# Do local isometries form a group? If so, what is the group for Euclidean space?

Burago, Burago, and Ivanov use the following definition on p.78, for metric spaces $(X,d)$, $(Y,\delta)$:

A map $f: X \to Y$ is called a local isometry at $x \in X$ if $x$ has a neighborhood $U_x$ such that (the restriction of) $f$ maps $U_x$ isometrically onto an open set $U_y$ in $Y$.

Based on this, I make the following definition:

A map $f: X \to X$ is called a local self-isometry at $x \in X$, or perhaps a local isometry anchored at $x \in X$, if $x$ has a neighborhood $U_x$ such that (the restriction of) $f$ maps $U_x$ isometrically onto a (possibly different) neighborhood $V_x$ of $x$, and if the map $f$ fixes $x$, i.e. $f(x) = x$.

EDIT: We could also say "pointed local isometries (autometries)", mirroring the terminology used in Theorem 10.10.1 on p.398 of Burago, Burago, Ivanov ("pointed homeomorphism"). /EDIT

If we don't require that $f$ fixes $x$, then $f$ could map onto a neighborhood of $x$ in a way which ("morally") is not "centered/focused" around $x$ (e.g. in Euclidean space a translation of $U_x$ to another neighborhood in which $x$ is closer or further from the boundary than before).

(The condition doesn't seem necessary to make the set a group, however, see below. Thus its purpose should be understood solely as a heuristic to avoid and exclude local isometries in which the presence of $x$ in the domain and range/image isn't an "afterthought", like for translations.)

Question: (a) Given a metric space $(X,d)$ and a point $x \in X$, does the set of all local self-isometries at $x$ form a group (which one might call the local isometry group at $x$)?

(b) If the answer to (a) is affirmative, what is this group for any point in Euclidean space? (It has to be the same at each point because of homogeneity.) Is it just $O(n)$?

Attempt: (a) Any isometry $f: X \to X$ is bijective, so has an inverse $f^{-1}: X \to X$, which is also clearly an isometry, since $f$ is. $d(f^{-1}(x_1),f^{-1}(x_2)) = d(f^{-1}(f(y_1)), f^{-1}(f(y_2))) = d(y_1, y_2)$, where $y_1$ and $y_2$ are the points such that $f(y_1) = x_1, f(y_2) = x_2$, and since $f$ is an isometry, $d(x_1, x_2) = d(f(x_1),f(x_2)) = d(y_1, y_2) = d(f^{-1}(x_1), f^{-1}(x_2))$, thus $f^{-1}$ is an isometry.

Now assume that $f$ is a local self-isometry at $x$. Then $f$ restricts to an isometry of $U_x$ onto $V_x$, for $U_x$ and $V_x$ neighborhoods of $x$. But that means that $f^{-1}$ restricts to an isometry of $V_x$ onto $U_x$, so $f^{-1}$ is a local self-isometry at $x$, so the set of all local self-isometries at $x$ is closed under inverses. (Since $f(x)=x$ implies that $f^{-1}(x) = f^{-1}(f(x)) = x$.)

The identity is obviously a local self-isometry at $x$ for any point $x \in X$ (because it maps any neighborhood $U_x$ of $x$ isometrically onto itself and fixes every point), and associativity follows from the associativity of function composition. So the set of local self-isometries at $x$ is a group.

(b) According to this document, exercise 40 on p. 9, every local isometry of Euclidean space extends to a global isometry. Therefore, it should be the case that any local self-isometry of Euclidean space extends to a global isometry which fixes $x$.

Perhaps another way to prove this is true is to note that every neighborhood of $x$ is homeomorphic to Euclidean space itself, thus in particular locally homeomorphic. Moreover, since Euclidean space is a length space, the metric coincides with the intrinsic metric, so every isometry is also an arcwise isometry. Then one might be able to use the result that every local homeomorphism which is an arcwise isometry is a local isometry to show the local isometry group is isomorphic to the group of (global) isometries for Euclidean space which fix $x$.

This question might also be relevant to finding an answer.

Anyway, since for any point $x$ in Euclidean space, the group of (global) isometries fixing $x$ is isomorphic to the orthogonal group $O(n)$, it would follow, if either of the above two arguments are correct, that the local isometry group at $x$ is isomorphic to $O(n)$.

Motivation: The notion of a "pointed local autometry group" is an attempt to fix this deficiencies of using "pointed (global) autometry groups", as pointed out here, for trying to define a notion of direction in arbitrary metric spaces, as well as of angle measure.

• Do you consider two local autometries $f,g: (X,x) \to (X,x)$ to be equal if there is a neighbourhood $U \ni x$ such that $f|_U = g|_U$? If so, I'm pretty sure your first argument in (b) is correct. Note that if you treat local autometries just as ordinary functions, they don't even need to be invertible: consider a function which is the identity on the open unit disc, but takes arbitrary chaotic discontinuous values elsewhere. That is a local autometry at 0. – Mees de Vries Jul 20 '17 at 10:45
• @MeesdeVries You are right, that does make a lot of sense. Both imposing such an equivalence relation, as well as considering local autometries as ordinary functions. Sort of like the concept of germs of functions maybe? (Also, do you think it is necessary to require explicitly that the autometries fix $x$ for such a germ-like definition? And without requiring that they fix $x$, does it follow that the local autometry groups are isomorphic to the Euclidean group instead of the orthogonal group?) – Chill2Macht Jul 20 '17 at 11:08
• It's like germs of functions, sure, but when you move to wider classes of functions the translation local to global doesn't work anymore. You've taken such a rigid class of functions that the local always determines the global; so moving to germs doesn't really do anything. Anchored local isometries become orthogonal transformations; local translations become translations; together these generate the Euclidean group. – Mees de Vries Jul 20 '17 at 11:23
• @MeesdeVries Well it also depends on the space too right? E.g. for a sphere I would expect that there is a difference between the local and global, since the local is more similar to Euclidean space than the entire sphere topologically, although on the other hand the curvature is still different. Also I'm not really sure what I'm asking anymore. – Chill2Macht Jul 20 '17 at 11:39
• Well yeah, but your question was about Euclidean spaces, right? (Anyway I think the same holds for spheres, since they are also highly uniform, but I suppose not in complete generality, no.) – Mees de Vries Jul 20 '17 at 11:40

(a) Yes, it is a group; your proof is correct. I would call it the local isometry group of pointed metric space $(X,x)$. (The term "pointed metric space" is fairly common; it means a metric space in which we decided to distinguish some point.)
(b) No. The global isometry group of pointed metric space $(\mathbb{R}^n, x)$ is indeed $O(n)$. But the local isometries of $(\mathbb{R}^n, x)$, the way you defined them, are not constrained by anything away from $x$. So for example the maps that fix the unit ball $B$ around $x$, and permute $\mathbb{R}^n\setminus B$ in whatever way would qualify. This is a huge group full of discontinuous maps.
• I will upvote your answer tomorrow -- I ran out of upvotes for today. Please remind me if I forget. As a follow-up, if we use an equivalence relation, whereby two pointed local isometries are equivalent if they agree on some neighborhood of $x$ (like in the definition of germs en.wikipedia.org/wiki/Germ_(mathematics) ) then would the group be the same as $O(n)$? Since the restrictions can always be extended to the global pointed isometries which form (a group isomorphic to) $O(n)$? – Chill2Macht Jul 20 '17 at 16:42
• Yes, isometric germs are in natural bijection with $O(n)$. – user357151 Jul 20 '17 at 18:34