Proof. Note that $x\in \Lambda _{(0,0)}(H_{>\nu \lambda e^M})$ implies that $x\in P(0,0)\times (M,\infty )$ . Let

$$ \begin{align*}y\in B(\mathcal{Z}_{\nu}(x),\alpha^{-1}R )\cap H_{>\nu\lambda e^M}.\end{align*} $$

Then by (2.1) we have that

$$ \begin{align*} |x-\Lambda_{(0,0)}(y)|=|\Lambda_{(0,0)}(\mathcal{Z}_{\nu}(x))-\Lambda_{(0,0)}(y)|\leq \alpha|\mathcal{Z}_{\nu}(x)-y|<R. \end{align*} $$

Hence, $\Lambda _{(0,0)}(y)\in B(x,R)\cap P(0,0)$ and thus $y=\mathcal {Z}_{\nu }(\Lambda _{(0,0)}(y))\in \mathcal {Z}_{\nu }(B(x,R)\cap H_{\geq M})$ .

Proof Proof of Proposition 1

Let us fix a point $x=(0,0,x_0)$ on the $x_3$ -axis and consider a neighbourhood U of that point. It is easy to see now that $\mathcal {Z}^{k}_{\nu }(x)=(0,0,E^k_{\nu \lambda }(x_0))$ , where $E_{\nu \lambda }^k$ denotes the kth iterate of the map $E_{\nu \lambda }(t)=\nu \lambda e^t$ . Since the $x_3$ -axis is invariant under our Zorich map and since $\nu \lambda> 1/e$ , we have that $E_{\nu \lambda }^k(x)\to \infty $ and thus we may assume that $x\in H_{\geq M}$ , for some $M>M_0$ . By repeatedly applying (2.2) we may now obtain a sequence $R_k\to \infty $ with

$$ \begin{align*} \mathcal{Z}_{\nu}^k(U)\supset B(\mathcal{Z}_{\nu}^k(x),R_k)\cap H_{E_{\nu\lambda}^k(M)}, \end{align*} $$

and we note that the intersection on the right-hand side always contains the upper half of the ball $B(\mathcal {Z}_{\nu }^k(x),R_k)$ . Hence, for large enough k, the set

$$ \begin{align*} V_k=\{(x_1,x_2)\in\mathbb{R}^2:|x_1|\leq 2\lambda,|x_2|\leq2\lambda\}\times [E_{\nu\lambda}^k(x_0),E_{\nu\lambda}^k(x_0)+R_k/2] \end{align*} $$

is contained in $B(\mathcal {Z}_{\nu }^k(x),R_k)\cap H_{E_{\nu \lambda }^k(M)}$ . Observe now that $\mathcal {Z}_{\nu }$ maps $V_k$ onto the shell

$$ \begin{align*} A_k=\{x\in\mathbb{R}^3:\nu\lambda \exp({E_{\nu\lambda}^k(x_0)})\leq|x|\leq \nu\lambda \exp({E_{\nu\lambda}^k(x_0)+R_k/2})\}. \end{align*} $$

It is easy to see now that this shell, for large enough k and since $R_k\to \infty $ , contains a set of the form

$$ \begin{align*} \{(x_1,x_2,x_3)\in\mathbb{R}^3:|x_1-2\lambda q_{k,1}|\leq 2\lambda,|x_2-2\lambda q_{k,2}|\leq2\lambda,|x_3|\leq t_k\}, \end{align*} $$

with $q_{k,1}, q_{k,2}\in \mathbb {Z}$ and $t_k\to \infty $ . This implies that

$$ \begin{align*} \{x\in\mathbb{R}^3:\nu\lambda e^{-t_k}\leq|x|\leq \nu\lambda e^{t_k}\}\subset \mathcal{Z}_{\nu}(A_k)\subset\mathcal{Z}_{\nu}^{k+2}(U). \end{align*} $$

Hence $\bigcup _{k=1}^{\infty }\mathcal {Z}_{\nu }^k(U)=\mathbb {R}^3\setminus \{(0,0,0)\}$ , which means that $x\in \mathcal {J}(\mathcal {Z}_{\nu })$ .

Proof. It is enough to find a lower bound for ${\operatorname {Im}}\ z\in [(4k-1)\sqrt 2,(4k+1)\sqrt 2]$ . This is true because for other z we have that $T(z)=\bar {z}+2\sqrt 2 i$ has imaginary part in $[(4k'-1)\sqrt 2,(4k'+1)\sqrt 2]$ for some $k'\in \mathbb {Z}$ and thus for those z we have $g(z)=g(T(z))=\psi (T(z))$ . Then by the chain rule we have that $Dg(z)=Dg(T(z))DT(z)$ . Since $DT(z)=-1$ this implies that $|\!\det Dg(z)|=|\!\det Dg(T(z))|$ .

With that in mind, if $z=x+iy$ we have that $Dg(z)$ is the linear transformation induced by the matrix

$$ \begin{align*} \begin{pmatrix} \nu\lambda e^{\lambda x}a(y)& \nu e^{\lambda x} \dfrac{da}{dy}({y})\\[12pt] \nu\lambda e^{\lambda x}b(y)& \nu e^{\lambda x} \dfrac{db}{dy}(y) \end{pmatrix}=\begin{pmatrix} \nu\lambda e^{\lambda x}\mathfrak{h}_3\bigg(\dfrac{y}{\sqrt2}\bigg)& \dfrac{\nu}{\sqrt2} e^{\lambda x} \dfrac{d\mathfrak{h}_3}{dy}\bigg(\dfrac{y}{\sqrt2}\bigg)\\[12pt] \nu\lambda\sqrt2 e^{\lambda x}\mathfrak{h}_1\bigg(\dfrac{y}{\sqrt2}\bigg)& \nu e^{\lambda x} \dfrac{d\mathfrak{h}_1}{dy}\bigg(\dfrac{y}{\sqrt2}\bigg) \end{pmatrix}, \end{align*} $$

$\hspace {1mm}\text {if}\hspace {1mm} y\in [(4k-1)\sqrt 2,(4k+1)\sqrt 2]$ .

Thus

$$ \begin{align*}|\!\det Dg(z)|&=\nu^2\lambda e^{2\lambda x}\bigg|\mathfrak{h}_3(y)\dfrac{d\mathfrak{h}_1}{dy}(y)-\mathfrak{h}_1(y)\dfrac{d\mathfrak{h}_3}{dy}(y)\bigg|\\ &=\nu^2\lambda e^{2\lambda x} \left|\det\begin{pmatrix} a(y)& \dfrac{da}{dy}({y})\\[8pt] b(y)& \dfrac{db}{dy}(y)\end{pmatrix}\right|\quad\text{for } y\in[(4k-1)\sqrt2,(4k+1)\sqrt2].\end{align*} $$

We now claim that

$$ \begin{align*} \left|\det\begin{pmatrix} a(y)& \dfrac{da}{dy}({y})\\[8pt] b(y)& \dfrac{db}{dy}(y)\end{pmatrix}\right|>\dfrac{1}{L}\ \text{a.e.}\end{align*} $$

Hence

$$ \begin{align*}|\!\det Dg(z)|\geq\dfrac{\nu^2\lambda e^{2\lambda x}}{L}\ \text{a.e.}\end{align*} $$

Indeed, because $a^2(y)+b^2(y)=1$ we obtain that the vectors $(a(y),b(y))$ and $({da}/{dy},{db}/{dy})$ are orthogonal and thus so is their matrix. This implies that

$$ \begin{align*} \left|\det\begin{pmatrix} a(y)& \dfrac{da}{dy}({y})\\[8pt] b(y)& \dfrac{db}{dy}(y)\end{pmatrix}\right|=|(a(y),b(y))|\bigg|\bigg(\dfrac{da}{dy}({y}), \dfrac{db}{dy}({y})\bigg)\bigg|=\bigg|\bigg(\dfrac{da}{dy}({y}),\dfrac{db}{dy}({y})\bigg)\bigg|.\end{align*} $$

Now because $\mathfrak {h}$ is a locally bi-Lipschitz map almost everywhere we have that

$$ \begin{align*}\bigg|\frac{d\mathfrak{h}}{dy}\bigg(\frac{y}{\sqrt2},\frac{y}{\sqrt2}\bigg)\bigg|\geq \frac{1}{L}\ \text{a.e.}\end{align*} $$

Since $|({da}/{dy})({y}),({db}/{dy})(y)|=|({d\mathfrak {h}}/{dy})({y}/{\sqrt 2},{y}/{\sqrt 2})|$ we obtain what we wanted.

Proof. We can assume that V lies on the right half-plane $\{z:{\operatorname {Re}}(z)>0\}$ , otherwise just consider $g(V)$ since g maps $\mathbb {C}$ to the right half-plane. We know that $m(g(V))=\int _{g(V)}\,dm$ . Since none of the iterates of V under g intersects the real axis, we have that those iterates also do not intersect any of the pre-images of the real axis, meaning the lines $y=2\sqrt 2k$ , $k\in ~\mathbb {Z}$ . Thus $g^n(V)$ is always inside strips of the form ${\{z\in \mathbb {C}:{\operatorname {Im}}\ z\in (2\sqrt 2k,2\sqrt 2(k+1))\}}$ . In each of those strips it is easy to see, by what we have said in the first paragraph of this section, that g is a two-to-one map. By using Lemma 2 and since ${\operatorname {Re}}\ z>0$ for all $z\in V$ we now obtain

$$ \begin{align*} \int_{g(V)}\,dm\geq\frac{1}{2}\int_V |\!\det (Dg)|\,dm\geq\frac{\nu^2\lambda}{2L} \int_V e^{2\lambda{\operatorname{Re}}(z)}\,dm\geq \frac{\nu^2\lambda}{2L}m(V). \end{align*} $$

This means that

$$ \begin{align*} m(g(V))\geq \frac{\nu^2\lambda }{2L}m(V). \end{align*} $$

Hence, since $\nu ^2\lambda>2L$ , if we iterate that inequality we have that $m(g^n(V))\to \infty $ .

Proof Proof of Theorem 5

Suppose, towards a contradiction, that there is a connected set V with $m(V)>0$ whose iterates never intersect the real axis. Then by Lemma 3 we have that $m(g^n(V))\to \infty $ as $n\to \infty $ . Since $g^n(V)$ never intersects the real axis it also does not intersect its pre-images, meaning the lines ${\operatorname {Im}}(z)=2\sqrt 2k$ , $k\in \mathbb {Z}$ . This means that $g^n(V)$ stays always inside strips of the form $\{ z\in \mathbb {C}:{\operatorname {Im}}(z)\in (2\sqrt 2k,2\sqrt 2(k+1)),k\in \mathbb {Z}\}$ . If we now take the pre-image of the lines ${\operatorname {Im}}(z)=2\sqrt 2m$ , $m\in \mathbb {Z}\setminus \{0\}$ , that lie inside all those strips we obtain curves $\gamma _m$ which the iterates $g^n(V)$ of our set must not cross (see Figure 1). By symmetry we can confine ourselves to the strip

$$ \begin{align*} S:=\{ z\in\mathbb{C}:{\operatorname{Im}}(z)\in(0,2\sqrt2)\}. \end{align*} $$

From now on let us write $\mathfrak {h}_i(y)$ instead of $\mathfrak {h}_i(y,y)$ , $i=1,2,3$ , for simplicity. We now have two cases to consider. Either S contains the pre-images of the lines ${\operatorname {Im}}(z)=2\sqrt 2m$ , $m>0$ , in which case $\mathfrak {h}_1(y)>0$ for $y>0$ , or S contains the pre-images of the lines for $m<0$ , in which case $\mathfrak {h}_1(y)<0$ for $y>0$ . We will only consider the first case here. The second one can be dealt with similarly. So suppose $m>0$ . We claim that the curves $\gamma _m$ partition the strip S into sets of uniformly bounded area. In fact if we call $A_m$ the area of the set defined by $\gamma _m$ and $\gamma _{m+1}$ and we call $A_0$ the area inside the strip between the imaginary axis and $\gamma _1$ , then we claim that $A_m$ is an eventually decreasing sequence. Clearly, since $m(g^n(V))\to \infty $ and $g^n(V)$ must stay inside those sets $A_m$ , this is impossible.

In order to prove those claims note that the curves $\gamma _m$ are given by the equations $\nu e^{\lambda x}\mathfrak {h}_1({y}/{\sqrt 2})=2m,$ when $y\in (0,\sqrt 2]$ , and $\nu e^{\lambda x}\mathfrak {h}_1(({2\sqrt 2-y})/{\sqrt 2})=2m$ , when ${y\in [\sqrt 2,2\sqrt 2)}$ . It is also easy to see that those curves do not have self-intersections and do not intersect with each other. The area we are looking for will be given by

$$ \begin{align*} A_m&=\int_{0}^{\sqrt2}\frac{1}{\lambda}\bigg(\log\frac{2(m+1)}{\nu \mathfrak{h}_1({y}/{\sqrt2})}-\log\frac{2m}{\nu \mathfrak{h}_1({y}/{\sqrt2})}\bigg)\,dy\\ &\quad+\int_{\sqrt2}^{2\sqrt2}\frac{1}{\lambda}\bigg(\log\frac{2(m+1)}{\nu \mathfrak{h}_1 (({2\sqrt2-y})/{\sqrt2})} -\log\frac{2m}{\nu \mathfrak{h}_1(({2\sqrt2-y})/{\sqrt2})}\bigg)\,dy. \end{align*} $$

Thus $A_m=({2\sqrt 2}/{\lambda })\log (({m+1})/{m})$ which proves what we wanted. We also need to find $A_0$ for which it is true that

$$ \begin{align*}A_0&=\int_{0}^{\sqrt2}\frac{1}{\lambda}\log \frac{2}{\nu \mathfrak{h}_1({y}/{\sqrt2})}\,dy+\int_{\sqrt2}^{2\sqrt2}\frac{1}{\lambda}\log\frac{2}{\nu \mathfrak{h}_1(({2\sqrt2-y})/{\sqrt2})}\,dy\\ &=2\int_{0}^{\sqrt2}\frac{1}{\lambda}\log \frac{2}{\nu \mathfrak{h}_1({y}/{\sqrt2})}\,dy,\end{align*} $$

if $\nu \mathfrak {h}_1({y}/{\sqrt 2})=2$ has no solution. If this equation has solutions then, although we can find the area again we do not need to since $A_0$ will be even smaller in this case (see Figure 2).

Figure 2

The curve $\gamma _0$ for the exponential map for two different values of $\nu $ . The situation is similar with our maps as well.

Notice that because $(\mathfrak {h}_1,\mathfrak {h}_2,\mathfrak {h}_3)$ is always a point on the unit sphere we have that

$$ \begin{align*}\mathfrak{h}_1(x,y)^2+\mathfrak{h}_2(x,y)^2&=\sin^2 \theta|\mathfrak{h}(x,y)-\mathfrak{h}(0,0)|^2\\ &=\sin^2 \theta (\mathfrak{h}_1(x,y)^2+\mathfrak{h}_2(x,y)^2+(\mathfrak{h}_3(x,y)-1)^2),\end{align*} $$

where $\theta $ is the angle between the $x_3$ -axis and the segment that connects $(0,0,1)$ with $(\mathfrak {h}_1,\mathfrak {h}_2,\mathfrak {h}_3)$ . Taking $x=y$ , the fact that $\mathfrak {h}$ is bi-Lipschitz on $[-1,1]^2$ and noticing that $\theta \geq \pi /4$ gives us $|\mathfrak {h}_1(y)|\geq ({|y|}/{\sqrt 2L}).$ Hence because $\mathfrak {h}_1(y)>0$ for $y\in (0,1)$ we have

$$ \begin{align*} A_0\leq \frac{2}{\lambda}\int_{0}^{\sqrt2}\log\bigg(\frac{4L}{\nu y}\bigg)\,dy, \end{align*} $$

which is finite.

Proof. First note that $|p(\mathcal {Z}_{\nu }^n(x))|=\nu e^{p_3(\mathcal {Z}_{\nu }^{n-1})}\cdot |(p\circ h\circ p\circ \mathcal {Z}_{\nu }^{n-1})(x) |$ . Also, using the fact that $h(0)=(0,0,\lambda )$ and $h(x_1,x_2)=(h_1(x_1,x_2),h_2(x_1,x_2),h_3(x_1,x_2))$ is a Lipschitz map, we have that

$$ \begin{align*} |p(h(x))|=|p(h(x)-h(0))|\leq|h(x)-h(0)|\leq L|x|. \end{align*} $$

Hence

(4.1) $$ \begin{align} |p(\mathcal{Z}_{\nu}^n(x))|&=\nu e^{p_3(\mathcal{Z}_{\nu}^{n-1})}|(p\circ h\circ p \circ \mathcal{Z}_{\nu}^{n-1})(x)|\nonumber\\ &\leq L\nu e^{p_3(\mathcal{Z}_{\nu}^{n-1})}|( p \circ \mathcal{Z}_{\nu}^{n-1})(x)|\leq\cdots\nonumber\\ &\leq (\nu L)^{n}e^{p_3(\mathcal{Z}_{\nu}^{n-1})}e^{p_3(\mathcal{Z}_{\nu}^{n-2})}\cdots e^{x_3}\sqrt{h_1^2 (x_1,x_2 )+h_2^2 (x_1,x_2 )}\nonumber\\ &\leq \lambda (\nu L)^{n}e^{p_3(\mathcal{Z}_{\nu}^{n-1})}e^{p_3(\mathcal{Z}_{\nu}^{n-2})}\cdots e^{x_3}. \end{align} $$

From the chain rule we know that

(4.2) $$ \begin{align}\det(D\mathcal{Z}_{\nu}^n(x))=\prod_{k=0}^{n-1}\det(D\mathcal{Z}_{\nu}({\mathcal{Z}_{\nu}^{k}(x)})).\end{align} $$

Now from the definition of $\mathcal {Z}_{\nu }$ we obtain

(4.3) $$ \begin{align}\det(D\mathcal{Z}_{\nu}(x))=\nu^3 e^{3x_3}\det H,\end{align} $$

where

$$ \begin{align*}H=\begin{pmatrix} \dfrac{\partial h_1}{\partial x_1}(p(x))&\dfrac{\partial h_1}{\partial x_2}(p(x))&h_1(p(x))\\ \dfrac{\partial h_2}{\partial x_1}(p(x))&\dfrac{\partial h_2}{\partial x_2}(p(x))&h_2(p(x))\\ \dfrac{\partial h_3}{\partial x_1}(p(x))&\dfrac{\partial h_3}{\partial x_2}(p(x))&h_3(p(x)) \end{pmatrix}.\end{align*} $$

We now set

$$ \begin{align*} A=\begin{pmatrix} \dfrac{\partial h_1}{\partial x_1}(p(x))\\[10pt] \dfrac{\partial h_2}{\partial x_1}(p(x))\\[10pt] \dfrac{\partial h_3}{\partial x_1}(p(x)) \end{pmatrix},\quad B=\begin{pmatrix} \dfrac{\partial h_1}{\partial x_2}(p(x))\\[10pt] \dfrac{\partial h_2}{\partial x_2}(p(x))\\[10pt] \dfrac{\partial h_3}{\partial x_2}(p(x))\end{pmatrix},\quad C=\begin{pmatrix} h_1(p(x))\\ h_2(p(x))\\h_3(p(x)) \end{pmatrix}. \end{align*} $$

Recall now that from linear algebra, the determinant of a matrix equals the scalar triple product. This means that

$$ \begin{align*} \det H=\langle A\times B, C\rangle, \end{align*} $$

where $\langle \cdot ,\cdot \rangle $ denotes the euclidean inner product. Since $\mathcal {Z}_{\nu }$ is sense preserving we will have that $\det H>0$ , and since A and B are orthogonal to C we will have that $A\times B$ is parallel to C. Remember that $|C|=\lambda $ , so

(4.4) $$ \begin{align} \det H= \lambda|A\times B| \bigg\langle \frac{A\times B}{|A\times B|}, \frac{C}{|C|}\bigg\rangle= \lambda |A\times B|.\end{align} $$

Now because $\mathfrak {h}$ is a locally bi-Lipschitz map we have that

$$ \begin{align*} |h(p(x)+tv)-h(p(x))|\geq\frac{|tv|}{L}, \end{align*} $$

for all small $t>0$ where $v=(v_1,v_2)\in \mathbb {R}^2$ . This implies that

(4.5) $$ \begin{align} |Dh(p(x))(v)|\geq \frac{|v|}{L}, \end{align} $$

and if we set $v=(|B|,({-\langle A,B\rangle })/{|B|})$ and square both sides we obtain

$$ \begin{align*} \bigg| |B|A-\frac{\langle A,B\rangle}{|B|}B\bigg|^2\geq \frac{1}{L^2}\bigg(|B|^2+\frac{\langle A,B\rangle^2}{|B|^2}\bigg)\geq\frac{|B|^2}{L^2}. \end{align*} $$

Simplifying, we obtain

$$ \begin{align*} |A|^2|B|^2-\langle A,B\rangle^2\geq\frac{|B|^2}{L^2}. \end{align*} $$

Now note that, by elementary properties of the cross product, $|A\times B|^2=|A|^2|B|^2-\langle A,B\rangle ^2$ and thus

$$ \begin{align*}|A\times B|^2\geq \frac{|B|^2}{L^2}\geq \frac{1}{L^4},\end{align*} $$

where the last inequality comes from (4.5) for $v=(0,1)$ . Hence, (4.4) becomes $\det H\geq ({\lambda }/{L^2})$ . Putting everything together in (4.3), we obtain

(4.6) $$ \begin{align}\det(D\mathcal{Z}_{\nu}(x))\geq \nu^3 e^{3x_3}\frac{\lambda}{L^2}\ \text{a.e.}\end{align} $$

Hence, by (4.1) and (4.6) we have that

$$ \begin{align*}|p(\mathcal{Z}_{\nu}^n(x))|^3&\leq\lambda^3 (L\nu)^{3n}e^{3p_3(\mathcal{Z}_{\nu}^{n-1})}e^{3p_3(\mathcal{Z}_{\nu}^{n-2})}\cdots e^{3x_3}\\ &\leq\frac{\lambda^3 L^{5n}}{\lambda^{n}}\det(D\mathcal{Z}_{\nu}(\mathcal{Z}_{\nu}^{n-1}(x)))\cdots\det(D\mathcal{Z}_{\nu}(x)) \ \text{a.e}.\end{align*} $$

By rearranging and (4.2) we now obtain the desired inequality.

Proof. (a) If $h(x_1,x_2)=(h_1(x_1,x_2),h_2(x_1,x_2),h_3(x_1,x_2))$ , then we have that $p_3(\mathcal {Z}_{\nu }(x))=\nu e^{x_3}h_3(x_1,x_2)$ . Now since $h(0,0)=(0,0,\lambda )$ and h is continuous, for all $\varepsilon>0$ there is a disk $D=D(0,\delta )$ of radius $\delta>0$ on which we have $h_3(x_1,x_2)>\lambda -\varepsilon $ . Hence if $x=(x_1,x_2,x_3)\in C_{\delta }=D\times \mathbb {R}$ , then

$$ \begin{align*}p_3(\mathcal{Z}_{\nu}(x))=\nu e^{x_3}h_3(x_1,x_2)> \nu e^{p_3(x)}(\lambda-\varepsilon)\geq p_3(x)+1+\log (\nu(\lambda-\varepsilon)),\end{align*} $$

where the last inequality follows by minimizing $\nu e^t(\lambda -\varepsilon )-t$ . Now notice that since $\nu \lambda>({1}/{e})$ we can find a small enough $\varepsilon>0$ such that $\nu (\lambda -\varepsilon )>{1}/{e}$ , which implies that $1+\log (\nu (\lambda -\varepsilon ))>0$ . Hence, $p_3(\mathcal {Z}_{\nu }(x))>p_3(x)+c$ with $c= 1+\log (\nu (\lambda -\varepsilon ))$ .

(b) For a $\delta $ as in (a) and $\delta <\lambda $ now assume that there is a point $x\in C_{\delta }$ such that $p(x)\not =0$ and $\mathcal {Z}_{\nu }^n(x)\in C_{\delta }$ for all $n\in \mathbb {N}$ . Then according to (a) we would have that $p_3(\mathcal {Z}_{\nu }^n(x))\to \infty $ when $n\to \infty $ . We know that

(4.7) $$ \begin{align}|(p\circ \mathcal{Z}_{\nu}^{n+1})(x)|=e^{p_3(\mathcal{Z}_{\nu}^n)}|(p\circ h\circ p\circ\mathcal{Z}_{\nu}^n)(x)|.\end{align} $$

Now its a simple geometric fact that, for each $y\in [-\lambda ,\lambda ]^2$ ,

(4.8) $$ \begin{align}|(p\circ h)(y)|=\sin \theta |h(y)-h(0)|,\end{align} $$

where $\theta $ is the angle between the line segment joining the point $h(y)$ , on the sphere, with the point $(0,0,\lambda )$ and the $x_3$ -axis. Moreover, $\theta \geq \pi /4$ for all such y. Also, by the fact that h is a bi-Lipschitz function, we have that $|h(y)-h(0)|\geq ({|y|}/{L}).$

Now taking $y=p(\mathcal {Z}_{\nu }^n(x))$ and combining this with (4.7) and (4.8) implies that

$$ \begin{align*}|(p\circ \mathcal{Z}_{\nu}^{n+1})(x)|\geq \frac{e^{p_3(\mathcal{Z}^n(x))}}{\sqrt2 L}|(p\circ\mathcal{Z}_{\nu}^n)(x)|.\end{align*} $$

Thus, since $p_3(\mathcal {Z}_{\nu }^n(x))\to \infty $ , for all large enough n we can say that

$$ \begin{align*}|(p\circ\mathcal{Z}_{\nu}^{n+1})(x)|\geq2|(p\circ\mathcal{Z}_{\nu}^n)(x)|.\end{align*} $$

This of course contradicts the fact that $\mathcal {Z}^n(x)\in C_{\delta }$ for all $n\in \mathbb {N}$ .

Proof. Since $\mathcal {Z}_{\nu }^{n_j}(V)$ stays outside the cylinder $C_a$ we have that, for $x\in V$ ,

$$ \begin{align*}|(p\circ \mathcal{Z}_{\nu}^{n_j})(x)|>a.\end{align*} $$

By using Lemma 4 we will now have that

$$ \begin{align*}\det(D\mathcal{Z}_{\nu}^{n_j})\geq{\bigg(\frac{\lambda }{L^5}\bigg)}^{n_j} \cdot \frac{a^3}{\lambda^{3}}\ \text{a.e. on} \ V.\end{align*} $$

Since all of the iterates of V do not intersect any of the planes $x_1=\pm x_2 + 2\lambda k$ , where $k\in \mathbb {Z}$ , and since the Zorich map is a homeomorphism in the square beams that remain if we remove those planes, we will have that $\mathcal {Z}^n$ is a homeomorphism in V. Hence, for all $n_j\in \mathbb {N}$ , we will have that

$$ \begin{align*}m(\mathcal{Z}_{\nu}^{n_j}(V))=\int_{V}|\!\det(D\mathcal{Z}_{\nu}^{n_j})|\,dm\geq{\bigg(\frac{\lambda }{L^5}\bigg)}^{n_j}{\bigg(\frac{a^3}{\lambda^{3}}\bigg)} m(V),\end{align*} $$

which tends to infinity as $j\to \infty $ since $\lambda>{L^{5}}$ .

Proof. Let us first find an implicit equation that describes each of these surfaces. We work on $B_{(0,0)}$ , but the same can be done on all other rectangle beams. Let us split $B_{(0,0)}$ into three different beams whose cross-sections with the $x_1x_2$ plane are the sets

$$ \begin{align*}Q_1:=h^{-1}(\{(x_1,x_2,x_3)\in S(0,\lambda):0\leq x_2\leq x_1\}),\end{align*} $$

$$ \begin{align*}\kern9pt Q_2:=h^{-1}(\{(x_1,x_2,x_3)\in S(0,\lambda): x_2\leq 0, x_1\geq 0\})\end{align*} $$

and

$$ \begin{align*}Q_3:=h^{-1}(\{(x_1,x_2,x_3)\in S(0,\lambda):x_2\leq x_1\leq 0\}),\end{align*} $$

where $S(0,\lambda )$ the sphere of centre $0$ and radius $\lambda $ . In the beam corresponding to the first cross-section, meaning $Q_1\times \mathbb {R}$ , the points on the surface $S_n$ satisfy

$$ \begin{align*}\nu e^{x_3}h_1(x_1,x_2)=-\nu e^{x_3}h_2(x_1,x_2)+2(n+1)\lambda.\end{align*} $$

On the beam $Q_2\times \mathbb {R}$ the surface points satisfy

$$ \begin{align*}\nu e^{x_3}h_1(x_1,x_2)=\nu e^{x_3}h_2(x_1,x_2)+2(n+1)\lambda,\end{align*} $$

while on the beam $Q_3\times \mathbb {R}$ that corresponds to the last cross-section the points satisfy

$$ \begin{align*}\nu e^{x_3}h_1(x_1,x_2)=-\nu e^{x_3}h_2(x_1,x_2)-2(n+1)\lambda.\end{align*} $$

Suppose now that the surfaces $S_n$ do not intersect the plane $x_3=0$ . Hence the volume that the surface $S_n$ encloses together with the plane $x_3=0$ and inside the beam $B_{(0,0)}$ is given by the integrals

$$ \begin{align*} I_n&=\int\int_{Q_1}\log \frac{2(n+1)\lambda}{\nu(h_2(x_1,x_2)+h_1(x_1,x_2))}\,dx_2\,dx_1\\[6pt] &\quad+\int\int_{Q_2}\log \frac{-2(n+1)\lambda}{\nu(h_1(x_1,x_2)-h_2(x_1,x_2))}\,dx_2\,dx_1\\[6pt] &\quad+\int\int_{Q_3}\log \frac{-2(n+1)\lambda}{\nu(h_2(x_1,x_2)+h_1(x_1,x_2))}\,dx_2\,dx_1. \end{align*} $$

It is not so hard to prove now that each of these integrals is finite since each is an integral of a bounded function except at a neighbourhood of $(0,0)$ and the points where $h_1\pm h_2 =0$ which are $(0,-2\lambda )$ , $(2\lambda ,0)$ . So in order to show that this sum is finite, it is enough to consider the integrals only in neighbourhoods of those points. On the other hand, when those surfaces $S_n$ intersect the plane $x_3=0$ the volumes are no longer given by the above integrals (the set where we integrate will change), but again we only have to consider them at a neighbourhood of $(0,0)$ as well as $(0,-2\lambda )$ , $(2\lambda ,0)$ , so that is what we do next. In fact, those volumes in that case are even smaller.

We will only treat here the second integral around an $\varepsilon $ -neighbourhood of $(0,0)$ , and the rest follows similarly. So we are looking at the integral

$$ \begin{align*}\int\int_{Q_2\cap B(0,\varepsilon)}\log \frac{-2(n+1)\lambda}{\nu(h_2(x_1,x_2)-h_1(x_1,x_2))}\,dx_2\,dx_1,\end{align*} $$

where $B(0,\varepsilon )$ is a ball centred at $0$ with radius $\varepsilon $ . Equivalently, we want to show that the integral

(4.9) $$ \begin{align}\int\int_{Q_2\cap B(0,\varepsilon)}-\log (h_1(x_1,x_2)-h_2(x_1,x_2))\,dx_2\,dx_1\end{align} $$

is finite. Now because h is a bi-Lipschitz map and because $h_2(x_1,x_2)\leq 0$ and $h_1(x_1,x_2)\geq 0$ in the set we are integrating we have that

$$ \begin{align*}(h_1-h_2)^2&=h_1^2+h_2^2+2h_1(-h_2)\geq h_1^2+h_2^2\\ &=\sin^2 \theta (h_1^2+h_2^2+(h_3-\lambda)^2)\geq \frac{c_{\varepsilon}}{L^2}(x_1^2+x_2^2),\end{align*} $$

where $\theta $ is the angle between the $x_3$ -axis and the segment that connects $(0,0,\lambda )$ with $(h_1,h_2,h_3)$ , and $c_{\varepsilon }>0$ is a constant that depends only on $\varepsilon $ . Hence

$$ \begin{align*}|h_1(x_1,x_2)-h_2(x_1,x_2)|\geq \frac{\sqrt{c_\varepsilon}}{L}\sqrt{x_1^2+x_2^2}.\end{align*} $$

Now since $h_1-h_2\geq 0$ in the set we are integrating, we will have that

$$ \begin{align*}&\int\int_{Q_2\cap B(0,\varepsilon)}\log (h_1(x_1,x_2)-h_2(x_1,x_2))\,dx_2\,dx_1\\ &\quad\geq \int\int_{Q_2\cap B(0,\varepsilon)}\log \bigg(\frac{\sqrt{c_\varepsilon}}{L}\sqrt{x_1^2+x_2^2}\bigg)\,dx_2\,dx_1.\end{align*} $$

Now since the last integral is finite we will have that the integral (4.9) is also finite.

Finally, let us show that the sequence $T_n$ is a decreasing one. Indeed,

$$ \begin{align*}T_n=I_{n+1}-I_n=\lambda^2\log\frac{n+2}{n+1}+\lambda^2 \log\frac{n+2}{n+1}+2\lambda^2\log\frac{n+2}{n+1}=4\lambda^2\log\frac{n+2}{n+1},\end{align*} $$

which can be easily seen to be a decreasing sequence.

Proof. Consider the iterates $V_i=\mathcal {Z}_{\nu }^i(V)$ of the set V. The sets $V_i$ always stay inside one of the rectangle beams by assumption and also they cannot intersect any of the surfaces in $\mathcal {S}$ that are in those beams since if they did on the next iterate they would intersect the boundary of one of the beams. Suppose now that we can find an $N\in \mathbb {N}$ such that $V_{i}\not \in B_{(0,0)}\cup B_{(0,-1)}$ for all $i>N$ . Then by Lemma 6 we have that $m(V_{i})\to \infty $ . This implies that our sets $V_{i}$ cannot lie between any two of the surfaces in $\mathcal {S}$ , for all large i since there is finite volume between them. Thus $V_{i}$ stays below the lowest surface in the relevant rectangle beam for all $i>N_1>N$ , where $N_1\in \mathbb {N}$ . This is a contradiction since being below that surface implies that $V_{i+1}$ is in either $B_{(0,0)}$ or $B_{(0,-1)}$ .

Proof. Either an $\varepsilon _0>0$ exists such that $d(\mathcal {Z}_{\nu }^n(V),x_3\text {-axis})>\varepsilon _0$ for all $n>N_0$ or such an $\varepsilon _0$ does not exist. In the first case we know from Lemma 6 that $m(\mathcal {Z}_{\nu }^n(V))\to \infty $ . We also know that since $\mathcal {Z}_{\nu }^n(V)\in B_{(0,0)}$ , for all $n>N_0$ , $\mathcal {Z}_{\nu }^n(V)$ must, by Lemma 7, lie below the surface $S_0$ . Since $m(\mathcal {Z}_{\nu }^n(V))\to \infty $ , it must be true that for all $M_1>0$ there exist a $n_0$ and a point $z_0$ in $\mathcal {Z}_{\nu }^{n_0}(V)$ with $p_3(z_0)<-M_1$ . The pre-image of that point inside $B_{(0,0)}$ is a point $z^{(1)}=(z_1^{(1)},z_2^{(1)},z_3^{(1)})$ for which $\nu e^{z_3^{(1)}}h_3(z_1^{(1)},z_2^{(1)})<-M_1$ and $h_3(z_1^{(1)},z_2^{(1)})<0$ . But since $h_3>-\lambda $ we have that

$$ \begin{align*}e^{z_3^{(1)}}>\frac{M_1}{\nu \lambda}\Rightarrow z_3^{(1)}>\log\frac{ M_1}{\nu \lambda}.\end{align*} $$

If we now take the pre-image in $B_{(0,0)}$ of that point, $z^{(2)}=(z_1^{(2)},z_2^{(2)},z_3^{(2)})$ then $0<h_3(z_1^{(2)},z_2^{(2)})<\lambda $ and

$$ \begin{align*} \nu e^{z_3^{(2)}}h_3(z_1^{(2)},z_2^{(2)})=z_3^{(1)}>\log \frac{M_1}{\nu\lambda},\end{align*} $$

which implies that

$$ \begin{align*}z_3^{(2)}>\log\frac{\log ({M_1}/{\nu\lambda})}{\nu\lambda}.\end{align*} $$

Thus we have shown that for any $M>0$ there is a point $z^{(2)}$ in $\mathcal {Z}_{\nu }^{n_0-2}(V)$ for which $z_3^{(2)}>M$ and also $|z_1^{(2)}|,|z_2^{(2)}|< \lambda $ . This leads to a contradiction. To see why, note that by our assumptions $\mathcal {Z}_{\nu }^n(V)$ is $\varepsilon _0$ away from the $x_3$ -axis and below the surface $S_0$ and thus all of its points, which are also inside the initial square beam $[-\lambda ,\lambda ]^2\times \mathbb {R}$ (pink in Figure 3), have a bounded $x_3$ -coordinate. This is true because the surface $S_0$ together with any cylinder around the $x_3$ -axis and the plane $x_3=0$ enclose a set inside $[-\lambda ,\lambda ]^2\times \mathbb {R}$ and outside the cylinder whose closure is compact.

For the second case, where such an $\varepsilon _0$ does not exist, there is a sequence $w_k\in \bigcup _{n>N_0}\mathcal {Z}_{\nu }^n(V)$ with $d(w_k,x_3\text {-axis})\to 0$ . If $p_3(w_k)\to \infty $ we are done. If, on the other hand, $p_3(w_k)\to -\infty $ then $y_k:=\mathcal {Z}_{\nu }(w_k)\to (0,0,0)$ and thus $\mathcal {Z}_{\nu }^n(y_k)\to \mathcal {Z}_{\nu }^n(0)=(0,0, E_{\nu \lambda }^n(0))$ , where $E_{\nu \lambda }^n(0))$ converges to $\infty $ and again we are done. The remaining case to consider is when there is a subsequence $w_{k_i}$ converging to some point $(0,0,a)$ in the $x_3$ -axis. By relabelling we may assume that $w_k\to (0,0,a)$ . Now choose an $N>0$ such that $E_{\nu \lambda }^N(a)>M$ , where $E_{\nu \lambda }$ denotes the map $x\mapsto \nu \lambda e^x$ . By continuity of $\mathcal {Z}_{\nu }^N$ , for all $\varepsilon>0$ we may find a $\delta $ such that if $|w_k-(0,0,a)|\leq \delta $ , then

$$ \begin{align*}|\mathcal{Z}_{\nu}^N(w_k)-\mathcal{Z}_{\nu}^N(0,0,a)|=|\mathcal{Z}^N_{\nu}(w_k)-(0,0,E_{\nu\lambda}^N(a))|\leq \varepsilon.\end{align*} $$

Hence, if we choose $\varepsilon $ small enough we get that $x_3(\mathcal {Z}^N_{\nu }(w_k))>M$ and $ \mathcal {Z}^N_{\nu }(w_k)$ is within $\varepsilon $ distance from the z-axis, when k is large enough.

Proof. The Zorich map is absolutely continuous on any line segment since it is locally Lipschitz. Using the fundamental theorem of calculus for the Lebesgue integral now, it is not too hard to prove that a version of the finite-increment theorem (see [Reference Zorich34, 10.4.1, Theorem 1]) is true for such functions. Specifically, we can prove that

$$ \begin{align*}|\mathcal{Z}_{\nu}^n(y_1)-\mathcal{Z}_{\nu}^n(y_2)|\leq \operatorname*{\mbox{ess sup}}_{x\in \gamma}|D\mathcal{Z}_{\nu}^n(x)||y_1-y_2|,\end{align*} $$

where $\gamma $ is the line segment that connects $y_1$ to $y_2$ . Remember that $|Df|$ denotes the operator norm of the total derivative. Hence, by the chain rule and elementary properties of linear maps we have that

$$ \begin{align*}|D\mathcal{Z}_{\nu}^n(x)|\leq |D\mathcal{Z}_{\nu}(\mathcal{Z}_{\nu}^{n-1}(x))|\cdots |D\mathcal{Z}_{\nu}(x)|.\end{align*} $$

Hence by the above inequalities and because $y_1,y_2 \in B(0,r)$ we have that

(4.10) $$ \begin{align}|\mathcal{Z}_{\nu}^n(y_1)-\mathcal{Z}_{\nu}^n(y_2)|\leq&\nonumber \operatorname*{\mbox{ess sup}}_{x\in \gamma}|D\mathcal{Z}_{\nu}(\mathcal{Z}_{\nu}^{n-1}(x))|\cdots \operatorname*{\mbox{ess sup}}_{x\in \gamma}|D\mathcal{Z}_{\nu}(x)||y_1-y_2|\\\leq& \operatorname*{\mbox{ess sup}}_{x\in B(0,r)}|D\mathcal{Z}_{\nu}(\mathcal{Z}_{\nu}^{n-1}(x))|\cdots \operatorname*{\mbox{ess sup}}_{x\in B(0,r)}|D\mathcal{Z}_{\nu}(x)||y_1-y_2|.\end{align} $$

We also know that $D\mathcal {Z}_{\nu }(x)= e^{x_3}D\mathcal {Z}_{\nu }(x_1,x_2,0)$ . Moreover, we will prove that

$$ \begin{align*}|D\mathcal{Z}_{\nu}(x_1,x_2,0)|\leq \nu\max\{L,\lambda\}.\end{align*} $$

Indeed, let $u=(u_1,u_2,u_3)\in \mathbb {R}^3$ . Then

$$ \begin{align*}|D\mathcal{Z}_{\nu}(x_1,x_2,0)|^2=\sup_{|u|=1}|D\mathcal{Z}_{\nu}(x_1,x_2,0)(u)|^2= \nu^2\sup_{|u|=1}|u_1A+u_2B+u_3C|^2, \end{align*} $$

where $A,B, C$ are as in the proof of Lemma 4. Remember now that C is orthogonal to A and B, and thus the above equation becomes

$$ \begin{align*} |D\mathcal{Z}_{\nu}(x_1,x_2,0)|^2&=\nu^2\sup_{|u|=1}(|u_1A+u_2B|^2+|u_3C|^2)\\ &=\nu^2\sup_{|u|=1}(|Dh(x)(u_1,u_2)|^2+|u_3|^2|C|^2)\\ &\leq\nu^2\sup_{|u|=1}(L^2|(u_1,u_2)|^2+\lambda^2|u_3|^2)\\ &\leq\nu^2\max\{L^2,\lambda^2\}, \end{align*} $$

where we have used the fact that h is locally bi-Lipschitz.

Hence, $|D\mathcal {Z}_{\nu }(x)|\leq \nu \max \{L,\lambda \}e^{x_3}$ and (4.10) becomes

$$ \begin{align*}|\mathcal{Z}_{\nu}^n(y_1)-\mathcal{Z}_{\nu}^n(y_2)| &\leq \nu^n\max\{L,\lambda\}^n\sup_{x\in B(0,r)}e^{p_3(\mathcal{Z}_{\nu}^{n-1}(x))}\cdots \sup_{x\in B(0,r)}e^{p_3(x)}|y_1-y_2|\\ &=\bigg(\frac{\max\{L,\lambda\}}{\lambda}\bigg)^n E_{\nu\lambda}(r)\cdots E_{\nu\lambda}^n(r)|y_1-y_2|, \end{align*} $$

where we have used the fact that $\nu \lambda \sup _{x\in B(0,r)}e^{p_3(\mathcal {Z}_{\nu }^{n}(x))}=E_{\nu \lambda }^{n+1}(r)$ which can be easily proved by induction on n.

Proof Proof of Theorem 1

First, let us note that by assumption $\nu>\sqrt {{2L}/{\lambda }}>{1}/{\lambda e}$ . Let V be any open and connected set in $\mathbb {R}^3$ . We want to show that $\mathcal {Z}_{\nu }^n(V)$ intersects one of the planes that belong to the Julia set for some n and thus V itself intersects the Julia set. Assume that this does not happen. By Lemma 8 we can consider two cases.

First case. Suppose first that the sequence of iterates $\mathcal {Z}_{\nu }^n(V)$ does not eventually stay inside the square beam $B_{(0,0)}\cup B_{(0,-1)}$ but also visits their complement infinitely often. Then we can find a subsequence $n_j$ such that $\mathcal {Z}_{\nu }^{n_j}(V)\in B_{(0,0)}\cup B_{(0,-1)}$ and $\mathcal {Z}_{\nu }^{n_j+1}(V)\in B_{(k,l)}$ for some $(k,l)\not =(0,0),(0,-1)$ . Without loss of generality we may assume that $\mathcal {Z}_{\nu }^{n_j}(V)\in B_{(0,0)}$ .

Consider now the sets

$$ \begin{align*}V_{n_j}^+=\bigg\{x\in V: |(p\circ \mathcal{Z}_{\nu}^{n_j})(x)|\geq \frac{\lambda}{2}\bigg\}\end{align*} $$

and

$$ \begin{align*}V_{n_j}^-=\bigg\{x\in V: |(p\circ \mathcal{Z}_{\nu}^{n_j})(x)|<\frac{\lambda}{2}\bigg\}\end{align*} $$

( $\lambda $ is the scale factor by which we scaled up the initial square). Notice that $V=V_{n_j}^+\cup V_{n_j}^-$ . Since $\mathcal {Z}_{\nu }^{n_j+1}(V)$ is outside of $B_{(0,0)}\cup B_{(0,-1)}$ we will have that $\mathcal {Z}_{\nu }^{n_j}(V)$ lies between two ‘level surfaces’ $S_k$ and $S_{k+1}$ , but we know, from Lemma 7, that those surfaces enclose a volume no greater than $M_0$ between them, where $M_0$ is a constant. Thus $m(\mathcal {Z}_{\nu }^{n_j}(V))\leq M_0$ , where m denotes the Lebesgue measure. Also, by Lemma 4, it is true that, for almost all points in $V_{n_j}^+$ ,

$$ \begin{align*}\det(D\mathcal{Z}_{\nu}^{n_j})\geq{\bigg(\frac{\lambda }{L^5}\bigg)}^{n_j}\frac{1}{\lambda^{3}}|(p\circ \mathcal{Z}_{\nu}^{n_j})(x)|^3\geq{\bigg(\frac{\lambda }{L^5}\bigg)}^{n_j}\frac{1}{8}. \end{align*} $$

Hence, because $\mathcal {Z}_{\nu }^{n_j}$ is a homeomorphism in V we have that

$$ \begin{align*}M_0\geq m(\mathcal{Z}_{\nu}^{n_j}(V_{n_j}^+))=\int_{V_{n_j}^+} |\!\det(D\mathcal{Z}_{\nu}^{n_j})|\,dm\geq{\bigg(\frac{\lambda }{L^5}\bigg)}^{n_j}\frac{1}{8}\cdot m(V_{n_j}^+).\end{align*} $$

This implies that $m(V_{n_j}^+)\to 0$ as $n_j\to \infty $ . On the other hand, the set $\mathcal {Z}_{\nu }^{n_j}(V_{n_j}^-)$ is inside the initial square beam $[-\lambda ,\lambda ]^2\times \mathbb {R}$ . Thus $\mathcal {Z}_{\nu }^{n_j+1}(V_{n_j}^-)$ lies in the half-space $x_3>0$ and outside the square beam $B_{(0,0)}\cup B_{(0,-1)}$ . This same set also lies between some level surfaces or below all of them. Moreover, as we proved in Lemma 7, the volume enclosed by those successive surfaces and the volume enclosed by the first one and the plane $x_3=0$ is smaller than some constant $M_0$ . Thus $m (\mathcal {Z}_{\nu }^{n_j+1}(V_{n_j}^-))\leq M_0$ . By arguing the same way as before now we get that

$$ \begin{align*}M_0\geq m(\mathcal{Z}_{\nu}^{n_j+1}(V_{n_j}^-))=\int_{V_{n_j}^-} |\!\det(D\mathcal{Z}_{\nu}^{n_j+1})|\,dm\geq{\bigg(\frac{\lambda }{L^5}\bigg)}^{n_j+1}\frac{1}{8}\cdot m(V_{n_j}^-).\end{align*} $$

This again implies that $ m(V_{n_j}^-)\to 0$ as $n_j\to \infty $ . But this is a contradiction since $m(V)=m(V_{n_j}^-)+m(V_{n_j}^+)$ .

Second case. Suppose now that $\mathcal {Z}_{\nu }^n(V)\in B_{(0,0)}\cup B_{(0,-1)}$ , for all $n>N_0$ . Observe that either $\mathcal {Z}_{\nu }(B_{(0,0)})\subset \{(x_1,x_2,x_3):x_2\leq x_1\}$ or $\mathcal {Z}_{\nu }(B_{(0,0)})\subset \{(x_1,x_2,x_3):x_2\geq x_1\}$ . In the first case $\mathcal {Z}_{\nu }^n(V)$ stays in $B_{(0,0)}$ for all large n or it stays in $B_{(0,-1)}$ , while in the second it alternates between $B_{(0,0)}$ and $B_{(0,-1)}$ . For simplicity we will assume that the first case holds and thus $\mathcal {Z}_{\nu }^n(V)\in B_{(0,0)}$ , for all $n>N_0$ .

Let us now consider the inverse image under $\mathcal {Z}_{\nu }$ of the boundary of $B_{(0,0)}$ , which lies inside $B_{(0,0)}$ , namely the surface $S_0$ we had in the proof of Lemma 7. Remember that this surface is defined as $S_0:=\{(x_1,x_2,x_3)\in B_{(0,0)}:x_3=f(x_1,x_2)\}$ , where f is continuous on $B_{(0,0)}\cap \{x_3=0\}$ and extends continuously on the boundary of this set except at the points $(0,0)$ , $(2\lambda ,0)$ , $(0,-2\lambda )$ where $f\to \infty $ . Notice then that all the iterates $\mathcal {Z}_{\nu }^n(V)$ stay below the surface $S_0$ .

Consider now a plane $x_3=c$ , with $c=E^{N}_{\nu \lambda }(0)-\lambda $ and N so large that this plane intersects this surface $S_0$ and also

(4.11) $$ \begin{align} (c+\lambda)^{\log (c+\lambda)+1}e^{c+\lambda}\nu^2\lambda^2 e^{-({\nu\lambda e^c}/{2})}\leq \lambda. \end{align} $$

We define sets $A_1$ , $A_2$ and $A_3$ as follows:

• • $A_1:=\{(x_1,x_2,x_3)\in B_{(0,0)}:c<x_3<f(x_1,x_2)\ \text {and}\ (x_1,x_2)\ \text {in a neighbourhood of} (0,0)\}$ ;

• • $A_2:=\{(x_1,x_2,x_3)\in B_{(0,0)}:c<x_3<f(x_1,x_2)\ \text {and}\ (x_1,x_2)\ \text {in a neighbourhood of} (2\lambda ,0)\}$ ;

• • $A_3:=\{(x_1,x_2,x_3)\in B_{(0,0)}:c<x_3<f(x_1,x_2)\ \text {and}\ (x_1,x_2)\ \text {in a neighbourhood of} (0,-2\lambda )\}$ .

Then the following assertions hold.

1. (i) All those sets lie below (in terms of the $x_3$ -coordinate) the surface $S_0$ . By the definitions and Lemma 7 it is easy to see that the sets $A_1$ , $A_2$ and $A_3$ are also of finite Lebesgue measure.

2. (ii) $\mathcal {Z}_{\nu }(A_2\cup A_3)\subset \{(x_1,x_2,x_3)\in \mathbb {R}^3:x_3< -({\nu \lambda e^c}/{2})\}$ and thus $\mathcal {Z}^2_{\nu }(A_2\cup A_3)\subset B(0,\delta ), $ where $\delta = \nu \lambda e^{-({\nu \lambda e^c}/{2})}$ . Note that $\delta <\nu \lambda =E_{\nu \lambda }(0)$ .

3. (iii) It is easy to show by induction on N and since $\nu \lambda>1/e$ that $E_{\nu \lambda }^N(0)\geq E_{\nu \lambda }(N-1)$ and thus

   (4.12) $$ \begin{align} E_{\nu\lambda}(N-1)\leq c+\lambda\Rightarrow N \leq \log(c+\lambda)+2. \end{align} $$

By Lemma 10 and since $\lambda> L^5$ we will now have that, for all $x\in B(0,\delta )$ ,

$$ \begin{align*}|\mathcal{Z}_{\nu}^N(x)-\mathcal{Z}_{\nu}^N(0)|&\leq E_{\nu\lambda}(\delta)\cdots E_{\nu\lambda}^N(\delta)|x|\\ &\leq E^2_{\nu\lambda}(0)\cdots E_{\nu\lambda}^{N+1}(0)\delta\\ &\leq(c+\lambda)^{N-1}E_{\nu\lambda}(\lambda+c)\delta\\ &\leq(c+\lambda)^{\log(c+\lambda )+1}\nu\lambda e^{\lambda+c}\delta.\end{align*} $$

Hence, by (4.11) we will have that

(4.13) $$ \begin{align} |\mathcal{Z}_{\nu}^N(x)-\mathcal{Z}_{\nu}^N(0)|\leq \lambda. \end{align} $$

Equation (4.13) together with (ii) implies that $\mathcal {Z}_{\nu }^{N+2}(A_2\cup A_3)\subset B(\mathcal {Z}_{\nu }^{N}(0),\lambda )$ , and by the choice of c this last ball is contained in $\{(x_1,x_2,x_3)\in \mathbb {R}^3:x_3>c\}$ . This implies that the part of $\mathcal {Z}_{\nu }^{N+2}(A_2\cup A_3)$ that lies below $S_0$ is contained in $A_1$ .

Recall that $\mathcal {Z}_\nu ^n(V)$ stays below $S_0$ for $n>N_0$ . By Lemma 9 we know that there is a point $x_0\in \mathcal {Z}_{\nu }^{n_0}(V)$ for some $n_0>N_0$ such that $x_0\in A_1$ . Take such a point $x_0$ . Let us carefully examine the behaviour of the iterates of this point. We can assume that $A_1$ is so close to the $x_3$ -axis that if $y\in A_1$ then $p_3(\mathcal {Z}_{\nu }(y))>p_3(y)$ by Lemma 5(a). Hence the points $\mathcal {Z}_{\nu }^n(x_0)$ go higher and higher up in the $x_3$ direction, while at the same time staying in $A_1$ , until at some point the iterate $\mathcal {Z}_{\nu }^k(x_0)$ , for some k, will lie in either $A_2$ or $A_3$ thanks to Lemma 5(b). Without loss of generality assume that $x_1:=\mathcal {Z}_{\nu }^{k}(x_0)\in A_2$ and take a small ball around $x_1$ , $B(x_1,r)$ with $B(x_1,r)\subset A_2\cap \mathcal {Z}^{n_0+k}_{\nu }(V)$ . By what we have said in the previous paragraph, we will have that $\mathcal {Z}_{\nu }^{N+2}(B(x_1,r))\subset ~A_1$ .

However, we know what happens in points inside $A_1$ when we iterate: they eventually leave $A_1$ . Thus for some $k>N+2$ we will have that $\mathcal {Z}_{\nu }^{k}(B(x_1,r))\subset A_2 \cup A_3$ since $B(x_1,r)$ is a connected set and the sets $A_1$ , $A_2$ and $A_3$ are disjoint. We can then repeat this whole argument, meaning that we take the set $\mathcal {Z}_{\nu }^{k}(B(x_1,r))$ which is now in $A_2$ or $A_3$ and thus will get mapped by $\mathcal {Z}_{\nu }$ to the lower half-space $x_3<-({\nu \lambda e^c}/{2})$ and by $\mathcal {Z}_{\nu }^{N+2}$ inside $A_1$ . Now continue as above and then repeat. Eventually we obtain a sequence $n_j\to \infty $ with

$$ \begin{align*}\mathcal{Z}_{\nu}^{n_j}(B(x_1,r))\subset A_2 \cup A_3.\end{align*} $$

Then by using Lemma 6 we will have that $m(\mathcal {Z}_{\nu }^{n_j}(B(x_1,r)))\to \infty $ , but this is impossible since $m(A_2\cup A_3)$ is finite.

Proof. Let $x_0\in \overline {\Gamma _0}$ , and let U be a neighbourhood of this point. We want to show that $\gamma _{k}\cap U\not =\emptyset $ for all sufficiently large k. We know, from the definition of $\overline {\Gamma _0}$ , that there is a point $x_1\in U\cap \Gamma _0 $ . This implies that $\mathcal {Z}_{\nu }^n(x_1)$ belongs to the $x_3$ -axis for all $n\geq N_0$ , for some $N_0\in \mathbb {N}$ , and in fact we can assume that $x_2:=\mathcal {Z}_{\nu }^{N_0}(x_1)\in H_{>M},$ where M is any positive number. Now taking $M>M_0$ , where $M_0$ is the constant we used in Lemma 1, and applying that lemma n times for a ball $B(x_2,R)\subset \mathcal {Z}_{\nu }^{N_0}(U)$ , we obtain

$$ \begin{align*}\mathcal{Z}_{\nu}^{n}(B(x_2,R)\cap H_{>M})\supset B(\mathcal{Z}^n_{\nu}(x_2),\alpha^{-n}R)\cap H_{>E^n_{\nu\lambda}(M)}.\end{align*} $$

For all large enough n the ball on the right-hand side, $B(\mathcal {Z}^n_{\nu }(x_2),\alpha ^{-n}R)$ , intersects the line $\gamma _1=\{(2\lambda ,0,t):t\in \mathbb {R}\}$ . Hence, for all large enough n, $\mathcal {Z}_{\nu }^n(U)$ intersects $\gamma _1$ . Thus for each n large enough there is a point $x_3\in \gamma _1$ whose backward orbit intersects U itself. This means that U contains a point in $\gamma _{k}$ for all large enough k, as we wanted.

Proof. Suppose $U\subset \mathbb {R}^3$ is an open set with $U\cap \Gamma _0\not =\emptyset $ and $\Gamma _0\cap \partial U=\emptyset $ . We show that $\Gamma _0\subset U$ .

By Lemma 11 we have that $\gamma _k\cap U\not =\emptyset $ , for all $k>k_0$ . Since $\gamma _k$ is a connected curve, this implies that $\gamma _k\subset U$ , for all $k> k_0$ . Thus

$$ \begin{align*}\Gamma_0\subset \overline{\Gamma_0}= \overline{\bigcup_{k\geq k_0}\gamma_k}\subset \overline{U}.\end{align*} $$

Hence, since $\Gamma _0\cap \partial U=\emptyset $ , we have that $\Gamma _0\subset U$ .

Proof. We will prove this by induction on j. Let us define the inverse branches of $\mathcal {Z}_{\nu }$ . Using the notation we introduced in the first paragraphs of this section, define $\Lambda _{k,l}:\mathbb {H}_p\to T_{(k,l)}$ , with $p=0,1,2,3$ , to be the inverse branches of $\mathcal {Z}_{\nu }$ that take values on the square beams $T_{(k,l)}$ . We also extend those maps to $\overline {\mathbb {H}_p}\setminus \{0\}$ and use the same symbol to denote those extensions. With that notation we have that

$$ \begin{align*}Y_{j+1}=\bigcup_{(k,l)\in\mathbb{Z}^2}\Lambda_{k,l}(Y_j)\cup Y_j.\end{align*} $$

By the inductive hypothesis now we know that $Y_j$ is connected and because $\Lambda _{k,l}$ is continuous the set $\Lambda _{k,l}(Y_j)$ is also connected. Observe now that the point $x_{n,m}=(2\lambda ,0,0)+n(2\lambda ,0,0)+m(0,2\lambda ,0)$ is inside $Y=Y_0$ and thus in $Y_j$ , for all $n,m\in \mathbb {Z}$ and for all $j\in \mathbb {N}$ . Also note that $\mathcal {Z}_{\nu }(x_{n,m})=(0,0,-\nu \lambda )$ or $(0,0,\nu \lambda )$ which are both points in $Y_j$ . This means that there are $m, n$ depending on $k,l$ such that $x_{n,m}\in \Lambda _{k,l}(Y_j)$ . Hence $\Lambda _{k,l}(Y_j)\cap Y_j\not =\emptyset $ . This implies that the set $\Lambda _{k,l}(Y_j)\cup Y_j$ is connected. Hence $Y_{j+1}$ is connected as a union of connected sets with non-empty intersections with each other, as we wanted.

Proof Proof of Theorem 3

Consider the set

$$ \begin{align*}\bigcup_{j\geq 0}\mathcal{Z}_{\nu}^{-j}((0,0,-1))\subset\bigcup_{j\geq 0}Y_j.\end{align*} $$

The Zorich map is bounded on $\{(x_1,x_2,x_3):x_3<0\}$ and thus it does not have the pits effect (see [Reference Bergweiler and Nicks5]). Hence, by [Reference Bergweiler and Nicks5, Theorem 1.8] we will have that the set $\bigcup _{j\geq 0}\mathcal {Z}_{\nu }^{-j}((0,0,-1))$ is dense in $\mathcal {J}(\mathcal {Z}_\nu )$ , which by Theorem 1 is $\mathbb {R}^3$ , and thus also dense in $I(\mathcal {Z}_{\nu })$ . Thus the set $\bigcup _{j\geq 0}Y_j $ is a connected dense subset of $I(\mathcal {Z}_{\nu })$ which implies that the escaping set itself is connected.

Proof Proof of Theorem 2

First let $U_0=B(x_0,r)$ be a ball centred at $x_0\in \mathbb {R}^3$ of radius $r>0$ . We seek a periodic point of $\mathcal {Z}_\nu $ in $U_0$ . Without loss of generality we may assume that $\overline {U_0}$ does not intersect any of the planes $x_1=\pm x_2+2\lambda k$ , $k\in \mathbb {Z}$ .

We will follow the method of [Reference Fletcher and Nicks16] where the authors prove that periodic points of a quasiregular version of the sine function are dense on $\mathbb {R}^3$ . We will do this by finding an $N\in \mathbb {N}$ and a finite sequence of open sets $U_j$ , $j=0,\ldots , N$ , such that:

1. (i) $U_{j+1}\subset \mathcal {Z}_{\nu }(U_j)$ , $0\leq j\leq N-1$ ;

2. (ii) $\mathcal {Z}_{\nu }$ is a homeomorphism on each $U_j$ for $j\leq N-1$ ;

3. (iii) $\overline {U_0}\subset U_N$ .

If these conditions are met then we can define a continuous inverse branch $\mathcal {Z}^{-N}_{\nu }:U_N\to U_0$ . Thus by the Brouwer fixed point theorem the map ${\mathcal {Z}^{-N}_{\nu }}|_{U_0}$ has a fixed point in $U_0$ .

We will now show how we can construct such a sequence. By Theorem 1 we know that $\mathcal {Z}^n_\nu (U_0)$ eventually covers $\mathbb {R}^3\setminus \{0\}$ . We set $U_j=\mathcal {Z}^j_\nu (U_0)$ for all j such that $\mathcal {Z}^j_\nu (U_0)$ does not intersect the set $P:=\bigcup _{k\in \mathbb {Z}}\{(x_1,x_2,x_3):x_1=\pm x_2+2k\lambda \}$ . Let $n_0$ be the biggest such j, so that we have defined $U_0,\ldots , U_{n_0}$ . Then take a point $y_1$ in $\mathcal {Z}^{n_0+1}(U_0)\cap P$ such that $y_1\not \in B_{\mathcal {Z}_\nu }$ and a ball $B(y_1,r)\subset \mathcal {Z}^{n_0+1}_\nu (U_0)\setminus B_{\mathcal {Z}_\nu }$ , where we recall here that $B_{\mathcal {Z}_\nu }$ is the branch set. Set $U_{n_0+1}=B(y_1,r)$ . We know that $\mathcal {Z}_\nu (U_{n_0+1})$ intersects one of the planes $x_1=\pm x_2$ and it is easy to see that $\mathcal {Z}_\nu $ is a homeomorphism on $U_{n_0+1}$ . Assume, without loss of generality that it intersects $x_1=x_2$ and take $y_2\in ( \mathcal {Z}_\nu (U_{n_0+1})\cap \{(x_1,x_2,x_3):x_1=x_2\})\setminus B_{\mathcal {Z}_\nu }$ . Set $U_{n_0+2}=B(y_2,r_2)$ , where $r_2>0$ is such that $B(y_2,r_2)\subset \mathcal {Z}_\nu (U_{n_0+1})\setminus B_{\mathcal {Z}_\nu }$ .

Consider the set $V_0=U_{n_0+2}\cap \{(x_1,x_2,x_3):x_1=x_2\}$ which is an open set of the plane $x_1=x_2$ in the subspace topology. We define the sets $V_n$ by induction as follows. Suppose that $V_n$ has been defined and that $V_n\cap B_{\mathcal {Z}_\nu }=\emptyset $ . We consider now two cases:

1. (1) $\mathcal {Z}_\nu (V_n)$ intersects one of the lines $\{(x_1,x_2,x_3):x_1=x_2=2\lambda k\}$ , $k\in \mathbb {Z}$ which are the pre-images of the $x_3$ -axis on the plane $x_1=x_2$ :

2. (2) $\mathcal {Z}_\nu (V_n)$ does not intersect any of those lines.

In the first case, let $y_3$ be a point in such an intersection. We define $V_{n+1}$ to be an open ball around $y_3$ in the subspace topology of $x_1=x_2$ of radius $r_3$ where $r_3$ is chosen in such a way that the ball does not contain branch points and such that $V_{n+1}\subset \mathcal {Z}_{\nu }(V_n)$ .

In the second case, we define $V_{n+1}:=\mathcal {Z}_\nu (V_n)\cap H_0$ , where $H_0$ is the whole plane $x_1=x_2$ if $\mathcal {Z}_\nu (V_n)\cap B_{\mathcal {Z}_\nu }=\emptyset $ , and it is an open half-plane on $x_1=x_2$ otherwise, which is defined as follows. Suppose that $\mathcal {Z}_\nu (V_n)$ intersects one of the lines of $B_{\mathcal {Z}_\nu }$ which we call  $\ell _1$ . We set $H_0$ to be the half-plane defined by this line and the property

$$ \begin{align*}m_2(\mathcal{Z}_\nu(V_n)\cap H_0)\geq \tfrac{1}{2}m_2(\mathcal{Z}_\nu(V_n)),\end{align*} $$

where $m_2$ is the two-dimensional Lebesgue measure on $x_1=x_2$ . Note now that we have inductively defined the sets $V_n$ .

We now claim that case (1) must occur for some n, otherwise notice that by construction $\mathcal {Z}_\nu $ is a homeomorphism on $V_n$ , for all $n\in \mathbb {N}$ . Hence,

(6.1) $$ \begin{align}m_2(V_{n+1})=m_2(\mathcal{Z}_\nu(V_n)\cap H_0)\geq\tfrac{1}{2}m_2(\mathcal{Z}_\nu(V_n)).\end{align} $$

Using the notation of §3, we now have that

$$ \begin{align*}m_2(\mathcal{Z}_\nu(V_n))=m_2((\phi^{-1}\circ g\circ \phi)(V_n))=Cm_2(g(\phi(V_n))), \end{align*} $$

where $C=|\!\det D\phi ^{-1}|$ , which is a constant since $\phi $ is linear. Combining with equation (6.1), this gives

$$ \begin{align*}m_2(V_{n+1})\geq\frac{C}{2} m_2(g(\phi(V_n))).\end{align*} $$

Thus by Lemma 3 we have that $m_2(V_n)\to \infty $ , as $n\to \infty $ . This implies, just like in the proof of Theorem 5, that there is an $m_0$ such that $V_{m_0}$ intersects the $x_3$ -axis.

We now set $U_{n_0+2+i}$ to be the open set

$$ \begin{align*}U_{n_0+2+i}:=\bigcup_{x\in V_i}B(x,r_x),\end{align*} $$

where $r_x$ is such that $B(x,r_x)\subset \mathcal {Z}_\nu (U_{n_0+1+i})$ and $B(x,r_x)\cap B_{\mathcal {Z}_\nu }=\emptyset $ , for all $1~\leq i~\leq ~m_0$ .

Notice that the sets $U_{n_0+2+i}$ satisfy properties (i) and (ii). We have that $U_{n_0+2+m_0}$ intersects the $x_3$ -axis, so let

$$ \begin{align*}U_{n_0+3+m_0}= \mathcal{Z}_\nu(U_{n_0+2+m_0})\cap B_{(0,0)}.\end{align*} $$

Define also

$$ \begin{align*}Q_{(0,0)}=\{(x_1,x_2):|x_1|+|x_2|<2\lambda \}\end{align*} $$

and

$$ \begin{align*}Q_{(k,l)}=Q_{(0,0)}+k(2\lambda,2\lambda)+l(2\lambda,-2\lambda), \quad k,l\in\mathbb{Z}.\end{align*} $$

We now set

$$ \begin{align*}U_{n_0+2+m_0+j}=\mathcal{Z}_\nu(U_{n_0+1+m_0+j})\cap B_{(0,0)} ,\end{align*} $$

for all $2 \leq j\leq m_1$ , where $m_1$ , depending on $M>0$ , is so large that $U_{n_0+2+m_0+m_1}$ contains a set of the form $Q_0\times [R,R+M]$ , where $Q_0=Q_{(0,0)}\cap \{(x_1,x_2):x_1<x_2\}$ and for some $R>0$ . We know that such an $m_1$ exists because the iterated image, under the Zorich map, of an open set that intersects the $x_3$ -axis eventually contains a ball of radius as large as we want (see Lemma 1).

If M is large enough then $\mathcal {Z}_\nu (U_{n_0+2+m_0+m_1})$ will contain a set of the form

$$ \begin{align*}U_{n_0+3+m_0+m_1}:=Q_{(k,l)}\times[-t_M,t_M],\end{align*} $$

for some $k,l\in \mathbb {Z}$ and $t_M\to \infty $ , as $M\to \infty $ . Note that $\mathcal {Z}_\nu $ is a homeomorphism on $U_{n_0+2+m_0+j}$ for all $2\leq j\leq m_1+1$ and that $\mathcal {Z}_\nu (U_{n_0+3+m_0+m_1})$ will be the set

$$ \begin{align*}U_N:=\{x\in\mathbb{R}^3:\nu\lambda e^{-t_M}< |x|< \nu\lambda e^{t_M}\}\setminus W ,\end{align*} $$

where $W=\{(x_1,x_2,x_3):x_1=\pm x_2,\ x_3\leq 0\}$ . If M is large enough $U_N$ will contain the closure of our initial set $U_0$ , since $U_0$ does not intersect any of the planes $x_1=\pm x_2 +2\lambda k$ , $k\in \mathbb {Z}$ , and we are done.

Proof. The only difference with the proof of Lemma 2 is in finding a lower bound for

$$ \begin{align*}\left|\det\begin{pmatrix} \mathfrak{a}(y)& \dfrac{d\mathfrak{a}}{dy}({y})\\ \mathfrak{b}(y)& \dfrac{d\mathfrak{b}}{dy}(y)\end{pmatrix}\right|.\end{align*} $$

This time we know that the absolute value of the determinant equals

$$ \begin{align*}|(\mathfrak{a}(y),\mathfrak{b}(y))|\bigg|\bigg(\dfrac{d\mathfrak{a}}{dy}({y}),\dfrac{d\mathfrak{b}}{dy}({y})\bigg)\bigg||\sin \theta(y)|,\end{align*} $$

where $\theta (y)$ is the angle between the vectors $(\mathfrak {a}(y),\mathfrak {b}(y))$ and $(({d\mathfrak {a}}/{dy})({y}),({d\mathfrak {b}}/{dy})(y))$ .

Hence, using the non-tangential position vector property and the fact that

$$ \begin{align*}\bigg|\bigg(\dfrac{d\mathfrak{a}}{dy}({y}),\dfrac{d\mathfrak{b}}{dy}({y})\bigg)\bigg|\geq \frac{1}{\sqrt2 L}\quad\text{and}\quad |(\mathfrak{a}(y),\mathfrak{b}(y))|\geq \frac{\min_{x\in Q} |\mathfrak{h}_{\mathrm{gen}}(x)|}{\sqrt2},\end{align*} $$

we get that

$$ \begin{align*}\left|\det\begin{pmatrix} \mathfrak{a}(y)& \dfrac{d\mathfrak{a}}{dy}({y})\\ \mathfrak{b}(y)& \dfrac{d\mathfrak{b}}{dy}(y)\end{pmatrix}\right|\geq \frac{\sin \theta_{\mathcal{S}}\min_{x\in Q} |\mathfrak{h}_{\mathrm{gen}}(x)|}{2L}\end{align*} $$

and thus

$$ \begin{align*}|\!\det D\hat{g}(z)|\geq\frac{ \sin \theta_{\mathcal{S}}\min_{x\in Q} |\mathfrak{h}_{\mathrm{gen}}(x)|\lambda e^{2\lambda {\operatorname{Re}}\ {z}}}{2 L}\ \text{a.e.}\\[-3.6pc] \end{align*} $$

Proof. Again the only difference is in obtaining a lower bound for the determinant $\det D\mathcal {Z}_{\mathrm {gen}}(x)$ . Define

$$ \begin{align*}\mathcal{H}=\begin{pmatrix} \dfrac{\partial h_{\mathrm{gen},1}}{\partial x_1}(p(x))&\dfrac{\partial h_{\mathrm{gen},1}}{\partial x_2}(p(x))&h_{\mathrm{gen},1}(p(x))\\[8pt] \dfrac{\partial h_{\mathrm{gen},2}}{\partial x_1}(p(x))&\dfrac{\partial h_{\mathrm{gen},2}}{\partial x_2}(p(x))&h_{\mathrm{gen},2}(p(x))\\[8pt] \dfrac{\partial h_{\mathrm{gen},3}}{\partial x_1}(p(x))&\dfrac{\partial h_{\mathrm{gen},3}}{\partial x_2}(p(x))&h_{\mathrm{gen},3}(p(x)) \end{pmatrix}\end{align*} $$

and set

$$ \begin{align*}\mathcal{A}=\begin{pmatrix} \dfrac{\partial h_{\mathrm{gen},1}}{\partial x_1}(p(x))\\ \dfrac{\partial h_{\mathrm{gen},2}}{\partial x_1}(p(x))\\ \dfrac{\partial h_{\mathrm{gen},3}}{\partial x_1}(p(x)) \end{pmatrix},\quad \mathcal{B}=\begin{pmatrix} \dfrac{\partial h_{\mathrm{gen},1}}{\partial x_2}(p(x))\\ \dfrac{\partial h_{\mathrm{gen},2}}{\partial x_2}(p(x))\\ \dfrac{\partial h_{\mathrm{gen},3}}{\partial x_2}(p(x))\end{pmatrix},\quad\mathcal{C}=\begin{pmatrix} h_{\mathrm{gen},1}(p(x))\\ h_{\mathrm{gen},2}(p(x))\\ h_{\mathrm{gen},3}(p(x)) \end{pmatrix}. \end{align*} $$

Then $\det \mathcal {H}=\langle \mathcal {A}\times \mathcal {B}, \mathcal {C}\rangle =|\mathcal {A}\times \mathcal {B}||\mathcal {C}| \cos \phi ,$ where $\phi $ is the angle between $\mathcal {A}\times \mathcal {B}$ and $\mathcal {C}$ . Using the fact that $|\mathcal {C}|\geq \lambda \min _{x\in Q}|\mathfrak {h}_{\mathrm {gen}}(x)|$ and $|\mathcal {A}\times \mathcal {B}|\geq {1}/{L^2}$ , together with the non-tangential position vector property, we can show that

$$ \begin{align*} \det\mathcal{H}\geq \frac{\lambda \min_{x\in Q}|\mathfrak{h}_{\mathrm{gen}}(x)|\sin \theta_{\mathcal{S}} }{L^2}. \end{align*} $$

Hence

$$ \begin{align*} \det D\mathcal{Z}_{\mathrm{gen}}(x)\geq e^{3x_3}\frac{\lambda \min_{x\in Q}|\mathfrak{h}_{\mathrm{gen}}(x)|\sin \theta_{\mathcal{S}}}{L^2}, \end{align*} $$

and the rest follows in exactly the same way as in the proof of Lemma 4.

Proof. The proof of (a) follows that of Lemma 5 word for word. For (b) again the proof is almost the same. The difference here is the lower bound for the angle $\theta $ used in the proof of Lemma 5: instead of ${\pi }/{4}$ it is now some constant greater than $0$ .

Proof. The proof is the same as in Lemma 6, only now we use Lemma 16 in place of Lemma 4.

Proof. The proof follows the proof of Lemma 10 almost word for word. Note, however, that $\mathcal {A}$ , $\mathcal {B}$ , $\mathcal {C}$ are not orthogonal. Still, when estimating $|D\mathcal {Z}_{\mathrm {gen}}(x_1,x_2,0)|$ (see the proof of Lemma 10) we can argue as follows:

$$ \begin{align*} |D\mathcal{Z}_{\mathrm{gen}}(x_1,x_2,0)|^2&=\sup_{|v|=1}(|v_1\mathcal{A}+v_2\mathcal{B}+v_3\mathcal{C}|^2)\\ &\leq\sup_{|v|=1}(|v_1\mathcal{A}+v_2\mathcal{B}|+|v_3\mathcal{C}|)^2\\ &\leq\sup_{|v|=1}(L|(v_1,v_2)|+\lambda|v_3|)^2\\ &\leq L^2+\lambda^2. \end{align*} $$

We also note that to argue here as in the last few lines of the proof of Lemma 10 we use the two conditions in equation (7.1).

Proof Proof of Theorem 4

Let V be any open and connected set of $\mathbb {R}^3$ . Assuming that

$$ \begin{align*} \lambda>C_{h_{\mathrm{gen}}}:=\frac{\max \{L^5,2 L\}}{\min_{x\in Q}|\mathfrak{h}_{\mathrm{gen}}(x)|\sin \theta_{\mathcal{S}}}, \end{align*} $$

we want to show that $\mathcal {Z}^n_{\mathrm {gen}}(V)$ intersects one of the planes that belong to the Julia set for some n and thus V itself intersects the Julia set.

The proof now proceeds in the same way as the proof of Theorem 1. We consider the same two cases.

1. (i) The iterates $\mathcal {Z}^n_{\mathrm {gen}}(V)$ do not eventually stay inside the beam $B_{(0,0)}\cup B_{(0,-1)}$ . In this case the proof is the same almost word for word.

2. (ii) The iterates $\mathcal {Z}^n_{\mathrm {gen}}(V)$ eventually stay inside $B_{(0,0)}\cup B_{(0,-1)}$ . The idea in this case will be the same. We leave the details, which will be slightly different, to the interested reader.