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Infinitesimally helicoidal motions with fixed pitch of oriented geodesics of a space form

Let L be the manifold of all (unparametrized) oriented lines of R^3. We study the controllability of the control system in L given by the condition that a curve in L describes at each instant, at the infinitesimal level, an helicoid with prescribed angular speed alpha. Actually, we pose the analogous more general problem by means of a control system on the manifold G_kappa of all the oriented complete geodesics of the three dimensional space form of curvature kappa: R^3 for kappa = 0, S^3 for kappa = 1 and hyperbolic 3-space for kappa = -1. We obtain that the system is controllable if and only if alpha ^2 not equal kappa. In the spherical case with alpha = (+/-) 1, an admissible curve remains in the set of fibers of a fixed Hopf fibration of S^3. We also address and solve a sort of Kendall's (aka Oxford) problem in this setting: Finding the minimum number of switches of piecewise continuous curves joininig two arbitrary oriented lines, with pieces in some distinguished families of admissible curves.

An helicoid in R 3 with angular speed α is a parametrized surface congruent to φ o by a rigid transformation of R 3 , that is, a map preserving the distance and the orientation.
Now we state vaguely the problem we are interested in: We fix α ∈ R. Given two oriented straight lines 1 and 2 in R 3 , can we move 1 to 2 in such a way that the swept surface resembles at each instant, at the infinitesimal level, an helicoid with angular speed α?
This is a control problem which does not arise from a linear or affine linear distribution. Thus, the convenient setting to pose it precisely is the following, that we learned of from [1] (see also Subsections 2.1 in [2] and 2.6 in [8]).
Definition 1 A control system on a smooth manifold N is a fiber subbundle of the tangent bundle T N , A smooth curve γ : (a, b) → N is said to be admissible if γ (t) ∈ ι (A) holds for each t ∈ (a, b). A control system in N is said to be controllable if for each pair of points in N there exists a piecewise admissible curve joining them.
Let L be the space of all oriented straight lines of R 3 . This is a four dimensional smooth manifold on which the group of rigid transformations of R 3 acts transitively. The problem above translates into defining a certain subbundle A of the tangent bundle T L. For the sake of generality, we study it for the three dimensional space forms, that is, we also consider curves in the manifolds of oriented lines in hyperbolic space H 3 and of oriented great circles in the sphere S 3 . We call G κ the manifold of all oriented geodesics of the three dimensional space form of curvature κ, in particular, L = G 0 . It is diffeomorphic to T S 2 for κ = 0, −1 and to S 2 × S 2 for κ = 1.
The fiber bundles involved are not trivial. Since the problem is global, this is another reason why we choose the above definition of control system.
Our main result, Theorem 8, asserts the following: For Euclidean space, the system is controllable if and only if α = 0. In the hyperbolic case, the system is controllable for all α, while in the spherical case it is controllable if and only if α = ±1; if α = ±1, an admissible curve consists of great circles in a Hopf fibration. The precise statement and the proof can be found in Section 2.
Section 3 addresses a related problem: Given a family F of distinguished curves in a manifold N , to find the minimum number of pieces in F of continuous curves in N joining two arbitrary points in N , which we call the Kendall number of F. In fact, this is a problem of the sort David Kendall used to pose to his students in Oxford in the mid-20th century for the system of a sphere rolling on the plane without slipping and spinning (that is, N is the five dimensional manifold of all positions of a sphere resting on a plane) and the family consists of curves in N determined by rolling along straight lines. It was solved by John Hammersley in [10], as a part of a book dedicated to Kendall for his sixty-fifth birthday (see also Section 4 of Chapter 4 in [13], where the problem is referred to as the Oxford problem and [4] for a more geometric approach).
In our context we can propose two analogues: for the family P α of curves in G 0 = L describing helicoids with angular speed α, and the family H α of α-admissible homogeneous curves in L, that is, those α-admissible curves which are orbits of monoparametric groups of rigid transformations. We find the Kendall numbers for both families.
We would like to thank Yamile Godoy and Eduardo Hulett for helpful suggestions.
2 The α-helicoidal control system For κ ∈ {0, 1, −1}, let M κ be the space form of dimension three with constant Gaussian curvature κ, that is, M 0 = R 3 , M 1 = S 3 and M −1 = H 3 . Let G κ be the space of all complete oriented geodesics in M κ up to parametrizations, i.e., The isometry group of M κ acts transitively on G κ and this induces on it a differentiable structure of dimension four, that renders it diffeomorphic to T S 2 for κ = 0, −1 [3], and S 2 × S 2 for κ = 1 (see Proposition 12). More precisely, for κ = 0, the map Before presenting the control system that concerns us, we need the following definitions. We denote by γ v the geodesic in M κ with initial velocity v.
Definition 2 Let κ ∈ {0, 1, −1} and α ∈ R. Given ∈ G κ , p ∈ and a unit vector A ∈ T p M κ orthogonal to , the α-helicoidal parametrized surface with initial ray and axis γ A , φ α ,p,A : where t → V t and t → B t are the parallel vector fields along γ A with initial values σ (0) and B, respectively. See Figure 1.
In other words, the axis begins at p ∈ with initial velocity A perpendicular to , and the rays rotate with constant angular speed α as they move along the axis with unit speed. Given , p and A as above, we define the α-helicoidal curve with initial ray and axis γ A as and the subset A α κ ⊂ T G κ by We call the elements of this set α-admissible tangent vectors. Now we are in the position of defining the α-helicoidal control system on G κ , that we present in the following proposition.
Proposition 4 Let κ ∈ {0, 1, −1} and α ∈ R. The canonical projection A α κ → G κ is a fiber bundle. Moreover, the inclusion ι α κ : A α κ → T G κ is a fiber subbundle and this gives the control system We will call the admissible curves of this system α-admissible curves.
Remark 5 Although the curves Γ α ,p,A are orbits of elements of G κ under monoparametric groups of isometries of M κ (conjugate one to another ), no vector field on G κ induced by elements of the Lie algebra of Iso (M κ ) is a section of the fiber bundle A α κ → G κ .
The following proposition reinforces the idea that the problem has a global nature and suggests the convenience of working in an invariant setting.
Example 7 a) The curves Γ α ,p,A , i.e. the α-helicoidal curves, are clearly α-admissible. b) The homogeneous α-admissible curves in the Euclidean case are characterized in Proposition 20. Among them, the curve of straight lines that sweeps a one-sheet hyperboloid is admissible for the control system (ι α 0 , A α 0 ), for suitable parameters (see the paragraph after that proposition). This also holds for analogous surfaces in H 3 and S 3 .
c) The curve in L associated with a circular helicoid with angular velocity α = 0 is not α-admissible. We recall that this parametrized surface can be built in an analogous manner as φ α ,p,A , but taking a unit speed circle c with initial velocity A, centered at a point on , instead of γ A , and using the normal connection of c to rotate along it, with angular velocity α. See Proposition 19.
Now we can state our main result. We recall that a submanifold of a vector space is said to be substantial if is not included in any affine subspace. The Hopf fibrations of S 3 are those fibrations by oriented great circles that are congruent by isometries of the sphere to the fibration whose fibers are intersections of S 3 with complex lines, identifying R 4 ≡ C 2 .
Theorem 8 Let α ∈ R. For κ ∈ {0, 1, −1}, the following assertions are equivalent: a) The control system (A α κ , ι α κ ) is controllable. b) It holds that α 2 = κ. c) For every ∈ G κ , the fiber of A α κ over is a substantial submanifold of T G κ . Moreover, in the Euclidean case, the image of a 0-admissible curve consists of parallel straight lines and in the spherical case, if α = ±1, the image of an admissible curve consists of great circles in a Hopf fibration.

Space of oriented geodesics
We begin by setting some notations for the three dimensional space forms. In general, we deal with the three cases simultaneously, but the spherical case will need partly a differentiated approach (see Subsection 2.3).
From now on, {e 0 , e 1 , e 2 , e 3 } denotes the canonical basis of R 4 . For κ ∈ {0, 1, −1}, let M κ be the three dimensional space form with Gaussian curvature κ, that is, M 0 = R 3 and for κ = ±1, M κ is the connected component of e 0 of {x ∈ R 4 : x, y κ = κ}, where x, y κ = κx 0 y 0 + x 1 y 1 + x 2 y 2 + x 3 y 3 , that induces a Riemannian metric on M κ . That is, M 1 is the sphere S 3 and M −1 is hyperbolic space H 3 . To handle the three cases simultaneously, sometimes it will be convenient to identify R 3 ≡ e 0 + R 3 = {p ∈ R 4 : p 0 = 1}. We denote G κ = Iso o (M κ ), the identity component of the isometry group on M κ . Let O (4) and O (1, 3) be the automorphism groups of the inner products , 1 and , −1 , respectively. With the identification R 3 ≡ e 0 + R 3 , it holds that Given an orthonormal subset {u, v} of T p M κ , the cross product u × v is defined as the unique unit vector w such that {u, v, w} is a positively oriented orthogonal basis of T p M κ , that is, {p, u, v, w} is a positively oriented orthogonal basis of (R 4 , , κ ). For instance, e 1 × e 2 = e 3 . It can be extended bilinearly to T p M κ × T p M κ .
Next we recall some properties of the space G κ of oriented geodesics in M κ . Their geometry for κ = 0, −1 has been studied for instance in [9,17,5,18]; for κ = 1 see Subsection 2.3. The isometry group G κ acts transitively on G κ through g · [σ] = [g • σ]. By abuse of notation, we say that a point p is in ∈ G κ if for some parametrization σ of there exists s o such that p = σ (s o ).
We introduce the notation sin 1 (r) = sin r, sin 0 (r) = r, sin −1 (r) = sinh r, cos κ = sin κ It will be convenient for us to present G κ explicitly as a homogeneous space. For B, C ∈ R 2×2 , we denote by diag (B, C) the 4 × 4 matrix with blocks A and B in the main diagonal. We have: We consider on G κ the differentiable structure induced by the bijection For κ ∈ {0, 1, −1} we denote by g κ the Lie algebra of G κ . Also from [7] we have The Lie algebra of K κ is For column vectors x, y ∈ R 2 we call The subspace p κ = {Z(x, y) ∈ g κ : x, y ∈ R 2 } of g κ is an Ad (K κ )-invariant complement of k κ and there exists a natural identification where : G κ → G κ /K κ is the canonical projection.

The fiber bundle
Now we consider a particular case of α-helicoidal curve as in (3), in good position. Let σ o and o be as in (6) and let We call Γ α o the curve in G κ defined by and denote by X α its initial velocity, that is, Proof of Proposition 4. Since G κ acts transitively on the positively oriented orthonormal frame bundle of M κ , given , p, A as in Definition 3, there exists g ∈ G κ such that g (e 0 ) = p, dg e 0 (e 3 ) = A and sends o to keeping the orientation. Since clearly G κ carries α-helicodal curves in α-helicoidal curves, it turns out that the group G κ acts transitively on A α κ . Thus, A α κ = {dg o (X α ) : g ∈ G κ }; in other words, it is the orbit of X α in T G κ under the action of G κ and therefore the inclusion is a fiber subbundle of T G κ .
Next we give an explicit homogeneous presentation of A α κ . We call where Z was defined in (8) and a α κ = 0 α κ 0 .
Proof. For any t ∈ R, let S t ∈ G κ given by We fix t and verify that both expressions are equal as functions of s. Since they are geodesics with the same initial value cos κ t e 0 + sin κ t e 3 , it suffices to see that they have the same initial velocity. We compute as desired. Finally, Proposition 11 a) Under the identification (9), the fiber of A α κ over 0 is Proof. a) We know from the proof of Proposition 4 that G κ acts transitively on A α κ via the differential. Hence, the fiber of The assertion follows now from the lemma above and the commutativity of the diagram b) By homogeneity, we may suppose that v is in the fiber over o . Hence v has the form Since R 1 (s + π) = R 1 (π) R 1 (s) = −R 1 (s), we have that −v = Ad (k (s + π, t)) (ξ α ) and so it belongs to the fiber over o . c) Let H κ (α) be the isotropy subgroup at X α of the action of If κ = 0, this implies that cos s = 1 and − sin s = αt, and so R κ (t) = R 1 (s) = I. If κ = −1, we have that cos s = cosh t = 1 and so we arrive at the same conclusion. In both cases, H κ (α) = {I}, as desired.

The α-helicoidal control system in the spherical case
Let H be the skew field of quaternions. We consider the sphere S 3 as the set of unit quaternions, that is, S 3 = {q ∈ H | |q| = 1}, which is a Lie group. It is well known that, identifying R 4 with H, the maps f : for x ∈ Im (H) ∼ = R 3 and y ∈ H ∼ = R 4 , are both surjective two-to-one morphisms. For brevity, we call C = G 1 the manifold of all oriented great circles of S 3 . We have that S 3 × S 3 acts transitively on C, since the action of SO (4) on C is transitive.
It is well known, for instance from [6] and [16], that C is diffeomorphic to S 2 ×S 2 . We include this assertion in the next proposition and write down the proof since it is different from the ones given in those articles and shorter; also, it contributes to establish the nomenclature used later. Note that Proof. Let (p, q) ∈ S 1 × S 1 . Then p = e it and q = e ir for some t, r ∈ R. Thus, s → pe is q = e i(s+t−r) belongs to the equivalence class [s → e is ] and so S 1 × S 1 is included in the isotropy subgroup. Now, we check the other inclusion. Let p, q ∈ S 3 such that pe is q = e i(s+so) for some s o and all s. Then, pe is = e is e iso q for all s and in particular, p = e iso q. Differentiating, we have pie is = ie is e iso q and so, pi = ie iso q = ip. Since p commutes with i, then p ∈ S 1 and so q = e −iso p ∈ S 1 as well. Therefore, the isotropy subgroup at . Then the expression for Φ follows from the fact that the morphism f in (17) induces a transitive action of S 3 on S 2 ⊂ Im H, given by (p, u) → pup, with isotropy subgroup at i equal to S 1 . Now, we describe in terms of the identification Φ above the curve Γ α o in C in good position defined in (11). Given β, τ ∈ R, we define the isometries R β (q) = e βk/2 qe −βk/2 and T τ (q) = e τ k/2 qe τ k/2 of S 3 (see (17)). The former is a rotation of R 4 fixing 1 and k, and rotating the i-j plane through the angle β. The latter is a transvection in τ along t → e tk (i.e. T τ e tk = e (t+τ )k and its differential realizes the parallel transport along t → e tk , see for instance Theorem 2 (3) in Note 7 of [14]). Notice that R β and T τ commute.

Proposition 13 a)
The α-helicoidal surface in S 3 with axis t → e tk and initial circle s → e si is given by In particular, c) The fiber of A α 1 over (x, y) ∈ S 2 × S 2 , via the identification Φ, is given by Proof. The first assertion follows from the properties of R β and T τ we mentioned when we introduced them above. It implies that Γ α o (t) = s → T t R αt e si = s → e tk/2 e αtk/2 e si e −αtk/2 e tk/2 = (p t , q t ) · c o ,

and ( 19) follows. A straightforward computation yields (20).
Now we verify (c). By homogeneity we may suppose x = y = i. As we saw in the proof of Proposition 4, the group G 1 = SO (4) acts transitively on A α 1 . Since S 3 × S 3 covers SO (4), we may write A α 1 = {pΓ o (0) q : p, q ∈ S 3 }. By Proposition 12, the isotropy subgroup of the action of S 3 × S 3 on S 2 × S 2 C is S 1 × S 1 . Thus, the fiber of A α 1 over c o (i, i) is {pΓ o (0) q : p, q ∈ S 1 } and using (20) we get that it equals Now, putting p = e it , we have pjp = e it je −it = cos (2t) j + sin (2t) k, which are exactly the unit elements on Im H orthogonal to i. Thus, (21) follows.
Next we recall the concept of Hopf fibration. The left multiplication by i in H induces on it a vector space structure over C. We have that {Cq ∩ S 3 | q ∈ S 3 }, the set formed by all the intersections of complex lines with the sphere, is the set of fibers of a fibration of S 3 by oriented great circles, which is known as the standard Hopf fibration. Any fibration congruent to it by an isometry of S 3 (which does not necessarily preserve the orientation) is called a Hopf fibration.
The following proposition is known, for instance, from [6]. For the reader's convenience we give a proof in the framework on this subsection. Proof. As above, let c o = [s → e is ]. The standard Hopf fibration has fibers c o q, with q ∈ S 3 . By (17), the elements of O (4) have either the form q → p 1 qp 2 (those preserving orientation) or the form q → p 1 q p 2 (those inverting orientation), with p 1 , p 2 ∈ S 3 . Then, the set of fibers of a Hopf fibration has the form H l or H r , where Proceeding as for H l , we have then that Φ (H r ) = S 2 × {z} with z = −p 2 ip 2 .
Proof. First we consider the Euclidean case with α = 0. Let t → t be a 0admissible curve in L = G 0 . For each t there exist p t ∈ t and A t such that d dt t = Γ t (0), with Γ t = Γ t,pt,At , that is, where v t ∈ S 2 is the direction of t , in particular, v t ⊥A t . Via the diffeomorphism ψ : T S 2 → L in (1) and recalling the expression for its inverse given afterwards, we have Comparing the first coordinates we obtain v t = 0. Therefore the curve t consists of parallel lines and in particular the system is not controllable.
In order to deal with the spherical case we use the identification C ∼ = S 2 × S 2 introduced in (18). Suppose that α = 1 and let γ = (γ 1 , γ 2 ) be a piecewise admissible curve in S 2 × S 2 . Then, the velocity γ (t) of each piece of γ is in the fiber of A 1 1 over γ (t), which by (21) is included in T γ 1 (t) S 2 × 0 γ 2 (t) . Thus, γ 2 = 0 and then γ 2 is constant, say γ 2 ≡ y o . So, the curve γ lies in S 2 × {y o }, that consists of the fibers of a Hopf fibration, as we saw in Proposition 14. Hence, two oriented circles cannot be joined by a piecewise 1-admissible curve if they do not share the projection onto the second factor. So, the system is not controllable. If α = −1 a similar argument applies, involving {x o } × S 2 .
Proof. Recall that a submanifold N of a vector space W is said to be substantial if it is not included in any proper affine subspace of W . If N is central symmetric, that is −N = N , we can substitute subspace for affine subspace, since the segment joining two opposite vectors in N contains the origin. If W is additionally endowed with an inner product , , then N is substantial if and only if q, u = 0 for every q ∈ N only when u = 0. Now we prove the statement of the proposition. By homogeneity, we may suppose that = o . By Proposition 11 (a) and (b), it suffices to show that Ad (K κ ) (ξ α ) is not contained in a proper subspace of p κ . On this vector space we consider the inner product Z (X, Y ) , Z (U, V ) = X, U + Y, V (see (8)). Let ζ = Z x z w y ∈ p κ and define f ζ : where k (s, t) ∈ K κ is as in (7). Suppose that f ζ ≡ 0. Then By the linear independence of cos and sin we obtain the linear system Now, if α 2 = κ, the system has only the trivial solution and so ζ = 0. Thus, in this case, the submanifold is substantial.
Finally, the submanifold is not substantial if α 2 = κ, since for ζ = Z α 1 −α 1 , a lengthy computation yields f ζ ≡ 0. Now we present the proof of the main result.
Proof of Theorem 8. By Proposition 15, we have that (a) implies (b) and that the last assertion of the theorem is true. The equivalence between (b) and (c) was proved in the previous proposition. Now we verify that (c) implies (a). We apply Sussmann's Orbit Theorem [19] (we also consulted [15]). We begin by showing the existence of a smooth vector field family D defined everywhere whose D-orbits are the whole manifold. Since A α κ → G κ is a fiber bundle with typical fiber F α κ we can take trivializations U i ×F α κ → π −1 (U i ) (i ∈ I) in such a way that the union of all U i covers G κ . Let which is a smooth vector field family defined everywhere. We have to show that its D-orbits are the whole manifold. Let ∆ D be the distribution on G κ defined as follows: ∆ D ( ) is the subspace of T G κ spanned by all v ( ) such that v ∈ D and v is defined on . Since α 2 = κ, we have by Proposition 16 that ∆ D ( ) = T G κ for all . Then the smallest D-invariant distribution containing ∆ D coincides with T G κ . By the Orbit Theorem, the D-orbit of any ∈ G κ is the whole G κ .
Finally, notice that if v ∈ D is as in (22), then −v is also in D by Proposition 11 (b). This implies that the system is controllable. Indeed, let 0 , ∈ G κ and v i ∈ D (i = 1, . . . , k) such that v k t k · · · v 1 t 1 ( 0 ) = , where t → v i t denotes the flow of v i . Call i = v i t i ( i−1 ) and suppose that t j < 0 and γ j : Proof of Proposition 6. We begin by describing the typical fibers. We consider first the cases κ = 0, −1. Since α 2 = κ, we know from the proof of Proposition 11 that the fiber over o can be identified with K κ . Hence, F α κ is homeomorphic to the cylinder by (7). When κ = 1 and α 2 = 1, we have by (21) that F α 1 is homeomorphic to To see that A α κ and G κ × F α κ are not homeomorphic we show that their fundamental groups do not coincide.

Examples
In this subsection we give examples of α-admissible curves. Since we have already dealt with various features of the spherical case, we concentrate on the Euclidean and hyperbolic cases. We relate α-admissible curves to Jacobi fields and use that to describe all the homogeneous α-admissible curves for κ = 0. This provides nontrivial examples, which, in their turn, constitute an interesting family to pose Kendall's problem.
Let σ be a unit speed geodesic of M κ , κ ∈ {0, 1, −1}. A Jacobi field along σ arises from geodesic variations as follows: Let ϕ : R × (−ε, ε) → M κ be a smooth map such that for each t ∈ R, s → ϕ (s, t) = def ϕ t (s) is a unit speed geodesic with ϕ 0 = σ. Then the associated Jacobi field J along σ is given by J (s) = d dt 0 ϕ t (s). We recall that Jacobi fields are the solutions of the equation D 2 J dt 2 + R κ (J, σ ) σ , where R κ is the curvature tensor of M κ , given by R κ (x, y) z = κ ( z, x κ y − z, y κ x) for x, y, z local vector fields on M κ . We have then that the Jacobi field along σ with initial conditions J (0) = u + aσ (0) and DJ dt (0) = v + bσ (0), with a, b ∈ R, u, v ∈ σ (0) ⊥ turns out to be J (s) = cos κ (s) U (s) + sin κ (s) V (s) + (a + sb) σ (s), where U, V are the parallel fields along σ with U (0) = u and V (0) = v. The Jacobi fields J arising from unit speed geodesic variations are exactly those with DJ dt ⊥σ (or equivalently, with b = 0 in the expression (23)). We call J σ the vector space consisting of all such Jacobi fields along σ. There is a canonical surjective linear morphism where σ t is any variation of σ by unit speed geodesics, associated with J (see Section Proposition 17 Fix α = 0 and let J ∈ J σ with J (0) ⊥J (0). If then T σ (J) is α-admissible. Moreover, the converse is true if κ = 0, −1. See Figure  2.
Suppose that T σ (J) is admissible, that is, T σ (J) ∈ A α κ . Since T σ (J) ∈ T M κ , there exist p ∈ and a unit vector A ∈ T p M κ orthogonal to such that T σ (J) = Next we focus on the Euclidean case. Let φ : be a ruled parametrized surface with V = 1 which is nowhere cylindrical, that is, V (t) = 0 for all t. It is said to be standard if β ⊥V . It is well-known that every nowhere cylindrical ruled surface admits such a parametrization; in this case β is called the striction line.
In the next proposition we present the details of Example (c) after Proposition 6.
Now we characterize the α-admissible homogeneous curves in L, that is, those which are orbits of monoparametric groups of rigid transformations. We exclude the trivial case α = 0. For s ∈ R, let R s be the rotation through the angle s around the z-axis and T s the translation given by T s (x) = x + se 3 .
Proposition 20 a) Any homogeneous curve in L is congruent, via an orientation preserving isometry, to the orbit under the one parameter group t → R θt T λt (for some θ, λ) of the oriented line for some ρ ≥ 0 and η.
Therefore, by the corollary, Γ (0) is α -admissible if and only if equations (29) hold. By the homogeneity of Γ and A α 0 , this is equivalent to Γ (t) being α-admissible for all t.

Kendall's problem for some families of α-admissible curves
This section addresses the analogue mentioned in the introduction of the well known rolling Kendall's problem. Given a family F of curves in a smooth manifold N , the Kendall number of F is the minimum number of pieces in F of continuous curves in N taking an initial point to a final point in N , both arbitrary and different. We consider N = G 0 = L and two families of distinguished α-admissible curves there: the family P α , consisting of all (pure) α-helicoidal curves, that is, all curves Γ α ,p,A as in Definition 3, and the family H α of all the α-admissible homogeneous curves in L. Note that this renders the result in Theorem 8 supefluous in the Euclidean case.
In the original Kendall's problem of a sphere rolling on the plane without slipping and spinning, the most difficult case was to roll along successive straight lines from a given position to another one over the same point, but rotated through some angle. In our problem, the most complex case will be to reach − from , two lines with the same image and opposite directions.

Kendall's problem for the family P α
Proposition 21 For α = 0, the Kendall number of the family P α is 3.
We begin by stating the following proposition, that implies that this number is greater than or equal to 3.

Proposition 22
Given α = 0, the oriented straight lines and − cannot be connected by a continuous curve of two α-helicoidal pieces.
Proof. Without loss of generality, we may suppose that = [s → se 1 ] (so, − = [s → −se 1 ]) and that the first piece is Γ ,0,e 2 , defined on the interval [0, t 0 ]. We call 1 = Γ ,0,e 2 (t 0 ) = − . We denote by v the direction of 1 , which is orthogonal to e 2 (the direction of the axis of the first piece), so we can write v = xe 1 + ze 3 with x 2 + z 2 = 1. Now we assume that there exist p ∈ 1 , a unit vector A orthogonal to 1 and t 1 such that Γ 1 ,p,A (t 1 ) = − . Since the axis of Γ 1 ,p,A is orthogonal to − and 1 , we have that A, e 1 = 0 = A, v . Hence z A, e 3 = 0.
If z = 0, then v = ±e 1 and so p = t 0 e 2 + s 0 e 1 for some s 0 . Since the axis t → p + tA of Γ 1 ,p,A intersects − at t 1 , we have that p + t 1 A = s 0 e 1 for some s 0 . Now, s o = t 0 e 2 + s 0 e 1 , e 1 = −t 1 A + s 0 e 1 , e 1 = s 0 .
We know from the previous proposition that the Kendall number of P α is greater than or equal to 3. Given and in L, we want to achieve from via the juxtaposition of three α-helicoidal curves in L. Without loss of generality we may assume that = [s → se 1 ] and = [s → de 2 + sv] for some d ≥ 0 and some unit vector v orthogonal to e 2 . We consider first the case α > 0.

Kendall's problem for the family H α
The elements of the family H α of all α-admissible homogeneous curves in L for α = 0 have been described in Proposition 20.
Proposition 23 Let α = 0. The Kendall number of the family H α is 2.
Let Γ be the curve in L determined by the orbit of under the monoparametric Figure 5: Standard position of and when they intersect. Also, (dh π/θ ) 0 (u) = (u ) group h t = def R θt T λt as in Proposition 20, with θ = α sin η and λ = 1 sin η (ρ = 0). The curve is α-admissible since the corresponding equations given in (29) are satisfied. One can also verify easily that Γ π θ = . Thus, and can be joined by one α-admissible homogeneous curve. See Figure 5. Now we consider two lines and in L that do not intersect. Let 1 ∈ L containing the shortest segment joining to , which is perpendicular to both of them. By the case above with 2η = π 2 , can be reached from via the juxtaposition of two curves in H α , the first joining to 1 and the second joining 1 to . If = − , one can take as 1 any curve orthogonal to . Then, the Kendall number is at most 2.
Finally, we show that the Kendall number is greater than 1. It suffices to see that for the monoporametric group t → g t = R θt T λt as in Proposition 20, if t → g t ( ) is α-admissible for some ∈ L, then g t ( ) = − for all t. We may suppose that = [s → ρe 2 + sv] is as in (28). The direction of g t ( ) is R t (v). If g t ( ) = − , then R t (v) = −v and equating the third components yields cos η = − cos η and so cos η = 0. In particular, is contained in the plane z = 0. Now, equations (29) imply that |θ| = |α| and αλ = θ.
Hence, g t = R λαt T λt with λ = ±1. Since g t is contained in the plane z = λt, we have that g t = − for all t (otherwise, we get λ = 0, a contradiction).