0

How to prove that $$g:\Bbb R^3 \to \Bbb R^3 \in G = \{g \, | \, \text{ for each } g \text{ exists } n\in\mathbb{Z} : r(g(x),g(y)=2^n r(x,y) \}$$ for each $x,y\in \Bbb R^3, r$ is an euclidean distance) is a bijection? Have no idea.

I have the task to prove that $G$ is a group or isn't. Here is the task : $r$ is a euclidean metric of space $L = \Bbb R^3$. Does $G$ - multiplicity of transformations of $L$ (for each $g\in G$ exist $n\in\mathbb Z : r(g(x),g(y))=2^n r(x,y)$ for each $x,y\in L$) form a group?

My teacher said that first of all it's necessary to prove if it is a bijection or not, and if it is closed or not.

Mark Fantini
  • 5,523
Phill
  • 21

1 Answers1

0

Let $g \in G$. First of all, $g$ is injective, since if $g(x) = g(y)$, then $$r(x,y) = 2^{-n}r(g(x),g(y)) = 0,$$ so $x = y$.

For surjectivity, define $f : \mathbb{R}^3 \to \mathbb{R}^3$ by $$f(x) = \frac{g(x)-g(0)}{2^n}.$$ To show that $g$ is surjective, it suffices to show that $f$ is. Now $f$ is constructed to satisfy $f(0) = 0$, and by homogeneity and translation invariance, $$\lVert f(x) -f(y) \rVert = r(f(x),f(y)) = 2^{-n}r(g(x),g(y)) = r(x,y) = \lVert x -y\rVert,$$ and so the claim follows by this answer to a similar question.

fuglede
  • 6,716
  • can't understand here r(f(x),f(0)) = 2^{-n}r(g(x),g(0)) =... how do you make this...and I don't understand why F is surjective – Phill May 27 '14 at 12:40
  • The first one follows from homogeneity and translation invariance, $$r(f(x),f(0)) = r(\frac{g(x)-g(0)}{2^n},\frac{g(0)-g(0)}{2^n}) = 2^{-n}r(g(x)-g(0),g(0)-g(0)) = 2^{-n}r(g(x),g(0)).$$ That $f$ is surjective is the content of the linked answer – fuglede May 27 '14 at 12:53
  • can't understand proof of 'similar question'. don't have much knowledge in this area. Could you explain in detail why F - surjective? – Phill May 27 '14 at 13:02
  • The links in this post give a couple of different proofs that may be useful. – fuglede May 27 '14 at 13:07
  • thanks, I'll try to sort out – Phill May 27 '14 at 13:23
  • maybe you know how to prove if G closed or not? – Phill May 27 '14 at 13:24
  • Closed under composition, you mean? – fuglede May 27 '14 at 13:31
  • yes, that's what i mean – Phill May 27 '14 at 13:41
  • If $g_1, g_2 \in G$ with corresponding integers $n_1$ and $n_2$, then $g_1 \circ g_2$ satisfies the condition defining $G$ with $n = n_1 + n_2$. This you can see by simply writing out the equation. – fuglede May 27 '14 at 13:42
  • Your trick allows you to look at maps from each ${x\in\Bbb R^3:|x|\le M}$ to itself, which gives you compactness, but you haven’t actually shown that $f$ is an isometry, so you can’t yet apply the linked result. – Brian M. Scott Oct 10 '14 at 18:19