Rotations of the three-sphere and symmetry of the Clifford Torus

John McCuan^{∗} and Lafe Spietz

October 5, 1998

Abstract

We describe decomposition formulas for rotations of ℝ^{3} and ℝ^{4} that have special properties with respect to stereographic projection. We use the lower dimensional decomposition to analyze stereographic projections of great circles in 𝕊^{2} ⊂ ℝ^{3}. This analysis provides a pattern for our analysis of stereographic projections of the Clifford torus C ⊂ 𝕊^{3} ⊂ ℝ^{4}. We use the higher dimensional decomposition to prove a symmetry assertion for stereographic projections of C which we believe we are the first to observe and which can be used to characterize the Clifford torus among embedded minimal tori in 𝕊^{3}—though this last assertion goes beyond the scope of this paper. An effort is made to intuitively motivate all necessary concepts including rotation, stereographic projection, and symmetry.

Introduction

It is known (and intuitively believable) that the spheres in ℝ^{3} are characterized by being the only compact surfaces with planes of symmetry whose normals exhaust all possible directions. That is, if S is a compact surface and, for each unit vector n in ℝ^{3}, there is some plane Π with normal n such that S is invariant under reflection in Π, then S is a sphere. Technically, S could be some collection of concentric spheres, but this can be fixed by requiring explicitly that S be connected, i.e., a single surface. The point is that spheres can be characterized by their (reflectional) symmetry.

In this paper we describe an analogous symmetry condition for a certain surface, the Clifford torus. Our task is complicated by the fact that this torus is located in the three-sphere, 𝕊^{3}, where our intuition from ℝ^{3} is of limited use. For this reason we will employ a certain transformation, stereographic projection, that allows us to realize 𝕊^{3} (at least most of it) in the Euclidean space ℝ^{3}. In fact, our symmetry condition will apply, more properly, to the stereographic projections of the Clifford torus. Furthermore, our symmetry condition is substantially more complicated than the simple one above for spheres, and we discuss at some length why it is a natural one.

∗Work supported in part by the University of California, Berkeley.

Another objective of the paper is to give what we consider novel, geometrically based, expositions of several well known topics. (Some of these are mentioned briefly below.) From this point of view, we offer an introduction to 𝕊^{3} that we hope is a geometric counterpart to the algebraic treatment in, for example, [12].

The paper is organized as follows. In the following §1 we review stereographic projection of 𝕊^{2} and discuss a decomposition formula for rotations of 𝕊^{2} ⊂ ℝ^{3}. We show, in particular, that projections of rotations of great circles are circles in ℝ^{2} whose size and position are given in terms of the parameters of the decomposition. This discussion is somewhat artificial because it is easy to show that the projection of essentially any circle in 𝕊^{2} is a circle in ℝ^{2} whose center and radius are easy to calculate. This is the case, however, owing to the fact that a circle in 𝕊^{2} is the intersection of an affine subspace (a plane) with 𝕊^{2}. This luxury is not afforded us by the Clifford torus, and the decomposition technique presented here will be used with considerable advantage in the more complicated higher dimensional case. Furthermore, our discussion of stereographic projection of 𝕊^{2} is used in §2 to give an exposition of symmetry for planar sets, and it provides intuition for the higher dimensional stereographic projection considered in §3.

We also give considerable attention to building up intuition about rotations, especially rotations of ℝ^{4} since we consider 𝕊^{3} as a subset of ℝ^{4}. Recall that rotations of ℝ^{2} (centered at zero) can be represented by matrices of the form

( \begin{matrix}cos θ & − sin θ \\ sin θ & cos θ \end{matrix} ) .

That is, given a rotation R : ℝ^{2} → ℝ^{2} there is a matrix M of the form given above such that for each vector x, we have that R(x) is given by the matrix multiplication Mx. Through such a representation we are immediately presented with two algebraic facts, namely that R is linear and that det M = 1. It is also easy to check from this representation that R is orthogonal, i.e., it preserves the orthonormality of bases. Many authors define a rotation of ℝ^{3} to be an orthogonal linear transformation corresponding to a matrix of determinant 1. This definition is concise and computationally convenient, but we find it unintuitive. It can perhaps be argued (via the parallelogram rule) that linearity is an intuitive assumption, but the role played by the determinant is difficult to see geometrically for rotations of ℝ^{3}, much less for rotations of ℝ^{4}. In an appendix to this paper we take the point of view that rotations are rigid motions (i.e., distance preserving transformations) that fix the origin and result from smooth “homogeneous” motions. We then prove linearity, orthogonality, and representation by matrices of determinant 1 (thereby showing the two definitions are equivalent). This appendix may be read at any time, but we recommend reading it after §1 and before §3.

In §2 we introduce circles of Apollonius and show that their symmetry as a family of planar curves is a natural generalization of the symmetry exhibited by concentric circles. We incorporate in this discussion an explanation of why symmetry and reflection about circles are natural generalizations of symmetry and reflection about lines. Furthermore, we observe that the generalized symmetry exhibited by circles of Apollonius is described naturally in terms of a line of centers l. More precisely, given a family of circles of Apollonius A, we can find, for each point x on l, an orthogonal Steiner circle C with center x and the property that every circle in A is symmetric with respect to C. Finally, we observe that this formulation of generalized symmetry can be easily extended to surfaces in ℝ^{3}. This sets the stage for our main (and we believe original) result which is roughly as follows. Given any stereographic projection Q of the Clifford torus, there is a line l in ℝ^{3}, and for each point x on l there is a sphere S centered at x so that Q is symmetric with respect to S. See Figure 5. A careful statement and proof, which include conditions on the radii of the spheres of symmetry, are given in §3. The proof follows, in outline, the discussion of §1 and §2.

From a wider perspective, the Clifford torus and its stereographic projections are interesting surfaces primarily due to curvature considerations that are beyond the scope of this paper. More primitively, they are interesting because they are critical points for certain functionals—which they are believed to minimize. For further information, see [3, 6, 10, 11]. In this framework, the Clifford torus plays a role in 𝕊^{3} similar to that of the plane and the sphere in ℝ^{3}. It has Gauss curvature and mean curvature zero like the plane, and it has constant mean curvature and is compact like the sphere. The symmetry properties shown in this paper may be considered as a first step in getting a feel for curvature of surfaces in 𝕊^{3}, and while the main result stated above may seem curious at first sight, it is completely analogous to the observation that any sphere in ℝ^{3} has a plane of reflective symmetry with any given normal direction.

We extend our thanks to Ed Bueler, David Hoffman, and Silvio Levy who gave us useful comments that greatly improved the exposition.

1 Stereographic Projection

In this section we consider the two-dimensional sphere 𝕊^{2} as a subset of ℝ^{3}:

𝕊^{2} = {x = (x, y, z) : |x| = 1}.

As usual |x| = \sqrt{x^{2} + y^{2} + z^{2}}.

We are interested in a map called stereographic projection that sends 𝕊^{2} (except for one point) into the x, y-plane, and we are interested in how the image of a certain set changes as the sphere is rotated. To be precise, stereographic projection π given by

π(x) = \frac{1}{1 − z} (x, y)  (1)

maps 𝕊^{2}\{(0, 0, 1)} onto ℝ^{2} = {(x, y, 0)} in a one-to-one fashion.

Exercise 1 Show that π is one-to-one and onto.

The map (1) has a convenient geometric interpretation:

Each point p ∈ 𝕊^{2}\{(0, 0, 1)} determines a unique line passing through p and (0, 0, 1). The line l, in turn, intersects the x, y-plane in a unique point q. We set π(p) = q.

[figure: p; q]

Figure 1: Stereographic Projection.

Exercise 2 (i) Using this geometric statement as a definition, derive formula (1) for π.

(ii) The geometric definition makes sense for any point in ℝ^{3} with third component not equal to 1. In this way π may be extended from 𝕊^{2}\{(0, 0, 1)} to ℝ^{3}\{(x, y, 1)}. We call this extended map π̄. Does formula (1) still apply?

Next we consider the equator circle

C = {x ∈ 𝕊^{2} : x^{2} + y^{2} = 1} = {x : x^{2} + y^{2} = 1, z = 0}

in the sphere. The stereographic projection π(C) of C is particularly simple—it is C itself.

What happens if we first rotate the sphere and then stereographically project? Say we rotate about the y-axis for example—by an angle ϕ. If we call this rotation R_{ϕ}^{y}, then R_{ϕ}^{y}(C) = {(x cos ϕ, y, x sin ϕ) : x^{2} + y^{2} = 1}, and the stereographic projection is

π ∘ R_{ϕ}^{y}(C) = \{(\frac{x cos ϕ}{1 − x sin ϕ}, \frac{y}{1 − x sin ϕ}) : x^{2} + y^{2} = 1\} .  (2)

This set in the plane is (perhaps) not so easy to recognize.

On page 19 of [1], however, Lars Ahlfors gave a nice way to look at it: R_{ϕ}^{y}(C) is the intersection of a plane P = {x : x sin ϕ − z cos ϕ = 0} with 𝕊^{2}. If we could find an inverse map π^{−1} : ℝ^{2} → 𝕊^{2}, then

π ∘ R_{ϕ}^{y}(C) = {a = (a, b) : π^{−1}(a) ∈ P }.

Stereographic projection on 𝕊^{2}\{(0, 0, 1)} does have an inverse (though the extended map π̄ does not—why?). The formula for the inverse is

π^{−1}(a) = \frac{1}{|a|^{2} + 1}(2a, 2b, |a|^{2} − 1).

Exercise 3 Derive this formula for π^{−1}.

The statement π^{−1}(a) ∈ P now translates into an equation:

−2a tan ϕ + a^{2} + b^{2} = 1.

Such an equation, as we know, represents a circle,

(a − tan ϕ)^{2} + b^{2} = sec^{2} ϕ,

with center (tan ϕ, 0) and radius r = | sec ϕ|.

If you think about what happens to C when the sphere is rotated, it is fairly clear that the simple rotation R_{ϕ}^{y} is typical. In particular, π ∘ R(C) should be a circle for any rotation R of 𝕊^{2}. One way to make this precise is to decompose an arbitrary rotation R into simple coordinate rotations like R_{ϕ}^{y}. The following theorem gives such a decomposition.

Theorem 4 Any rotation R is the composition of three rotations—one about the zaxis, one about the y-axis, and another about the z-axis. Thus, there are angles θ, ϕ, and ψ such that

R = R_{ψ}^{z} ∘ R_{ϕ}^{y} ∘ R_{θ}^{z}.

Proof. The simple rotations R_{θ}^{z}, R_{ϕ}^{y}, and R_{ψ}^{z}, like rotations of ℝ^{2}, can be represented by matrices. For example,

R_{ψ}^{z}(x) = ( \begin{matrix}cos ψ & − sin ψ & 0 \\ sin ψ & cos ψ & 0 \\ 0 & 0 & 1 \end{matrix} ) x

for every x ∈ ℝ^{3}. Consequently, these rotations are linear transformations of ℝ^{3}. Moreover, we show in the appendix that all rotations are linear. Because of this, we only need to find θ, ϕ, and ψ such that R_{ψ}^{z} ∘ R_{ϕ}^{y} ∘ R_{θ}^{z} and R agree on an orthonormal basis.

Let e_{1} = (1, 0, 0), e_{2} = (0, 1, 0) and e_{3} = (0, 0, 1). Notice that for any θ we have R_{θ}^{z}(e_{3}) = e_{3} so that R_{ψ}^{z} ∘ R_{ϕ}^{y} ∘ R_{θ}^{z}(e_{3}) = R_{ψ}^{z} ∘ R_{ϕ}^{y}(e_{3}). Thus, we want to find ϕ and ψ such that, in particular,

R(e_{3}) = R_{ψ}^{z} ∘ R_{ϕ}^{y}(e_{3}) = ( \begin{matrix}− cos ψ sin ϕ \\ − sin ψ sin ϕ \\ cos ϕ \end{matrix} )  (3)

where the second equality is by direct calculation. On the other hand, R(e_{3}) is some unit vector (u_{1}, u_{2}, u_{3}), and since |u_{3}| ≤ 1, we can find an angle ϕ with cos ϕ = u_{3}. If |u_{3}| = 1, then sin ϕ = 0 and (3) holds. Otherwise, sin ϕ ≠ 0 and (u_{1}/ sin ϕ)^{2} + (u_{2}/ sin ϕ)^{2} = 1. Thus, for some angle ψ, cos ψ = −u_{1}/ sin ϕ, sin ψ = −u_{2}/ sin ϕ, and again (3) holds.

We have then for any θ, R_{ψ}^{z} ∘ R_{ϕ}^{y} ∘ R_{θ}^{z}(e_{3}) = R(e_{3}). It remains to choose θ so that R_{ψ}^{z} ∘ R_{ϕ}^{y} ∘ R_{θ}^{z}(e_{j}) = R(e_{j}) for j = 1 and 2.

Let S = R_{ψ}^{z} ∘ R_{ϕ}^{y} and note that S^{−1} ∘ R(e_{3}) = e_{3}. Therefore, S^{−1} ∘ R induces a rotation R_{θ} (by some angle θ) on ℝ^{2}. That is, S^{−1} ∘ R(e_{j}) = R_{θ}(e_{j}) for j = 1 and 2. Since R_{θ} extends to a rotation R_{θ}^{z} on ℝ^{3} we are done. □

We can now apply Theorem 4 as follows.

π ∘ R(C) = π ∘ R_{ψ}^{z} ∘ R_{ϕ}^{y} ∘ R_{θ}^{z}(C)
= R_{ψ}^{z} ∘ π ∘ R_{ϕ}^{y}(C).

This set is the circle (a − tan ϕ)^{2} + b^{2} = sec^{2} ϕ rotated by an angle ψ, i.e., it is the circle with center (cos ψ tan ϕ, sin ψ tan ϕ) and radius r = | sec ϕ|.

Exercise 5 Show that any circle in 𝕊^{2} stereographically projects to a circle in ℝ^{2}, and any circle in ℝ^{2} is the projection of a circle in 𝕊^{2}. Hint: A circle in 𝕊^{2} is the intersection of a plane in ℝ^{3} with 𝕊^{2}.

We will need one other fact about the stereographic projection π : 𝕊^{2} → ℝ^{2}:

Lemma 6 If C_{1} and C_{2} are two (smooth) curves in 𝕊^{2} that intersect at a point p in an angle γ, then the image curves π(C_{1}) and π(C_{2}) intersect at π(p) in the same angle γ.

This property is expressed by saying that π is conformal. Henry Wente told us a short proof of Lemma 6 which we have included in an appendix. For another proof, we refer the reader to the classic book [5] by David Hilbert.

2 Symmetry

When we described the stereographic projection of a rotation of the equator circle (π ∘ R(C)) in the last section, we ignored the unpleasant possibility that C had been rotated onto (0, 0, 1)—where stereographic projection is not defined. This happens, of course, when the angle ϕ of the preceding section is π/2. Our analysis in that case is flawed since tan ϕ and sec ϕ are not defined, and in fact the image of R(C) (aside from the point (0, 0, 1)) is then a line in ℝ^{2}. It is one of our objectives in this section to address this apparent difficulty. Our second and main objective is to generalize in a natural way our intuitive notion of symmetry, so that we can introduce the symmetry assertion of the main theorem.

The analysis of stereographic projection of circles that pass through (0, 0, 1) is simple. For example, if we take ϕ = π/2, then in place of (2) we have

π ∘ R_{π/2}^{y}(C\\{(1, 0, 0)\}) = \{(0, \frac{y}{1 − x}) : x^{2} + y^{2} = 1, x ≠ 1\}

= \{(0, ±\sqrt{\frac{1 + x}{1 − x}}) : −1 ≤ x < 1\} .

The last set is clearly the line x = 0, since (1 + x)/(1 − x) maps [−1, 1) monotonely onto [0, ∞). The more general cases can be handled similarly. What we really wish to emphasize, however, is the following: because circles in 𝕊^{2} that pass through (0, 0, 1) are geometrically identical to other circles that do not, it is natural for our purposes to view straight lines in ℝ^{2} as circles with (infinite radius and) one point at ∞.

For starters, this viewpoint allows Exercise 5 to make sense as stated. More importantly it illustrates how sets and structures in 𝕊^{2} can provide insight for terminology and constructions in ℝ^{2}. We proceed further along this line presently.

The symmetry of a circle in ℝ^{2} is perhaps most easily described in terms of its center. Given a point p in a circle A with center a, A is generated by rotating p about a. The same circle can also be generated by reflecting p about each of the lines through a. This latter characterization will be the one of interest to us.

Definition 7 A set A ⊂ ℝ^{2} has Euclidean reflectional symmetry with respect to a point a ∈ ℝ^{2} if, for each line E passing through a, we have ψ_{E}(A) = A, where ψ_{E} : ℝ^{2} → ℝ^{2} is the reflection about E.

Exercise 8 Show that any such set (with Euclidean reflectional symmetry) is a union of concentric circles with center a.

If, as we have suggested, lines should be considered simply as circles with infinite radius, then it is natural to ask for a definition of symmetry in which reflection about lines is replaced with reflection about circles. Extrapolating directly from the definition above we might try to replace the family of symmetry lines passing through a with a family of symmetry circles passing through a common point a. Unfortunately, the full geometric situation is not completely evident from considering the plane alone. Again we turn to stereographic projection. The inverse image of each line through a is a circle in 𝕊^{2} passing through π^{−1}(a) and (0, 0, 1). (There are two points of intersection.) If one then rotates slightly this family of circles in 𝕊^{2} and stereographically projects, a family of circles in ℝ^{2} is obtained that pass through two distinct points a_{1} and a_{2}, as shown on the left in Figure 4. Note that this family also contains the line determined by a_{1} and a_{2}. We call the circles that pass through two given points a_{1} and a_{2} Steiner symmetry circles. We use these circles to generalize the symmetry lines in Definition 7.

It remains to specify, for each circle S passing through a_{1} and a_{2}, a transformation ψ_{S} : ℝ^{2} → ℝ^{2} which we will call reflection about S. Again we look to the sphere for intuition. What transformation of 𝕊^{2} corresponds to reflection about a line in ℝ^{2}? If E is a line in ℝ^{2}\{0}, then the inverse image of E is some circle C in 𝕊^{2} passing through (0, 0, 1). Furthermore, there is a unique point c = (c_{1}, c_{2}, 1) such that the segments connecting c to C form a right circular cone (tangent to 𝕊^{2}), n.b., Figure 2. The point c can now be used to define a transformation Ψ_{C} : 𝕊^{2} → 𝕊^{2}:

For each point p ∈ 𝕊^{2}, the line determined by p and c intersects 𝕊^{2} in a set {p, q}. We set Ψ_{C}(p) = q.

Note that this definition is much like the geometric definition of stereographic projection. The reader can check (and we will show below) that Ψ_{C} corresponds to the reflection ψ_{E} in the sense that ψ_{E} = π ∘ Ψ_{C} ∘ π^{−1}. This construction also works if

[figure: p; q; c]

Figure 2: Reflection on 𝕊^{2}.

E passes through 0—though in that case the transformation Ψ_{C} is simply given by reflection about the plane determined by C.

Notice that the fact C passes through (0, 0, 1) is not required for the geometric definitions of Ψ_{C} to make sense. That is, for any circle C in 𝕊^{2}, if C is not a great circle, it defines a cone point^{1} c, and the definition above gives a transformation Ψ_{C} of 𝕊^{2} that is geometrically identical (modulo rotation) to one that corresponds to Euclidean reflection. For great circles we use the alternative construction.

Going back to ℝ^{2}, we may start with any circle (or straight line) S, take C = π^{−1}(S) and apply the construction described above to obtain a transformation ψ_{S} = π ∘ Ψ_{C} ∘ π^{−1} of ℝ^{2}\{a = π̄(c)}. This is the transformation we call reflection about S.

We proceed to derive a formula for ψ_{S} : ℝ^{2}\{a} → ℝ^{2}. Notice first that if C′ is a circle in 𝕊^{2} that meets C in right angles at points p_{1} and p_{2}, then C′ lies in the plane determined by c, p_{1} and p_{2}, and one sees from this that Ψ_{C}(C′) = C′. It follows moreover, since π is an angle and circle preserving transformation, that any circle S′ in ℝ^{2} which is orthogonal to S is mapped by ψ_{S} into itself. Thus, consider a point p inside S as shown in Figure 3. Let S′ be the line determined by the center of S and p. There is also a circle S′′ which passes through p, meets S orthogonally, and has its center on S′. Since ψ_{S}(S′\{a}) = S′\{a}, ψ_{S}(S′′) = S′′, and ψ_{S}(p) ≠ p, it follows that ψ_{S}(p) = p′ is the other point of intersection of S′ and S′′. Moreover, if a is the center of S and q ∈ S ∩ S′′, then triangles aqp and aqp′ are similar. It follows that |p − a||p′ − a| = ρ^{2} where ρ is the radius of S. The same reasoning applies if p lies outside of S, and we obtain the formula

p′ = ψ_{S}(p) = ρ^{2} \frac{p − a}{|p − a|^{2}} + a  (4)

for reflection about the circle S in ℝ^{2} with center a and radius ρ. This same discussion applied to the case when C passes through (0, 0, 1) and S = E is a straight line provides a proof of the assertion made above that Ψ_{C} corresponds to ψ_{E}.

1Or horizon point. See [5] for other interesting properties of this point. In particular, π̄(c) is the center of π(C).

S a  S′′  S′ p  p′ q

Figure 3: Mapping Circles

Finally we have the following

Definition A set A ⊂ ℝ^{2} has generalized reflectional symmetry if there are two distinct points a_{1} and a_{2} such that for each Steiner circle S passing through a_{1} and a_{2}, ψ_{S}(A) = A where ψ_{S} is reflection about S.

We may allow one of the points a_{1} or a_{2} to be at ∞, in which case the circles S are all the lines passing through the other point. Furthermore, it can be shown (in analogy to Exercise 8) that if A has generalized reflectional symmetry, then A is a union of circles, each of which is orthogonal to all the circles through a_{1} and a_{2}. We will not need this fact, but it is a special case of an assertion proved in [8]. These circles are called circles of Apollonius, and once it is known that they are circles, it is easy to show the following. (See Figure 4.)

Lemma 9 The circles of Apollonius determined by the Steiner circles through a_{1} and a_{2} in ℝ^{2} are disjoint and are in one to one correspondence with their centers which comprise the line E passing through a_{1} and a_{2} except for the segment between a_{1} and a_{2}. Let m_{0} = (a_{1} + a_{2})/2. The circle of Apollonius with center a ∈ ℝ^{2} has radius r = \sqrt{d^{2} − ρ_{0}^{2}} where d = |a − m_{0}| and ρ_{0} = |a_{1} − a_{2}|/2.

The centers of the Steiner circles also form a line l, and once we know the point m_{0} = (a_{1} + a_{2})/2 on l and the reference distance ρ_{0} = |a_{1} − a_{2}|/2, we can express generalized reflectional symmetry without reference to a_{1} or a_{2} as follows.

Definition 10 A set Q has generalized reflectional symmetry along a line l if, for each point a ∈ l, we have ψ_{S}(Q) = Q, where ψ_{S} is given by (4) with ρ = \sqrt{d^{2} + ρ_{0}^{2}} and d = |a − m_{0}|.

[figure: a_{1}; a_{2}; l]

Figure 4: Steiner circles and circles of Apollonius.

Notice finally that formula (4) makes sense for points a and p in ℝ^{3} and gives a generalization of reflection about circles to reflection about spheres. Thus, this statement of generalized symmetry can be applied to sets Q ⊂ ℝ^{3}.

In the next section we use this formulation and take Q to be a stereographic projection of the Clifford torus.

3 Stereographic projection of the Clifford Torus

The unit sphere in ℝ^{4} is the three-dimensional space

𝕊^{3} = {x = (x, y, z, w) : |x| = 1}.

Because we are used to visualizing things that are described by three Euclidean coordinates (i.e., things in ℝ^{3}), it is often difficult to see what objects look like in 𝕊^{3}. For this reason, it is convenient to use a stereographic projection π : 𝕊^{3}\{(0, 0, 0, 1)} → ℝ^{3}. The formula for such a map is similar to the one for 𝕊^{2}:

π(x) = \frac{1}{1 − w}(x, y, z),

and a similar geometric description applies as well.

We are interested in a particular geometric object in 𝕊^{3} called the Clifford torus:

C = {x : x^{2} + y^{2} = 1/2 = z^{2} + w^{2}}.

The stereographic projection π(C) of the Clifford torus is particularly nice because of its symmetry.

Exercise 11 (i) Show that π(C) is rotationally symmetric with respect to the z-axis in ℝ^{3} = {(x, y, z, 0)}.

(ii) What is the intersection of π(C) with the x, z-plane?

From Exercise 11 it is clear that π(C) is described by its intersection with the half planes Π_{θ} = {(r cos θ, r sin θ, z) : r > 0}. In fact, it is enough to know only π(C) ∩ Π_{0}.

As with circles in 𝕊^{2}, rotating the three-sphere (i.e., moving C around in 𝕊^{3}) changes the stereographic projection. Since the rotated surface is geometrically identical to C however, one might expect that some kind of symmetry of the projection is preserved. In fact, we show the following.

Theorem 12 Let R be any rotation of 𝕊^{3} ⊂ ℝ^{4}. The surface Q = π ∘ R(C) has generalized symmetry in the sense described in the last section.

We will, as we did in §1, consider first a particular rotation and then show that that rotation is typical via a decomposition formula (Theorem 14) for general rotations. With hindsight from the decomposition formula, we consider the rotation R_{ψ}^{xw} of the x, w-plane corresponding to the matrix

( \begin{matrix}cos ψ & 0 & 0 & − sin ψ \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ sin ψ & 0 & 0 & cos ψ \end{matrix} ) .

Proposition 13 Let Q = Q(ψ) = π ∘ R_{ψ}^{xw}(C).

(i) If ψ = 0, then Q = Q_{0} has generalized reflectional symmetry along the z-axis.

(ii) If 0 < |ψ| ≤ π/2, then Q has generalized reflectional symmetry along the vertical line (x, y) = (tan ψ, 0) and along the horizontal line (x, z) = (− cot ψ, 0).

(iii) Let R be the rotation of ℝ^{3} about the x-axis by π/2. Then

(a) Q(ψ + π/2) = R(Q(ψ)), and

(b) Q(ψ + π) = Q(ψ), for all ψ.

In particular, Q = Q(ψ) has generalized reflectional symmetry for all ψ.

Theorem 14 Any rotation R of 𝕊^{3} ⊂ ℝ^{4} is a composition

R = R_{0} ∘ R_{ψ}^{xw} ∘ R_{ϕ}^{zw} ∘ R_{θ}^{xy}  (5)

where R_{0} is the trivial extension^{2} to ℝ^{4} of a rotation of ℝ^{3} = {(x, y, z, 0)} and R_{θ}^{xy} is the trivial extension^{2} to ℝ^{4} of a rotation of ℝ^{2} = {(x, y, 0, 0)} etc..

2For a precise definition see the appendix on rotations.

Q(0)  Q(π/8)  Q(π/4)

Figure 5: Stereographic projections and lines of centers.

Before we prove Proposition 13 and Theorem 14 we will show that they imply Theorem 12. It is easy to check (see Lemma 17 below) that rotations of the x, y and z, w-planes leave C invariant. Furthermore, if S is any set in 𝕊^{3}\{(0, 0, 0, 1)} and R_{0} is a rotation of {(x, y, z, 0)} as above, then π ∘ R_{0}(S) = R_{0} ∘ π(S). To see this, let R_{0}(x, y, z, 0) = (x′, y′, z′, 0) and note that

π ∘ R_{0}(x, y, z, w) = π(x′, y′, z′, w)
= \frac{1}{1 − w}(x′, y′, z′, 0)
= R_{0} (\frac{1}{1 − w}(x, y, z, 0))
= R_{0} ∘ π(x, y, z, w).

Thus, using the decomposition formula (5) we have

π ∘ R(C) = π ∘ R_{0} ∘ R_{ψ}^{xw} ∘ R_{ϕ}^{zw} ∘ R_{θ}^{xy}(C)
= R_{0} ∘ π ∘ R_{ψ}^{xw}(C).

The set π ∘ R_{ψ}^{xw}(C) is described in Proposition 13. The rotation R_{0} only changes that description by a rigid rotation in ℝ^{3}. In particular, π ∘ R(C) has generalized reflectional symmetry. □

Proof of Proposition 13. Part (i) follows from Exercise 11 where one finds that Definition 10 is satisfied with m_{0} = 0 ∈ ℝ^{3} and ρ_{0} = 1. Therefore, the sphere of symmetry with center a = (0, 0, c) which we denote by S = S_{ρ}(a) has radius

ρ = \sqrt{1 + c^{2}}.  (6)

For 0 < |ψ| ≤ π/2 we will show that C = π^{−1}(S) determines a transformation Ψ_{C} of 𝕊^{3}. This construction is analogous to the discussion in §2 of circles C ⊂ 𝕊^{2}. Moreover, we will again have the correspondence ψ_{S} = π ∘ Ψ_{C} ∘ π^{−1} where ψ_{S} is the reflection about S. Furthermore, a geometrically identical transformation ΨC̃ will be determined by C̃ = R_{ψ}^{xw}(C), and ΨC̃ will correspond to reflection about the sphere S̃ = π(C̃). The spheres S̃ thus corresponding to the spheres S = S_{ρ}(a) will be symmetry spheres for Q = Q(ψ) that satisfy Definition 10. Of course, at this point we do not even know that S̃ is a sphere. We give now a precise higher dimensional version of Exercise 5 which will establish this fact. For the statement we use the notation \underline{n} = (n_{1}, n_{2}, n_{3}) when n = (n_{1}, n_{2}, n_{3}, n_{4}) and the notation (n, n_{4}) = (n_{1}, n_{2}, n_{3}, n_{4}) when n = (n_{1}, n_{2}, n_{3}).

Lemma 15 Let Π = {x = (x, y, z, w) : n · x = e} be a three-plane in ℝ^{4}. If (0, 0, 0, 1) ∉ Π, then

π(Π ∩ 𝕊^{3}) = \{a = (a, b, c) : |a − \frac{\underline{n}}{n_{4} − e}|^{2} = \frac{n_{4} + e}{n_{4} − e} + \frac{|\underline{n}|^{2}}{(n_{4} − e)^{2}} \}  (7)

(which is a sphere). If (0, 0, 0, 1) ∈ Π, then

π(Π ∩ 𝕊^{3}\\{(0, 0, 0, 1)\}) = \{a : \underline{n} · a = n_{4}\}  (8)

(which is a plane).

On the other hand, let S = {a : |a − a_{0}|^{2} = ρ^{2}} be a sphere in ℝ^{3}. Then π^{−1}(S) = Π ∩ 𝕊^{3} where

Π = {x : (−2a_{0}, ρ^{2} − |a_{0}|^{2} + 1) · x = ρ^{2} − |a_{0}|^{2} − 1}.  (9)

Let P = {a : n · a = e} be a plane in ℝ^{3}. Then π^{−1}(P) ∪ {(0, 0, 0, 1)} = Π ∩ 𝕊^{3} where

Π = {x : (n, e) · x = e}.  (10)

Since we have given explicit equations, Lemma 15 follows from simple substitution using the formulas for π and π^{−1}, and we omit the proof. Note that equations (7) and (8) allow degenerate cases corresponding to Π∩𝕊^{3} ⊂ {(0, 0, 0, 1)}. In our applications below however, we will know that Π ∩ 𝕊^{3} is nontrivial.

Recall that S = S_{ρ}(a) is a sphere of symmetry for Q_{0}. One sees from (9) and (6) that C = π^{−1}(S) = Π ∩ 𝕊^{3} where Π = {x : (a, −1) · x = 0}. Note that Π passes through 0 ∈ ℝ^{4}, i.e., C is a great sphere in 𝕊^{3}. Let n = (a, −1) be the normal to Π. We consider the reflection Ψ_{C} of 𝕊^{3} about Π defined by

Ψ_{C}(x) = x − \frac{2x · n}{|n|^{2}}n.  (11)

To see that ψ_{S} = π ∘ Ψ_{C} ∘ π^{−1}, we extend the discussion of Figure 3 in §2. Let S′ = {a′ : n′ · (a′ − a) = 0} be a plane orthogonal to S. According to (10) we have π^{−1}(S′) ∪ {(0, 0, 0, 1)} = Π′ ∩ 𝕊^{3} where Π′ = {x : (n′, n′ · a) · x = n′ · a}. Since (n′, n′ · a) · n = (n′, n′ · a) · (a, −1) = 0, we see that Ψ_{C}(Π′) = Π′. Consequently, π ∘ Ψ_{C} ∘ π^{−1}(S′\{a}) = S′\{a}. It follows similarly that π ∘ Ψ_{C} ∘ π^{−1}(S′′) = S′′ for any sphere S′′ orthogonal to S.

It then follows that formula (4) gives the value of π ∘ Ψ_{C} ∘ π^{−1}(p), and hence that ψ_{S} = π ∘ Ψ_{C} ∘ π^{−1}. To see this, we can apply the discussion of Figure 3 in §2 where we interpret S, S′ and S′′ as S_{ρ}(a), a plane (orthogonal to the paper), and a sphere respectively. Technically, we should also introduce the plane of the paper S′′′ which can be used to show that p′ ∈ S′′′.

There is nothing special about S = S_{ρ}(a) in this reasoning, except that its inverse image is a great sphere. We have actually shown the following.

Lemma 16 Let S be any sphere or plane in ℝ^{3} whose inverse image C = π^{−1}(S) is determined by a three-plane Π = {x · n = 0} through 0 ∈ ℝ^{4}. Then the reflection ψ_{S} about S is given by ψ_{S} = π ∘ Ψ_{C} ∘ π^{−1} where Ψ_{C} is the reflection about Π given by (11).

We are now in a position to finish the proof of Proposition 13. The rotation C̃ = R_{ψ}^{xw}(C) is also a great sphere in 𝕊^{3}, and for 0 < |ψ| ≤ π/2, the plane Π̃ = R_{ψ}^{xw}(Π) = {x : (sin ψ, 0, c, − cos ψ) · x = 0} does not contain (0, 0, 0, 1). Thus, we have from (7) that π(C̃) is the sphere

S̃ = {ã : |ã − (tan ψ, 0, c/ cos ψ)|^{2} = (1 + c^{2})/ cos^{2} ψ}.

If we set m̃_{0} = (tan ψ, 0, 0) and ρ̃_{0} = 1/ cos^{2} ψ, we see that Q = π ∘ R_{ψ}^{xw}(C) satisfies Definition 10 since

ψS̃(Q) = π ∘ ΨC̃ ∘ π^{−1} ∘ π ∘ R_{ψ}^{xw}(C)
= π ∘ R_{ψ}^{xw} ∘ Ψ_{C} ∘ R_{−ψ}^{xw} ∘ R_{ψ}^{xw}(C)
= π ∘ R_{ψ}^{xw} ∘ π^{−1} ∘ π ∘ Ψ_{C} ∘ π^{−1} ∘ π(C)
= π ∘ R_{ψ}^{xw} ∘ π^{−1} ∘ ψ_{S} ∘ π(C)
= π ∘ R_{ψ}^{xw}(C) = Q.

We have established that Q has a vertical line (x, y) = (tan ψ, 0) of generalized reflectional symmetry for 0 < |ψ| ≤ π/2.

The planes P_{ϕ} = {a : (− sin ϕ, cos ϕ, 0) · a = 0} are also planes of reflective symmetry for Q_{0}. Applying Lemma 16 and Lemma 15 much as we have done above with the spheres S_{ρ}(a) we find that for 0 < |ψ| ≤ π/2 the spheres

S̃ = π ∘ R_{ψ}^{xw} ∘ π^{−1}(P_{ϕ})
= {a : |a − (− cot ψ, cot ϕ/ sin ψ, 0)|^{2} = (1 + cot^{2} ϕ)/ sin^{2} ψ}

are spheres of symmetry along the horizontal line (x, z) = (− cot ψ, 0) which satisfy Definition 10 with m̃_{0} = (− cot ψ, 0, 0) and ρ̃_{0} = 1/ sin^{2} ψ. This finishes the proof of statement (ii).

The first identity in statement (iii) follows from explicit calculation and the following observation.

Lemma 17 For any function f : ℝ^{4} → ℝ^{4} we have {f(x, y, z, w) : x ∈ C} = {f(x′, y′, z′, w′) : x ∈ C} where x′ is ±x and y′ is ±y (or possibly x′ is ±y and y′ is ±x), and similarly z′ is ±z and w′ is ±w (or possibly z′ is ±w and w′ is ±z).

Statement (iiib) follows from (iiia). □

Proof of Theorem 14. We use again the fact that rotations are precisely those linear transformations that correspond to orthogonal matrices of determinant 1. Let M be the matrix representing R. Let N = N(θ, ϕ, ψ) be the unknown matrix representing R_{ψ}^{xw} ∘ R_{ϕ}^{zw} ∘ R_{θ}^{xy}, and let N_{0} be the unknown matrix representing R_{0}. We then need to show M = N_{0}N.

We know that N_{0} has the form

N_{0} = ( \begin{matrix}& & & 0 \\ & ⋆ & & 0 \\ & & & 0 \\ 0 & 0 & 0 & 1 \end{matrix} )  (12)

where ⋆ represents a rotation matrix for ℝ^{3}. From this we see that the last row of N_{0}N and the last row of N are the same. This last row is given by N^{T}e_{4} (where T indicates the transpose), and we need to have M^{T}e_{4} = N^{T}e_{4}.

Lemma 18 There exist angles θ, ϕ, and ψ such that M^{T}e_{4} = N^{T}e_{4}.

Proof. Since M is orthogonal, so is M^{−1} = M^{T}. Therefore, the columns of M^{T} (i.e., the rows of M) form an orthonormal basis—see [4, pp. 127–129]. In particular, M^{T}e_{4} = (m_{41}, m_{42}, m_{43}, m_{44}) is some unit vector.

On the other hand, by direct calculation we see that

N^{T}e_{4} = (sin ψ cos θ, − sin ψ sin θ, cos ψ sin ϕ, cos ψ cos ϕ).

Since m_{41}^{2} + m_{42}^{2} ≤ 1, there is some angle ψ with sin^{2} ψ = m_{41}^{2} + m_{42}^{2}. If sin ψ ≠ 0, then we can find θ with cos θ = m_{41}/ sin ψ and sin θ = −m_{42}/ sin ψ. If sin ψ = 0, then m_{41} = m_{42} = 0. In either case, our choice of θ and ψ implies that the first two coordinates of M^{T}e_{4} and N^{T}e_{4} agree. Since m_{43}^{2} + m_{44}^{2} = 1 − m_{41}^{2} + m_{42}^{2} = cos^{2} θ, we can choose ϕ, much as we chose θ, and have the last two coordinates match. □

To prove Theorem 14 it remains to specify N_{0}. Let m_{j} = (m_{j1}, m_{j2}, m_{j3}, m_{j4}) be the jth row of M and n_{j} be the jth row of N for j = 1, 2, 3, 4, and consider

MN^{−1} = ( \begin{matrix}m_{1} \\ m_{2} \\ m_{3} \\ n_{4} \end{matrix} ) (n_{1}, n_{2}, n_{3}, m_{4})

= ( \begin{matrix}& & & m_{1} · m_{4} \\ & ⋆ & & m_{2} · m_{4} \\ & & & m_{3} · m_{4} \\ n_{4} · n_{1} & n_{4} · n_{2} & n_{4} · n_{3} & n_{4} · m_{4} \end{matrix} )

= ( \begin{matrix}& & & 0 \\ & ⋆ & & 0 \\ & & & 0 \\ 0 & 0 & 0 & m_{4} · m_{4} \end{matrix} )

which is of the form (12). Thus, we let N_{0} be the rotation matrix on the right and clearly we have M = N_{0}N. □

Epilogue

We observed in §2 that stereographic projections of rotations of the equator circle in 𝕊^{2} are circles. Exercise 5 points out that other circles in 𝕊^{2} have this property as well. That is, we have not characterized the rotations of the equator circle C. Nevertheless, C and its rotations (the great circles) are “balanced” on the surface of 𝕊^{2} in a way that the other circles are not. This “balance” is expressed precisely by saying the geodesic curvature is zero or simply that these curves are geodesics. It is this balance that justifies the specific attention we have given to the great circles.

In a similar way, the Clifford torus is “balanced” in the three-sphere 𝕊^{3}, because its mean curvature is zero, i.e., it is a minimal surface. While it can be shown that every geodesic curve in 𝕊^{2} is (part of) a great circle, there is a great variety of minimal surfaces in 𝕊^{3}. There are spheres and tori and surfaces of genus two (two holed tori), etc.. In fact, Lawson [7] has given examples of closed minimal surfaces in 𝕊^{3} of every topological genus, i.e., tori with any number of holes. On the other hand, Bryant [3] has shown that every embedded (non self-intersecting) minimal surface in 𝕊^{3} that is topologically spherical stereographically projects to a standard Euclidean sphere. A similar characterization for minimal tori is not known, but it is believed that, up to a rigid rotation of 𝕊^{3}, the Clifford torus C is the unique embedded minimal torus.

For this reason, symmetry properties of C as shown above are of great interest. We remark finally that the only closed surfaces in ℝ^{3} possessing the symmetry shown above for stereographic projections of C are topologically spherical or toroidal [8]. This symmetry, moreover, has geometric consequences as well, and we hope to give, in another paper, an elementary introduction to the curvature of surfaces in 𝕊^{3} and prove that C is the unique minimal torus possessing such symmetry.

Appendix: Rotations

We seek below to give an intuitive introduction to the family of rotations of Euclidean space ℝ^{n}, n ≥ 3. Our starting point is with the distance preserving transformations T : ℝ^{n} → ℝ^{n} which satisfy

|T (x) − T (y)| = |x − y|.  (13)

We refer to all such transformations as rigid motions, and our objective is to determine which rigid motions should be called rotations.

Notice first of all that translations are rigid motions. That is, for any fixed vector a, the transformation defined by T (x) = x + a satisfies (13). It is a fundamental algebraic fact that up to a translation every rigid motion is linear.

Theorem 19 If T_{0} : ℝ^{n} → ℝ^{n} is a rigid motion, then T defined by T (x) = T_{0}(x) − T_{0}(0) is a linear transformation.

Proof.^{3} Recall that T is linear if T (ax) = aT (x) and T (x) + T (y) = T (x + y) for all x, y ∈ ℝ^{n} and a ∈ ℝ^{1}. Recall also the triangle inequality in ℝ^{n}:

|x + y| ≤ |x| + |y| with equality only if x = λy for some λ ≥ 0.

See [9, Exercise 1-2] for a proof.

Because T preserves distance and fixes the origin,

|T (x)| = |x|, |T (ax)| = |a| |x|, |T (ax) − T (x)| = |a − 1| |x|.  (14)

On the other hand, by the triangle inequality

|T (x) − T (ax)| + |T (ax)| ≥ |T (x)|.  (15)

3This discussion is considered in a more general setting in [2, pp. 1–6].

If 0 ≤ a ≤ 1, one can use (14) to check that equality holds in (15). Consequently, for some λ ≥ 0

T (x) − T (ax) = λT (ax),

or T (x) = (1 + λ)T (ax). Taking the norm of both sides we see that T (ax) = aT (x).

By exchanging |T (ax)| and |T (x)| in (15), and following the same line of reasoning, one sees that T (ax) = aT (x) also for 1 < a.

Finally, if a < 0, the same reasoning applied to the inequality

|T (ax)| + |T (x)| ≥ |T (x) − T (ax)|

yields again that T (ax) = aT (x).

Next consider T (x + y). In fact, let a < 0 and note that

|T (x + ay) − T (x)| + |T (x) − T (x + y)| ≥ |T (x + ay) − T (x + y)|.

The left side is |ay| + |y| = (1 − a)y, and the right side is |ay − y| = (1 − a)y. Since they are equal, there exists λ ≥ 0 with T (x + ay) − T (x) = λ(T (x) − T (x + y)). It is easy to see that λ = −a, so

T (x + ay) = T (x) + a(T (x + y) − T (x)).

Subtracting aT (y) = T (ay) from both sides and rearranging we get

a[T (x + y) − (T (x) + T (y))] = T (x + ay) − T (ay) − T (x).

Therefore,

|a| |T (x + y) − (T (x) + T (y))| ≤ |T (x + ay) − T (ay)| + |T (x)|
= 2|x|.

Notice that the right side is a fixed value, but |a| on the left may be taken as large as we like. The only way the inequality can continue to hold is if

|T (x + y) − (T (x) + T (y))| = 0,

i.e., T (x + y) = T (x) + T (y). □

From now on, we assume our rigid motions satisfy

T (0) = 0.  (16)

Thus, we are only going to consider rotations about the origin—since other rotations only differ from these by a translation. Furthermore, we can refer to the matrix M which corresponds to a rigid motion T (which will be the matrix of T with respect to the standard basis unless stated otherwise).

From Theorem 19 we can easily prove Corollary 20 If e_{1}, …, e_{n} form an orthonormal basis, then T (e_{1}), …, T (e_{n}) form an orthonormal basis as well.

Proof. Once linearity is established, the preservation of orthonormal bases follows by expressing the inner product in terms of the norm |x| = \sqrt{x · x}. In fact,

x · y = \frac{1}{2}(|x + y|^{2} − |x|^{2} − |y|^{2}).

It follows, using the linearity, that T (x) · T (y) = x · y. In particular, T (e_{i}) · T (e_{j}) = e_{i} · e_{j} = δ_{ij} where e_{1}, …, e_{n} form an orthonormal basis. □

Any linear transformation that preserves orthonormality of bases is called an orthogonal transformation. The main properties of orthogonal transformations and the matrices that represent them may be found in [4, pp. 127–129]. In particular, the inverse matrix M^{−1} that represents T^{−1} is the transpose matrix M^{T} of M. It follows from this and the product formula for determinants that (det M)^{2} = 1, or since M is a real matrix, that det M = ±1. As pointed out in the introduction, the additional condition

det M = 1  (17)

is often used to distinguish T as a rotation.

So far we have used the intuitive condition (13) and the normalization (16) to derive some algebraic facts. We now return to our intuition concerning rotations and ask for a precise condition which, in conjunction with (13), will express our intuitive idea of what defines a rotation. This is not so easy, but we should keep in mind that such a condition is likely to be equivalent to (17).

Our first approach might be to give an intuitive (yet precise) definition of orientation, and then try to connect a condition concerning orientation preserving transformations with (17). The reader may be surprised to find, as we were, that our physical intuition concerning this approach is limited to two dimensions. To see this, take a piece of paper and draw an orthonormal basis on it dark enough so that you can see it from the back side of the paper (see Figure 6). Now draw a dot on the desk to represent the origin 0. Any rigid motion of the plane (i.e., the paper) which fixes the origin must map e_{1} to some unit vector (cos θ, sin θ)—which you can draw on the desk. Now there are two obvious ways to rigidly move the paper so that it lies flat on the desk and e_{1} lies on top of (cos θ, sin θ). Intuitively, if the paper is facing up, the motion is a rotation. If the paper is facing down, it is not. That is, whether or not a rigid motion is a rotation is determined by how we “orient” the paper before placing it on the desk.

It is very difficult however (if not impossible) for us to rigidly move a physical representation of ℝ^{3} (like a wooden block) so as to change its orientation.

It turns out that the only way to define orientation of bases for ℝ^{3} is, one way or another, to append an additional dimension. One feels, however, that we do have an intuitive idea of what constitutes a rotation of ℝ^{3} independent of additional dimensions.

A second approach might be based on the idea of a rotation axis. Indeed, every rotation of ℝ^{3} has 1 as an eigenvalue so that it does have a fixed vector, x = R(x), which can be used to define an axis of rotation (See [4, pg. 291, Corollary 33.3]). Unfortunately, there is no such rotation axis for non-trivial rotations of ℝ^{2}, and there need not be one for rotations of ℝ^{4}.

A third approach (since we are getting frustrated) could be to use the idea of decomposition as in Theorems 4 and 14—except in reverse. That is, we could define an elementary rotation to be a rotation of just one coordinate two-plane, i.e., a transformation R_{ψ}^{kj} corresponding to a matrix of the form

\begin{matrix}k → \\ j → \end{matrix} ( \begin{matrix}I_{k−1} \\ & cos ψ & & − sin ψ \\ & & I_{j−k−1} \\ & sin ψ & & cos ψ \\ & & & & I_{n−j} \end{matrix} )

where I_{m} denotes an m × m identity matrix and there are zeros filling all the spaces. Then we could define a rotation to be a composition of elementary rotations. After pondering this, however, it is not at all clear that a composition of rotations should be a rotation. In fact, the decomposition in Theorem 4 is not intuitively a rotation of ℝ^{3}—it is the composition of three rotations, one executed after another in time. And here is the key. A rotation is a transformation which can be realized as a physical rigid motion (parameterized by time) that is the same motion at each instant of time. To make this statement precise is fairly easy.

0 ^{}e_{2} e_{1}

Figure 6: Moving orthonormal bases.

Definition 21 A rigid motion R is a rotation if there is a smoothly parameterized family of rigid motions R_{0}(t) such that R_{0}(0) = id_{ℝ^{n}} and, for each m = 1, 2, 3, …,

R_{0}(1/m)^{m} = \underbrace{R_{0}(1/m) ∘ ⋯ ∘ R_{0}(1/m)}_{m times} = R.

Notice that R_{0}(t) for t ∈ (0, 1) is not explicitly required to be a rotation (so the definition is not circular). One might be worried however that the definition allows transformations of determinant −1 which we don’t want as rotations in ℝ^{2}. It is easy to show that this does not happen, but it turns out that the most difficult thing to see is that condition (17) does not allow transformations that Definition 21 excludes. Nevertheless, we have the following.

Proposition 22 If R : ℝ^{n} → ℝ^{n} is a rigid motion represented by the matrix M, then the following are equivalent.

(i) R is a rotation.

(ii) There exist elementary rotations R_{1}, R_{2}, …, R_{k} such that R = R_{1} ∘ ⋯ ∘ R_{k}.

(iii) det M = 1.

(iv) With respect to some basis R is represented by a matrix of the form

( \begin{matrix}I_{k} \\ & R_{θ_{1}} \\ & & . & . & . \\ & & & & & R_{θ_{(n−k)/2}} \end{matrix} )

where I_{k} is a k × k identity matrix and R_{θ_{1}}, …, R_{θ_{(n−k)/2}} are 2 × 2 rotation matrices.

Notice that condition (ii) was mentioned above as a condition that came somewhat short in expressing our intuitive idea of a rotation. It is however a useful condition, of which Theorems 4 and 14 are particular instances, so we have included it. We will use the following exercise and lemma to show that it is part of the equivalence.

Exercise 23 Show that elementary rotations are rotations and have determinant 1.

Lemma 24 Given any two vectors v, w ∈ ℝ^{n} of the same length, there is a rotation Q, which is a composition of elementary rotations R_{θ}^{kl}, such that Qv = w.

Proof. We first note that it is enough to prove the lemma for v = e_{n} and w = u an arbitrary unit vector. To see this simply note that ṽ = v/|v| and w̃ = w/|w| are unit vectors. Thus, if we can find Q_{1} and Q_{2} (compositions of elementary rotations) with Q_{1}(e_{n}) = ṽ and Q_{2}(e_{n}) = w̃, we can take Q = Q_{2} ∘ Q_{1}^{−1} and it is easily checked that Q(v) = w.

We prove that e_{n} can be “coordinate rotated” to u by induction. The initial case, n = 2 follows from the fact that in ℝ^{2} any unit vector u can be represented by (cos θ, sin θ) for some angle θ.

For n > 2, let u = (u_{1}, …, u_{n}) and u_{n} = cos ϕ. It follows that for some v = (v_{1}, …, v_{n−1}) ∈ ℝ^{n−1}

R_{ϕ}^{1n}(e_{n}) = (v_{1}, …, v_{n−1}, u_{n}).

Also, by Exercise 23, R_{ϕ}^{1n} preserves length, so |v| = |w| where w = (u_{1}, …, u_{n−1}). By induction, there is a composition Q of elementary rotations of ℝ^{n−1} such that Q(v) = w. Notice that Q extends to a composition of elementary rotations of ℝ^{n}, and we have Q ∘ R_{ϕ}^{1n}(e_{n}) = u. This completes the induction and the proof of Lemma 24. □

Proof of Proposition 22. That (i) implies (iii) follows from the product formula for determinants applied to R = R_{0}(1/2) ∘ R_{0}(1/2). That (iii) implies (iv) is Theorem 30.5 in [4, pg. 270] where rotations are viewed as orthogonal transformations of determinant 1. Condition (i) follows from (iv) by taking R_{0}(t) to be the transformation corresponding (in the same basis) to the matrix

( \begin{matrix}I_{k} \\ & R_{tθ_{1}} \\ & & . & . & . \\ & & & & & R_{tθ_{(n−k)/2}} \end{matrix} ) .

Condition (ii) we deal with separately. It is clear from the product formula for determinants and Exercise 23 that (ii) implies (iii). We obtain the reverse implication by induction. As discussed above, if n = 2, then R(e_{1}) must be some unit vector (cos θ, sin θ). By orthogonality R(e_{2}) = ±(− sin θ, cos θ). Only the + sign is compatible with (iii).

For n ≥ 3, let M be the orthogonal matrix representing R. Let v = R(e_{n}). According to Lemma 24, there is a composition Q of elementary rotations such that Qv = e_{n}. Thus, Q ∘ R, fixes e_{n}, and if N is the matrix representing Q we have

NM = ( \begin{matrix}& & & 0 \\ & U & & ⋮ \\ & & & 0 \\ 0 & ⋯ & 0 & 1 \end{matrix} )  (18)

where U is an (n−1)×(n−1) rotation matrix. By Exercise 23 and the multiplication formula for determinants det U = det (NM) = det M = 1. Therefore, by induction U = U_{1} ⋯ U_{k} for some elementary rotation matrices U_{1}, …, U_{k}. These rotations extend linearly to elementary rotations of ℝ^{n} represented by matrices N_{j} of the form (18) with U_{j} in place of U. Hence, M = N^{−1}N_{1} ⋯ N_{k} is a product of elementary rotation matrices. □

We note that the construction just used to extend the elementary rotation matrices can be used to extend any rotation to a rotation on a higher dimensional space. To be precise, if R is a rotation of ℝ^{k} with standard basis elements e_{1}, …, e_{k}, k < n, and J = {j_{1} < ⋯ < j_{k}} is a subset of k indices from {1, …, n}, then one obtains a rotation R̃ of ℝ^{n} with standard basis elements ẽ_{1}, …, ẽ_{n} by setting R̃(ẽ_{j}) = ẽ_{j} if j ∉ J and R̃(ẽ_{j_{l}}) = ∑ b_{m}ẽ_{j_{m}} where R(e_{l}) = ∑ b_{m}e_{m}. The rotation R̃ is called a trivial extension of R.

Appendix: Conformality of Stereographic Projection

Here we give a short proof of Lemma 6. Let us first assume that the two curves C_{1} and C_{2} intersect at the point p = (0, 0, −1) in an angle γ(C_{1}, C_{2}, p). Let T_{j} be a unit tangent vector to C_{j} at p_{j} for j = 1 and 2, and let C̃_{j} be the intersection (circle) of the plane Π_{j} containing T_{j} and (0, 0, 1) with 𝕊^{2}. Clearly we have γ(C_{1}, C_{2}, p) = γ(C̃_{1}, C̃_{2}, p) is the angle between Π_{1} and Π_{2}.

On the other hand, π(p) = (0, 0), and π(C̃_{j}) is the intersection (line) of Π_{j} with the x, y-plane for j = 1 and 2. Thus, the angle of intersection of π(C̃_{1}) and π(C̃_{2}) is again the angle between Π_{1} and Π_{2}. We have shown that π is conformal at p = (0, 0, −1).

If the point of intersection p is any point other than (0, 0, −1) in 𝕊^{2}\{(0, 0, 1)}, we again take tangents T_{j} to C_{j} at p and let C̃_{j} be the intersection (circle) determined by the plane Π_{j} containing T_{j} and (0, 0, −1) for j = 1 and 2. Here π(C̃_{1}) and π(C̃_{2}) are circles that intersect in two points q_{1} = π(p) and q_{2} = (0, 0). Since circles (on the sphere and in the plane) intersect in equal angles at their two points of intersection, we have

γ(π(C_{1}), π(C_{2}), π(p)) = γ(π(C̃_{1}), π(C̃_{2}), π(p))
= γ(π(C̃_{1}), π(C̃_{2}), (0, 0))
= γ(C̃_{1}, C̃_{2}, (0, 0, −1))
= γ(C̃_{1}, C̃_{2}, p)
= γ(C_{1}, C_{2}, p).□

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John McCuan Mathematics Department University of California, Berkeley Berkeley, CA 94720 johnm@math.berkeley.edu