# "Visualizing'' rotation in even dimensions

So as we know, a linear transformation $\Bbb R^n\to\Bbb R^n$ must have an even number of non-real complex eigenvalues. One consequence of this is that, in $4$ dimensions, we cannot talk about rotation about a line — the only non-trivial rotation fixes a plane.

Since we can't visualize $4$ dimensions, I was trying to think of a way to interpret these rotations. One useful way is to imagine a $3$ dimensional space where each point has a fourth coordinate, which we can interpret as something like "temperature" of the point.

After playing around for a bit, I realized that rotations around a plane in this world look like stretching along one axis. For example, if we fix the $x$-$y$ plane, then the point $(0,0,1)$ might become $(0,0,2)$. Then the temperature of that point would have to decrease correspondingly. It is probably a coincidence, but in that case, a rotation looks like compressing/decompressing a gas.

As far as I can tell, nothing about $6$ dimensions makes this any more interesting. You just have to add in some other properties (density, color) and you have the same basic picture.

What other methods do you know for interpreting rotation about a plane? Without getting too much into physics, what is the connection between rotation in $4$ dimensions and our apparently $3$ dimensional world?

• You get something particularly interesting if your T rotations use imaginary angles. – eyeballfrog May 22 '17 at 22:23
• I feel like it's more useful to think of rotation as a codimension-2 operation, and this picture still 'works' in 4 dimensions (though in 6 dimensions things are decidedly more complicated). That is, a rotation is just the standard $\left(\begin{smallmatrix}\cos\theta & \sin\theta \\ -\sin\theta & \cos\theta\end{smallmatrix}\right)$ operation, 'conjugated' appropriately so that the invariant is a specific codimension-2 object. In three dimensions this is a line (the axis of rotation), in four dimensions it's a plane; in two it's just the origin. – Steven Stadnicki May 22 '17 at 22:28
• @RobArthan I think OP is suggesting that the point $(0,0, 1, 2)$ might rotate into $(0, 0, 2, 1)$, for instance - but omitting that fourth coordinate doesn't paint the whole picture. – Steven Stadnicki May 22 '17 at 22:31
• @ElliotG I'll say that one problem with your analogy is that it's, for want of a better phrase, 'non-isometric': a rotation no longer maintains 'distances' between points, which is arguably one of the most fundamental properties of a rotation. – Steven Stadnicki May 22 '17 at 23:04
• @Rahul I didn't really know enough about the geometry of higher-dimensional rotations to be able to say with any surety, but from reading up on SO(n) a little bit it looks like that fact (the decomposition into 'block diagonal' form with 2x2 rotation matrices along the diagonal) holds true in any even dimension (and in the odd dimensions, one just needs to add a final 1). – Steven Stadnicki May 22 '17 at 23:21

$\newcommand{\Reals}{\mathbf{R}}\newcommand{\Basis}{\mathbf{e}}$To visualize $\Reals^{4}$, I find it helpful to think of the orthogonal planes \begin{align*} \Reals^{2} \times \{0\} &= \{(x, y, 0, 0) : x, y \in \Reals\}, \\ \{0\} \times \Reals^{2} &= \{(0, 0, z, w) : z, w \in \Reals\}. \end{align*} These planes obviously intersect in a single point, the origin. It is clear how to travel from $(0, 0, 1, 0)$ to $(0, 0, -1, 0)$ without passing through the first plane: Travel along a semi-circle in the $(0, 0, z, w)$-plane, such as $(0, 0, \cos\theta, \sin\theta)$ for $0 \leq \theta \leq \pi$. That is, a plane fails to separate $\Reals^{4}$ into two pieces just as a line fails to separate $\Reals^{3}$ into two pieces.

Now, as Steven Stadnicki says, a "fundamental" rotation of $\Reals^{n}$ should be viewed as choosing some oriented plane $\Pi$, fixing its $(n - 2)$-dimensional orthogonal complement, and rotating $\Pi$ through an angle $\theta$ according to ordinary two-dimensional intuition, i.e., fixing an oriented orthonormal basis $\Basis_{1}$, $\Basis_{2}$ for $\Pi$, mapping \begin{align*} \Basis_{1} &\mapsto \phantom{-}(\cos\theta)\Basis_{1} + (\sin\theta)\Basis_{2}, \\ \Basis_{2} &\mapsto -(\sin\theta)\Basis_{1} + (\cos\theta)\Basis_{2}, \end{align*} and extending by linearity. Generally, a rotation of a finite-dimensional space $\Reals^{n}$ about a point $p$ acts by fixing some collection of mutually-orthogonal oriented planes through $p$, rotating each by some angle $\theta$, and extending by linearity.

This 6MB animation loop (too large to upload to Math.SE) shows the effect of rotation on the Riemann surface $x + iy = (z + iw)^{2}$ of the complex square root, projected into the three-dimensional Cartesian space with coordinates $(x, y, z, 0)$. The "imaginary" $(0, y, 0, w)$-plane is being rotated, so that the "real" $(x, 0, z, 0)$-plane containing the parabola is fixed pointwise.

Despite the fixed plane, the result does strongly resemble the shadow of a rotating object in ordinary space.

In four dimensions and higher, incidentally, there's a new phenomenon: Non-compact one-parameter subgroups. Rotation about an arbitrary axis in $\Reals^{3}$ returns to the identity after an angle of $2\pi$. In $\Reals^{n}$ with $n \geq 4$, rotation in a single plane has the same property, but a rotation can act on a pair of orthogonal planes with angular velocities whose ratio is irrational.

Concretely, the one-parameter group of rotations acting with unit angular speed on the $(x, y, 0 ,0)$-plane and with angular speed $\alpha$ on the $(0, 0, z, w)$-plane carries the point $(1, 0, 1, 0)$ to $$\gamma(t) = (\cos t, \sin t, \cos \alpha t, \sin \alpha t)$$ at time $t$. If $\alpha$ is irrational, then $\gamma(t) = (1, 0, 1, 0)$ if and only if $t = 0$.