Isometric map in an open set of Euclidean space In classical geometry, it is known that isometric maps in $\mathbb{R}^3$ 
(i.e. a map $T:\mathbb{R}^3\rightarrow\mathbb{R}^3$ s.t. $d(Tx,Ty)=d(x,y)$ for all points $x,y$ in $\mathbb{R}^3$) 
are orthogonal transformations with translations.
$T$ is continuous by Lipschitz. Is it possible that $T$ is not $C^1$?
And I wonder what will happen when restricting to a given open set $U\subset\mathbb{R}^3$. How to decide the isometry group of $U$?
A special case is the open ball. It preserves most symmetries of the entire space. But I can't see if $T(O)\ne O$ could happen.
Any help will be appreciated.
 A: Any isometry is obviously continuous since it is Lipschitz continuous.
The group of symmetries of $U$ is clearly a subgroup of the group of all symmetries of $\mathbb{R^3}$. Usually one finds the isometry group of such $U$ by looking at $U$ really hard, then writing down what appears to be its symmetry group, then proving that the group you wrote down is correct. There is no general procedure.
A: Here is a step-by-step sketch of the proof, the details you should be able to fill in:
Let $T: U\to E^3$ be an isometric map of an open connected subset of $R^3$. 


*

*Show that $T$ sends geodesic (straight line) segments in $U$ to geodesic segments. (Hint: define geodesic segments in terms of triangle inequalities.)

*Show that $T$ preserves angles $\angle(ABC)$ between geodesic segments $AB, BC$ in $U$. 

*Consider three pairwise orthogonal geodesic segments $OA, OB, OC$ in $E^3$ (I will call such a triple a "base" with the center at $O$). Show that every geodesic segment $OD$ is determined by its length and the angles that it makes with the segments $OA, OB, OC$. (Hint: Use linear algebra if you have hard time proving this via elementary geometry.) 

*Assume that $T: U\to E^3$ fixes a base at $O$ in $U$ (sends $OA$ to itself, etc). Show (using 2 and 3) that $T$ restricts to the identity on a sufficiently small  open ball $B(O,r)$  centered at $O$. 

*Let ${\mathcal B}_{\delta}$ denote the set of bases in $E^3$ such that each segment in the base has length $\delta$. Show that the group of rigid motions of $E^3$ acts transitively on ${\mathcal B}_{\delta}$. 

*Use 5 to show that given any isometric map $T: U\to E^3$ and a base $b\in {\mathcal B}_\delta$ which is entirely contained in $U$, there exists a rigid motion $g$ of $E^3$ such that $T\circ g$ fixes the base $b$. (A rigid motion is a composition of reflections.) Conclude using 4 that $T\circ g$ is the identity on the ball $B_\delta(O,r)$ where $O$ is the center of the base $b$. 

*Conclude that each isometric map $T: U\to E^3$ is locally a rigid motion: For every $x\in U$ there exists $\delta>0$ such that $T|B(x,\delta)$ equals the restriction of a rigid motion of $E^3$. 

*Use connectivity of $U$ to conclude that $T: U\to E^3$ is the restriction of a rigid motion. Here you will use the fact that if two rigid motions agree on a nonempty open set then they are equal. 
Bonus 1: Prove the same for isometric maps of domains in the Euclidean $n$-space $E^n$.
Bonus 2: Prove the same for isometric maps of domains in the rough sphere (of any dimension) and in the hyperbolic space (of any dimension). 
A: Euclidean space has this amazing property: an isometry defined on a subset may be extended to an isometry of the whole space.  Most metric spaces do not have this property.
