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I am considering rotations in 4D space. We use $x, y, z, w$ as coordinates in a Cartesian basis. I have found sources that give a parameterization of the rotation matrices as \begin{align} &R_{yz}(\theta) = \begin{pmatrix} 1&0&0&0\\0&\cos\theta&-\sin\theta&0\\0&\sin\theta&\cos\theta&0\\0&0&0&1 \end{pmatrix}, R_{zx}(\theta) = \begin{pmatrix} \cos\theta&0&\sin\theta&0\\0&1&0&0\\-\sin\theta&0&\cos\theta&0\\0&0&0&1 \end{pmatrix},\\ &R_{xy}(\theta) = \begin{pmatrix} \cos\theta&-\sin\theta&0&0\\\sin\theta&\cos\theta&0&0\\0&0&1&0\\0&0&0&1 \end{pmatrix}, R_{xw}(\theta) = \begin{pmatrix} \cos\theta&0&0&-\sin\theta\\0&1&0&0\\0&0&1&0\\\sin\theta&0&0&\cos\theta \end{pmatrix},\\ &R_{yw}(\theta) = \begin{pmatrix} 1&0&0&0\\0&\cos\theta&0&-\sin\theta\\0&0&1&0\\0&\sin\theta&0&\cos\theta \end{pmatrix}, R_{zw}(\theta) = \begin{pmatrix} 1&0&0&0\\0&1&0&0\\0&0&\cos\theta&-\sin\theta\\0&0&\sin\theta&\cos\theta \end{pmatrix}, \end{align} where the subscript labels a plane that is being rotated. This seems to be a very intuitive extension of lower dimensional rotations. However, I would really like to see a proof that these are correct, and I'm not sure how I could go about doing that. By correct, I mean that these 6 matrices can generate any 4D rotation.

My initial attempt was to construct a set of transformations from the definition of the transformations (as matrices) that define a 4D rotation, \begin{align} \{R|RR^T = I\}, \end{align} where $I$ is the identity matrix (4D), but this has 16 (constrained) parameters and I thought that there must be an easier way.

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  • $\begingroup$ You might not be interested, but your six matrices are overkill. You could generate every rotation by just 4, in principle, noting S(4)~SO(3)xSO(3), and that a presentation of SO(3) rotations requires only two generators, not 3, as the 3rd can be obtained by the group commutator of just two: so you have azimuth and altitude on a globe. $\endgroup$ Commented Jul 21, 2021 at 15:42
  • $\begingroup$ This is very interesting actually, can you recommend any resources that go into some more detail? $\endgroup$
    – Bedge
    Commented Jul 21, 2021 at 16:19
  • $\begingroup$ Sorry, not in compact form. The SO(4)~SO(3)xSO(3) split is in all discussions of the Lorentz group (which adds i's for the noncompact form) representations. The sufficiency of two angles, hence generators to parameterize a sphere, is in many discussions of the rotation group and astronomy. The obvious point, Lie Algebraically, is that $[L_x,L_y]\propto L_z$, so you may go everywhere with $L_x$ and $L_y$, combining them. It's a different question. Books and reviews on presentations are rare. $\endgroup$ Commented Jul 21, 2021 at 16:31
  • $\begingroup$ @CosmasZachos see de Guise H, Di Matteo O, Sánchez-Soto LL. Simple factorization of unitary transformations. Physical review A. 2018 Feb 20;97(2):022328 (or the arXiv version arxiv.org/pdf/1708.00735.pdf) and replace the unitaries by real $2\times 2$ rotation matrices. You then only need in fact the correct sequence of "adjacent" rotations $R_{12}\cdot R_{23}\cdot R_{34}\cdot R_{12}\cdot R_{23} \cdot R_{12}$ each with its own rotation angle. $\endgroup$ Commented Jul 21, 2021 at 18:11
  • $\begingroup$ Nice, thanks!... $\endgroup$ Commented Jul 21, 2021 at 18:14

3 Answers 3

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An indirect and sneaky way of doing this is to construct the generators $L_{ij}=-i d R_{ij}/d\theta \vert_{\theta=0}$ and verify that the resulting matrices span the Lie algebra $\mathfrak{so}(4)$. This avoids having to construct a general rotation matrix as a product of your 6 elements. You can then use the result that the exponential of any linear combo of your generators is guaranteed to generate an element in the group.

A more direct way is to check that $R_{ij}R_{ij}^T=\mathbb{I}$, and then check that $(R_{ij}R_{ab})^T (R_{ij}R_{ab})=\mathbb{I}$ and then by induction that any product $R=R_{ab}R_{cd}R_{ef}...$ satisfies $RR^T=\mathbb{I}$.

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  • $\begingroup$ A more direct approach would have been to take a vector $(w,x,y,z)^T$, and impose the rotations plane by plane, because it is just the rotations in 2D with two unchanged coordinates. Then you can trivially reconstruct the rotations matrices. $\endgroup$ Commented Jul 21, 2021 at 15:11
  • $\begingroup$ It is true that the Lie algebra is $\mathfrak{so}(4)$, but is it not the case that the Lie algebra is not uniquely associated to the group, so that does not seem sufficient to me. Also, perhaps my question is not clear enough, but I am happy that the elements I have listed are at least a subset of the possible 4D rotations. What I really want is to ensure that I haven't 'missed' any, i.e. prove that there does not exist a 4D rotation that cannot be constructed as a combination of the elements I have listed. $\endgroup$
    – Bedge
    Commented Jul 21, 2021 at 15:18
  • $\begingroup$ I'm not sure I understand your comment about uniqueness: are you worried about spinor representations? $\endgroup$ Commented Jul 21, 2021 at 15:21
  • $\begingroup$ As in multiple groups can have the same Lie algebra. So showing that the Lie algebra is equivalent does show equivalence at the level of groups, unless I'm mistaken. $\endgroup$
    – Bedge
    Commented Jul 21, 2021 at 15:33
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    $\begingroup$ @Bedge It's true in the abstract sense that you can have different groups with isomorphic Lie algebras, but if you're working in a specific representation (in this case $GL(4)$) then there is a one to one correspondence between groups and their Lie algebras. $\endgroup$
    – Carmeister
    Commented Jul 22, 2021 at 12:23
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Rotation matrices are orthogonal matrices, i.e., $$R^{-1}=R^T.$$ An orthogonal n-by-n matrix has $n(n-1)/2$ independent parameters (see here for an elegant proof), which in our case is 6 parameters. Linear algebra tells us that a 6-component vector can be represented in terms of 6 linearly independent vectors. The rest is to prove that the six given matrices are linearly independent, i.e., that there are no such coefficients $c_j$ that $$ \sum_jc_jR_j(\theta_j)=0 $$ for arbitrary combination of $\theta_j$, which is easily done by hand.

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  • $\begingroup$ Thank you, I tried this and it was immediately clear. Just to clarify, is the vector space in question the $\mathbb{R}$-vector space on the set of special orthogonal maps $\phi: \mathbb{R}^4 \longmapsto \mathbb{R}^4$? $\endgroup$
    – Bedge
    Commented Jul 21, 2021 at 16:14
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    $\begingroup$ hold on a minute here. You cannot take combinations of group elements: $R_{xz}+R_{yw}$ doesn’t make sense as a group element. Your argument about the combos is at the algebra level, and elements in the algebra will NOT depend on the $\theta_i$, i.e. $R_i(\theta_i)$ doesn’t make sense as an element in the algebra. Or am I missing something obvious (like strong coffee)? $\endgroup$ Commented Jul 21, 2021 at 16:26
  • $\begingroup$ Thank you for pointing this out, it's not clear. I'll un-accept the answer for now. $\endgroup$
    – Bedge
    Commented Jul 21, 2021 at 16:52
  • $\begingroup$ @ZeroTheHero in this argument thetas are merely a matter of parametrization. Btw, I think that your answer is correct . I merely wanted to provide a pedestrian alternative. $\endgroup$
    – Roger V.
    Commented Jul 21, 2021 at 17:34
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For each of your 4D rotation matrix $~\mathbf R~$ if this equation

$$\mathbf Z^T\, \mathbf Z= \left(\mathbf R\,\mathbf Z\right)^T\,\left(\mathbf R\,\mathbf Z\right)$$

is fulfilled the rotation matrix $~\mathbf R~$ is orthonormal .$~\mathbf R^T\,\mathbf R=\mathbf I_4$

where

$$\mathbf Z= \begin{bmatrix} x \\ y \\ z \\ w \\ \end{bmatrix}$$

Edit

you can also check the determinate of the Rotation matrix ,if the determinate of the Rotation matrix is equal one the matrix is orthonormal ?

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    $\begingroup$ not quite. If the determinant is +1 then you are in SO(4) (special orthogonal group). If the determinant is -1, you are in O(4) (orthogonal group), which contains in addition to rotations reflections and still satisfy $R^T\cdot R=\textbf{I}_4$. $\endgroup$ Commented Jul 21, 2021 at 23:45
  • $\begingroup$ @ZeroTheHero thank you $\endgroup$
    – Eli
    Commented Jul 22, 2021 at 6:51

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