throughout the following question, whenever I'm wrong please correct me!

Recently I came across the notions of symmetry and isometry. Though, there is something obscure (definitely in my head) concerning the distinction of those two things. For instance, let's say that $\Sigma \subset \mathbb{R}^{n},$ is an arbitrary geometric object (I'm looking for the general point view of the notion of symmetries and isometries, hence am going to assume that at the moment no further structure has been assumed on this object). Then the group of symmetries of $\Sigma$, is $$Symm(\Sigma)= \{ \sigma \in Isom(\mathbb{R}^n) \thinspace | \thinspace \sigma(\Sigma) = \Sigma \},$$ whilst isometries of $\mathbb{R}^n$ are defined as usually, being distance-preserving maps $\sigma : \mathbb{R}^n \rightarrow \mathbb{R}^n,$ which turn out to be continuous, one-to-one and onto (hence homeomorphisms of the underlying topological structure induced by the metric). Now, let's say that a geometric figure $\Sigma \subset \mathbb{R}^{n},$ is given on its own again and someone asks, "What's the group of isomotries and symmetries of $\Sigma$?". Then there are two possibilities:

  • $\Sigma$, inherits the metric by $\mathbb{R}^n,$ as a subspace.
  • $\Sigma$, becomes a metric space with some other metric and we examine it by forgetting any ambient space.

Now, for the first one, I think the symmetries and isometries coincide, right (if no, a counterexample suffices)? It's just another name for the same map $\sigma: \Sigma \rightarrow \Sigma$ which preserves distances. But what happens for the second case?

For instance, for the last question I have in my mind the distinctive case $\Sigma= \mathbb{S}^{n-1},$ the $(n-1)$-dimensional sphere which naturally inherits a metric by $\mathbb{R}^n$, but it can be equipped with another metric too, hence the isometries change in those two cases since different type of measurment is being applied each time. What happens if moreover we assume some differentiable structure and someone asks for the isomotries/symmetries of Riemmanian manifolds instead of subsets of $\mathbb{R}^n$? What about the symmetries and isometries in that case?

Thank you, I hope haven't done something wrong, because is my first post! If yes, do let me know.


The word "symmetry" in mathematics is quite overloaded and really is context dependent so let me try and work with your definition of symmetry. To fix some notation, let $(X,d_X)$ be a metric space. The group of isometries of $(X,d_X)$ is defined as

$$ \operatorname{Iso}(X,d_X) := \{ \phi \colon X \rightarrow X \, | \, \phi \text{ is one-to-one and onto and } d_X(\phi(x),\phi(y)) = d_X(x,y) \,\,\forall x,y \in X \}. $$

Given a set $\Sigma \subseteq X$, you suggest defining

$$ \operatorname{Sym}(\Sigma) = \operatorname{Sym}(\Sigma,(X,d_X)) = \{ \phi \in \operatorname{Iso}(X,d_X) \, | \, \phi(\Sigma) = \Sigma \}. $$

In order to discuss the isometries of $\Sigma$, we need to put on it some metric $d_{\Sigma}$. In general, if the metric $d_{\Sigma}$ we put on $\Sigma$ has nothing to do with the metric $d_X$ on $X$, there won't be any relation between $\operatorname{Iso}(\Sigma,d_{\Sigma})$ and $\operatorname{Symm}(\Sigma,(X,d_x))$. However, like you noted, we can restrict $d$ to $\Sigma$ and define $d_{\Sigma}(x,y) = d_X(x,y)$ and then ask what is the relation between $\operatorname{Iso}(\Sigma,d_{\Sigma})$ and $\operatorname{Sym}(\Sigma, (X,d_X))$.

We have a natural map $T \colon \operatorname{Sym}(\Sigma) \rightarrow \operatorname{Iso}(\Sigma)$ defined by restriction. Namely, $T(\phi) = \phi|_{A}$. However, this map is in general not one-to-one nor onto. Here are some examples to think about:

  1. Let $X = \mathbb{R}^n$ with the standard metric and $\Sigma = \{ 0 \}$. Then $\operatorname{Sym}(\Sigma) = \operatorname{O}_n(\mathbb{R})$ while $\operatorname{Iso}(\Sigma) = \{ \operatorname{id} \}$. The problem here is that while any symmetry of $\Sigma$ gives us an isometry of $(\Sigma, d_{\Sigma})$, different symmetries (in this case, all the symmetries) might give us the same isometry.
  2. Let $(X,d)$ be any metric space with $|X| \geq 2$ and $\operatorname{Iso}(X,d) = \{ \operatorname{id} \}$ (a space with no isometries). Choose two points $p \neq q$ in $X$ and set $\Sigma = \{p, q \}$. Then $\operatorname{Sym}(\Sigma) = \{ \operatorname{id} \}$ while $\operatorname{Iso}(\Sigma) = \mathbb{Z}_2$. The problem here is that an isometry of $(\Sigma,d_{\Sigma})$ might not extend to an isometry of $(X,d)$.

Finally, let $(X,g)$ be a Riemannian manifold and $i \colon \Sigma \rightarrow X$ be a submanifold of $X$. Using $g$, one can define a distance function $d_g$ which turns $X$ into a metric space and then the group of isometries of $(X,g)$ (as a Riemannian manifold) is the same as the group $(X,d_g)$ (as a metric space). Now we can actually do two things:

  1. We can restrict $d_g$ to $\Sigma$ to get a metric $d_1$ and ask your question above.
  2. We can endow $\Sigma$ with the Riemannian metric $i^{*}(g)$, consider the metric $d_2 = d_{i^{*}(g)}$ and ask your question above.

To see the difference, let's take your example of $X = \mathbb{R}^3$ with the standard Riemannian metric and $\Sigma = S^{2}$ to be the unit sphere. If we perform the first procedure, we get a metric $d_1$ on $S^{2}$ such that $d_1(x,y) = |x - y|$. For example, $d_1(N,S) = 2$ where $N,S \in S^{2}$ are the north pole and the south pole. If we perform the second procedure, we get a metric $d_2$ for which $d_2(N,S) = \pi$ because $d_2(N,S)$ is the length of the distance minimizing geodesic of $S^2$ which is a great circle.

In general, $(\Sigma,d_1)$ and $(\Sigma,d_2)$ are not isometric (as you can see from this example). Like before, you have maps $\operatorname{Sym}(\Sigma) \rightarrow \operatorname{Iso}(\Sigma,d_i)$ but they don't have to be one-to-one or onto.

  • 2
    $\begingroup$ Thank you for your answer. I'll read carefully what you've written and will come up with some questions in a bit I guess. Cheers! $\endgroup$ – user430191 Mar 28 '17 at 18:42
  • $\begingroup$ @user430191: No problem. Welcome to the site and kudos for a well-written and interesting question! $\endgroup$ – levap Mar 28 '17 at 18:45
  • $\begingroup$ Very good answer it's definitely something to mull over and your examples are very beautiful indeed. The Riemannian manifold example is great too, though the only point I don't really get is for $\Sigma = \{ 0 \}$ the part where $Isom(\Sigma)=\{ id \}$. It is true that any orthogonal matrix is an isomerty of $\mathbb{R}^n$ I think (simple rotations and reflections). So you mean that the restriction of all those matrices on the "zero" metric space just coincide with the identity of that metric space, right? $\endgroup$ – user430191 Mar 28 '17 at 19:18
  • $\begingroup$ And also, is generally wrong that the words isometry and symmetry can be used interchangeably (perhaps for some "good" spaces only) right? $\endgroup$ – user430191 Mar 28 '17 at 19:18
  • $\begingroup$ @user430191: The set $\Sigma = \{ \vec{0} \}$ consists only of one point - the zero vector. The isometry group of a metric space with one point is then always the trivial group. Regarding your second question, I would say yes. People sometimes call replace isometries with symmetries but in a context where you don't have an object sitting inside a bigger object. $\endgroup$ – levap Mar 28 '17 at 19:26

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