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In context of geometry and points in a plane Wikipedia describes symmetry as a type of invariance - the property that something does not change under a set of transformations.

Isn't isometry the exact same thing? A type of invariance that preserves relative distances between points.

This definition from wikipedia adds to my confusion:

If the object X is a set of points in the plane with its metric structure or any other metric space, a symmetry is a bijection of the set to itself which preserves the distance between each pair of points (an isometry).

So from this I can conclude that every symmetry is an isometry, but not every isometry is a symmetry. And which type of invariance, in addition to ones it already has, does an isometry need to have to be considered a symmetry? Coud I apply a rigid motion to any figure in a plane and call that symmetry as well? I have not seen this explicitly stated anywhere.

Travis Willse
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Michael Munta
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    That definition of symmetry is not as universal as the definition of isometry. I wouldn't use the word symmetry without clarification as to the meaning. You can use symmetry in that manner but you should be just clarify what you mean. – Clara M Johnson Sep 02 '19 at 15:13

2 Answers2

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An isometry is a set bijection $\Phi : (X, d) \to (X', d')$ between metric spaces that identifies $d, d'$, that is, that satisfies $$d(x, y) = d'(\Phi(x), \Phi(y)) \qquad \textrm{for all $x, y \in X$} .$$ A symmetry (as defined in the excerpt), then, is just an isometry from a metric space $(X, d)$ to itself.

Note every metric space admits at least one symmetry, namely the identity map. Checking the axioms directly that the set of symmetries of a fixed metric space $(X, d)$ form a group under composition, which is called the isometry group and is sometimes denoted $\operatorname{Iso}(X, d)$.

As Clara points out in the comments, the term symmetry does not usually have this restricted meaning. More generally, given a set $X$ equipped with some structure $\mathcal S$, we can consider the bijections $X \to X$ that preserve $\mathcal S$ in an appropriate sense. Generically these maps are called automorphisms rather than symmetries, and again for any $(X, \mathcal S)$ the set of such maps form a group under composition called the automorphism group of $(X, \mathcal S)$, and it is often denoted $\operatorname{Aut}(X, \mathcal S)$. In some contexts, often when $(X, \mathcal S)$ has some geometric interpretation, this group is sometimes called the symmetry group of $(X, \mathcal S)$.

Travis Willse
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  • Is it required that $d$ and $d'$ are equivalent functions or just that they evaluate to the same value? – Michael Munta Sep 04 '19 at 14:49
  • I'm not sure what difference you have in mind---but by definition a symmetry is an isometry $(X, d) \to (X, d)$. – Travis Willse Sep 04 '19 at 14:54
  • Talking about isometry now and the above equation you have specified. $d(x, y) = d'(\Phi(x), \Phi(y)) \qquad \textrm{for all $x, y \in X$}$ – Michael Munta Sep 04 '19 at 15:07
  • I'm not sure I understand the question still, but maybe this will answer it: The notion of isometry is a notion of equivalence (in the sense of equivalence relations). In other words, $d$ and $d'$ are equivalent in the sense that there is an isometry $\Phi : (X, d) \to (X', d')$. – Travis Willse Sep 04 '19 at 15:40
  • ...or do you have in mind some other notion of equivalence (e.g., that the topological spaces $(X, \tau_d), (X', \tau_{d'})$ underlying $(X, d), (X', d')$ are homeomorphic)? – Travis Willse Sep 04 '19 at 15:41
  • Well $d$ and $d'$ are functions on their metric spaces respectively. Are they equivalent in the sense for example if $d(x) = x^2$ and $d'(x)= x^2$ or they have different expressions that evaluate to the same value? – Michael Munta Sep 04 '19 at 16:54
  • If the first function were one that squares its input, does the second function also square its input or it can arrive to the equivalent output of the first function via something other than squareing. – Michael Munta Sep 04 '19 at 17:20
  • A metric $d$ is not a function on the underlying topological space $X$ but rather a function $X \times X \to \Bbb R_{\geq 0}$, and on a general space $X$ the expression $x^2$ has no meaning. But, speaking informally, isometric metric spaces can have metrics $d, d'$ that appear rather different in standard notation. – Travis Willse Sep 04 '19 at 18:29
  • One can produce many examples of isometries by picking a metric space $(X', d')$ and a homeomorphism $\Phi : X \to X'$. Then, if we define $d(x, y) := d'(\Phi(x), \Phi(y))$, by construction $\Phi$ is an isometry $(X, d) \to (X', d')$. (In fact all isometries arise this way.) For example, if we take: $X' = \Bbb R$, $d'$ to be the standard metric $$d'(x', y') = |x' - y'| ,$$ $X = (-1, 1) ,$ and $\Phi(x) = \operatorname{artanh} x$, then $$d(x, y) = |\operatorname{artanh} x - \operatorname{artanh} y| .$$ – Travis Willse Sep 04 '19 at 18:37
  • Hmm, I have just started learning abstract algebra and it is a bit overwhelming so just speaking informally is the following true? $d(x,y)=d'(x,y)$ – Michael Munta Sep 04 '19 at 20:16
  • I wouldn't say that even informally. After all, $d$ and $d'$ are respectively defined on sets $X$ and $X'$ that may be different, so if $x, y \in X$ then $d'(x, y)$ doesn't make sense (unless $x, y$ happen to be in $X'$, too). – Travis Willse Sep 04 '19 at 21:19
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The answer Travis provided is right. I just want to emphasize that in your particular case not all isometries of the plane are symmetries of $X$. Suppose $I$ is an isometry of the plane. We say that $I$ is a symmetry of the set of points $X$ if $I(X) = X$.

Just as Travis says, generically, a symmetry of a set $A$ with structure $\mathcal{S}$ is a bijection $A \to A$ such that $\mathcal{S}$ is preserved. In this case, the set $A$ is the plane, and the structure $\mathcal{S}$ is both the metric structure on the plane and the set of points $X$. Thus, $I : A \to A$ is a symmetry of $(A, \mathcal{S})$ if $d(x,y) = d(I(x), I(y))$ (where $d : A \times A \to \mathbb{R}$ is the distance function) and $I(X) = X$, i.e. $I|_X : X \to X$ is a bijection.

Charles Hudgins
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  • This is a well-put addition to my answer, and it answers OP's last question as well as possible. Thanks for writing this up! (And +1.) – Travis Willse Sep 03 '19 at 03:56