# Why is the space of all frames in $\mathbb{R}^3$ the space $\mathbb{R}^3 \times \operatorname{SO}(3)$?

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.

• It would be helpful to have some more definitions: how are frames defined, how about configurations or rigid bodies? Commented May 13, 2022 at 1:59
• I've edited my answer to include them :) Commented May 13, 2022 at 8:46
• 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. Commented May 13, 2022 at 9:13

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 $$\color{blue}{\text{blue}}$$.

• 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? Commented May 13, 2022 at 8:26
• 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. Commented May 13, 2022 at 9:17
• Let's see if I understood this correctly. Would it be correct to say that the configuration space is the space $\{(b,A):b \in \mathbb{R}^3, A \in SO(3)\}$? Commented May 13, 2022 at 9:28
• Yes. Choosing one of each type of object is the same thing as choosing a pair from the product. Commented May 13, 2022 at 9:32
• Uh, just to be clear, what does "each type of object" refers to? Do you mean $b$ and $A$? Commented May 13, 2022 at 9:39