# Two questions about isometry group of Riemannian manifolds

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)

• If $M$ is the sphere with its standard metric, what do you mean by a "translation" of $M$? Sep 22, 2019 at 18:17
• Does that map the unit sphere to itself? Sep 22, 2019 at 18:35
• 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. Sep 22, 2019 at 18:43
• I think you're confusing "an isometry of $M$" with "an isometry from one manifold to another." Sep 22, 2019 at 18:43
• A translation is an isometry of $\mathbb R^{n+1}$ (with its standard metric), but not of $\mathbb S^n$. Sep 22, 2019 at 18:47

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.

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 $$\mathbb S^n$$.
• What about isometries of a manifold $N\supseteq M$ which also preserve the submanifold $M$? Sep 22, 2019 at 19:19
• @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. Sep 22, 2019 at 19:20
• @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. Oct 8, 2020 at 18:53