathrm{Par}}(\sigma,xb) + R(\sigma,xab)+R(\sigma,xbb) ,\end{align*}$$

then in all three cases, we have $\sigma (\sqrt {2}) = (-1)^{R_{r,1}(\sigma ,x)}\sqrt {2}$ , as desired.

Proof Let X be the set of nodes of $T_{\infty }$ . Each of the subgroups G in question is defined by relations involving functions $P:\operatorname {\mathrm {Aut}}(T_{\infty })\times X\to \mathbb {Z}/2\mathbb {Z}$ for which $P(\sigma ,x)$ depends only on $\operatorname {\mathrm {Par}}(\sigma ,y)$ for nodes y lying a bounded number of levels above x. (Here, P is any of $P^a_{r,s}$ or $P^b_{r,s}$ or $R_{3,2}$ or $R_{r,1}$ .) By definition of the profinite topology, then, for each $x\in X$ , the function $\sigma \mapsto P(\sigma ,x)$ is continuous, and thus each of the equations defining the subgroup G (for a single $x\in X$ or a pair $x_1,x_2\in X$ ) defines a closed subset of $\operatorname {\mathrm {Aut}}(T_{\infty })$ . The subgroup G is therefore an intersection of closed sets and hence is closed.

Proof By Lemma 6.1, it suffices to prove that each of the generators $\boldsymbol {\alpha }_1,\ldots ,\boldsymbol {\alpha }_r$ of Pink’s group belongs to $B_{r,s,\infty }$ . As in equation (57), let $\ell :=r-s\geq 1$ , and let $w_{\ell }:= b a^{\ell -1}$ .

Fix a node x of the tree, and an integer $i\in \{ 1,\ldots , r\}$ . Recalling Definition 1.1, we must show that $P_{r,s}^t(\boldsymbol {\alpha }_i,x)\equiv 0\pmod {2}$ for both $t=a$ and $t=b$ . We consider three cases.

First, suppose $i\leq s$ . Then $\operatorname {\mathrm {Par}}(\boldsymbol {\alpha }_i,w)=0$ for all words w of length at least s. In particular, all of the terms appearing in the definition of $P_{r,s}^a(\boldsymbol {\alpha }_i,x)$ and $P_{r,s}^b(\boldsymbol {\alpha }_i,x)$ are zero, as desired.

Second, suppose that $i=s+1+j$ for some $0\leq j \leq \ell -1$ , and that there is some node y exactly s levels above x for which $\operatorname {\mathrm {Par}}(\boldsymbol {\alpha }_i, y) =1$ . Then according to equation (57), we must have $y=a^j w_{\ell }^n a^s$ for some $n\geq 0$ , and hence $x=a^j w_{\ell }^n$ . Thus, $y=xaa^{s-1}$ is the only node s levels above x for which $\operatorname {\mathrm {Par}}(\boldsymbol {\alpha }_i, y) =1$ , and $y':=xw_{\ell }a^s = xb a^{r-1}$ is the only node r levels above x for which $\operatorname {\mathrm {Par}}(\boldsymbol {\alpha }_i, y') =1$ . Therefore,

$$\begin{align*}P_{r,s}^a(\boldsymbol{\alpha}_i,x) = 0 + 0 = 0, \quad\text{and}\quad P_{r,s}^b(\boldsymbol{\alpha}_i,x) = 1 + 1 \equiv 0 \quad\pmod{2}, \end{align*}$$

as desired.

Third, suppose that $i=s+1+j$ for some $0\leq j \leq \ell -1$ , and that $\operatorname {\mathrm {Par}}(\boldsymbol {\alpha }_i, y) =0$ for all nodes y that are s levels above x. Again by equation (57), x cannot be of the form $a^j w_{\ell }^n$ for any $n\geq 0$ , and therefore we must have $\operatorname {\mathrm {Par}}(\boldsymbol {\alpha }_i, y') =0$ for all nodes $y'$ that are r levels above x, since the word $w_{\ell } a^s$ has length r. Thus, all of the terms appearing in the definition of $P_{r,s}^a(\boldsymbol {\alpha }_i,x)$ and $P_{r,s}^b(\boldsymbol {\alpha }_i,x)$ are zero, as desired.

Proof For each integer $n\geq 1$ , define

$$\begin{align*}B^{\prime}_{r,s,n}:= \{ \sigma\in \operatorname{\mathrm{Aut}}(T_n) : \forall m<n-r \text{ and } \forall x\in\{a,b\}^m, \, P^a_{r,s}(\sigma,x)=P^b(\sigma,x)=0\},\end{align*}$$

where we recall $P^a_{r,s}$ and $P^b_{r,s}$ take values in $\mathbb {Z}/2\mathbb {Z}$ . By Proposition 6.2 and Definition 1.1, we have