What is the topological properties of this set Let $x_1,\dots,x_n$ be $n$ points forming a rigid body in $\mathbb{R}^3$. The distance between each pair of points is constant. Let $R\in SO(3)$ be a rotation and $T\in\mathbb{R}^3$ be a translation. Then $\{R,T\}$ actually is a rigid body transformation. If I apply $\{R,T\}$ to $\{x_i\}_{i=1}^N$, then I obtain a $\{x'_i=Rx_i+T\}_{i=1}^N$, which is different from the original one by a rigid body transformation. 
My question is: Since $\{R,T\}$ can be arbitrary, all these $\{x'_i\}_{i=1}^N$ form a set. If I define $x'=[x'_1^T,\dots,x'_n^T]^T\in\mathbb{R}^{3n}$, all these $\{x'_i\}_{i=1}^N$ form a set $\Omega$ in $\mathbb{R}^{3n}$. Then what does this set $\Omega$ look like? Is it connected? compact? or a manifold? I'm not very familiar with the set theory or differential geometry, can someone give me a hint on how to analyze the problem? Thanks.

The usual way to model a rigid body is to attach an orthonormal frame at some point of the rigid body.  Thus the configuration space becomes the trivial frame bundle $\mathbb{R}^3 \times SO(3)$.
OK, reading your question again my answer wasn't very helpful.  May I ask what do you need it for?
Some simple observations can be done:


*

*It's connected:  Rotations and translations are connected.  Use this to construct a continuous path between two points.

*It's closed: It's complete.  Each convergent sequence converges towards another point from that set.

*It's not bounded (you can move the rigid body arbitrary far away with a translation) and thus not compact.


I think more interesting would be what the topology of the set of equivalence classes is.
