Mappings from R^n to F^n which preserve unit Euclidean distance, where F is a field of characteristic 0
Abstract
Description
Let F be a commutative field of characteristic 0, G_n: F^n \times F^n -> F, G_n((x_1,...,x_n),(y_1,...,y_n))=(x_1-y_1)^2+...+(x_n-y_n)^2. We say that g:R^n->F^n preserves distance d>=0 if for each x,y \in R^n |x-y|=d implies G_n(g(x),g(y))=d^2. Let f:R^n->F^n preserve unit distance. We prove: (1) if n>=2, x,y \in R^n and x \neq y, then G_n(f(x),f(y)) \neq 0, (2) if A,B,C,D \in R^2, r \in Q and \overrightarrow{CD}=r\overrightarrow{AB}, then \overrightarrow{f(C)f(D)}=r\overrightarrow{f(A)f(B)}, (3) if A,B,C,D \in R^2 and \overrightarrow{AB} and \overrightarrow{CD} are linearly dependent, then \overrightarrow{f(A)f(B)} and \overrightarrow{f(C)f(D)} are linearly dependent, (4) if A,B,C,D \in R^2 and \overrightarrow{AB} is perpendicular to \overrightarrow{CD}, then \overrightarrow{f(A)f(B)} is perpendicular to \overrightarrow{f(C)f(D)}, (5) if A,B,C,D \in R^2 and |AB|=|CD|, then G_2(f(A),f(B))=G_2(f(C),f(D)). Let D_n(F) denote the set of all positive numbers d with the property: if x,y \in R^n and |x-y|=d then there exists a finite set S(x,y) with {x,y} \subseteq S(x,y) \subseteq R^n such that any map g:S(x,y)->F^n that preserves unit distance preserves also the distance between x and y. Obviously, {1} \subseteq D_n(F) \subseteq A_n(F). We prove: (6) A_n(C) \subseteq {d>0: d^2 \in Q}, (7) {d>0: d^2 \in Q} \subseteq D_2(F).
enlarged version, 20 pages
enlarged version, 20 pages