THE CONVERSE OF THE FOUR VERTEX THEOREM

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1 PROCEEDINGS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 133, Number 7, Pages S (5) Article electronically published on January 31, 25 THE CONVERSE OF THE FOUR VERTEX THEOREM BJÖRNE.J.DAHLBERG (Communicated by Ronald A. Fintushel) Abstract. We establish the converse to the four vertex theorem without the positivity condition. 1. Introduction Let T denote the unit circle of the complex plane C and let γ : T C be the parametrisation of a smooth, simple and closed curve such that γ and such that γ is also regular. Here the smoothness condition means that γ is infinitely many times continuously differentiable. If κ : T R denotes the curvature function of γ, then the four vertex theorem asserts that if κ is not a constant, then κ has at least four critical points p i T, i =1,...,4, ordered counterclockwise such that p 1,p 3 are local maxima and p 2,p 4 are local minima and that furthermore κ(p 1 ) >,κ(p 3 ) > and max(κ(p 2 ),κ(p 4 )) < min(κ(p 1 ),κ(p 3 )). This result was apparently first proved for the case of closed convex curves by Mukhopadhyaya [4]. For proofs in the case of simple closed curves see Fog [1], Jackson [3] and Vietoris [5]. The converse of this was studied by Gluck [2], who proved that if κ is a smooth and strictly positive function satisfying the above four vertex property, then κ is the curvature function of a smooth, simple and closed curve. The purpose of this note is to establish the converse to the four vertex theorem without the positivity condition. Theorem 1.1. Suppose κ : T R is not a constant and assume that κ has at least four critical points ordered counterclockwise p i T, i =1,...,4, such that p 1,p 3 are local maxima and p 2,p 4 are local minima with κ(p 1 ) >,κ(p 3 ) > and max(κ(p 2 ),κ(p 4 )) < min(κ(p 1 ),κ(p 3 )). If in addition κ is smooth, then κ is the curvature function of a smooth, simple and closed curve. Received by the editors March 1, 23 and, in revised form, March 29, Mathematics Subject Classification. Primary 53A4. The author was supported by a grant from the NFR, Sweden. Björn Dahlberg died on the 3th of January The results of this paper appeared in his posthumous work. The final version was prepared by Vilhelm Adolfsson and Peter Kumlin, Department of Mathematics, Chalmers University of Technology and Göteborg University, SE Göteborg, Sweden; vilhelm@math.chalmers.se c 25 American Mathematical Society

2 2132 BJÖRNE.J.DAHLBERG We remark that a smooth function K : T R represents the curvature of a smooth, simple and closed curve parametrised by the arc length if and only if the following three conditions are satisfied: (1.1) (1.2) (1.3) τ t Kds=2, e iα(s) ds =, e iα(s) ds if t<τ<2. Here α(s) = s K(t) dt and the parametrisation of the associated curve γ is given by t γ K (t) = e iα(s) ds. The above three conditions all have a geometric interpretation. The first condition expresses that the curve γ K has a well-determined tangent at s =, and the second condition expresses that γ K is a closed curve. Finally, the third condition expresses that γ K is simple, that is, without self-intersections. We will say that κ is a non normalised curvature function if { I = (1.4) κdt and K = 2 I κ satisfies (1.2) and (1.3). In order to prove the theorem it is enough to show the existence of a smooth diffeomorphism ϕ : T T such that κ ϕ is a non normalised curvature function. For letting ψ denote the inverse of ϕ and setting K = 2 I κ ϕ, I = κ ϕdt,then the curve Γ parametrised by Γ(t) = 2 I γ K(ψ(t)) has κ as its curvature function. 2. Preliminary results The construction of the diffeomorphisms required for the proof of Theorem 1.1 will be based on the following observations. Proposition 2.1. Let E j T,j=1,...,4, be non-empty, pairwise disjoint open intervals that are ordered counterclockwise on the unit circle with E j = T. Let a, b be two positive numbers with a b and define the function k by { a on E1 E k = 3, b on E 2 E 4. Suppose furthermore that kds=2. Then k is the curvature function of a closed convex curve parametrised by the arc length if and only if E 3 = { w : w E 1 } and E 4 = { w : w E 2 }. Proof. We begin by establishing the necessity. Let γ be a curve of length 2 whose curvature is given by k = k(s). Since k is strictly positive it is well known that γ is convex. Let κ = κ(ϑ) represent the curvature of γ as a function of the angle ϑ that the tangent forms with the positive x-axis. Then it is easily seen that (2.1) e iϑ 1 dϑ =. κ(ϑ)

3 THE CONVERSE OF THE FOUR VERTEX THEOREM 2133 Let the function F = F (ϑ) parametrise γ with respect to the angle ϑ that the tangent forms with the x-axis. Now set e j = {ϑ : F (ϑ) γ(e j )}, j =1,...,4. Then the e j s are non empty, pairwise disjoint open intervals in T that are ordered counterclockwise and for which e j = T. Then κ = a on U = e 1 e 3 and κ = b on V = e 2 e 4. Since a b and e iϑ dϑ = it follows from (2.1) that (2.2) e iϑ dϑ = e iϑ dϑ =. U Denote by 2l j the length of the interval e j and let c j denote its centre. From (2.2) it follows that e ic1 sin l 1 + e ic3 sin l 3 =ande ic2 sin l 2 + e ic4 sin l 4 =. Since l j > for j =1,...,4 it is easily seen that c 3 = c 1 +, c 2 = c 4 +, l 3 = l 1 and l 4 = l 2, which yields the necessity part of the proposition. The sufficiency part is an easy consequence of (1.1), (1.2) and (1.3). Proposition 2.2. For α C with α < 1, letg α denote the restriction to T of the Möbius transformation g α (w) = w α 1 ᾱw. Let E j T, j =1,...,4, be non-empty, pairwise disjoint open intervals that are ordered counterclockwise on the unit circle with E j = T. Suppose E 3 = { w : w E 1 } and E 4 = { w : w E 2 }. If g α (E 1 )= g α (E 3 ) and g α (E 2 )= g α (E 4 ), then α =. Proof. The proof will be carriedout by a contradiction argument. We assume therefore that α. Let z,ζ denote the end points of the interval E 1. The assumptions on g α imply that g α ( z) = g α (z) andg α ( ζ) = g α (ζ). A straightforward computation shows that therefore ᾱz 2 = α and ᾱζ 2 = α. Under the assumption that α it follows therefore that z 2 = ζ 2, which is impossible since E 1 is non-empty with length strictly less than. This contradiction establishes the proposition. We will need an infinitesimal version of the above propositions. Proposition 2.3. For α C with α < 1, letg α denote the restriction to T of the Möbius transformation g α (w) = w α 1 ᾱw. Let f : T R denote the function defined by f(e iθ )=1whenever θ 2 < 4 or θ 3 2 < 4 and zero elsewhere. Let a and b be two positive numbers such that a b and set κ = a(1 f) +bf. Furthermore, let a and b be normalised so that κdt =2 and define I α = κ g α. Let A α be defined by A α () = and A α = κ g α. Suppose that α depends smoothly on a real parameter t such that α() =. Letting α denote the derivative with respect to the t-variable evaluated at t =we have that I =and V Ȧ(t)e ia(t) dt = z (ξz 1 + ηz 2 ). Here ξ and η are defined by α = ξ + iη, and z =2 2i(b a)e iw, w =(a +2b)/4, z 1 =(e ib 2 1)/b and z2 =(e ia 2 1)/a. Furthermore the vectors z 1 and z 2 are linearly independent over R.

4 2134 BJÖRNE.J.DAHLBERG Proof. We begin by selecting a smooth branch of the argument for points in a neighbourhood of {e iθ : θ θ θ 1 },where<θ <θ 1 < 2. Now let G α denote the argument of the inverse of g α. It is easily seen that Ȧ(t) = if <t< 4, (a b)ġ( 4 ) if 4 <t< 3 4, (a b)(ġ( 4 ) Ġ( 3 4 )) if 3 4 <t< 5 4, (a b)(ġ( 4 ) Ġ( 3 4 )+Ġ( 5 4 )) if 5 4 <t< 7 4, (a b)(ġ( 4 ) Ġ( 3 4 )+Ġ( 5 4 ) Ġ( 7 4 )) if 7 4 <t<2. Since the inverse of g α is given by g α it is also easily seen that Ġ(θ) =2Im( αe iθ ), where Im(w) denotes the imaginary part of w. Hence Ġ(θ + ) = Ġ(t), so that if <t< 4 or 7 4 <t<2, (a Ȧ(t) = b)ġ( 4 ) if 4 <t< 3 4, (a b)(ġ( 4 ) Ġ( 3 4 )) if 3 4 <t< 5 4, (b a)ġ( 3 4 ) if 5 4 <t< 7 4. In particular, we see that since I α = A α (2) wehavethati =. By using that A (t + ) = + A (t) and integrating it is easy to verify that the expression for Ȧ(t) holds. It remains to verify that the vectors z 1 and z 2 are linearly independent over R. We first note that the normalisation κdt =2 means that a + b =2 so that a, b (, 2). Assuming that z 1 and z 2 are not linearly independent, there would exist real numbers α, β such that αz 1 + βz 2 =wherenotbothα and β equals zero. In fact, both α and β have to be non-zero in this case since otherwise z 1 or z 2 has to equal zero in which case 1 = e iw 2 for w = a or w = b, whichin turn implies that w =4nfor some n Z and this contradicts that a, b (, 2). Thus, since both α and β have to be non-zero we get that there is a real number c suchthat c(1 e ib 2 )=e ia 2 1 e ib 4 c(e ib 4 e ib 4 )=e ia 4 (e ia 4 e ia 4 ) c sin(b 4 )=ei(a+b) 4 sin(a 4 ) c sin(b 4 )=ei 2 sin(a 4 ) whereweusedthata + b =2. Consequently, c sin(b 4 )=Re(c sin(b 4 )) = Re(i sin(a )) =. 4 Since c wegetsin(b 4 ) =, which implies that b 4 = n for some n Z, which contradicts b (, 2). Thus, the assumption of linear dependence must be wrong and this yields the proposition. 3. Proof of the plane case Let f : T R denote the function defined by f(e iθ ) = 1 whenever θ 2 < 4 or θ 3 2 < 4 and zero elsewhere. The proof of the theorem will be carried out in two steps. First one chooses a diffeomorphism η of the circle such that for some positive numbers a, b we have that κ = κ η a(1 f)+bf on T. This step is an easy consequence of the four vertex condition. The second step consists in showing the existence of a complex

5 THE CONVERSE OF THE FOUR VERTEX THEOREM 2135 number β with β < 1 such that if g β denotes the fractional transformation g β (w) = w β,w T, 1 βw then K = κ g β satisfies (1.4). We need some definitions in order to construct the first preliminary diffeomorphism. For j =1,...,4, let θ j =(j 1) 2 and set q j = e iθj,a j = {e iθ : θ θ j < 4 ɛ} for a small positive number ɛ. By continuity there are points r j T, j = 1,...,4, also ordered counterclockwise and positive numbers a, b such that <a<band κ(r 1 )=κ(r 3 )= a, κ(r 2 )=κ(r 4 )=b. Pick intervals B j T,j =1,...,4, such that r j B j and κ(w) κ(r j ) <ɛfor all w B j. Let η be any smooth orientation preserving C diffeomorphism of the circle such that η(a j ) B j for j =1,...,4. It is now easily seen that κ η = a(1 f)+bf + e, where e is bounded with e dt < Cɛ and C is independent of ɛ. We now claim that if ɛ has been chosen small enough, then there is a β C, β 1 2 such that κ g β is a non normalised curvature function. To see this set k = κ g β, k = a(1 f g β )+bf g β and let I = kdt, I = k dt, K = 2 I k and K = 2 I k. If ɛ has been chosen small enough, then clearly I whenever β 1 2. Set F (β,ɛ) = e iα (s) ds and F (β) = e iα(s) ds, where α, α are defined by α() = α () = and α = K, α = K. It follows from Propositions 2.1 and 2.2 that F (β) whenever β, β < 1. Noticing that F () = we see from Proposition 2.3 that the restriction of F to a sufficiently small circle centred at has non-vanishing winding number around. By a standard topology argument and the fact that lim ɛ F (β,ɛ) =F (β) it follows that if ɛ has been chosen sufficiently small, then for all ɛ (,ɛ )thereisaβ, β < 1 2 such that F (β,ɛ) =. Since the curve γ(t) = t eiα(u) du is simple and closed and α α Cɛ, it follows that τ (s) ds whenever <t<τ<2, which completes the proof of the t eiα theorem. References [1] D. Fog, Über den Vierscheitelsatz und seine Verallgemeinerungen, Sitzungsbereiche der Berlin Akademie der Wissenschaft (1933), [2] H. Gluck, The converse of the four vertex theorem, L Enseignement Mathématique, II e Serie, Tome XVII (1971), MR (49:9737) [3] S.B.Jackson,Vertices for plane curves, Bulletin of the American Mathematical Society 5 (1944), MR1992 (6:1e) [4] S. Mukhopadhyaya, New methods in the geometry of a plane arc, Bull. Calcutta Math. Soc. 1 (199), [5] L. Vietoris, Ein einfacher Beweis des Vierscheitelsatzes der ebenen Kurven, Archivder Mathematik 3 (1952), MR52137 (14:577h) Department of Mathematics, ChalmersUniversityofTechnology, SE Göteborg, Sweden