Duality for curves in equiaffine geometry

Jan Dymara

0. Introduction

The purpose of this article is to describe an analogue of the classical dualities for plane curves in the equiaffine (special affine) context. Let us first rewiev some of the main features of the classical cases.

To construct the projective dual of a plane curve, one takes the curve of lines tangent to it, and thus gets a dual curve in the dualprojective space (RP^{2})^{∗}. Any curve can be obtained as a dual. This duality helps to recognize inflexion points and double tangents of plane curves—they result in singular points of the dual: cusps and nodes (crossings).

There is also a duality corresponding in plane euclidean geometry: one associates to a plane curve its evolute—the curve γ̂ of centres of circles osculating γ. The vertices (extremal points of euclidean curvature) of γ correspond to singular points (cusps) of γ̂. It is not possible to recover γ from γ̂: we should also encode the radii of the osculating circles, say as a third coordinate of γ̂. Thus γ̂ becomes a curve in R^{3} (the space of circles in R^{2}), and γ can be obtained as the intersection of R^{2} × {0} ⊂ R^{3} with lines tangent to γ̂.

If we have a duality ˆ that relates curves in R^{n} and curves in R^{k}, then it is a diffeomorphism of the spaces of such curves, which are C_{n(k)} := C^{∞}(I, R^{n(k)}) (I is an interval) divided by the ’gauge’ group of reparametrisations G := Diff (I, I). The differential of the duality at a curve γ ∈ C_{n} should be an isomorphism of the tangent spaces T_{γ}(C_{n}/G) and Tγ̂(C_{k}/G), i.e. an elliptic differential operator between the spaces of sections of the normal bundles of γ and γ̂. In particular these normal bundles should have the same dimension, which means k = n. This is rarely satisfied (e.g. not in the plane euclidean duality or in equiaffine duality). As a result, there are restrictions on the curve γ̂.

Coming back to plane euclidean duality, it turns out that γ̂ is always isotropic in the standard Minkowski structure x_{1}^{2} + x_{2}^{2} − x_{3}^{2} on R^{3}, and, conversely, every isotropic curve (with no inflexion points) is dual to a unique plane curve. Using elementary properties of isotropic curves one deduces Kneser’s lemma: circles osculating a vertex–free segment of a plane curve form a nested family. Note also that singular points of γ̂ are SO(2, 1)–invariant, hence euclidean vertices of plane curves are Möbius invariant. These corrollaries of euclidean duality have been applied successfully in the study of 4–vertex problem ([Tab],[Pin],[KU],[CS]).

In analogy with the above classical dualities, we describe a duality for curves in RP^{2} associated to the equiaffine (special affine) geometry. Out of a curve in RP^{2} we produce the curve γ̂ of its osculating conics, which is a curve in the space P(SM_{3})= RP^{5} of conics in RP^{2}. Actually, we get a curve in an open subset GC ⊂ P(SM_{3}) consisting of non–degenerate conics. The dimension of GC being bigger then 2, we can’t expect all curves in GC to be dual to curves in RP^{2}. Instead, we characterize γ̂ as follows. Let T 3_{p} be the space of conics tangent to c ∈ GC with order 3. Let C_{c} be the subset of T_{c}GC consisting of directions tangent to T 3_{c}. In fact, C_{c} is a 2–dimensional cone. A curve in GC turns out to be dual to a curve in RP^{2} iff its tangents lie in in the cones C_{c} (theorem 3). The equiaffine duality transforms equiaffine vertices of plane curves (extremal points of equiaffine curvature) into singular points of the dual curve (proposition 3). We prove also Kneser’s lemma in the equiaffine context.

One might resonably hope that equiaffine duality will be useful in considerations concerning equiaffine vertices.

I would like to thank Tadeusz Januszkiewicz, Pierre Pansu and Jacek Świa‘tkowski for many helpful conversations.

1. Basic properties of higher tangency

Definition 1 Two curves γ_{1}, γ_{2}: (−ϵ, ϵ) → R^{2} with non–vanishing tangent vector are called tangent of order n (at the point γ_{1}(0) = γ_{2}(0)) if there exist reparametrisations s_{1}, s_{2}: (−δ, δ) → (−ϵ, ϵ) , such that s′_{1} ≠ 0, s′_{2} ≠ 0, s_{1}(0) = s_{2}(0) = 0 and for 0 ≤ k ≤ n we have

(1.1)  \frac{d^{k}}{dt^{k}} [γ_{1} ∘ s_{1}(t)]|_{t=0} = \frac{d^{k}}{dt^{k}} [γ_{2} ∘ s_{2}(t)]|_{t=0} .

It follows from the chain rule that diffeomorphic images of curves tangent of order n are also tangent of order n.

Lemma 1 Suppose γ, η: (−ϵ, ϵ) → R^{2} have non–vanishing tangent vector, and γ(0) = η(0). Let F: R^{2} → R be such that F ∘ η ≡ 0 and dF_{p} ≠ 0 for p in the image of η. Then γ and η are tangent of order n at γ(0) = η(0) iff

(1.2)  \frac{d^{k}}{dt^{k}} [F ∘ γ(t)]|_{t=0} = 0 for 0 ≤ k ≤ n.

Proof.

(⇒) The chain rule implies that (1.2) does not depend on the parametrisation of γ. Fix the parametrisations of γ,η as in definition 1. Now expand (using the chain rule) \frac{d^{k}}{dt^{k}} [F ∘ γ(t)]|_{t=0} to get an expression depending on the derivatives of F and the derivatives of γ at 0 of order not exceeding k. But for l ≤ n we have \frac{d^{l}γ}{dt^{l}} (0) = \frac{d^{l}η}{dt^{l}} (0), and, consequently \frac{d^{k}}{dt^{k}} [F ∘ γ(t)]|_{t=0} = \frac{d^{k}}{dt^{k}} [F ∘ η(t)]|_{t=0} = 0.

(⇐) Choose a coordinate system standardizing F and η, ie such that F (x, y) = y, η(t) = (t, 0) for t in some neighbourhood of 0. We will modify (step by step) the parametrisation of γ, changing its higher and higher derivatives (at 0) into those of η.

1)

(F ∘ γ)′(0) = 0 is equivalent to γ′_{2}(0) = (β, 0) for some β ∈ R. Put s(t) = t/β (β ≠ 0 since γ′(0) ≠ 0). Then (γ ∘ s)′(0) = (1, 0) = η′(0).

2)

Suppose γ^{(l)}(0) = 0 = η^{(l)}(0) for 1 < l < k. (F ∘ γ)^{(k)}(0) = 0 implies γ_{2}^{(k)}(0) = 0, ie γ^{(k)}(0) = (β, 0). Let s(t) = t − \frac{β}{k!}t^{k}; then (γ ∘ s)^{(l)}(0) = 0 for l ≤ k.

QED

2 The dual curve

Definition 2 We call γ: (−ϵ, ϵ) → RP^{2} non–degenerate if it is a smooth curve with non–vanishing tangent vector, satisfying det(γ′, γ′′) ≠ 0 (in some nonhomogeneous coordinate system).

Lemma 2 In the above definition, the condition det(γ′, γ′′) ≠ 0 does not depend on the choice of a nonhomogeneous coordinate system.

Proof. For any nonhomogeneous coordinate system there is a canonical lift of γ to R^{3}\{0} :

(2.1)  ξ(t) = (γ_{1}(t), γ_{2}(t), 1)

(written in the corresponding coordinate system on R^{3} ). The lift ζ of γ attached to some other coordinate system can be expressed in terms of ξ as

(2.2)  ξ(t) = f(t)ξ(t)

where f: (−ϵ, ϵ) → R\{0}. Differentiating (2.2) we get

(2.3)  det(ζ, ζ′, ζ′′) = f^{3}det(ξ, ξ′, ξ′′).

Thus det(ζ, ζ′, ζ′′) ≠ 0 iff det(ξ, ξ′, ξ′′) ≠ 0 QED

Let γ be non-degenerate. Fix some nonhomogeneous coordinate system . We can reparametrize γ in such a way, that det(γ′, γ′′) = 1 (this parametrization is called equiaffine). Then γ′′′(t) = k(t)γ(t). The function k is called the equiaffine curvature of γ.

Theorem 1 Let det(γ′, γ′′) = 1.Then for any t ∈ (−ϵ, ϵ) there exists a unique non–degenerate conic Γ(t) tangent of order 4 to γ at γ(t) (the so–called osculating conic). The map t ↦ Γ(t) from (−ϵ, ϵ) to the space of conics is smooth. Γ(t) and γ are tangent of order 5 at γ(t) iff k′_{γ}(t) = 0 , and of order 6 iff k′_{γ}(t) = k′′_{γ}(t) = 0 .

Proof.(cf. [GMO]) We can assume t = 0. Consider the Taylor series of γ in a neighbourhood of γ(0):

(2.4)  γ(s) = sγ′(0) + \frac{1}{2}s^{2}γ′′(0) + \frac{1}{6}s^{3}γ′′′(0) + ...

Suppose our parametrisation is equiaffine. Then

(2.5)  \begin{aligned}γ′′′ & = kγ′ \\ γ^{′ν} & = k′γ′ + kγ′′ \\ γ^{ν} & = (k′′ + k^{2})γ′ + 2k′γ′′ \\ γ^{ν}′ & = (k′′′ + 4kk′)γ′ + (k^{2} + 3k′′)γ′′ \end{aligned}

Putting (2.5) into (2.4) we get

(2.6)  \begin{matrix}γ(s) = & (s + k(0)\frac{s^{3}}{6} + k′(0)\frac{s^{4}}{24} + (k(0)^{2} + k′′(0))\frac{s^{5}}{120} + O(s^{6}))γ′(0) \\ + & (\frac{s^{2}}{2} + k(0)\frac{s^{4}}{24} + 2k′(0)\frac{s^{5}}{120} + (k(0)^{2} + 3k′′(0))\frac{s^{6}}{720} + O(s^{7}))γ′′(0) \end{matrix}

Now consider a coordinate system such that γ′(0) = (1, 0), γ′′(0) = (0, 1) and the conic defined by the following polynomial:

(2.7)  F_{0}(x, y) = x^{2} − 2y − k(0)y^{2}

Putting (2.6) into (2.7) we get

(2.8)  F_{0}(x(s), y(s)) = k′(0)\frac{s^{5}}{20} + k′′(0)\frac{s^{6}}{120}(s^{6}) + O(s^{7})

Thus, the conic defined by F_{0} is (by lemma 1) Γ(0). The uniqueness of Γ(0) is a consequence of the following lemma:

Lemma 3 Two conics tangent of order 4 at some point are equal.

Lemma 3 is a consequence of formulae (4.3) (to be obtained independently of our present considerations). Returning to the proof of theorem 1, note that the conic defined by (2.7) depends smoothly on γ(0),

γ′(0), γ′′(0) and k(0). However, γ(t), γ′(t), γ′′(t) and k(t) are smooth in t. Hence Γ(t) is smooth in t.

The last claim of the theorem follows immediately from (2.8).

QED

Remarks 1. Theorem 1 implies the projective invariance of k′ ≠ 0 (since the osculating conic and the order of tangency are projectively invariant)

2. Γ(t) depends only on the 4–jet of γ at γ(t).

Thus, to a curve γ in RP^{2} we have assigned a curve Γ in the space of conics.

3 Dual curves are not arbitrary

The space of conics is the projectivisation P(SM_{3}) of the space of symmetric matrices 3×3 with real coefficients; the matrix

( \begin{matrix}a & b & d \\ b & c & e \\ d & e & f \end{matrix} )

corresponds to the conic ax^{2} + 2bxy + cy^{2} + 2dx + 2ey + f = 0. To any curve γ(t) in RP^{2} satisfying det(γ′, γ′′) ≠ 0 we assigne a curve Γ(t) in P(SM_{3}) – the curve of conics osculating γ.

Consider any lifts of γ and Γ to curves in R^{3}\{0} and SM_{3}\{0}; for convenience of notation call these lifts also γ and Γ.

Osculation means

(3.1)  \frac{d^{k}}{ds^{k}} [^{t}γ(s)Γ(t)γ(s)]|_{s=t} = 0 for k = 0, 1, 2, 3, 4

for any t in the domain of γ. Expand all these derivatives:

w_{0}(Γ, γ) = ^{t}γΓγ
w_{1}(Γ, γ) = ^{t}γ′Γγ +^{t}γΓγ′
w_{2}(Γ, γ) = ^{t}γ′′Γγ + 2 ^{t}γ′Γγ′ +^{t}γΓγ′′
w_{3}(Γ, γ) = ^{t}γ′′′Γγ + 3 ^{t}γ′′Γγ′ + 3 ^{t}γ′Γγ′′ +^{t}γΓγ′′′
w_{4}(Γ, γ) = ^{t}γ^{′ν}Γγ + 4 ^{t}γ′′′Γγ′ + 6 ^{t}γ′′Γγ′′ + 4 ^{t}γ′Γγ′′′ +^{t}γΓγ^{′ν}

Each w_{k}(γ, Γ) is the everywhere–zero function of t. Now differentiate w_{0}, w_{1}, w_{2} i w_{3} with respect to t:

0 = w′_{0}(Γ, γ) = w_{0}(Γ′, γ) + w_{1}(Γ, γ)
0 = w′_{1}(Γ, γ) = w_{1}(Γ′, γ) + w_{2}(Γ, γ)
0 = w′_{2}(Γ, γ) = w_{2}(Γ′, γ) + w_{3}(Γ, γ)
0 = w′_{3}(Γ, γ) = w_{3}(Γ′, γ) + w_{4}(Γ, γ)

Consequently w_{0}(Γ′, γ) = w_{1}(Γ′, γ) = w_{2}(Γ′, γ) = w_{4}(Γ′, γ) = 0. However, this means

(3.2)  \frac{d^{k}}{ds^{k}} [^{t}γ(s)Γ′(t)γ(s)]|_{s=t} = 0 for k = 0, 1, 2, 3

Coming back to (not lifted) curves in RP^{2} we translate it as:

Theorem 2 Let γ be a non–degenerate curve in RP^{2}, and Γ the curve of conics osculating γ . Then Γ′(t) and γ are tangent of order 3 at γ(t). In other words: the conics Γ(t) and Γ′(t) are tangent of order 3 at γ(t).

In fact, an obvious iteration of the above argument proves the following:

Lemma 4 Let γ be a non–degenerate curve in RP^{2}, and Γ the curve of conics osculating γ . If Γ(t) is tangent of order k to γ at γ(t) then Γ^{(l)}(t) is tangent of order k − l to γ at γ(t) . If the exact orders of tangency of Γ(t) and γ at γ(t) is k (ie they are tangent of order k but not k + 1),

then the exact order of tangency of Γ^{(l)}(t) and γ at γ(t) is k − l (as long as k − l ≥ 0).

4 The space of conics

The relation of higher–order tangency of conics is intimately related to the geometry of the P GL(3)– action on the space of conics P(SM_{3}) (induced by the action of P GL(3) on RP^{2}). We begin by describing the orbits of this action.

A) Matrices (modulo scaling) of quadratic forms with signature (2,1) or (1,2). Each of them is P GL(3)– equivalent to the matrix

( \begin{matrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} )

Such matrices correspond to non–degenerate nonempty conics in RP^{2}. We denote this orbit GC (general conics). The images of the dual curves Γ are contained in GC.

B) Definite matrices (signature (3,0) or (0,3)), equivalent to

( \begin{matrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & −1 \end{matrix} )

define empty conics. We call this orbit Z (like zero).

The set D = {A | detA = 0} of matrices that are degenerate, ie equivalent to

( \begin{matrix}a & b & 0 \\ b & c & 0 \\ 0 & 0 & 0 \end{matrix} )

splits into three orbits.

C) V = {A | rank A = 0} (Veronese variety). Such matrices define double lines in RP^{2}. D) V_{+} = {A | signature A = (1, 1)}. Such matrices define pairs of lines in RP^{2}. E) V_{−} = {A | signature A = (2, 0) or (0, 2)}. Such matrices define points in RP^{2} (with a conformal

structure on the tangent space).

Remark V is a embedded submanifold of P(SM_{3}), diffeomorphic to RP^{2}. D is made of lines in P(SM_{3}) passing through two points of V (or tangent to V). Z is an open ball with boundary V_{+} ∪ V_{−}.

These properties of the orbits are illustrated by the following two–dimensional picture: Now we will describe the set of conics tangent with a given order to a fixed conic C ∈ GC at a point

p ∈ C.

Since P GL(3) acts transitively on pairs (C, p) (p ∈ C ∈ GC), we can choose (for calculation) any conic and any point on it. Choose C = {x^{2} + y^{2} = 1} and p = (1, 0) (in some nonhomogeneous coordinate system ). Parametrize C by ϕ: s ↦ (sin(s), cos(s)). Consider the conic

(4.1)  A(x, y) = ax^{2} + 2bxy + cy^{2} + 2dx + 2ey + f

To calculate the order of its tangency to C at p we should determine how many consecutive derivatives \frac{d^{n}}{ds^{n}} [A(ϕ(s))]|_{s=0} are zero. Using (4.1) we get:

(4.2)  \begin{aligned}\frac{d^{0}}{ds^{0}} A(ϕ(s))|_{s=0} & = a + 2d + f \\ \frac{d}{ds} A(ϕ(s))|_{s=0} & = 2b + 2e \\ \frac{d^{2}}{ds^{2}} A(ϕ(s))|_{s=0} & = −2a + 2c − 2d \\ \frac{d^{3}}{ds^{3}} A(ϕ(s))|_{s=0} & = −8b − 2e \\ \frac{d^{4}}{ds^{4}} A(ϕ(s))|_{s=0} & = 8a − 8c + 2d \end{aligned}

Let T i_{p}(A) = {A ∈ P(SM_{3}) | A is tangent of order i to C at p}. By (4.2) we get

(4.3)  \begin{aligned}T 0_{p}(C) & = \{( \begin{matrix}a & b & d \\ b & c & e \\ d & e & −a − 2d \end{matrix} )\} \\ T 1_{p}(C) & = \{( \begin{matrix}a & b & d \\ b & c & −b \\ d & −b & −a − 2d \end{matrix} )\} \\ T 2_{p}(C) & = \{( \begin{matrix}a & b & d \\ b & a + d & −b \\ d & −b & −a − 2d \end{matrix} )\} \\ T 3_{p}(C) & = \{( \begin{matrix}a & 0 & d \\ 0 & a + d & 0 \\ d & 0 & −a − 2d \end{matrix} )\} \\ T 4_{p}(C) & = \{( \begin{matrix}a & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & −a \end{matrix} )\} \end{aligned}

Let T i(C) = ⋃{T i_{p}(C) | p ∈ C}. Now we will describe

• T 3_{p}(C) in a P GL(3)–invariant manner — using no coordinate system ,

• T 3(C) in terms of the P GL(3) action on P(SM_{3}), without mentioning RP^{2}.

Proposition 1 T 3_{p}(C) is the pencil of conics (ie a line in P(SM_{3})) containing C and the double line tangent to C at p. This double line is the only degenerate conic in T 3_{p}(C).

Proof. The proposition being P GL(3)–invariant, we are reduced to the case C = {x^{2} + y^{2} = 1} and p = (1, 0). From (4.3) we see that T 3_{p}(C) is a pencil, and obviously C ∈ T 3_{p}(C). Calculate the determinant of the matrix in (4.3):

(4.4)  det ( \begin{matrix}a & 0 & d \\ 0 & a + d & 0 \\ d & 0 & −a − 2d \end{matrix} ) = −(a + d)^{3}

Thus, our pencil contains only one degenerate conic (a = −d). That conic is the double line tangent to C at p:

(4.5)  ( \begin{matrix}−d & 0 & d \\ 0 & 0 & 0 \\ d & 0 & d \end{matrix} ) ↔ −d(x − 1)^{2}

QED

Proposition 2 Take the boundary of {P ∈ V | the pencil through P and C intersects Z} in the set V of double lines; take the cone with apex C over that boundary; the result is T 3(C).

Proof. Consider all pencils through C and double lines.

If a double line D intersects C, there are no emty conics in the pencil through D and C ( any element of that pencil contains the points of intersection of D and C).

If a double line D does not intersect C, the pencil through D and C contains empty conics (we change coordinates making D the line at infinity and C = {x^{2} + y^{2} = 1}; then x^{2} + y^{2} + 1 = 0 is in the pencil).

T 3(C) is the C–apexed cone over the set of double lines tangent to C — this set is the common boundary of the above two sets: {P ∈ V | P ∩ C ≠ ∅} and {P ∈ V | P ∩ C = ∅} = {P ∈ V | the pencil through P and C intersects Z}

QED

5 Duality

Define the incidence variety

I = {(p, C) ∈ RP^{2} × GC | p ∈ C}.

Let p_{1}: I → RP^{2} and p_{2}: I → GC be the projections. On I we have a 2–dimensional distribution F. The subspace F_{(p,C)} of T_{(p,C)}I is spanned by vectors tangent to the following curves:

• (p_{t}, C), p_{t} parametrizing the conic C, p_{0} = p

• (p, C_{t}), C_{t} parametrizing the pencil of conics tangent of order 3 to C at p, C_{0} = C. We say that a curve ξ : (−ϵ, ϵ) → I is F–controlled if ξ′(s) ∈ F_{ξ(s)} for every s ∈ (−ϵ, ϵ) Any non–

degenerate curve γ in RP^{2} has a natural lift to I:

γ̃(t) = (γ(t), the conic tangent of order 4 to γ at γ(t)). Similarily, any curve Γ in GC, such that Γ′(t) ∈ T 3(Γ(t))\{0} (we will call such curve T 3–controlled),

has a natural lift to I:

Γ̃(t) = (p | Γ′(t) ∈ T 3_{p}(Γ(t)), Γ(t)). Points A) and C) of the following lemma are basic for the duality.

Lemma 5 A) Let η be a curve in I, such that \frac{d}{dt}(p_{1} ∘ η) ≠ 0. Then

p_{1} ∘ η is non–degenerate and η = p̃r_{1} ∘ η ⟺ η is controlled by F.

B) Let γ be a non–degenerate curve in RP^{2}. Then

\frac{d}{dt}(pr_{2} ∘ γ̃)(t) ≠ 0 ⟺ k′(t) ≠ 0

(k is the equiaffine curvature in an arbitrary nonhomogeneous coordinate system ). C) Let η be a curve in I, such that \frac{d}{dt}(p_{2} ∘ η) ≠ 0. Then

p_{2} ∘ η is T 3–controlled and η = p̃r_{2} ∘ η ⟺ η is controlled by F

.

D) Let Γ be a curve in GC with non–vanishing tangent vector, controlled by T 3. Then

\frac{d}{dt}(pr_{1} ∘ Γ̃)(t) ≠ 0 ⟺ Γ′(t) i Γ′′(t) are linearly independent

.

Remark Some readers may prefer to interpret points A and C as follows: I can be identified with the space of non–parametrised non–degenerate 4–jets of curves in RP^{2}, or

with the space of non–parametrised nonzero 1–jets of curves in GC; these spaces posess their holonomy distributions, which, carried onto I, both coincide with F.

Proof of lemma 5. 5.A) Let η be a curve in I, \frac{d}{dt}(p_{1} ∘ η) ≠ 0. Fix a coordinate system in RP^{2}. Knowing a pair (p, C) ∈ I is equivalent to knowing the first two derivatives of the conic C at p (in the equiaffine parametrisation) and its equiaffine curvature.

Hence the curve η can be reparametrised (locally) as (γ, v_{1}, v_{2}, k), where γ = p_{1} ∘ η, v_{1}, v_{2} are some vector fields along γ, k : (−ϵ, ϵ) → R. Change the parametrisation η, so that γ is parametrised equiaffinly. Then the condition η = γ̃ can be expressed as follows:

(5.1)  \{ \begin{matrix}γ′ = v_{1} \\ v′_{1} = v_{2} \\ v′_{2} = kv_{1} \end{matrix}

Returning to an arbitrary parametrisation, (5.1) can be written as

(5.2)  η′ = (γ′, v′_{1}, v′_{2}, k′) = t(v_{1}, v_{2}, kv_{1}, 0) + s(0, 0, 0, 1)

for some t, s ∈ R.

But (v_{1}, v_{2}, kv_{1}, 0) and (0, 0, 0, 1) are exactly the velocity vectors of the curves spanning F_{(γ,v_{1},v_{2},k)} (as in the definition of F). Thus η = γ̃ ⟺ η′ ∈ F_{η}.

5.B) Parametrize γ equiaffinly. Then γ̃′ = 1(v_{1}, v_{2}, kv_{1}, 0) + k′(0, 0, 0, 1) [notation as in the proof of 5.A]. Note that Dp_{2}(v_{1}, v_{2}, kv_{1}, 0) = 0 and Dp_{2}(0, 0, 0, 1) ≠ 0. Because of this Dp_{2}(γ̃′) ≠ 0 ⟺ k′ ≠ 0.

5.C) η – a curve in I, \frac{d}{dt}(p_{2} ∘η) ≠ 0. Fix some nonhomogeneous coordinate system on P(SM_{3}). Knowing a pair (p, C) ∈ I is equivalent to knowing the pencil of conics tangent of order 3 to C at p, for which it is enough to know a vector tangent to that pencil at C. Thus, η can be represented (locally) as (Γ, v), where Γ = p_{2} ∘ η and v is a unit–length vector field along Γ, v(t) ∈ T 3(Γ(t)). Reparametrize η, so that |Γ′| = 1. Then η = Γ̃ can be written as Γ′ = v. This is equivalent to

(5.3)  η′ = (Γ′, v′) = t(v, 0) + s(0, v^{⊥})

in an arbitrary parametrisation. Here v^{⊥} is a vector tangent to {u ∈ T 3(Γ(t)) | |u| = 1} at v. However, (0, v^{⊥}) and (v, 0) are exactly the velocity vectors of the curves spanning F_{(Γ,v)} (cf the definition of F).

Thus η = Γ̃ ⟺ η′ ∈ F_{η}

5.D) In the notation of the proof of 5.C: Γ̃′ = t(Γ′, 0) + s(0, Γ^{′⊥}), Dp_{1}(v, 0) = 0,

Dp_{1}(0, v^{⊥}) ≠ 0. As a result Dp_{1}(Γ̃′) ≠ 0 ⟺ s ≠ 0 ⟺ Γ′, Γ′′ are linearly independent.

QED

Note that γ ↦ Γ = p_{2} ∘ γ̃ assigns the curve of osculating conics to a curve γ. Lemma 5 implies that Γ ↦ γ = p_{1} ∘ Γ̃ is the inverse map, if we restrict ourselves to certain classes of curves γ, Γ. We will describe one (“the smallest”) possible choice of these classes.

Theorem 3(duality theorem) The mappings γ ↦ p_{2} ∘ γ̃ and Γ ↦ p_{1} ∘ Γ̃ are mutually inverse bijections of the following classes of curves:

K_{1} = {γ: (−ϵ, ϵ) → RP^{2} | det(γ′, γ′′) ≠ 0, k′_{γ} ≠ 0}
K_{2} = {Γ: (−ϵ, ϵ) → GC | Γ is controlled by T 3, Γ′, Γ′′ are linearly independent }

.

Proof. Let

K = \{η: (−ϵ, ϵ) → I | \frac{d}{dt}(p_{1} ∘ η)(t) ≠ 0, \frac{d}{dt}(p_{2} ∘ η)(t) ≠ 0, η is controlled by F\}.

Then γ ↦ Γ and Γ ↦ γ are mutually inverse bijections of K_{1} = {p_{1} ∘ η | η ∈ K} and K_{2} = {p_{2} ∘ η | η ∈ K}. K_{1} and K_{2} can be written as follows:

K_{1} = \{γ: (−ϵ, ϵ) → RP^{2} | γ′ ≠ 0, γ non–degenerate, \frac{d}{dt}(p_{2} ∘ γ̃)(t) ≠ 0\}
K_{2} = \{Γ: (−ϵ, ϵ) → GC | Γ′ ≠ 0, Γ controlled by T 3, \frac{d}{dt}(p_{1} ∘ Γ̃)(t) ≠ 0\}

In view of the lemma 5.B,D we can write K_{1} and K_{2} as in the theorem. QED

Proposition 3 In theorem 3 we can extend the classes of curves K_{1} , K_{2} to L_{1} , L_{2} as follows: we allow γ have non–degenerate vertices (ie points where k = 0 , k′ ≠ 0 ); the corresponding extension of K_{2} is allowing Γ to have points where Γ′ = 0 , Γ′′, Γ′′′ are linearly independent.

Proof. Take Γ in L_{2} with Γ′(0) = 0. Write Γ(t) = tα(t) , α ∈ C^{∞} , α(0) ≠ 0 [Mil]. Since α(t) ∈ T 3(Γ(t)) , we have α(0) ∈ T 3(Γ(0)). We put Γ̃(t) = (p|α(t) ∈ T 3_{p}(Γ(t)), Γ(t)) ; for t ≠ 0 this agrees with Γ̃ defined at the begining of chapter 5 using Γ′ instead of α. Our new definition shows that Γ̃ is C^{∞} and F controlled even at t = 0. We have to show that γ = pr_{1} ∘ Γ̃ has a non–degenerate vertex at 0.

First we show that γ is non–degenerate at 0. This follows from Lemma 5A if we show that γ′(0) ≠ 0. To prove it we use Lemma 5D, with Γ′ , Γ′′ replaced by α , α′ , which are linearly independent because α(0) = Γ′′(0) , α′(0) = \frac{1}{2}Γ′′′(0) (the proof of 5D carries over).

Now, since Γ̃ is F-controlled, we have (by 5A) Γ̃ = p̃r_{1} ∘ Γ̃ = γ̃. So Γ = pr_{2} ∘γ̃, and from 5B we conclude k′_{γ}(0) = 0.

It remains to prove k′′(0) ≠ 0. However, if k′′(0) = 0 then theorem 1 and lemma 4 imply that Γ(0) is tangent to Γ′(0) with order 5, and to Γ′′(0) with order 4. Hence from lemma 3 we get Γ′′(0) = cΓ(0)(= 0 in P(SM_{3}))—contradiction with linear independence of Γ′′(0), Γ′′′(0).

Now suppose we are given a curve γ with a non–degenerate vertex at 0. γ is non–degenerate everywhere, so pr_{2} ∘γ̃ is well defined and C^{∞}; it coincides with Γ for t ≠ 0, hence also for t = 0. From 5B we get Γ′(0) = 0 .

Knowing k′′(0) ≠ 0 we look again at theorem 1 and lemma 3; we conclude that the exact orders of tangency of Γ(0), Γ′′(0) and of Γ(0), Γ′′′(0) are 3 and 2 respectively. This means that Γ′′(0) ∈ T 3_{γ(0)}(Γ(0))\{0} , Γ′′′(0) ∈ T 2_{γ(0)}(Γ(0))\T 3_{γ(0)}(Γ(0)) . In suitable coordinates T 3, T 2 are as in (4.3), which implies that Γ′′(0), Γ′′′(0) are linearly independent.

QED

6 Kneser’s Lemma

Let Z̄ be a compact convex subset of R^{n}, 0 ∈ intZ̄. For x ∈ R^{n} let C_{x} denote the set of all vectors v ∈ R^{n}, such that the half-line starting at x with direction v intersects Z̄:

(5.1)  C_{x} = {v ∈ R^{n} | (∃t ∈ R^{+})(x + tv ∈ Z̄)}.

Also, put

(5.2)  C_{x}^{a} = {x} ∪ (x + C_{x} \ Z̄ + C_{x}).

Lemma 6 (obvious) y ∈ C_{x}^{a} ⇒ C_{y}^{a} ⊆ C_{x}^{a}, C_{x} ⊆ C_{y}

Lemma 7 Let Γ: (−L, L) → R^{n}\Z̄ be a C^{1} curve with non–vanishing tangent vector, such that Γ′(t) ∈ C_{Γ(t)} for any t. Then t > t_{0} implies Γ(t) ∈ C_{Γ(t_{0})}^{a}.

Proof. 1) Suppose that for any t Γ′(t) ∈ intC_{Γ(t)}.

Then for each t there exists a δ, such that

(6.3)  Γ([t, t + δ)) ⊆ C_{Γ(t)}^{a}.

For t = t_{0} choose the largest δ = δ_{0} satisfying (6.3). If t_{0} + δ_{0} < L, then Γ(t_{0} + δ_{0}) lies on/in the boundary of C_{Γ(t_{0})}^{a} (in R^{n}\Z̄). In view of this, (6.3) for t = t_{0} + δ_{0} implies the existence of a δ_{1}, such that

(6.4)  Γ([t_{0} + δ_{0}, t_{0} + δ_{0} + δ_{1})) ⊆ C_{Γ(t_{0}+δ_{0})}^{a} ⊆ C_{Γ(t_{0})}^{a}

(the last inclusion follows from lemma 6). Then, however, δ_{0} can be enlarged — contradiction. Thus t_{0} + δ_{0} = L 2) Without additional assumptions: We can assume t_{0} = 0. Let v be an open neighbourhood of 0 contained in Z̄. Put

(6.5)  r_{ϵ} = −ϵ\frac{Γ(0)}{|Γ(0)|} , Γ_{ϵ}(t) = Γ(t) + tr_{ϵ}.

Fix ϵ_{0}, l > 0, in such a way that Γ_{ϵ}(t) ∈ Γ(0) + V holds for ϵ < ϵ_{0}, 0 ≤ t ≤ l. Then

(6.6)  Γ_{ϵ}(t) = Γ(t) + tr_{ϵ} ∈ C_{Γ(t)}^{a}

Furthermore, Γ′(t) ∈ C_{Γ(t)} implies

(6.7)  Γ′_{ϵ}(t) = Γ′(t) + r_{ϵ} ∈ intC_{Γ(t)} ⊆ intC_{Γ(t)+tr_{ϵ}} = intC_{Γ_{ϵ}(t)}

Now we applying the first part of the proof to Γ_{ϵ}(t) we get

(6.8)  Γ_{ϵ}(t) ∈ C_{Γ(0)}^{a}

for ϵ < ϵ_{0}, 0 ≤ t ≤ l. Making ϵ go to 0 we obtain

(6.9)  Γ(t) ∈ C_{Γ(0)}^{a}

for 0 ≤ t ≤ l. In this way we have shown that for any t there is a δ(= l), such that (6.3) holds. Now we argue exactly as in the first part of the proof. QED

Applying lemma 7 to dual curves in the space of conics we will prove the following

Theorem 4(Kneser’s Lemma). The conics osculating a non–degenerate curve in RP^{2} having no extremal points of equiaffine curvature form a nested family of disjoint curves.

Remark A family of conics Γ(s) is called nested if for s_{0} < s_{1} the contractible part of the complement of Γ(s_{0}) (in RP^{2}) is contained in the contractible part of the complement of Γ(s_{1}).

Proof of theorem 4. In the proof we will use the notation of chapter 4.

Consider the trace functional tr: SM_{3} → R.

Lemma 8 P(tr^{−1}(0)) ∩ (Z ∪ V ∪ V_{−}) = ∅

Proof. Matrices representing elements of Z ∪ V ∪ V_{−} are semi–definite and non–zero, so their traces are positive. QED

In nonhomogeneous coordinates sending P(tr^{−1}(0)) to the hyperplane at infinity Z ∪ V ∪ V_{−} is convex compact and has nonempty boundary (Z is convex and open and V ∪ V_{−} is its boundary). Making perhaps a translation we can assume 0 ∈ Z. We will understand (6.1) and (6.2) in such a coordinate system .

Consider a curve Γ(t) in P(SM_{3}) dual to some non–degenerate curve γ(t) in RP^{2}. We can assume that tr(Γ(t)) > 0 for t ∈ (−ϵ, ϵ) (If this is not the case we change the hyperplane at infinity, which is easy since Z̄ is compact). By theorem 3 and proposition 2 (4), Γ′(t) has the same direction as a line from Γ(t) to some point in V ⊂ ∂Z̄. In particular, Γ′(t) ∈ C_{Γ(t)}, and, by lemma 7, Γ(t) ∈ C_{Γ(t_{0})}^{a} for t > t_{0}. Now, the nondegeneracy of γ implies a stronger condition:

Lemma 9 Γ(t) ∈ intC_{Γ(t_{0})}^{a}

Proof.Suppose not. Suppose Γ(t) ∈ ∂C_{Γ(t_{0})}^{a} for some t > t_{0}. Then for s ∈ [t, t_{0}] we have Γ(t) ∈ ∂C_{Γ(s)}^{a}, hence Γ(s) ∈ ∂C_{Γ(t_{0})}^{a} (otherwise we would have C_{Γ(s)}^{a} ⊆ intC_{Γ(t_{0})}^{a}). Moreover, for any pair s_{0} < s_{1} in [t_{0}, t] we have Γ(s_{1}) ∈ ∂C_{Γ(s_{0})}^{a}, which means that Γ(s_{0}) and Γ(s_{1}) are tangent at some point and do not intersect transversally (a small deformation moving Γ(s_{1}) inside C_{Γ(s_{0})} removes all common points of Γ(s_{0}) and Γ(s_{1})). In an appropriate coordinate system on RP^{2} {Γ(s)} is a family of ellipses continuously depending on s, and such that any two its elements are tangent to each other. In a neighbourhood of the tangency point any of the two ellipses lies on one side of the other; hence

s_{0} < s_{1} ⟺ Γ(s_{0}) is contained in the convex hull of Γ(s_{1}).

In particular all the ellipses Γ(s) are contained between Γ(t) and Γ(t_{0}), hence are all tangent (at p) to the double line B tangent to Γ(t) and Γ(t_{0}) at their point of tangency p . Then, however, Γ|_{[t_{0},t]} takes its values in

W = {C ∈ SM_{3} | C is tangent to B at p}.

W is a three–dimensional projective subspace of SM_{3}, so we have Γ′ ∈ W. But the pencil through Γ and Γ′ passes also through the double line tangent to Γ and lying in W. The double line that are in W are exactly those that pass by p; among them, only B is tangent to non–degenerate conics from W, in particular to Γ. Thus the pencil through Γ and Γ′ passes also through B, hence Γ|_{[t_{0},t]} runs (in SM_{3}) along the pencil determined by Γ(t_{0}) and B. This means γ|_{[t_{0},t]} ≡ p, which contradicts γ′ ≠ 0. QED

Lemma 9 implies that the pencil through Γ(t) and Γ(t_{0}) intersects Z, ie the conics Γ(t) and Γ(t_{0}) are disjoint. QED

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