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The following regards the textbook "Mathematical Methods of Classical Mechanics" by V.I. Arnold.

Definition. The configuration space of a system of $n$ points is the direct product of $n$ copies of $\mathbb{R}^3.$

Definition. Rigid bodies are defined as a system of point masses constrained by the fact that the distance between them is constant, i.e. $| \mathbf{x}_i - \mathbf{x}_j | = r_{ij}$.

Theorem. The configuration space of a rigid body is the 6-dimensional manifold $\mathbb{R}^3 \times \operatorname{SO}(3)$ (the direct product of a three-dimensional space $\mathbb{R}^3$ and the group $\operatorname{SO}(3)$ of its rotations), as long as there are three points in the body not in a straight line.

Proof. Let $\mathbf{x}_1$, $\mathbf{x}_2$, and $\mathbf{x}_3$ be three points of the body which do not lie in a straight line. Consider the right-handed orthonormal frame whose first vector is in the direction of $\mathbf{x}_2-\mathbf{x}_1$, and whose second is on the $\mathbf{x}_3$ side in the $\mathbf{x}_1 \mathbf{x}_2 \mathbf{x}_3$-plane (Figure). It follows from the conditions $|\mathbf{x}_i - \mathbf{x}_j|=r_{ij}\; (i=1,2,3)$, that the positions of all the points of the body are uniquely determined by the positions of $\mathbf{x}_1$, $\mathbf{x}_2$, and $\mathbf{x}_3$, which are given by the position of the frame. Finally, the space of frames in $\mathbb{R}^3$ is $\mathbb{R}^3 \times \operatorname{SO}(3)$, since every frame is obtained from a fixed one by a rotation and a translation.

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Question 1. Why do we identify space of all frames in $\mathbb{R}^3$ with $\mathbb{R}^3 \times \operatorname{SO}(3)$?

Question 2. Why is every frame obtained from a fixed one by a rotation and a translation?

Question 3. Why is the configuration space of a rigid body identified with the space of all frames in $\mathbb{R}^3$?

Question 4. I do not see how the definition of the configuration space relates with the configuration space in question. How did we get from "the direct product of $n$ copies of $\mathbb{R}^3$" to $\mathbb{R}^3 \times \operatorname{SO}(3)$?

Unfortunately, I wasn't able to find the definition of a frame in the book.

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  • $\begingroup$ It would be helpful to have some more definitions: how are frames defined, how about configurations or rigid bodies? $\endgroup$ Commented May 13, 2022 at 1:59
  • $\begingroup$ I've edited my answer to include them :) $\endgroup$ Commented May 13, 2022 at 8:46
  • $\begingroup$ A frame is usually just a basis consisting of orthonormal vectors (unit length, orthogonal to one another). You can assemble them as column vectors into a (nonsingular) square matrix. All frames based at the origin are just the standard frame (identity matrix) acted on by a special orthogonal matrix $A \in \operatorname{SO}(3)$, so you can just identify the frame with the matrix. $\endgroup$ Commented May 13, 2022 at 9:13

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The choice of a frame is sometimes referred to as the choice of coordinate axes. You have to specify where your origin is—that's a vector in $\mathbb{R}^3$—and which way each of your axes point—that's an orientation-preserving orthogonal transformation, i.e. a matrix in $\operatorname{SO}(3)$. The combination of these independent choices forms the special euclidean group $\operatorname{SE}(3) \cong \mathbb{R}^3 \rtimes \operatorname{SO}(3)$.

You can think of the standard frame, anchored at the origin $\mathbf{0} \in \mathbb{R}^3$ with the standard axes $(\mathbf{e}_1, \mathbf{e}_2, \mathbf{e}_3)$. To uniquely identify any other frame, you need to rotate the axes, which is equivalent to applying a transformation $A \in \operatorname{SO}(3)$. Then you need to translate the origin, which is equivalent to adding a vector $\mathbf{b} \in \mathbb{R}^3$. The space of all such transformations happens to be $3$-dimensional, hence the the space of frames is $3+3 = 6$-dimensional.

By the way, if you also allow left-handed frames, then you identify a frame by an element of the orthogonal group $\operatorname{O}(3)$, which includes reflections and is a double cover of $\operatorname{SO}(3)$, so it's also $3$-dimensional. Together with translations of the origin, these form the entire euclidean group $\operatorname{E}(3) \cong \mathbb{R}^3 \rtimes \operatorname{O}(3)$.

Finally, if you allow axes that aren't necessarily perpendicular (just linearly independent), then the linear transformation will be an element of the general linear group $\operatorname{GL}(3)$, which is $9$-dimensional. Together with translations of the origin, these form the group of all affine transformations in $\mathbb{R}^3$, which is $\mathbb{R}^3 \rtimes \operatorname{GL}(3)$. You can think of an affine transformation as a pair $(\mathbf{b}, A)$ that acts on a vector $\mathbf{x} \in \mathbb{R}^3$ by $$ \mathbf{x} \mapsto A\mathbf{x} + \mathbf{b}. $$ All of these Lie groups are nested: \begin{array}{*5{c}} \operatorname{SO}(3) & \subset & \operatorname{O}(3) & \subset & \operatorname{GL}(3) \\ \cap & & \cap & & \cap \\ \color{blue}{\mathbb{R}^3 \rtimes \operatorname{SO}(3)} & \subset & \mathbb{R}^3 \rtimes \operatorname{O}(3) & \subset & \mathbb{R}^3 \rtimes \operatorname{GL}(3) \end{array} The space of frames is parametrized by the special euclidean group in $\color{blue}{\text{blue}}$.

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    $\begingroup$ Thank you for the answer! However, I still have trouble seeing why the configuration space of a rigid body is identified with the space of all frames in $\mathbb{R}^3$. Could you please comment on that? $\endgroup$ Commented May 13, 2022 at 8:26
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    $\begingroup$ Relative to your coordinate axes, where is the rigid body $\bigl( \mathbf{b} \in \mathbb{R}^3 \bigr)$ and which direction is it facing $\bigl( A \in \operatorname{SO}(3) \bigr)$? That pair $\bigl( \mathbf{b}, A \bigr)$ is a frame. $\endgroup$ Commented May 13, 2022 at 9:17
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    $\begingroup$ Let's see if I understood this correctly. Would it be correct to say that the configuration space is the space $\{(b,A):b \in \mathbb{R}^3, A \in SO(3)\}$? $\endgroup$ Commented May 13, 2022 at 9:28
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    $\begingroup$ Yes. Choosing one of each type of object is the same thing as choosing a pair from the product. $\endgroup$ Commented May 13, 2022 at 9:32
  • $\begingroup$ Uh, just to be clear, what does "each type of object" refers to? Do you mean $b$ and $A$? $\endgroup$ Commented May 13, 2022 at 9:39

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