On the geometric concomitant equation

Abstract

We solve in some cases the equation in the definition of the concomitant of a geometric object for each of the four functions in this equation. Remarks and examples are given, too.

1 Introduction

For $\hbox {i}=1,2$ let $\hbox {S}_{\mathrm{i}}$ be a set and let ($\hbox {G}_{\mathrm{i}},+$) be a groupoid. Let the function $\hbox {F}_{\mathrm{i}}:\hbox {S}_{\mathrm{i}}\times \hbox {G}_{\mathrm{i}}\rightarrow \hbox {S}_{\mathrm{i}}$ be the solution of the translation equation, i.e., such that

$$\begin{aligned} \hbox {F}_{\mathrm{i}}(\hbox {F}_{\mathrm{i}}(\hbox {x},\hbox {t}_{1}),\hbox {t}_{2})=\hbox {F}_{\mathrm{i}}(\hbox {x},\hbox {t}_{1}+\hbox {t}_{2}) \hbox { for } \hbox {x}\upepsilon \hbox {S}_{\mathrm{i}},\hbox {t}_{1},\hbox {t}_{2}\upepsilon \hbox {G}_{\mathrm{i}} \quad \hbox { for } \hbox {i}=1,2. \end{aligned}$$

The equation

$$\begin{aligned} \hbox {f}[\hbox {F}_{1}(\hbox {x,t})]=\hbox {F}_{2}[\hbox {f(x),h(t)}]\quad \hbox { for } (\hbox {x,t})\upepsilon \hbox {S}_{1}\times \hbox {G}_{1}, \end{aligned}$$

(1)

where $\hbox {f}:\hbox {S}_{1}\rightarrow \hbox {S}_{2}$ and h is a homomorphism from $\hbox {G}_{1}$ to $\hbox {G}_{2}$, is said to be the geometric concomitant equation.

This equation means that the diagram

is commutative.

If $\hbox {G}_{1}=\hbox {G}_{2}$, $\hbox {h}(\hbox {t})=\hbox {t}$ for $\hbox {t}\upepsilon \hbox {G}_{1}$, $\hbox {G}_{1}$ has the neutral element 0 and $\hbox {F}_{\mathrm{i}}(\hbox {x},0)=\hbox {x}$ for $\hbox {x}\upepsilon \hbox {S}_{\mathrm{i}}$, $\hbox {i}=1,2$, then in the theory of geometric objects [1] the function $\hbox {F}_{1}$ is the transformation law of the object x, f(x) is the concomitant of this object x [4] and $\hbox {F}_{2}$ is the transformation law of the object f(x).

If f is a bijection, then $\hbox {F}_{1}$ and $\hbox {F}_{2}$ are said to be equivalent (e.g., in the theory of abstract automata [11]).

If G is a group, then $\hbox {G/G}^{*}$ means the family of the right cosets of the group G for the subgroup $\hbox {G}^{*}$ throughout the paper.

We have four functions in Eq. (1), thus there are four possibilities for the unknown function.

2 Unknown function f

Let $\hbox {G}_{\mathrm{i}}$ be a group and let $\hbox {F}_{\mathrm{i}}(\hbox {x},0)=\hbox {x}$ for $\hbox {x}\upepsilon \hbox {S}_{\mathrm{i}}$ and $\hbox {i}=1,2$. The solution $\hbox {F}_{\mathrm{i}}$ of the translation equation has the form

$$\begin{aligned} \hbox {F}_{\mathrm{i}}(\hbox {x,t})= \text{ b }_{\mathrm{i}}^{-1} [\hbox {b}_{\mathrm{i}}(\hbox {x})+\hbox {t}] \quad \hbox { for } (\hbox {x,t})\upepsilon \hbox {S}_{\mathrm{i}}\times \hbox {G}_{\mathrm{i}}, \end{aligned}$$

(2)

where $\hbox {b}_{\mathrm{i}}=\cup _{\mathrm{k}\upepsilon K_{i}} :\hbox {b}_{\mathrm{ik}}$, $\hbox {S}_{\mathrm{i}}=\cup _{\mathrm{k}\upepsilon K_{i}} :\hbox {S}_{\mathrm{ik}}$, $\hbox {S}_{\mathrm{ik}}$ for $\hbox {k}\upepsilon \hbox {K}_{\mathrm{i}}$ are non-empty disjoint sets such that for every $\hbox {k}\upepsilon \hbox {K}_{\mathrm{i}}$ there exists a subgroup $\hbox {G}_{\mathrm{ik}}$ of the group $\hbox {G}_{\mathrm{i}}$ for which $\hbox {cardS}_{\mathrm{ik}}=\hbox {cardG}_{\mathrm{i}}/\hbox {G}_{\mathrm{ik}}$ and $\hbox {b}_{\mathrm{ik}}:\hbox {S}_{\mathrm{ik}}\rightarrow \hbox {G}_{\mathrm{i}}/\hbox {G}_{\mathrm{ik}}$ is a bijection ([5, 7]).

From here the above function $\hbox {F}_{\mathrm{i}}$ and the solution of the translation equation $F_{i}(\hbox {u,t})= \hbox {u}+\hbox {t}:(\cup _{\mathrm{k}\upepsilon K_{i}} \hbox {G}_{\mathrm{i}}/\hbox {G}_{\mathrm{ik}})\times \hbox {G}_{\mathrm{i}}\rightarrow \cup _{\mathrm{k}\upepsilon K_{i}} \hbox {G}_{\mathrm{i}}/\hbox {G}_{\mathrm{ik}}$ are equivalent since $\hbox {b}_{\mathrm{i}}(\hbox {F}_{\mathrm{i}}(\hbox {x,t}))=\hbox {b}_{\mathrm{i}}(\hbox {x})+\hbox {t}=F_{i}(\hbox {b}_{\mathrm{i}}(\hbox {x}),\hbox {t})$. This implies that the solution $\hbox {F}:\hbox {S}\times \hbox {G} \rightarrow \hbox {S}$ such that G is a group and $\hbox {F}(\hbox {s},0)=\hbox {s}$ for $\hbox {s}\upepsilon \hbox {S}$ may be identified with the family of the right cosets of the group G.

Theorem 1

If $\hbox {F}_{\mathrm{i}}$ are of the form (2), then a function $\hbox {f}:\hbox {S}_{1}\rightarrow \hbox {S}_{2}$ is a solution of (1) if and only if there exist functions $\upalpha :\hbox {K}_{1}\rightarrow \hbox {G}_{2}$ and $\upbeta :\hbox {K}_{1}\rightarrow \hbox {K}_{2}$ such that

$$\begin{aligned} \hbox {h}(\hbox {G}_{\mathrm{1k}})\subset [-\upalpha (\hbox {k})+\hbox {G}_{2\upbeta (\mathrm{k})}+\upalpha (\hbox {k})] \,for\, \hbox {k}\upepsilon \hbox {K}_{1} \end{aligned}$$

(3)

and f is of the form $\hbox {f}=\text{ b }_{2}^{-1} \upvarphi \hbox {b}_{1}$, where

$$\begin{aligned} \upvarphi (\hbox {C})=\upvarphi (\hbox {G}_{\mathrm{1k}}+\hbox {t})=\hbox {G}_{2\upbeta (\mathrm{k})}+\upalpha (\hbox {k})+\hbox {h}(\hbox {t}) \,for \,\hbox {C}=\hbox {G}_{\mathrm{1k}}+\hbox {t}\upepsilon \hbox {G}_{1}/\hbox {G}_{\mathrm{1k}}\,and\, \hbox {k}\upepsilon \hbox {K}_{1}.\nonumber \\ \end{aligned}$$

(4)

Proof

Note that the function $\upvarphi$ defined in (4) is well-defined. In fact, if $\hbox {G}_{\mathrm{1k}}+\hbox {t}_{1}=\hbox {G}_{\mathrm{1k}}+\hbox {t}_{2}$, then $\hbox {t}_{2}\hbox {--}\hbox {t}_{1}\upepsilon \hbox {G}_{\mathrm{1k}}$, thus the function

$$\begin{aligned} \hbox {h}(\hbox {t}_{2}\hbox {--t}_{1})\upepsilon [\hbox {--}\upalpha (\hbox {k})+\hbox {G}_{2\upbeta (\mathrm{k})}+\upalpha (\hbox {k})]=:\hbox {A}. \end{aligned}$$

Since A is a group, $\hbox {A}+\hbox {h}(\hbox {t}_{2}\hbox {--}\hbox {t}_{1})=\hbox {A}$. This yields that $\hbox {A}+\hbox {h}(\hbox {t}_{2})=\hbox {A}+\hbox {h}(\hbox {t}_{1})$.

Let f be a solution of (1). Then the function $\upvarphi =\hbox {b}_{2}\hbox {f}\text{ b }_{1}^{-1}$ is a solution of the equation

$$\begin{aligned} \upvarphi (\hbox {C}+\hbox {t})=\upvarphi (\hbox {C})+\hbox {h}(\hbox {t}), \end{aligned}$$

(5)

where $\upvarphi :\cup _{\mathrm{k}\upepsilon K_{1}} :\hbox {G}_{1}/\hbox {G}_{\mathrm{1k}}\rightarrow \cup _{\mathrm{k}\upepsilon K_{2}} :\hbox {G}_{2}/\hbox {G}_{\mathrm{2k}}$. Indeed, for $\hbox {C}\upepsilon \cup _{\mathrm{k}\upepsilon K_{1}} :\hbox {G}_{1}/\hbox {G}_{\mathrm{1k}}$ there exists $\hbox {x}\upepsilon \hbox {S}_{1}$ such that $\hbox {b}_{1}(\hbox {x})=\hbox {C}$ thus $\hbox {x}=\text{ b }_{1}^{-1} (\hbox {C})$ and

$$\begin{aligned} \hbox {b}_{2}\text{ fb }_{1}^{-1} (\hbox {C}+\hbox {t})= & {} \hbox {b}_{2}\hbox {fb}_{1}^{-1}(\hbox {b}_{1}(\hbox {x})+\hbox {t})= \hbox {b}_{2}\hbox {f}(\hbox {F}_{1}(\hbox {x,t}))=\hbox {b}_{2}\hbox {F}_{2}(\hbox {f}(\hbox {x}),\hbox {h}(\hbox {t}))\\= & {} \hbox {b}_{2}\text{ b }_{2}^{-1} [\hbox {b}_{2}(\hbox {f}(\hbox {x}))+\hbox {h}(\hbox {t})]=\hbox {b}_{2}\text{ fb }_{1}^{-1} (\hbox {C})+\hbox {h}(\hbox {t}). \end{aligned}$$

Since $\upvarphi (\hbox {G}_{\mathrm{1k}})\upepsilon \cup _{\mathrm{k}\upepsilon K_{2}} :\hbox {G}_{2}/\hbox {G}_{\mathrm{2k}}$ there exist functions $\upalpha :\hbox {K}_{1}\rightarrow \hbox {G}_{2}$ and $\upbeta :\hbox {K}_{1}\rightarrow \hbox {K}_{2}$ such that $\upvarphi (\hbox {G}_{\mathrm{1k}})=\hbox {G}_{2\upbeta (\mathrm{k})}+\upalpha (\hbox {k})$. We have thus for $\hbox {t}\upepsilon \hbox {G}_{\mathrm{1k}}$ that $\upvarphi (\hbox {G}_{\mathrm{1k}}+\hbox {t})=\upvarphi (\hbox {G}_{\mathrm{1k}})=\upvarphi (\hbox {G}_{\mathrm{1k}})+\hbox {h}(\hbox {t})$. This yields that $\hbox {G}_{ 2\upbeta (\mathrm{k})}+\upalpha (\hbox {k})=\hbox {G}_{2\upbeta (\mathrm{k})}+\upalpha (\hbox {k})+\hbox {h}(\hbox {t})$. From here $\hbox {h}(\hbox {t})\upepsilon [\hbox {--}\upalpha (\hbox {k})+\hbox {G}_{2\upbeta (\mathrm{k})}+\upalpha (\hbox {k})]$ if $\hbox {t}\upepsilon \hbox {G}_{\mathrm{1k.}}$. This implies $\hbox {h}(\hbox {G}_{\mathrm{1k}})\subset [\hbox {--}\upalpha (\hbox {k})+\hbox {G}_{2\upbeta (\mathrm{k})}+\upalpha (\hbox {k})]$. Moreover for $\hbox {C}=\hbox {G}_{\mathrm{1k}}+\hbox {t}$ we obtain

$$\begin{aligned} \upvarphi (\hbox {C})=\upvarphi (\hbox {G}_{\mathrm{1k}}+\hbox {t})=\upvarphi (\hbox {G}_{\mathrm{1k}})+\hbox {h}(\hbox {t})= \hbox {G}_{2\upbeta (\mathrm{k})}+\upalpha (\hbox {k})+\hbox {h}(\hbox {t}). \end{aligned}$$

Assume now that $\hbox {f}=\text{ b }_{2}^{-1} \upvarphi \hbox {b}_{1}$ with $\upvarphi$ as in (4). The function $\upvarphi$ is a solution of the Eq. (5) since for $\hbox {C}=\hbox {G}_{\mathrm{1k}}+\hbox {t}_{1}$

$$\begin{aligned} \upvarphi (\hbox {C}+\hbox {t})= & {} \upvarphi (\hbox {G}_{\mathrm{1k}}+\hbox {t}_{1}+\hbox {t})=\hbox {G}_{2\upbeta (\mathrm{k})}+\upalpha (\hbox {k})+\hbox {h}(\hbox {t}_{1}+\hbox {t})\\ {}= & {} \hbox {G}_{2\upbeta (\mathrm{k})}+\upalpha (\hbox {k})+\hbox {h}(\hbox {t}_{1})+\hbox {h}(\hbox {t}) = \upvarphi (\hbox {G}_{\mathrm{1k}}+\hbox {t}_{1})+\hbox {h}(\hbox {t})=\upvarphi (\hbox {C})+\hbox {h}(\hbox {t}). \end{aligned}$$

From here we have

$$\begin{aligned} \hbox {f}[\hbox {F}_{1}(\hbox {x,t})]= & {} \text{ b }_{2}^{-1} \upvarphi \hbox {b}_{1}[\text{ b }_{1}^{-1} (\hbox {b}_{1}(\hbox {x})+\hbox {t})]= \text{ b }_{2}^{-1} \upvarphi (\hbox {b}_{1}(\hbox {x})+\hbox {t})= \text{ b }_{2}^{-1} [\upvarphi (\hbox {b}_{1}(\hbox {x}))+\hbox {h}(\hbox {t})]\\= & {} \text{ b }_{2}^{-1} [\hbox {b}_{2}\text{ b }_{2}^{-1} \upvarphi (\hbox {b}_{1}(\hbox {x}))+\hbox {h}(\hbox {t})]=\text{ b }_{2}^{-1} [\hbox {b}_{2}(\hbox {f}(\hbox {x}))+\hbox {h}(\hbox {t})]=\hbox {F}_{2}(\hbox {f}(\hbox {x}),\hbox {h}(\hbox {t})). \end{aligned}$$

$\square$

Remark

The solution f of (1) is a function from $\hbox {S}_{1}$ to $\hbox {S}_{\mathrm{2}}$ but only the stability groups $\hbox {G}_{\mathrm{ik}}$ and the homomorphism h decide about the existence of this solution.

Examples

1.:

Put

and $\hbox {h}(\hbox {t})=\hbox {t}$.

We have $\hbox {G}_{1}=\hbox {G}_{2}=\mathbb {R}$, $\hbox {S}_{1}=(\hbox {--} \frac{\pi }{2}, \frac{\pi }{2})$, $\hbox {K}_{1}=\{ 1\}$ , $\hbox {S}_{{11}}=\hbox {S}_{1}$, $\hbox {b}_{11}=\hbox {tn}$, $\hbox {G}_{{11}}=\{ 0\}$, $\hbox {S}_{2}=\mathbb {R}$, $\hbox {K}_{2}=\{ 1,2,3\}$, $\hbox {S}_{{21}}=(0,+\infty )$, $\hbox {S}_{{22}}=\{ 0\}$, $\hbox {S}_{{23}}=(\hbox {--}\infty ,0)$, $\hbox {b}_{{21}}=\hbox {ln}$, $\hbox {b}_{{22}}=0$, $\hbox {b}_{{23}}(\hbox {x})=\hbox {ln}(\hbox {-x})$, $\hbox {G}_{{21}}=\{ 0\}$, $\hbox {G}_{{22}}=\mathbb {R}$, $\hbox {G}_{{23}}=\{ 0\}$.

The condition (3) has the form $\hbox {G}_{11}\subset \hbox {G}_{\mathrm{2\upbeta (1)}}$ and this is true for $\upbeta (1)=1,2,3$ and for every $\upalpha (1)=:\hbox {a}\upepsilon \mathbb {R}$.

If $\upbeta (1)=1$, then $\upvarphi (\hbox {G}_{11}+\hbox {t})=\upvarphi (\{ 0\} +\hbox {t})=\{ 0\} +\hbox {a}+\hbox {t}$. For $\hbox {C}(\hbox {t}):=\{ 0\} +\hbox {t}$ we have $\upvarphi (\hbox {C}(\hbox {t}))= \hbox {C}(\hbox {a}+\hbox {t})$. From here $\upvarphi (\hbox {b}_{11}(\hbox {x}))=\upvarphi (\hbox {C}(\hbox {tnx}))=\hbox {C}(\hbox {a}+\hbox {tnx})$.This yields that

$$\begin{aligned} \text{ f }\left( {\text{ x }} \right) =\text{ b }_{21}^{-1} (\hbox {C}(\hbox {a}+\hbox {tnx}))=\hbox {exp}(\hbox {a}+\hbox {tnx}). \end{aligned}$$

Analogously if $\upbeta (1)=2$, then $\hbox {f}(\hbox {x})=0$ and if $\upbeta (1)=3$, then $\hbox {f}(\hbox {x})=\hbox {-exp}(\hbox {a}+\hbox {tnx})$.

From here the functions $\hbox {f}(\hbox {x})=\upvarepsilon \hbox {exp}(\hbox {a}+\hbox {tnx})$ for $\hbox {x}\upepsilon (\hbox {--} \frac{\pi }{2}, \frac{\pi }{2})$, $\hbox {a}\upepsilon \mathbb {R}$, $\upvarepsilon \upepsilon \{ \hbox {--}1,0,1\}$ are the only solutions of (1) in the case of the above functions $\hbox {F}_{1}$ and $\hbox {F}_{2}$.

2.:

The equation

$$\begin{aligned} \hbox {f}[\hbox {F}_{2}(\hbox {x,t})]=\hbox {F}_{1}(\hbox {f}(\hbox {x}),\hbox {h}(\hbox {t})), \end{aligned}$$

where $\hbox {F}_{1},\hbox {F}_{2},\hbox {h}$ are as in Example 1, does not have a solution. In fact, there does not exist a function $\upbeta :\hbox {K}_{2}\rightarrow \hbox {K}_{1}$ for which $\mathbb {R}=\hbox {G}_{22}\subset \hbox {G}_{1\upbeta (\mathrm{k})}=\hbox {G}_{11}=\{ 0\}$.

We prove this fact directly, too. We have $\hbox {f}[\hbox {F}_{2}(0,\hbox {t})]=\hbox {f}(0)\ne \hbox {F}_{1}(\hbox {f}(0),\hbox {t})$ since $\hbox {F}_{1}(\hbox {f}(0),\mathbb {R})= (\hbox {--}\frac{\pi }{2}, \frac{\pi }{2})$.

• 3. Let the function F be of the form

  $$\begin{aligned} \hbox {F}(\hbox {x,t})=\left\{ \begin{array}{ll} \text{ b }_{\mathrm{n}}^{-1} [\hbox {b}_{\mathrm{n}}(\hbox {x})+\hbox {t}] &{} \quad \hbox { for } \hbox {x}\upepsilon \hbox {S}_{\mathrm{n}}, \quad \hbox {t}\upepsilon \hbox {G},\\ \hbox {x} &{}\quad \hbox { for } \hbox {x}\upepsilon \hbox {S}\backslash \cup \hbox {S}_{\mathrm{n}}=:\hbox {N}_{2}, \hbox {t}\upepsilon \hbox {G}, \end{array}\right. \end{aligned}$$

  (6)

where G is an abelian group, $\hbox {S}_{\mathrm{n}}$ are disjoint subsets of S such that $\hbox {cardS}_{\mathrm{n}}=\hbox {cardG}$ for n in a set $\hbox {N}_{1}$ of indices which is disjoint with $\hbox {N}_{2}$ and $\hbox {h}_{\mathrm{n}}:\hbox {S}_{\mathrm{n}}\rightarrow \hbox {G}$ are bijections.

Consider the equation

$$\begin{aligned} \hbox {f}[\hbox {F}(\hbox {x,t})]=\hbox {F}(\hbox {f}(\hbox {x}),\hbox {t}). \end{aligned}$$

(7)

Put $\hbox {N}=\hbox {N}_{1}\cup \hbox {N}_{2}$. The formula

$$\begin{aligned} \hbox {f}(\hbox {x})=\left\{ \begin{array}{ll} \upgamma (\hbox {x}) &{}\quad \hbox { for } \hbox {x}\upepsilon \hbox {N}_{2},\\ \upgamma (\hbox {n}) &{}\quad \hbox { for } \hbox {x}\upepsilon \hbox {S}_{\mathrm{n}}, \upgamma (\hbox {n})\upepsilon \hbox {N}_{2},\\ \text{ b }_{\upgamma \left( \text{ n } \right) }^{-1} [\hbox {b}_{\mathrm{n}}(x)+\updelta (\hbox {n})] &{}\quad \hbox { for } \hbox {x}\upepsilon \hbox {S}_{\mathrm{n}}, \upgamma (\hbox {n})\upepsilon \hbox {N}_{1}, \end{array}\right. \end{aligned}$$

(8)

where the function $\upgamma :\hbox {N}\rightarrow \hbox {N}$ is such that $\upgamma (\hbox {N}_{2})\subset \hbox {N}_{2}$ and $\updelta$ is an arbitrary function from $\hbox {N}_{1}$ to G, gives all the solutions of (7).

For the proof assume that the function $\hbox {f}:\hbox {S}\rightarrow \hbox {S}$ is a solution of (7). If $\hbox {x}\upepsilon \hbox {S}\backslash \cup \hbox {S}_{\mathrm{n}}= \hbox {N}_{2}$, then $\hbox {f}(\hbox {x})=\hbox {f}[\hbox {F}(\hbox {x,t})]=\hbox {F}(\hbox {f}(\hbox {x}),\hbox {t})$, thus $\hbox {f}(\hbox {x})\upepsilon \hbox {N}_{2}$ and we put $\upgamma (\hbox {x})=\hbox {f}(\hbox {x})$.

If $\hbox {x}\upepsilon \hbox {S}_{\mathrm{n}}$ and $\hbox {f}(\hbox {x})\upepsilon \hbox {N}_{2}$, then we have $\hbox {f}[\hbox {F}(\hbox {x,t})\} =\hbox {F}(\hbox {f}(\hbox {x}),\hbox {t})=\hbox {f}(\hbox {x})$. This yields that f is a constant function on $\hbox {S}_{\mathrm{n}}$ and this constant, denoted $\upgamma (\hbox {n})$, is in $\hbox {N}_{2}$. From here $\hbox {f}(\hbox {x})=\upgamma (\hbox {n})$ for $\hbox {x}\upepsilon \hbox {S}_{\mathrm{n}}$.

If $\hbox {x}\upepsilon \hbox {S}_{\mathrm{n}}$ and $\hbox {f}(\hbox {x})\upepsilon \cup \hbox {I}_{\mathrm{n}}$, then $\hbox {f}(\hbox {x})\upepsilon \hbox {S}_{\mathrm{m}}$ for some $\hbox {m}\upepsilon \hbox {N}_{1}$. Since $\hbox {f}[\hbox {F}(\hbox {x,t})]=\hbox {F}(\hbox {f}(\hbox {x}),\hbox {t})\upepsilon \hbox {S}_{\mathrm{m}}$, thus the values of the function f on $\hbox {S}_{\mathrm{n}}$ are in $\hbox {S}_{\mathrm{m}}$. From here m is determined by n, thus m is a function of $\hbox {n}: \hbox {m}=\upgamma (\hbox {n})\upepsilon \hbox {N}_{1}$. Let $\hbox {x}_{0}$ be a fixed point of $\hbox {S}_{\mathrm{n}}$, then $\hbox {f}(\hbox {x}_{0})\upepsilon \hbox {S}_{{\upgamma (\mathrm{n})}}$. Since $\hbox {x}\upepsilon \hbox {S}_{\mathrm{n}}$, for $\uptau =\hbox {b}_{\mathrm{n}}(\hbox {x})\hbox {--b}_{\mathrm{n}}(\hbox {x}_{0})$ we have $\hbox {F}(\hbox {x}_{0},\uptau )=\text{ b }_{\mathrm{n}}^{-1} (\hbox {b}_{\mathrm{n}}(\hbox {x}_{0})+\uptau )=\hbox {x}$. This implies that

$$\begin{aligned} \hbox {f}(\hbox {x})&=\hbox {f}[\hbox {F}(\hbox {x}_{0},\uptau )]=\hbox {F}(\hbox {f}(\hbox {x}_{0}),\uptau )=\text{ b }_{{\upgamma }\left( {\text{ n }} \right) }^{-1} [\hbox {b}_{{\upgamma (\mathrm{n})}}(\hbox {f}(\hbox {x}_{0}))+ \hbox {b}_{\mathrm{n}}(\hbox {x})\hbox {--b}_{\mathrm{n}}(\hbox {x}_{0})]\\&=\text{ b }_{{\upgamma }\left( {\text{ n }} \right) }^{-1} (\hbox {b}_{\mathrm{n}}(\hbox {x})+\updelta (\hbox {n})), \end{aligned}$$

where $\updelta (\hbox {n})= \hbox {b}_{{\upgamma (\mathrm{n})}}(\hbox {f}(\hbox {x}_{0}))\hbox {--b}_{\mathrm{n}}(\hbox {x}_{0})$ is a function from $\hbox {N}_{1}$ to G.

It is easily to verify that a function of the form (8) is the solution of the Eq. (7).

This result is proven in [9] if F: $\mathbb {R}\times \mathbb {R}\rightarrow \mathbb {R}$ is the continuous solution of the translation equation. Note that the formula (8) gives all the functions f which are commutative with the functions of the family $\{ \hbox {F}(\mathbf . , \hbox {t})\} _{\mathrm{t}\upepsilon }\hbox {G}$ of functions from S to S.

If a solution $\hbox {F}:\mathbb {R}\times \mathbb {R}\rightarrow \mathbb {R}$ of the translation equation, such that $\hbox {F}(\hbox {x},0)=\hbox {x}$ for $\hbox {x}\upepsilon \mathbb {R}$, is continuous with respect to the second variable for every $\hbox {x}\upepsilon \mathbb {R}$ and for which at least one of the functions $\hbox {F}(\mathbf ., \hbox {t}): \mathbb {R}\rightarrow \mathbb {R}$ is continuous, then the function F has the form (6). In this case $\hbox {S}_{\mathrm{n}}$ are open intervals and $\hbox {b}_{\mathrm{n}}$ are homeomorphisms and the function F is continuous [10]. This solution F of the translation equation is continuous, too, if F is Carathéodory, i.e., the function $\hbox {F}(\hbox {x}, \mathbf . ): \mathbb {R}\rightarrow \mathbb {R}$ is measurable for every $\hbox {x}\upepsilon \mathbb {R}$ and function $\hbox {F}(\mathbf . ,\hbox {t}):\mathbb {R}\rightarrow \mathbb {R}$ is continuos for every $\hbox {t}\upepsilon \mathbb {R}$ [2].

Recall that $\hbox {x}_{0}$ is said to be a fixed point of the function $\hbox {F}:\hbox {S}\times \hbox {G}\rightarrow \hbox {G}$ if $\hbox {F}(\hbox {x}_{0},\hbox {t})=\hbox {x}_{0}$ for $\hbox {t}\upepsilon \hbox {G}$.

Theorem 2

Let $\hbox {F}:\mathbb {R}\times \mathbb {R}\rightarrow \mathbb {R}$ be a continuous solution of the translation equation such that $\hbox {F}(\hbox {x},0)=\hbox {x}$ for $\hbox {x}\upepsilon \mathbb {R}$. Every solution f of (7) is continuous if and only if F does not have any fixed points.

Proof

Assume that f is a solution of (7). If F does not have any fixed points, then the set $\hbox {N}_{1}$ in the form (8) of F is a singelton $\hbox {N}_{1}=\{ \hbox {m}\}$ and $\hbox {S}_{\mathrm{m}}=\mathbb {R}$. From here $\upgamma (\hbox {m})=\hbox {m}$ and $\hbox {f}(\hbox {x})=\text{ b }_{\mathrm{m}}^{-1} (\hbox {b}_{\mathrm{m}}(\hbox {x})+\updelta (\hbox {m}))$ for $\hbox {x}\upepsilon \mathbb {R}$, where $\hbox {b}_{\mathrm{m}}:\mathbb {R}\rightarrow \mathbb {R}$ is a homeomorphism and $\updelta (\hbox {m})\upepsilon \mathbb {R}$. The function f is thus continuous.

If a is the only fixed point of F, i.e., $\hbox {N}_{2}=\{ \hbox {a}\}$ in (6), then $\hbox {N}_{\mathrm{1}}$ has two elements, which we denote as 1,2, so that $\hbox {S}_{1}=(\hbox {--}\infty ,\hbox {a})$ and $\hbox {S}_{2}=(\hbox {a},+\infty )$. If $\upgamma (1)=2$, $\upgamma (2)=2$, $\upgamma (\hbox {a})=\hbox {a}$, $\updelta :\{ 1,2\} \rightarrow \mathbb {R}$ is an arbitrary function and $\hbox {b}_{1}:\hbox {S}_{1}\rightarrow \mathbb {R}$ and $\hbox {b}_{2}:\hbox {S}_{2}\rightarrow \mathbb {R}$ are arbitrary increasing bijections, then the function f defined as in (8) is a solution of (7). For $\hbox {x}\upepsilon \hbox {S}_{1}$ we obtain $\hbox {f}(\hbox {x})=\text{ b }_{2}^{-1} (\hbox {b}_{1}(\hbox {x})+\updelta (1))$. We have for $\hbox {x}\rightarrow \hbox {a}$ that $\hbox {b}_{1}(\hbox {x})+\updelta (1) \rightarrow +\infty$, thus $\text{ b }_{2}^{-1} (\hbox {b}_{1}(\hbox {x})+\updelta (1))\rightarrow +\infty$ and from here f is not continuous.

Let a and b be two different fixed point of F. We consider two cases.

• a/ Every point of the interval (a,b) is a fixed point of F.

• b/ The opposite case to a/.

In case a/ let the function $\upgamma$ in (8) be such that $\upgamma (\hbox {a})=\hbox {b}$ and $\upgamma (\hbox {x})=\hbox {x}$ for $\hbox {x}\upepsilon (\hbox {a,b})$. From here we have for the function f defined as in (8) that

$$\begin{aligned} \hbox {lim}_{\mathrm{x}\rightarrow \mathrm{a}+} \hbox {f}(\hbox {x})=\hbox {a}\ne \hbox {b}=\upgamma (\hbox {a})=\hbox {f}(\hbox {a}), \end{aligned}$$

thus f is discontinuous.

In case b/ there exists at least one n$\upepsilon \hbox {N}_{1}$ in the form (6) of the function F such that $\hbox {S}_{\mathrm{n}}\subset (\hbox {a,b})$. Let $\hbox {c}=\hbox {infS}_{\mathrm{n}}$. Then $\hbox {c}\upepsilon \hbox {N}_{2}$. Let $\upgamma$ in (8) be such that $\upgamma (\hbox {c})=\hbox {b}$, $\upgamma (\hbox {n})=\hbox {n}$ and let $\updelta (\hbox {n})=0$ in (8). From here $\hbox {f}(\hbox {x})=\hbox {x}$ for $\hbox {x}\upepsilon \hbox {S}_{\mathrm{n}}$. This yields that

$$\begin{aligned} \hbox {lim}_{\mathrm{x}\rightarrow \mathrm{c}+}\hbox {f}(\hbox {x})=\hbox {c}\ne \hbox {b}=\upgamma (\hbox {c})=\hbox {f}(\hbox {c}), \end{aligned}$$

thus f is discontinuous. $\square$

If in Eq. (7) we take the function $\hbox {F}_{1}$ of Example 1 as the function F, then the solution f has the form $\hbox {f}(\hbox {x})=\hbox {tn}^{-1}(\hbox {tnx}+\updelta )$ for $\updelta \upepsilon \mathbb {R}$. If in this equation we have $\hbox {F}_{2}$, then

$$\begin{aligned} \hbox {f}(\hbox {x})=\left\{ \begin{array}{ll} 0 &{}\quad \hbox { for } \hbox {x}=0,\\ 0 &{}\quad \hbox { for } \hbox {x}\upepsilon \hbox {S}_{\mathrm{n}}, \upgamma (\hbox {n})=0,\\ \text{ b }_{\upgamma \left( \mathrm{n} \right) }^{-1} (\hbox {b}_{\mathrm{n}}(\hbox {x})+\updelta (\hbox {n})) &{}\quad \hbox { for } \hbox {x}\upepsilon \hbox {S}_{\mathrm{n}}, \upgamma (\hbox {n})\ne 0, \end{array}\right. \end{aligned}$$

where $\hbox {S}_{1}=(0,+\infty )$, $\hbox {S}_{2}=(\hbox {--}\infty ,0)$, $\upgamma :\{ 0,1,2\} \rightarrow \{ 0,1,2\}$, $\upgamma (0)=0$, $\updelta :\{ 1,2\} \rightarrow \mathbb {R}$, $\hbox {b}_{1}(\hbox {x})=\hbox {lnx}$, $\hbox {b}_{2}(\hbox {x})=\hbox {ln}(\hbox {-x})$.

4.:

The constant function $\hbox {f}(\hbox {x})=\hbox {c}$ for $\hbox {x}\upepsilon \hbox {S}_{1}$ is a solution of (1) if and only if $\hbox {F}_{2}(\hbox {c,h}(\hbox {t}))=\hbox {c}$ for $\hbox {t}\upepsilon \hbox {G}_{1}$. We verify condition (3) directly in this case. If $\hbox {f}(\hbox {x})=\hbox {c}\upepsilon \hbox {S}_{2}$ is a solution of (1), then there exists a $\hbox {l}\upepsilon \hbox {K}_{2}$ such that $\hbox {c}\upepsilon \hbox {S}_{\mathrm{2l}}$. From here $\hbox {F}_{2}(\hbox {c,s})=\text{ b }_{21}^{-1} [\hbox {b}_{\mathrm{2l}}(\hbox {c})+\hbox {s}]$ thus $\hbox {c}= \hbox {F}_{2}(\hbox {c,h}(\hbox {t}))=\text{ b }_{21}^{-1} [\hbox {b}_{\mathrm{2l}}(\hbox {c})+\hbox {h}(\hbox {t})]$ for $\hbox {t}\upepsilon \hbox {G}_{1}$. There exists $\hbox {s}_{0}\upepsilon \hbox {G}_{2}$ for which $\hbox {b}_{\mathrm{2l}}(\hbox {c})=\hbox {G}_{\mathrm{2l}}+\hbox {s}_{0}$. This yields that $\hbox {G}_{\mathrm{2l}}+\hbox {s}_{0}=\hbox {b}_{\mathrm{2l}}(\hbox {c})=\hbox {G}_{\mathrm{2l}}+\hbox {s}_{0}+\hbox {h}(\hbox {t})$. From here $-\hbox {s}_{0}+ \hbox {G}_{\mathrm{2l}}+\hbox {s}_{0}=(\hbox {--s}+\hbox {G}_{\mathrm{2l}}+\hbox {s}_{0})+\hbox {h}(\hbox {t})$. Since $-\hbox {s}_{0}+ \hbox {G}_{\mathrm{2l}}+\hbox {s}_{0}$ is a group, $\hbox {h}(\hbox {G}_{1})\subset [-\hbox {s}_{0}+ \hbox {G}_{\mathrm{2l}}+\hbox {s}_{0}]$. We have $\hbox {h}(\hbox {G}_{\mathrm{1k}})\subset \hbox {h}(\hbox {G}_{1})$ for $\hbox {k}\upepsilon \hbox {K}_{1}$, thus the condition (3) is satisfied with $\upbeta (\hbox {k})=\hbox {s}_{0}$ for $\hbox {k}\upepsilon \hbox {K}_{1}$.

5.:

Consider the equation

$$\begin{aligned} \hbox {f}(\hbox {g}(\hbox {x}))=\hbox {g}(\hbox {f}(\hbox {x})), \end{aligned}$$

(9)

where $\hbox {g}:\hbox {S}\rightarrow \hbox {S}$ is a given idempotent function and $\hbox {f}:\hbox {S}\rightarrow \hbox {S}$ is the unknown function. This is an equation of the form (1) for $\hbox {F}_{1}(\hbox {x,t})=\hbox {F}_{2}(\hbox {x,t})=\hbox {g}(\hbox {x})$.

If g is an injective function on the set $\hbox {S}\backslash \hbox {g}(\hbox {S})$, then every function f of of the form

$$\begin{aligned} \hbox {f}(\hbox {x})=\left\{ \begin{array}{ll} \hbox {f}_{1}(\hbox {x}) &{}\quad \hbox { for } \hbox {x}\upepsilon \hbox {g}(\hbox {S}),\\ \text{ g }_{1}^{-1} [\hbox {f}_{1}(\hbox {g}(\hbox {x})]&{}\quad \hbox { for } \hbox {x}\upepsilon \hbox {S}\backslash \hbox {g}(\hbox {S}), \end{array}\right. \end{aligned}$$

(10)

where $\hbox {f}_{1}$ is an arbitrary function from g(S) to g(S) and $\hbox {g}_{1}=\hbox {g}_{\vert \mathrm{S}\backslash \mathrm{g}(\mathrm{S})}$, is a solution of (9).

Indeed, if $\hbox {x}\upepsilon \hbox {g}(\hbox {S})$, then there exists $\hbox {y}\upepsilon \hbox {S}$ such that $\hbox {x}=\hbox {g}(\hbox {y})$. From here $\hbox {f}(\hbox {g}(\hbox {x}))=\hbox {f}[\hbox {g}(\hbox {g}(\hbox {y}))]= \hbox {f}(\hbox {g}(\hbox {y}))= \hbox {f}(\hbox {x})=\hbox {f}_{1}(\hbox {x})=\hbox {g}(\hbox {f}_{1}(\hbox {x}))=\hbox {g}(\hbox {f}(\hbox {x}))$ since g is the identity on g(S) as the idempotent. If $\hbox {x}\upepsilon \hbox {S}\backslash \hbox {g}(\hbox {S})$, then $\hbox {g}(\hbox {f}(\hbox {x}))=\hbox {f}_{1}(\hbox {g}(\hbox {x}))=\hbox {f}(\hbox {g}(\hbox {x}))$ since $\hbox {g}(\hbox {x})\upepsilon \hbox {g}(\hbox {S})$.

If $\hbox {S}=\hbox {g}(\hbox {S})$, i.e., if $\hbox {g}(\hbox {x})=\hbox {x}$ for $\hbox {x}\upepsilon \hbox {S}$, then every function $\hbox {f}:\hbox {S}\rightarrow \hbox {S}$ is a solution of (9).

If $\hbox {S}\ne \hbox {g}(\hbox {S})$, then a solution f of (9) is of the form (10) if and only if $\hbox {f}(\hbox {S}\backslash \hbox {g}(\hbox {S}))\subset \hbox {S}\backslash \hbox {g}(\hbox {S})$. In fact, if f is a solution of (9) and $\hbox {f}(\hbox {S}\backslash \hbox {g}(\hbox {S}))\subset \hbox {S}\backslash \hbox {g}(\hbox {S})$, then

• a/ for $\hbox {x}\upepsilon \hbox {g}(\hbox {S})$ we have $\hbox {f}(\hbox {x})=\hbox {f}(\hbox {g}(\hbox {x}))=\hbox {g}(\hbox {f}(\hbox {x}))\upepsilon \hbox {g}(\hbox {S})$, thus $\hbox {f}_{1}=\hbox {f}_{\vert \mathrm{g(S)}}$ is a function from g(S) to g(S) and $\hbox {f}=\hbox {f}_{1}$ on g(S),

• b/ for $\hbox {x}\upepsilon \hbox {S}\backslash \hbox {g}(\hbox {S})$ we have $\hbox {f}(\hbox {x})\upepsilon \hbox {S}\backslash \hbox {g}(\hbox {S})$ and $\hbox {f}(\hbox {x})=\text{ g }_{1}^{-1} [\hbox {f}_{1}(\hbox {g}(\hbox {x})]$ since

  $$\begin{aligned} \hbox {g}_{1}(\hbox {f}(\hbox {x}))=\hbox {g}(\hbox {f}(\hbox {x}))= \hbox {f}(\hbox {g}(\hbox {x}))=\hbox {f}_{1}(\hbox {g}(\hbox {x})). \end{aligned}$$

  If f is of the form (10), then for $\hbox {x}\upepsilon \hbox {S}\backslash \hbox {g}(\hbox {S})$ we obtain $\hbox {f}(\hbox {x})=\text{ g }_{1}^{-1} [\hbox {f}_{1}(\hbox {g}(\hbox {x})]\upepsilon \hbox {S}\backslash \hbox {g}(\hbox {S})$.

Note in the end that if $\hbox {S}\ne \hbox {g}(\hbox {S})$, then the function $\hbox {f}(\hbox {x})=\hbox {c}\upepsilon \hbox {g}(\hbox {S})$ for $\hbox {x}\upepsilon \hbox {S}$ is a solution of (9) and f is not of the form (10) by the above.

6.:

Let G be the groupoid of $\hbox {n}\times \hbox {n-matrices}$ with the usual multiplication. We give all the solutions of Eq. (7) in which $\hbox {F}(\hbox {x,t})=\hbox {xt}:\mathbb {R}^{\mathrm{n}}\times \hbox {G}\rightarrow \mathbb {R}^{\mathrm{n}}$ (the transformation law of the vector). Equation (7) is in this case of the form $\hbox {f}(\hbox {xt})=\hbox {f}(\hbox {x})\hbox {t}$. Assume that a function $\hbox {f}:\mathbb {R}^{\mathrm{n}}\rightarrow \mathbb {R}^{\mathrm{n}}$ is a solution of this equation. Put in this equation $\hbox {x}=(1,0,\ldots ,0)$ and $t= \vert \vert \hbox {t}_{\mathrm{ik}}\vert \vert _{\mathrm{i,k}=1,\ldots ,\mathrm{n}}$, where $\hbox {t}_{\mathrm{1k}}=\hbox {x}_{\mathrm{k}}$ for $\hbox {k}=1,\ \ldots ,\hbox {n}$ and the other $\hbox {t}_{\mathrm{ik}}$ are arbitrary in $\mathbb {R}$, we have $\hbox {f}(\hbox {x}_{1},\ \ldots ,\hbox {x}_{\mathrm{n}})= (\hbox {y}_{1},\ \ldots ,\hbox {y}_{\mathrm{n}})\hbox {t}$, where $\hbox {y}:= (\hbox {y}_{1},\ \ldots ,\hbox {y}_{\mathrm{n}})= \hbox {f}(1,0,\ \ldots ,0)$. Assume that there exists $\hbox {m}\ne 1$ for which $\hbox {y}_{\mathrm{m}}\ne 0$. Let $\hbox {a}_{1}=\vert \vert \hbox {a}_{\mathrm{ik}}\vert \vert _{\mathrm{i,k}=1,\ldots ,\mathrm{n}}$, where $\hbox {a}_{\mathrm{1k}}=\hbox {x}_{\mathrm{k}}$ for $\hbox {k}=1,\ \ldots ,\hbox {n}$, $\hbox {a}_{\mathrm{i1}}=0$ for $\hbox {i}\ne 1$ and the other $\hbox {a}_{\mathrm{ik}}$ are arbitrary in $\mathbb {R}$. Let $\hbox {a}_{2}=\vert \vert \hbox {b}_{\mathrm{ik}}\vert \vert _{\mathrm{i,k}=1,\ldots ,\mathrm{n}}$, where $\hbox {b}_{\mathrm{1k}}=\hbox {x}_{\mathrm{k}}$ for $\hbox {k}=1,\ \ldots ,\hbox {n}$, $\hbox {b}_{\mathrm{m1}}=1$, $\hbox {b}_{\mathrm{i1}}=0$ for $\hbox {i}\ne 1,\hbox {m}$ and the other $\hbox {b}_{\mathrm{ik}}$ are arbitrary in $\mathbb {R}$. Puting $\hbox {t}=\hbox {a}_{\mathrm{l}}$ in our equation we have $\hbox {f}(\hbox {x}_{1},\ \ldots ,\hbox {x}_{\mathrm{n}})=\hbox {ya}_{\mathrm{l}}$. From here $\hbox {ya}_{1}=\hbox {ya}_{2}$ and in particular

$$\begin{aligned} \hbox {y}_{1}\hbox {x}_{1}&=\hbox {y}_{1}\hbox {a}_{11}+\hbox {y}_{2}\hbox {a}_{\mathrm{21}}+\ \cdots +\hbox {y}_{\mathrm{n}}\hbox {a}_{\mathrm{n1}}=\hbox {y}_{1}\hbox {b}_{11}+\ \cdots +\hbox {y}_{\mathrm{m}}\hbox {b}_{\mathrm{m1}}+\ \cdots \\&+\hbox {y}_{\mathrm{n}}\hbox {b}_{\mathrm{n1}}= \hbox {y}_{1}\hbox {x}_{1}+\hbox {y}_{\mathrm{m}}, \end{aligned}$$

thus $\hbox {y}_{\mathrm{m}}=0$. We have a contradiction thus $\hbox {y}_{\mathrm{i}}=0$ for $\hbox {i}=2,\ \ldots ,\hbox {n}$. This yields, by the equation, that $\hbox {f}(\hbox {x}_{1},\ \ldots ,\hbox {x}_{\mathrm{n}})=\hbox {y}_{1}(\hbox {x}_{1},\ \ldots ,\hbox {y}_{\mathrm{n}})$. These functions are evidently solutions of our equation.

We have the same result if G is the group of nonsingular $\hbox {n}\times \hbox {n-matrices}$. We prove this for $\hbox {n}=2$. If we put in the above

• i/ $\hbox {a}_{1}=\left\| {\begin{array}{cc} {x_{1} } &{}\quad {x_{2} } \\ 0 &{}\quad 1 \\ \end{array}}\right\|$ and $\hbox {a}_{2}=\left\| {\begin{array}{cc} {x_{1} } &{}\quad {x_{2} } \\ 1 &{}\quad {x_{2} x_{1}^{-1} +1} \\ \end{array}}\right\|$ for $\hbox {x}_{1}\ne 0$ and

• ii/ $\hbox {a}_{1} = \left\| {\begin{array}{cc} {x_{1} } &{}\quad {x_{2} } \\ 1 &{}\quad 0 \\ \end{array}}\right\|$ and $\hbox {a}_{2}=\left\| {\begin{array}{cc} {x_{1} } &{}\quad {x_{2} } \\ {x_{1} x_{2}^{-1} +1} &{}\quad 1 \\ \end{array}}\right\|$ for $\hbox {x}_{2}\ne 0$,

then we obtain as above $\hbox {y}_{2}=0$. Thus by the equation for t nonsingular (e,g., for $\hbox {t}= \left\| {\begin{array}{cc} {x_{1} } &{}\quad {x_{2} } \\ {-x_{2} } &{}\quad {x_{1} } \\ \end{array} }\right\| )$ we have $\hbox {f}(\hbox {x}_{1},\hbox {x}_{2})=\hbox {y}_{1}(\hbox {x}_{1},\hbox {x}_{2})$ for $x_{1}^{2} +x_{2}^{2} \ne 0$.

For $(\hbox {x}_{1},\hbox {x}_{2})=(0,0)$ put in our equation $\hbox {t}=\left\| {\begin{array}{cc} 1 &{}\quad 1 \\ 1 &{}\quad 0 \\ \end{array}}\right|$ and $\hbox {f}((0,0))=(\upalpha , \upbeta )$. Then we have $(\upbeta , \upalpha \hbox {--}\upbeta )= (0,0)$, thus $\upalpha =\upbeta =0$ and from here $\hbox {f}((0,0))=(0,0)=\hbox {y}_{1}(0,0)$, too.

It is possible to obtain this result also by the method as in Theorem 1 [9]. The above result means in the theory of geometric objects that the multiplication of a vector by a scalar is the only geometric concomitant of the vector.

3 Unknown function $\hbox {F}_{2}$

We only need to define the function $\hbox {F}_{2}$ on $\hbox {S}_{2}\times \hbox {h}(\hbox {G}_{1})$ due to the form of the Eq. (1)

Theorem 3

• 1/ Equation (1) has a solution for $\hbox {F}_{2}:\hbox {S}_{2}\times \hbox {h}(\hbox {G}_{1})\rightarrow \hbox {S}_{2}$ if and only if

  $$\begin{aligned} \hbox {f}(\hbox {x})=\hbox {f}(\hbox {y}) \,and\, \hbox {h}(\hbox {t})=\hbox {h}(\hbox {s}) \Rightarrow \hbox {f }[\hbox {F}_{1}(\hbox {x,t})]=\hbox {f }[\hbox {F}_{1}(\hbox {y,s})]\,for\, \hbox {x,y}\upepsilon \hbox {S}_{1}, \hbox {t,s}\upepsilon \hbox {G}_{1}.\quad \end{aligned}$$

  (11)

• 2/ Let functions $\hbox {f},\hbox {h},\hbox {F}_{1}$ satisfy (11). Then a function of the form

  $$\begin{aligned} \hbox {F}_{2}(\hbox {u,v})=\left\{ \begin{array}{ll} \hbox {f }[\hbox {F}_{1}(\hbox {x,t})] &{} \,for\, \hbox {u}=\hbox {f}(\hbox {x}), \hbox {v}=\hbox {h}(\hbox {t})\\ \hbox {H}(\hbox {u,v}) &{}\, for\, \hbox {u}\upepsilon \hbox {S}_{2}\backslash \hbox {f}(\hbox {S}_{1}),\hbox {v}\upepsilon \hbox {h}(\hbox {G}_{1}), \end{array}\right. \end{aligned}$$

  (12)

where $\hbox {H}(\hbox {u,v}):(\hbox {S}_{2}\backslash \hbox {f}(\hbox {S}_{1}))\times \hbox {G}_{1}\rightarrow \hbox {S}_{2}\backslash \hbox {f}(\hbox {S}_{1})$ is an arbitrary solution of the translation equation, is a solutions of (1).

Proof

The proof of the “only if” part is evident.

The function $\hbox {F}_{2}$ defined by (12) is well-defined due to (11) and it is a solution of (1). Moreover, for $\hbox {u}=\hbox {f}(\hbox {x})$, $\hbox {v}=\hbox {h}(\hbox {t})$, $\hbox {w}=\hbox {h}(\hbox {s})$, we have

$$\begin{aligned} \hbox {F}_{2}(\hbox {F}_{2}(\hbox {u,v}),\hbox {w})= & {} \hbox {F}_{2}[\hbox {F}_{2}(\hbox {f}(\hbox {x}),\hbox {h}(\hbox {t})),\hbox {h}(\hbox {s})]= \hbox {F}_{2}[\hbox {f}(\hbox {F}_{1}(\hbox {x,t})),\hbox {h}(\hbox {s})]\\= & {} \hbox {f}[\hbox {F}_{1}(\hbox {F}_{1}(\hbox {x,t}),\hbox {s})]=\hbox {f}[\hbox {F}_{1}(\hbox {x,t}+\hbox {s})]= \hbox {F}_{2}(\hbox {u,v}+\hbox {w}). \end{aligned}$$

$\square$

Example

7.:

The condition (11) is satisfied for $\hbox {h}(\hbox {t})=\hbox {t}$ and by every $\hbox {f}(\hbox {x})=\upvarepsilon \hbox {exp}(\hbox {a}+\hbox {tnx})$ ($\upvarepsilon =\hbox {--}1,0,1)$ in the equation in Example 1

• a/ since the function h and the functions f for $\upvarepsilon \ne 0$ are injective and for $\hbox {f}=0$ ($\upvarepsilon =0$) the condition (11) is evident or

• b/ since there exist function $\hbox {f},\hbox {h},\hbox {F}_{1},\hbox {F}_{2}$ which satisfy equation (7).

If we want to have $\hbox {F}_{2}$ defined on $\hbox {S}_{2}\times \hbox {G}_{2}$, then we must extend the solution $\hbox {F}_{2}$ of the translation equation on $\hbox {S}_{2}\times \hbox {h}(\hbox {G}_{1})$ to the solution of the translation equation on $\hbox {S}_{2}\times \hbox {G}_{2}$. This extension is not always possible (see Example 8 below). For the condition when this extension is possible see [6] and [8].

Examples

8.:

Let $\hbox {S}_{1}=\hbox {S}_{2}=\mathbb {Z}$, $\hbox {G}_{1}=\hbox {G}_{2}=\mathbb {Z}$ (the additive group of integers), $\hbox {f}(\hbox {x})=\hbox {x}, \hbox {h}(\hbox {t})=2\hbox {t}$, $\hbox {F}_{1}(\hbox {x,t})=\hbox {x}+\hbox {t}$. These functions evidently satisfy (11). Assume that there exists a solution $\hbox {F}_{2}:\mathbb {Z}\times \mathbb {Z}\rightarrow \mathbb {Z}$ of the translation equation such that $\hbox {F}_{2}(\hbox {x,u})=\hbox {F}(\hbox {x,t})$ for $\hbox {u}=2\hbox {t}$, i.e., such that $\hbox {F}_{2}(\hbox {x},2\hbox {t})= \hbox {F}_{1}(\hbox {x,t})= \hbox {x}+\hbox {t}$ for $(\hbox {x,t})\upepsilon \mathbb {Z}\times \mathbb {Z}$. From here we have for $\hbox {x}\upepsilon \mathbb {Z}$

$$\begin{aligned} \mathbb {Z} =\hbox {F}_{1}(\hbox {x},\mathbb {Z})=\hbox {F}_{2}(\hbox {x},2\mathbb {Z})\subset \hbox {F}_{2}(\hbox {x},\mathbb {Z})\subset \mathbb {Z}. \end{aligned}$$

Thus $\hbox {F}_{2}(\hbox {x},\mathbb {Z})=\mathbb {Z}$. Since $\hbox {F}_{2}$ is the solution of the translation equation and $\hbox {F}_{2}(\hbox {x},0)=\hbox {x}$ for $\hbox {x}\upepsilon \mathbb {Z}$, $\hbox {F}_{2}(\hbox {x,t})=\text{ b }_{1}^{-1} (\hbox {b}_{1}(\hbox {x})+\hbox {t})$, where $\hbox {b}_{1}$ is a bijection from $\mathbb {Z}$ to $\mathbb {Z}/\mathbb {Z}^{*}$ for a subgroup $\mathbb {Z}^{*}$ of $\mathbb {Z}$ [7]. We have $\hbox {b}_{1}(0)=\mathbb {Z}^{*}+\hbox {a}$ for some $\hbox {a}\upepsilon \mathbb {Z}$. We obtain for $\hbox {b}(\hbox {x}):=\hbox {b}_{1}(\hbox {x})\hbox {--a}$ that

$$\begin{aligned} \hbox {F}_{2}(\hbox {x,t})=\text{ b }_{1}^{-1} (\hbox {b}_{1}(\hbox {x})+\hbox {t})= \hbox {b}^{-1}(\hbox {b}(\hbox {x})+\hbox {t}) \hbox { and } \hbox {b}(0)=\mathbb {Z}^{*}. \end{aligned}$$

This yields that $\hbox {b}^{-1}(\hbox {b}(\hbox {x})+2)=\hbox {F}_{2}(\hbox {x},2)=\hbox {F}_{1}(\hbox {x},1)=\hbox {x}+1$. From here $\hbox {b}(\hbox {x})+2=\hbox {b}(\hbox {x}+1)$, thus $\hbox {b}(1)=\hbox {b}(0)+\mathbb {Z}^{*}$. This implies that $\hbox {b}(\hbox {x})=\mathbb {Z}^{*}+2\hbox {x}$ for $\hbox {x}\upepsilon \mathbb {Z}$. Let $\hbox {c}\ge 0$ be the generator of $\mathbb {Z}^{*}$, i.e., $\mathbb {Z}^{*}=\{ \hbox {nc:n}\upepsilon \mathbb {Z}\}$.

• i/ If $\hbox {c}=0$, i.e., $\mathbb {Z}^{*}=\{0\}$ , then $\hbox {b}(\hbox {x})=\{0\} +2\hbox {x}$. From here b is not a function onto $\mathbb {Z}/\mathbb {Z}^{*}$.

• ii/ If $\hbox {c}> 0$, then $\hbox {b}(\hbox {c})=\mathbb {Z}^{*}+2\hbox {c}=\mathbb {Z}^{*}=\mathbb {Z}^{*}+4\hbox {c}=\hbox {b}(2\hbox {c})$. Thus the function b is not injective.

9.:

Let $\hbox {S}_{1}=\hbox {S}_{2}=\hbox {G}_{1}=\hbox {G}_{2}=\hbox {G}$ be a semigrup and $\hbox {f}(\hbox {x})=\hbox {x}$ for $\hbox {x}\upepsilon \hbox {G}$. Let $\hbox {h:G}\rightarrow \hbox {G}$ be a homomorphism such that $\hbox {h}(\hbox {G})\ne \hbox {G}$ and $\hbox {F}_{1}(\hbox {x,t})=\hbox {x}+\hbox {h}(\hbox {t})$ for $\hbox {x,t}\upepsilon \hbox {G}$. The function $\hbox {F}_{1}$ is the solution of the translation equation and $\hbox {F}_{1}$ satifies (11). The function $\hbox {F}_{2}$, definied on $\hbox {G}\times \hbox {h}(\hbox {G})$ as in (12) (with $\hbox {f}(\hbox {x})=\hbox {x})$, is of the form $\hbox {F}_{2}(\hbox {x,h}(\hbox {t}))=\hbox {F}_{1}(\hbox {x,t})= \hbox {x}+\hbox {h}(\hbox {t})$. Thus the function $\hbox {x}+\hbox {t}$ is an extention of $\hbox {F}_{2}$ on the set $\hbox {G}\times \hbox {G}$.

Remarks

It is possible that the functions of the form (12) are not all the solutions of Eq. (1). E.g., if $\hbox {S}_{1}=\hbox {S}_{2}=\mathbb {R}$, $\hbox {G}_{1}=\hbox {G}_{2}=\mathbb {R}$ with the usual addition, $\hbox {h}(\hbox {t})=\hbox {t}$, $\hbox {f}(\hbox {x})=\hbox {expx}$, $\hbox {F}_{1}(\hbox {x,t})=\hbox {x}$, $\hbox {F}_{2}(\hbox {u,v})=\hbox {u}$ for $\hbox {u}> 0,\hbox {v}\upepsilon \mathbb {R}$ and $\hbox {F}_{2}(\hbox {u,v})=1$ for $\hbox {u}\le 0,\hbox {v}\upepsilon \mathbb {R}$, then the solution of the translation equation $\hbox {F}_{2}$ is the solution of (1), too. This function $\hbox {F}_{2}$ is not of the form (12) since $0\upepsilon \hbox {S}_{2}\backslash \hbox {f}(\hbox {S}_{1})=(\hbox {--}\infty , 0]$ and $\hbox {F}_{2}(0,\hbox {v})=1\notin \hbox {S}_{2}\backslash \hbox {f}(\hbox {S}_{1})$.

If $\hbox {f}(\hbox {S}_{1})=\hbox {S}_{2}$, then $\hbox {F}_{2}$ must be of the form $\hbox {F}_{2}(\hbox {u,v})=\hbox {f}[\hbox {F}_{1}(\hbox {x,t})]$ for $\hbox {u}=\hbox {f}(\hbox {x})$, $\hbox {v}=\hbox {h}(\hbox {t})$, thus Eq. (1) has only one solution. In this case if there exists $\hbox {t}_{0}$ such that $\hbox {F}_{1}(\hbox {x},\hbox {t}_{0})=\hbox {x}$ for $\hbox {x}\upepsilon \hbox {S}_{1}$, then $\hbox {F}_{2}(\hbox {u,h}(\hbox {t}_{0}))=\hbox {f}(\hbox {x})=\hbox {u}$ for $\hbox {u}\upepsilon \hbox {S}_{2}$. From here if $\hbox {F}_{1}$ is the transformation law for a geometric object, then $\hbox {F}_{2}$ is the same.

If $\hbox {S}_{1}=\hbox {S}_{2}$, then for the function

$$\begin{aligned} \hbox {F}_{2}(\hbox {u,v})=\left\{ \begin{array}{ll} \hbox {f}[\hbox {F}_{1}(\hbox {x,t})] &{}\quad \hbox { for } \hbox {u}=\hbox {f}(\hbox {x}), \hbox {v}=\hbox {h}(\hbox {t}),\\ \hbox {u} &{}\quad \hbox { for } \hbox {u}\upepsilon \hbox {S}_{2}\backslash \hbox {f}(\hbox {S}_{1}), \hbox {v}\upepsilon \hbox {G}_{2} \end{array}\right. \end{aligned}$$

we have the same situation as above.

The condition (11) and the equation

$$\begin{aligned} \hbox {f}[\hbox {F}_{1}(\hbox {x,t})]=\hbox {F}_{1}(\hbox {f}(\hbox {x}),\hbox {h}(\hbox {t})) \end{aligned}$$

are not equivalent. A solution of this equation is a solution of (11) but not vice versa. E.g., every constant function f is a solution of (11) and $\hbox {f}(\hbox {x})=\hbox {c}$ is a solution of the above equation if and only if $\hbox {F}_{1}(\hbox {c,h}(\hbox {t}))=\hbox {c}$ for $\hbox {t}\upepsilon \hbox {G}$.

4 Unknown function $\hbox {F}_{1}$

If the function f is injective, then the function $\hbox {F}_{1}(\hbox {x,t})= \hbox {f}^{-1}[\hbox {F}_{2}(\hbox {f}(\hbox {x}),\hbox {h}(\hbox {t}))]$ is the only solution of (1).

There exists solutions of (1) for a non-injective function f.

Examples

10.:

If $\hbox {f}(\hbox {x})=\hbox {c}$, $\hbox {F}_{2}$ is a solution of the translation equation such that c is a fixed point of $\hbox {F}_{2}$ ($\hbox {F}_{2}(\hbox {c,v})= \hbox {c}$ for $\hbox {v}\upepsilon \hbox {G}_{2})$ and h is an arbitrary homomorphism, then every function $\hbox {F}_{1}:\hbox {S}_{1}\times \hbox {G}_{1}\rightarrow \hbox {S}_{1}$ is a solution of (1).

A function $\hbox {F}_{1}:\hbox {S}_{1}\times \hbox {G}_{1}\rightarrow \hbox {S}_{1}$ is the solution of (1) only if

$\hbox {f}(\hbox {x})=\hbox {f}(\hbox {y}) \Rightarrow \hbox {f}(\hbox {F}_{1}(\hbox {x,t}))=\hbox {f}(\hbox {F}_{1}(\hbox {y,t}))$ for $\hbox {x,y}\upepsilon \hbox {S}_{1}$, $\hbox {t}\upepsilon \hbox {G}_{1}$.

Note that the solution $\hbox {F}_{1}(\hbox {x,t})=\hbox {c}$ for $(\hbox {x,t})\upepsilon \hbox {S}_{1}\times \hbox {G}_{1}$ of the above condition is a solution of (1) only if $\hbox {F}_{2}(\hbox {f}(\hbox {x}),\hbox {h}(\hbox {t}))=\hbox {f}(\hbox {c})$ for $(\hbox {x,t})\upepsilon \hbox {S}_{1}\times \hbox {G}_{1}$.

11.:

Let $\hbox {F}_{\mathrm{i}}:\hbox {S}_{\mathrm{i}}\times \hbox {G}_{\mathrm{i}}\rightarrow \hbox {S}_{\mathrm{i}}$ for $\hbox {i}=1,2$ be arbitrary functions not necessarily a solutions of the translation equation. Consider a special case of equation (11) namely if $\hbox {F}_{2}$ is such that $\hbox {F}_{2}(\hbox {u,v})=\hbox {u}$ on $\hbox {f}(\hbox {S}_{1})\times \hbox {h}(\hbox {G}_{1})$. Then Eq. (11) has the form

$$\begin{aligned} \hbox {f}(\hbox {F}_{1}(\hbox {x,t}))=\hbox {f}(\hbox {x}). \end{aligned}$$

(13)

In this case the function $\hbox {F}_{1}$, such that $\hbox {x}\upepsilon \hbox {F}_{1}(\hbox {x},\hbox {G}_{1})$ for $\hbox {x}\upepsilon \hbox {S}_{1}$, is a solution of (11) if and only if the set $\hbox {F}_{1}(\hbox {x},\hbox {G}_{1})$ is included in some level of the function f for every $\hbox {x}\upepsilon \hbox {S}_{1}$.

From here the function $\hbox {F}_{1}$ of the form as in (2) satisfisies Eq. (13) if and only if for every $\hbox {k}\upepsilon \hbox {K}_{1}$ the set $\hbox {S}_{\mathrm{1k}}$ is included in some level of f.

E.g., let $\hbox {S}_{1}=\mathbb {R}$ and $\hbox {f}(\hbox {x})=\vert \hbox {x}\vert$. The levels of f have the form $\{ \hbox {--a,a}\}$ for a$\upepsilon \mathbb {R}$. Assume that the group $\hbox {G}_{1}$ is non-orientable, i.e., there exists no subgroup of $\hbox {G}_{1}$ with index 2 (e.g., the additive group $\mathbb {R}$ is non-orientable). Thus it is impossible that $\hbox {S}_{\mathrm{1k}}=\{ \hbox {--a,a}\}$ for $\hbox {a}\ne 0$. From here every $\hbox {S}_{\mathrm{1k}}$ in the form $\hbox {F}_{1}$ in (2) is a singelton. This yields that the constant functions $\hbox {F}_{1}(\hbox {x,t})=\hbox {c} \upepsilon \mathbb {R}$ are the only solutions of (13) in this case.

Let $\hbox {S}_{1}$ and f be as above. Assume that $\hbox {G}_{\mathrm{1}}$ is orientable by the subgroup $G_{1}^{*}$ (like the additive group of integers by even integers). We have in this case, except for the constant solutions, the non-constant solutions of (13), too, e.g., the solution of the form

$$\begin{aligned} \hbox {F}_{1}(\hbox {x,t})=\left\{ \begin{array}{ll} \hbox {--}1&{}\quad \hbox { for } \hbox {x}=\hbox {--}1, \hbox {t}\upepsilon \,G_{1}^{*} \hbox { or or } \hbox {x}=1, \hbox {t}\upepsilon \hbox {G}_{1}\backslash \,G_{1}^{*},\\ 1 &{}\quad \hbox { for } \hbox {x}=\hbox {--}1, \hbox {t}\upepsilon \hbox {G}_{1}\backslash \,G_{1}^{*} \hbox { or } \hbox {x}=1, \hbox {t}\upepsilon \, G_{1}^{*},\\ \hbox {c}\upepsilon \mathbb {R}&{}\quad \hbox { for } \hbox {x}\upepsilon \mathbb {R}\backslash \{\hbox {--}1,1\} , \hbox {t}\upepsilon \hbox {G}_{1}. \end{array}\right. \end{aligned}$$

5 Unknown homomorphism h

Theorem 4

If a function $\hbox {F:S}\times \hbox {G} \rightarrow \hbox {S}$ is of the form (6) and function f is of the form (8), then the equation

$$\begin{aligned} \hbox {f}[\hbox {F}(\hbox {x,t})]=\hbox {F}(\hbox {f}(\hbox {x}),\hbox {h}(\hbox {t})) \end{aligned}$$

(14)

has $\hbox {h}(\hbox {t})=\hbox {t}$ as the only solution if $\upgamma (\hbox {N}_{1})\not \subset \hbox {N}_{2}$.

If $\upgamma (\hbox {N}_{1})\subset \hbox {N}_{2}$, then every function $\hbox {h:G}\rightarrow \hbox {G}$ is a solution of (14).

Proof

If $\upgamma (\hbox {N}_{1})\not \subset \hbox {N}_{2}$, then there exists $\hbox {m}\upepsilon \hbox {N}_{1}$ such that $\upgamma (\hbox {m})\upepsilon \hbox {N}_{1}$. Assume that the function h is a solution of (1). Then for $\hbox {x}\upepsilon \hbox {I}_{\mathrm{m}}$ we have $\hbox {f}(\hbox {x})=\text{ b }_{{\upgamma }\left( {\text{ m }} \right) }^{-1} [\hbox {b}_{\mathrm{m}}(\hbox {x})+\updelta (\hbox {m})]\upepsilon \hbox {S}_{{\upgamma (\mathrm{m})}}$, $\hbox {F}(\hbox {x,t})= \text{ b }_{\mathrm{m}}^{-1} [\hbox {b}_{\mathrm{m}}(\hbox {x})+\hbox {t}]\upepsilon \hbox {S}_{\mathrm{m}}$. From here (14) has the form

$$\begin{aligned} \text{ b }_{{\upgamma }\left( {\text{ m }} \right) }^{-1} [\hbox {b}_{\mathrm{m}}(\hbox {x})+\hbox {t} +\updelta (\hbox {m})]= \text{ b }_{{\upgamma }\left( {\text{ m }} \right) }^{-1} [\hbox {b}_{\mathrm{m}}(\hbox {x})+\updelta (\hbox {m})+ \hbox {h}(\hbox {t})], \end{aligned}$$

thus $\hbox {h}(\hbox {t})=\hbox {t}$. This homomorphism is a solution of (1) by Example 3.

If $\upgamma (\hbox {N}_{1})\subset \hbox {N}_{2}$, then $\upgamma (\hbox {n})\upepsilon \hbox {N}_{1}$ for every $\hbox {n}\upepsilon \hbox {N}$. From here for every function $\hbox {h:G}\rightarrow \hbox {G}$

• 1/ if $\hbox {x}\upepsilon \hbox {N}_{2}$, then $\hbox {f}(\hbox {x})=\upgamma (\hbox {x})$ and we obtain $\hbox {f}[\hbox {F}(\hbox {x,t})]=\hbox {f}(\hbox {x})=\upgamma (\hbox {x})=\hbox {F}(\upgamma (\hbox {x}),\hbox {h}(\hbox {t}))$,

• 2/ if $\hbox {x}\upepsilon \hbox {I}_{\mathrm{n}}$, then $\hbox {f}(\hbox {x})=\upgamma (\hbox {n})$ and since $\hbox {F}(\hbox {x,t})\upepsilon \hbox {I}_{\mathrm{n}}$ we obtain $\hbox {f}[\hbox {F}(\hbox {x,t})]=\upgamma (\hbox {x})=\hbox {F}(\upgamma (\hbox {x}),\hbox {h}(\hbox {t}))= \hbox {F}(\hbox {f}(\hbox {x}),\hbox {h}(\hbox {t}))$.

$\square$

Proposition

Let functions $\hbox {F}_{\mathrm{i}}$ be as in (2). If there exists $\hbox {x}_{0}\upepsilon \hbox {S}_{1}$ such that $\hbox {F}_{1}(\hbox {x}_{0}, \mathbf . ):\hbox {G}_{1}\rightarrow \hbox {S}_{1}$ is a constant function and $\hbox {F}_{2}(\hbox {f}(\hbox {x}_{0}), \mathbf . ):\hbox {G}_{2}\rightarrow \hbox {S}_{2}$ is an injection, then $\hbox {h}(\hbox {t})=0$ for $\hbox {t}\upepsilon \hbox {G}_{1}$ is the only solution of (1).

The function $\hbox {h}(\hbox {t})=0$ for $\hbox {t}\upepsilon \hbox {G}_{1}$ is a solution of (1) if and only if the function f is constant on every $\hbox {S}_{\mathrm{1k}}$ for $\hbox {k}\upepsilon \hbox {K}_{1}$. In this case other solutions h can exist.

Proof

We have $\hbox {f}[\hbox {F}_{1}(\hbox {x}_{0},\hbox {t})]=\hbox {f}[\hbox {F}_{1}(\hbox {x}_{0},0)]=\hbox {f}(\hbox {x}_{0})$. For $\hbox {f}(\hbox {x}_{0})\upepsilon \hbox {S}_{\mathrm{2k}}$ for some $\hbox {k}\upepsilon \hbox {K}_{2}$ we obtain $\hbox {F}_{2}(\hbox {f}(\hbox {x}_{0}),\hbox {h}(\hbox {t}))=\text{ b }_{2{\mathrm{k}}}^{-1} [\hbox {b}_{\mathrm{2k}}(\hbox {f}(\hbox {x}_{0})+\hbox {h}(\hbox {t})]$, where $\hbox {b}_{\mathrm{2k}}$ is a bijection from $\hbox {S}_{\mathrm{2k}}$ onto $\hbox {G}_{2}$. From here $\hbox {b}_{\mathrm{2k}}(\hbox {f}(\hbox {x}_{0}))=\hbox {b}_{\mathrm{2k}}(\hbox {f}(\hbox {x}_{0}))+\hbox {h}(\hbox {t})$. This yields that $\hbox {h}(\hbox {t})=0$ for $\hbox {t}\upepsilon \hbox {G}_{1}$.

If $\hbox {h}(\hbox {t})=0$, then $\hbox {f}[\hbox {F}_{1}(\hbox {x,t})]=\hbox {F}_{2}(\hbox {f}(\hbox {x}),0)=\hbox {f}(\hbox {x})$. Since $\hbox {F}_{1}(\hbox {x},\hbox {G}_{1})=\hbox {S}_{\mathrm{1k}}$ for $\hbox {x}\upepsilon \hbox {S}_{\mathrm{1k}}$, f is constant on $\hbox {S}_{\mathrm{1k}}$. The inverse implication is evident. If $\hbox {f}(\hbox {x})=\hbox {c} \upepsilon \hbox {S}_{2}$ for $\hbox {x}\upepsilon \hbox {S}_{1}$ and $\hbox {F}_{2}(\hbox {c,t})=\hbox {c}$ for $\hbox {t}\upepsilon \hbox {G}_{2}$, the every function $\hbox {h:G}_{1}\rightarrow \hbox {G}_{2}$ is a solution of (1). $\square$

Lemma 1

Let $\hbox {F:S}\times \hbox {G} \rightarrow \hbox {S}$, where G is a group, be a solution of the translation equation such that $\hbox {F}(\hbox {x},0)=\hbox {x}$ for $\hbox {x}\upepsilon \hbox {S}$. A function $\hbox {g:G}\rightarrow \hbox {G}$ (not necessarly a homomorphism) is a solution of the equation

$$\begin{aligned} \hbox {F}(\hbox {x,t})=\hbox {F}(\hbox {x},\hbox {g}(\hbox {t})) \end{aligned}$$

(15)

if and only if $\hbox {g}(\hbox {t})\hbox {--t}\upepsilon \cap _{\mathrm{x}\upepsilon \mathrm{S}}\{ \hbox {s}\upepsilon \hbox {G:F}(\hbox {x,s})=\hbox {x}\}$ for $\hbox {t}\upepsilon \hbox {G}$.

Proof

If $\hbox {g}(\hbox {t})\hbox {--t}\upepsilon \cap _{\mathrm{x}\upepsilon \mathrm{S}}\{ \hbox {s}\upepsilon \hbox {G:F}(\hbox {x,s})=\hbox {x}\}$, then

$$\begin{aligned} \hbox {F}(\hbox {x,g}(\hbox {t}))=\hbox {F}(\hbox {F}(\hbox {x},\hbox {g}(\hbox {t})\hbox {--t}),\hbox {t})= \hbox {F}(\hbox {x,t}). \end{aligned}$$

If $\hbox {F}(\hbox {x,g}(\hbox {t}))=\hbox {F}(\hbox {x,t})$, then $\hbox {g}(\hbox {t})\hbox {--t}\upepsilon \cap _{\mathrm{x}\upepsilon \mathrm{S}}\{ \hbox {s}\upepsilon \hbox {G:F}(\hbox {x,s})=\hbox {x}\}$ since

$$\begin{aligned} \hbox {F}(\hbox {x,g}(\hbox {t})\hbox {--t})= \hbox {F}(\hbox {F}(\hbox {x,g}(\hbox {t})),\hbox {--t})=\hbox {F}(\hbox {F}(\hbox {x,t}),\hbox {--t})=\hbox {F}(\hbox {x},0)=\hbox {x}. \end{aligned}$$

$\square$

Lemma 2

Let a function F be of the form $\hbox {F}(\hbox {x,t})= b_{k}^{-1} (\hbox {b}_{\mathrm{k}}(\hbox {x})+\hbox {t})$ for $\hbox {x}\upepsilon \hbox {S}_{\mathrm{k}}$, $\hbox {k}\upepsilon \hbox {K}$, $\hbox {t}\upepsilon \hbox {G}$, where $\hbox {S}=\cup _{\mathrm{k}\upepsilon \mathrm{K}}\hbox {S}_{\mathrm{k}}$, $\hbox {S}_{\mathrm{k}}$ are disjoint and $\hbox {b}_{\mathrm{k}}$ is a bijection from $\hbox {S}_{\mathrm{k}}$ to $\hbox {G/G}_{\mathrm{k}}$ (see (2)). Then

$$\begin{aligned} \cap _{\mathrm{x}\upepsilon \mathrm{S}}\{ \hbox {t}\upepsilon \hbox {G:F}(\hbox {x,t})=\hbox {x}\} =\cap _{\mathrm{u}\upepsilon \mathrm{G}}\cap _{\mathrm{k}\upepsilon \mathrm{K}}(\hbox {--u}+\hbox {G}_{\mathrm{k}}+\hbox {u}). \end{aligned}$$

Proof

Let $\hbox {s}\upepsilon \{ \hbox {t}\upepsilon \hbox {G:F}(\hbox {x,t})=\hbox {x}\}$ for every $\hbox {x}\upepsilon \hbox {S}$. We have $\hbox {G}_{\mathrm{k}}+\hbox {u}\upepsilon \hbox {G/G}_{\mathrm{k}}$ for every $\hbox {u}\upepsilon \hbox {G}$, $\hbox {k}\upepsilon \hbox {K}$. There exists $\hbox {y}\upepsilon \hbox {S}_{\mathrm{k}}$ such that $\hbox {b}_{\mathrm{k}}(\hbox {y})=\hbox {G}_{\mathrm{k}}+\hbox {u}$. Since $\hbox {s}\upepsilon \{ \hbox {t}\upepsilon \hbox {G:F}(\hbox {y,t})=\hbox {y}\}$, $\hbox {y}=\hbox {F}(\hbox {y,s})=b_{k}^{-1} (\hbox {b}_{\mathrm{k}}(\hbox {y})+\hbox {s}$). From here $\hbox {G}_{\mathrm{k}}+\hbox {u}=\hbox {G}_{\mathrm{k}}+\hbox {u}+\hbox {s}$. This yields that $\hbox {--u}+\hbox {G}_{\mathrm{k}}+\hbox {u}=[\hbox {--u}+\hbox {G}_{\mathrm{k}}+\hbox {u}]+\hbox {s}$, thus $\hbox {s}\upepsilon [\hbox {--u}+\hbox {G}_{\mathrm{k}}+\hbox {u}]$ for every $\hbox {u}\upepsilon \hbox {G}$, $\hbox {k}\upepsilon \hbox {K}$.

Let $\hbox {s}\upepsilon (\hbox {--u}+\hbox {G}_{\mathrm{k}}+\hbox {u})$ for every $\hbox {u}\upepsilon \hbox {G}$, $\hbox {k}\upepsilon \hbox {K}$ and $\hbox {x}\upepsilon \hbox {S}_{\mathrm{k}}$. Then $\hbox {b}_{\mathrm{k}}(\hbox {x})=\hbox {G}_{\mathrm{k}}+\hbox {a}(\hbox {x})$ for some $\hbox {a}(\hbox {x})\upepsilon \hbox {G}$. From here $\hbox {b}_{\mathrm{k}}(\hbox {x})+\hbox {s}=\hbox {a}(\hbox {x})+[\hbox {--a}(\hbox {x})+\hbox {G}_{\mathrm{k}}+\hbox {a}(\hbox {x})]+\hbox {s}=\hbox {b}_{\mathrm{k}}(\hbox {x})$, thus $\hbox {F}(\hbox {x,s})= b_{k}^{-1} (\hbox {b}(\hbox {x})+\hbox {s})=\hbox {x}$. $\square$

Note that if $\cap _{\mathrm{u}\upepsilon \mathrm{G}}\cap _{\mathrm{k}\upepsilon \mathrm{K}}(\hbox {--u}+\hbox {G}_{\mathrm{k}}+\hbox {u})=\{ 0\}$, then $\hbox {g}(\hbox {t})=\hbox {t}$ for $\hbox {t}\upepsilon \hbox {G}$ is the only solution of (15).

If the group G is abelian, then the function g in Lemma 1 is a homomorphism if and only if the function g(t)–t is a homomorphism. We have thus by Lemmas 1 and 2 the following theorem.

Theorem 5

Let $\hbox {F}$ be as in Lemma 2. Assume that the group $\hbox {G}$ is abelian. The homomorphism $\hbox {h:G}\rightarrow \hbox {G}$ is a solution of Eq. (15) if and only if $\hbox {h(t)}=\hbox {h}_{1}(\hbox {t})+\hbox {t}$ for $\hbox {t}\upepsilon \hbox {G}$, where $\hbox {h}_{1}$ is an arbitrary homomorphism from $\hbox {G}$ to $\cap _{\mathrm{k}\upepsilon \mathrm{K}}\hbox {G}_{\mathrm{k}}$.

Remarks

1/ If the group G is not abelian, then Theorem 5 is not true since the difference of two homomorphism may not be a homomorphism. E.g., for the smallest non-abelian group given in the table below and for $\hbox {h}(\hbox {t})=\hbox {--a}+\hbox {t}+\hbox {a}$ the function $\hbox {h}_{1}(\hbox {t})=\hbox {h}(\hbox {t})\hbox {--t}$ is not a homomorphism. Indeed, $\hbox {h}_{1}(\hbox {a}+\hbox {b})=\hbox {c}\ne \hbox {d}=\hbox {h}_{1}(\hbox {a})+\hbox {h}_{1}(\hbox {b})$.

The assumption that G is abelian in the above Theorem 5 can not be replaced by the assumption that $\hbox {G}_{\mathrm{k}}$ are abelian in F for $\hbox {k}\upepsilon \hbox {K}$. We give an example below.

Let $\hbox {S}=\{ 1,2\}$ and $\hbox {K}=\{ 1\}$. Let G be a group as in the table above. Put $\hbox {b}_{1}(1)=\{ 0,\hbox {c,d}\}$, $\hbox {b}_{1}(2)= \{ \hbox {a,b,e}\}$ and $\hbox {F}(\hbox {x,t})= b_{1}^{-1} (\hbox {b}_{1}(\hbox {x})+\hbox {t})$ for $\hbox {x}\upepsilon \hbox {S}$, $\hbox {t}\upepsilon \hbox {G}$. Then $\hbox {G}_{1}=\{ 0,\hbox {c,d}\}$ is an abelian group, the homomrphism $\hbox {h}(\hbox {t})= \hbox {--a}+\hbox {x}+\hbox {a}$ is a solution of (15) and h(t)–t is not a homomorphism.

The tables of these functions F(x,t), h(t) and F(x,h(t)) are shown below.

2/ If a homomorphism $\hbox {h:G}\rightarrow \hbox {G}$ is a solution of (15), then

$$\begin{aligned} \hbox {h}(\hbox {t})=\hbox {h}(\hbox {s}) \Rightarrow \hbox {F}(\hbox {x,t})=\hbox {F}(\hbox {x,s}) \hbox { for } \hbox {x}\upepsilon \hbox {S}, \hbox {t},\hbox {s}\upepsilon \hbox {G}. \end{aligned}$$

(16)

Examples

12.:

An equation of the form (15) can have a solution which is not the identity function. In fact, let $\hbox {S}=\hbox {G}=\mathbb {R}$ and let $\mathfrak {b}_{1},\mathfrak {b}_{2}$ be different elements of the Hamel base H of $\mathbb {R}$ over $\mathbb {Q}$. Let $\hbox {a,h}:\mathbb {R}\rightarrow \mathbb {R}$ be additive functions such that $\hbox {a}(\mathfrak {b}_{1})=\hbox {a}(\mathfrak {b}_{2})= \mathfrak {b}_{1}$, $\hbox {h}(\mathfrak {b}_{1})= \mathfrak {b}_{2}$, $\hbox {h}(\mathfrak {b}_{2})= \mathfrak {b}_{1}$, $\hbox {a}(\hbox {t})=\hbox {h}(\hbox {t})=0$ for $\hbox {t}\upepsilon \hbox {H}\backslash \{ \mathfrak {b}_{1}, \mathfrak {b}_{2}\}$. We have for $\hbox {t}=\hbox {q}_{1}\mathfrak {b}_{1}+\hbox {q}_{2}\mathfrak {b}_{2}+\hbox {q}_{3}\mathfrak {b}_{3}+\ \cdots + \hbox {q}_{\mathrm{n}}\mathfrak {b}_{\mathrm{n}}$, where $\hbox {q}_{\mathrm{i}}\upepsilon \mathbb {Q}$, $\mathfrak {b}_{\mathrm{i}}\upepsilon \hbox {H}$ for $\hbox {i}=1,\ \ldots ,\hbox {n}$, that

$$\begin{aligned} \hbox {a}(\hbox {h}(\hbox {t}))&=\hbox {a}(\hbox {q}_{1}\hbox {h}(\mathfrak {b}_{1}){+}\hbox {q}_{2}\hbox {h}(\mathfrak {b}_{2})) =\hbox {a}(\hbox {q}_{1}\mathfrak {b}_{2}{+}\hbox {q}_{2}\mathfrak {b}_{1}) =\hbox {q}_{1}\hbox {a}(\mathfrak {b}_{2}){+}\hbox {q}_{2}\hbox {a}(\mathfrak {b}_{1}) =(\hbox {q}_{1}{+}\hbox {q}_{2})\mathfrak {b}_{1}\\&=\hbox {q}_{1}\hbox {a}(\mathfrak {b}_{1})+\hbox {q}_{2}\hbox {a}(\mathfrak {b}_{2})=\hbox {a}(\hbox {q}_{1}\mathfrak {b}_{1}+\hbox {q}_{2}\mathfrak {b}_{2}+\hbox {q}_{3}\mathfrak {b}_{3}+\ \cdots + \hbox {q}_{\mathrm{n}}\mathfrak {b}_{\mathrm{n}})=\hbox {a}(\hbox {t}). \end{aligned}$$

From here for $\hbox {F}(\hbox {x,t})=\hbox {x}+\hbox {a}(\hbox {t})$ we obtain $\hbox {F}(\hbox {x,h}(\hbox {t}))=\hbox {x}+\hbox {a}(\hbox {h}(\hbox {t}))=\hbox {x}+\hbox {a}(\hbox {t})=\hbox {F}(\hbox {x,t})$. The function $\hbox {h}_{1}$ of Theorem 6 is here of the form $\hbox {h}_{1}(\hbox {t})=(\hbox {q}_{2}\hbox {--}\hbox {q}_{1})\mathfrak {b}_{1}+(\hbox {q}_{1}\hbox {--}\hbox {q}_{2})\mathfrak {b}_{2}\hbox {--}\hbox {q}_{3}\mathfrak {b}_{3}\hbox {--}\ \cdots \hbox {--}\hbox {q}_{\mathrm{n}}\mathfrak {b}_{\mathrm{n}}$ for t as above.

13.:

If $\hbox {F}_{\mathrm{i}}(\hbox {x,t})=\hbox {x}$ for $(\hbox {x,t})\upepsilon (\hbox {S}_{\mathrm{i}},\hbox {G}_{\mathrm{i}})$ and $\hbox {i}=1,2$ and $\hbox {f}:\hbox {S}_{1}\rightarrow \hbox {S}_{2}$ is an arbitrary function, then the arbitrary function $\hbox {h:G}_{1}\rightarrow \hbox {G}_{2}$ is a solution of (1).

14.:

If functions $\hbox {F}_{1},\hbox {F}_{\mathrm{2}}$,f are as in Example 1, then Eq. (1) has the form

$$\begin{aligned} \upvarepsilon \hbox {exp}(\hbox {a}+\hbox {tnx}+\hbox {t})=\upvarepsilon \hbox {exp}(\hbox {a}+\hbox {tnx}+\hbox {h}(\hbox {t})). \end{aligned}$$

Thus if $\upvarepsilon \ne 0$, then $\hbox {h}(\hbox {t})=\hbox {t}$ is the only solution of (1) and if $\upvarepsilon =0$, then every function $\hbox {h}:\mathbb {R}\rightarrow \mathbb {R}$ is a solution of this equation.

15.:

The homomorphism $\hbox {h}(\hbox {t})=\hbox {t}$ is not a solution of (1) in Example 2.

16.:

If $\hbox {S}_{1}=\hbox {S}_{2}=\hbox {G}_{1}=\hbox {G}_{2}=\mathbb {Z}$, $\hbox {F}_{1}(\hbox {x,t})=\hbox {F}_{2}(\hbox {x,t})=\hbox {x}+\hbox {t}$ and $\hbox {f}(\hbox {x})=2\hbox {x}$, then Eq. (1) has the form $2(\hbox {x}+\hbox {t})=2\hbox {x}+\hbox {h}(\hbox {t})$. From here $\hbox {h}(\hbox {t})=2\hbox {t}$ is the only homomorphism which is a solution of (1).

17.:

Let $\hbox {G}_{1}$ be a semigroupe, let $\hbox {G}_{2}$ be a semigroupe with the left cancellation property and let $\hbox {S}_{1}=\hbox {G}_{1}$, $\hbox {S}_{2}=\hbox {G}_{2}$, $\hbox {F}_{1}(\hbox {x,t})=\hbox {x}+\hbox {t} :\hbox {G}_{1}\times \hbox {G}_{1}\rightarrow \hbox {G}_{1}$, $\hbox {F}_{2}(\hbox {u,v})=\hbox {u}+\hbox {v} :\hbox {G}_{2}\times \hbox {G}_{2}\rightarrow \hbox {G}_{2}$. If a function $\hbox {f}:\hbox {G}_{1}\rightarrow \hbox {G}_{2}$ is a homomorphism, then $\hbox {h}=\hbox {f}$ is the only solution of Eq. (1). In fact, Eq. (1) has the form $\hbox {f}(\hbox {x}+\hbox {t})=\hbox {f}(\hbox {x})+\hbox {h}(\hbox {t})$.

6 Conditional equation (11)

The conditional equation (11) is said to be the equation associated with the geometric concomitant equation (1). If functions $\hbox {f},\hbox {F}_{1},\hbox {F}_{2},\hbox {h}$ satisfy (1), then $\hbox {f},\hbox {F}_{1},\hbox {h}$ are evidently solutions of (11) and if $\hbox {f},\hbox {F}_{1},\hbox {h}$ satisfy (11), then there exists $\hbox {F}_{2}:\hbox {S}_{2}\times \hbox {h}(\hbox {G}_{1})$ such that $\hbox {f},\hbox {F}_{1},\hbox {h},\hbox {F}_{2}$ are solutions of (1) by Theorem 3. A solution of the translation equation $\hbox {F}_{2}$ with this property and defined on $\hbox {S}_{2}\times \hbox {G}_{2}$ may not exist (see Example 8). We consider the conditional equation (11) (without the index ,,1” in $\hbox {F}_{1})$ in the cases if

• (i) the homomorphism h is injective and $\hbox {G}_{1}=\hbox {G}_{2}$

  or

• (ii) the function f is injective and $\hbox {S}_{1}=\hbox {S}_{2}$.

In case (i) the functions f,h,F are solutions of (11) if and only if

$$\begin{aligned} \hbox {f}(\hbox {x})=\hbox {f}(\hbox {y}) \Rightarrow \hbox {f}[\hbox {F}(\hbox {x,t})]=\hbox {f}[\hbox {F}(\hbox {y,t})] \hbox { for } \hbox {x,y,t}\upepsilon \hbox {G}. \end{aligned}$$

(17)

In case (ii) the functions f,h F are solutions of (11) if and only if

$$\begin{aligned} \hbox {h}(\hbox {t})=\hbox {h}(\hbox {s}) \Rightarrow \hbox {F}(\hbox {x,t})=\hbox {F}(\hbox {x,s}) \hbox { for } \hbox {x}\upepsilon \hbox {S}, \hbox {t},\hbox {s}\upepsilon \hbox {G}. \end{aligned}$$

(18)

Theorem 6

Let G be a group. A function $\hbox {f:S}\rightarrow \hbox {S}$ is a solution of the conditional equation (17), where $\hbox {F}(\hbox {x,t})=\hbox {b}^{-1}(\hbox {b}(\hbox {x})+\hbox {t})$ and $\hbox {b:S}\rightarrow \hbox {G}$ is a bijection, if and only if $\hbox {f}(\hbox {x})=\hbox {b}^{-1}\Phi ([\hbox {b}(\hbox {x})])$ for $\hbox {x}\upepsilon \hbox {S}$, where $[\hbox {x}]\upepsilon \hbox {G/G}^{*}$ for some subgroup $\hbox {G}^{*}$ of $\hbox {G}$ and $\Phi$ is an injection from $\hbox {G/G}^{*}$ to $\hbox {G}$.

Proof

If $\hbox {b}^{-1}\Phi ([\hbox {b}(\hbox {x})])=\hbox {b}^{-1}\Phi ([\hbox {b}(\hbox {y})])$, then $[\hbox {b}(\hbox {x})]=[\hbox {b}(\hbox {y})]$. From here $(\hbox {b}(\hbox {x})+\hbox {t})\hbox {--}(\hbox {b}(\hbox {y})+\hbox {t})= \hbox {b}(\hbox {x})\hbox {--b}(\hbox {y})\upepsilon \hbox {G}^{*}$ for every $\hbox {t}\upepsilon \hbox {G}$. This yields that $[\hbox {b}(\hbox {x})+\hbox {t}]=[\hbox {b}(\hbox {y})+\hbox {t}]$. This implies that

$$\begin{aligned} \hbox {b}^{-1}\Phi ([\hbox {b}(\hbox {F}(\hbox {x,t}))])=\hbox {b}^{-1} \Phi ([\hbox {b}(\hbox {x})+\hbox {t}])=\hbox {b}^{-1}\Phi ([\hbox {b}(\hbox {y})+\hbox {t}])=\hbox {b}^{-1}\Phi ([\hbox {b}(\hbox {F}(\hbox {y,t}))]). \end{aligned}$$

Analogously, if f is a solution of (15), then $\upvarphi (\hbox {u})=\hbox {bfb}^{-1}(\hbox {u})$ for $\hbox {u}\upepsilon \hbox {G}$ is a solution of the implication

$$\begin{aligned} \upvarphi (\hbox {u})=\upvarphi (\hbox {v}) \Rightarrow \upvarphi (\hbox {u}+\hbox {t})=\upvarphi (\hbox {v}+\hbox {t}). \end{aligned}$$

We define a relation R on $\hbox {G: uRv}\Leftrightarrow [\upvarphi (\hbox {u})=\upvarphi (\hbox {v})]$. It is an equivalence relation and uRv implies $(\hbox {u}+\hbox {t})\hbox {R}(\hbox {v}+\hbox {t})$. From here the family of equivalence classes of R is the same as the family $\hbox {G/G}^{*}$ for some subgroup $\hbox {G}^{*}$ [3]. Let S be a selector of this family. Put $\Phi ([\hbox {u}])=\upvarphi (\hbox {u}_{0})$, where $\hbox {u}_{0}\upepsilon \hbox {S}\cap [\hbox {u}]$. Since $\hbox {u}_{0}\upepsilon [\hbox {u}]$, $[\hbox {u}]=[\hbox {u}_{0}]$. For $\hbox {u}\upepsilon [\hbox {u}]=[\hbox {u}_{0}]$ we have $\upvarphi (\hbox {u})=\upvarphi (\hbox {u}_{0})$, thus $\upvarphi (\hbox {u})= \Phi ([\hbox {u}])$. From here $\hbox {f}(\hbox {x})=\hbox {b}^{-1}\Phi ([\hbox {b}(\hbox {x})])$. $\square$

Example

18.:

Consider the conditional equation (16) where F is a function as in (2). The function $\hbox {f}(\hbox {x})= \Phi (\hbox {k})$ for $\hbox {x}\upepsilon \hbox {S}_{\mathrm{1k}}$, where $\Phi$ is a function from $\hbox {K}_{1}$ to $\hbox {S}_{1}$, is a solution of this condition. In fact, if $\hbox {f}(\hbox {x})=\hbox {f}(\hbox {y})$, $\hbox {f}(\hbox {x})=\Phi (\hbox {k})$ for $\hbox {x}\upepsilon \hbox {S}_{\mathrm{1k}}$ and $\hbox {f}(\hbox {y})=\Phi (\hbox {l})$ for $\hbox {y}\upepsilon \hbox {S}_{\mathrm{1l}}$, then $\Phi (\hbox {k})=\Phi (\hbox {l})$. Since $\hbox {F}_{1}(\hbox {x,t})\upepsilon \hbox {S}_{\mathrm{1k}}$ for $\hbox {t}\upepsilon \hbox {G}_{1}$, $\hbox {f}[\hbox {F}_{1}(\hbox {x,t})]= \Phi (\hbox {k})$. Analogously $\hbox {f}[\hbox {F}_{1}(\hbox {y,t})]=\Phi (\hbox {l})$. From here $\hbox {f}[\hbox {F}_{1}(\hbox {x,t})]=\hbox {f}[\hbox {F}_{1}(\hbox {y,t})]$.

Example

19.:

Let $\hbox {F}(\hbox {x,t})=\hbox {x}+\hbox {t}$ be a function from $\mathbb {R}\times \mathbb {R}$ to $\mathbb {R}$, let G be the additive group of real numbers. The function $\Phi ([\hbox {x}])=\hbox {M}(\hbox {x})$ for $[\hbox {x}]\upepsilon \mathbb {R}/\mathbb {Z}$, where M(x) is the mantissa of $\hbox {x}\upepsilon \mathbb {R}$, is an injection from $\mathbb {R}/\mathbb {Z}$ to $\mathbb {R}$. From here M(x) is a solution of (15) and it is not a solution of Eq.  (7).

Proposition

Let $\hbox {G}$ be a group and let $\hbox {F:S}\times \hbox {G} \rightarrow \hbox {S}$ be a solution of the translation equation such that $\hbox {F}(\hbox {x},0)=\hbox {x}$ for $\hbox {x}\upepsilon \hbox {S}$. A homomorphism $\hbox {h:G}\rightarrow \hbox {G}$ is a solution of the conditional equation (18) if and only if

$$\begin{aligned} \hbox {Kern}(\hbox {h})\subset \cap _{\mathrm{x}\upepsilon \mathrm{S}}\{ \hbox {s}\upepsilon \hbox {G:F}(\hbox {x,s})=\hbox {x}\}. \end{aligned}$$

(19)

Proof

If h is a solution of (18) and $\hbox {t}\upepsilon \hbox {Kern}(\hbox {h})$, then $\hbox {h}(\hbox {t})=0=\hbox {h}(0)$, thus $\hbox {F}(\hbox {x,t})=\hbox {F}(\hbox {x},0)=\hbox {x}$ for every $\hbox {x}\upepsilon \hbox {S}$. Fom here $\hbox {t}\upepsilon \cap _{\mathrm{x}\upepsilon \mathrm{S}}\{ \hbox {s}\upepsilon \hbox {G:F}(\hbox {x,s})=\hbox {x}\}$.

Assume that (19) is true. If $\hbox {h}(\hbox {t})=\hbox {h}(\hbox {s})$, then $\hbox {h}(\hbox {t--s})=0$, thus $\hbox {t--s}\upepsilon \hbox {Kern}(\hbox {h})$. This yields that $\hbox {F}(\hbox {x,t--s})=\hbox {x}$ for $\hbox {x}\upepsilon \hbox {S}$. From here $\hbox {F}(\hbox {x,t})=\hbox {F}(\hbox {F}(\hbox {x,t--s}),\hbox {s})=\hbox {F}(\hbox {x,s})$. $\square$

Example

20.:

Let S,G,H be as in Example 10. Let $\hbox {F}(\hbox {x}+\hbox {t})=\hbox {x}+\hbox {t}$. For the homomorphism

$$\begin{aligned} \hbox {h}(\hbox {q}_{1}\mathfrak {b}_{1}+\hbox {q}_{2}\mathfrak {b}_{2}+\hbox {q}_{3}\mathfrak {b}_{3}+\ \cdots +\hbox {q}_{\mathrm{n}}\mathfrak {b}_{\mathrm{n}})= \hbox {q}_{2}\mathfrak {b}_{1}+\hbox {q}_{1}\mathfrak {b}_{2}+\hbox {q}_{3}\mathfrak {b}_{3}+\ \cdots +\hbox {q}_{\mathrm{n}}\mathfrak {b}_{\mathrm{n}} \end{aligned}$$

we have $\hbox {Kern}(\hbox {h})=\{ 0\}$, h satisfies (19). This implies that h is a solution of (18). This function h is evidently not a solution of (17).

Example

21.:

Injection from $\mathbb {R}$ to $\mathbb {R}$ and constant function are the only solutions of (17) for the geometric object $\hbox {x}=\hbox {v}_{2}/\hbox {v}_{1}$, where $\hbox {v}_{1},\hbox {v}_{\mathrm{2}}$ are the coordinates of the 2-dimensional contravariant vector. Indeed, this object has the transformation law of the form $\frac{A_{1}^{1} +A_{2}^{1} x}{A_{1}^{2} +A_{2}^{2} x}$ [1]. Thus (17) is of the form

$$\begin{aligned} \hbox {f}(\hbox {x})=\hbox {f}(\hbox {y}) \Rightarrow \hbox {f}\left( \frac{A_{1}^{1} +A_{2}^{1} x}{A_{1}^{2} +A_{2}^{2} x}\right) =\hbox {f}\left( \frac{A_{1}^{1} +A_{2}^{1} y}{A_{1}^{2} +A_{2}^{2} y}\right) . \end{aligned}$$

(20)

For $A_{2}^{1} =A_{1}^{2} =1$ and $A_{2}^{2} =0$ we have

$$\begin{aligned} \hbox {f}(\hbox {x})=\hbox {f}(\hbox {y}) \Rightarrow \hbox {f}(A_{1}^{1}+x)=\hbox {f}(A_{1}^{1}+y) \end{aligned}$$

for every $\hbox {x}, \hbox {y}, A_{1}^{1} \upepsilon \mathbb {R}$. From here the family F of the levels of f is invariant under arbitrary translations. This yields that this family $F=\mathbb {R}/\mathbb {R}^{*}$ for a subgroupe $\mathbb {R}^{*}$ of $\mathbb {R}$. Assume that $\{ 0\} \ne \mathbb {R}^{*}\ne \mathbb {R}$. This implies that there exist $\hbox {x}_{1},\hbox {x}_{2}$ such that $\hbox {x}_{1}\ne 0$ and $\hbox {f}(0)= \hbox {f}(\hbox {x}_{1})\ne \hbox {f}(\hbox {x}_{2})$. We have by (20) that $\hbox {f}(\hbox {x})=\hbox {f}(\hbox {y}) \Rightarrow \hbox {f}(\hbox {ax})=\hbox {f}(\hbox {ay})$ for every x,y,a$\upepsilon \mathbb {R}$. From here

$$\begin{aligned} \hbox {f}(0)=\hbox {f}\left( \frac{x_{2} }{x_{1} } \mathbf . 0\right) =\hbox {f}\left( \frac{x_{2} }{x_{1} } \hbox {x}_{1}\right) =\hbox {f}(\hbox {x}_{2}), \end{aligned}$$

thus a contradiction. If $\mathbb {R}^{*}=\{ 0\}$, then f is an injection. If $\mathbb {R}^{*}=\mathbb {R}$, then f is a constant function.