# Why are the generators of rotation in the 4-dimensional Euclidean space correspond to rotations in a plane?

In three-dimensions, the rotation generators are represented by $J_1$, $J_2$ and $J_3$ where $1,2,3$ respectively stands for the generator of rotation about $x,y,z$ axes respectively. In general, in the rotation about the direction $\hat{\textbf{n}}$ is generated by $\textbf{J}\cdot\hat{\textbf{n}}$.

However, for even dimensional rotation groups such as $SO(4)$, the generators are labelled by $J_{ij}$ where $i,j=1,2,3,4$ which corresponds to rotations in $12$, $23$, $34$, $14$, $24$ and $13$ planes. Similar situation happens in case of the Lorentz group $SO(3,1)$ in special relativity.

Does it mean that a given rotation in 4-dimensional Euclidean space cannot be associated with a unique axis ($\hat{\textbf{n}}$) of rotation? If yes, why is that the case?

• Not only for even-dimensional rotation groups. For all rotation groups $\mathrm O(n)$ for $n\neq 3$, the generators are (skew) tensors. The three dimensional case is special because of the vector product $\times$. – AccidentalFourierTransform Dec 29 '17 at 10:52
• Is it not possible to associate a unique axis with the rotation in $\mathbb{R}^4$ in the $23$ plane (for example)? @AccidentalFourierTransform – SRS Dec 29 '17 at 10:57
• No. Rotations in $n\ge4$ dimensions fix a plane, not a line. – AccidentalFourierTransform Dec 29 '17 at 11:05
• Possible duplicates: How to define angular momentum in other than three dimensions? , Rotation in Higher Dimensions and links therein. – Qmechanic Dec 29 '17 at 11:14

Does it mean that a given rotation in 4-dimensional Euclidean space cannot be associated with a unique axis ($\hat{\textbf{n}}$) of rotation? If yes, why is that the case?

Yes, this is absolutely true. The notion of a one dimensional axis is an "accident" of three dimensions. Rotations transform planar (dimension 2) linear subspaces of Euclidean space and so one needs to specify the transformed plane and the rotation angle to specify the rotation.

In 3D dimensions we can cheat a little: a plane is uniquely defined by a unit normal vector, and the rotation angle can be encoded as the length of this vector. This is what we mean by an axis. The axis is the untransformed space of the rotation; the 3D space splits into two orthogonal, invariant spaces, the former being the plane of rotation, which is invariant but transformed (i.e. nontrivially bijectively mapped to itself) and the latter the axis, which is both invariant and untransformed. In 4 and higher dimensions, the invariant spaces are of 2 or higher dimensions.

A member of the Lie algebra of a rotation group (with the algebra written as a faithful matrix representation) is a skew-symmetric matrix, i.e. an entity of the form $\sum\limits_i X_i \wedge Y_i$ where the $X_i$ and $Y_i$ are 1D vectors in the Euclidean space. A general rotation matrix is then of the form $\exp\left(\sum\limits_i X_i \wedge Y_i\right)$. Things get kind of complicated in 4 and higher dimensions; the most general thing one can say is that a general proper orthogonal transformation on $N$ dimensional space can be decomposed as $R_1\circ R_2\circ\,\cdots R_{N\,\mathrm{div}\, 2}$ where each of the $R_i$ is a rotation that bijectively transforms a plane into itself and leaves the plane's complement invariant. However, the planes for each of the $R_i$ are not in general the same plane.

Further Questions and Useful Rotation Properties

User John Dvorak points out:

I would think that $R_1\circ R_2\circ\,\cdots R_{N\,\mathrm{div}\, 2}$ would always be pairwise orthogonal. Is that not the case?

This is indeed absolutely true and it is worth sketching the proof to get more insight into a higher dimensional rotation.

Let our rotation matrix be $R=\exp(H)$ with $H=\sum\limits_i X_i \wedge Y_i\in \mathfrak{so}(N)$ as above. Then there exists another orthogonal transformation $\tilde{R}$ (i.e. $\tilde{R}\in \mathrm{SO}(N)$) that, through similarity transformation, reduces the skew symmetric $H\in \mathfrak{so}(N)$ to block diagonal form:

$$H = \tilde{R}\,\mathrm{diag}(\Lambda_1,\,\Lambda_2,\,\cdots)\,\tilde{R}^T=\tilde{R}\,\mathrm{diag}(\Lambda_1,\,\Lambda_2,\,\cdots)\,\tilde{R}^{-1}$$

where each of the blocks is of the form:

$$\Lambda_j=\left(\begin{array}{cc}0&-\theta_j\\\theta_j&0\end{array}\right)$$

with $\theta_j\in\mathbb{R}$ being a rotation angle and that, if $N$ is odd, there is also a $1\times1$ zero block left over.

Therefore, if we put:

$$H_j = \tilde{R}\,\mathrm{diag}(0,\,0,\,\cdots,\,\Lambda_j,\cdots)\,\tilde{R}^T$$

then $R_j=\exp(H_j)$ with $R_1\circ R_2\circ\,\cdots R_{N\,\mathrm{div}\, 2}$ are then readily seen to make up the decomposition with the properties that John claims, to wit:

1. The $R_j$ are each rotations, each which transforms one plane only and each also has a dimension $N-2$ invariant and untransformed space (the analogue of the "axis");
2. The planes transformed by the $R_j$ are mutually orthogonal and indeed the planes spanned by the unit vectors $\tilde{R}_j\,\hat{e}_{2\,j}$ and $\tilde{R}_j\,\hat{e}_{2\,j+1}$, where the $\hat{e}_j$ are the orthonormal basis in which all the operators discussed have matrices as written above;
3. (as a consequence of 2.) the $R_j$ are mutually commuting.

Thus we can easily see that:

1. If the dimension $N$ is odd, there is always a dimension 1 invariant, untransformed space, corresponding to the 1D zero block cited above, further to the invariant spaces described below;
2. If the dimension is even, a nontrivial proper orthogonal transformation's untransformed space can be any of the dimensions $0,\,2,\,4,\,\cdots N-2$. The invariant spaces are of dimensions $0,\,2,\,4,\,\cdots,\,N$

This decomposition is about one particular rotation operator and is not to be confused with the notion of Canonical Co-ordinates of the Second Kind (see Chapter 1, Proposition 3.3 of V.V. Gorbatsevich, E.B. Vinberg, "Lie Groups and Lie Algebras I: Foundations of Lie Theory and Lie Transformation Groups", Springer, 2013), which are a generalized notion of Euler Angles. Here, a set of $H_j\in\mathfrak{so}(N)$ for $j=1,\,\cdots,\,N$ (note, there are now $N$ of them, not $N\,\mathrm{div}\,2$ of them) is chosen as a basis, i.e. to span $\mathfrak{so}(N)$. The the following are true:

1. The set $\mathbf{G}=\left\{\left.\prod\limits_{j=1}^N\,\exp(\theta_j\,H_j)\,\right|\,\theta_j\in\mathbb{R}\right\}$ contains a neighborhood of the identity in $\mathrm{SO}(N)$;
2. If, further, the $H_j$ are orthogonal with respect to the Killing form $\langle X,\,Y\rangle=\mathrm{tr}(\mathrm{ad}(X)\,\mathrm{ad}(Y))$, then the set $\mathbf{G}$ above is the whole of $\mathrm{SO}(N)$.

Property 1, as shown in the Gorbatsevich & Vinberg reference cited above, is a general and fundamental property of all Lie groups (if we replace $\mathfrak{so}(N)$ by the group's Lie algebra and $\mathrm{SO}(N)$ by the group); property 2 holds for compact semisimple ones only.

If the similarity transformation I have here pulled out of thin air seems mysterious, readers may be more familiar with the a re-ordered version of the similarity transformation $\tilde{R}$ above where we decompose a skew-symmetric, closed 2-form $\omega$ in an even dimension case so that its matrix $\Omega$ is:

$$\Omega = \tilde{R}\; \left(\begin{array}{cc}0&-\mathrm{id}_{\frac{N}{2}}\\\mathrm{id}_{\frac{N}{2}}&0\end{array}\right)\;\tilde{R}^T$$

which we implicitly do whenever we label a symplectic space with (in general nonunique) "canonical co-ordinates" so that $\omega$ then has the matrix:

$$\Omega = \left(\begin{array}{cc}0&-\mathrm{id}_{\frac{N}{2}}\\\mathrm{id}_{\frac{N}{2}}&0\end{array}\right)$$

Here we have a different usage of the word "canonical", this time as used in Hamiltonian mechanics. The word "canonical" well and truly needs a well pensioned retirement as it has worked so hard in Physics!

• I didn't understand any of this until I read, "The axis is the invariant space of the rotation. In 4 and higher dimensions, the invariant space is of 2 or higher dimensions.", and I have a degree in pure mathematics! (shame) Excellent explanation. +1 – Todd Wilcox Dec 29 '17 at 20:32
• I would think that $R_1\circ R_2\circ\,\cdots R_{N\,\mathrm{div}\, 2}$ would always be pairwise orthogonal. Is that not the case? – John Dvorak Dec 29 '17 at 20:38
• @JohnDvorak I think you might be thinking of conditions for the product of the $R_j$ to reach all of $SO(N)$ by adjusting the rotation angles (i.e. you're thinking generalized Euler angles). In any particular case, the decomposition is not unique and the planes of rotation may not be orthogonal. Let me look into this some more. The following is certainly true: Let $R_j = \exp(\alpha_j\,X_j)$ with $\alpha_j\in\mathbb{R}$ and $X_j\in \mathfrak{so}(N)$. If we have $N$ of them (rather than $N\,\mathrm{div}\, 2$) and if the $X_j$ span $\mathfrak{so}(N)$ then the product can reach any ..... – Selene Routley Dec 29 '17 at 22:52
• @JohnDvorak I've thought about it some more in the shower and you're absolutely correct. The $R_j$ are mutually commuting. There is an orthogonal similarity transformation that reduces the skew-symmetric Lie algebra member to block diagonal form, where each of the blocks is of the form $\left(\begin{array}{cc}0&-\theta_j\\\theta_j&0\end{array}\right)$ where $\theta_j$ is a rotation angle. If $N$ is odd, there is also a $0$ on the diagonal left over. ... – Selene Routley Dec 30 '17 at 0:15
• To be a bit picky, 3-d rotations have two invariant subspaces: the plane of rotation that all rotations have, and a line normal to that plane—the “axis.’ – amd Dec 30 '17 at 0:40

It's simply because 3-2 = 1 but 4 - 2 = 2. A rotation consists of the exchange of two axes. Since it involves two dimensions, it occurs in a plane, and there are n-2 dimensions left over. In one dimensional space, there aren't enough dimensional to do a rotation (unless you consider flipping the space a rotation). In two dimensional space, all of the dimensions are involved in a rotation (although the origin is a fixed point). In three dimensional space, there's one dimension left over, and this dimension can be treated as being an axis, and we can represent rotations in three dimensional space with three dimensional vectors. While the sign is arbitrary (right hand rule is a convention, not an inherent property three dimensional space), the line along which the vector lies is not. In four dimensional space, there are two dimensions left over, so the fixed points of a rotation are a plane, and choosing a direction to represent the rotation would be arbitrary. (We could represent a rotation among indices i,i+1 with a vector in the direction i+2, but that would require an arbitrary ordering of dimensions. Also note that if we have a rotation among nonconsecutive indices, such as 1 and 3, that can be composed of rotation among consecutive indices, in this case 1,2 and 2,3.)