Smooth Plane Curves

The euclidean plane $\mathbb{R}^2$ has the special property that there is a distinguished  linear map $J:\mathbb{R}\to\mathbb{R}^2$, the rotation by $90^\circ$ degrees. As a matrix $J$ has the form

\[J=\left(\begin{array}{cc}0 & -1 \\ 1 & 0\end{array}\right).\]

$J$ is distinguished by the fact that it squares to minus the identity and for all $X,Y\in \mathbb{R}^2$ we have

\[ \langle J X,Y \rangle =\mbox{det}(X,Y) =-\langle X, JY\rangle.\]

This unique feature of the two-dimensional case allows us to define for an arclength parametrization $\gamma: [0,L] \to \mathbb{R}^2$ its curvature as follows: Because $T:=\gamma’$ has unit length there must be a real-valued function   $\kappa: [0,L] \to \mathbb{R}$ such that

\[T’=\kappa JT.\]


Define $\alpha: [0,L] \to \mathbb{R}$ as

\[\alpha(s) = \int_0^s \kappa.\]

It will be very convenient to identify $\mathbb{R}^2$ with the complex plane $\mathbb{C}$. After this identification $J$ just becomes multiplication with the imaginary unit $i=\sqrt{-1}$. Define

\begin{align*}\tilde{T}: [0,L] &\to \mathbb{C}\\ \tilde{T}&=e^{i\alpha}.\end{align*}

Then also $\tilde{T}’=i\kappa \tilde{T}$ and from


we conclude that $T$ is a constant multiple of $\tilde{T}$. Evaluating this at $s=0$ reveals the constant to be $T(0)$ and we obtain

\[T(s) = e^{i\alpha(s)}\,T(0).\]

So $\kappa$ determines $T$ uniquely up to a multiplicative constant $a=T(0)$ of norm one. $T$ in turn determines $\gamma$ up to an additive (translational) constant $b=\gamma(0)$:

\[\gamma(s)=\gamma(0)+\int_0^s T.\]

Thus for each curvature function $\kappa$ there exists a curve $\gamma$ with curvature $\kappa$. Every other such curve $\tilde{\gamma}$ differs from $\gamma$ only by a euclidean motion:

\[\tilde{\gamma}=a \gamma+b\]

with $|a|=1$.

For a closed curve we must have $T(L)=T(0)$ which means that there has to be an integer $m\in \mathbb{Z}$ such that

\[\int_0^L \kappa = \alpha(L)=2\pi m.\]

$m$ is called the tangent winding number of $\gamma$.

Leave a Reply