Appendix
1.1 Appendix 1: Calculations to Show that the Zorich Map is Quasiregular
Theorem 6.1
If \(g:D\rightarrow \mathbb {R}^n\), where \(D\subset \mathbb {R}^{n-1}\times \{0\}\) is a \(n-1\) regular polytope with g(D) being the upper unit sphere is infinitesimally bi-Lipschitz, then
$$\begin{aligned} \mathcal {Z}(x)=e^{x_n}h(x_1,\ldots ,x_{n-1},0) \end{aligned}$$
is quasiregular in \(\mathbb {R}^n\), where \(h:\mathbb {R}^{n-1}\times \{0\}\rightarrow \mathbb {R}^n\) is the extension of g by reflections as defined earlier.
Proof
Since h is extended by reflections in \((n-2)\)-faces of D, we can restrict our attention to \(h|_D=g\). Note that we can see that since g is infinitesimally bi-Lipschitz that
$$\begin{aligned} \mathcal {Z}|_D(x)=e^{x_n}g(x_1,\ldots ,x_{n-1},0) \end{aligned}$$
is absolutely continuous on lines. Also, since we are multiplying each coordinate in the image of g by \(e^{x_n}\) we can see that \(\mathcal {Z}|_D\) must also be locally \(L^n\)-integrable. All that is left to show is that \(\mathcal {Z}_g\) has bounded distortion
Since g is infinitesimally bi-Lipschitz, there is a \(L\ge 1\) such that
$$\begin{aligned} \frac{1}{L}\le \liminf _{\epsilon \rightarrow 0}\frac{|g(x+\epsilon )-g(x)|}{|\epsilon |} \le \limsup _{\epsilon \rightarrow 0}\frac{|g(x+\epsilon )-g(x)|}{|\epsilon |}\le L, \end{aligned}$$
for all \(x\in D\), \(\epsilon =(\epsilon _1,\ldots ,\epsilon _n)\). The linear distortion function from Iwaniec and Martin, [9, Sect. 6.4], of \(\mathcal {Z}\) is defined to be
$$\begin{aligned} H(x,\mathcal {Z})&=\limsup _{r\rightarrow 0}\frac{\max _{|\epsilon |=r}|\mathcal {Z}(x+\epsilon ) -\mathcal {Z}(x)|}{\min _{|\epsilon |=r}|\mathcal {Z}(x+\epsilon )-\mathcal {Z}(x)|}\\&=\limsup _{r\rightarrow 0}\frac{\max _{|\epsilon |=r}|e^{x_n}\left( e^{\epsilon _n}g(x_1 +\epsilon _1,\ldots ,x_{n-1}+\epsilon _{n-1},0)-g(x_1,\ldots ,x_{n-1},0) \right) |}{\min _{|\epsilon |=r}|e^{x_n}\left( e^{\epsilon _n}g(x_1+\epsilon _1,\ldots , x_{n-1}+\epsilon _{n-1},0)-g(x_1,\ldots ,x_{n-1},0)\right) |}. \end{aligned}$$
Note that
$$\begin{aligned} \lim _{x\rightarrow 0}\frac{e^x-1}{x}=1, \end{aligned}$$
so there is \(a>0, a\in \mathbb {R}\), such that \(|e^{\epsilon _n}-1|=a|\epsilon _n|\), where \(a\rightarrow 1\) as \(\epsilon _n\rightarrow 0\). For notation, let \(\bar{x}=(x_1,\ldots ,x_{n-1},0)\). Also note that \(r^2=|\epsilon |^2=|\bar{\epsilon }|^2+|\epsilon _n|^2\), so that \(|\epsilon _n|^2=r^2-|\bar{\epsilon }|^2\). This leads to
$$\begin{aligned} |\mathcal {Z}(x+\epsilon )-\mathcal {Z}(x)|&=e^{x_n}|e^{\epsilon _n}g(\bar{x} +\bar{\epsilon })-g(\bar{x})|\\&=e^{x_n}|e^{\epsilon _n}\left( g(\bar{x}+\bar{\epsilon })-g(\bar{x}) \right) +g(\bar{x})\left( e^{\epsilon _n}-1 \right) |. \end{aligned}$$
Notice that \(g(\bar{x}+\bar{\epsilon })-g(\bar{x})\) describes how the first \(n-1\) coordinates map onto the unit sphere. In particular for a point A on the unit sphere, \(g(\bar{x}+\bar{\epsilon })-g(\bar{x})\) moves point A to point B, still on the unit sphere, by a distance of \(c|\bar{\epsilon }|\) where \(\frac{1}{L}\le c\le L\), since h is bi-Lipschitz. Then \(e^{\epsilon _n}-1\) will move point B orthogonally from the unit sphere to a point C by a distance of \(|e^{\epsilon _n}-1|=a|\epsilon _n|\). Let \(L'\) be the distance from point A to point C, in particular
$$\begin{aligned} L'=|e^{\epsilon _n}\left( g(\bar{x}+\bar{\epsilon })-g(\bar{x}) \right) +g(\bar{x})\left( e^{\epsilon _n}-1 \right) |. \end{aligned}$$
One can also notice that \(\angle ABC=\pi /2+\delta \) with \(\delta >0\) where \(\delta \rightarrow 0\) as \(r\rightarrow 0\). The linear distance \(L'\) is then
$$\begin{aligned} L'^2&=\left( c|\bar{\epsilon }| \right) ^2+a^2|\epsilon _n|^2-2ac|\bar{\epsilon }||\epsilon _n|\cos \left( \frac{\pi }{2} +\delta \right) \\&=c^2|\bar{\epsilon }|^2+a^2r^2-a^2|\bar{\epsilon }|^2+2ac|\bar{\epsilon }||\epsilon _n |\delta . \end{aligned}$$
Since \(c\le L\), we have that
$$\begin{aligned} L'^2\le L^2|\bar{\epsilon }|^2+a^2r^2+2L|\bar{\epsilon }||\epsilon _n|\delta \le r^2(L^2+a^2+2La\delta ). \end{aligned}$$
For \(\epsilon \) sufficiently small, we can have a close enough to 1 and \(\delta \) small enough so that
$$\begin{aligned} L'\le r\sqrt{L^2+a^2+2La\delta }\le 2r\sqrt{L^2+1}. \end{aligned}$$
We also have that
$$\begin{aligned} L'^2&\ge \frac{1}{L^2}|\bar{\epsilon }|^2+a^2r^2-a^2|\bar{\epsilon }|^2 +\frac{2}{L}a|\bar{\epsilon }||\epsilon _n|\delta \\&\ge \frac{r^2}{L^2}-\frac{|\epsilon _n|^2}{L^2}+a^2|\epsilon _n|^2\\&=\frac{r^2}{L^2}+\frac{(a^2L^2-1)|\epsilon _n|^2}{L^2}. \end{aligned}$$
Since \(L^2\ge 1\) we have \(a^2L^2\ge a^2\) which gives us \(a^2L^2-1\ge a^2-1\). We have that
$$\begin{aligned} L'^2\ge \frac{r^2}{L^2}+\frac{(a^2-1)|\epsilon _n|^2}{L^2}. \end{aligned}$$
Since \(r^2=|\bar{\epsilon }|^2+|\epsilon _n|^2\), we know that \(|\epsilon _n|\in [0,r]\). If \(a^2-1\ge 0\), then
$$\begin{aligned} L'^2\ge \frac{r^2}{L^2} \end{aligned}$$
which means
$$\begin{aligned} L'\ge \frac{r}{L}>\frac{r}{2\sqrt{L^2+1}}. \end{aligned}$$
If \(a^2-1<0\) we have
$$\begin{aligned} L'^2\ge \frac{r^2}{L^2}+\frac{(a^2-1)r^2}{L^2}=\frac{a^2r^2}{L^2}. \end{aligned}$$
Since \(a\rightarrow 1\) as \(r\rightarrow 0\), we can find r small enough so that \(a>\frac{1}{2}\). Then we have
$$\begin{aligned} L'^2\ge \frac{(1/2)^2r^2}{L^2}=\frac{1}{4L^2}. \end{aligned}$$
Again, we get
$$\begin{aligned} L'\ge \frac{r}{2L} >\frac{r}{2\sqrt{L^2+1}}. \end{aligned}$$
Then we have that our linear distortion
$$\begin{aligned} H(x,\mathcal {Z})&=\limsup _{r\rightarrow 0}\frac{\max _{|\epsilon |=r}|e^{x_n} \left( e^{\epsilon _n}g(x_1+\epsilon _1,\ldots ,x_{n-1}+\epsilon _{n-1},0) -g(x_1,\ldots ,x_{n-1},0)\right) |}{\min _{|\epsilon |=r}|e^{x_n}\left( e^{\epsilon _n} g(x_1+\epsilon _1,\ldots ,x_{n-1}+\epsilon _{n-1},0)-g(x_1,\ldots ,x_{n-1},0)\right) |}\\&\le \frac{2r\sqrt{L^2+1}}{\frac{r}{2\sqrt{L^2+1}}}=4(L^2+1)\le 8L^2. \end{aligned}$$
From (2.1) and (2.2) we have that the distortion K of \(\mathcal {Z}\) is bounded by
$$\begin{aligned} \left( H(x,\mathcal {Z}) \right) ^{n-1}=(8L^2)^{n-1}. \end{aligned}$$
Therefore \(\mathcal {Z}\) is quasiregular. \(\square \)
Appendix 2: Our Particular Function For Zorich Map is Infinitesimally Bi-Lipschitz
Recall that we defined
$$\begin{aligned} g(x_1,\ldots ,x_{n-1},0)= & {} \left( \frac{x_1\sin M(x_1,\ldots ,x_{n-1})}{\sqrt{x_1^2+\cdots +x_{n-1}^2}},\ldots ,\frac{x_{n-1}\sin M(x_1,\ldots ,x_{n-1})}{\sqrt{x_1^2+\cdots +x_{n-1}^2}}\right. ,\\&\quad \left. \,\cos M(x_1,\ldots ,x_{n-1}) \right) , \end{aligned}$$
where \(M(x_1,\ldots ,x_{n-1})=\max \{|x_1|,\ldots ,|x_{n-1}|\}\), which maps the \([-\pi /2,\pi /2]^{n-1}\) cube to the half unit sphere in \(\mathbb {R}^n\) where \(y_n\ge 0\) in the image. The calculations for \(n>3\) are very similar, but even more tedious than the calculations for \(n=3\). We will show that for \(n=3\) that for \(g:[-\pi /2,\pi /2]^2\rightarrow \mathbb {R}^3\) defined by
$$\begin{aligned} g(x,y,0)=\left( \frac{x\sin M(x,y)}{\sqrt{x^2+y^2}},\frac{y\sin M(x,y)}{\sqrt{x^2+y^2}},\cos M(x,y) \right) \end{aligned}$$
where \(M(x,y)=\max \{|x|,|y| \},\) is infinitesimally bi-Lipschitz, and then note that by similarity we can conclude that all other g functions for \(n>3\) are also infinitesimally bi-Lipschitz.
Without loss of generality, since g is symmetric in the square, we will restrict ourselves to
$$\begin{aligned} A:=\{(x,y,z)\in [-\pi /2,\pi /2]^2\times \{0 \}:x\ge |y|\}, \end{aligned}$$
so that \(M(x,y)=x\) for \((x,y,z)\in A\). Note that when we take \((x,y,z)\in A\) we can omit the origin, a single point has Lebesgue measure zero, and so our map will still have bounded distortion and will be quasiregular. First we will note some useful Taylor series expansions:
$$\begin{aligned} \cos \epsilon&=1-\frac{\epsilon ^2}{2}+o(\epsilon ^2),\\ \sin \epsilon&=\epsilon +o(\epsilon ^2),\text { and }\\ \left( (x+\epsilon )^2+(y+\delta )^2\right) ^{-1/2}&=(x^2+y^2)^{-1/2}\left( 1-\frac{\epsilon x+\delta y}{x^2+y^2}+o(|(\epsilon ,\delta )|^2) \right) . \end{aligned}$$
Here we will take \(\epsilon \) and \(\delta \) to be small enough so that \((x+\epsilon ,y+\delta )\in A\) for our calculations. One can ask about how we handle the distortion about the boundary of A. The following calculations will be similar with same final estimates when we consider the other triangle quadrants, which will give us our infinitesimally bi-Lipschitz result for h. We have
$$\begin{aligned} |g(x,y)-g(x+\epsilon ,y+\delta )|^2=|(u,v,w)|^2, \end{aligned}$$
where
$$\begin{aligned} u&=\frac{x\sin x}{\sqrt{x^2+y^2}}-\frac{(x+\epsilon )\sin (x+\epsilon )}{\sqrt{(x+\epsilon )^2+(y+\delta )^2}},\\ v&=\frac{y\sin x}{\sqrt{x^2+y^2}}-\frac{(y+\delta )\sin (x+\epsilon )}{\sqrt{(x+\epsilon )^2+(y+\delta )^2}},\text { and}\\ w&=\cos x-\cos (x+\epsilon ). \end{aligned}$$
Using the Taylor series above, we have the following calculations,
$$\begin{aligned} u^2&=\left( \frac{x\sin x}{\sqrt{x^2+y^2}}-\frac{(x+\epsilon )(\sin x\cos \epsilon +\cos x\sin \epsilon )}{\sqrt{x^2+y^2}}\left( 1-\frac{\epsilon x+\delta y}{x^2+y^2} \right) \right) ^2\\&\quad +\,o(|(\epsilon ,\delta )|^2)\\&=\frac{1}{x^2+y^2}\left( \frac{-\epsilon x^2\sin x}{x^2+y^2}-\frac{\delta yx\sin x}{x^2+y^2}+\epsilon x\cos x+\epsilon \sin x \right) ^2+ o(|(\epsilon ,\delta )|^2)\\&=\frac{1}{x^2+y^2}\left( \frac{\epsilon ^2x^4\sin ^2x}{(x^2+y^2)^2}+\frac{2\epsilon \delta x^3y\sin ^2x}{(x^2+y^2)^2}-\frac{2\epsilon ^2x^3\sin x\cos x}{x^2+y^2}-\frac{2\epsilon ^2x^2\sin ^2x}{x^2+y^2}\right. \\&\left. \quad +\,\frac{\delta ^2x^2y^2\sin ^2x}{(x^2+y^2)^2}\right) +\frac{1}{x^2+y^2}\\&\quad \times \,\left( \frac{-2\epsilon \delta x^2y\sin x\cos x}{x^2+y^2}-\frac{2\epsilon \delta xy\sin ^2x}{x^2+y^2}+\epsilon ^2x^2\cos ^2x+2\epsilon ^2x\sin x\cos x\right. \\&\left. \quad +\,\epsilon ^2\sin ^2x \right) +o(|(\epsilon ,\delta )|^2),\\ v^2&=\left( \frac{y\sin x}{\sqrt{x^2+y^2}}-\frac{(y+\delta )(\sin x\cos \epsilon +\cos x\sin \epsilon )}{\sqrt{x^2+y^2}}\left( 1-\frac{\epsilon x+\delta y}{x^2+y^2}\right) \right) ^2\\&\quad +\,o(|(\epsilon ,\delta )|^2)\\&=\frac{1}{x^2+y^2}\left( \frac{-\epsilon xy\sin x}{x^2+y^2}-\frac{\delta y^2\sin x}{x^2+y^2}+\epsilon y\cos x+\delta \sin x \right) ^2+o(|(\epsilon ,\delta )|^2)\\&=\frac{1}{x^2+y^2}\left( \frac{\epsilon ^2x^2y^2\sin ^2x}{(x^2+y^2)^2}+\frac{2\epsilon \delta xy^3\sin ^2x}{(x^2+y^2)^2}-\frac{2\epsilon ^2xy^2\sin x\cos x}{x^2+y^2}\right. \\&\left. \quad -\,\frac{2\epsilon \delta xy\sin ^2x}{x^2+y^2}+\frac{\delta ^2y^4\sin ^2x}{(x^2+y^2)^2} \right) \\&\quad +\frac{1}{x^2+y^2}\left( \frac{-2\epsilon \delta y^3\sin x\cos x}{x^2+y^2}-\frac{2\delta ^2y^2\sin ^2x}{x^2+y^2}+\epsilon ^2y^2\cos ^2x \right. \\&\left. \quad +\,2\epsilon \delta y\sin x\cos x+\delta ^2\sin ^2x \right) +o(|(\epsilon ,\delta )|^2), \end{aligned}$$
and
$$\begin{aligned} w^2&=\left( \cos x-(\cos x\cos \epsilon -\sin \epsilon \sin x) \right) ^2+o(|(\epsilon ,\delta )|^2)\\&=\left( \cos x-\cos x+\frac{\epsilon ^2}{2}\cos x+\epsilon \sin x\right) ^2+o(|(\epsilon ,\delta )|^2)\\&=\epsilon ^2\sin ^2x+o(|(\epsilon ,\delta )|^2). \end{aligned}$$
Then separating into \(\epsilon ^2\), \(\delta ^2\), and \(\epsilon \delta \) terms we have
$$\begin{aligned} |(u,v,w)|^2&=u^2+v^2+w^2\\&=\epsilon ^2+\frac{\epsilon ^2\sin ^2x}{(x^2+y^2)^3}\left( x^4-2x^4-2x^2y^2+2x^2y^2+y^4+x^2y^2 \right) \\&\quad +\frac{\delta ^2\sin ^2x}{(x^2+y^2)^3}\left( x^2y^2+y^4-2x^2y^2-2y^4+x^4+2x^2y^2+y^4 \right) \\&\quad +\frac{\epsilon \delta \sin ^2x}{(x^2+y^2)^3}\left( 2x^3y-2x^3y-2xy^3+2xy^3-2x^3y-2xy^3 \right) \\&\quad +\,o(|(\epsilon ,\delta )|^2)\\&=\epsilon ^2\left( 1+\frac{y^2\sin ^2x}{(x^2+y^2)^2} \right) +\delta ^2\left( \frac{x^2\sin ^2x}{(x^2+y^2)^2} \right) -2\epsilon \delta \left( \frac{xy\sin ^2x}{(x^2+y^2)^2}\right) \\&\quad +\,o(|(\epsilon ,\delta )|^2). \end{aligned}$$
Here we have that
$$\begin{aligned} |g(x,y)-g(x+\epsilon ,y+\delta )|^2&=\epsilon ^2\left( 1+\frac{y^2\sin ^2x}{(x^2+y^2)^2} \right) +\delta ^2\left( \frac{x^2\sin ^2x}{(x^2+y^2)^2} \right) \\&\quad -\,2\epsilon \delta \left( \frac{xy\sin ^2x}{(x^2+y^2)^2}\right) +o(|(\epsilon ,\delta )|^2). \end{aligned}$$
We can notice that the term
$$\begin{aligned} \epsilon ^2\left( 1+\frac{y^2\sin ^2x}{(x^2+y^2)^2} \right) +\delta ^2\left( \frac{x^2\sin ^2x}{(x^2+y^2)^2} \right) -2\epsilon \delta \left( \frac{xy\sin ^2x}{(x^2+y^2)^2}\right) \end{aligned}$$
is a quadratic form in \((\epsilon , \delta )\) with corresponding matrix
$$\begin{aligned} B=\begin{pmatrix} 1+\frac{y^2\sin ^2x}{(x^2+y^2)^2}&{}\frac{-xy\sin ^2x}{(x^2+y^2)^2}\\ \frac{-xy\sin ^2x}{(x^2+y^2)^2}&{} \frac{x^2\sin ^2x}{(x^2+y^2)^2} \end{pmatrix}. \end{aligned}$$
Since we are in quadratic form, the eigen-values and eigen-vectors tell us how much and in what direction we have distortion. If the eigen-values are bounded above and below by positive constants, then we have that our map h is infinitesimally bi-Lipschitz. That is, if the eigen-values \(\lambda \) have the bounding \(\frac{1}{L}\le \lambda \le L\) for some \(L\ge 1\), then we have
$$\begin{aligned} \frac{1}{L}(\epsilon ^2+\delta ^2)\le & {} \epsilon ^2\left( 1+\frac{y^2\sin ^2x}{(x^2+y^2)^2} \right) +2\epsilon \delta \left( \frac{-xy\sin ^2x}{(x^2+y^2)^2}\right) +\delta ^2\left( \frac{x^2\sin ^2x}{(x^2+y^2)^2}\right) \\\le & {} L(\epsilon ^2+\delta ^2), \end{aligned}$$
which when we consider the small error term gives us an \(\tilde{L}\ge 1\) such that
$$\begin{aligned} \frac{1}{\tilde{L}}(\epsilon ^2+\delta ^2)\le |g(x,y)-g(x+\epsilon ,y+\delta )|^2 \le \tilde{L}(\epsilon ^2+\delta ^2). \end{aligned}$$
To find our eigen-values, we have
$$\begin{aligned} \det (\lambda I-B)&=\lambda ^2-\lambda \left( \frac{x^2\sin ^2x}{(x^2+y^2)^2}+1 +\frac{y^2\sin ^2x}{(x^2+y^2)^2} \right) -\frac{x^2y^2\sin ^4x}{(x^2+y^2)^4}\\&\quad +\,\left( 1+\frac{y^2\sin ^2x}{(x^2+y^2)^2} \right) \left( \frac{x^2\sin ^2x}{(x^2+y^2)^2} \right) \\&=\lambda ^2-\lambda \left( 1+\frac{\sin ^2x}{x^2+y^2}\right) +\frac{x^2\sin ^2x}{(x^2+y^2)^2}, \end{aligned}$$
so that \(\det (\lambda I-B)=0\) when
$$\begin{aligned} \lambda =\frac{1}{2}\left( 1+\frac{\sin ^2x}{x^2+y^2}\pm \sqrt{\frac{(x^2-\sin ^2x)^2 +2x^2y^2+y^4+2y^2\sin ^2x}{(x^2+y^2)^2} } \right) . \end{aligned}$$
For the rest of the calculations, we will use facts about \(\sin x/x\), that is
Lemma 7.1
If \(f(x)=\frac{\sin x}{x}\), then f is decreasing on \((0,\pi /2)\) and \(f:[0,\pi /2]\rightarrow [2/\pi ,1].\)
Here, we will show that \(\lambda >0\), first we are assuming that \(x\ne 0\), so that we are not at the origin. Also note that
$$\begin{aligned} \lambda =\frac{1}{2a}(-b\pm \sqrt{b^2+4ac})=\frac{1}{2}\left( -b\pm \sqrt{b^2-4c} \right) \end{aligned}$$
where
$$\begin{aligned} a&=1,\\ b&=-\left( 1+\frac{\sin ^2x}{x^2+y^2} \right) ,\text { and}\\ c&=\frac{x^2\sin ^2x}{(x^2+y^2)^2}. \end{aligned}$$
First note that \(-b>0\) and \(c>0\). We also have
$$\begin{aligned} b^2-4c=\frac{(x^2-\sin ^2x)^2+2x^2y^2+y^4+2y^2\sin ^2x}{(x^2+y^2)^2}>0, \end{aligned}$$
then \(|b|>\sqrt{b^2-4c}.\) This gives us
$$\begin{aligned} \lambda =-b\pm \sqrt{b^2-4c}>0. \end{aligned}$$
Since we have \(x\ge |y|\), with \(x\ne 0\) since we are not at the origin, then
$$\begin{aligned} \lambda&\le \frac{1}{2}\left( 1+\frac{\sin ^2x}{x^2}+\sqrt{\frac{(x^2)^2}{x^4}+\frac{2x^4}{x^4}+\frac{x^4}{x^4}+\frac{2x^2\sin ^2x}{x^4} } \right) \\&\le \frac{1}{2}\left( 1+1+\sqrt{1+2+1+2}\right) =1+\frac{\sqrt{6}}{2}. \end{aligned}$$
Let \(p=-b\) and \(q=\sqrt{b^2-4c}\), then \(\lambda =p\pm q\). We showed that
$$\begin{aligned} \lambda \le p+q<1+\frac{\sqrt{6}}{2}. \end{aligned}$$
Also, note
$$\begin{aligned} p^2-q^2&=(-b)^2-\left( \sqrt{b^2-4c} \right) ^2=b^2-b^2+4c=4c\\&=4\frac{x^2\sin ^2x}{(x^2+y^2)^2}\ge 4\frac{x^2\sin ^2x}{(x^2+x^2)^2} =4\frac{x^2\sin ^2x}{4x^4}\\&=\frac{\sin ^2x}{x^2}\ge \frac{4}{\pi ^2}, \end{aligned}$$
since we have \((x,y)\in A\). Here we want to show that \(\lambda \ge p-q\) is bounded from below. We know that
$$\begin{aligned} p+q&\le 1+\frac{\sqrt{6}}{2},\text { and} \\ p-q&\ge \frac{4}{\pi ^2}. \end{aligned}$$
This leads to the following calculation,
$$\begin{aligned} \lambda&\ge p-q=\frac{p^2-q^2}{p+q}\ge \frac{\frac{4}{\pi ^2} }{p+q}\\&\ge \frac{\frac{4}{\pi ^2} }{1+\frac{\sqrt{6}}{2}}=\frac{8}{\pi ^2(2+\sqrt{6})}. \end{aligned}$$
Since \(\frac{\pi ^2(2+\sqrt{6})}{8}>1+\frac{\sqrt{6}}{2}\), then we can let \(L=\frac{\pi ^2(2+\sqrt{6})}{8}\), so that g(x, y) is infinitesimally bi-Lipschitz.
1.1 Appendix 3: Derivative Calculations For the Radial Stretch with Spiraling Map
The Zorich transform of \(R_\mathrm{{s}}\) is defined by
$$\begin{aligned} \tilde{R_\mathrm{{s}}}(x_1,\ldots ,x_n)=(u_1,\ldots ,u_n) \end{aligned}$$
with
$$\begin{aligned} u_i={\left\{ \begin{array}{ll} Mm(x_1\cos (\alpha x_n)-x_2\sin (\alpha x_n))&{} \text { for }i=1\\ Mm(x_1\sin (\alpha x_n)+x_2\cos (\alpha x_n))&{} \text { for }i=2\\ Mmx_i&{}\text { for } 3\le i\le n-1\\ x_n+\ln K-\frac{1}{2}\ln \left( K^2+(1-K^2)\frac{x_1^2\sin ^2 M}{x_1^2+\cdots +x_{n-1}^2}\right) &{} \text { for }i=n \end{array}\right. }, \end{aligned}$$
where
$$\begin{aligned} M&=M(x_1,\ldots ,x_{n-1})=\max \{|x_1|,\ldots ,|x_{n-1}| \}\text { and,}\\ m&=m(x_1,\ldots ,x_{n-1})=\min \left\{ \frac{1}{|x_1\cos (\alpha x_n)-x_2\sin (\alpha x_n)|},\right. \\&\left. \quad \,\frac{1}{|x_1\sin (\alpha x_n)+x_2\cos (\alpha x_n)|},\frac{1}{|x_3|},\ldots ,\frac{1}{|x_{n-1}|} \right\} . \end{aligned}$$
We will discuss bounding \(\tilde{R_\mathrm{{s}}}'\) and \(J_{\tilde{R_\mathrm{{s}}}}\) where the derivatives exist. Also, for \(x_{1},\ldots ,x_{n-1}\) not all zero,
$$\begin{aligned} M(u_1,\ldots ,u_{n-1})=M(x_1,\ldots ,x_{n-1}) \end{aligned}$$
since m will cancel with one of the following, \((x_1\cos (\alpha x_n)-x_2\sin (\alpha x_n)), (x_1\sin (\alpha x_n)+x_2\cos (\alpha x_n)),x_3,\ldots ,x_{n-1}\) leaving one of the \(u_i\) as \(\pm M\). By definition of m we have that \(|u_j|\le |u_i|\) for \(1\le j\le n-1\). We indeed have that \(\mathcal {Z}\circ \tilde{R_\mathrm{{s}}}=R_s\circ \mathcal {Z}\). A useful calculation is that if either \(x_1\) or \(x_2\) are not zero, then
$$\begin{aligned} \frac{\sqrt{x_1^2+x_2^2}}{\sqrt{2}}\le & {} \max \{|x_1\cos (\alpha x_n)-x_2\sin (\alpha x_n)|,|x_1\sin (\alpha x_n)+x_2\cos (\alpha x_n)| \}\\\le & {} \sqrt{x_1^2+x_2^2}, \end{aligned}$$
which gives us
$$\begin{aligned} \frac{1}{\sqrt{x_1^2+x_2^2}}\le & {} \min \left\{ \frac{1}{|x_1\cos (\alpha x_n)-x_2\sin (\alpha x_n)|},\frac{1}{|x_1\sin (\alpha x_n)+x_2\cos (\alpha x_n)|}\right\} \nonumber \\\le & {} \frac{\sqrt{2}}{\sqrt{x_1^2+x_2^2}}. \end{aligned}$$
(7.1)
For \(\tilde{R_\mathrm{{s}}}\) to be quasiregular we want \(\tilde{R_\mathrm{{s}}}'\) and \(\left( \tilde{R_\mathrm{{s}}}'\right) ^{-1}\) to be bounded. For \(1\le i\le n-1\), let
$$\begin{aligned} A_{i}:= & {} \left\{ (x_1,\ldots ,x_{n-1},x_n)\in \left[ -\frac{\pi }{2},\frac{\pi }{2}\right] ^{n-1}\times \mathbb {R}: x_i>|x_j|\text { for }j\ne i, 1\right. \\\le & {} \left. j\le n-1 \right\} . \end{aligned}$$
We will break our calculations into three cases, when \((x_1,\ldots ,x_n)\in A_1\), \((x_1,\ldots ,x_n)\in A_2\), and \((x_1,\ldots ,x_n)\in A_j\) for \(3\le j\le n-1\).
Case I Suppose that \((x_1,\ldots ,x_n)\in A_1\), so that \(M=x_1\). First note that the solution sets of the equations \(x_j=x_1\cos (\alpha x_n)-x_2\sin (\alpha x_n)\), \(x_j=x_1\sin (\alpha x_n)+x_2\cos (\alpha x_n)\), and \(x_1\cos (\alpha x_n)-x_2\sin (\alpha x_n)=x_1\sin (\alpha x_n)+x_2\cos (\alpha x_n)\) are closed, \(\sigma \)-finite \((n-1)\)-dimensional Hausdorff measurable sets. Note that our function is not differentiable on these sets as well. The following three sub-cases address the different regions we can be in \(A_1\) which are bounded by the solution sets described, or the boundary of \(A_1\).
Sub-case a Suppose that
$$\begin{aligned} m=\frac{1}{x_1\cos (\alpha x_n)-x_2\sin (\alpha x_n)}, \end{aligned}$$
if we had \(-m\) the derivative calculations will just have opposite signs and the bounding would work the same. Also if \(M=-x_1\), the following calculations would also just be of opposite sign and will not significantly change.
For this case we have
$$\begin{aligned} u_1&=x_1,\\ u_2&=(x_1^2\sin (\alpha x_n)+x_1x_2\cos (\alpha x_n))(x_1\cos (\alpha x_n)-x_2\sin (\alpha x_n))^{-1},\\ u_i&=x_1x_i(x_1\cos (\alpha x_n)-x_2\sin (\alpha x_n))^{-1}\text { for }3\le i\le n-1\text {, and}\\ u_n&=x_n+\ln (K)-\frac{1}{2}\ln \left( K^2+\left( 1-K^2\right) \frac{x_1^2\sin ^2x_1}{x_1^2+\cdots +x_{n-1}^2}\right) . \end{aligned}$$
We now have the derivative matrix
$$\begin{aligned} \tilde{R_\mathrm{{s}}}'=\begin{pmatrix} 1 &{} 0&{}0&{}0&{}\cdots &{}0&{}0\\ (u_2)_{x_1}&{} (u_2)_{x_2}&{}0&{}0&{}\cdots &{}0&{}(u_2)_{x_n}\\ (u_3)_{x_1}&{} (u_3)_{x_2}&{}(u_3)_{x_3}&{}0&{}\cdots &{}0&{}(u_3)_{x_n}\\ \vdots &{} \ddots &{}\cdots &{}\vdots &{}\ddots &{}\cdots &{}\vdots \\ (u_{n-1})_{x_1}&{} (u_{n-1})_{x_2}&{}0 &{}0&{}\cdots &{}(u_{n-1})_{x_{n-1}}&{}(u_{n-1})_{x_n}\\ (u_n)_{x_1}&{}(u_n)_{x_2}&{}(u_n)_{x_3}&{}(u_n)_{x_4}&{} \cdots &{}(u_{n})_{x_{n-1}}&{}1 \end{pmatrix}, \end{aligned}$$
where
$$\begin{aligned} (u_2)_{x_i}={\left\{ \begin{array}{ll} \frac{(x_1^2-x_2^2)\sin (\alpha x_n)\cos (\alpha x_n)-2x_1x_2\sin ^2(\alpha x_n)}{(x_1\cos (\alpha x_n)-x_2\sin (\alpha x_n))^2}&{}i=1\\ x_1^2/(x_1\cos (\alpha x_n)-x_2\sin (\alpha x_n))^2&{}i=2\\ 0 &{}3\le i\le n-1\\ \left( \alpha x_1^3+\alpha x_1x_2^2\right) /(x_1\cos (\alpha x_n)-x_2\sin (\alpha x_n))^2&{}i=n \end{array}\right. }, \end{aligned}$$
for \(3\le j\le n-1\) we have
$$\begin{aligned}&(u_j)_{x_i}\\&={\left\{ \begin{array}{ll} -x_2x_j\sin (\alpha x_n)/(x_1\cos (\alpha x_n)-x_2\sin (\alpha x_n))^2&{}i=1\\ x_1x_j\sin (\alpha x_n)/(x_1\cos (\alpha x_n)-x_2\sin (\alpha x_n))^2&{}i=2\\ 0&{}i\ne 1,2,j,n\\ \left( x_1^2\cos (\alpha x_n)-x_1x_2\sin (\alpha x_n)\right) /(x_1\cos (\alpha x_n)-x_2\sin (\alpha x_n))^2&{}i=j\\ \left( \alpha x_1^2x_j\sin (\alpha x_n)+\alpha x_1x_2x_j\cos (\alpha x_n) \right) /(x_1\cos (\alpha x_n)-x_2\sin (\alpha x_n))^2&{}i=n \end{array}\right. }, \end{aligned}$$
and
$$\begin{aligned}&(u_n)_{x_1}\\= & {} \frac{(K^2-1)\left[ \left( x_1\sin ^2x_1+x_1^2\sin x_1\cos x_1 \right) (x_1^2+\cdots +x_{n-1}^2)^{-1}-\frac{x_1^3\sin ^2x_1}{(x_1^2+\cdots +x_{n-1}^2)^2} \right] }{K^2+(1-K^2)\frac{x_1^2\sin ^2x_1}{x_1^2+\cdots +x_{n-1}^2}}, \end{aligned}$$
for \(2\le i\le n-1\) we have
$$\begin{aligned} (u_n)_{x_i}=\frac{(1-K^2)x_ix_1^2\sin ^2x_1/(x_1^2+\cdots +x_{n-1}^2)^2}{K^2+(1-K^2)\frac{x_1^2\sin ^2x_1}{x_1^2+\cdots +x_{n-1}^2}}, \end{aligned}$$
and
$$\begin{aligned} (u_n)_{x_n}=1. \end{aligned}$$
Here we will give bounding for the partial derivatives. Using the fact that \((x_1,\ldots ,x_{n})\in A_1\) and (7.1), and the fact that we are in sub-case a), we have
$$\begin{aligned} |(u_2)_{x_1}|&=\left| \frac{(x_1^2-x_2^2)\sin (\alpha x_n)\cos (\alpha x_n)-2x_1x_2\sin ^2(\alpha x_n)}{(x_1\cos (\alpha x_n)-x_2\sin (\alpha x_n))^2}\right| \\&\le \frac{2\left( x_1^2+x_2^2+2x_1^2\right) }{x_1^2+x_2^2}\\&\le 2\left( 1+2\right) =6. \end{aligned}$$
Using similar methods we have the following bounds:
$$\begin{aligned} |(u_2)_{x_i}|\le {\left\{ \begin{array}{ll} 6&{} i=1\\ 2&{}i=2\\ 0&{} 3\le i\le n-1\\ 8|\alpha |&{} i=n \end{array}\right. }, \end{aligned}$$
and
$$\begin{aligned} |(u_j)_{x_i}|\le {\left\{ \begin{array}{ll} 2&{}i=1\\ 2&{} i=2\\ 0&{} 3\le i\le n-1,i\ne j\\ 4&{} i=j\\ 8|\alpha |&{} i=n \end{array}\right. }. \end{aligned}$$
We need to use slightly different tactics to calculate a bound for the partial derivative \((u_n)_{x_1}\). We know that
$$\begin{aligned} 0\le \frac{ x_1^2\sin ^2 x_1}{x_1^2+\cdots +x_{n-1}^2}\le 1, \end{aligned}$$
so that
$$\begin{aligned} 1\le K^2+(1-K^2)\frac{x_1^2\sin ^2x_1}{x_1^2+\cdots +x_{n-1}^2}\le K^2. \end{aligned}$$
We will also use the fact that \(x_1\ge \sin (x_1)\) for \(x_1\ge 0\). From here, we have
$$\begin{aligned}&|(u_n)_{x_1}|\\= & {} \left| \frac{(K^2-1)\left[ \left( x_1\sin ^2x_1+x_1^2\sin x_1\cos x_1 \right) \left( x_1^2+\cdots +x_{n-1}^2 \right) ^{-1}-x_1^3\sin ^2x_1\left( x_1^2+\cdots +x_{n-1}^2\right) ^{-2}\right] }{K^2+(1-K^2)\frac{x_1^2\sin ^2x_1}{x_1^2+\cdots +x_{n-1}^2}}\right| \\\le & {} \left| (K^2-1)\left[ \left( x_1\sin ^2x_1+x_1^2\sin x_1\cos x_1 \right) \left( x_1^2+\cdots +x_{n-1}^2 \right) ^{-1}\right. \right. \\&\left. \left. \quad -\,x_1^3\sin ^2x_1\left( x_1^2+\cdots +x_{n-1}^2\right) ^{-2}\right] \right| \\\le & {} \left| K^2-1\right| \left( 1+1+|x_1|^5(x_1^2+\cdots +x_{n-1}^2)^{-2} \right) \\\le & {} |K^2-1|(2+\pi /2)\le 4(K^2-1). \end{aligned}$$
We also have for \(2\le i\le n-1\) that
$$\begin{aligned} |(u_n)_{x_i}|&=\left| \frac{(1-K^2)x_1^2x_i\sin ^2x_1}{\left( x_1^2+\cdots +x_{n-1}^2\right) ^2\left( K^2+(1-K^2)\frac{x_1^2\sin ^2x_1}{x_1^2+\cdots +x_{n-1}^2} \right) } \right| \\&\le (K^2-1)\frac{|x_1|^3}{(x_1^2+\cdots +x_{n-1}^2)^2}\\&\le (K^2-1). \end{aligned}$$
In conclusion, we have the bounds
$$\begin{aligned} |(u_n)_{x_i}|\le {\left\{ \begin{array}{ll} 4(K^2-1)&{} i=1\\ (K^2-1)&{} 2\le i\le n-1\\ 1 &{} i=n \end{array}\right. }. \end{aligned}$$
The above bounds are not sharp, but for our result of \(\tilde{R_\mathrm{{s}}}\) to be quasiregular, all we need to know is that these partial derivatives are bounded above by some constant value, so that \(\Vert \tilde{R_\mathrm{{s}}}'\Vert \) is bounded. We also want \(J_{\tilde{R_\mathrm{{s}}}}\) to be bounded from below, so that we can use Theorem 2.3. To the end of bounding \(J_{\tilde{R_\mathrm{{s}}}}\) from below, we can notice that the only terms that appear without an \(\alpha \) multiplying them occur when we multiply the diagonal of \(\tilde{R_\mathrm{{s}}}'\) together. That is, the non-alpha term of the Jacobian is
$$\begin{aligned} Q:=\frac{x_1^2(x_1^2\cos (\alpha x_n)-x_1x_2\sin (\alpha x_n))^{n-3}}{(x_1\cos (\alpha x_n)-x_2\sin (\alpha x_n))^{2(n-2)}}. \end{aligned}$$
Notice that using (7.1), we have that
$$\begin{aligned} Q&\ge \frac{x_1^{n-1}}{(x_1\cos (\alpha x_n)-x_2\sin (\alpha x_n))^{n-1}}\\&\ge \frac{x_1^{n-1}}{(\sqrt{x_1^2+x_2^2})^{n-1}}\\&\ge \frac{x_1^{n-1}}{(\sqrt{2x_1^2})^{n-1}}=2^{-(n-1)/2}. \end{aligned}$$
For here, we can choose \(\alpha \) so that \(|\alpha |>0\) is sufficiently small so that the alpha terms have absolute value less than \(\frac{1}{2}Q\). That is,
$$\begin{aligned} J_{\tilde{R_\mathrm{{s}}}}>\frac{1}{2}Q\ge 2^{-(n+1)/2}. \end{aligned}$$
Sub-case b For the case when
$$\begin{aligned} m=\frac{1}{x_1\sin (\alpha x_n)+x_2\cos (\alpha x_n)} \end{aligned}$$
we have
$$\begin{aligned} u_1&=(x_1^2\cos (\alpha x_n)-x_1x_2\sin (\alpha x_n))(x_1\sin (\alpha x_n)+x_2\cos (\alpha x_n))^{-1},\\ u_2&=x_1,\\ u_i&=x_1x_i(x_1\sin (\alpha x_n)+x_2\cos (\alpha x_n))^{-1}\text { for }3\le i\le n-1\text {, and}\\ u_n&=x_n+\ln (K)-\frac{1}{2}\ln \left( K^2+\left( 1-K^2\right) \frac{x_1^2\sin ^2x_1}{x_1^2+\cdots +x_{n-1}^2}\right) . \end{aligned}$$
This gives us the derivative matrix
$$\begin{aligned} \tilde{R_\mathrm{{s}}}'=\begin{pmatrix} (u_1)_{x_1}&{} (u_1)_{x_2}&{}\cdots &{}(u_1)_{x_n}\\ 1 &{} 0&{}\cdots &{}0\\ (u_3)_{x_1}&{}(u_3)_{x_2}&{}\cdots &{}(u_3)_{x_n}\\ \vdots &{} \ddots &{}\cdots &{}\vdots \\ (u_n)_{x_1}&{}\cdots &{} (u_n)_{x_{n-1}}&{}1 \end{pmatrix}, \end{aligned}$$
where
$$\begin{aligned} (u_1)_{x_i}={\left\{ \begin{array}{ll} \frac{(x_1^2-x_2^2)\sin (\alpha x_n)\cos (\alpha x_n)+2x_1x_2\cos ^2(\alpha x_n)}{(x_1\sin (\alpha x_n)+x_2\cos (\alpha x_n))^2}&{}i=1\\ -x_1^2/(x_1\sin (\alpha x_n)+x_2\cos (\alpha x_n))^2&{}i=2\\ 0&{}3\le i\le n-1\\ \left( -\alpha x_1^3-\alpha x_1x_2^2\right) /(x_1\sin (\alpha x_n)+x_2\cos (\alpha x_n))^2&{}i=n \end{array}\right. }, \end{aligned}$$
and for \(3\le j\le n-1\) we have
$$\begin{aligned} (u_j)_{x_i}={\left\{ \begin{array}{ll} x_2x_j\cos (\alpha x_n)/(x_1\sin (\alpha x_n)+x_2\cos (\alpha x_n))^2&{}i=1\\ -x_1x_j\cos (\alpha x_n) /(x_1\sin (\alpha x_n)+x_2\cos (\alpha x_n))^2&{}i=2\\ 0&{}i\ne 1,2, j,n\\ \left( x_1^2\sin (\alpha x_n)+x_1x_2\cos (\alpha x_n)\right) /(x_1\sin (\alpha x_n)+x_2\cos (\alpha x_n))^2&{}i=j\\ \left( -\alpha x_1^2x_j\cos (\alpha x_n)+\alpha x_1x_2x_j\sin (\alpha x_n) \right) /(x_1\sin (\alpha x_n)+x_2\cos (\alpha x_n))^2&{} i=n \end{array}\right. }. \end{aligned}$$
The partial derivatives of \(u_n\) are the same as sub-case a. Also, by looking at the similarities we can see that all of these derivatives are bounded from above, and that we can choose \(\alpha \) small enough so that \(J_{\tilde{R_\mathrm{{s}}}}>2^{-(n+1)/2}\).
Sub-case c Let \(m=x_j^{-1}\) for \(3\le j\le n-1\), we have
$$\begin{aligned} u_1&=(x_1^2\cos (\alpha x_n)-x_1x_2\sin (\alpha x_n))x_j^{-1},\\ u_2&=(x_1^2\sin (\alpha x_n)+x_1x_2\cos (\alpha x_n))x_j^{-1},\\ u_i&=x_1x_ix_j^{-1}\text { for }3\le i\le n-1,i\ne j,\\ u_j&=x_1\text {, and}\\ u_n&=x_n+\ln (K)-\frac{1}{2}\ln \left( K^2+\left( 1-K^2\right) \frac{x_1^2 \sin ^2x_1}{x_1^2+\cdots +x_{n-1}^2}\right) . \end{aligned}$$
We have the derivative matrix
$$\begin{aligned} \tilde{R_\mathrm{{s}}}'=\begin{pmatrix} (u_1)_{x_1}&{}(u_1)_{x_{2}}&{} \cdots &{}(u_1)_{x_n}\\ \vdots &{} \ddots &{}\cdots &{}\vdots \\ (u_n)_{x_1}&{}\cdots &{} (u_n)_{x_{n-1}}&{}1 \end{pmatrix}, \end{aligned}$$
where
$$\begin{aligned} (u_1)_{x_i}= & {} {\left\{ \begin{array}{ll} \left( 2x_1\cos (\alpha x_n)-x_2\sin (\alpha x_n)\right) x_j^{-1}&{}i=1\\ -x_1\sin (\alpha x_n)x_j^{-1}&{}i=2\\ 0&{}3\le i\le n-1,i\ne j\\ \left( -x_1^2\cos (\alpha x_n)+x_1x_2\sin (\alpha x_n) \right) x_j^{-2}&{} i=j,\\ \left( -\alpha x_1^2\sin (\alpha x_n)-\alpha x_1x_2\cos (\alpha x_n)\right) x_j^{-1}&{}i=n \end{array}\right. },\\ (u_2)_{x_i}= & {} {\left\{ \begin{array}{ll} \left( 2x_1\sin (\alpha x_n)+x_2\cos (\alpha x_n)\right) x_j^{-1}&{}i=1\\ x_1\cos (\alpha x_n)x_j^{-1}&{}i=2\\ 0&{}3\le i\le n-1,i\ne j\\ \left( -x_1^2\sin (\alpha x_n)-x_1x_2\cos (\alpha x_n) \right) x_j^{-2}&{} i=j,\\ \left( \alpha x_1^2\cos (\alpha x_n)-\alpha x_1x_2\sin (\alpha x_n)\right) x_j^{-1}&{}i=n \end{array}\right. },\\ (u_j)_{x_i}= & {} {\left\{ \begin{array}{ll} 1 &{} i=1\\ 0&{} i\ne 1 \end{array}\right. }, \end{aligned}$$
for \(3\le k\le n-1\), \(k\ne j\), we have
$$\begin{aligned} (u_k)_{x_i}={\left\{ \begin{array}{ll} x_kx_j^{-1}&{}i=1\\ 0&{}i\ne 1,k,j \\ x_1x_j^{-1}&{}i=k\\ -x_1x_kx_j^{-2}&{}i=j \end{array}\right. }. \end{aligned}$$
Note that the partial derivatives of \(u_n\) are the same as in the previous two cases and are bounded. Since we are assuming that \((x_1,\ldots ,x_n)\in A_1\) where the point at the origin is not included, and that \(M=x_1\ne 0\), this means that \(x_1>|x_i|\) for all \(2\le i\le n-1\). For \(m=1/|x_j|\) for some j, the definition of m and (7.1) give
$$\begin{aligned} \frac{1}{|x_j|}\le & {} \min \left\{ \frac{1}{|x_1\cos (\alpha x_n)-x_2\sin (\alpha x_n)|},\frac{1}{|x_1\sin (\alpha x_n)+x_2\cos (\alpha x_n)|} \right\} \nonumber \\\le & {} \frac{\sqrt{2}}{\sqrt{x_1^2+x_2^2}}. \end{aligned}$$
(7.2)
Using the fact that \((x_1,\ldots ,x_{n})\in A_1\) and (7.2), for \(l=1,2\) we have
$$\begin{aligned} |(u_l)_{x_i}|\le {\left\{ \begin{array}{ll} 6&{} i=1\\ 2&{}i=2\\ 0&{} 3\le i\le n-1,i\ne j\\ 4&{}i=j\\ 8|\alpha |&{} i=n \end{array}\right. }, \end{aligned}$$
and for \(k\ne j\), \(3\le k\le n-1\), we have
$$\begin{aligned} |(u_k)_{x_i}|\le {\left\{ \begin{array}{ll} 2&{}i=1,j,k\\ 0&{} i\ne 1,j,k \end{array}\right. }. \end{aligned}$$
In this case, we can calculate the Jacobian by first taking the determinate across jth row, so that
$$\begin{aligned} J_{\tilde{R_\mathrm{{s}}}}=(-1)^{j+1}\det \begin{pmatrix} (u_1)_{x_2}&{}\cdots &{} (u_1)_{x_n}\\ \vdots &{}\ddots &{} \vdots \\ (u_{j-1})_{x_2}&{}\cdots &{}(u_{j-1})_{x_n}\\ (u_{j+1})_{x_2}&{}\cdots &{}(u_{j+1})_{x_n}\\ \vdots &{}\ddots &{}\vdots \\ (u_n)_{x_2}&{}\cdots &{}(u_n)_{x_n} \end{pmatrix}. \end{aligned}$$
Now take the determinate down the column where we take the partial derivative with respect to \(x_n\), then the Jacobian is
$$\begin{aligned} J_{\tilde{R_\mathrm{{s}}}}&=(u_1)_{x_2}(u_2)_{x_j} \left( \prod _{\begin{array}{c} 3\le i\le n-1\\ i\ne j \end{array}}(u_i)_{x_i} \right) -(u_1)_{x_j}(u_2)_{x_2}\left( \prod _{\begin{array}{c} 3\le i\le n-1\\ i\ne j \end{array}} (u_i)_{x_i} \right) \\ {}&\qquad +(-1)^{n+1}(u_1)_{x_n}\det \begin{pmatrix} (u_2)_{x_1}&{}\cdots &{} (u_2)_{x_{n-1}}\\ \vdots &{}\ddots &{}\vdots \\ (u_n)_{x_1}&{} \cdots &{} (u_n)_{x_{n-1}} \end{pmatrix}\\&\qquad +\,(-1)^n(u_2)_{x_n}\begin{pmatrix} (u_1)_{x_1}&{} \cdots &{}(u_1)_{x_{n-1}}\\ (u_3)_{x_1}&{}\cdots &{} (u_3)_{x_{n-1}}\\ \vdots &{}\ddots &{}\vdots \\ (u_n)_{x_1}&{} \cdots &{} (u_n)_{x_{n-1}} \end{pmatrix} , \end{aligned}$$
so that the term without being multiplied by \(\alpha \) will be
$$\begin{aligned}&\frac{(x_1^2\cos (\alpha x_n)-x_1x_2\sin (\alpha x_n))x_1\cos (\alpha x_n) x_1^{n-4}+(-x_1^2-x_1x_1\cos (\alpha x_n))(-x_1\sin (\alpha x_n))x_1^{n-4}}{x_j^{n-1}}\\&\qquad =\frac{x_1^{n-1}(x_1^{3}\sin ^2(\alpha x_n)+x_1^2\cos ^2(\alpha x_n))}{x_j^{n-1}}=\frac{x_1^{n-1}}{x_j^{n-1}}>1. \end{aligned}$$
We need \(\alpha \) to be sufficiently small where we have
$$\begin{aligned} J_{\tilde{R_\mathrm{{s}}}}&>\frac{1}{2}\frac{(x_1^2\cos (\alpha x_n)-x_1x_2\sin (\alpha x_n))x_1\cos (\alpha x_n) x_1^{n-4}}{x_j^{n-1}}\\&\qquad +\,\frac{(-x_1^2-x_1x_1\cos (\alpha x_n))(-x_1\sin (\alpha x_n))x_1^{n-4}}{x_j^{n-1}}\\&>\frac{1}{2}>2^{-(n+1)/2}. \end{aligned}$$
The last inequality shows that all we need do is to choose \(\alpha \) in finitely many cases, so that the Jacobian is bounded from below by \(2^{-(n+1)/2}\). In other words, we can let \(\alpha \) be the minimal in size from sub-cases a, b, and c, then we obtain \(\Vert \tilde{R_\mathrm{{s}}}'\Vert \) bounded in each region.
Case II We have the case where \(M=x_2\), i.e., \((x_1,\ldots ,x_n)\in A_2\), which is similar to the case when \(M=x_1\). Running through similar calculations as in case I we can show that \(\tilde{R_\mathrm{{s}}}\) has bounded derivative matrix, where the derivative matrix is invertible. Moreover, we show that the Jacobian is bounded from below giving us that the inverse derivative matrix is bounded as well in the corresponding regions.
Case III Let \((x_1,\ldots ,x_n)\in A_j\) for some \(3\le j\le n-1\), so that \(M=x_j\), with \(x_j>|x_i|\) for \(1\le i\le n-1,\) \(i\ne j\), and that \(x_j\ne 0\). Here we will also break this case into three sub-cases for the same reasoning as in case I.
Sub-case a Suppose that
$$\begin{aligned} m=\frac{1}{x_1\cos (\alpha x_n)-x_2\sin (\alpha x_n)}, \end{aligned}$$
which means, by definition of m that
$$\begin{aligned} \frac{1}{x_1\cos (\alpha x_n)-x_2\sin (\alpha x_n)}\le \frac{1}{x_j}. \end{aligned}$$
(7.3)
This means that
$$\begin{aligned} x_1\cos (\alpha x_n)-x_2\sin (\alpha x_n)\ge x_j>0, \end{aligned}$$
which also implies that either \(x_1\ne 0\) or \(x_2\ne 0\). Since either \(x_1\) or \(x_2\) are not zero we have that (7.1) holds. We also have the inequality
$$\begin{aligned} \sqrt{x_1^2+x_2^2}\ge x_1\cos (\alpha x_n)-x_2\cos (\alpha x_n)\ge x_j. \end{aligned}$$
(7.4)
For this case we have
$$\begin{aligned} u_1&=x_j,\\ u_2&=(x_1x_j\sin (\alpha x_n)+x_2x_j\cos (\alpha x_n))(x_1\cos (\alpha x_n)-x_2\sin (\alpha x_n))^{-1},\\ u_i&=x_ix_j(x_1\cos (\alpha x_n)-x_2\sin (\alpha x_n))^{-1}\text { for }3\le i\le n-1, i\ne j\\ u_j&=x_j^2(x_1\cos (\alpha x_n)-x_2\sin (\alpha x_n))^{-1}\text { and,}\\ u_n&=x_n+\ln (K)-\frac{1}{2}\ln \left( K^2+\left( 1-K^2\right) \frac{x_1^2\sin ^2x_j}{x_1^2+\cdots +x_{n-1}^2}\right) . \end{aligned}$$
Define \(\Omega :=\{1,2,j,n\}\). We have the corresponding derivative matrix
$$\begin{aligned} \tilde{R_\mathrm{{s}}}'=\begin{pmatrix} (u_1)_{x_1}&{}(u_1)_{x_{2}}&{} \cdots &{}(u_1)_{x_n}\\ \vdots &{} \ddots &{}\cdots &{}\vdots \\ (u_n)_{x_1}&{}\cdots &{} (u_n)_{x_{n-1}}&{}1 \end{pmatrix}, \end{aligned}$$
where
$$\begin{aligned} (u_1)_{x_i}&={\left\{ \begin{array}{ll} 0&{} i\ne j\\ 1&{} i=j \end{array}\right. },\\ (u_2)_{x_i}&={\left\{ \begin{array}{ll} -x_2x_j/(x_1\cos (\alpha x_n)-x_2\sin (\alpha x_n))^2&{}i=1\\ x_1x_j/(x_1\cos (\alpha x_n)-x_2\sin (\alpha x_n))^2&{}i=2\\ 0&{}i\notin \Omega \\ \frac{\left( x_1^2-x_2^2\right) \sin (\alpha x_n)\cos (\alpha x_n)+x_1x_2\left( \cos ^2(\alpha x_n)-\sin ^2(\alpha x_n)\right) }{(x_1\cos (\alpha x_n)-x_2\sin (\alpha x_n))^2} &{}i=j\\ \left( \alpha x_1^2x_j+\alpha x_2^2x_j\right) /(x_1\cos (\alpha x_n)-x_2\sin (\alpha x_n))^2&{}i=n \end{array}\right. },\\ (u_j)_{x_i}\\&={\left\{ \begin{array}{ll} -x_j^2\cos (\alpha x_n)/\left( x_1\cos (\alpha x_n)-x_2\sin (\alpha x_n)\right) ^2 &{}i=1\\ x_j^2\sin (\alpha x_n)/\left( x_1\cos (\alpha x_n)-x_2\sin (\alpha x_n)\right) ^2 &{} i=2\\ 0 &{} i\notin \Omega \\ 2x_j/\left( x_1\cos (\alpha x_n)-x_2\sin (\alpha x_n)\right) &{}i=j\\ \left( \alpha x_1x_j^2\sin (\alpha x_n)+\alpha x_2x_j^2\cos (\alpha x_n) \right) /\left( x_1\cos (\alpha x_n)-x_2\sin (\alpha x_n)\right) ^2&{} i=n \end{array}\right. }, \end{aligned}$$
for \(3\le k\le n-1\), \(k\ne j\) we have
$$\begin{aligned} (u_k)_{x_i}\\&={\left\{ \begin{array}{ll} -x_kx_j\cos (\alpha x_n) /(x_1\cos (\alpha x_n)-x_2\sin (\alpha x_n))^2&{}i=1\\ x_kx_j\sin (\alpha x_n)/(x_1\cos (\alpha x_n)-x_2\sin (\alpha x_n))^2&{}i=2\\ 0&{}i\notin \Omega \cup \{k\} \\ x_k/(x_1\cos (\alpha x_n)-x_2\sin (\alpha x_n))&{}i=j\\ x_j/(x_1\cos (\alpha x_n)-x_2\sin (\alpha x_n))&{}i=k\\ \left( \alpha x_1x_kx_j\sin (\alpha x_n)+\alpha x_2x_kx_j\cos (\alpha x_n) \right) /(x_1\cos (\alpha x_n)-x_2\sin (\alpha x_n))^2&{}i=n \end{array}\right. }, \end{aligned}$$
and
$$\begin{aligned} (u_n)_{x_1}= & {} \frac{(K^2-1)\left[ \left( x_1\sin ^2x_j\right) \left( x_1^2+\cdots +x_{n-1}^2 \right) ^{-1}-x_1^3\sin ^2x_j\left( x_1^2+\cdots +x_{n-1}^2\right) ^{-2}\right] }{K^2+(1-K^2)\frac{x_1^2\sin ^2x_j}{x_1^2+\cdots +x_{n-1}^2}},\\ (u_n)_{x_i}= & {} \frac{(1-K^2)x_1^2x_i\sin ^2x_j}{\left( x_1^2+\cdots +x_{n-1}^2\right) \left( K^2+(1-K^2)\frac{x_1^2\sin ^2x_j}{x_1^2+\cdots +x_{n-1}^2} \right) }, \end{aligned}$$
for \(2\le i\le n-1\), \(i\ne j\),
$$\begin{aligned} (u_n)_{x_j}=\frac{(K^2-1)\left[ \left( x_1^2\sin x_j\cos x_j\right) \left( x_1^2+\cdots +x_{n-1}^2 \right) ^{-1}-x_1^2x_j\sin ^2x_j\left( x_1^2+\cdots +x_{n-1}^2\right) ^{-2}\right] }{K^2+(1-K^2)\frac{x_1^2\sin ^2x_j}{x_1^2+\cdots +x_{n-1}^2}}, \end{aligned}$$
and
$$\begin{aligned} (u_n)_{x_n}=1. \end{aligned}$$
Note that the partial derivatives of \(u_n\) are bounded using similar calculations as in case I. Using the fact that \((x_1,\ldots ,x_n)\in A_j\) and (7.3) we have the following bounds for \(l=2,j\)
$$\begin{aligned} |(u_l)_{x_i}|\le & {} {\left\{ \begin{array}{ll} 1&{} i=1,2\\ 0&{}i\ne 1,2,j,n\\ 6&{} i=j\\ 4|\alpha |&{} i=n \end{array}\right. }, \\ |(u_1)_{x_i}|\le & {} {\left\{ \begin{array}{ll} 1&{}i=1\\ 0&{}i\ne 1 \end{array}\right. }, \end{aligned}$$
by similar methods from Case I sub-case a) the bounds for partial derivatives of \(u_n\) are
$$\begin{aligned} |(u_n)_{x_i}|\le {\left\{ \begin{array}{ll} 3|K^2-1|&{} i=1,j\\ 2|K^2-1|&{}2\le i\le n-1, i\ne j\\ 1&{} i=n \end{array}\right. }, \end{aligned}$$
and for \(k\ne 1,2,j,n\) we have
$$\begin{aligned} |(u_k)_{x_i}|\le {\left\{ \begin{array}{ll} 1&{}i=1,2\\ 0&{} i\ne 1,2,j,n,k\\ 4&{} i=j\\ 2&{} i=k\\ 4|\alpha |&{} i=n \end{array}\right. }. \end{aligned}$$
To compute the Jacobian of \(\tilde{R_\mathrm{{s}}}\) for this case, first let
$$\begin{aligned} M=\begin{pmatrix}(u_2)_{x_1}&{}\cdots &{}(u_2)_{x_{j-1}}&{}(u_2)_{x_{j+1}}&{}\cdots &{}(u_2)_{x_n}\\ \vdots &{}\cdots &{}\ddots &{}\cdots &{}\ddots &{}\vdots \\ (u_n)_{x_n}&{}\cdots &{}(u_n)_{x_{j-1}}&{}(u_n)_{x_{j+1}}&{}\cdots &{}(u_n)_{x_n} \end{pmatrix}. \end{aligned}$$
Taking the determinate first row, we have
$$\begin{aligned} J_{\tilde{R_\mathrm{{s}}}}=(-1)^{j+1}\det M. \end{aligned}$$
Define \(M_i\) to be the square matrix of order \(n-2\) derived from removing the \((i-1)\)th row, \(2\le i\le n\) and \((n-1)\)th column from M. Taking the determinate of M first along the column where the partial derivatives are taken with respect to \(x_n\), we have that
$$\begin{aligned} J_{\tilde{R_\mathrm{{s}}}}&=\left( \prod _{\begin{array}{c} 3\le i\le n-1\\ i\ne j \end{array}}(u_i)_{x_i} \right) \left( (u_j)_{x_2}(u_2)_{x_1}-(u_j)_{x_1}(u_2)_{x_2}\right) +\sum _{i=2}^{n-1}\left[ (-1)^{n+i+j+1}(u_n)_{x_i}\det (M_i)\right] , \end{aligned}$$
so that the non-alpha term in \(J_{\tilde{R_\mathrm{{s}}}} \) is
$$\begin{aligned} Q&=\frac{-(-x_j^2)\cos (\alpha x_n)(x_1x_j)x_j^{n-4}(x_1\cos (\alpha x_n)-x_2\sin (\alpha x_n))^{n-4}}{\left( (x_1\cos (\alpha x_n)-x_2\sin (\alpha x_n))^2\right) ^{n-2}}\\ {}&\quad +\,\frac{x_j^2\sin (\alpha x_n)(-x_2x_j)x_j^{n-4}(x_1\cos (\alpha x_n)-x_2\sin (\alpha x_n))^{n-4}}{\left( (x_1\cos (\alpha x_n)-x_2\sin (\alpha x_n))^2\right) ^{n-2}}. \end{aligned}$$
Using (7.3) and simplifying equations we have the following lower bound for Q,
$$\begin{aligned} Q&=\frac{x_j^{n-1}(x_1\cos (\alpha x_n)-x_2\sin (\alpha x_n))^{n-3}}{(x_1\cos (\alpha x_n)-x_2\sin (\alpha x_n))^{2n-4}}\\&=\frac{x_j^{n-1}}{(x_1\cos (\alpha x_n)-x_2\sin (\alpha x_n))^{n-1}}\ge \frac{x_j^{n-1}}{x_j^{n-1}}=1. \end{aligned}$$
Then we need \(\alpha \) to be sufficiently small so that
$$\begin{aligned} J_{\tilde{R_\mathrm{{s}}}}>\frac{1}{2}Q\ge \frac{1}{2}. \end{aligned}$$
Then \(J_{\tilde{R_\mathrm{{s}}}}\) is bounded below and the norms of \(\tilde{R_\mathrm{{s}}}'\) and \(\left( \tilde{R_\mathrm{{s}}}'\right) ^{-1}\) are bounded above.
Sub-case bThe case when
$$\begin{aligned} m=\frac{1}{x_1\sin (\alpha x_n)+x_2\cos (\alpha x_n)} \end{aligned}$$
is very similar to sub-case a. Using similar calculations we have that \(J_{\tilde{R_\mathrm{{s}}}}\) is bounded below, and that \(\Vert \tilde{R_\mathrm{{s}}}'\Vert \) is bounded from above.
Sub-case c Finally, we are left with our last case when we let
$$\begin{aligned} m=\frac{1}{x_j}. \end{aligned}$$
We have that
$$\begin{aligned} u_1&=x_1\cos (\alpha x_n)-x_2\sin (\alpha x_n),\\ u_2&=x_2\sin (\alpha x_n)+x_2\sin (\alpha x_n),\\ u_i&=x_i\text { for }3\le i\le n-1\text {, and}\\ u_n&=x_n+\ln (K)-\frac{1}{2}\ln \left( K^2+\left( 1-K^2\right) \frac{x_1^2\sin ^2x_j}{x_1^2+\cdots +x_{n-1}^2}\right) . \end{aligned}$$
We have the corresponding derivative matrix
$$\begin{aligned} \tilde{R_\mathrm{{s}}}'=\begin{pmatrix} (u_1)_{x_1}&{}(u_1)_{x_{2}}&{} (u_1)_{x_3}&{}(u_1)_{x_4}&{} (u_1)_{x_5}&{}\cdots &{}(u_1)_{x_{n-1}}&{} (u_1)_{x_n}\\ (u_2)_{x_1}&{}(u_2)_{x_{2}}&{} (u_2)_{x_3}&{}(u_2)_{x_4}&{}(u_2)_{x_5} &{}\cdots &{}(u_2)_{x_{n-1}}&{}(u_2)_{x_n}\\ 0 &{} 0 &{}1 &{} 0 &{} 0 &{} \cdots &{} 0 &{}0 \\ 0 &{} 0 &{}0 &{} 1 &{} 0 &{} \cdots &{} 0 &{}0 \\ \vdots &{} \ddots &{}\cdots &{} \ddots &{}\cdots &{}\ddots &{}\cdots &{}\vdots \\ 0 &{} 0 &{}0 &{} 0 &{} 0 &{} \cdots &{} 1 &{}0 \\ (u_n)_{x_1}&{}(u_n)_{x_{2}}&{} (u_n)_{x_3}&{}(u_n)_{x_4}&{} (u_n)_{x_5}&{} \cdots &{} (u_n)_{x_{n-1}}&{}1 \end{pmatrix}, \end{aligned}$$
where
$$\begin{aligned} (u_1)_{x_i}= & {} {\left\{ \begin{array}{ll} \cos (\alpha x_n)&{} i=1\\ -\sin (\alpha x_n) &{} i=2\\ 0&{} 3\le i\le n-1\\ -\alpha x_1\sin (\alpha x_n)-\alpha x_2\cos (\alpha x_n) &{} i=n \end{array}\right. },\\ (u_2)_{x_i}= & {} {\left\{ \begin{array}{ll} \sin (\alpha x_n)&{} i=1\\ \cos (\alpha x_n)&{} i=2\\ 0&{}3\le i\le n-1\\ \alpha x_1\cos (\alpha x_n)-\alpha x_2\sin (\alpha x_n) &{} i=n \end{array}\right. }, \end{aligned}$$
and the partial derivatives for \(u_n\) are the same as in the sub-case a which we already remarked were all bounded from above.
We have the following bounds for the partial derivatives corresponding to \(l=1,2\)
$$\begin{aligned} |(u_l)_{x_i}|\le {\left\{ \begin{array}{ll} 1 &{} i=1,2\\ 0&{} 3\le i\le n-1\\ 6|\alpha |&{} i=n \end{array}\right. }. \end{aligned}$$
The term without \(\alpha \) in \(J_{\tilde{R_\mathrm{{s}}}}\) is
$$\begin{aligned} \cos ^2(\alpha x_n)+\sin ^2(\alpha x_n)=1. \end{aligned}$$
We can find an \(\alpha \) sufficiently small so that
$$\begin{aligned} J_{\tilde{R_\mathrm{{s}}}}\ge \frac{1}{2}. \end{aligned}$$
Therefore, the norm of \(\tilde{R_\mathrm{{s}}}'\) is bounded from above in the regions where \(\tilde{R_\mathrm{{s}}}\) is differentiable.
From cases I, II, and III, we have that the linear distortion of \(\tilde{R_\mathrm{{s}}}\) is bounded from above where \(\tilde{R_\mathrm{{s}}}\) is differentiable.
