Curves in $\mathbb {R}^2$

Abstract

Curves in the plane $\mathbb {R}^2$ are special in several respects: For a closed plane curve $\gamma$ an enclosed area $\mathcal {A}(\gamma )$ can be defined, providing another geometric functional in addition to length and bending energy.

Curves in the plane $\mathbb {R}^2$ are special in several respects: For a closed plane curve $\gamma$ an enclosed area $\mathcal {A}(\gamma )$ can be defined, providing another geometric functional in addition to length and bending energy. Unlike the situation in higher dimensions, the geometry of an arbitrary unit speed plane curve $\gamma \colon [a,b] \to \mathbb {R}^2$ is captured in a smooth real-valued curvature function $\kappa \colon [a,b]\to \mathbb {R}$. We prove our first theorem in Global Differential Geometry: The integral of the curvature of a closed plane curve is $2\pi n$ where n is an integer, called the tangent winding number of $\gamma$. Two closed plane curves can be smoothly deformed into each other if and only if they have the same tangent winding number.

1 Plane Curves

The case of curves $\gamma \colon [a,b]\to \mathbb {R}^2$ is special because $\mathbb {R}^2$ comes with a distinguished linear map $J\colon \mathbb {R}^2 \rightarrow \mathbb {R}^2$, the $90^\circ$-rotation in the counterclockwise (positive) direction:

$$\displaystyle \begin{aligned} J\colon \mathbb{R}^2 \rightarrow \mathbb{R}^2, \qquad J \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}\end{aligned}$$

Here are some properties of J that are easy to check: We have $J^2=-I$ and J is orthogonal as well as skew-adjoint, i.e. for all vectors $X,Y\in \mathbb {R}^2$ we have

$$\displaystyle \begin{aligned} \langle JX, JY \rangle &= \langle X, Y \rangle \\ \langle J X, Y \rangle &= -\langle X, J Y \rangle. \end{aligned} $$

Furthermore, the determinant function $\det$ on $\mathbb {R}^2$ can be expressed in terms of J and the scalar product:

$$\displaystyle \begin{aligned} \langle JX, Y\rangle = \det(X,Y).\end{aligned}$$

If $\gamma \colon [a,b]\to \mathbb {R}^2$ is a curve and $T\colon [a,b]\to \mathbb {R}^2$ is its unit tangent, then $\frac {dT}{ds}$ is orthogonal to T and therefore proportional to JT:

Definition 3.1

Let $\gamma \colon [a,b]\to \mathbb {R}^2$ be a curve and $T \colon [a, b] \to \mathbb {R}^2$ its unit tangent. Then the unique function $\kappa \colon [a,b]\to \mathbb {R}$ such that

$$\displaystyle \begin{aligned} \frac{dT}{ds}=\kappa JT\end{aligned}$$

is called the curvature of $\gamma$.

More explicitly,

$$\displaystyle \begin{aligned} \kappa=\left\langle JT, \frac{dT}{ds}\right\rangle = \left\langle \frac{1}{v}J\gamma',\frac{1}{v}\left(\frac{1}{v}\gamma'\right)'\right\rangle=\frac{\det(\gamma^\prime,\gamma^{\prime\prime})}{|\gamma'|{}^3}.\end{aligned}$$

We convince ourselves that the curvature is independent of the parametrization: Let $\tilde \gamma =\gamma \circ \varphi$ be a reparametrization of a plane curve $\gamma$. Then

$$\displaystyle \begin{aligned} \tilde\gamma'=\varphi'\cdot \gamma'\circ\varphi\end{aligned}$$

and, since $\varphi '>0$, $|\tilde \gamma '|=\varphi '|\gamma '\circ \varphi |$. Hence

$$\displaystyle \begin{aligned} \tilde\gamma''=\varphi''\cdot\gamma'\circ\varphi + \varphi'(\varphi'\cdot\gamma''\circ\varphi)\end{aligned}$$

and therefore

$$\displaystyle \begin{aligned} \tilde\kappa = \frac{\det(\tilde\gamma',\tilde\gamma'')}{|\tilde\gamma'|{}^3} = \frac{(\varphi')^3\cdot\det(\gamma',\gamma'')}{(\varphi')^3|\gamma'|{}^3} = \kappa\,.\end{aligned}$$

The curvature $\kappa$ of a straight line segment vanishes since we can parametrize the segment going from a point $p\in \mathbb {R}^2$ to a point $q\in \mathbb {R}^2$ by

$$\displaystyle \begin{aligned} \gamma\colon[a,b]\to\mathbb{R}^2,\ x\mapsto p + \frac{x-a}{b-a}(q-p)\end{aligned}$$

and therefore $\gamma ''=0$, so that $\kappa =0$. A circular arc

$$\displaystyle \begin{aligned} \gamma\colon[a,b]&\to \mathbb{R}^2,\ x \mapsto \begin{pmatrix} r\cos x \\ r\sin x \end{pmatrix}\end{aligned} $$

of radius r has constant curvature $\kappa =\frac {1}{r}$. If we restrict attention to unit speed curves $\gamma \colon [0,L]\to \mathbb {R}^2$, the curvature function $\kappa \colon [0,L]\to \mathbb {R}$ determines $\gamma$ up to orientation-preserving congruence:

Theorem 3.2 (Fundamental Theorem of Plane Curves)

• (i) For everysmooth function$\kappa \colon [0,L]\to \mathbb {R}$there is a unit speed curve$\gamma :[0,L]\to \mathbb {R}^2$with curvature$\kappa$.

• (ii) If$\gamma ,\tilde {\gamma }\colon [0,L]\to \mathbb {R}^2$are unit speed curves with the same curvature function$\kappa$, then there is an orthogonal$(2\times 2)$-matrix A with$\det A=1$and a vector$\mathbf {b}\in \mathbb {R}^2$such that

  $$\displaystyle \begin{aligned} \tilde{\gamma}=A\gamma +\mathbf{b}. \end{aligned}$$

Proof

For (ii), denote by $T,\tilde {T}$ the unit tangent fields of $\gamma$ and $\tilde {\gamma }$ and take for A the orthogonal $(2\times 2)$-matrix A with determinant one for which $A T(0)=\tilde {T}(0)$. Then both $\tilde {T}$ and

$$\displaystyle \begin{aligned} \hat{T}:=AT\end{aligned}$$

solve the linear initial value problem

$$\displaystyle \begin{aligned} Y(0)&=\tilde{T}(0) \\ Y'&=\kappa JY\end{aligned} $$

and therefore, by the uniqueness part of the Picard-Lindelöf theorem, we must have $\hat {T}=\tilde {T}$. Then

$$\displaystyle \begin{aligned} (\tilde{\gamma}-A\gamma)'=\kappa J(\tilde{T}-\hat{T})=0,\end{aligned}$$

which proves (ii). For (i), define $\alpha \colon [0,L]\to \mathbb {R}$ and $T,\gamma \colon [0,L]\to \mathbb {R}^2$ by

$$\displaystyle \begin{aligned} \alpha(x)&:=\int_0^x \kappa \\ T&:=\begin{pmatrix}\cos \alpha \\ \sin\alpha\end{pmatrix} \\ \gamma(x)&:=\int_0^x T.\end{aligned} $$

Then $|T|=1$ and $\gamma '=T$, so $\gamma$ is a curve and T is its unit tangent field. Furthermore, $T'=\kappa JT$ and therefore $\gamma$ has curvature $\kappa$. □

2 Area of a Plane Curve

Let $\gamma \colon [a,b]\to \mathbb {R}^2$ be a curve such that $\det (\gamma ,\gamma ')>0$ and the map

$$\displaystyle \begin{aligned} f\colon(0,1]\times [a,b]\to \mathbb{R}^2,\ f(t,x)=t\gamma(x)\end{aligned} $$

is a bijective map onto a subset $S\subset \mathbb {R}^2$. Then the derivative $f'(t,x)$ at the point $(t,x)\in (0,1)\times [a,b]$ satisfies

$$\displaystyle \begin{aligned} \det f'(t,x)=t\det(\gamma(x),\gamma'(x))>0\end{aligned}$$

and using the transformation formula of integrals it is not difficult to show that the area of S is given by

$$\displaystyle \begin{aligned} \mathrm{area}(S)=\int_S1 = \int_{f((0,1]\times[a,b])}1=\int_a^b\int_0^1 \det f'=\frac{1}{2}\int_a^b\det(\gamma,\gamma').\end{aligned}$$

For the curve $\gamma$ shown on the left of Fig. 3.1, the above formula correctly yields the area enclosed by $\gamma$ and the line segments from the origin to $\gamma (a)$ and $\gamma (b)$. It therefore seems reasonable to use this formula in order to define an area for arbitrary curves $\gamma \colon [a,b]\to \mathbb {R}^2$:

Fig. 3.1

For the curve on the right, the position vector from the origin to $\gamma (x)$ covers some areas multiple times

Definition 3.3

The sector area of a curve $\gamma \colon [a,b]\to \mathbb {R}^2$ is defined as

$$\displaystyle \begin{aligned} \mathcal{A}(\gamma)=\frac{1}{2}\int_a^b\det(\gamma,\gamma').\end{aligned}$$

The curve on the right of Fig. 3.1 illustrates the consequences of this definition. There the position vector from the origin to $\gamma (x)$ covers some areas multiple times. However, for some of these times (where $\gamma$, as seen from the origin, moves clockwise) the contribution to the covered area, as it is computed by the above formula, is negative.

The sector area $\mathcal {A}(\gamma )$ depends on the origin in $\mathbb {R}^2$, which means that it changes if we apply a translation $\mathbf {p}\mapsto \mathbf {p}-\mathbf {v}$ to $\gamma$. Therefore, at first sight the sector area does not look like a good geometric invariant for curves. However, this dependence disappears as soon as we restrict attention to closed curves, or consider differences between the sector areas of curves that share the same endpoints (see Fig. 3.2):

Fig. 3.2

Independent of the choice of origin in $\mathbb {R}^2$, the sector area of the curve on the left of the above picture equals the area of the blue region minus the area of the orange region. Similarly, the difference of the sector areas of the two curves on the right equals the area of the blue region between them

Let $\mathbf {v}\in \mathbb {R}^2$ be a vector and $\gamma \colon [a,b]\to \mathbb {R}^2$ a curve. Then we define a modified sector area $\mathcal {A}_{\mathbf {v}}(\gamma )$ as the sector area of the curve $\gamma$ translated by the vector $\mathbf {v}$:

$$\displaystyle \begin{aligned} \mathcal{A}_{\mathbf{v}}(\gamma):=\mathcal{A}(\gamma+\mathbf{v}).\end{aligned}$$

Theorem 3.4

Let$\gamma \colon [a,b]\to \mathbb {R}^2$be a closed curve and$\gamma _1,\gamma _2\colon [a,b]\to \mathbb {R}^2$two curves with$\gamma _1(a)=\gamma _2(a)$and$\gamma _1(b)=\gamma _2(b)$. Then, for any vector$\mathbf {v}\in \mathbb {R}^2$we have

$$\displaystyle \begin{aligned} \mathcal{A}_{\mathbf{v}}(\gamma)&=\mathcal{A}(\gamma) \\ \mathcal{A}_{\mathbf{v}}(\gamma_2)-\mathcal{A}_{\mathbf{v}}(\gamma_1) &=\mathcal{A}(\gamma_2)-\mathcal{A}(\gamma_1).\end{aligned} $$

Proof

Because $\gamma$ is closed, we have

$$\displaystyle \begin{aligned} \mathcal{A}_{\mathbf{v}}(\gamma)-\mathcal{A}(\gamma) &=\frac{1}{2}\int_a^b \det(\gamma+\mathbf{v},\gamma')-\frac{1}{2}\int_a^b \det(\gamma,\gamma') \\&= \frac{1}{2}\int_a^b \det(\mathbf{v},\gamma')\\&= \frac{1}{2}\int_a^b \det(\mathbf{v},\gamma)' \\&= \frac{1}{2}\left.\det(\mathbf{v},\gamma)\right|{}_{a}^b \\&=0.\end{aligned} $$

By the same arguments we obtain

$$\displaystyle \begin{aligned} &\quad \left(\mathcal{A}_{\mathbf{v}}(\gamma_2)-\mathcal{A}_{\mathbf{v}}(\gamma_1)\right)-\left(\mathcal{A}(\gamma_2)-\mathcal{A}(\gamma_1)\right)\\ &=\frac{1}{2}\int_a^b \det(\mathbf{v},\gamma_2)'-\frac{1}{2}\int_a^b \det(\mathbf{v},\gamma_1)'\\ &= \left.\det(\mathbf{v},\gamma_2-\gamma_1)\right|{}_{a}^b \\&=0.\end{aligned} $$

□

In particular, we expect that for variations with support in the interior of $[a,b]$ of a curve $\gamma \colon [a,b]\to \mathbb {R}^2$, the corresponding variation of sector area is independent of the choice of origin:

Theorem 3.5

Let$t\mapsto \gamma _t$be a variation with support in the interior of$[a,b]$of a curve$\gamma \colon [a,b]\to \mathbb {R}^2$. Then

$$\displaystyle \begin{aligned} \left.\frac{d}{dt}\right|{}_{t=0}\mathcal{A}(\gamma_t)=-\int_a^b \langle Y, J\gamma'\rangle.\end{aligned}$$

Proof

Since Y  vanishes at the endpoints, we have

$$\displaystyle \begin{aligned} \left.\frac{d}{dt}\right|{}_{t=0}\mathcal{A}(\gamma_t)&= \frac{1}{2}\int_a^b \det(Y,\gamma')+\frac{1}{2}\int_a^b \det(\gamma,{\left(\gamma'\right)\,}^{\mbox{\bfseries \hspace{-0.35ex}.}})\\&= \frac{1}{2}\int_a^b \det(Y,\gamma')+\frac{1}{2}\int_a^b \det(\gamma,Y')\\&= \frac{1}{2}\int_a^b \det(Y,\gamma')-\frac{1}{2}\int_a^b \det(\gamma',Y)\\&=\int_a^b \det(Y,\gamma')\\&=-\int_a^b \langle Y, J\gamma'\rangle.\end{aligned} $$

□

As a consequence, the sector area functional by itself does not have any critical points. On the other hand, minimizing length among all curves with the same endpoints and the same sector area is possible:

Theorem 3.6

A curve$\gamma \colon [a,b]\to \mathbb {R}^2$is a critical point of length under the constraint of fixed sector area if and only if its curvature$\kappa$is constant, i.e. if and only if its image lies on a circle or a straight line.

Proof

By Theorems 2.9 and 2.20$\gamma$ is a critical point of length under the constraint of fixed sector area if and only if there is a constant $\lambda \in \mathbb {R}$ such that

$$\displaystyle \begin{aligned} \lambda(-J\gamma')=-T'=-\kappa J\gamma'.\end{aligned}$$

□

3 Planar Elastic Curves

For a unit speed curve $\gamma \colon [0,L]\to \mathbb {R}^2$ with unit tangent T and curvature $\kappa$ we have

$$\displaystyle \begin{aligned} T'&=\kappa JT \\ T^{\prime\prime}&=-\kappa^2 T +\kappa' JT \\ T^{\prime\prime\prime}&=-3\kappa\kappa'T+(\kappa^{\prime\prime}-\kappa^3)JT\end{aligned} $$

The bending energy of a plane curve is also called its total squared curvature. This is because for a unit speed plane curve $\gamma$ as above we have

$$\displaystyle \begin{aligned} \mathcal{B}(\gamma)=\frac{1}{2}\int_a^b \langle T',T'\rangle \,ds= \frac{1}{2}\int_a^b \kappa^2 \,ds.\end{aligned}$$

By Theorem 2.23, $\gamma$ is an elastic curve with tension $\lambda$ if and only if

$$\displaystyle \begin{aligned} 0&= T^{\prime\prime\prime}+3\langle T',T^{\prime\prime}\rangle T+\frac{3}{2}\langle T',T'\rangle T' -\lambda T' \\ &=(\kappa^{\prime\prime}+\frac{\kappa^3}{2}+\lambda \kappa)JT\end{aligned} $$

which means

$$\displaystyle \begin{aligned} \kappa^{\prime\prime}+\frac{\kappa^3}{2}+\lambda \kappa=0.\end{aligned}$$

This differential equation can be interpreted as the equation of motion

$$\displaystyle \begin{aligned} \kappa^{\prime\prime}+\frac{\partial V}{\partial\kappa}(\kappa)=0\end{aligned}$$

for a particle with unit mass moving on the real line subject to the potential energy

$$\displaystyle \begin{aligned} V(\kappa)=\tfrac{1}{8}\kappa^4+\tfrac{\lambda}{2}\kappa^2.\end{aligned}$$

As expected (and as is easy to verify by taking the derivative) the total energy

$$\displaystyle \begin{aligned} E:=\frac{1}{2}(\kappa')^2+ V(\kappa)\end{aligned}$$

is constant. In particular, we see that along for each solution the potential energy is bounded from above by E. In Fig. 3.3 we see examples that should be compared to the shapes of the corresponding curves that were shown in Sect. 2.5.

Fig. 3.3

The potential wells for various values of $\lambda$. For a solution $\kappa$ of the equation of motion, $V(\kappa )-E$ is always non-positive. The values of $\kappa$ that satisfy this condition are indicated in blue

If we look for critical points of the total squared curvature while constraining not only the length but also the sector area, by Theorem 3.6 we arrive at the differential equation for $\kappa$:

$$\displaystyle \begin{aligned} \kappa^{\prime\prime}+\frac{\kappa^3}{2}+\lambda \kappa +\mu =0.\end{aligned}$$

The closed curve in Fig. 3.4 is such a critical point.

Fig. 3.4

A curve which is a critical points of the total squared curvature with constrained length and the sector area

4 Tangent Winding Number

Definition 3.7

For a curve $\gamma \colon [a,b]\to \mathbb {R}^2$ with curvature $\kappa$ the integral

$$\displaystyle \begin{aligned} \int_a^b \kappa\,ds\end{aligned}$$

is called the total curvature of $\gamma$.

In this section we will prove that for a closed curve in $\mathbb {R}^2$ the total curvature is an integer multiple of $2\pi$:

Theorem 3.8

If$\gamma \colon [a,b]\to \mathbb {R}^2$is a closed curve with curvature$\kappa$, then there is an integer$n\in \mathbb {Z}$such that

$$\displaystyle \begin{aligned} \int_a^b \kappa \,ds=2\pi n.\end{aligned}$$

n is called thetangent winding numberof$\gamma$.

Proof

Define $\alpha \colon [a,b]\to \mathbb {R}$ by

$$\displaystyle \begin{aligned} \alpha(x) := \alpha_0 + \int_a^x \kappa\,ds\end{aligned}$$

where $\alpha _0$ is chosen in such a way that

$$\displaystyle \begin{aligned} T(a) = (\cos \alpha_0, \sin \alpha_0).\end{aligned}$$

As in the proof of Theorem 3.2, we conclude

$$\displaystyle \begin{aligned} T= (\cos \alpha, \sin \alpha).\end{aligned}$$

Since $\gamma$ is closed, we have $T(b) = T(a)$, which means

$$\displaystyle \begin{aligned} (\cos \alpha(b), \sin \alpha(b)) = (\cos \alpha(a), \sin \alpha(a)).\end{aligned}$$

Therefore, there is an integer $n\in \mathbb {Z}$ such that

$$\displaystyle \begin{aligned} \int_a^b \kappa \,ds = \alpha(b) - \alpha(a) = 2\pi n.\end{aligned}$$

□

As is clear from the above proof, the tangent winding number counts how often the unit tangent $T(x)$ turns around the unit circle $S^1$ as x runs from a to b (see Fig. 3.5). Figure 3.6 shows that all integers $n\in \mathbb {Z}$ arise as the tangent winding number of some curve in $\mathbb {R}^2$.

Fig. 3.5

The path on $S^1$ of the unit tangent can be visualized more clearly if it is drawn slightly outside of the unit circle

Fig. 3.6

A list of representatives for every homotopy class of plane curves

5 Regular Homotopy

The following two sections will deal with the question: “Given two curves $\gamma ,\tilde {\gamma }$ in $\mathbb {R}^n$, is it always possible to smoothly deform $\gamma$ into $\tilde {\gamma }$ through intermediate curves?” For convenience, we assume that $\gamma$ and $\tilde {\gamma }$ have the same parameter interval.

Definition 3.9

A regular homotopy between two curves $\gamma ,\tilde {\gamma }\colon [a,b]\to \mathbb {R}^n$ is a one-parameter family $t\mapsto \gamma _t$ of curves $\gamma _t\colon [a,b]\to \mathbb {R}^n$, defined for $t\in [0,1]$, such that $\gamma _0=\gamma$ and $\gamma _1=\tilde {\gamma }$. If there exists such a regular homotopy, $\gamma$ and $\tilde {\gamma }$ are called regularlyhomotopic.

Regular homotopy is an equivalence relation on the set of curves $\gamma \colon [a,b]\to \mathbb {R}^n$: Reflexivity and symmetry are easy and for transitivity we make use (see Appendix A.2) of a smooth function $h\colon [0,1]\to [0,1]$ such that

$$\displaystyle \begin{aligned} h(x)=\begin{cases} 0, &\text{for } x\in[0,\epsilon] \\ 1, &\text{for } x\in [1-\epsilon,1].\end{cases}\end{aligned}$$

If now $t\mapsto \gamma _t$ is a regular homotopy between $\gamma$ and $\hat {\gamma }$ and $t\mapsto \tilde {\gamma }_t$ a regular homotopy between $\hat {\gamma }$ and $\tilde {\gamma }$ then

$$\displaystyle \begin{aligned} t\mapsto \begin{cases} \gamma_{h(2t)}, &\text{for } t\in \left[0,\frac{1}{2}\right]\\ \tilde{\gamma}_{h(2t-1)}, &\text{for } t\in \left(\frac{1}{2},1\right]\end{cases}\end{aligned}$$

is a regular homotopy from $\gamma$ to $\tilde {\gamma }$. One can think of regular homotopies as smooth paths in the space of all curves $\gamma \colon [a,b]\to \mathbb {R}^n$, the equivalence classes under regular homotopy are the path-connected components of this space. Indeed, this space is connected, as we will prove below for the case $n=2$. Using the curvature function for curves in $\mathbb {R}^n$ that will be introduced in Sect. 4.2, it would not be difficult to modify the proof and show that any two curves $\gamma \colon [a,b]\to \mathbb {R}^n$ are regularly homotopic.

Theorem 3.10

Any two curves$\gamma ,\tilde {\gamma }\colon [a,b]\to \mathbb {R}^2$are regularly homotopic.

Proof

By transitivity, we can construct the desired regular homotopy in steps. As a first step we use a regular homotopy to achieve that $\gamma$ has length $b-a$:

$$\displaystyle \begin{aligned} \gamma_t=\left(1-t+t\frac{b-a}{\mathcal{L}(\gamma)}\right)\gamma\end{aligned}$$

Therefore, without loss of generality we may assume that the original curve $\gamma$ already has length b-a. Then we can use a regular homotopy to achieve that $\gamma$ has unit speed: using the inverse of the arclength function $s\colon [a,b]\to [0,L]$ of $\gamma$ (Definition 1.13), we define a regular homotopy

$$\displaystyle \begin{aligned} \gamma_t(x) = \gamma((1-t)x + t\,s^{-1}(x)).\end{aligned}$$

So we can assume without loss of generality that $\gamma$ has unit speed. Now we use a regular homotopy in order to translate the starting point of $\gamma$ to the origin and achieve $\gamma (a)=0$:

$$\displaystyle \begin{aligned} \gamma_t = (1-t)\gamma(a)+\gamma\,.\end{aligned}$$

Similarly, we can rotate $\gamma$ to achieve that the unit tangent

$$\displaystyle \begin{aligned} T(a)=\begin{pmatrix}\cos \beta \\ \sin\beta\end{pmatrix}\end{aligned}$$

of $\gamma$ at the starting point becomes the first standard basis vector ${\mathbf {e}}_1$ of $\mathbb {R}^2$ (Fig. 3.7):

$$\displaystyle \begin{aligned} \gamma_t = \begin{pmatrix}\cos ((1-t)\beta) & \sin{}((1-t)\beta) \\ -\sin{}((1-t)\beta) &\cos{}((1-t)\beta)\end{pmatrix}\gamma\end{aligned}$$

Fig. 3.7

The initial regular homotopy that brings the curve into a standard position and size

We apply the same normalizations to $\tilde {\gamma }$. Now we consider the linear interpolation

$$\displaystyle \begin{aligned} \kappa_t=(1-t)\kappa +t\tilde{\kappa}\end{aligned}$$

between the curvature functions $\kappa$ of $\gamma$ and $\tilde {\kappa }$ of $\tilde {\gamma }$ and define the desired regular homotopy from $\gamma$ to $\tilde {\gamma }$ by

$$\displaystyle \begin{aligned} \alpha_t(x) &:= \int_0^x \kappa_t \\ T_t &:=\begin{pmatrix}\cos \alpha_t \\ \sin\alpha_t\end{pmatrix} \\ \gamma_t(x) &:=\int_0^x T_t.\end{aligned} $$

□

6 Whitney-Graustein Theorem

Definition 3.11

A regular homotopy through closed curves between two closed curves $\gamma ,\tilde {\gamma }\colon [a,b]\to \mathbb {R}^2$ is a regular homotopy $t\mapsto \gamma _t$ between $\gamma$ and $\tilde {\gamma }$ such that for all $t\in [0,1]$ the curve $\gamma _t$ is closed. If there exists such a regular homotopy, $\gamma$ and $\tilde {\gamma }$ are called regularly homotopic through closed curves.

Let us start with an example that will be needed below. Recall that $\gamma \colon [a,b]\to \mathbb {R}^n$ is called closed if $\gamma =\hat {\gamma }|{ }_{[a,b]}$ for some periodic smooth map $\hat {\gamma }\colon \mathbb {R}\to \mathbb {R}^n$ with period $b-a$. A simple way to make a new closed curve $\tilde {\gamma }\colon [a,b]:\to \mathbb {R}^n$ out of such a curve $\gamma$ is by a so-called parameter shift, which depends on a number $\tau \in \mathbb {R}$:

$$\displaystyle \begin{aligned} \tilde{\gamma}(x):=\hat{\gamma}(x-\tau).\end{aligned}$$

This closed curve $\tilde {\gamma }$ is regularly homotopic through closed curves to $\gamma$, a suitable regular homotopy being $t\mapsto \gamma _t$ with

$$\displaystyle \begin{aligned} \gamma_t(x)=\hat{\gamma}(x-t\tau).\end{aligned}$$

Like regular homotopy in Sect. 3.5, regular homotopy as closed curves is an equivalence relation on the set of closed curves in $\mathbb {R}^2$ and the equivalence classes can be thought of as the connected components of this space. This time however, the whole space is not connected:

Theorem 3.12 ([45])

Two closed curves$\gamma ,\tilde {\gamma }\colon [a,b]\to \mathbb {R}^2$are regularly homotopic through closed curves if and only if they have the same tangent winding number.

In Fig. 3.8 we see an example of a regular homotopy through closed curves.

Fig. 3.8

A sequence of curves from a regular homotopy between the elastic figure eight curve traversed twice (top left) and the elastic figure eight curve traversed only once (bottom right)

Proof

Suppose there is a regular homotopy as closed curves $t\mapsto \gamma _t$ between $\gamma$ and $\tilde {\gamma }$. Denote by $ds_t=|\gamma _t^{\prime }|$ and $\kappa _t$ the speed and the curvature of $\gamma _t$. Then the tangent winding number

$$\displaystyle \begin{aligned} n_t=\frac{1}{2\pi}\int_a^b \kappa_t \,ds_t\end{aligned}$$

is an integer for all $t\in [0,1]$ and it depends continuously on t. Therefore it is constant and $n_0=n_1$ means that $\gamma$ and $\tilde {\gamma }$ they have the same tangent winding number.

Conversely, suppose that $\gamma$ and $\tilde {\gamma }$ they have the same tangent winding number. As in the proof of Theorem 3.10 we can assume without loss of generality that $\gamma$ and $\tilde {\gamma }$ both have unit speed. By Lemma 3.13 below and the fact that parameter shifts can be accomplished by regular homotopy as closed curves, we may also assume that the curvature functions $\kappa$ and $\tilde {\kappa }$ are either constant or linearly independent. As in the proof of Theorem 3.10, we can apply another regular homotopy through closed curves to achieve $\gamma (a)=0$ and $\gamma '(a)={\mathbf {e}}_1$ (see Fig. 3.7). The same can be assumed for $\tilde {\gamma }$.

Let then $t\mapsto \gamma _t$ be the regular homotopy between $\gamma$ and $\tilde {\gamma }$ constructed at the end of the proof of Theorem 3.10. The only problem is that for the intermediate curves $\gamma _t$ might not be closed. We are going to repair this by modifying $\gamma _t$ to a closed curve $\tilde {\gamma }_t$ as follows:

$$\displaystyle \begin{aligned} \tilde{\gamma}_t(x)=\gamma_t(x)-\frac{x-a}{b-a}\int_a^b T_t.\end{aligned}$$

The only fact that needs to be checked is that $\tilde {\gamma }_t^{\prime }(x)\neq 0$ for all $x\in [a,b]$.

Suppose we would have

$$\displaystyle \begin{aligned} 0=\tilde{\gamma}_t^{\prime}(x)=T(x)-\frac{1}{b-a}\int_a^b T_t,\end{aligned}$$

and therefore

$$\displaystyle \begin{aligned} 1=|T(x)| &=\frac{1}{b-a}\left|\int_a^b T_t\right| \leq \frac{1}{b-a}\int_a^b |T_t|=1.\end{aligned} $$

The inequality sign in the above formula must be an equality, and this implies that $T_t$ is constant, i.e.

$$\displaystyle \begin{aligned} 0=\kappa_t=(1-t)\kappa +t\tilde{\kappa}.\end{aligned}$$

This would imply that $\kappa$ and $\tilde {\kappa }$ are linearly dependent as functions, which by our assumptions means that $\kappa$ and $\tilde {\kappa }$ are constant. Since both coefficients in the previous equation are positive, this would imply $\kappa =\tilde {\kappa }=0$, which is impossible for closed curves. □

As a consequence of Theorem 3.12, every closed curve in $\mathbb {R}^2$ is regularly homotopic through closed curves to one of the curves in the following list in Fig. 3.6.

We conclude this chapter with the Lemma that was needed in the proof of the Whitney-Graustein theorem:

Lemma 3.13

Let$\kappa \colon \mathbb {R}\to \mathbb {R}$be a non-zero periodic function such that for all$\tau \in \mathbb {R}$the functions$\kappa$and$x\mapsto \kappa (x-\tau )$are linearly dependent. Then$\kappa$is constant.

Proof

Given our assumptions, there is a smooth function $\lambda \colon \mathbb {R}\to \mathbb {R}$ such that for all $x\in \mathbb {R}$ we have

$$\displaystyle \begin{aligned} \kappa(x-\tau)=\lambda(\tau) \kappa(x).\end{aligned}$$

Differentiation with respect to $\tau$ at $\tau =0$ yields

$$\displaystyle \begin{aligned} \kappa'(x)=\lambda'(0)\kappa(x)\end{aligned}$$

The only non-zero periodic functions that satisfy such a differential equations are the constant functions. □

Suppose we have a diffeomorphism $g\colon M\to \mathbb {R}^2$ where

$$\displaystyle \begin{aligned} M:= \{x\in\mathbb{R}^2\,|\, |x|\leq 1\}\end{aligned}$$

is the unit disk in $\mathbb {R}^2$. Then we can define a closed curve $\gamma \colon [a,b]\to \mathbb {R}^2$ in such a way that the Fig. 3.9 becomes a commutative diagram.

Fig. 3.9

A diffeomorphism g from the unit disk M into $\mathbb {R}^2$ and the corresponding boundary loop $\gamma$

If a closed curve bounds a region in $\mathbb {R}^2$ that can be mapped onto the unit disk by a diffeomorphism, its tangent winding number is one or minus one:

Theorem 3.14

In the setup of Fig.3.9, the tangent winding number of$\gamma$is$\pm 1$, where the plus sign applies if and only if g preserves orientation, i.e. if$\det g'(x)>0$for all$x\in M$.

Proof

Already in the proof of Theorem 3.12 we saw that applying a scale or a rotation to $\gamma$ does not change the regular homotopy class of $\gamma$. Therefore, without loss of generality we may assume

$$\displaystyle \begin{aligned} g'(0){\mathbf{e}}_1={\mathbf{e}}_1.\end{aligned}$$

For $t\in [0,1]$ let $A_t$ be the $2\times 2$-matrix such that

$$\displaystyle \begin{aligned} A_t {\mathbf{e}}_1&={\mathbf{e}}_1 \\ A_t\, g'({\mathbf{e}}_2)&= (1-t)g'({\mathbf{e}}_2)\pm t{\mathbf{e}}_2 \end{aligned} $$

where the plus sign is chosen if and only if $\det g'(0)>0$. Then the matrix $A_t$ is invertible for all t and the one-parameter family $t\mapsto \gamma _t=A_t \gamma$ of closed curves is a regular homotopy, so after replacing g with $A_1\circ g$ we can assume without loss of generality that

$$\displaystyle \begin{aligned} g'(0)=I.\end{aligned}$$

Now define for $r\in (0,1]$ closed curves $\gamma _r\colon [0,2\pi ]\to \mathbb {R}^2$ by

$$\displaystyle \begin{aligned} \gamma_r(x)=g\left(\!\begin{pmatrix}\cos{}(rx)\\ \sin{}(rx)\end{pmatrix}\!\right).\end{aligned}$$

For small $\epsilon >0$ the curve $\gamma _\epsilon$ is close the parametrization

$$\displaystyle \begin{aligned} x\mapsto \begin{pmatrix}\cos{}(x)\\ \pm\sin{}(x)\end{pmatrix}\end{aligned}$$

of to the unit circle, so the tangent winding number of $\gamma _\epsilon$ is $\pm 1$. On the other hand, $\gamma =\gamma _1$ is regularly homotopic to $\gamma _\epsilon$, and therefore also the tangent winding number of $\gamma$ is $\pm 1$ (Fig. 3.8). □