Mappings from Rn to Fn which preserve unit Euclidean distance, where F is a field of characteristic 0
Abstract
Let F be a commutative field of characteristic 0, Gn: Fn × Fn -> F, Gn((x1,...,xn),(y1,...,yn))=(x1-y1)2+...+(xn-yn)2. We say that g:Rn->Fn preserves distance d>=0 if for each x,y ∈ Rn |x-y|=d implies Gn(g(x),g(y))=d2. Let f:Rn->Fn preserve unit distance. We prove: (1) if n>=2, x,y ∈ Rn and x ≠ y, then Gn(f(x),f(y)) ≠ 0, (2) if A,B,C,D ∈ R2, r ∈ Q and CD=rAB, then f(C)f(D)=rf(A)f(B), (3) if A,B,C,D ∈ R2 and AB and CD are linearly dependent, then f(A)f(B) and f(C)f(D) are linearly dependent, (4) if A,B,C,D ∈ R2 and AB is perpendicular to CD, then f(A)f(B) is perpendicular to f(C)f(D), (5) if A,B,C,D ∈ R2 and |AB|=|CD|, then G2(f(A),f(B))=G2(f(C),f(D)). Let Dn(F) denote the set of all positive numbers d with the property: if x,y ∈ Rn and |x-y|=d then there exists a finite set S(x,y) with x,y ⊂eq S(x,y) ⊂eq Rn such that any map g:S(x,y)->Fn that preserves unit distance preserves also the distance between x and y. Obviously, 1 ⊂eq Dn(F) ⊂eq An(F). We prove: (6) An(C) ⊂eq d>0: d2 ∈ Q, (7) d>0: d2 ∈ Q ⊂eq D2(F).
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