Anti-orthotomics of frontals and their applications

TL;DR

This paper introduces the concept of anti-orthotomics for frontals, establishes their properties, and explores applications including a generalized vector formula, optical interpretations, and criteria for frontals.

Contribution

It defines the anti-orthotomic of a frontal relative to a point, proves its properties, and applies these results to geometric and optical problems.

Findings

Anti-orthotomic of a frontal is uniquely defined and retains frontal properties.

The anti-orthotomic satisfies specific geometric relations with the original frontal.

Applications include a generalized Cahn-Hoffman vector formula, optical interpretations, and criteria for frontals.

Abstract

Let $f: N^n\to \mathbb{R}^{n+1}$ be a frontal with its Gauss mapping $\nu: N\to S^n$ and let $P\in \mathbb{R}^{n+1}$ be a point such that $(f(x)-P)\cdot \nu(x) \ne 0$ for any $x\in N$. In this paper, for the mapping $\widetilde{f}: N\to \mathbb{R}^{n+1}$ defined by $$ \widetilde{f}(x)=f(x)-\frac{||f(x)-P||^2}{2(f(x)-P) \cdot \nu(x)}\nu(x), $$ the following four are shown. (1) $\widetilde{f}$ is a frontal with its Gauss mapping $\widetilde{\nu}(x)=\frac{f(x)-P}{||f(x)-P||}$ at $\widetilde{f}(x)$. (2) $\widetilde{f}$ is the unique anti-orthotomic of $f$ relative to $P$. (3) The property $(\widetilde{f}(x)-P)\cdot \widetilde{\nu}(x)\ne 0 $ holds for any $x\in N$. (4) The equality $||\widetilde{f}(x)-P||=||\widetilde{f}(x)-f(x)||$ holds for any $x\in N$. Moreover, three applications of the main result are given. As the first application, a generalization of Cahn-Hoffman vector formula is given. The second application is to clarify an optical meaning of anti-orthotomics. The third application gives a criterion to be a front for a given frontal.

Figures

Equations

$$\tilde{f}(x)=f(x)-\frac{||f(x)-P|{|}^2}{2(f(x)-P)\cdot \nu (x)}\nu (x),$$

$$f(x)=2\left(\left(\tilde{f}(x)-P\right)\cdot \tilde{\nu }(x)\right)\tilde{\nu }(x)+P.$$

$$g(x)=\left(\left(\tilde{f}(x)-P\right)\cdot \tilde{\nu }(x)\right)\tilde{\nu }(x)+P.$$

$$\left(\tilde{f}(x)-P\right)\cdot \tilde{\nu }(x)\ne 0.$$

$$f(x)=2\left((\tilde{f}(x)-P)\cdot \tilde{\nu }(x)\right)\tilde{\nu }(x)+P$$

$$\left(f(x)-P\right)\cdot \nu (x)\ne 0$$

$$\left(\tilde{f}(x)-P\right)\cdot \tilde{\nu }(x)\ne 0.$$

$$g(x)=\left((\tilde{f}(x)-P)\cdot \tilde{\nu }(x)\right)\tilde{\nu }(x)+P$$

$$\left(g(x)-P\right)\cdot \nu (x)\ne 0$$

$${\mathcal{N}\mathcal{S}}_{\tilde{f}}=\left\{P\in {\mathbb{R}}^2|{\mathbb{R}}^2-\bigcup _{x\in S^1}\left(\tilde{f}(x)+d{\tilde{f}}_x(T_x{S}^1)\right)\right\}.$$

$${\mathcal{N}\mathcal{S}}_{\tilde{f}}=\left\{P\in {\mathbb{R}}^{n+1}|\left(\tilde{f}(x)-P\right)\cdot \tilde{\nu }(x)\ne 0\text{mbox}foranyx\in N\right\}.$$

$$f(x)=2\left(\left(\tilde{f}(x)-P\right)\cdot \tilde{\nu }(x)\right)\tilde{\nu }(x)+P.$$

$$g(x)=\left(\left(\tilde{f}(x)-P\right)\cdot \tilde{\nu }(x)\right)\tilde{\nu }(x)+P.$$

$$\Phi (X,Y,\theta )=\nu (\theta )\cdot ((X,Y)-g(\theta ))$$

$$\left\{\begin{array}[]{cc}x(t)=1&(\mbox{if }\;8n\leq t\leq 8n+3),\\ 1\geq x(t)\geq-1&(\mbox{if }\;8n+3\leq t\leq 8n+4),\\ x(t)=-1&(\mbox{if }\;8n+4\leq t\leq 8n+7),\\ -1\leq x(t)\leq 1&(\mbox{if }\;8n+7\leq t\leq 8n+8).\end{array}\right.$$

$$\left\{\begin{array}[]{cc}y(t)=-1&(\mbox{if }\;8n-2\leq t\leq 8n+1),\\ -1\leq y(t)\leq 1&(\mbox{if }\;8n+1\leq t\leq 8n+2),\\ y(t)=1&(\mbox{if }\;8n+2\leq t\leq 8n+5),\\ 1\geq x(t)\geq-1&(\mbox{if }\;8n+5\leq t\leq 8n+6).\end{array}\right.$$

$$\cdots ,[-2,-1],[0,1],[2,3],[4,5],[6,7],[8,9],\cdots .$$

$$(1,-1)=\tilde{f}([8n,8n+1]),$$

$$(-1,1)=\tilde{f}([8n+4,8n+5]),$$

$$\left\{\begin{array}[]{cc}0\leq{\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}n_{1}}(t)\leq 1&(\mbox{if }\;8n\leq t\leq 8n+1),\\ {\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}n_{1}}(t)=1&(\mbox{if }\;8n+1\leq t\leq 8n+2),\\ 1\geq{\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}n_{1}}(t)\geq 0&(\mbox{if }\;8n+2\leq t\leq 8n+3),\\ {\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}n_{1}}(t)=0&(\mbox{if }\;8n+3\leq t\leq 8n+4),\\ 0\geq{\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}n_{1}}(t)\geq-1&(\mbox{if }\;8n+4\leq t\leq 8n+5),\\ {\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}n_{1}}(t)=-1&(\mbox{if }\;8n+5\leq t\leq 8n+6),\\ -1\leq{\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}n_{1}}(t)\leq 0&(\mbox{if }\;8n+6\leq t\leq 8n+7),\\ {\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}n_{1}}(t)=0&(\mbox{if }\;8n+7\leq t\leq 8n+8).\\ \end{array}\right.$$

$$\left\{\begin{array}[]{cc}-1\leq{\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}n_{2}}(t)\leq 0&(\mbox{if }\;8n\leq t\leq 8n+1),\\ {\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}n_{2}}(t)=0&(\mbox{if }\;8n+1\leq t\leq 8n+2),\\ 0\leq{\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}n_{2}}(t)\leq 1&(\mbox{if }\;8n+2\leq t\leq 8n+3),\\ {\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}n_{2}}(t)=1&(\mbox{if }\;8n+3\leq t\leq 8n+4),\\ 1\geq{\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}n_{2}}(t)\geq 0&(\mbox{if }\;8n+4\leq t\leq 8n+5),\\ {\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}n_{2}}(t)=0&(\mbox{if }\;8n+5\leq t\leq 8n+6),\\ 0\geq{\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}n_{2}}(t)\geq-1&(\mbox{if }\;8n+6\leq t\leq 8n+7),\\ {\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}n_{2}}(t)=-1&(\mbox{if }\;8n+7\leq t\leq 8n+8).\\ \end{array}\right.$$

$$({\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}n_{1}}(t),{\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}n_{2}}(t))\neq(0,0)\quad\mbox{ and }\quad\left(\frac{dx}{dt}(t),\frac{dy}{dt}(t)\right)\cdot({\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}n_{1}}(t),{\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}n_{2}}(t))=0.$$

$$\widetilde{\nu}(t)=\frac{1}{\sqrt{({\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}n_{1}}(t))^{2}+({\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}n_{2}}(t))^{2}}}({\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}n_{1}}(t),{\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}n_{2}}(t)).$$

$$\tilde{f}(x)=f(x)-\frac{||f(x)-P|{|}^2}{2(f(x)-P)\cdot \nu (x)}\nu (x).$$

$$\tilde{f}(x)=2g(x)-P-\frac{||g(x)-P|{|}^2}{(g(x)-P)\cdot \nu (x)}\nu (x).$$

$$\tilde{f}(x_0)-P=2\left((\tilde{f}(x_0)-P)\cdot \tilde{\nu }(x_0)\right)\tilde{\nu }(x_0).$$

$$\nu (x)=\frac{f(x)-\tilde{f}(x)}{||f(x)-\tilde{f}(x)||}.$$

$$(f\circ \xi {)}^{\prime }(0)\cdot \nu (x)=0$$

$$||f(x)-\tilde{f}(x)||\left({\left(f\circ \xi \right)}^{\prime }(0)\cdot \nu (x)\right)$$

$$||f(x)-\tilde{f}(x)||\left(f(x)-P\right)\cdot \nu (x)$$

$$\tilde{\nu }(x)=\frac{f(x)-P}{||f(x)-P||}.$$

Full text

[TABLE]

the following four are shown. (1) $\widetilde{f}$ is a frontal with its Gauss mapping $\widetilde{\nu}(x)=\frac{f(x)-P}{||f(x)-P||}$ at $\widetilde{f}(x)$. (2) $\widetilde{f}$ is the unique anti-orthotomic of $f$ relative to $P$. (3) The property $(\widetilde{f}(x)-P)\cdot\widetilde{\nu}(x)\neq 0$ holds for any $x\in N$. (4) The equality $||\widetilde{f}(x)-P||=||\widetilde{f}(x)-f(x)||$ holds for any $x\in N$.

Moreover, three applications of the main result are given. As the first application, a generalization of Cahn-Hoffman vector formula is given. The second application is to clarify an optical meaning of anti-orthotomics. The third application gives a criterion to be a front for a given frontal.

1. Introduction

Throughout this paper, let $n$, $N$ be a positive integer and an $n$-dimensional $C^{\infty}$ manifold without boundary respectively. Moreover, all mappings in this paper are of class $C^{\infty}$ unless otherwise stated.

A mapping $f:N\to\mathbb{R}^{n+1}$ is called a frontal if there exists a mapping $\nu:N\to\mathbb{R}^{n+1}$ such that the following two conditions are satisfied, where the dot in the center stands for the scalar product of two vectors in $\mathbb{R}^{n+1}$ and two vector spaces $T_{f(x)}\mathbb{R}^{n+1}$ and $\mathbb{R}^{n+1}$ are identified:

1. (1)

$\nu(x)\cdot\nu(x)=1$, i.e. $\nu(x)\in S^{n}$ for any $x\in N$. 2. (2)

$df_{x}({\bf v})\cdot\nu(x)=0$ for any $x\in N$ and any ${\bf v}\in T_{x}N$.

By the above conditions $(1)$ and $(2)$, it is natural to call $\nu:N\to S^{n}$ the Gauss mapping of $f$. In this paper, sometimes, even the mapping $(f,\nu):N\to\mathbb{R}^{n+1}\times S^{n}$ is called a frontal. The notion of frontal was independently introduced in several literature (e.g. [6, 12, 23, 27]) and it has been rapidly and intensively investigated (see [14]).

Definition 1**.**

Let $\widetilde{f}:N\to\mathbb{R}^{n+1}$ be a frontal with its Gauss mapping $\widetilde{\nu}:N\to S^{n}$ and let $P$ be a point of $\mathbb{R}^{n+1}$.

1. (1)

A mapping ${f}:N\to\mathbb{R}^{n+1}$ is called the orthotomic of $\widetilde{f}$ relative to $P$ if the following equality holds for any $x\in N$.

[TABLE] 2. (2)

A mapping ${g}:N\to\mathbb{R}^{n+1}$ is called the pedal of $\widetilde{f}$ relative to $P$ if the following equality holds for any $x\in N$.

[TABLE]

Figure 1 clearly illustrates the relation between the orthotomic $f$ and the pedal $g$ of $\widetilde{f}$ relative to $P$. The following Proposition 1 guarantees that the orthotomic of a given frontal is a frontal.

Proposition 1**.**

Let $(\widetilde{f},\widetilde{\nu}):N\to\mathbb{R}^{n+1}$ be a frontal and let $P$ be a point of $\mathbb{R}^{n+1}$ such that the following condition is satisfied for any $x\in N$.

[TABLE]

Then, the orthotomic of $\widetilde{f}$ relative to $P$ defined by

[TABLE]

is a frontal with its Gauss mapping ${\nu}(x)=\frac{f(x)-\widetilde{f}(x)}{||f(x)-\widetilde{f}(x)||}$. Moreover, the condition

[TABLE]

holds for any $x\in N$.

Proposition 1 will be proved in Section 2. By definition, it is clear that for the pedal $g$, $f=2g-P$ is the orthotomic. Thus, it is clear that Proposition 1 yields the following corollary which is a generalization of [18].

Corollary 1**.**

Let $(\widetilde{f},\widetilde{\nu}):N\to\mathbb{R}^{n+1}$ be a frontal and let $P$ be a point of $\mathbb{R}^{n+1}$ such that the following condition is satisfied for any $x\in N$.

[TABLE]

Then, the pedal of $\widetilde{f}$ relative to $P$ defined by

[TABLE]

is a frontal with its Gauss mapping ${\nu}(x)=\frac{2g(x)-P-\widetilde{f}(x)}{||2g(x)-P-\widetilde{f}(x)||}$. Moreover, the condition

[TABLE]

holds for any $x\in N$.

Notice that in the case that $\widetilde{f}$ is a plane regular curve, it is well-known that $\widetilde{f}(x)-f(x)$ is a normal vector to ${f}$ at ${f}(x)$ (for instance, see [4]). Therefore, a part of Proposition 1 may be regarded as just a generalization of the classical result to frontals of general dimension.

Notice also that even if $\widetilde{f}:N\to\mathbb{R}^{n+1}$ is non-singular, the condition “$(\widetilde{f}(x)-P)\cdot\widetilde{\nu}(x)\neq 0$ for any $x\in N$”  seems not so mild. In other words, even when $\widetilde{f}:N\to\mathbb{R}^{n+1}$ is an embedding, if the image of Gaussian curvature function of $\widetilde{f}(N)$ is a large interval containing zero as an interior point, then there are no points $P$ satisfying the condition “$(\widetilde{f}(x)-P)\cdot\widetilde{\nu}(x)\neq 0$ for any $x\in N$”. On the other hand, if $\widetilde{f}:N\to\mathbb{R}^{n+1}$ is an embedding and the Gaussian curvature of $\widetilde{f}(N)$ is always positive, then the set $\{P\in\mathbb{R}^{n+1}\;|\;(\widetilde{f}(x)-P)\cdot\widetilde{\nu}(x)\neq 0\mbox{ for any }x\in N\}$ is a non-empty open set. Moreover, the assumption “$\widetilde{f}:N\to\mathbb{R}^{n+1}$ is an embedding and the Gaussian curvature of $\widetilde{f}(N)$ is always positive”  seems to be common for the study of orthotomics and pedals. (for instance, see [1, 5]). Therefore, the assumption given in Proposition 1 generalizes the common assumption for the study of orthotomics and pedals.

The same condition as the assumption of Proposition 1 has been already introduced by J. W. Bruce and P. J. Giblin in [4] 7.14 in the case that $\widetilde{f}:I\to\mathbb{R}^{2}$ is regular; and also

by the second author ([21] in the case that $\widetilde{f}:N\to S^{n+1}$ is an immersion, [22] in the case that $\widetilde{f}:S^{n}\to\mathbb{R}^{n+1}$ is a Legendrian map and [16] in the case that $n=1$ and $\widetilde{f}:S^{1}\to\mathbb{R}^{2}$ is an embedding). Namely, in [16] the following set, called no-silhouette of $\widetilde{f}$ and denoted by $\mathcal{NS}_{\widetilde{f}}$, is defined.

[TABLE]

For a frontal $\widetilde{f}:N\to\mathbb{R}^{n+1}$ with its Gauss mapping $\widetilde{\nu}$, the notion of no-silhouette $\mathcal{NS}_{\widetilde{f}}$ can be naturally generalized as follows. The optical meaning of no-silhouette is illustrated by Figure 2.

[TABLE]

Definition 2**.**

1. (1)

Let ${f}:N\to\mathbb{R}^{n+1}$ be a $C^{\infty}$ mapping and let $P$ be a point of $\mathbb{R}^{n+1}$. A frontal $\widetilde{f}:N\to\mathbb{R}^{n+1}$ with its Gauss mapping $\widetilde{\nu}:N\to S^{n}$ is called the anti-orthotomic of $f$ relative to $P$ if the following equality holds for any $x\in N$.

[TABLE] 2. (2)

Let $g:N\to\mathbb{R}^{n+1}$ be a $C^{\infty}$ mapping and let $P$ be a point of $\mathbb{R}^{n+1}$. A frontal $\widetilde{f}:N\to\mathbb{R}^{n+1}$ with its Gauss mapping $\widetilde{\nu}:N\to S^{n}$ is called the negative pedal of $g$ relative to $P$ if the following equality holds for any $x\in N$.

[TABLE]

By definition, if $f=2g-P$, then the anti-orthotomic of $f$ relative to $P$ is exactly the same as the negative pedal of $g$ relative to $P$. Depending on situations, sometimes, the negative pedal is also called the primitive (for example, see [2]) or the Cahn-Hoffman map (for instance, see [10, 15, 17]). In Geometric Optics, the notions of anti-orthotomic is very important (for example, see [1, 4, 5]), and for the study of Wulff shape, the notion of negative pedal is the core concept (for instance, see [7, 15, 17, 19, 26]).

By definition, it is clear that an anti-orthotomic (resp., a negative pedal) is a solution frontal for a given orthotomic equation (resp., pedal equation). Therefore, obtaining anti-orthotomics or negative pedals may be considered as inverse problems. It seems that, except for the case that the Gauss mapping $\widetilde{\nu}$ of $\widetilde{f}$ is non-singular (i.e. the case that $\widetilde{f}:N\to\mathbb{R}^{n+1}$ is an embedding and the Gaussian curvature of $\widetilde{f}(N)$ is always positive), such inverse problems have been usually investigated only by solving simultaneous function equations for the envelopes.

Example 1**.**

Let $g:\mathbb{R}\to\mathbb{R}^{2}$ be the mapping defined by $g(\theta)=\left(\cos\left(\theta^{3}\right),\sin\left(\theta^{3}\right)\right)$. Define $\nu:\mathbb{R}\to\mathbb{R}^{2}$ by $\nu=g$. Then, $(g,\nu):\mathbb{R}\to\mathbb{R}^{2}\times S^{1}$ is a frontal. Set $P=(0,0)$ and $\widetilde{f}=\widetilde{\nu}=g$. Then, clearly, $(\widetilde{f},\widetilde{\nu}):\mathbb{R}\to\mathbb{R}^{2}\times S^{1}$ is the negative pedal of $g$ relative to $P$ and $\widetilde{f}(\mathbb{R})=S^{1}$.

On the other hand, the function $\Phi:\mathbb{R}^{2}\times\mathbb{R}\to\mathbb{R}$ given by

[TABLE]

defines the one-parameter family of affine tangent lines to $g(\mathbb{R})=S^{1}$. And the solution figure of the simultaneous equation $\Phi=0,\frac{\partial\Phi}{\partial\theta}=0$ is $S^{1}\cup\{(1,Y)\;|\;Y\in\mathbb{R}\}$.

Example 2**.**

Let $x:\mathbb{R}\to\mathbb{R}$ be a $C^{\infty}$ periodic function of period $8$ satisfying the following condition for any $n$.

[TABLE]

Let $y:\mathbb{R}\to\mathbb{R}$ be a $C^{\infty}$ periodic function of period $8$ satisfying the following condition for any integer $n$.

[TABLE]

Define $\widetilde{f}:\mathbb{R}\to\mathbb{R}^{2}$ by $\widetilde{f}(t)=(x(t),y(t))$. Then, $\widetilde{f}$ is a $C^{\infty}$ periodic mapping of period $8$ and the set of singular points of $\widetilde{f}$ contains infinitely many closed intervals

[TABLE]

From Figure 3, it is clear that even if $\widetilde{f}$ has other singular points, the image of $\widetilde{f}$ is always the square with the following $4$ vertexes

[TABLE]

Next, in order to assert that $\widetilde{f}$ is a frontal, we construct a non-zero normal vector ({\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}n_{1}}(t),{\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}n_{2}}(t)) to $\widetilde{f}$ at $\widetilde{f}(t)$. Let {\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}n_{1}}:\mathbb{R}\to\mathbb{R} be a $C^{\infty}$ periodic function of period $8$ satisfying the following condition for any integer $n$.

[TABLE]

Let {\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}n_{2}}:\mathbb{R}\to\mathbb{R} be a $C^{\infty}$ periodic function of period $8$ satisfying the following condition for any integer $n$.

[TABLE]

From Figure 3 and Figure 4, it is easily seen that the following two properties hold for any $t\in\mathbb{R}$.

[TABLE]

Set

[TABLE]

Then, $(\widetilde{f},\widetilde{\nu}):\mathbb{R}\to\mathbb{R}^{2}\times S^{1}$ is actually a frontal.

For the frontal $(\widetilde{f},\widetilde{\nu})$, the envelope of the one-parameter family of lines $\ell_{t}$ perpendicular to the unit vector $\widetilde{\nu}(t)$ and passing through the point $\widetilde{f}(t)$ does not restore the square $\widetilde{f}(\mathbb{R})$.

These two examples show that, for frontals, the classical notion of envelope cannot restore the original figure in general. In order to eliminate the influence of singularities of frontals, Masatomo Takahashi has succeeded to improve the notion of envelopes ([24, 25]). The improvement due to Takahashi is nice, and thus for Example 1, the original figure $g(\mathbb{R})=S^{1}$ can be actually obtained as the envelope of Takahashi’s sense. However, unfortunately, the variability condition defined in [24, 25] is not satisfied for the frontal $(\widetilde{f},\widetilde{\nu})$ given in Example 2. Thus, even Takahashi’s envelope cannot restore the original square $\widetilde{f}(\mathbb{R})$ of Example 2. In Ishikawa’s words, a frontal satisfying Takahashi’s variability condition is called a proper frontal ([14], §6). The frontal $(\widetilde{f},\widetilde{\nu}):\mathbb{R}\to\mathbb{R}^{2}\times S^{1}$ of Example 2 is not a proper frontal. To the best of authors’ knowledge, except for Example 2.5 given in [14], all frontals investigated in detail so far are proper frontals. We would like to assert that non-proper frontals, too, are useful especially in application to surface science (see §5).

The following Example 3 shows that the uniqueness of anti-orthotomic (resp., negative pedal) does not hold in general even when the given mapping $f$ (resp., $g$) is a frontal.

Example 3**.**

Let $f,g:\mathbb{R}\to\mathbb{R}^{2}$ be the constant mappings defined by $f(t)=(0,-1)$, $g(t)=(0,0)$. Let $\nu:\mathbb{R}\to S^{1}$ be the constant mapping defined by $\nu(t)=(-1,0)$. Set $P=(0,1)$ and define the constant mapping $\widetilde{\nu}:\mathbb{R}\to S^{1}$ by $\widetilde{\nu}(t)=(0,1)$. Then, for any $C^{\infty}$ mapping $\widetilde{f}:\mathbb{R}\to\mathbb{R}^{2}$ with the form $\widetilde{f}(t)=(\widetilde{f}_{1}(t),0)$, the frontal $(\widetilde{f},\widetilde{\nu})$ is an anti-orthotomic of $f$ relative to $P$ and a negative pedal of $g$ relative to $P$.

The main purpose of this paper is to obtain the unique solution of the inverse problem for a given orthotomic $f$ relative to $P$ such that $P\in\mathcal{NS}_{f}$.

Theorem 1**.**

Let $(f,\nu):N\to\mathbb{R}^{n+1}\times S^{n}$, $P$ be a frontal and a point of $\mathbb{R}^{n+1}$ such that $P\in\mathcal{NS}_{f}$ respectively. Let $\widetilde{f}:N\to\mathbb{R}^{n+1}$ be the mapping defined by

[TABLE]

Then, the following four holds:

1. (1)

The mapping $\widetilde{f}$ is a frontal with its Gauss mapping $\widetilde{\nu}(x)=\frac{f(x)-P}{||f(x)-P||}$. 2. (2)

The mapping $\widetilde{f}$ is the unique anti-orthotomic of $f$ relative to $P$. 3. (3)

The point $P$ belongs to $\mathcal{NS}_{\widetilde{f}}$. 4. (4)

*The equality $||\widetilde{f}(x)-P||=||\widetilde{f}(x)-f(x)||$ holds for any $x\in N$. *

In the case that $P=O$ and $f:I\to\mathbb{R}^{2}$ is a regular curve such that $f(x)\cdot\nu(x)\neq 0$ for any $x\in I$, the same formula for $\widetilde{f}$ has been already given in [4] 7.14 as the envelope of perpendicular bysectors of segments joining $f(x)$ and $P$. On the other hand, by Proposition 1, $\widetilde{f}(x)$ must be in the normal line $\{f(x)+a\nu(x)\;|\;a\in\mathbb{R}\}$. Therefore, in Theorem 1, just by solving simultaneous linear equations, the same formula for $\widetilde{f}$ can be obtained easily as the intersections of the perpendicular bysectors and the normal lines; and the no-silhouette condition $P\in\mathcal{NS}_{f}$ guarantees that each simultaneous linear equation must have the unique solution.

The following corollary clearly follows from Theorem 1

Corollary 2**.**

Let $(g,\nu):N\to\mathbb{R}^{n+1}\times S^{n}$, $P$ be a frontal and a point of $\mathbb{R}^{n+1}$ such that $P\in\mathcal{NS}_{g}$ respectively. Let $\widetilde{f}:N\to\mathbb{R}^{n+1}$ be the mapping defined by

[TABLE]

Then, the following four holds:

1. (1)

The mapping $\widetilde{f}$ is a frontal with its Gauss mapping $\widetilde{\nu}(x)=\frac{g(x)-P}{||g(x)-P||}$. 2. (2)

The mapping $\widetilde{f}$ is the unique negative pedal of $g$ relative to $P$. 3. (3)

The point $P$ belongs to $\mathcal{NS}_{\widetilde{f}}$. 4. (4)

*The equality $||\widetilde{f}(x)-P||=||\widetilde{f}(x)-2g(x)+P||$ holds for any $x\in N$. *

This paper is organized as follows. In Section 2, Proposition 1 is proved. Theorem 1 is proved in Section 3. In the case that the Gauss mapping $\widetilde{\nu}$ of $\widetilde{f}$ is the identity mapping, there is the famous Cahn-Hoffman vector formula for $\widetilde{f}$ ([9]). In Section 4, as the first application of Theorem 1, Cahn-Hoffman formula is shown. In Section 5, as the second application of Theorem 1, the optical meaning of the anti-orthotomic $\widetilde{f}$ is clarified even at a singular point of the Gauss mapping $\widetilde{\nu}$ of $\widetilde{f}$. Moreover, in order to show how the clarified optical meaning is useful, it is applied to construct the exact shape of the orthotomic $f$ for the frontal $\widetilde{f}$ in Example 2 and a given point $P\in\mathcal{NS}_{\widetilde{f}}$. Finally, in Section 6, as the third application of Theorem 1, it is given a criterion that a given frontal is actually a front.

2. Proof of Proposition 1

2.1. Proof that $f$ is a frontal

with its Gauss mapping $\nu(x)=\frac{f(x)-\widetilde{f}(x)}{||f(x)-\widetilde{f}(x)||}$

Recall that $f$ is defined by $f(x)=2\left((\widetilde{f}(x)-P)\cdot\widetilde{\nu}(x)\right)\widetilde{\nu}(x)+P$.

Lemma 2.1**.**

For any $x\in N$, $f(x)-\widetilde{f}(x)$ is a non-zero vector.

Proof.

Suppose that $f(x_{0})=\widetilde{f}(x_{0})$ for some $x_{0}\in N$. Then, for the $x_{0}$, the following holds:

[TABLE]

This implies $(\widetilde{f}(x_{0})-P)\cdot\widetilde{\nu}(x_{0})=0,$ which contradicts the assumption $P\in\mathcal{NS}_{\widetilde{f}}$. $\Box$

Set

[TABLE]

Then, it is sufficient to show that $d{f}_{x}({\bf v})\cdot{\nu}(x)=0$ for any $x\in N$ and any ${\bf v}\in T_{x}N$. In other words, it is sufficient to show that

[TABLE]

for any curve $\xi:(-\varepsilon,\varepsilon)\to N$ such that $\xi(0)=x$. The following lemma clearly holds:

Lemma 2.2**.**

1. (1)

$(\widetilde{f}\circ\xi)^{\prime}(0)\cdot\widetilde{\nu}(x)=0$. 2. (2)

$(\widetilde{\nu}\circ\xi)^{\prime}(0)\cdot\widetilde{\nu}(x)=0$. 3. (3)

$\left(f\circ\xi\right)^{\prime}(0)=2\left(\left(\widetilde{f}(x)-P\right)\cdot\left(\widetilde{\nu}\circ\xi\right)^{\prime}(0)\right)\widetilde{\nu}(x)+2\left(\left(\widetilde{f}(x)-P\right)\cdot\widetilde{\nu}(x)\right)\left(\widetilde{\nu}\circ\xi\right)^{\prime}(0).$**

By using Lemma 2.2, we have the following:

[TABLE]

$\Box$

2.2. Proof that

$\left(f(x)-P\right)\cdot\nu(x)\neq 0$ for any $x\in N$

For any $x\in N$, we have the following:

[TABLE]

By the assumption $P\in\mathcal{NS}_{\widetilde{f}}$, it follows that $\left(f(x)-P\right)\cdot\nu(x)\neq 0$ for any $x\in N$. $\Box$

3. Proof of Theorem 1

3.1. Proof that $\widetilde{f}$ is a frontal

with Gauss mapping $\widetilde{\nu}(x)=\frac{f(x)-P}{||f(x)-P||}$

From the assumption that $(f(x)-P)\cdot\nu(x)\neq 0$ for any $x\in N$, it follows that $f(x)\neq P$ for any $x\in N$. Thus,

[TABLE]

is well-defined. Then, it is sufficient to show that

[TABLE]

for any curve $\xi:(-\varepsilon,\varepsilon)\to N$ such that $\xi(0)=x$. Since $\widetilde{f}$ has the form

[TABLE]

we have the following:

[TABLE]

$\Box$

3.2. Proof that $\widetilde{f}$ is

the unique anti-orthotomic of $f$ relative to $P$

The proof is essentially given in the paragraph next to Theorem 1. Thus, in this subsection, just a confirmation by definition is given.

Recall that $\widetilde{\nu}(x)=\frac{f(x)-P}{||f(x)-P||}$. We have the following:

[TABLE]

$\Box$

3.3. Proof that

$\left(\widetilde{f}(x)-P\right)\cdot\widetilde{\nu}(x)\neq 0$ for any $x\in N$

For any $x\in N$ we have the following:

[TABLE]

$\Box$

3.4. Proof that the equality

$||\widetilde{f}(x)-P||=||\widetilde{f}(x)-f(x)||$ holds for any $x\in N$

Since $\widetilde{f}(x)-P=(f(x)-P)-\frac{||f(x)-P||^{2}}{2(f(x)-P)\cdot\nu(x)}\nu(x)$, the following holds for any $x\in N$:

[TABLE]

$\Box$

4. Application 1:

Generalization of Cahn-Hoffman vector formula

Let $(g,\nu):N\to\mathbb{R}^{n+1}\times S^{n}$ be a frontal. We assume that $\mathcal{NS}_{g}$ is not empty. Let $P$ be a point of $\mathcal{NS}_{g}$. Then, by Corollary 2, the mapping $\left(\widetilde{f},\widetilde{\nu}\right):N\to\mathbb{R}^{n+1}\times S^{n}$ defined by

[TABLE]

is a frontal and the unique negative pedal of $g$ relative to $P$. Set $\gamma(x)=||g(x)-P||$. Then, by using $\widetilde{\nu}:N\to S^{n}$ and $\gamma:N\to\mathbb{R}_{+}$, $g(x)-P$ can be expressed as follows.

[TABLE]

In [9], under the assumption that $N$ is the unit sphere $S^{n}$ and $\widetilde{\nu}:S^{n}\to S^{n}$ is the identity mapping and under the identification $\mathbb{R}^{n+1}=T_{\widetilde{\nu}(x)}\mathbb{R}^{n+1}=T_{\widetilde{\nu}(x)}S^{n}\oplus\mathbb{R}\widetilde{\nu}(x)$, D. W. Hoffman and J. W. Cahn showed the following.

Theorem 2** (Cahn-Hoffman vector formula [9]).**

For any $x\in S^{n}$, the following equality holds.

[TABLE]

Here, $\nabla\gamma(x)$ stands for the gradient vector of $\gamma$ at $x$ with respect to the normal coordinate system of $S^{n}$ around $x$. In this section, as an application of Theorem 1, we generalize Theorem 2 as follows.

Theorem 3**.**

Let $(g,\nu):N\to\mathbb{R}^{n+1}\times S^{n}$ be a frontal and let $P$ be a point of $\mathcal{NS}_{g}$. Suppose that $\widetilde{\nu}$ is non-singular at $x$. Then, the following equality holds.

[TABLE]

Here, $\left(\left(J\widetilde{\nu}(x)\right)^{-1}\right)^{t}$ stands for the transposed matrix of the inverse of Jacobian matrix of $\widetilde{\nu}$ with respect to an arbitrary local coordinate system around $x\in N$ and the normal coordinate system around $\widetilde{\nu}(x)\in S^{n}$. Theorem 3 yields not only Theorem 2 but also the following.

Corollary 3**.**

*Let $(g,\nu):N\to\mathbb{R}^{n+1}\times S^{n}$ be a frontal and let $P$ be a point of $\mathcal{NS}_{g}$. Suppose that $\widetilde{\nu}$ is non-singular at $x$. Then, $x$ is a singular point of $\gamma$ if and only if $\widetilde{f}(x)=g(x)$ is satisfied. *

Proof of Theorem 3.

Since

[TABLE]

it follows

[TABLE]

where $\nu(x)=\nu_{1}(x)\oplus\nu_{2}(x)$ and $\nu_{1}(x)\in T_{\widetilde{\nu}(x)}S^{n},\nu_{2}(x)\in\mathbb{R}\widetilde{\nu}(x)$.

In order to represent $\nu_{1},\nu_{2}$ in terms of $\gamma$ and $\widetilde{\nu}$, the same technique as in [20] is used. Since $f(x)-P=2\gamma(x)\widetilde{\nu}(x)$, for any $v\in T_{x}N$,

[TABLE]

where $\gamma(x)d\widetilde{\nu}_{x}(v)\in T_{\widetilde{\nu}(x)}S^{n}$ and $d\gamma_{x}(v)\widetilde{\nu}(x)\in\mathbb{R}\widetilde{\nu}(x)$. Thus, the Jacobian matrix $Jf$ of $f$ at $x$ with respect to an arbitrary local coordinate system around $x\in N$ and the direct product of the normal coordinate system and $\mathbb{R}$ around $f(x)=2\gamma(x)\widetilde{\nu}(x)$ has the following form, where $J\widetilde{\nu}(x)$ stands for the Jacobian matrix of $\widetilde{\nu}$ at $x$ and $\left(\nabla\gamma(t)\right)^{t}$ stands for the transposed vector of the gradient of $\gamma$ at $x$.

[TABLE]

Let $\widetilde{J}\widetilde{\nu}(x)$ and $|J\widetilde{\nu}|(x)$be the cofactor matrix of the Jacobian matrix $J\widetilde{\nu}(x)$ and the Jacobian determinant of $\widetilde{\nu}$ at $x$ respectively. Moreover, let ${O}$ be the $n\times 1$ zero vector. Multiplying the matrix

[TABLE]

to the Jacobian matrix $Jf(x)$ from the left side yields the following, where $E_{n}$ stands for the $n\times n$ unit matrix.

[TABLE]

Hence we have the following.

Lemma 4.1**.**

Suppose that $|J\widetilde{\nu}|(x)\neq 0$. Then, we may put as follows:

[TABLE]

Notice that in order to show that $||\nu(x)||\neq 0$, the assumption “$|J\widetilde{\nu}|(x)\neq 0$”  is used .

Set $h(x)=2g(x)-\widetilde{f}(x)$. Then, by elementary geometry, we have

[TABLE]

Since $\widetilde{f}(x)-g(x)=g(x)-h(x)={(g(x)-P)}-{(h(x)-P)}$, we have

[TABLE]

$\Box$

5. Application 2: Opening of Gauss mapping of

anti-orthotomic

The application in Section 4 is a result only at a non-singular point of $\widetilde{\nu}$. In this section, as the second application of Theorem 1, we investigate what can be asserted even at a singular point of $\widetilde{\nu}$. Several definitions are needed for the investigation of this section.

Definition 3** ([13]).**

Let $f=(f_{1},\ldots,f_{n}):(\mathbb{R}^{n},0)\to(\mathbb{R}^{n},0)$ be an equidimensional $C^{\infty}$ map-germ.

1. (1)

Let $\Omega_{n}^{1}$ denote the $\mathcal{E}_{n}$-module of $1$-forms on $(\mathbb{R}^{n},0)$. Then, the $\mathcal{E}_{n}$-module generated by $df_{i}$ $(i=1,\ldots,n)$ in $\Omega_{n}^{1}$ is called the Jacobi module of $f$ and is denoted by $\mathcal{J}_{f}$, where $dh$ for a function-germ $h:(\mathbb{R}^{n},0)\to\mathbb{R}$ stands for the exterior differential of $h$. 2. (2)

The ramification module of $f$ (denoted by $\mathcal{R}_{f}$) is defined as the $f^{*}\left(\mathcal{E}_{n}\right)$ -module consisting of all function-germs $\gamma$ such that $d\gamma$ belongs to $\mathcal{J}_{f}$.

Definition 4** ([13]).**

Let {\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}\varphi}:(\mathbb{R}^{n},0)\to(\mathbb{R}^{n},0) be an equidimensional $C^{\infty}$ map-germ and let {\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}\delta}:(\mathbb{R}^{n},0)\to(\mathbb{R},0) a $C^{\infty}$ function-germ. Then, the map-germ ({\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}\varphi,\delta}):(\mathbb{R}^{n},0)\to(\mathbb{R}^{n}\times\mathbb{R},(0,0)) is called an opening of ${\varphi}$ if $\delta\in\mathcal{R}_{\varphi}$.

Theorem 4**.**

Let $(f,\nu):N\to\mathbb{R}^{n+1}\times S^{n}$ be a frontal. Let $x_{0}\in N$ and $P\in\mathbb{R}^{n+1}$ satisfy $(f(x_{0})-P)\cdot\nu(x_{0})\neq 0$. Then, $(f-P):(N,x_{0})\to\mathbb{R}^{n+1}$ is an opening of the Gauss map-germ $\widetilde{\nu}=\frac{f-P}{||f-P||}:(N,x_{0})\to S^{n}$ of its anti-orthotomic $\widetilde{f}:(N,x_{0})\to\mathbb{R}^{n+1}$.

The optical meaning of anti-orthotomic is straightforward from Theorem 4. By definition, we have the following corollary.

Corollary 4**.**

Let $(g,\nu):N\to\mathbb{R}^{n+1}\times S^{n}$ be a frontal. Let $x_{0}\in N$ and $P\in\mathbb{R}^{n+1}$ satisfy $(g(x_{0})-P)\cdot\nu(x_{0})\neq 0$. Then, $(g-P):(N,x_{0})\to\mathbb{R}^{n+1}$ is an opening of the Gauss map-germ $\widetilde{\nu}=\frac{g-P}{||g-P||}:(N,x_{0})\to S^{n}$ of its negative pedal $\widetilde{f}:(N,x_{0})\to\mathbb{R}^{n+1}$.

Proof of Theorem 4.

Let $V\subset S^{n}$ be a sufficiently small open neighbourhood of $\widetilde{\nu}(x_{0})$ and let $h=(h_{1},\ldots,h_{n}):V\to T_{\widetilde{\nu}(x_{0})}S^{n}$ be a normal coordinate system at $\widetilde{\nu}(x_{0})$. Let $U$ be a sufficiently small open neighbourhood of $x_{0}$ such that $U\subset\widetilde{\nu}^{-1}(V)$ and set $\nu_{1}(x)=\left(\nu_{1,1}(x),\ldots,\nu_{1,n}(x)\right)$ for any $x\in U$. Moreover, for any $i$ $(1\leq i\leq n)$, set $\widetilde{\nu}_{i}=h_{i}\circ\widetilde{\nu}$. Since $\nu:N\to S^{n}$ is the Gauss mapping of $f:N\to\mathbb{R}^{n+1}$ and $f(x)-P=2\gamma(x)\widetilde{\nu}(x)$, we have

[TABLE]

Since $(f(x_{0})-P)\cdot\nu(x_{0})\neq 0$, it follows $\nu_{2}(x_{0})\neq 0$. Thus, we have

[TABLE]

$\Box$

Consider again the frontal $(\widetilde{f},\widetilde{\nu}):\mathbb{R}\to\mathbb{R}^{2}\times S^{1}$ given in Example 2. Recall that the image $\widetilde{f}(\mathbb{R})$ is the square $S$ with vertexes $(1,1),(-1,1),(1,-1),(-1,-1)$. Let $P=(p_{1},p_{2})$ be a point such that $-1<p_{1},p_{2}<1$. Then, $P$ belongs to $\mathcal{NS}_{\widetilde{f}}$. Let $f:\mathbb{R}\to\mathbb{R}^{2}$ be the orthotomic of $\widetilde{f}$ relative to $P$. Theorems 1 and 4 reduce the construction of the image of $f$ to elementary geometry, which is explained as follows (see Figure 5).

By the construction of $\widetilde{\nu}$, if $t$ belongs to the union of closed intervals

[TABLE]

then $t$ must be a singular point of $\widetilde{\nu}$. Thus, by Theorem 4, $t$ must be a singular point of $f$, and therefore each of $f([8n+1,8n+2]),f([8n+3,8n+4]),f([8n+5,8n+6]),f([8n+7,8n+8])$ must be one point. By definition, the one point must be the mirror image of $P$ as follows.

Lemma 5.1**.**

[TABLE]

By the construction of $\widetilde{f}$,

[TABLE]

By the assertion (4) of Theorem 1, the following holds.

Lemma 5.2**.**

[TABLE]

By the construction of $\widetilde{\nu}$, we have the following.

Lemma 5.3**.**

1. (1)

$f([8n,8n+1])$* is exactly the hemicircle centered at $\widetilde{f}([8n,8n+1])=(1,-1)$ with boundary $f([8n+7,8n+8])=(p_{1},-2-p_{2})$ and $f([8n+1,8n+2])=(2-p_{1},p_{2})$ which does not contain $P$.* 2. (2)

$f([8n+2,8n+3])$* is exactly the hemicircle centered at $\widetilde{f}([8n+2,8n+3])=(1,1)$ with boundary $f([8n+1,8n+2])=(2-p_{1},p_{2})$ and $f([8n+3,8n+4])=(p_{1},2-p_{2})$ which does not contain $P$.* 3. (3)

$f([8n+4,8n+5])$* is exactly the hemicircle centered at $\widetilde{f}([8n+4,8n+5])=(-1,1)$ with boundary $f([8n+3,8n+4])=(p_{1},2-p_{2})$ and $f([8n+5,8n+6])=(-2-p_{1},p_{2})$ which does not contain $P$.* 4. (4)

$f([8n+6,8n+7])$* is exactly the hemicircle centered at $\widetilde{f}([8n+6,8n+7])=(-1,-1)$ with boundary $f([8n+5,8n+6])=(-2-p_{1},p_{2})$ and $f([8n+7,8n+8])=(p_{1},-2-p_{2})$ which does not contain $P$.*

For the precise shape of the pedal $g:\mathbb{R}\to\mathbb{R}^{2}$ of $\widetilde{f}$ relative to $P$, just shrink $f(\mathbb{R})$ to $50$ percent with respect to $P$.

It seems that the method of C. Herring explained in [8] is similar as our method. However, his method seems to rely on a thermodynamical consideration of atoms. Our method needs no physical consideration. Once the given shape is realized as the image of frontal $\widetilde{f}(\mathbb{R})$, by applying Theorem 1 and Theorem 4, only elementary geometry is needed. In other words, under any physical situation, if the same square is given, then the $\gamma$-plot for the square must have the same shape.

6. Application 3: A criterion for fronts

Definition 5**.**

A germ of frontal $(f,\nu):(N,x_{0})\to\mathbb{R}^{n+1}\times S^{n}$ is called a germ of front (or front-germ) if $(f,\nu)$ is non-singular at $x_{0}$.

Given a frontal $(f,\nu):N\to\mathbb{R}^{n+1}\times S^{n}$, if $(f,\nu):(N,x)\to\mathbb{R}^{n+1}\times S^{n}$ is a germ of front for any $x\in N$, then $(f,\nu)$ is called a front. A front is also called a wave-front. For details on fronts, see for example [2, 3].

Theorem 5**.**

Let $(f,\nu):N\to\mathbb{R}^{n+1}\times S^{n}$ be a frontal and let $x_{0}$ be a point of $N$. Then, for any point $P\in\mathbb{R}^{n+1}$ such that $(f(x_{0})-P)\cdot\nu(x_{0})\neq 0$, the following are equivalent, where $\left(\widetilde{f},\widetilde{\nu}\right):(N,x_{0})\to\mathbb{R}^{n+1}\times S^{n}$ is the anti-orthotomic germ of $f$ relative to $P$.

1. (1)

$(f,\nu):(N,x_{0})\to\mathbb{R}^{n+1}\times S^{n}$* is a front-germ.* 2. (2)

$\left(\widetilde{f},\widetilde{\nu}\right):(N,x_{0})\to\mathbb{R}^{n+1}\times S^{n}$* is a front-germ.* 3. (3)

$\left(f,\widetilde{f}\right):(N,x_{0})\to\mathbb{R}^{n+1}\times\mathbb{R}^{n+1}$* is non-singular.*

Theorem 5 answers the question communicated by A. Honda and K. Teramoto ([11]). Theorem 5 yields the following corollaries.

Corollary 5**.**

Let $(g,\nu):N\to\mathbb{R}^{n+1}\times S^{n}$ be a frontal and let $x_{0}$ be a point of $N$. Then, for any point $P\in\mathbb{R}^{n+1}$ such that $(g(x_{0})-P)\cdot\nu(x_{0})\neq 0$, the following are equivalent, where $\left(\widetilde{f},\widetilde{\nu}\right):(N,x_{0})\to\mathbb{R}^{n+1}\times S^{n}$ is the negative pedal germ of $g$ relative to $P$.

1. (1)

$(g,\nu):(N,x_{0})\to\mathbb{R}^{n+1}\times S^{n}$* is a front-germ.* 2. (2)

$\left(\widetilde{f},\widetilde{\nu}\right):(N,x_{0})\to\mathbb{R}^{n+1}\times S^{n}$* is a front-germ.* 3. (3)

$\left(g,\widetilde{f}\right):(N,x_{0})\to\mathbb{R}^{n+1}\times\mathbb{R}^{n+1}$* is non-singular.*

Corollary 6**.**

Let $(f,\nu):N\to\mathbb{R}^{n+1}\times S^{n}$ be a frontal. Let two points $x_{0}\in N$ and $P\in\mathbb{R}^{n+1}$ satisfy $(f(x_{0})-P)\cdot\nu(x_{0})\neq 0$. Let $(\widetilde{f},\widetilde{\nu}):N\to\mathbb{R}^{n+1}\times S^{n}$ be the anti-orthotomic of $f$ relative to $P$. If $x_{0}$ is not contained in Sing$(\widetilde{f})\cap$Sing$(\widetilde{\nu})$, then the map-germ $f:(N,x_{0})\to\mathbb{R}^{n+1}$ is a front-germ; where for a $C^{\infty}$ mapping $\varphi:X\to Y$, Sing$(\varphi)$ stands for the singular set of $\varphi$. In particular, if $\widetilde{f}:(N,x_{0})\to\mathbb{R}^{n+1}$ is non-singular, then $f:(N,x_{0})\to\mathbb{R}^{n+1}$ must be a front-germ.

Corollary 7**.**

Let $(g,\nu):N\to\mathbb{R}^{n+1}\times S^{n}$ be a frontal. Let two points $x_{0}\in N$ and $P\in\mathbb{R}^{n+1}$ satisfy $(g(x_{0})-P)\cdot\nu(x_{0})\neq 0$. Let $(\widetilde{f},\widetilde{\nu}):N\to\mathbb{R}^{n+1}\times S^{n}$ be the negative pedal of $g$ relative to $P$. If $x_{0}$ is not contained in Sing$(\widetilde{f})\cap$Sing$(\widetilde{\nu})$, then the map-germ $g:(N,x_{0})\to\mathbb{R}^{n+1}$ is a front-germ. In particular, if $\widetilde{f}:(N,x_{0})\to\mathbb{R}^{n+1}$ is non-singular, then $g:(N,x_{0})\to\mathbb{R}^{n+1}$ must be a front-germ.

Corollary 8**.**

*Let $(\widetilde{f},\widetilde{\nu}):N\to\mathbb{R}^{n+1}\times S^{n}$ be a frontal. Let two points $x_{0}\in N$ and $P\in\mathbb{R}^{n+1}$ satisfy $(\widetilde{f}(x_{0})-P)\cdot\widetilde{\nu}(x_{0})\neq 0$. Let $({f},{\nu}):N\to\mathbb{R}^{n+1}\times S^{n}$ be the orthotomic of $\widetilde{f}$ relative to $P$. If $x_{0}$ is not contained in Sing$({f})\cap$Sing$({\nu})$, then the map-germ $\widetilde{f}:(N,x_{0})\to\mathbb{R}^{n+1}$ is a front-germ. *

Corollary 9**.**

*Let $(\widetilde{f},\widetilde{\nu}):N\to\mathbb{R}^{n+1}\times S^{n}$ be a frontal. Let two points $x_{0}\in N$ and $P\in\mathbb{R}^{n+1}$ satisfy $(\widetilde{f}(x_{0})-P)\cdot\widetilde{\nu}(x_{0})\neq 0$. Let $({g},{\nu}):N\to\mathbb{R}^{n+1}\times S^{n}$ be the pedal of $\widetilde{f}$ relative to $P$. If $x_{0}$ is not contained in Sing$({g})\cap$Sing$({\nu})$, then the map-germ $\widetilde{f}:(N,x_{0})\to\mathbb{R}^{n+1}$ is a front-germ. *

Proof of Theorem 5.

$(2)\Leftrightarrow(3)$ is just a corollary of Theorem 4. $(1)\Rightarrow(3)$ is trivial. Thus, in order to complete the proof, it is sufficient to show $(3)\Rightarrow(1)$. Suppose that $(f,\widetilde{f})$ is non-singular. Then, $(f,f-\widetilde{f})$ is non-singular. By the assumption $({f}(x_{0})-P)\cdot{\nu}(x_{0})\neq 0$, it follows that the projection $\pi:d_{x_{0}}(f-\widetilde{f})(\mbox{\rm Ker}(d_{x_{0}}f))\to T_{{\nu(x_{0})}}S^{n}$ is injective. Therefore, $\left(f,{\nu}\right)=\left(f,\frac{f-\widetilde{f}}{||f-\widetilde{f}||}\right):(N,x_{0})\to\mathbb{R}^{n+1}\times S^{n}$ is non-singular. $\Box$

Acknowledgement

The authors thank Masatomo Takahashi for teaching them his improvement of classical envelope. The second author was supported by JSPS KAKENHI Grant Number 17K05245.