The Hopf fibration is a map from the three-sphere onto the two-sphere whose preimages are circles, written $S^1 \hookrightarrow S^3 \xrightarrow{\ h\ } S^2$. Every point of $S^2$ pulls back to a full circle in $S^3$, no two of those circles meet, and together they fill $S^3$ exactly once. It matters here because it is not an abstraction invented for its own sake: a single qubit state is a point of $S^3$, global phase is the circle, and the Bloch sphere is what is left over.
A fiber bundle is a space that looks locally like a product but need not be one globally. Three pieces: a total space $E$, a base $B$, and a fiber $F$, with a projection $\pi : E \to B$ such that every point of $B$ has a neighbourhood $U$ where $\pi^{-1}(U)$ looks like $U \times F$.
The familiar case is the cylinder, $S^1 \times [0,1]$, which is a bundle over $S^1$ with fiber $[0,1]$ and is a product globally. The Möbius band is the same base and the same fiber and is not a product, which is the whole point of the definition: local triviality is cheap, global triviality is not. Draw both as a square whose left and right edges get glued, with the fibers as the vertical lines, and the entire difference is which way round the gluing goes.
The Hopf fibration is the same story one dimension up, and it is the interesting case in exactly the same way the Möbius band is: $S^3$ is not $S^2 \times S^1$, though every small patch of it is.
Write $S^3$ as the unit sphere in $\CC^2$ rather than in $\RR^4$, which costs nothing and buys everything:
$$S^3 = \left\{ (\alpha, \beta) \in \CC^2 \ :\ |\alpha|^2 + |\beta|^2 = 1 \right\}$$
The Hopf map sends such a pair to a point of $\RR^3$:
$$h(\alpha,\beta) = \Big(\, 2\,\Re(\bar\alpha\beta),\ \ 2\,\Im(\bar\alpha\beta),\ \ |\alpha|^2 - |\beta|^2 \,\Big)$$
That the image lands on $S^2$ is a one-line check. With $u = 2\bar\alpha\beta$ and $v = |\alpha|^2-|\beta|^2$,
$$|u|^2 + v^2 = 4|\alpha|^2|\beta|^2 + \big(|\alpha|^2-|\beta|^2\big)^2 = \big(|\alpha|^2+|\beta|^2\big)^2 = 1$$
In real coordinates $\alpha = x_1 + ix_2$, $\beta = x_3 + ix_4$ it is four quadratics:
$$h(x) = \big(\,2(x_1x_3 + x_2x_4),\ \ 2(x_1x_4 - x_2x_3),\ \ x_1^2 + x_2^2 - x_3^2 - x_4^2\,\big)$$
Multiply both coordinates by the same unit complex number $e^{i\chi}$. Then $\bar\alpha\beta$ is unchanged, because $\overline{e^{i\chi}\alpha}\,e^{i\chi}\beta = e^{-i\chi}e^{i\chi}\bar\alpha\beta = \bar\alpha\beta$, and the moduli are unchanged too. So
$$h(e^{i\chi}\alpha,\ e^{i\chi}\beta) = h(\alpha,\beta) \quad \text{for every } \chi$$
The set $\{(e^{i\chi}\alpha, e^{i\chi}\beta) : \chi \in [0,2\pi)\}$ is a circle in $S^3$, and it is exactly one fiber. The converse takes slightly more work and is also true: if $h(\alpha,\beta) = h(\alpha',\beta')$ then the two differ by a common phase. So the fibers are precisely the orbits of the $U(1)$ action, and
$$S^3 / U(1) \;\cong\; S^2$$
This is the group-theoretic statement of the same fact, and it is worth carrying: $S^3$ is the group $SU(2)$, the fiber is its $U(1)$ subgroup, and the base is the coset space. The unit quaternions are the same object again, with the fiber the circle of unit complex numbers sitting inside them.
Parametrise a point of $S^3$ by three angles:
$$\alpha = \cos\tfrac{\theta}{2}\,e^{i\chi}, \qquad \beta = \sin\tfrac{\theta}{2}\,e^{i(\chi + \varphi)}$$
with $\theta \in [0,\pi]$, and $\varphi, \chi \in [0, 2\pi)$. Feeding this through $h$ gives
$$h = \big(\sin\theta\cos\varphi,\ \sin\theta\sin\varphi,\ \cos\theta\big)$$
which is ordinary spherical coordinates on $S^2$. The base is $(\theta, \varphi)$ and $\chi$ is the fiber coordinate: it moves the point along its circle and never shows up downstairs. Anyone who has written $\lvert\psi\rangle = \cos\frac{\theta}{2}\lvert 0\rangle + e^{i\varphi}\sin\frac{\theta}{2}\lvert 1\rangle$ has already used these coordinates, with $\chi$ silently set to zero.
The two letters are worth keeping apart, since this article is about the difference between them: $\chi$ moves within a fiber and is invisible, $\varphi$ moves the base point and is not.
$S^3$ does not fit in $\RR^3$, so the standard move is stereographic projection from one point. Remove the pole $(0,0,0,1)$ and send
$$\sigma(x_1,x_2,x_3,x_4) = \frac{1}{1 - x_4}\,(x_1,\ x_2,\ x_3)$$
Everything except the removed pole lands somewhere in $\RR^3$, circles stay circles, and angles are preserved. The one cost is that a circle through the removed pole becomes a straight line, which is a circle through infinity if that is a comfortable thing to say and an annoying special case if it is not.
Two fibers are worth knowing by heart, because they anchor every picture below. Take the projection point to be on the fiber over the south pole. Then:
A line threading a circle. They are linked, once, and they cannot be pulled apart.
Each base point $(\theta,\varphi)$ gives a circle, and here it is. Press play to set it moving, and drag to turn the picture, which works whether it is running or not.
Every sketch on this page carries controls="play", so they start paused. Five of them looping at sixty frames a second is five cores' worth of nothing much, and most of what there is to see holds still.
<p5sketch width="440" height="360" controls="play"> // a point of S^3 on the fiber over (theta, phi), then projected to R^3 function fiber(theta, phi, chi) { const c = cos(theta / 2), s = sin(theta / 2); const x1 = c * cos(chi), x2 = c * sin(chi); const x3 = s * cos(chi + phi), x4 = s * sin(chi + phi); const k = R / (1 - x4); return [k * x1, k * x2, k * x3]; } </p5sketch>
The grey circle and the grey line are the two anchor fibers. The orange one belongs to a base point at fixed latitude $\theta = 1.1$ going once around in $\varphi$, and it threads the grey circle every time.
This is the part that makes the Hopf fibration more than a bookkeeping device. Take two distinct points of $S^2$, however close together. Their fibers are disjoint circles in $S^3$, and after projection they are two circles in $\RR^3$ with linking number one. Not merely tangled: they cannot be separated without cutting.
Two fibers, side by side. Turn it, or press play and let it turn itself, and the linking shows from every angle:
The two circles never touch, they are the preimages of two different points, and no amount of rotating finds an angle where they come apart.
Fix $\theta$ and let $\varphi$ run all the way around. That is a circle of latitude on $S^2$, and its preimage is a circle of fibers, which is a torus in $S^3$. Projected into $\RR^3$ it is an ordinary torus of revolution, and the fibers lie on it as Villarceau circles: neither the tube circles nor the circles of revolution, but the diagonal ones that wind once around each way.
Two facts fall out of the picture. Every one of those circles is linked with every other, so a torus here is not a bundle of parallel curves but a fully interlocked family. And the two anchor fibers from earlier are the degenerate ends: as $\theta \to 0$ the torus collapses onto the unit circle, and as $\theta \to \pi$ it collapses onto the $z$-axis.
Stack several latitudes and the tori nest, one inside the next, filling $\RR^3$ and therefore filling $S^3$.
Notice what is not happening: no fiber crosses another, and none is redundant. Every point of $\RR^3$ in the picture lies on exactly one circle. That is the fibration.
Now the reason this sits in a wiki about quantum computing rather than one about topology.
A pure single-qubit state is $\lvert\psi\rangle = \alpha\lvert 0\rangle + \beta\lvert 1\rangle$ with $\alpha,\beta \in \CC$ and $|\alpha|^2+|\beta|^2 = 1$. That is not like a point of $S^3$. It is a point of $S^3$, the same equation written with different letters.
Global phase is unobservable: $\lvert\psi\rangle$ and $e^{i\chi}\lvert\psi\rangle$ give identical probabilities for every measurement, because every expectation value is a sandwich $\langle\psi\rvert A \lvert\psi\rangle$ and the two phases cancel. So the physically distinct states are not the points of $S^3$ but the circles, and the space of them is $S^3/U(1) = S^2$. The Bloch sphere is the base space of the Hopf fibration, and nothing else.
The Bloch vector is the Hopf map spelled out in physics notation:
$$\vec r = \big(\langle X\rangle, \langle Y\rangle, \langle Z\rangle\big) = \big(2\Re(\bar\alpha\beta),\ 2\Im(\bar\alpha\beta),\ |\alpha|^2 - |\beta|^2\big) = h(\alpha,\beta)$$
Which is to say the three Pauli expectation values are the three components of $h$. The correspondence is not an analogy:
| Fibration | Qubit |
|---|---|
| total space $S^3$ | normalised state vector $(\alpha,\beta)$ |
| fiber $S^1$ | global phase $e^{i\chi}$ |
| base $S^2$ | Bloch sphere, the observable state |
| projection $h$ | $\vec r = (\langle X\rangle,\langle Y\rangle,\langle Z\rangle)$ |
| $\theta,\varphi$ | polar and azimuthal Bloch angles |
| structure group $U(1)$ | phase freedom |
A global phase gate slides the state along its own fiber and leaves the Bloch vector exactly where it was. Below, the left panel is the fiber circle with the state travelling around it, and the right panel is the Bloch vector that results. Press play: the left panel moves and the right one does not.
The same statement in a circuit. A phase applied to the whole register is not a gate anyone can detect:
A relative phase is a different animal entirely. It moves $\varphi$, which is a base coordinate, so it moves the Bloch vector and is measurable. That is the difference between $S\lvert\psi\rangle$ mattering and $e^{i\chi}\lvert\psi\rangle$ not mattering, and the fibration is where the difference lives: one is motion in the base, the other is motion in the fiber.
$S^3 \ne S^2 \times S^1$, and the linking is the reason. In a product $S^2 \times S^1$ the fibers $\{p\}\times S^1$ can be pulled apart: pick two points of $S^2$, connect them by nothing at all, and the two circles sit in separate slices, unlinked. In $S^3$ every pair of fibers is linked once, so no such splitting exists.
The invariant that detects this is the Hopf invariant, and it is why the map is famous beyond geometry. Before it, $\pi_n(S^m)$ for $n > m$ was expected to vanish. The Hopf map is an element of
$$\pi_3(S^2) \cong \ZZ$$
of infinite order, generating the group, and the linking number of two fibers is precisely the invariant that measures it. A map from a higher sphere onto a lower one that cannot be contracted to a point was a genuine surprise in 1931 and is the origin of a large amount of what followed.
Fibrations of spheres by spheres of this kind exist for each of the four normed division algebras and, by a theorem of Adams, nowhere else:
| Algebra | Fibration | Dimensions |
|---|---|---|
| $\RR$ | $S^0 \to S^1 \to S^1$ | $0, 1, 1$ |
| $\CC$ | $S^1 \to S^3 \to S^2$ | $1, 3, 2$ |
| $\mathbb{H}$ | $S^3 \to S^7 \to S^4$ | $3, 7, 4$ |
| $\mathbb{O}$ | $S^7 \to S^{15} \to S^8$ | $7, 15, 8$ |
The complex row is the one this page is about. The real row is the double cover of the circle by itself and is not very interesting.
The quaternionic one, $S^3 \to S^7 \to S^4$, is the two-qubit story in the same way the complex one is the one-qubit story, and it is where entanglement enters the picture.
The mark is a fiber, is a base point, and the whole subject is the claim that you cannot comb the first into the second without something interesting happening.