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In a world of three spatial dimensions plus time, every atom rotates around a line, the axis of rotation.

In a world of $N$ spatial dimensions where $N$ is greater than 3, must every atom rotate, and if so does it rotate around a line, a plane, or a subspace of smaller number of dimensions?

Qmechanic
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2 Answers2

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  1. One may show that a general rotation $R\in SO(N)$ in $N\geq 2$ spatial dimensions can be composed
    $$R ~=~ R_1\circ \ldots\circ R_{k} $$ of at most $k=[\frac{N}{2}] $ pairwise commuting rotations $$R_1,\ldots, R_{k}~\in~ SO(N)$$ that each leaves a co-dimension-2 subspace invariant (although not necessarily the same subspace).

  2. More explicitly, given a rotation $R\in SO(N)$ there exists an orthonormal basis $(e_1, \ldots, e_N)$ [which may depend on $R$] such that the rotation $R$ is represented by a block-diagonal matrix of the form $$ \begin{pmatrix} \cos\theta_1 & \sin\theta_1 & \cr -\sin\theta_1 & \cos\theta_1 & \cr && \cos\theta_2 & \sin\theta_2 & \cr &&-\sin\theta_2 & \cos\theta_2 & \cr &&&& \ddots \cr &&&&&\cos\theta_k & \sin\theta_k & \cr &&&&&-\sin\theta_k & \cos\theta_k & \cr &&&&&&&1\cr &&&&&&&&1\cr &&&&&&&&& \ddots \cr &&&&&&&&&&1\end{pmatrix}. $$

  3. The rotation $R$ itself is only guaranteed to leave invariant a dimension-1 subspace (=a line through the origin) if the space dimension $N$ is odd.

Qmechanic
  • 201,751
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In 2d, a rotation matrix has the form $$ r(\theta)=\left(\begin{array}{cc} \cos\theta&-\sin\theta\\ \sin\theta&\cos\theta\end{array}\right):= \left(\begin{array}{cc} c(\theta)&-s(\theta)\\ s(\theta)&c(\theta)\end{array}\right) $$ and rotates vector in a plane.

In 3d a rotation matrix can be written as a product $$ r_{12}(\psi)r_{13}(\theta)r_{12}(\varphi) $$ where \begin{align} r_{12}(\psi)&=\left(\begin{array}{ccc} c(\psi)&-s(\psi)&0\\ s(\psi)&c(\psi)&0\\ 0&0&1 \end{array}\right)\\ r_{13}(\theta)&=\left(\begin{array}{ccc} c(\psi)&0&-s(\psi)\\ 0&1&0\\ s(\psi)&0&c(\psi) \end{array}\right) \end{align} leaving one axis invariant. This axis can be identified by the row or column containing $0$s everywhere except for one entry.

In SO(4), one can write a rotation matrix as a sequence or $r_{ij}$ matrices. $r_{12}$ would have the form $$ r_{12}(\psi)=\left(\begin{array}{cccc} c(\psi)&-s(\psi)&0&0\\ s(\psi)&c(\psi)&0&0\\ 0&0&1&0\\ 0&0&0&1 \end{array}\right) $$ and so leaves a 2-dimensional subspace invariant. An SO(4) matrix can be written in the factored form $$ r_{34}(\beta_1)r_{23}(\beta_2)r_{12}(\beta_3) r_{34}(\beta_4)r_{23}(\beta_5)r_{34}(\beta_6) $$ by restricting to real values the entries of the $SU(4)$ matrix factored as as done here. This is not by any means the only possible factorization.

Obviously, an SO(5) rotation can be written in terms of matrices leaving a 3-dimensional subspace invariant etc.

ZeroTheHero
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  • What about double rotations in 4D? The simultaneous rotations in 2 orthogonal planes intercepting in a point at the origin (replacing units in your matrix with sines and cosines). – safesphere Dec 01 '17 at 13:33
  • @safesphere I’m not sure I understand your question. It’s still a 4d rotation, but clearly it can be realized as a sequence of two commuting SO(2) rotations: $r_{12} r_{34}$. Take $r_{23}(0)=1$ etc. – ZeroTheHero Dec 01 '17 at 13:38
  • As a clarification of what I mean, please see this https://math.stackexchange.com/q/2543122/ and also see Double Rottions under Geometry here https://en.wikipedia.org/wiki/Rotations_in_4-dimensional_Euclidean_space – safesphere Dec 01 '17 at 13:58
  • @safesphere I really need more coffee... I still don’t get what you ask, but the link I will read with interest later. – ZeroTheHero Dec 01 '17 at 14:04
  • No problem :) To clarify again, in 3D, an object can rotate only around one axis. If we try rotating an object around two axes at the same time, this only turns the rotation axis, but it remains a single one and the rotation still has a single rate. In 4D however an object can rotate in one plane (single rotation) and also in another orthogonal plane at the same time (dual rotation) independently and with a different rate. – safesphere Dec 01 '17 at 14:12
  • @safesphere let me (eventually) read the material you linked. – ZeroTheHero Dec 01 '17 at 14:13
  • @safesphere The 'rigid object' can in principle be rotated any number of times, and the product of the associated rotation matrices is always a rotation matrix (which does not distort the distances and shapes of the 'rigid object'). That rotation matrix does not describe how many steps and what axes are used, it just relates the final state to the initial one. – Whit3rd Dec 30 '17 at 03:00
  • @Whit3rd True, but this view is limited by seeing rotation as a one-time result of some undescribed steps. This question was about rotation not as a one-time result, but as a continuous process and also about the geometric meaning of this process. As such, rotation in 2D is always in the same plane, in 3D it is in any single plane, but in 4D it can be on one or it can be in two planes at the same time that are orthogonal to each other and intersecting in the center of rotation. – safesphere Dec 30 '17 at 04:11
  • @ZeroTheHero So... is "eventually" before or after the end of the year? :) Your answer states that in higher dimensions, a rotation leaves a $k-2$ subspace invariant. This is true only for a single rotation in one plane. However, in higher dimensions you get multiple rotations in different orthogonal planes at the same time. So, as Qmechanic stated, in odd dimensions you only get in general one 1D axis invariant while in even dimensions in general you get only one zero-D point invariant. So higher dimensions translate to more complex rotations while the invariance remains a point or a line. – safesphere Dec 30 '17 at 04:38
  • @safesphere Mon! I forgot about this... – ZeroTheHero Dec 30 '17 at 04:39
  • @ZeroTheHero Rotation is always in a 2D plane. So in 1D - no rotation, but you have one invariant axis: in 2D - one rotation, no axis; in 3D - one rotation plus one axis; in 4D - two rotations (2Dx2=4D), no axis; in 5D - two rotations plus one axis, and so on. The matrix has diagonal pairs of sines and cosines, except for unity in odd dimensions in the lower right corner as in your 3D example (axis). However in your 4D example in general there should be no unities, only sines and cosines (no exis). – safesphere Dec 30 '17 at 09:27