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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.

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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:

  1. It's connected: Rotations and translations are connected. Use this to construct a continuous path between two points.
  2. It's closed: It's complete. Each convergent sequence converges towards another point from that set.
  3. 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.

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    Well, I was still tired. Actually I'm pretty sure the set is $\mathbb{R}^3 \times SO(3)$. I fear the proof will be tedious though. As a start: glue and orthonormal frame at $x_1$. At least intuitively the coordinate $x_1$ and the frame uniquely define all other points of the rigid body. If you drop translations you are left with just $SO(3)$ thus the set is compact. Additional symmetries of the rigid body may make the set smaller though. And yes, it's a manifold, it's a regular point of the distances of the rigid body. Maybe Arnolds mechanics book gives you some more details.2012-03-24