Beiträge zur Algebra und Geometrie Contributions to Algebra and Geometry Volume 43 (2002), No. 2, 407-410.

Osculating Plane Preserving Diffeomorphisms

F. J. Craveiro de Carvalho

Departamento de Matemática, Universidade de Coimbra, Portugal e-mail: fjcc@mat.uc.pt

For someone familiar with the notion of self-parallel group for an immersion into euclidean space [3], [6] it is only natural to wonder what happens if, in the case of space curves, normal planes are replaced by, say, osculating planes. We give here a necessary and sufficient condition for the non-triviality of the osculating group of a simple space curve. This is, in the new situation, the group corresponding to the self-parallel group.

Problems of a similar nature have also been considered in [1] and [2].

1.

In what follows X will stand for R or S^{1} and we will be dealing with smooth space curves, that is, C^{∞} immersions f : X → R^{3}. In the case X = S^{1}, we will write f for both f : S^{1} → R^{3} and f ∘ exp, with exp (t) = (cos 2πt, sin 2πt). It will be clear from the context which one we are considering.

The curvature k_{f} of f is assumed not to vanish and (T_{f}, N_{f}, B_{f}) will denote the Frenet-Serret frame. Also we do not assume parametrization by arc-length and denote by v_{f} the velocity of f.

Definition 1. The osculating group O(f) of f : X → R^{3} is the subgroup of Diff (X) formed by the diffeomorphisms δ : X → X such that, for x ∈ X, the osculating planes of f at x and δ(x) coincide.

If f is a plane curve then O(f) is Diff (X) precisely.

Proposition 1. Let f : X → R^{3} be a smooth curve with non-vanishing curvature and torsion. Then O(f) is

0138-4821/93 $ 2.50 \begin{matrix}c \\ ◯ \end{matrix} 2002 Heldermann Verlag

a) cyclic of finite order if X = S^{1},

b) trivial or not finite if X = R.

Proof. The proof is almost a duplicate of proofs given in [1], [2]. It is only included for completeness.

Denote by A_{2}^{3} the open Grassmannian of affine planes in R^{3} and define Õ : X → A_{2}^{3}, where Õ(x) is the osculating plane at x. Since we are assuming non-vanishing torsion Õ is an immersion. This fact implies that the action ϕ : O(f) × X → X, with ϕ(δ, x) = δ(x), is properly discontinuous.

In fact let δ ∈ O(f) and suppose that x ∈ X is such that δ(x) = x. Since Õ is an immersion there is an open neighbourhood U of x such that Õ | U is injective. Then, for y ∈ U ∩ δ^{−1}(U), δ(y) = y because Õ(δ(y)) = Õ(y). Therefore the fixed point set Δ of δ is open. Since Δ is also closed it follows that either δ has no fixed points or is the identity. Consequently the action of O(f) on X is free.

Furthermore if U ∩ δ(U) ≠ ∅ then δ is the identity and O(f) acts in a properly discontinuous way. Hence the projection p : X → X/O(f) is a covering projection and π_{1}(X/O(f), p(x))/p_{∗}(π_{1}(X, x)) ≈ O(f) [4].

If X = S^{1} then X/O(f) is diffeomorphic to S^{1} and it follows that O(f) is cyclic of finite order.

Assume now that X = R and that O(f) is finite. Then X/O(f) is either R or S^{1}. Since O(f) ≈ π_{1}(X/B(f)) it follows that it must be trivial. □

We will also use the tangent group T (f) formed by the diffeomorphisms δ : X → X such that, for x ∈ X, the tangent lines of f at x and δ(x) coincide. As above one can show that if the curvature k_{f} never vanishes the natural action of T (f) on X is properly discontinuous.

2. Non-vanishing torsion

We start by recalling that a simple point for f is a point y ∈ X such that f^{−1}(f(y)) = {y}.

Proposition 1. Let f : X → R^{3} be a smooth curve with non-vanishing curvature. If f has a simple point then T (f) is trivial.

Proof. Let δ ∈ T (f). Then, for x ∈ X,

< f(x) − f(δ(x)) | N_{f}(x) > = < f(x) − f(δ(x)) | B_{f}(x) > = 0.

It then follows that k_{f}(x) v_{f}(x) < f(x) − f(δ(x)) | T_{f}(x) > = 0. Consequently f(x) − f(δ(x)) = 0, for x ∈ X.

If y ∈ X is a simple point then y = δ(y) and, since the action of T (f) is properly discontinuous, δ = id_{X}. □

Proposition 2. Let f : X → R^{3} be a smooth curve with non-vanishing curvature and torsion. If f has a simple point then O(f) is trivial.

Proof. Let δ ∈ O(f). From

f(x) − f(δ(x)) = α(x) T_{f}(x) + β(x) N_{f}(x)

one can conclude, by differentiation, that β(x) = 0, for x ∈ X. That is, f(δ(x)) belongs to the tangent line of f at x.

Since B_{f}(x) = ± B_{f}(δ(x)), for x ∈ X, we can conclude that also T_{f}(x) = ± T_{f}(δ(x)), for x ∈ X. Therefore δ ∈ T (f). By Proposition 1, δ = id_{X} and O(f) is trivial. □

3. Plane arcs

From now on we will assume that τ_{f} vanishes but that the curve is not plane.

Lemma 1. Let f : X → R^{3} be a smooth, simple curve with non-vanishing curvature. If x_{0} is a point such that τ_{f}(x_{0}) ≠ 0 and δ ∈ O(f) then δ(x_{0}) = x_{0}.

Proof. There is an open interval I containing x_{0} where τ_{f} does not vanish. Then, for x ∈ I, f(δ(x)) = f(x) + α(x) T_{f}(x).

Since B_{f}(x) = ± B_{f}(δ(x)), for x ∈ X, we also have T_{f}(x) = ± T_{f}(δ(x)). By differentiation, α(x) = 0, for x ∈ I, and, due to the injectivity of f, δ | I = id_{I}. □

Proposition 1. Let f : X → R^{3} be a smooth, simple curve with non-vanishing curvature and such that τ_{f} vanishes but not everywhere. Then O(f) is non-trivial if and only if f has a plane arc.

Proof. Assume that f : [a, b] → R^{3} is a plane arc for f. Without loss of generality we assume 0 < a < b < 1.

Let δ be a diffeomorphism of [a, b] such that δ(a) = a, δ(b) = b, δ′(a) = δ′(b) = 1 and δ^{(k)}(a) = δ^{(k)}(b) = 0, for k ≥ 2. Any such δ can be extended to a diffeomorphism of R by letting the extension be the identity outside [a, b] if X = R or, in the case X = S^{1}, by letting the extension \overline{δ} be the identity in [0, 1] \ [a, b] and satisfy \overline{δ}(x + 1) = \overline{δ}(x) + 1. The resulting diffeomorphism or the diffeomorphism that it induces for S^{1}, in the case X = S^{1}, is then an element of O(f).

Assume now that there are no plane arcs for f and that δ ∈ O(f). If x_{0} is such that τ_{f}(x_{0}) = 0 then x_{0} belongs to the topological closure of A = {x ∈ X | τ_{f}(x) ≠ 0}. Therefore there exists a sequence (x_{n}), x_{n} ∈ A, which converges to x_{0}. The sequence (δ(x_{n})) converges to δ(x_{0}). Since by Lemma 1 δ(x_{n}) = x_{n}, n ∈ N, it follows that δ(x_{0}) = x_{0}. Using Lemma 1 again, δ must be id_{X}. □

Examples of curves with plane arcs can be constructed using convenient bump functions [5].

References

[1] d’Azevedo Breda, A. M.; Craveiro de Carvalho, F. J.; Wegner, Bernd: On the existence of Bertrand pairs. Pré-publicação 01-06, Departamento de Matemática da Universidade de Coimbra.

[2] Craveiro de Carvalho, F. J.; Robertson, S. A.: The parallel group of a plane curve. Proceedings of the 1^{st} International Meeting on Geometry and Topology, Braga (1997), 57–61. Zbl 0942.53002

[3] Farran, H. R.; Robertson, S. A.: Parallel immersions in euclidean space. J. London Math. Soc. 35 (1987), 527–538. Zbl 0623.53022

[4] Kosniowski, C.: A first course in algebraic topology. Cambridge University Press 1980. Zbl 0441.55001

[5] Spivak, Michael: A comprehensive introduction to differential geometry. Vol. 1, Publish or Perish Inc., 1979. Zbl 0439.53001

[6] Wegner, B.: Self-parallel and transnormal curves. Geom. Dedicata 38 (1991), 175–191. Zbl 0697.53006

Received September 16, 2001