Four Vertex Theorem

Simple Closed Curves

A parameterized curve \(\gamma\) is closed with a period of \(T\), if \(\forall t\in \mathbb R, \forall k\in\mathbb N. \gamma(t) = \gamma(t+kT)\)

A parameterized curve \(\gamma\) is simple if it has no self-intersections.

Jordan's Theorem

Claim Any simple closed curve in the plane has a connected bounded "interior" and an connected unbounded "exterior".

Hopf's Umlaufsatz Theorem

The total signed curvature of a simple closed curve in \(\mathbb R^2\) is \(\pm 2\pi\).

Vertex

A vertex of a plane curve \(\gamma\) is a point where \(\frac{d\kappa_s}{dt} = 0\). Note that this definition is independent of parameterizations.

Example Every point on the circle \(\gamma(t) = (R \cos t, R\sin t)\) is a vertex.

proof. We known that \(\kappa_s = R^{-1}\) for the circle \(\gamma\), so that \(\frac{d\kappa_s}{dt} = 0\) everywhere.

Circumscribed circle

For a simple closed curve \(\gamma:\mathbb R\rightarrow\mathbb R^2\), we can define a circumscribed circle \(C(c, R) = \{x\in\mathbb R^2: \|x - c\| = R\}\) that contains \(\gamma(\mathbb R)\) and attains the smallest possible radius.

Claim existence

proof. By Jordon's Theorem, since \(\gamma\) is simple, closed. Its interior is bounded. There exists some circles \(C(c,R)\) that bounds \(\gamma(\mathbb R)\).

Then, we need to consider whether \(\text{inf}\{R\}\) exists.

Take a sequence \(c_n, R_n\) s.t. \(R_n\searrow \text{inf}\{R\} = R_\infty\).
Note that \(\|c-\gamma(s)\| \leq R_n\), take a convergent subsequence \(c_{n_k}\) s.t. \(\lim_{k\rightarrow\infty} c_{n_k} = c_\infty\).

We claim that circle \(C(c_\infty, R_\infty)\) contains \(\gamma(\mathbb R)\).
The idea of the claim is that if we assume \(C_\infty = C(c_\infty, R_\infty)\) does not contain \(\gamma(\mathbb R)\), then there are some part of \(\gamma(\mathbb R)\) "sticks out" of \(C_\infty\), i.e.

\[\exists s_0\in\mathbb R. \|\gamma(s_0) - c_\infty\| > R_\infty\]

By triangle inequality we have

\[\|\gamma(s_0) - c_\infty\| \leq \|\gamma(s_0) - c_{n_k}\| \leq \|\gamma(s_0) - c_\infty\| + \|c_\infty - c_{n_k}\|\]

Let \(D:= \|\gamma(s_0) - c_\infty\| + \|c_\infty - c_{n_k}\|, \epsilon_k := \|c_\infty - c_{n_k}\|\). so that

\[\|\gamma(s_0) - c_{n_k}\| \geq D - \epsilon_{k}\]

Take \(k\) large enough s.t. \(\epsilon_k < \frac{D-R_\infty}{2}\) and \(R_{n_k} - R_\infty < \frac{D-R_\infty}{2}\). Therefore,

\[\|\gamma(s_0) - c_{n_k}\| \geq D - \frac{D-R_\infty}{2} = \frac{D-R_\infty}{2} + R_\infty > R_{n_k}\]

Then, by the construction of our sequence, this is a contradiction.

Claim. uniqueness

Assume not unique, then exists \(C_1 = (c_1, R), C_2 = (c_2, R)\) (\(R\) must be the same since minimum) s.t. \(\|\gamma(s) - c_1\| < R\) and \(\|\gamma(s) - c_2\| < R\). Therefore, \(\gamma(\mathbb R) \in \text{int}(C_1)\cap \text{int}(C_2)\). Then, we can construct new circle according to the picture, where \(c = \frac{c_1 + c_2}{2}\) and by triangle inequality, \(R' < R\)

Lemmas

Fact 1 \(C(c, R)\) and \(\gamma(\mathbb R)\) has at least one intersection.
Assume not, then we can keep \(c\) unmoved, note that \(\forall s. \|\gamma(s) - c\| < R\) so that we can shrink \(R' < R\).

Fact 2 \(C(c, R)\) and \(\gamma(\mathbb R)\) has at least two intersections.
Assume not, by fact 1, we take \(s_0\) to be the only intersection. Then move \(c\) along the direction of \(\gamma(s_0)-c\) by arbitrarily small \(\epsilon\), then it has no intersection, which violates Fact 1.

Fact 3 If \(C(c, R)\) and \(\gamma(\mathbb R)\) has exactly two intersections, then the line segment between intersection points is a diameter.
If not, then move \(c\) along the direction normal to the line.

Fact 4 \(\gamma(\mathbb R)\) and \(C\) has the same tangent at their point of intersections.
Assume there are at least two intersections \(p_1,...,p_n\).
First, we want to show that at all intersections, the orientation is the same. Assume \(\gamma\) is oriented the same as \(C\) as intersection point \(p_k\), to show that all intersection points have the same intersection, it's sufficient to show that of \(p_{k+1}\). Note that the curve \(\gamma([p_k, p_{k+1}])\) and the arc of circle between \(p_k\) and \(p_{k+1}\) will form a simple closed curve, by Jordon's theorem, it's bounded and connected.
Then, consider any simple curve within the enclosed region from \(p_{k+1}\) to \(p_k\), and it forms a simple closed curve with \(\gamma([p_k, p_{k+1}])\). Therefore, they must oriented the same.

Fact 5.1 If \(\gamma\)'s image is within \(\text{int}(C)\) in a neighborhood of the intersection point, then the curvature of \(\gamma, \kappa_s(s_i) \geq R^{-1}\). Wlog assume \(c = (0,0)\). At point of intersection, we have that \(p_k = \gamma(s_k)\), then note that \(\|\gamma(s_k)\|^2 = \gamma(s_k)\cdot\gamma(s_k)= R^2\) must attain its maximum (since other points are \(\|\gamma(s)\| \leq R\) by circumscribed circle). Then, we have

\[\frac{d}{ds}(\gamma(s_k)\cdot \gamma(s_k)) = 2\gamma'(s_k)\cdot\gamma(s_k) = 0\]

Therefore, we have that \(\mathbf t(s_k)\) is perpendicular to the radius at \(p_k\) and is the same tangent vector.

In addition, we have that

\[\begin{align*} \frac{d^2}{ds^2}(\gamma(s_k)\cdot \gamma(s_k)) &\leq 0\\ 2(\gamma''(s_k)\cdot\gamma(s_k) + \gamma'(s_k)\cdot\gamma'(s_k))&\leq 0\\ 2(\kappa_s(s_k)p_k \cdot \mathbf n_s(s_k) + 1) &\leq 0 \end{align*}\]

Then note \(\mathbf n_s(s_k)\) is parallel and directly opposite to \(p_k\), since \(\|p_k\| = R, p_k \cdot \mathbf n_s(s_k) = -R\)

\[\begin{align*} \frac{d^2}{ds^2}(\gamma(s_k)\cdot \gamma(s_k)) &\leq 0\\ (-\kappa_s(s_k)R + 1) &\leq 0\\ \kappa_s(s_k) &\geq R^{-1} \end{align*}\]

Fact 5.2 If \(\gamma\)'s image is within \(\text{ext}(C)\) in a neighborhood of the intersection point, then the curvature of \(\gamma, \kappa_s(s_i) \leq R^{-1}\).

The idea is similar to Fact 5.1 while here \(\gamma(s_k)\cdot \gamma(s_k)\) attains its local minimum, so that

\[\begin{align*} \frac{d^2}{ds^2}(\gamma(s_k)\cdot \gamma(s_k)) &\geq 0\\ \kappa_s(s_k) &\leq R^{-1} \end{align*}\]

Theorem (Four Vertex Theorem)

For a simple closed curve \(\gamma:\mathbb R\rightarrow\mathbb R^2\), it has at least 4 vertex.

Let \(\gamma\) be a simple closed curve, and \(C\) be its circumscribed curve, \(p_k = \gamma(s_k)\) and \(p_{k+1} = \gamma(s_{k+1})\) be two conservative intersection points. Connecting the two points with the line segment \(l = \overline{p_k p_{k+1} }\) and move the the center of the circle along the direction normal to \(l\) until the last moment it touches \(\gamma\). Then, the \(\gamma([s_k, s_{k+1}])\) is all in \(\text{ext}(C)\), by Fact 5.2, there exists at least one point \(q_k\) s.t. \(\kappa_s(q_k) \leq R^{-1}\). Note that there are two opposite directions to move \(c\).

Therefore, we have that the curve passing through \(p_{k}, q_k, p_{k+1}\) must have a local minimum since it goes down from \(\geq R^{-1}\) to \(\leq R^{-1}\) and then back to \(\geq R^{-1}\). Therefore, we have at least \(n\) local minimum. Then, consider the curve passing through \(q_k, p_k, q_{k+1}\), similarly we can obtain \(n\) local maximum.

Note that the only possible case is that \(\kappa_s\) at \(p_k, q, p_{k+1}\) are all \(R^{-1}\). In this case, there is infinitely many points of \(\gamma\) in the neighborhood of \(p_k\), that is coincided with the arc of the circle. Since \(C\) is the circumscribed circle, either \(\gamma([p_k, p_{k+1]}\) coincides with the arc of the circle all the way, or \(\kappa(q_k) < R^{-1}\)

Example : Limacon

Let \(\gamma:\mathbb R\rightarrow\mathbb R^2\) be defined as

\[\gamma(t) = ((1+2\cos t)\cos t, (1 + 2\cos t)\sin t)\]

Note that \(\gamma\) is closed (but not simple) with a period of \(2\pi\).

Claim Limacon only has 2 vertices.

proof.

\[\begin{align*} \gamma'(t) &= (-\sin t - 2\sin 2t, \cos t + 2\cos 2t)\\ \|\gamma'(t) \| &= \sqrt{\sin^2 t + 4\sin t\sin 2t + 4\sin^2 2t + \cos^2 t + 4\cos t\cos 2t + 4\cos^2 2t}\\ &= \sqrt{5 + 8\cos t \sin^2 t + 4\cos t - 8\cos t\sin^2 t)}\\ &= \sqrt{5 + 4\cos t} \end{align*}\]

Then, let \(\gamma'(t) = \frac{1}{\sqrt{5 + 4\cos t} }(-\sin t - 2\sin 2t, \cos t + 2\cos 2t)\) and define the turning angle by

\[\cos\varphi = \frac{-\sin t - 2\sin 2t}{\sqrt{5 + 4\cos t} }. \sin\varphi = \frac{\cos t + 2\cos 2t}{ {\sqrt{5 + 4\cos t} } }\]

Because we only interested in \(\kappa_s = \varphi'\), observe that

\[\begin{align*} \frac{d}{dt}\sin \varphi &= \frac{d}{d\varphi}\sin\varphi\frac{d\varphi}{dt}\\ \frac{d}{dt}\sin \varphi &= \frac{(-\sin t - 4\sin 2t)\sqrt{5+4\cos t} - (\cos t + 2\cos 2t)(-4\sin t)\frac{1}{2\sqrt{5+4\cos t} } }{5+4\cos t}\\ &= \frac{-\sin t(24\cos^2 t + 42\cos t +9)}{(5+4\cos t)^{3/2} }\\ \varphi'(t) &= \frac{\frac{d}{dt}\sin \varphi}{\cos \varphi} \\ &= \frac{\frac{-\sin t(24\cos^2 t + 42\cos t +9)}{(5+4\cos t)^{3/2} } }{\frac{-\sin t - 2\sin 2t}{\sqrt{5 + 4\cos t} } }\\ &= \frac{9+6\cos t}{5+4\cos t}\\ \kappa_s &= \frac{d\varphi}{ds} = \frac{9+6\cos t}{5+4\cos t} \frac{1}{\sqrt{5+4\cos t} }\\ &= \frac{9+6\cos t}{(5+4\cos t)^{3/2} }\\ \frac{d\kappa_s}{dt} &= \frac{12\sin t (2+\cos t)}{(5+4\cos t)^{5/2} } \end{align*}\]

Therefore, \(\frac{d\kappa_s}{dt} = 0\) only when \(t=0, \pi\) for \(t\in [0, 2\pi)\)