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I was going through Hall's book about Lie groups.While presenting Euler groups $E(n)$ and on the way to prove that they form a matrix Lie group hee made a proposition that Every one one onto distance preserving map from $\mathbb{R}^{n}$ to $\mathbb{R}^{n}$ which fixes origin will be linear.He has not proved it.For $n=1$ it is clear.But how to prove it generally ?

Gerry Myerson
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Madhu
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  • see http://en.wikipedia.org/wiki/Orthogonal_matrix – Emilio Novati Mar 01 '15 at 07:24
  • This is a special case of the Mazur-Ulam Theorem: http://en.wikipedia.org/wiki/Mazur%E2%80%93Ulam_theorem , which is discussed some here: http://math.stackexchange.com/questions/146232/norm-preserving-map-is-linear – Travis Willse Mar 01 '15 at 07:27

2 Answers2

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Let $\varphi: {\mathbb R}^n\to{\mathbb R}^n$ be an isometry with $\varphi(0)=0$. Then, for any $v\in {\mathbb R}^n$ and any $\lambda\in [0,1]$, we have $$(\dagger)\qquad \|\varphi(\lambda v)\| = \|\varphi(\lambda v) - \varphi(0)\|=\|\lambda v - 0\| = \lambda \| v\|=\lambda\|\varphi(v)\|\\(\ddagger)\qquad\|\varphi(v)-\varphi(\lambda v)\| = \|v - \lambda v\| = (1-\lambda)\|v\|=(1-\lambda)\|\varphi(v)\|$$ In particular $\|\varphi(v) - \varphi(\lambda v)\| + \|\varphi(\lambda v)\| = \|\varphi(v)\|$, implying that $0$, $\varphi(\lambda v)$ and $\varphi(v)$ are colinear, say $\varphi(\lambda v)=\mu\varphi(v)$. Then $(\dagger)$ and $(\ddagger)$ give $|\mu|=\lambda$ and $|1-\mu|=(1-\lambda)$, so $\mu=\lambda$ and hence $\varphi(\lambda v)=\lambda \varphi(v)$.

More generally, you can check that the argument generalizes to show that $$\varphi(\lambda v + (1-\lambda)w)=\lambda \varphi(v) + (1-\lambda)\varphi(w)$$ for $v,w\in{\mathbb R}^n$ and $\lambda\in [0,1]$; applying this with $\lambda=\frac{1}{2}$ and using $\varphi(\lambda\cdot -)=\lambda\cdot\varphi(-)$ gives the linearity.

Hanno
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  • @Madhu: Do you have any questions? – Hanno Mar 03 '15 at 07:23
  • it is clear to me now...I have started reading Hall's in last month only .Moving very slow reached Poincare groups yesterday..soon will come with more doubts.... – Madhu Mar 05 '15 at 00:22
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Let $T: \mathbb R^n \to \mathbb R^n$ be a bijective isometry, and let $\ell \subset \mathbb R^n$ be a line in $\mathbb R^n$. Take $v, w\in \ell$, and let $$ c_t = (1-t)v + tw. $$ Now, let $S$ and $S'$ be the spheres of radius $\Vert v - c_{1/2}\Vert$ around $v$ and $w$ respectively. Then $S \cap S' = \{c_{1/2}\}$, and since $T$ maps spheres bijectively to spheres of the same radius, $T(S) \cap T(S') = \{T(c_{1/2})\}$. Thus, $T(v)$, $T(w)$, and $T(c_{1/2})$ are all colinear. By repeating this argument, we see that $T(v)$ and $T(w)$ are colinear to every $T(c_t)$ where $t\in[0,1]$ is a dyadic rational.

By continuity, $T(\ell)$ is a line, so $T$ is affine. The linear maps are precisely the affine maps that fix the origin, so $T$ is linear.

William Stagner
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