We are used to rotations having an axis. A top spins about its axis, the Earth about its own, a door about its hinges. The habit is so strong that “what axis does a tesseract rotate about?” sounds like a sensible question. It has no answer: in four dimensions rotation works differently. To see why, it helps to notice that even in our world the axis is not the point.
Rotation in the plane
Start with two dimensions. Rotating the plane by an angle $\theta$ about the origin sends a point $(x, y)$ to $$x' = x\cos\theta - y\sin\theta,\qquad y' = x\sin\theta + y\cos\theta.$$ As a matrix: $$R(\theta) = \begin{pmatrix}\cos\theta & -\sin\theta\\ \sin\theta & \cos\theta\end{pmatrix}.$$ A rotation keeps distances and does not flip the plane over: the determinant is $\cos^2\theta + \sin^2\theta = 1$. One point stays put, the centre. Rotations of the plane compose in the simplest way: turning by $\alpha$ and then by $\beta$ is the same as turning by $\alpha + \beta$, in either order.
The axis is whatever stays put
In three dimensions a rotation “about the $z$ axis” is written $$\begin{pmatrix}\cos\theta & -\sin\theta & 0\\ \sin\theta & \cos\theta & 0\\ 0 & 0 & 1\end{pmatrix}.$$ Look at what actually happens. The coordinates $x$ and $y$ mix exactly as in the plane, and $z$ does not change. What moves is the $XY$ plane: it turns within itself by $\theta$. The $z$ axis is simply everything perpendicular to that plane, which is why it stays where it is.
In three dimensions a plane and the line perpendicular to it determine each other, so “about an axis” is a convenient and harmless way to talk. In 1775 Leonhard Euler proved that every rotation of space about a point has a fixed axis. The same coincidence lets us describe rotation by a vector: angular velocity points along the axis. But this is a coincidence of three-dimensional space, not a law of nature. The right object is the plane of rotation, the plane in which points travel along circles.
In four dimensions a whole plane stays put
Add a coordinate $w$ and rotate the $XW$ plane: $$R_{XW}(\theta) = \begin{pmatrix}\cos\theta & 0 & 0 & -\sin\theta\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ \sin\theta & 0 & 0 & \cos\theta\end{pmatrix}.$$ Now $x$ and $w$ mix, while $y$ and $z$ do not change. So what stays in place is not a line but the entire $YZ$ plane. The pattern is clear: a rotation in one plane fixes a subspace of dimension $n - 2$. In the plane that is a point, in space a line, in 4D a plane. There is no axis of rotation in 4D; what stays put is two-dimensional.
How many coordinate planes are there? A plane is picked by a pair of axes, so there are $$\binom{n}{2} = \frac{n(n-1)}{2}:$$ one in the plane, three in space and six in 4D: $XY$, $XZ$, $XW$, $YZ$, $YW$, $ZW$. That is the number of independent “knobs” for turning a four-dimensional object, the dimension of the rotation group $\mathrm{SO}(4)$. In three dimensions there are three planes and three axes, and this second coincidence is what the language of axes rests on. In 4D there are six planes and only four axes, so there is nothing to build an angular-velocity vector from. Rotation is described by a skew-symmetric $4\times4$ matrix with six independent entries; physicists call such an object a bivector.
Press “Reset” and move only $YZ$: this is an ordinary 3D rotation, and the “cube within a cube” simply turns as a whole. Now move only $XW$: the inner cube grows, the outer one shrinks, and only the four corner entries of the matrix change. The white dot is always the same vertex, $(\tfrac12, \tfrac12, \tfrac12, \tfrac12)$; keep an eye on it.
Rotation matrices and their order
The matrix of a rotation in the plane of axes $i$ and $j$ is the identity except for four entries: where rows and columns $i$ and $j$ cross stand $\cos\theta$, $-\sin\theta$, $\sin\theta$, $\cos\theta$. In numerical linear algebra these are called Givens rotations. Doing two rotations in a row means multiplying their matrices; if $A$ comes first and $B$ second, the result is $BA$.
Matrix multiplication is not commutative in general, and rotations notice. Take $90^\circ$ rotations in the $XW$ and $XY$ planes and follow the unit vector $e_x$:
- $XW$ first, then $XY$: the $XW$ turn sends $e_x$ to $e_w$, and $XY$ leaves $w$ alone. Result: $e_w$;
- $XY$ first, then $XW$: $XY$ sends $e_x$ to $e_y$, and $XW$ leaves $y$ alone. Result: $e_y$.
Different answers, so the order matters. Rotations in $XW$ and $YZ$, on the other hand, commute: the first touches only $x$ and $w$, the second only $y$ and $z$, and their matrices are made of disjoint blocks. The general rule: rotations in planes with no common axis commute; rotations in planes that share an axis usually do not. In three dimensions any two coordinate planes share an axis, so there are no commuting pairs at all. In 4D there are three: $XY$ and $ZW$, $XZ$ and $YW$, $XW$ and $YZ$. Such planes are called completely orthogonal: every vector of one is perpendicular to every vector of the other, and they meet in a single point, the origin.
Incidentally, the composition of the two $90^\circ$ turns in $XW$ and $XY$ is a simple rotation by $120^\circ$. Neither turn touches $z$, so everything happens inside the three-dimensional space $XYW$, where this is a familiar fact: two quarter turns about perpendicular axes make a third of a turn about a diagonal.
Switch the widget to “XW and YZ”: the white tesseract ($XW$ first) and the pink one ($YZ$ first) coincide at every angle, and so do their matrices. In “XW and XY” they come apart, and the matrix entries that differ light up pink.
Canonical form: two planes and two angles
Six angles describe a rotation redundantly: different sets of angles give the same matrix, and they say little about what the rotation looks like. There is a shorter and clearer description; the widget shows it in the line “Canonical angles”.
For every rotation $A$ of four-dimensional space (an orthogonal matrix with determinant $1$) there are two completely orthogonal planes $P$ and $Q$ and angles $\alpha$, $\beta$ such that $A$ turns $P$ within itself by $\alpha$ and $Q$ by $\beta$. In an orthonormal basis made of two vectors of $P$ and two of $Q$, $$A = \begin{pmatrix}\cos\alpha & -\sin\alpha & 0 & 0\\ \sin\alpha & \cos\alpha & 0 & 0\\ 0 & 0 & \cos\beta & -\sin\beta\\ 0 & 0 & \sin\beta & \cos\beta\end{pmatrix}.$$
The characteristic polynomial of $A$ has degree 4 and real coefficients, so its non-real roots come in pairs $\lambda$, $\bar\lambda$. Every root has modulus 1: if $Az = \lambda z$ for a complex vector $z \ne 0$, then $|\lambda|^2\,\bar z^{\mathsf T} z = \overline{(Az)}^{\mathsf T}(Az) = \bar z^{\mathsf T}A^{\mathsf T}\!A\,z = \bar z^{\mathsf T} z$.
Step 1: a non-real root gives an invariant plane. Let $\lambda = \cos\alpha + i\sin\alpha$ with $\sin\alpha \ne 0$, and $z = u + iv$ with real vectors $u, v$. Comparing real and imaginary parts of $Az = \lambda z$ gives $Au = u\cos\alpha - v\sin\alpha$ and $Av = u\sin\alpha + v\cos\alpha$: the plane $P$ spanned by $u$ and $v$ is mapped to itself. Moreover $u \perp v$ and $|u| = |v|$. Indeed, for the unconjugated expression $z^{\mathsf T}z$ we have $z^{\mathsf T}z = (Az)^{\mathsf T}(Az) = \lambda^2 z^{\mathsf T}z$, and $\lambda^2 \ne 1$, so $z^{\mathsf T}z = |u|^2 - |v|^2 + 2i\,u\cdot v = 0$. Normalising $u$ and $v$ gives an orthonormal basis of $P$ in which $A$ acts as the rotation by $\alpha$ (up to the order of the basis vectors).
Step 2: the complement is invariant too. Let $Q = P^\perp$. If $x \in Q$ and $p \in P$, then $p = Ap'$ for some $p' \in P$ ($A$ is invertible on $P$), and $Ax\cdot p = Ax\cdot Ap' = x\cdot p' = 0$. So $A$ maps $Q$ to $Q$ and acts on it orthogonally. The determinant of $A$ is the product of its determinants on $P$ and on $Q$; on $P$ it is 1, hence on $Q$ it is 1 as well, so on $Q$ the map is a rotation by some angle $\beta$.
Step 3: all roots are real. Then they are $\pm 1$. Take a real unit eigenvector; its orthogonal complement is invariant for the same reason as in step 2, and we pass to a three-dimensional space, which again contains a real eigenvector. This produces an orthonormal basis of eigenvectors with eigenvalues $\pm 1$. Their product is $\det A = 1$, so $-1$ occurs an even number of times. A pair of vectors with $-1$ spans a plane turned by $180^\circ$, a pair with $+1$ a plane turned by $0^\circ$. ∎
A handy consequence: the trace (the sum of the diagonal entries) does not depend on the basis, so $\operatorname{tr}A = 2\cos\alpha + 2\cos\beta$. Together with the trace of $A^2$, which is $2\cos 2\alpha + 2\cos 2\beta$, this yields both angles, and that is how the widget finds them from the matrix. For example, the composition of $60^\circ$ turns in $XW$ and $XY$ has trace $2.25$ and $\beta = 0$, so $\cos\alpha = 1/8$ and $\alpha \approx 82.8^\circ$: again a simple rotation, though no longer in a coordinate plane.
The theorem sorts all rotations of 4D into three kinds.
- A simple rotation: $\beta = 0$. The plane $Q$ is fixed point by point while $P$ turns. This is the direct analogue of a 3D rotation about an axis, except that the “axis” is two-dimensional.
- A double rotation: $\alpha$ and $\beta$ are both non-zero. Only one point stays put, the centre. This never happens in three dimensions, where there is always an axis.
- An isoclinic rotation: $|\alpha| = |\beta|$. The most symmetric case, and the most surprising.
Double rotation and Clifford parallels
Take a point of the unit three-sphere $S^3$ and write it through its projections onto $P$ and $Q$: $$p = (\cos\eta\cos\varphi,\ \cos\eta\sin\varphi,\ \sin\eta\cos\psi,\ \sin\eta\sin\psi).$$ A rotation with angles $\alpha$ and $\beta$ adds $\alpha$ to $\varphi$ and $\beta$ to $\psi$ and leaves $\eta$ alone. So the point stays on the surface where $\eta$ is constant. That surface is a torus, the product of two circles of radii $\cos\eta$ and $\sin\eta$, lying in the three-sphere (for $\eta = 45^\circ$ it is called the Clifford torus). If $\beta/\alpha$ is rational, the point's path closes up into a knot on the torus; if it is irrational, the path winds around the torus forever without closing and comes arbitrarily close to every point of it.
By what angle $\delta$ does a point move in one turn? The dot product of $p$ with its image is $$\cos\delta = \cos^2\eta\,\cos\alpha + \sin^2\eta\,\cos\beta,$$ so the displacement lies between $\alpha$ (for points of $P$) and $\beta$ (for points of $Q$). When $\alpha = \beta$ the formula gives $\cos\delta = \cos\alpha$ for every $\eta$: every point of the sphere moves by the same angle. Hence the name: “isoclinic” means “equally inclined”.
Each point then travels along a great circle, a “straight line” of spherical geometry. There are no longer two invariant planes but infinitely many: the plane spanned by any point $p$ and its image is mapped to itself. The whole three-sphere falls apart into great circles, the paths of the points, and the distance from a point of one of them to another is the same wherever you measure it. On an ordinary sphere any two great circles meet, and there are no parallel “lines”. On $S^3$ there are: they are called Clifford parallels, after William Kingdon Clifford, who described them in 1873. What is more, any two of these circles are linked, like the links of a chain. It is the same decomposition of the sphere that underlies the Hopf fibration, which we meet in the chapter on the hypersphere.
The widget shows the paths of a dozen points of $S^3$ in stereographic projection: the sphere is mapped into ordinary space from the “north pole” $(0,0,0,1)$, and circles go to circles (or to straight lines if they pass through the pole). The blue ring is the great circle in the $XY$ plane; the orange line is the circle in the $ZW$ plane, which passes through the pole and has therefore become the $z$ axis.
In “Simple” mode the paths are horizontal rings around the $z$ axis, none threaded through another, and the $ZW$ circle does not move. In “Isoclinic” every path is a circle and any two are linked. Set the $\beta : \alpha$ slider to $0.5$: the paths become knots wrapped around tori. The line under the widget says how far the points moved while the $XY$ plane turned by $60^\circ$.
An isoclinic rotation “combs” the three-sphere: at every point the velocity is non-zero and tangent to the sphere. This is impossible for the ordinary sphere: any rotation of $S^2$ fixes two poles, and indeed every continuous tangent field on $S^2$ vanishes somewhere (the hairy ball theorem). Spheres of odd dimension, like $S^1$ and $S^3$, have no such obstruction.
Quaternions: rotation as multiplication
On 16 October 1843 William Rowan Hamilton, walking past Broom Bridge in Dublin, saw how to multiply quadruples of numbers, and cut into a stone of the bridge the formula $$i^2 = j^2 = k^2 = ijk = -1.$$ A quaternion $q = a + bi + cj + dk$ is a point $(a, b, c, d)$ of four-dimensional space. The key property of the product is that moduli multiply: $|pq| = |p|\,|q|$. Quaternions of modulus 1 form the three-sphere $S^3$, and multiplying by one of them preserves lengths — that is, it is a rotation of $\R^4$.
Multiplying on the left by $l = \cos\theta + u\sin\theta$, where $u$ is a unit pure imaginary quaternion, moves every point by the same angle: the dot product of $p$ and $lp$ is $\operatorname{Re}(\bar p\, l\, p) = |p|^2\cos\theta$. This is an isoclinic rotation; call it a left one. Multiplying on the right gives right isoclinic rotations. In coordinates, left multiplication by $\cos\theta + i\sin\theta$ turns $XY$ and $ZW$ by $\theta$ in the same sense ($\alpha = \beta$), right multiplication in opposite senses ($\alpha = -\beta$): two mirror-image varieties.
Quaternions are better known for rotations in three dimensions. The imaginary quaternions $bi + cj + dk$ form our space, and the map $v \mapsto q\,v\,\bar q$ with $q = \cos\tfrac{\theta}{2} + u\sin\tfrac{\theta}{2}$ rotates it by $\theta$ about the axis $u$. Hamilton and, independently, Arthur Cayley published this formula in the mid-1840s. The half angle is no accident: $q$ and $-q$ give the same rotation, so the sphere $S^3$ covers the rotation group $\mathrm{SO}(3)$ twice. Quaternions rotate objects in computer graphics and keep track of spacecraft attitude.
Every rotation of $\R^4$ has the form $p \mapsto l\,p\,\bar r$ for unit quaternions $l$ and $r$, and the pair $(l, r)$ is unique up to a common sign. Therefore $$\mathrm{SO}(4) \cong (S^3 \times S^3)/\{\pm(1, 1)\}.$$
The map $p \mapsto l p \bar r$ preserves lengths, and its determinant is 1: it varies continuously, can only be $\pm 1$, and the pair $(l, r)$ can be moved continuously to $(1, 1)$. Conversely, let $A$ be a rotation and $a = A(1)$, a unit quaternion. The map $B(p) = \bar a\,A(p)$ is also a rotation, and it fixes $1$. So $B$ maps the space of imaginary quaternions, perpendicular to $1$, to itself and is a three-dimensional rotation there. By the 3D theorem $B(p) = q p \bar q$ for some unit $q$ (and $q\cdot 1\cdot\bar q = 1$, so the formula holds on all of $\R^4$). Hence $A(p) = a q\, p\, \bar q$, that is $l = aq$, $r = q$. Uniqueness: if $l p \bar r = p$ for all $p$, then $p = 1$ gives $l = r$, and $l p \bar l = p$ for all $p$ means that $l$ commutes with every quaternion, so it is real: $l = \pm 1$. ∎
The formula $p \mapsto l\,p\,\bar r$ is a left isoclinic rotation followed by a right one. The two commute because quaternion multiplication is associative: $(l p)\bar r = l(p \bar r)$. So every rotation of 4D is the product of two commuting isoclinic rotations. The angles add up too: if $l$ moves points by $\alpha'$ and $r$ by $\beta'$, the two invariant planes turn by $\alpha' + \beta'$ and $|\alpha' - \beta'|$. Cayley found this representation in 1854–1855. On the showcase the “flow” rotation works exactly this way: the object is multiplied on the right by a quaternion, and each vertex drifts along its own Hopf circle.
Does the tesseract turn inside out?
The most famous animation of the fourth dimension is a rotating tesseract. The small cube inside grows, passes through the walls of the big one and becomes the outside, while the outer cube shrinks and ends up inside. It looks as if the figure turns inside out and gets squashed, its edges changing length.
In fact the tesseract is rigid: a rotation in 4D keeps every edge at its length. Only the picture is distorted. It is drawn by perspective projection from a four-dimensional “eye” on the $w$ axis at distance $d$ from the centre: a point $(x, y, z, w)$ lands at $$X = \frac{(x,\, y,\, z)}{1 - w/d}.$$ The cell at $w = -\tfrac12$ is further from the eye and appears as the small cube inside; the cell at $w = +\tfrac12$ is nearer and appears as the big cube outside. A rotation in the $XW$ plane changes the points' $w$ coordinate: after half a turn the far cell has become the near one, and a quarter of the way through it faces the eye sideways and shows as a truncated pyramid. The centre of the highlighted cell moves along the circle $\bigl(\tfrac12\sin\theta,\ 0,\ 0,\ -\tfrac12\cos\theta\bigr)$, and its on-screen scale $1/(1 - w/d)$ grows smoothly from the smaller to the larger value.
The same happens to an ordinary cube drawn in “square within a square” perspective, looking along an axis. Turn it about a vertical axis and the back face comes forward, passing through the position of a side face; in the drawing the small square “turns inside out” into the big one. Nobody thinks the cube gets squashed: we know how perspective works. The tesseract is exactly the same, except that depth is the fourth coordinate.
Pause the animation and slowly move the angle from $0^\circ$ to $180^\circ$. In the “side view” inset the square is the tesseract seen in the $XW$ plane (the other two coordinates collapsed to a point), the orange side is the highlighted cell, and the eye is at the top. Nothing turns inside out: the side simply travels around the centre and ends up closer to the eye.
Rotation happens in a plane, not about an axis. 4D has six coordinate planes, and every rotation comes down to two completely orthogonal planes turned by $\alpha$ and $\beta$. When the angles are equal, all points of the sphere move alike, along linked great circles.
Summary
- A rotation turns a plane; the “axis” in three dimensions is just what is perpendicular to that plane and stays put. In 4D a rotation in one plane leaves a whole plane fixed.
- There are $n(n-1)/2$ coordinate planes in $n$ dimensions: six in 4D, and the rotation group $\mathrm{SO}(4)$ is six-dimensional.
- Rotations in planes with no common axis commute; with a common axis they usually do not: after $90^\circ$ turns in $XW$ and $XY$ the vector $e_x$ ends up at $e_w$ or at $e_y$ depending on the order.
- Every rotation of 4D turns two completely orthogonal planes by angles $\alpha$ and $\beta$. Simple: $\beta = 0$; double: both angles non-zero; isoclinic: $|\alpha| = |\beta|$.
- In an isoclinic rotation all points of the sphere move by the same angle along great circles — Clifford parallels — and any two of these circles are linked.
- Quaternions write every rotation of 4D as $p \mapsto l p \bar r$, a product of two commuting isoclinic rotations; $\mathrm{SO}(4) \cong (S^3 \times S^3)/\{\pm 1\}$.
- The tesseract “turning inside out” is a perspective effect: a rotation in $XW$ brings the far cell near.
Next come cross-sections: what a three-dimensional being sees when a four-dimensional body passes through its space.
Sources
- E. B. Vinberg. A Course in Algebra. AMS, 2003 — the canonical form of an orthogonal operator.
- J. H. Conway, D. A. Smith. On Quaternions and Octonions. A K Peters, 2003 — quaternions and rotations in three and four dimensions.
- Rotations in 4-dimensional Euclidean space — Wikipedia: simple, double and isoclinic rotations; left and right isoclinic rotations commute.
- Euler's rotation theorem — Wikipedia: Euler's paper, presented to the St Petersburg Academy on 9 October 1775 and published in 1776.
- History of quaternions — Wikipedia: 16 October 1843, Broom Bridge.
- Lorentz transformation (Quaternions) — Wikiversity: excerpts from and references to Hamilton (1844–1845) and Cayley (Phil. Mag., 1845 and 1854; J. reine angew. Math. 50, 1855).
- Clifford parallel — Wikipedia: Clifford parallels (1873) and their link to isoclinic rotations.