Difference between revisions of "Isometry"

 
 
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An '''isometry''' is a map which preserves distances between points.
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An '''isometry''' is a map which preserves [[distance]]s between [[point]]s. Isometries exist in any space in which a distance function is defined, i.e. an arbitrary abstract [[metric space]].  In the particular case where we take our space to be the usual Euclidean plane or Euclidean 3-space (<math>\mathbb{R}^2</math> or <math>\mathbb{R}^3</math> with the standard [[Euclidean metric]]), the isometries are known as ''rigid motions'' and two sets which can be transformed onto each other by an isometry are said to be [[congruent (geometry) | congruent]].
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==Isometries are injective==
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Since for any metric we have <math>d(x, y) > 0</math> whenever <math>x \neq y</math>, it follows that every isometry must be an [[injection]]. 
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===Proof===
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Suppose otherwise.  Then there is some non-injective isometry <math>\phi: S \to T</math>.  Since <math>\phi</math> is not injective, we know <math>\exists x, y \in S</math> such that <math>x \neq y</math> and <math>\phi(x) = \phi(y)</math>.  But <math>x \neq y \Longrightarrow d_S(x, y) > 0</math> while <math>\phi(x) = \phi(y) \Longrightarrow d_T(\phi(x), \phi(y)) = 0</math>, and this contradicts the fact that <math>\phi</math> is an isometry.
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Note that this does ''not'' mean that isometries are necessarily [[bijection]]s.  Consider, for example, the [[discrete metric]] on the integers, <math>(\mathbb{Z}, d)</math> such that <math>d(x, x) = 0</math> and <math>x\neq y \Longrightarrow d(x, y) = 1</math>.  It is simple to verify that this is a metric space.  The map <math>\phi: \mathbb{Z} \to \mathbb{Z}</math> given by <math>\phi(n) = 2n</math> is an isometry but it is not [[surjection | surjective]].
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Latest revision as of 17:17, 23 September 2006

An isometry is a map which preserves distances between points. Isometries exist in any space in which a distance function is defined, i.e. an arbitrary abstract metric space. In the particular case where we take our space to be the usual Euclidean plane or Euclidean 3-space ($\mathbb{R}^2$ or $\mathbb{R}^3$ with the standard Euclidean metric), the isometries are known as rigid motions and two sets which can be transformed onto each other by an isometry are said to be congruent.

Isometries are injective

Since for any metric we have $d(x, y) > 0$ whenever $x \neq y$, it follows that every isometry must be an injection.

Proof

Suppose otherwise. Then there is some non-injective isometry $\phi: S \to T$. Since $\phi$ is not injective, we know $\exists x, y \in S$ such that $x \neq y$ and $\phi(x) = \phi(y)$. But $x \neq y \Longrightarrow d_S(x, y) > 0$ while $\phi(x) = \phi(y) \Longrightarrow d_T(\phi(x), \phi(y)) = 0$, and this contradicts the fact that $\phi$ is an isometry.


Note that this does not mean that isometries are necessarily bijections. Consider, for example, the discrete metric on the integers, $(\mathbb{Z}, d)$ such that $d(x, x) = 0$ and $x\neq y \Longrightarrow d(x, y) = 1$. It is simple to verify that this is a metric space. The map $\phi: \mathbb{Z} \to \mathbb{Z}$ given by $\phi(n) = 2n$ is an isometry but it is not surjective.


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