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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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Can you explain a little more about "configuration space" and "frame bundle"? –  Shiyu Mar 24 '12 at 13:00
    
The term "configuration space" is used by physicist as the space where the dynamic is happening. I don't think it's a well defined term as in most cases you can factor out some symmetries to get a smaller space (think of the Kepler problem, you would start by writing down the equations of motions in $\mathbb{R}^3$ and reduce it then to a plane by using conservation of angular momentum). –  David Mar 24 '12 at 13:18
    
Forget about the exact definition of "frame bundle" (it's quite technical and requires some differential geometry), intuitively a point in the frame bundle is a point in some base manifold (in this case $\mathbb{R}^3$) and an orthonormal frame attached to this point. You can describe the frame by the rows or columns of a matrix $R \in SO(3)$. –  David Mar 24 '12 at 13:22
    
What I need it for: suppose one point $x$ in $\mathbb{R}^{3n}$ is an equilibrium. And all $x'$ in $\mathbb{R}^{3n}$ are also equilibria, which are obtained by transforming $x$. So the set $\Omega$ is an equilibrium set. What I want is to identify the structure or properties of this set. But I'm not familiar with related mathematics. –  Shiyu Mar 24 '12 at 14:54
    
Thanks for the editing. It is helping. Regarding the boundness of the set, if there is only rotation and no translation, would the set be bounded? And is the set a manifold? In fact, I think I am quite familiar with linear algebra such as matrix and subspace, is it possible to somehow check the structure of the set from a linear algebra perspective? For example, let's only consider rotation transformation. Let $x\in\mathbb{R}^{3n}$ be an original point and $x'=(I_n\otimes R) x$. So the whole set is $\Omega=\{x' | x'=(I_n\otimes R) x, \forall R\in SO(3)\}$. But I don't know how to analyze next. –  Shiyu Mar 24 '12 at 15:07
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