VISUALIZING THE HOPF FIBRATION

Rick Kreminski

Besides a novel visual rendering of an elusive mathematical object, fundamental concepts in abstract algebra are depicted in an important example The Hopf fibration is a versatile mathematical veteran, arising in numerous and seemingly unrelated situations. Even if you have never heard of it, you may still have encountered it. For instance, the Hopf fibration surfaces in particle physics: it underlies the mathematics of the Dirac monopole (see [Ryder 1996]). It also appears in general relativity, for instance in the Robinson congruence (see [PenRin 1984]). And of course it is found in several pure math contexts, such as in algebraic topology: roughly, there are infinitely many (topologically inequivalent) ways to map the three-dimensional sphere onto the two-dimensional sphere. 2 More precisely, in the language of homotopy groups, π3(S ) = Z (see almost any text in algebraic topology, for instance [Whitehead 1978]). Even with all its applications, the Hopf fibration is a mathematical Kilroy: it’s been all over the place, but it’s difficult to literally “see”. The Hopf fibration usually arises algebraically, rather than visually. The typical formulation involves S3, the unit sphere in C2 (or R4), and this in turn is usually visualized by taking ordinary space, R3, and artificially adding a point out at infinity. This is the approach in [Berger 1987 chapters 4 and 18]. While this method of visualization is valuable and interesting and does lead to an actual picture of the Hopf fibration, it is not the approach of this paper. Instead, we present what we think is a new way, and more importantly an elementary way, to honestly visualize this elusive character, emphasizing that we do this without having to venture into C2, R4 or adding infinity to R3. In fact, we only use some linear algebra, and a basketball. We postpone discussion of the Hopf fibration until after we examine a close relative. Indeed, the object in Figure 1 is our main interest in this article; the Hopf fibration will arise almost as an afterthought. We therefore begin by explaining what the cut-away ball in Figure 1 really is, and then explain what those strings or trajectories inside it represent. Solid ball model of the set of all rotations in R3: The “ball” in Figure 1 really depicts something called a projective space, which is obtained as follows. Begin with a solid ball in R3 that is centered at the origin and which has radius π, i.e. all points in ordinary space that are at most π units from the origin. Then for each point p on the ball’s boundary, identify it with −p, its antipodal (diametrically opposite) point. Do this identification procedure only for the points on the boundary, i.e. only for those points in the ball that are exactly π units from the origin. The resulting space is precisely how we are to interpret the ball-like object in the figure. [This situation is somewhat analogous to that which arises in various video games: if a curve or trajectory strikes the boundary of the solid ball at some point, it reappears at the antipodal point.] The space obtained by taking a solid ball in Rn and identifying antipodal boundary points is called a (real) projective space and denoted RP n. Thus, the curves in Figure 1 represent curves in the space denoted RP 3. To understand precisely how the curves in RP 3 that appear in Figure 1 are obtained, we first have to un- derstand the connection between RP 3 and the 3-dimensional rotation group. Every (orientation preserving) rotation of R3 can be completely characterized by two pieces of information, namely: (1) an axis of rotation given by some ray, l, and (2) a real number θ between 0 and π describing the amount of (counterclockwise) rotation about that axis. (All rotations in this article are considered to keep the origin fixed.) Then to each rotation (with given l and θ) we associate a point in RP 3 as follows: simply go out θ units from the origin, along the ray l, making sure that when peering back at the origin the rotation would be counterclockwise by θ radians1. The origin itself depicts the identity transformation (or, the trivial rotation by 0 about any

1As an example, the point (0, 0, π/2) in RP 3 corresponds to the rotation of R3 about the z-axis by π/2 in the counterclockwise sense when viewed from above. This would rotate the x-axis to the former location of the y-axis, and the y-axis to the former

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axis). Any point r in the solid ball model of RP 3 from Figure 1 therefore corresponds to a unique rotation R, and vice versa. Note that antipodal boundary points indeed correspond to identical rotations of R3 (since both points correspond to rotations about the same axis, with angle of rotation π). In this way, we see how the set of rotations of R3 corresponds to the solid ball model of RP 3 depicted in Figure 1. The Figure 1 trajectories: Here is where the basketball mentioned in the introduction truly proves useful. Figure 2 depicts our virtual basketball. This ball has its center at the origin in R3; its radius is irrelevant. Imagine holding the ball with some pre-determined point on it pointing upwards, for instance the pinhole where the ball is inflated. For the rest of the article “pinhole” refers to this point that is originally on top of the ball. Let q be any other point on the surface of the basketball. Now imagine a rotation of R3 that rotates the basketball in such a way that the pinhole has moved to q’s former location. Remember that rotations must keep the center of the ball fixed. The reader’s duty is to try and find all such rotations, that send the pinhole to q’s former location. In demonstrations, people first come up with the rotation that seems the “quickest”; one simply rotates the pinhole down along the line of longitude connecting it to q. In other words, rotate the ball so that the pinhole stays in the plane that contains the origin as well as the original postions of the pinhole and q. This rotation will be referred to below as the “obvious” rotation. Trouble sometimes arises when it is pointed out that there are infinitely many other rotations that take the pinhole to q’s former location. (The next rotation typically found is the one where one rotates by π about a point halfway down the line of longitude from the pinhole to q.) The reader equipped with a basketball is urged to find all rotations that take the pinhole to q, and characterize them, before reading on. A glance back at Figure 2 reveals the answer: all rotations of R3 that move the pinhole to q’s former location have the following form. They have axis of rotation that lies in the plane τ that is the perpendicular bisector of the chord that connects the pinhole to q. [In fact, it is easy to physically see how much of a rotation angle is needed for any axis of rotation l that lies in τ, if one is actually holding a basketball: simply grip the basketball tightly at the two points where l pierces the basketball, and swivel about l until indeed the pinhole reaches q’s former location.] We can now state how the trajectories in Figure 1 come about: Fix a point q on the basketball, and consider the collection of all rotations of R3 which rotate the pinhole to q’s former location. Plot these rotations in the solid ball model of RP 3. The curve obtained is one of the trajectories. Moreover, all trajectories in Figure 1 are obtained this way. Since for a given q the relevant rotations all have axes that lie in a plane (called τ above), one sees that any one trajectory in Figure 1 should lie in a plane, namely τ 2. Using q as depicted in Figure 2, the plot of the trajectory obtained from it is indicated in Figure 3. Symmetry in Figure 1: We have stated that each trajectory in Figure 1 represents a certain set of rotations of R3, namely: each trajectory represents all rotations which rotate the top point on a given sphere centered at the origin to a fixed given point q on the same sphere. Many observations follow almost immediately from this: (1) Each trajectory in the solid ball model of RP 3 (depicted in Figure 1) lies in a plane. (This was discussed in the previous section.) (2) The set of all rotations that leave the top point, or “pinhole”, fixed are precisely the rotations about the z-axis; and these rotations correspond to the vertical trajectory along the z-axis in Figure 1. (3) Each trajectory is topologically a circle. (This is true because when a trajectory reaches the boundary of the solid ball, it has reached a point that gets identified with its antipodal point.) (4) One of the trajectories is the equator. (This corresponds to all rotations that rotate the top point, or “pinhole”, to the point q which is the bottommost point.) (5) The trajectories are disjoint. (6) The space of all trajectories can be identified with a 2-dimensional sphere. There are several ways to see this. (a) First, note that each trajectory, with the exception of the equator, strikes the upper hemisphere in exactly one point. (That point represents the “maximal” rotation, by π about the

location of the negative x-axis. By contrast, the rotation corresponding to the point (0, 0, −π/2) will rotate the x-axis to the former location of the negative y-axis, and the y-axis to the former location of the x-axis. 2Or really in the plane in the solid ball model of RP 3 that’s parallel to τ, since τ itself cuts through the basketball. VISUALIZING THE HOPF FIBRATION 3

axis that points halfway down the line of longitude from the pinhole to the given point q.) Thus the collection of all trajectories can be identified with the upper hemisphere, with the proviso that the equator is to be identified to a single point. But the space one gets by doing this is topologically equivalent to a 2-dimensional sphere. (b) For a second approach, note that each trajectory, with the exception of the equator, strikes the equatorial plane in exactly one point. (That point represents the “obvious” rotation, namely rotating the pinhole “straight down” the line of longitude from the pinhole to its destination q.) Thus the collection of all trajectories can be identified with the disk of radius π lying in the equatorial plane, with the proviso that the boundary circle of the disk (the equator) is to be identified to a single point. Once again, the result is topologically a 2-dimensional sphere. (c) The last way we mention is the shortest: each trajectory is constructed from some fixed point q on the surface of the basketball. There is clearly a one-to-one correspondence, therefore, between trajectories and points on S2. (7) The collection of all trajectories, if it could be plotted, would be symmetric about the z-axis. That is, if Figure 1 depicted all trajectories, it would be rotationally symmetric about the z-axis. (Note that there is no x-axis or y-axis symmetry.) (8) Each trajectory is symmetric about its midpoint, in the plane that it lies in. Its midpoint in fact is where the trajectory strikes the plane z = 0 (which again represents the “obvious” rotation). (9) Denote the angle between the “pinhole” and the endpoint of a given trajectory (i.e. the point where the trajectory strikes the boundary of the solid ball in Figure 1) by φ³. Then in polar´ coordinates, −1 sin2θ−tan2φ the equation of the trajectory in its plane is given by r(θ) = −cos sin2θ+tan2φ . We leave the direct (but somewhat tedious) computation as an exercise in linear algebra. The reader is challenged to understand why the above statements hold true; detailed proofs can be obtained from the author, including proofs that use only linear algebra. Figure 1 as a multiplication table; or how to compose two rotations3 : One can use Figure 1 (and a bas- ketball) to determine the RP 3 point that represents the outcome when two given rotations of R3, say R and S, are combined by first doing rotation R and then rotation S. If R and S have the same axis of symmetry, it is easy to see how to find the point corresponding to SR: just add the vectors corresonding to R and S in the solid ball model of RP 3. The interesting case is, instead, when R and S have different axes of symmetry, as in Figure 4a. First, track where the topmost point of a given sphere in R3 (i.e. the basketball’s “pinhole”) is rotated by R, and then where that result is rotated by S. Denote the overall outcome by q0, i.e. q0 = S(R(pinhole)). Then rotation SR must be given by one point on the relevant trajectory in Figure 1; that is, SR must lie on the (unique) trajectory, say α, which lies in the plane that is the perpendicular bisector of the chord connecting the pinhole to q0. See Figure 4b. It remains to deduce which point on α corresponds to the rotation SR. To find the point on α corresponding to SR, consider the point n, one of the points where the axis of rotation for R passes through the basketball. Let m denote the outcome when SR rotates n, i.e. m = S(R(n)). (Note that R(n) = n, so we one only need examine where S rotates n.) Find the plane that is the perpendicular bisector of the chord from n to m, and see where it slices α. The point on α so determined is the point in RP 3 that corresponds to SR. This situation is illustrated in Figure 4c. Mathematical aside: This therefore puts a multiplicative structure on RP 3, using Figure 1. Some readers may already have realized that finding inverses in RP 3 is straightforward: if r is a point representing a rotation R, then the point representing R−1 is simply −r, i.e. the inversion of r in the origin. We thus have a geometric way to put a non-abelian (i.e. non-commutative) group structure on the solid ball with antipodal boundary points identified. This gives an example of a non-abelian group structure on a geometric object with group operations defined purely geometrically. Mathematical comments (including How to see a left coset, and The Hopf fibration): This section assumes some familiarity with the notions of group, coset and topological homeomorphism. Let Sn denote the n- dimensional sphere, i.e. the set of all points in Rn+1 that are exactly one unit from the origin. Let SO(n) denote the set of n by n orthogonal matrices of determinant 1, which is equivalently the matrices in Rn that correspond to rotations. SO(3) is then the set of all rotations of R3 we have been examining4. We

4It is a fact, not immediately obvious (but surprisingly easy to show), that every element in SO(3) does indeed represent a counterclockwise rotation about some axis L by some amount θ. This is often taught in a linear algebra course, and we leave 4 RICK KREMINSKI

have discussed above how SO(3) ' RP 3, where ' denotes topological homeomorphism. Rotations about the z-axis are given by block diagonal 3x3 matrices, with a ‘1’ in the lower right entry and a 2x2 block consisting of an element of SO(2) in the upper left. Abusing notation, we denote the rotations about the z-axis by SO(2). These are precisely the rotations that preserve the point (0, 0, 1) (the pinhole). SO(2) is known as the isotropy group of (0, 0, 1). Note that a (left) coset of SO(2) in SO(3) is exactly the set of all rotation matrices that send the point (0, 0, 1) to a common point. That is, all elements of a coset [gSO(2)] map (0, 0, 1) to the same point, call it q; and no other elements of SO(3) send (0, 0, 1) to q. (For a more pedestrian way to consider this, due to the block-diagonal form of the group we call SO(2), all elements of a left coset of SO(2) must have the same third column; but this is precisely the point that (0, 0, 1) gets sent to by the action of a 3x3 matrix.) This implies that each left coset of SO(2) in SO(3) directly corresponds to one of the trajectories, and vice versa. Since we saw that the collection of all trajectories could be identified with S2, we realize that part of our discussion above yields an elementary way of understanding that SO(3)/SO(2) ' S2. We now briefly discuss some key properties of the concepts fibration and fibre bundle, skipping the formal (and unfortunately technical) definitions. Fibrations are useful in algebraic topology, for instance in consid- ering homotopic maps between spaces (and hence arising in computations of higher homotopy groups, like 2 3 π3(S ).) Fibre bundles give rise to fibrations, and our discussion of SO(3) and RP leads to two nontrivial fiber fundles. Let X, Y and Z be connected topological spaces. One consequence of a fibre bundle (and hence fibration) X → Y → Z is that space Y is composed of lots of pieces, each space looking like (or homeomorphic to) space X. Space Z is essentially a bookkeeping device for how many copies of space X to use: a different version of space X is used for each point in space Z. This process, of chopping Y into a Z’s worth of disjoint copies of X, must be done “continuously”. Earlier discussions imply that we have the fibre bundle (and hence fibration) S1 → RP 3 → S2, and this fibration is exactly what Figure 1 illustrates. In words, RP 3 is comprised of disjoint pieces, each of which is an S1 or circle, and there are an S2’s worth of these circles. This fibration is also exactly double covered by the fibration S1 → S3 → S2, which is the Hopf fibration. It is this nontrivial cutting up of S3 into a disjoint collection of S1’s that arises in the situations cited in the introduction. To “see” the Hopf fibration, take two copies of Figure 1 and place them side by side; but slightly change the identification rules so that when trajectories reach the boundary of one solid ball they actually appear at the corresponding point in the other ball (instead of appearing at the antipodal point of the same ball). This is our (possibly never-before-actually-seen) visualization of the Hopf fibration. As claimed in the introduction, this rendition does not require visualization of R4 or adding a point at infinity to R3. Rather, S3 here is visualized as the result obtained when two solid balls are glued together on their boundary. (How gluing two solid balls together on their boundary truly yields S3 is most readily understood by analogy. If one takes two solid 2-dimensional “balls”, which would really be two ordinary disks, and glues them togther on their boundary, one gets an ordinary 2-dimensional sphere, at least topologically speaking. For instance, think of one of the solid 2-dimensional “balls” or disks as slightly stretched so as to be the lower hemisphere, and the other ball or disk as the upper hemisphere. The gluing occurs along their boundaries, and forms the equator 2 of the sphere.) In this way, for instance, we can directly “see” the generator of the group π3(S ). Right cosets, double cosets, and nested tori: Finally, we comment on Figure 5. Let H denote the sub- group of rotations fixing the z-axis, i.e. the subgroup we have been denoting SO(2). Figure 5 illustrates one typical right coset of H in SO(3), along with a fairly random sample of the left cosets it intersects, in the RP 3 model of SO(3). One can show that the obvious observed symmetry is real, namely that the trajectory corresponding to a right coset [Hg] will: (a) lie in a plane; (b) only intersect left cosets’ trajectories which are members of a certain “family”, i.e. only intersect left cosets whose corresponding trajectories are all rotated versions of one another; and (c) be the mirror image of the left coset trajectory that lies in its plane. From a more purely algebraic viewpoint, (c) states: the elements of SO(3) comprising the right coset are point-by-point the group-theoretic inverses of the elements of the left coset whose trajectory shares the plane

it as an exercise for the reader. VISUALIZING THE HOPF FIBRATION 5 of the right coset’s trajectory. [Hint: Recall that group theoretic inversion corresponds to geometric inversion in the origin, and note that (Hg)−1 = g−1H.] This in turn illustrates a situation arising in introductory abstract algebra: the natural bijection between the collection of left cosets of a subgroup H of a group G and the right cosets is given by [gH] 7→ [Hg−1]. Here, the geometric interpretation is simply: map a left coset to its inversion in the origin. Note in passing that the only right cosets that are also left cosets are (a) the coset of the identity element of the group and (b) the coset of the element representing rotation by π about any axis in the x − y plane. Geometrically, in Figure 1 the trajectory along the z-axis is both a left coset and a right coset, as is the equator. Lastly, consider the double cosets SO(2)\SO(3)/SO(2), i.e. H\SO(3)/H. The reader might think that this would be difficult to “see”! However, one double coset is essentially depicted in Figure 5. A moment’s thought indicates that HgH must be precisely the result of taking one trajectory and rotating about the z-axis. [Hint: hgH rotates the pinhole to wherever g did, followed by a rotation about the z-axis; the end result is a point on the same latitude line as g’s end result. The trajectory associated with this final resting place for the pinhole is the rotation by h of the trajectory associated to g.] Recalling that antipodal points on the boundary of the ball in Figure 5 are identified, we see that such double cosets are topologically torii - with two exceptions: (a) the vertical trajectory is a double coset and (b) the equator is a double coset. These form “degenerate” torii, namely circles. Note that the torii are nested. Compare with the figure of the Hopf fibration given in chapter 18.9 of [Berger 1987] (who cites work of Penrose from the 1960’s).

References BERGER M., Geometry, Springer-Verlag, New York, 1987. PENROSE R. and RINDLER W., Spinors and Spacetime, Cambridge University Press, Cambridge, 1984. RYDER L., Quantum Field Theory, second edition, Cambridge University Press, Cambridge, 1996. WHITEHEAD G., Elements of Homotopy Theory, Springer-Verlag, New York, NY, 1978. About the author Rick Kreminski is a mathematical physicist at Texas A & M University - Commerce. He is interested in interdisciplinary applied mathematics, in the general (and literal) sense of applying results from some branch of pure mathematics to physics, biology, chemistry, or economics, or even to other branches of mathematics.

Department of Mathematics, Texas A & M University - Commerce, Commerce, TX 75429