Writhe - Four Vertex Theorem.14 June, 2006

THE FOUR VERTEX THEOREM AND ITS CONVERSE in honor of Björn Dahlberg

The Four Vertex Theorem says that a simple closed curve in the plane, other than a circle, must have at least four "vertices", that is, at least four points where the curvature has a local maximum or local minimum.

There must be at least two local maxima, separated by two local minima.

The Converse to the Four Vertex Theorem says that any continuous real-valued function on the circle which has at least two local maxima and two local minima is the curvature function of a simple closed curve in the plane.

Preassign curvature ..... then find the curve

That is, given such a function κ defined on the circle S1 , there is an embedding α: S1 → R2 whose image has 1 curvature κ(t) at the point α(t) for all t ∊ S .

History.

In 1909, the Indian mathematician S. Mukhopadhyaya proved the Four Vertex Theorem for strictly convex curves in the plane.

Syamadas Mukhopadhyaya (1866 - 1937)

In 1912, the German mathematician Adolf Kneser proved it for all simple closed curves in the plane, not just the strictly convex ones.

Adolf Kneser (1862 - 1930)

In 1971, we proved the converse for strictly positive preassigned curvature, as a special case of a general result about the existence of embeddings of Sn into Rn+1 with strictly positive preassigned Gaussian curvature.

In 1997, the Swedish mathematician Björn Dahlberg proved the full converse to the Four Vertex Theorem without the restriction that the curvature be strictly positive.

Björn E. J. Dahlberg (1949 - 1998)

Publication was delayed by Dahlberg's death in January 1998, but his paper was edited afterwards by his student Vilhelm Adolfsson in collaboration with Peter Kumlin, and appeared in 2005 in the Proceedings of the AMS.

The work of Dahlberg completes the almost hundred year long thread of ideas begun by Mukhopadhyaya.

I. Why is the Four Vertex Theorem true?

A counter-example would have a nonconstant curvature function with one max and one min, and be weakly monotonic on the two arcs between them.

Let's try to build such a counter-example from a few arcs of circles. Start with the largest circle at the bottom, giving the minimum curvature there.

Then go part way up to the right, switch to a smaller circle as shown, go up a little more and switch to a yet smaller circle, and go to its top.

As we increase curvature by switching to smaller circles, their vertical diameters move to the right.

Hence the top point lies to the right of the vertical diameter of the original circle.

When we try the same construction, but this time move up to the left from the bottom of the original circle, the top of the third circular arc will lie to the left of the vertical diameter of the original circle.

Thus the convex curve that we are trying to build does not close up at the top...and so does not exist.

If we permit one self-intersection, then monotonicity of curvature is easy to realize.

r = — 1 — 2 sin θ

Recommended reading.

Robert Osserman

"The Four-or-More Vertex Theorem" American Mathematical Monthly Vol. 92, No. 5 (May 1985), 332 - 337.

II. The Converse to the Four Vertex Theorem for strictly positive curvature.

The Generalized Minkowski Problem.

Does there exist an embedding of the n-sphere Sn into Euclidean space Rn+1 whose curvature has been preassigned as a strictly positive continuous function?

In the early 1970's, we proved the following result.

GENERALIZED MINKOWSKI THEOREM. Let K: Sn → R , for n ≥ 2 , be a continuous, strictly positive function. Then there exists an embedding α: Sn → Rn+1 onto a closed convex hypersurface whose Gaussian curvature at the point α(p) is K(p) for all n p ∊ S .

When n = 1, the Four Vertex Theorem places a constraint on the curvature, so the above theorem says that by contrast, there are no constraints for n ≥ 2 .

The proof of this theorem was by a degree argument in the diffeomorphism group of the n-sphere Sn .

The Converse to the Four Vertex Theorem.

The proof of the Generalized Minkowski Theorem, when rewritten for n = 1 , shows that the required embedding α exists if and only if the curvature is either a nonzero constant or else has at least two local maxima and two local minima.

This gives us the converse to the Four Vertex Theorem in the case of strictly positive curvature. The spirit of the argument is shown in the following picture.

A snake searches unsuccessfully for its tail

What story does this picture tell?

Each of the eight little figures shows a map of an interval into the plane, which begins at the origin, moves off in the positive x-direction, ends up pointing the same way, and is positively curved throughout.

We imagine that the snake is following a concealed loop of instructions, from which the above pictures are samples.

The error vector in each picture extends from the snake's tail to its head, and the pictures are arranged so that the one at location θ on the circle has an error vector pointing in the direction θ .

The resulting loop of error vectors circles once around the origin.

If the loop of instructions is contractible in the space of all instructions, then some instruction in that space will have a corresponding error vector which is zero, and hence will tell the snake how to close up.

A specific example of the snake dance.

The most elementary curvature function which can illustrate these ideas is a step-function, and the simplest case among these leads to curves built from four circular arcs, two cut from a circle of one size, alternating with two cut from a circle of another size, as shown below.

Bicircle

Since our curvatures are strictly positive, we can express the preassigned curvature as a function of the angle of inclination ϕ of the outward pointing normal vector. This has the advantage of making total curvature 2π automatic.

Given a curvature function κ(ϕ) , there will then be a unique map of the interval [0, 2π] into the plane which realizes this curvature function, subject to the initial condition that it begin at the origin, and start moving in the positive x-direction.

The figures on the left report the curvature κ as a step function of the angle ϕ with the heavier markings indicating the larger circle.

The figures on the right show the resulting curve in the plane, built from arcs of these two circles.

Preassign κ0(ϕ) Get this bicircle

Preassign κ(ϕ) Get this curve

A two-curvature-snake searches unsuccessfully for its tail

Explanation of this picture.

The inner group of eight circles reports the preassigned curvature step-functions κ(ϕ) , while the outer group shows the corrresponding curves.

Each curvature step function κ(ϕ) is a distortion, via a diffeomorphism of the circle, of the step function κ0(ϕ) which corresponds to the bicircle.

The eight diffeomorphisms are a sample from a loop of diffeomorphisms, which is easily seen to be contractible in the space of all diffeomorphisms of the circle.

Each error vector is parallel to the position of the curve in the group of eight, and therefore the error vectors turn once around the origin as we go once around the loop of diffeomorphisms.

It follows that there is some diffeomorphism whose corresponding error vector is zero, and so the corresponding curve closes up.

Of course we knew this, since a bicircle closes up.

But this argument is robust, and so will apply to curvature functions which are close to κ0(ϕ) .

Analogy with standard topological proof of the Fundamental Theorem of Algebra.

This is analogous to proving that the polynomial n p0(z) = an z has a zero by noting that it takes any circle in the complex plane to a curve which winds n times around the origin, even though we already know that p0(z) has a zero at the origin.

Again the point is that this argument is robust, and will apply just as well to the polynomial

n n—1 p(z) = an z + an—1 z + ... + a1 z + a0 , once we pay attention to a suitably large circle.

Proving the converse to the Four Vertex Theorem for strictly positive curvature.

Given a preassigned strictly positive curvature funtion κ: S1 → R having at least two relative maxima and two relative minima, we first agree to parametrize our proposed curve by the inverse of the Gauss map.

Then we find a diffeomorphism h: S1 → S1 so that the curvature function κ ◦ h: S1 → R is ε-close in measure to the two-valued step function of some bicircle, meaning that the function κ ◦ h is within ε of this step function on almost all of S1 , and violates this only on a subset of measure less than ε .

The robustness of the winding number argument for the two-valued step function then implies that, for sufficiently small ε > 0 , there is another diffeomorphism h1 of the circle so that the curve built with curvature function κ ◦ h ◦ h1 closes up smoothly.

Since the curvature is positive, this closed curve is convex, and since its normal vector rotates by just 2π as we go once around, the curve is simple.

Reparametrizing this curve, it realizes the preassigned curvature function κ .

III. Dahlberg's proof of the Full Converse to the Four Vertex Theorem.

When asked to draw a simple closed curve in the plane with strictly positive curvature, and another one with mixed positive and negative curvature, a typical response might be...

But in response to the second question, Dahlberg envisioned the following curve.

Its four major subarcs are almost circular. They are connected by four small wiggly arcs, each of which has an almost constant tangent direction, but largely varying curvature, including negative curvature.

This curve "marginalizes" its negative curvature and emphasizes its positive curvature, and from a distance looks like a bicircle.

Dahlberg's Key Idea. You can construct such a curve with any preassigned curvature which has at least two local maxima and two local minima. You can use the winding number argument to get it to close up smoothly, and you can also make it C1 close to a fixed convex curve, which will imply that it is simple.

Dahlberg's proof plan.

(1) Choose as common domain the unit circle S1 , with arc length s as parameter.

Since most of the curves will fail to close up, you can also think of the domain as the interval [0, 2π] .

All the curves will have length 2π , and at the end can be scaled up or down to modify their curvature.

(2) Given a preassigned curvature function κ: S1 → R 2π κ which is not identically zero, evaluate ∫0 (s) ds .

If this is zero, precede κ by a preliminary diffeomorphism of S1 so as to make the integral nonzero.

Then rescale κ by a constant c so that

2π κ π ∫0 c (s) ds = 2 .

When we later modify this curvature by a diffeomorphism h: S1 → S1 , we will immediately rescale the new curvature κ ◦ h by a constant ch so that

2π κ π (*) ∫0 ch h(s) ds = 2 .

(3) Construct an arc-length parametrized curve 2 αh: [0, 2π] → R with curvature ch κ h in the usual way, starting at the origin and heading to the right.

In general, the curve αh will not close up, and the gap is measured by the error vector

I(h) = αh(2π) — αh(0) .

If the error vector is zero, then the curve closes up, and does so smoothly, because its total curvature is 2π .

(4) Use the winding number argument to solve the equation I(h) = 0 for the unkown diffeomorphism h: S1 → S1 , with the search for h restricted to a certain 2-cell D (yet to be defined) within Diff(S1) .

Finding h will give us a smooth closed curve with curvature function ch κ h , which we then rescale to realize the curvature function κ h , and then reparametrize to realize the curvature function κ .

We can find such a curve arbitrarily C1-close to a bicircle, and conclude that it is simple, finishing the proof.

Configuration space.

A central role in the proof is played by curvature step- functions with just two values, and these correspond to curves built from arcs of two different size circles.

To deal with this kind of data, let CS denote the configuration space of ordered 4-tuples (p1 , p2 , p3 , p4) of distinct points on the unit circle S1 , arranged in counterclockwise order, as shown below.

The configuration space CS is homeomorphic to S1 × R3 .

The diffeomorphism group Diff(S1) acts on CS in the natural way:

h(p1 , p2 , p3 , p4) = (h(p1) , h(p2) , h(p3) , h(p4)) .

The core of configuration space.

If the ordered 4-tuple of points (p1 , p2 , p3 , p4) guides the construction of a curve composed of arcs cut alternately from circles of two different sizes, then the curve will close up if and only if opposite arcs are equal in length, equivalently, if and only if p1 and p3 are antipodal, and also p2 and p4 are antipodal.

These conditions are easily seen to be equivalent to the equation

p1 — p2 + p3 — p4 = 0 in the complex plane.

We will call the set of such points (p1 , p2 , p3 , p4) the core of the configuration space CS.

This core is homeomorphic to S1 × R1 .

Reduced configuration space.

To aid in visualization, we can reduce the dimension of the configuration space from four to three, without losing any essential information.

Define the reduced configuration space RCS ⊂ CS to be the subset where p1 = 1 . Then RCS is homeomorphic to R3 and we can use the group structure on S1 to express the homeomorphism S1 × RCS → CS by

(eiθ, (1 , p , q , r)) → (eiθ, eiθp , eiθq, eiθr) .

Notice that this homeomorphism preserves cores, that is,

1 — p + q — r = 0 iff eiθ — eiθp + eiθq — eiθr = 0 .

Now change coordinates by writing

p = e2πix , q = e2πiy and r = e2πiz .

Then RCS ≅ {(x, y, z) : 0 < x < y < z < 1}.

The reduced configuration space appears as an open solid tetrahedron

A point (1, p, q, r) in RCS is in the core if and only if 1 and q are antipodal, and also p and r are antipodal.

In the x, y, z coordinates for RCS , this means that

0 < x < y = ½ < z = x + ½ < 1.

In the open tetrahedron in xyz-space which represents RCS , the core corresponds to the open line segment connecting the point (0, ½, ½) to the point (½, ½, 1) .

The core of the reduced configuration space.

Special Möbius transformations.

Dahlberg's choice of 2-cell D ⊂ Diff(S1) consists of the special Möbius transformations

gβ(z) = (z — β) / (1 —β z) , where |β| < 1 and β is the complex conjugate of β .

These special Möbius transformations are all isometries of the Poincaré disk model of the hyperbolic plane. The transformation g0 is the identity, and if β ≠ 0 , then gβ is a hyperbolic translation of the line through 0 and β which takes β to 0 and 0 to —β . The point β/|β| and its antipode —β/|β| on S1 (the circle at infinity) are the only fixed points of gβ on the unit disk.

Mapping Dahlberg's 2-cell into configuration space.

Start with Dahlberg's 2-cell D ⊂ Diff(S1) , and map it into the configuration space CS by picking any point (p1 , p2 , p3 , p4) in the core of CS , and sending

gβ → (gβ(p1), gβ(p2), gβ(p3), gβ(p4)) .

KEY PROPOSITION. The above map of D into CS is a smooth embedding which meets the core transversally at the point p , and nowhere else.

We omit the proof, which makes good use of the fact that the transformations gβ are isometries of the Poincaré disk model of the hyperbolic plane.

Finding the image of Dahlberg's disk in the reduced configuration space.

Take the embedding of Dahlberg's disk D into the configuration space CS , and then project to the reduced configuration space RCS .

Dahlberg's disk D mapped into the tetrahedron

In this tetrahedral picture of the reduced configuration space RCS , the midpoint (1/4 , 1/2 , 3/4) of the core corresponds to the point (1 , i , —1 , —i) .

Dahlberg's disk is mapped into RCS by

—1 —1 —1 gβ → (1, gβ(1) gβ(i) , gβ(1) gβ(—1) , gβ(1) gβ(—i)) , and the answer is then converted to xyz-coordinates.

Completion of Dahlberg's proof.

We start with a continuous, preassigned curvature function κ: S1 → R which has at least two local maxima and two local minima, and must find an embedding α: S1 → R2 with curvature κ(t) at each point α(t) .

Changing the sign of κ if necessary, there are real numbers 0 < a < b and four points on S1 in counterclockwise order where κ takes the values a , b , a , b in succession.

The points 1 , i , —1 , —i divide the circle S1 into four equal arcs of length π/2 . Let κ0 by the step function which takes the values a , b , a , b on these arcs. The value of κ0 at the four division points is irrelevant.

Given any ε > 0 , it is easy to find a diffeomorphism h: S1 → S1 such that κ ◦ h is "ε-close in measure" to the step function κ0 .

For notational simplicity, replace κ ◦ h by κ .

Since κ is bounded, we can choose ε sufficiently small that the total curvatures of κ and κ0 are arbitrarily close. We then rescale both of these to achieve total curvature 2π, and they will again be ε-close in measure, for a new small ε .

Suppose now that h = gβ is one of the diffeomorphisms selected from Dahlberg's disk D , and consider the curvature functions κ h and κ0 h .

If β is close to 0 , then h is close to the identity, so κ h and κ0 h have total curvatures close to 2π . Hence they can be rescaled to have total curvatures exactly 2π :

2π κ σ σ π 2π κ σ σ ∫0 ch h( ) d = 2 = ∫0 c0h 0 h( ) d .

Then in the usual way we construct curves

2 αh and α0h : [0, 2π] → R , with these curvatures, and consider their error vectors

I(h) = αh(2π) — αh(0) and I0(h) = α0h(2π) — α0h(0) .

The finale.

Let the diffeomorphism h = gβ circle around the identity in Dahlberg's disk D , keeping |β| small and fixed.

Since the image of the disk D is transverse to the core, the error vector I0(h) will circle once around the origin in the complex plane.

If the curvature function κ is sufficiently close in measure to the step function κ0 , and |β| is sufficiently small, then the error vector I(h) will also circle once around the origin in the complex plane.

Hence there must be a diffeomorphism h close to the identity in the Dahlberg disk, for which the error vector I(h) = 0 .

This means that the corresponding curve αh closes up, and since the total curvature is 2π , it does so smoothly.

For ε and |β| sufficiently small, this curve αh is C1- close to the bicircle corresponding to the original step function κ0 , and hence must itself be simple.

The simple closed curve αh realizes the curvature function ch κ h , rescaling it realizes the curvature function κ h , and reparametrizing this realizes the curvature function κ .

This completes Dahlberg's proof of the Converse to the Four Vertex Theorem.