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In Riemannian geometry, the term "isometry" can refer to two different concepts. Firstly, for Riemannian manifolds $(M_1, g_1)$ and $(M_2, g_2)$, an isometry from $\boldsymbol{M_1}$ to $\boldsymbol{M_2}$ is a smooth map $F \colon M_1 \to M_2$ that satisfies $F^*g_2 = g_1$. Secondly, for a fixed Riemannian manifold $(M, g)$, an isometry of $\boldsymbol{M}$ is a smooth map $F \colon M \to M$ such that $F^*g = g$. The set of all such isometries forms a group under composition, known as the isometry group of $M$.

If $M = \mathbb{R}^2$ with its Euclidean metric, it is well-known that translations, rotations, reflections, and glide reflections are isometries. My question is: Are these transformations the only isometries of $\mathbb{R}^2$ with the Euclidean metric?

I am looking for a proof based in the context of Riemannian geometry. I mean, we don't have an inner product defined on $\mathbb R^2$ as a vector space, but a inner product for every tangent space $T_p\mathbb R^2$.

A. J. Pan-Collantes
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  • If $F:\Bbb R^n\to\Bbb R^n$ preserves the metric (on the tangent space at each point), won't it preserve the distance? https://en.wikipedia.org/wiki/Riemannian_manifold#Metric_space_structure – Anne Bauval Jul 27 '24 at 10:54
  • Yes, of course. So in the proof it is necessary to define the notion of distance, right? – A. J. Pan-Collantes Jul 27 '24 at 10:59
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    The list here contains in addition srew motions (concerning your boldface question). – Dietrich Burde Jul 27 '24 at 11:02
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    Yes, screw motions. But all this is for $\Bbb R^3$. – Anne Bauval Jul 27 '24 at 11:43
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    Did you study orthogonal transformations of $\mathbb R^n$ in your linear algebra class? Did they cover the normal (block-diagonal) form of orthogonal transformations? – Moishe Kohan Jul 27 '24 at 13:14
  • (i) Those classes represent all isometries of the Euclidean plane. (ii) They do not describe all isometries of Euclidean $3$-space. In addition to screw motions, there are compositions of rotations and reflections in a plane orthogonal to the axis, which have no widely-used name as far as I know. (iii) Every isometry of Euclidean $n$-space is an orthogonal affine transformation (non-trivial, but idiomatic proof); Moishe's comment should be viewed as decisive. – Andrew D. Hwang Jul 27 '24 at 14:30
  • Considering $\mathbb R^2$ as an inner product space, it is clear that the group of maps preserving the inner product is just the classical rigid motions, that is not my question. In the context I have described you don't have a "global" inner product defined on $\mathbb R^2$; instead, there is an inner product defined on every tangent space. – A. J. Pan-Collantes Jul 27 '24 at 14:44

1 Answers1

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According to the comments above, and with my own thinking, I conclude that to prove that the only isometries of the Euclidean plane (in the sense of Riemannian geometry) are translations, rotations, reflections and glide reflections, we have to follow these steps:

  1. The geodesics of this Riemannian manifold can be shown to be straight lines.
  2. Given two points, there is only one straight line joining them.
  3. The geodesic distance can be explicitly proven to be $$d(p,q)=\sqrt{(q_1-p_1)^2+(q_2-p_2)^2}$$ which is the the Euclidean norm formula.
  4. Isometries preserve geodesics, so they preserve this distance. We are now in the context of an Euclidean affine space.
  5. In the context of an affine space, it is well-known that the distance-preserving maps are only translations, rotations, and reflections. Is basic linear algebra.

Please, let me know if I am wrong, or if there is an easier way to prove this.

A. J. Pan-Collantes
  • 2,545