The Four-Vertex Theorem and
its converse
In memoriam Björn Dahlberg
Dennis DeTurck, University of Pennsylvania
with Herman Gluck, Dan Pomerleano, Shea Vick
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
curvature κ(t) at the point α(t) for all t ∊ S1 .
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, Herman Gluck 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)
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 it's easy to get
just one max and one min for the curvature.
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, Herman Gluck 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
φ: Sn  Rn+1 onto a closed convex hypersurface whose
Gaussian curvature at the point φ(p) is K(p) for all
p ∊ Sn .
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 Converse to the Four Vertex Theorem.
The proof of the Generalized Minkowski Theorem, when
rewritten for n = 1 , shows that the required embedding a
exists if and only if the curvature is either a nonzero constant or
else has at least two local maxima and two local minima.
The spirit of the argument is shown in the following picture.
What story does this picture tell?
A snake searches unsuccessfully for its tail
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
normal angle φ with the
heavier markings indicating
the larger circle (hence the
smaller curvature).
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
Explanation of this picture.
The coloring of the left circle reports the preassigned curvature stepfunction κ(φ) , while the corrresponding curves are shown on the right.
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 diffeomorphisms comprise a loop of diffeomorphisms, which is easily
seen to be contractible in the space of all diffeomorphisms of the circle.
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
p0(z) = an zn 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
p(z) = an zn + an—1 zn—1 + ... + 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
which is not identically zero, evaluate ∫02π κ(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
∫02π c κ(s) ds = 2π .
When we later modify this curvature by a diffeomorphism
h: S1  S1 , we will rescale the new curvature κ ◦ h by
a constant ch so that the total curvature is 2π:
∫02π ch κ ◦ h(s) ds = 2π .
(3) Construct an arc-length parametrized curve
ψh: [0, 2π]  R2 with curvature ch κ ◦ h in the usual
way, starting at the origin and heading to the right.
In general, the curve ψh will not close up, and the gap
is measured by the error vector
I(h) = ψh(2π) — ψh(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 finally
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.
Dahlberg – using arclength as the parameter
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 here.
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 — p 2 + p 3 — 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
The core of the reduced configuration space.
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) .
Special Möbius transformations.
Dahlberg's choice of 2-cell D  Diff(S1) consists of thespecial 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 essential 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 .
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
gβ  (1, gβ(1)—1gβ(i) , gβ(1)—1gβ(—1) , gβ(1)—1gβ(—i)) ,
and the image 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 φ: S1  R2 with curvature κ(t) at each point φ(t) .
Changing the sign of κ if necessary, there are real numbers
0 < a < b and four points on S1 in counter-clockwise 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 small enough so 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 :
∫02π ch κ ◦ h(σ) dσ = 2π = ∫02π c0h κ0 ◦ h (σ) dσ .
Then in the usual way we construct curves
φh and φ0h : [0, 2π]  R2 ,
with these curvatures, and consider their error vectors
I(h) = φh(2π) — φh(0) and I0(h) = φ0h(2π) — φ0h(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 φh closes up, and since the
total curvature is 2π , it does so smoothly.
For ε and |β| sufficiently small, this curve φh is C1-close to the
bicircle corresponding to the original step function κ0 , and hence must
itself be simple.
The simple closed curve φh 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.