A description/proof of that rotation in \(n\)-dimensional Euclidean vectors space is same \(2\)-dimensional rotations along \((n - 2)\)-dimensional subspace axis
Topics
About: Euclidean vectors space
The table of contents of this article
Starting Context
- The reader knows a definition of Euclidean vectors space.
Target Context
- The reader will have a description and a proof of the proposition that any rotation in the \(n\)-dimensional Euclidean vectors space is any same \(2\)-dimensional rotations along any \((n - 2)\)-dimensional subspace axis.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Description
For any Euclidean vectors space, \(\mathbb{R}^n\), any rotation in \(\mathbb{R}^n\) is any same \(2\)-dimensional rotations along any \((n - 2)\)-dimensional subspace axis.
2: Note
This is not any rigorous proof but an intuitive understanding what rotation in any higher-dimensional Euclidean vectors space is.
3: Proof
Any rotation is a kind of map, \(f: \mathbb{R}^n \to \mathbb{R}^n\).
Let us always think of rotations centered at the origin, which means that \(f (0) = 0\). Any rotation centered at another point, \(p_0 \in \mathbb{R}^n\), can be regarded to be \(f'^{-1} \circ f \circ f'\), where \(f': \mathbb{R}^n \to \mathbb{R}^n\), \(p \mapsto p - p_0\), the translation.
Any rotation in \(\mathbb{R}^3\) is a rotation around an axis, which is a line (that passes the origin, which we will not mention here after, because our lines are always so, because we are thinking only about rotations centered at the origin).
As we tend to imagine a rotation as in \(\mathbb{R}^3\), we tend to imagine a line as the axis.
But what is the axis for any rotation in \(\mathbb{R}^2\)? The line perpendicular to \(\mathbb{R}^2\)? But the line is not in \(\mathbb{R}^2\), which seems a problem: why do we need to introduce the ambient \(\mathbb{R}^3\) when we are thinking of the \(\mathbb{R}^2\) space? At least, we did not need to introduce the ambient \(\mathbb{R}^4\) in order to think of rotations in \(\mathbb{R}^3\), which seems not symmetric.
What is 'axis' indeed? The axis of any rotation is the set of points that are fixed by the rotation.
In fact, the axis for any rotation in \(\mathbb{R}^2\) is not the line in \(\mathbb{R}^3\), but the origin, the point.
What is the axis for any rotation in \(\mathbb{R}^4\)? \((x^1, x^2, x^3, x^4) \mapsto (cos \theta x^1 - sin \theta x^2, sin \theta x^1 + cos \theta x^2, x^3, x^4)\) is a rotation, which fixes the \(x^3, x^4\) coordinates, which means that the fixed set is \(\{(0, 0)\} \times \mathbb{R}^2\), which is the axis. So, the axis for any rotation in \(\mathbb{R}^4\) is a \(2\)-dimensional subspace of \(\mathbb{R}^4\).
Generally, the axis for any rotation in \(\mathbb{R}^n\) is an \((n - 2)\)-dimensional subspace of \(\mathbb{R}^n\).
Let us think of a rotation in \(\mathbb{R}^3\). The rotation has the axis as a \((3 - 2)\)-dimensional subspace, a line. At each point on the axis, there is the \((3 - (3 - 2) = 2)\)-dimensional plane perpendicular to the axis. Do the rotation for such each plane centered at the intersection with the axis by any same angle, which is nothing but the rotation in \(\mathbb{R}^3\). The reason why the axis of any rotation in \(\mathbb{R}^3\) is a line is that \(3 - 2 = 1\), not that axis in general is a line, which is wrong.
Let us think of a rotation in \(\mathbb{R}^n\). The rotation has the axis as an \((n - 2)\)-dimensional subspace. At each point on the axis, there is the \((n - (n - 2) = 2)\)-dimensional plane perpendicular to the axis. Do the rotation for such each plane centered at the intersection with the axis by any same angle, which is nothing but the rotation in \(\mathbb{R}^n\).
