About one composition of partial mapping of Euclidean space $E_{5}$

Abstract

In the domain  $ \Omega\subset E_{5} $ we consider a set of smooth lines such that through each point   $X\in\Omega$ there passes exactly one line  $\omega^{1}$ from the given set. The moving frame of the domain   $\Omega$ is a Frenet frame \cite{Rashevsky} associated with the line  $\omega^{1}$. The integral lines of the coordinate vector fields form a Frenet net \cite{Rashevsky}. We define the point  $F_{1}^{5}$ on the tangent of the line  $\omega^{1}$ in an invariant manner. As the point $X$ moves within the domain  $\Omega$, the point  $F_{1}^{5}$ traces out a new domain $\Omega_{1}^{5}\subset E_{5}$. This defines the partial mapping $f_{1}^{5}:\Omega \rightarrow \Omega_{1}^{5}$ such that $f_{1}^{5}(X)=F_{1}^{5}$.

Similarly, we define another partial mapping $f_{5}^{4}:\Omega \rightarrow \Omega_{5}^{4}$ Next, we consider the composition of these two partial mappings, specifically the inverse mapping $(f_{1}^{5})^{-1}$ and $f_{5}^{4}$ given by:

 $f_{5}^{4}\circ (f_{1}^{5})^{-1} :\Omega_{1}^{5} \rightarrow \Omega_{5}^{4}$ such that $f_{5}^{4}\circ (f_{1}^{5})^{-1}(F_{1}^{5})=F_{5}^{4}$, where $(f_{1}^{5})^{-1}$ - is the inverse mapping $f_{1}^{5}$.

 Let the line $\gamma$, which belongs to the distribution $\Delta_{4}=(X,\overrightarrow{e}_{2},\overrightarrow{e}_{3},\overrightarrow{e}_{4},\overrightarrow{e}_{5})$ be a quasi-double line of the pair of distributions $(\Delta_{4},\Delta'_{4})$ in the partial mapping $f_{1}^{5}$ (where $\Delta'_{4}=f_{1}^{5}(\Delta_{4})$).

We establish necessary and sufficient conditions for the line $f_{5}^{4}\circ (f_{1}^{5})^{-1}(\gamma)$ to be a quasi-double line of the pair $(\Delta_{4},\Delta'_{4})$ of distributions $\Delta_{4},\Delta'_{4}$
in the partial mapping  $f_{5}^{4}\circ (f_{1}^{5})^{-1}$.