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Beside the algebraic definition of isometry:

Question 1. Is it correct that the isometry group of a Riemannian manifold depends on its embedding? e.g. I think isometry group of a circle in plan and in 3-space are different?

Question 2. Why we can talk about trivial isometry group while always a translation is a nontrivial member of isometry group?

After these, I think it would be clear for me that why an isometry of a Riemannian manifold, also called a symmetry. (The strange point for me is: members of isometry group like translation)

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    $\begingroup$ If $M$ is the sphere with its standard metric, what do you mean by a "translation" of $M$? $\endgroup$
    – Jack Lee
    Sep 22, 2019 at 18:17
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    $\begingroup$ Does that map the unit sphere to itself? $\endgroup$
    – Jack Lee
    Sep 22, 2019 at 18:35
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    $\begingroup$ As I mentioned in a comment to this answer, an isometry of a Riemannian manifold $(M,g)$, is a smooth map $F:M\to M$ sattisfying $F^∗g=g$. A translation of $\mathbb R^{n+1}$ does not map the sphere to itself, so it is not an isometry of the sphere. $\endgroup$
    – Jack Lee
    Sep 22, 2019 at 18:43
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    $\begingroup$ I think you're confusing "an isometry of $M$" with "an isometry from one manifold to another." $\endgroup$
    – Jack Lee
    Sep 22, 2019 at 18:43
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    $\begingroup$ A translation is an isometry of $\mathbb R^{n+1}$ (with its standard metric), but not of $\mathbb S^n$. $\endgroup$
    – Jack Lee
    Sep 22, 2019 at 18:47

1 Answer 1

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OK, let me see if I can answer your two questions.

But first, let's clear up a common misunderstanding. In Riemannian geometry, the word "isometry" is used with two different meanings. First, if $(M_1,g_1)$ and $(M_2,g_2)$ are Riemannian manifolds, an isometry from $\boldsymbol {M_1}$ to $\boldsymbol {M_2}$ is a smooth map $F\colon M_1\to M_2$ that satisfies $F^*g_2 = g_1$. For a given pair of Riemannian manifolds, there may or may not be any isometries between them.

Second, if $(M,g)$ is a fixed Riemannian manifold, an isometry of $\boldsymbol M$ is an isometry from $(M,g)$ to itself, that is, a smooth map $F\colon M\to M$ such that $F^*g = g$. In this case, the set of isometries of $M$ is a group under composition. It always contains at least the identity map, and it might or might not contain others. If $M=\mathbb R^{n+1}$ with its Euclidean metric, the isometry group contains all translations, rotations, reflections, and glide reflections. If $M=\mathbb S^n$ with its standard round metric, the group contains only (restrictions of) rotations and reflections.

To address your questions:

Question 1: The isometry group of a given Riemannian manifold $(M,g)$ depends only on the manifold $M$ and the metric $g$, not on any particular isometric embedding into a Euclidean space. This is immediate from the definition I gave above. (Of course, if you choose an embedding that is not an isometric embedding, then you will induce a different metric on $M$, and it will very likely have a different group of isometries.)

Question 2: The term "translation" only makes sense if we're talking about a vector space (or more generally an affine space). All translations of $\mathbb R^{n+1}$ are isometries of its standard metric, but there are also metrics on $\mathbb R^{n+1}$ that are not translation-invariant. You can't talk about "translations of $\mathbb S^n$" because there are no translations that take $\mathbb S^n$ to itself.

Of course, any translation of $\mathbb R^{n+1}$ takes $\mathbb S^n$ to another unit sphere, let's call it $S'$, and the restriction of that translation becomes an isometry from $\mathbb S^n$ to $S'$. But that's not an isometry of $\mathbb S^n$.

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  • $\begingroup$ +1.5. Why you are expert in finding my faults? $\endgroup$
    – C.F.G
    Sep 22, 2019 at 19:13
  • $\begingroup$ @C.F.G: I don't consider myself an expert at that. But I've seen the same or similar mistakes hundreds of times, so they start to be somewhat recognizable. This will all seem very easy and natural eventually. $\endgroup$
    – Jack Lee
    Sep 22, 2019 at 19:17
  • $\begingroup$ What about isometries of a manifold $N\supseteq M$ which also preserve the submanifold $M$? $\endgroup$
    – mr_e_man
    Sep 22, 2019 at 19:19
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    $\begingroup$ @mr_e_man: What about them? If $M$ is isometrically embedded in $N$, they restrict to isometries of $M$. But there might be other isometries of $M$ that are not of this form. $\endgroup$
    – Jack Lee
    Sep 22, 2019 at 19:20
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    $\begingroup$ @ArvinRasoulzadeh: I'm not sure what you mean by "wrapping it around one of the main axes." But the isometry group of $U$ consists of exactly two elements: the identity and the reflection across the plane containing the removed semicircle. One way to see this is to note that for each $p\in U$, the injectivity radius at $p$ is an intrinsically defined number, and the set $J$ of points whose injectivity radius is $\pi/2$ is an isometry-invariant subset. Every isometry of $U$ is the restriction of an isometry of $S^2$, and the only such isometries that preserve $J$ are the two I mentioned. $\endgroup$
    – Jack Lee
    Oct 8, 2020 at 18:53

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