Seeing the 3-Sphere

A lamp at the north pole

Take the unit sphere $S^2$, put a lamp at its north pole $N=(0,0,1)$, and follow the ray from the lamp through a point $(x,y,z)$ of the sphere until it meets the equatorial plane. That is stereographic projection,

It is very old. The oldest treatise on it to survive is Ptolemy's Planisphere from the second century, and it is the mathematics behind the flat plates of the astrolabe, which carried a map of the sky for more than a thousand years [1]. The southern hemisphere lands inside the unit circle, the equator on it, and the northern hemisphere outside. The lamp itself has no image: it becomes a single point at infinity, and the plane is the sphere with one point missing.

Two properties make the map special. It preserves angles, so a small square on the sphere stays a square in the plane, only scaled by the factor $(1+\rho^2)/2$: half size at the centre and larger without bound far out, which is why the regions near the lamp look so enlarged. And it sends every circle on the sphere to a circle in the plane, or to a straight line when the circle runs through the lamp. The parallel at angle $\theta$ from the north pole becomes a circle of radius $\cot(\theta/2)$ around the origin, and the meridians, which all pass through the lamp, become straight lines through the origin. The left pane of the demo shows the plain shadow for comparison: light from far above, each point simply dropped to $(x,y)$. There the two hemispheres land on top of each other, and tilted circles turn into ellipses.

Writing the plane as the complex numbers, $\zeta = (x+iy)/(1-z)$, makes the map even more useful. Every rotation of the sphere becomes a fraction

A turn by the angle $\alpha$ about the $x$ axis, for example, has $a=\cos(\alpha/2)$ and $b=i\sin(\alpha/2)$. Two remarks prepare the rest of the page. Fractions of the same kind, with more general coefficients, describe how the night sky looks to a traveller moving close to the speed of light, which is why a sphere flying past still looks round [2]. And the pair $(a,b)$, four real numbers with squares adding up to one, is a point of the 3-sphere. Because the angle enters halved, a full turn changes $(a,b)$ into $(-a,-b)$, the opposite point. The rotations of an ordinary sphere already live on the 3-sphere.

One dimension up

The 3-sphere is the set of points of four-dimensional space at distance one from the origin,

The natural coordinates on it are Hamilton's quaternions $q = w + x\,i + y\,j + z\,k$, with $i^2=j^2=k^2=ijk=-1$ [3]. Hamilton found these rules on a walk along the Royal Canal in Dublin in October 1843 and carved them into the stone of Broom Bridge [4]. The quaternions of length one are exactly the points of $S^3$, and each of them turns ordinary space:

a rotation by the angle $\alpha$ about the axis $\mathbf n$. The demo shows $S^3$ in two ways at once. On the left is the shadow, the analogue of the plain shadow of the sphere: the 3-sphere is first turned in four dimensions with the three sliders for the $(w,x)$, $(w,y)$ and $(w,z)$ planes, and then $w$ is dropped. The two halves $w>0$ and $w<0$ then fill the same solid ball on top of each other. On the right is the stereographic image of the same turned 3-sphere, from the pole $N=(1,0,0,0)$,

It fills all of ordinary space. The half with $w<0$ lands inside the unit ball, the equator $w=0$ on the unit sphere, and the pole $N$ goes to infinity, exactly as on the sphere one dimension down. Angles are preserved again, and circles on $S^3$ become circles or straight lines.

The red, green and blue circles are the great circles $\cos t + i\sin t$, $\cos t + j\sin t$ and $\cos t + k\sin t$, rotations by the angle $2t$ about the $x$, $y$ and $z$ axes. All three pass through the pole, so on the right they become the three coordinate axes, meeting at the origin, the image of $-1$, and again at infinity. Follow the red circle from $1$: halfway round, at $t=\pi$, the rotation has made a full turn of 360°, yet the point has reached $-1$, the opposite end of $S^3$, not $1$. Only after two full turns does the circle close. Since $q$ and $-q$ turn every vector in the same way, the difference is invisible for a rotating object, but not for a spin-½ particle, whose wavefunction is a point of this $S^3$ and changes sign after a full turn. Neutron interferometers measured that sign in 1975 [5][6].

A space woven from circles

Group the four coordinates into two complex numbers, $z_1 = w + i x$ and $z_2 = y + i z$, with $|z_1|^2+|z_2|^2=1$, and write them with three angles,

These are the three sliders of the demo, and the yellow point is $q(\mu,\xi,\eta)$. In 1931 Heinz Hopf found a map from the 3-sphere onto the ordinary sphere with a remarkable property [7]. In the demo it is $h(q)=\bar q\,i\,q$, and in the three angles it reads

It depends on $\mu$ and on the difference $\xi-\eta$ only. Slide $\xi$ and $\eta$ forward together, which multiplies $q$ by $e^{it}$, and $q$ runs around a great circle while its image on $S^2$ stays put. Every point of the ordinary sphere therefore has a whole circle of $S^3$ sitting above it, its fibre. Different points have different fibres, the fibres never meet, and together they fill the 3-sphere completely. This is the Hopf fibration. The highlighted circle in the demo is the fibre through $q$, and the readout below the sliders shows its base point $h(q)$.

Keeping $\mu$ fixed and letting $\xi$ and $\eta$ run gives a torus, and on the right it becomes a ring torus around the $x$ axis,

It lies over one circle of latitude of $S^2$, the one at height $\cos 2\mu$, which is the colour of its fibres in the demo. As $\mu$ grows the tori shrink: near $\mu=0$ a huge torus wraps tightly around the red line, at $\mu=\pi/4$ sits the Clifford torus, with radii $\sqrt2$ and 1, which cuts $S^3$ into two equal solid tori, and as $\mu\to\pi/2$ the torus thins down to the unit circle in the $yz$ plane. The fibres on each torus are tilted circles that wind once around the hole and once around the tube. A ring torus carries exactly such slanted circles, discovered by Yvon Villarceau in 1848 [8], and the Hopf fibres are precisely those.

Now look at two of them. The fibre with $\mu=0$ is the red line, which passes through infinity, and the fibre with $\mu=\pi/2$ is the unit circle in the $yz$ plane. The line runs straight through the middle of the circle: the two are linked. Every other fibre winds once around the hole of its torus and once around the tube, and with that, any two fibres of the Hopf fibration are linked exactly once, like two rings of a chain. This linking number, the Hopf invariant, is what Hopf used to prove that the map cannot be shrunk to a point, the first sign that spheres of higher dimension can wrap around lower ones in ways that cannot be undone [7].

Where physics lives on the 3-sphere

The 3-sphere and its Hopf fibration turn up all over physics [9]. Three places stand out, each more surprising than the one before.

A quantum bit

The most direct case is a single quantum bit. Its state is a pair of complex amplitudes, $|\psi\rangle = z_1|0\rangle + z_2|1\rangle$, with $|z_1|^2+|z_2|^2=1$: a point of $S^3$. Multiplying the state by an overall phase $e^{it}$ changes no prediction of quantum mechanics, so all the points of one Hopf fibre describe the same physical state, and the physical states are the points of the base sphere. That base sphere is the Bloch sphere, and the Hopf map is the Bloch vector [10]. With the amplitudes of the demo,

so the colour of each torus is the difference between the probabilities of finding the qubit in $|0\rangle$ and in $|1\rangle$, and sliding the yellow point along its fibre changes the global phase and nothing else.

The Kepler problem

A planet around the Sun moves under $H = p^2/2m - k/r$, with $k = GMm$. Besides the energy and the angular momentum $\mathbf L = \mathbf r\times\mathbf p$, it keeps a third, less famous quantity fixed, the Laplace–Runge–Lenz vector

It points from the Sun to the perihelion, the point of closest approach, and its length is $mke$, with $e$ the eccentricity of the orbit [11]. Because $\mathbf A$ never moves, the orbit closes on itself: an ellipse that comes back to the same place forever. Any small extra force sets $\mathbf A$ turning, and the ellipse slowly rotates. For Mercury the corrections of general relativity turn it by 43 arcseconds per century, the number Einstein computed in 1915 [12].

At a fixed negative energy $E$, the angular momentum and the rescaled vector $\mathbf D = \mathbf A/\sqrt{-2mE}$ close among themselves,

the Poisson brackets of the six rotations of four-dimensional space [11]. Three of them are the ordinary rotations. The other three turn an orbit into another of the same energy but a different eccentricity, a circle into a long thin ellipse, which is why all orbits of one energy share the same period whatever their shape, as Kepler's third law says. A symmetry made of four-dimensional rotations needs a 3-sphere to turn, and there is one.

Hamilton noticed in 1847 that the momentum of a body on a Kepler orbit runs around a circle, which he called the hodograph [13]. For an orbit with angular momentum $L$ and eccentricity $e$, the circle has radius $mk/L$, and its centre lies a distance $mke/L$ from the origin. Now treat momentum space as the space of the right-hand picture, measured in units of $p_0=\sqrt{-2mE}$, and run the projection of the demo backwards, $q=\sigma^{-1}(\mathbf p/p_0)$. Stereographic projection turns circles into circles, so every hodograph becomes a circle on $S^3$. It is even a great circle, because a circle becomes a great circle exactly when it cuts the unit sphere, the image of the equator, at two opposite points, and for every Kepler orbit

The Kepler orbits of one energy are the great circles of a 3-sphere, and the hidden symmetry is simply the rotation group of that sphere [14]. Moser made the picture exact: after a change of the time variable, Kepler motion becomes uniform motion along these great circles [15]. A body falling straight into the Sun has a momentum that grows without bound at the collision, so its great circle runs through the pole. The point at infinity of the demo is the moment of collision, and on the sphere it is an ordinary point that the motion passes straight through. Kustaanheimo and Stiefel reached the same goal another way. They lifted three-dimensional space to four with the Hopf map itself, stretched out radially, and the Kepler problem became a four-dimensional harmonic oscillator, which runs smoothly through every collision [16]. Checked numerically for eccentricities up to 0.99, every momentum along an orbit lands on its great circle to within $10^{-15}$. Add a third body and all of this is lost: no Runge–Lenz vector is conserved any more, and the motion turns chaotic, as the three-body page shows.

The hydrogen atom

The hydrogen atom has the same Hamiltonian, with $k = e^2/4\pi\varepsilon_0$ for the pull between electron and proton, and the hidden symmetry survives quantum mechanics. The energy levels depend on the principal quantum number $n$ alone, and level $n$ holds $n^2$ states, more than ordinary rotations can explain. In early 1926 Pauli derived the spectrum from the quantum version of the Runge–Lenz vector, before wave mechanics had solved it [17]. The combinations $\mathbf J_\pm = (\mathbf L\pm\mathbf D)/2$ behave like two independent angular momenta of the same size $j$, because $\mathbf L\cdot\mathbf A = 0$, so the states of one level come in $(2j+1)^2$ copies, and the algebra fixes the energy,

Nine years later Fock found the 3-sphere behind the algebra [18]. He applied the same map as above, momentum space onto a 3-sphere at the scale $\sqrt{-2mE}$, to the wavefunction in momentum space, and the Schrödinger equation became the equation for the spherical harmonics of $S^3$. The $n^2$ states of level $n$ are the $n^2$ harmonics of degree $n-1$, and Bargmann showed the following year that Pauli's algebra and Fock's sphere are one and the same symmetry [19]. The orbits of the planets and the levels of hydrogen are two views of one rotating 3-sphere.

Real hydrogen breaks the symmetry a little, just as real planets do. Relativity and the spin of the electron split each level according to its total angular momentum, and in 1947 Lamb and Retherford found that the 2s₁/₂ and 2p₁/₂ states, which Dirac's theory puts at exactly the same energy, differ by about a gigahertz [20]. Like the slow turning of Mercury's orbit, the small splitting measures how far nature departs from a perfect $1/r$ force.

Inside the code

The projection itself is a single division. The only care needed is at the pole, which has no image, and near it, where images run off far beyond the screen:

// Stereographic image of a point u of S³ (after the 4D view rotation).
// At the pole there is no image: that point has gone to infinity.
function stereoInto(u, out, o) {
  const d = 1 - pole * u[0];               // 1 − w from N, 1 + w from S
  if (!(d > POLE_EPS)) return false;
  const x = u[1] / d, y = u[2] / d, z = u[3] / d;
  if (x * x + y * y + z * z > CLIP * CLIP) return false;
  out[o] = x; out[o + 1] = y; out[o + 2] = z;
  return true;
}

Every curve, a great circle, a fibre or a coordinate circle, is evaluated once per point and drawn in both panes. A circle through the pole has one point at infinity, and connecting the samples on either side of it would draw a false chord across the picture, so such segments are skipped:

// One evaluation of each curve feeds both pictures. On the right a segment
// is drawn only if both of its ends have an image, so a circle through the
// pole opens into a line instead of a chord across the whole picture.
for (let k = 0; k <= segs; k++) {
  param((2 * Math.PI * k) / segs, P);       // a point of S³
  const u = view(P);                        // turned in four dimensions
  shadow[o] = u[1]; shadow[o + 1] = u[2]; shadow[o + 2] = u[3];  // drop w
  const ok = stereoInto(u, stereo, o);
  if (k > 0) {
    left.segment(shadow, prev, shadow, o);
    if (ok && okBefore) right.segment(stereo, prev, stereo, o);
  }
  okBefore = ok;
}

The fibres are drawn straight from their definition, as the circles along which $\xi$ and $\eta$ advance together:

// The Hopf fibres on the torus μ: ξ runs once around and η = ξ + d follows,
// one circle for each offset d. Each is the orbit e^{it} q of a point q.
for (let j = 0; j < perTorus; j++) {
  const d = (2 * Math.PI * j) / perTorus;
  family.curve((t, P) => {
    P[0] = c * Math.cos(t);      P[1] = c * Math.sin(t);       // z1 = cos μ e^{iξ}
    P[2] = s * Math.cos(t + d);  P[3] = s * Math.sin(t + d);   // z2 = sin μ e^{iη}
  }, segs, color);
}

The option Random points on $S^3$ scatters up to 20,000 points uniformly over the 3-sphere, using the fact that a vector of independent Gaussians points in a uniformly random direction:

// Uniform random points on S³: four independent Gaussians, normalized.
// The Gaussian looks the same in every direction, so the direction is
// uniform. Normalizing points of a cube would crowd them toward its corners.
for (let k = 0; k < 4; k++) { v[k] = randn(); len += v[k] * v[k]; }
len = Math.sqrt(len);
for (let k = 0; k < 4; k++) out[4 * i + k] = v[k] / len;

The cloud is a good check on the geometry. On the right exactly half of the points land inside the unit ball, the image of the half $w<0$. On the left they crowd toward the rim of the ball, where the 3-sphere is seen edge on: 54% of them fall in the outer shell beyond radius 0.9, which holds only 27% of the volume. Both numbers are checked in the tests of the site.

Try it yourself

Both demos open full screen with the buttons above, and each has a guided tour that sets the scene for every step. On the sphere, drag θ toward 0 and watch the point race off to infinity as it nears the lamp, then switch the projection to the south pole and see the inside and the outside of the unit circle swap. Tilt the sphere with the (z, x) slider: in the shadow the circles squash into ellipses, while on the right they stay round.

On the 3-sphere, change ξ and then η by the same amount. The yellow point slides along its highlighted fibre, and the base point $h(q)$ below the sliders does not change. Drag μ to 0 and the fibre through the point straightens into the red line, drag it to π/2 and it becomes the unit circle in the yz plane, and at π/4 it lies on the Clifford torus, the middle one of the three tori drawn. Then press Run with the (w, x) rotation ticked. The whole picture turns in four dimensions, and the tori stream out through the point at infinity and come back from the other side, something no rotation of ordinary space could do. Finally, add 20,000 random points and compare the two panes, the dense core fading outward on the right and the bright rim on the left.

References

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