# Darboux Transformations of Discrete Space Curves

Let $\gamma_1,\ldots,\gamma_n\in \mathbb{R}^3$ be a regular discrete space curve. We are looking for another discrete space curve $\tilde{\gamma}$ that satisfies the first two conditions we had in the planar case: The distance between corresponding points $|\tilde{\gamma}_j-\gamma_j|$ is a constant $\ell$ and corresponding edges of $\gamma$ and $\tilde{\gamma}$ have the same length. Then (with a notation as in the previous post) we see that compared two the planar case we have a whole one-parameter family of possible choices for $S_{j+1}$: The vector $S_{j+1}$ can be the result of rotating $S_j$ around $\ell S_j + v_j T_j$ by any chosen angle.

One can imagine to start with a parallelogram with sides $v_j T_j$ and $\ell S_j$. The parallelogram is then folded around the diagonal whose direction is $\ell S_j-v_jT_j$.

Folding by 180 degrees corresponds to the Darboux-transformations we studied in the two-dimensional setting. It is given by the map

$y\mapsto (\ell S_j-v_jT_j)y(\ell S_j-v_jT_j)^{-1}.$

Any other folding angle (except for angle zero that would leave us with the parallelogram) can be realized as

$y \mapsto qyq^{-1}$

with $q$ of the form

$q=r+\ell S_j-v_jT_j.$

Here $r$ is an arbitrary real number. This leads to

$S_{j+1}=(r+\ell S_j-v_jT_j)S_j(r+\ell S_j-v_jT_j)^{-1}$.

Definition: A discrete space curve $\tilde{\gamma}$ is called a Darboux transform of $\gamma$ with rod length $\ell$ and twist $r$ if

$\gamma_j=\gamma_j+\ell S_j$

and $S_1,\ldots, S_n$ satisfy the above equation.

Let us omit the subscript $j$ and use a subscript “+” instead of $j+1$. Combining then in the above formula the $S$ in the middle with the left bracket we see that the map $S\mapsto S_+$ is fractional linear:

$S_+=((r -v T) S-\ell)((r -vT)+\ell S)^{-1}.$

To investigate this map further we write $S$ as the stereographic projection with center $-T$ of some point $Z\in \mathbb{R}^3$ orthogonal to $T$:

\begin{align*} S &=(1-ZT)T(1-ZT)^{-1} \\\\ &=(T+Z)(1-ZT)^{-1} \\\\ &= T(1-ZT)(1+ZT)^{-1}.\end{align*}

The first line shows that $S\in S^2$ for $Z\perp T$. The second line shows that $Z\mapsto S$ is fractional linear. The last line shows that $S=Z$ in case $|Z|=1$. Let us plug this into the last formula for $S_+$:

\begin{align*}S_+&= ((r-vT)(T+Z)-\ell (1-ZT)((r-vT)(1-ZT)+\ell(T+Z))^{-1} \\\\ &=(((v-\ell)+rT)+(r-(v+\ell)T)Z)((r+(\ell-v)T)+((\ell+v)+rT)Z)^{-1} \\\\ &= \left(T+Z\frac{(\ell+v)-rT}{(\ell-v)-rT}\right)\left(1+TZ\frac{(\ell+v)-rT}{(\ell-v)-rT}\right)^{-1}.\end{align*}

Here the fractional notation is justified because the quaternions involved all commute. Thus in terms of the sterographic coordinate $Z$ the transition from $S$ to $S_+$ is just multiplication

$Z+=MZ=Z\overline{M}$

by a quaternion

$M = \frac{(\ell+v)+rT}{(\ell-v)+rT}.$

In the case $r=0$ this is in perfect accordance with the results of an earlier post.

As in the last post let $\gamma$ be a discrete curve in the $\mathbb{C}$ and $\tilde{\gamma}= \gamma+ \ell S$ be a Darboux transformation. Denote by $\alpha_j$ and $\beta_j$ the angles from the edges $T_J$ of $\gamma$ to the rod directions $S_j$ and S_{j+1} respectively. Then \begin{align*}S_j &= e^{i \alpha_j} T_j \\\\ S_{j+1}&= e^{i \beta_j} T_j.\end{align*} We know thatSsatisfies $S_{j+1}=\frac{\ell S_j- v_j T_j}{\ell-v_j \bar{T}_j S_j}.$ This yields $e^{i\beta_j}=\frac{\ell e^{i\alpha_j}- v_j}{\ell-v_j e^{i\alpha_j}}.$ Using the formula $e^{i\alpha}=\frac{1+it}{1-it}$ where $t:=\tan \frac{\alpha}{2}$ we then have \begin{align*} e^{i\beta_j}&=\left(\ell \frac{1+it_j}{1-it_j}- v_j \right) \left(\ell-v_j \frac{1+it_j}{1-it_j} \right)^{-1} \\\\ &=\left(\frac{(\ell -v_j)+(\ell +v_j)it_j}{1-it_j}\right) \left( \frac{(\ell-v_j )-(\ell+v_j )it_j}{1-it_j} \right)^{-1} \\\\ &= \frac{1+i \mu_j t_j}{1-i \mu_j t_j} \,.\end{align*} where $\mu_j= \frac{\ell+v_j}{\ell-v_j}.$ From this we conclude $\tan \frac{\beta_j}{2}=\mu_j \tan \frac{\alpha_j}{2}.$ With the notation \begin{align*}t_j&:=\tan \frac{\alpha_j}{2}\\\\ \tau_j&:=\tan \frac{\kappa_j}{2}\end{align*} we obtain \begin{align*}\tau_j&=\tan \frac{\beta_{j-1}-\alpha_j}{2}\\\\ &= \frac{\mu_{j-1} t_{j-1}-t_j}{1+\mu_{j-1} t_{j-1} t_j}.\end{align*} The corresponding quantities for the Darboux transform\tilde{\gamma}are \begin{align*}\tilde{\alpha}_j &=\pi-\beta_j \\\\ \tilde{\beta}_j &= \pi- \alpha_j \\\\ \tilde{\tau}_j&=\tan \frac{\beta_j-\alpha_{j-1}}{2}\\\\ &= \frac{\mu_j t_j-t_{j-1}}{1+\mu_j t_j t_{j-1}}.\end{align*} Note that we arrive at\tilde{\tau}_j$starting from$\tau_j$if we interchange the roles of$t_{j-1}$and$t_j$and replace$\mu_{j-1}$by$\mu_j. If we apply this rule to \begin{align*}1+\tau_j^2 &= \frac{(1+2\mu_{j-1} t_{j-1} t_j +\mu_{j-1}^2 t_{j-1}^2 t_j^2) + (t_j^2 -2 \mu_{j-1} t_{j-1} t_j +\mu_{j-1}^2 t_{j-1}^2)}{(1+\mu_{j-1} t_{j-1} t_j)^2} \\\\ &= \frac{(1+t_j^2)(1+\mu_{j-1}^2 t_{j-1}^2)}{(1+\mu_{j-1} t_{j-1} t_j)^2} \end{align*} we obtain $1+\tilde{\tau}_j^2=\frac{(1+t_{j-1}^2)(1+\mu_j^2 t_j^2)}{(1+\mu_j t_j t_{j-1})^2}$ Now the following is easy to prove: Theorem: If\gamma$is a closed polygon in$\mathbb{R}^2$with equal edge lengths and$\tilde{\gamma}$is a closed Darboux transform of$\gamma$then $\prod_{j=1}^n (1+\tau^2) = \prod_{j=1}^n (1+\tilde{\tau}^2).$ We can interpret this as follows: The bending energy $\sum_{j=1}^n \log (1+\tau_j^2)=-\sum_{j=1}^n \log (1+\cos \kappa_j)$ is the same for$\gamma$and$\tilde{\gamma}$. # Discrete Darboux Transforms Let$\gamma_1,\ldots,\gamma_n$be a regular polygon in$\mathbb{R}^2=\mathbb{C}$. A second polygon$\tilde{\gamma}_1,\ldots,\tilde{\gamma}_n$is said to be a Darboux transform of$\gamma$if three conditions are satisfied. The first two conditions are: 1) The distance between corresponding points is a fixed number$\ell$. This means for all$j$we have $|\tilde{\gamma}_j-\gamma_j|=\ell.$ 2) Corresponding edges have the same length. This means $|\tilde{\gamma}_{j+1}-\tilde{\gamma}_j|=|\gamma_{j+1}-\gamma_j|.$ These two conditions are trivially met by a polygon$\tilde{\gamma}$that is just a translate of$\gamma$, i.e.$\tilde{\gamma}_j=\gamma_j+a$for some fixed$a\in \mathbb{R}^2$. In this case for all$j$the four points$\gamma_j, \gamma_{j+1}, \tilde{\gamma}_j,\tilde{\gamma}_{j+1}$form a parallelogram. To eliminate this trivial solution we add a third condition: 3) Given the points$\gamma_j,\gamma_{j+1}$and$\tilde{\gamma}_j$the next point$\tilde{\gamma}_{j+1}$is chosen to be different from$\tilde{\gamma}_j+(\gamma_{j+1}-\gamma_j)$, if possible. Here the word “if possible” refer to the case where$\gamma_j,\gamma_{j+1},\tilde{\gamma}_j$lie on a straight line. In this case the parallelogram solution is the only one that complies with the first two conditions. We denote the edge lengths by $v_j=|\gamma_{j+1}-\gamma_j|$ and introduce unit vectors$T_j$and$S_jsuch that \begin{align*}\gamma_{j+1}-\gamma_j&=v_j T_j \\\\ \tilde{\gamma}_j -\gamma_j &= \ell S_j.\end{align*} Now it is easy to see that the three conditions above amount to saying thatS_{j+1}$is obtained from$S_j$(the parallelogram solution) by reflecting it in that diagonal of the parallelogramm that does not contain$\gamma_j$. The direction of this diagonal is given by the vector$\ell S_j – v_j T_j$. Reflection in a vector$a\in \mathbb{C}$is the map $z\mapsto \frac{a}{\bar{a}}\bar{z}.$ Using this we obtain $S_{j+1}=\frac{\ell S_j – v_j T_j}{\ell \bar{S}_j – v_j \bar{T}_j}\bar{S}_j=\frac{\ell S_j -v_j T_j}{\ell -v_j \bar{T}_j S_j}.$ Obviously the four points$\gamma_j, \gamma_{j+1}, \tilde{\gamma}_j,\tilde{\gamma}_{j+1}$lie on a circle. One might formulate this by saying that the polygons$\gamma$and$\tilde{\gamma}$are “enveloped” by a sequence of circles in such a way that the arclength on the two envelopes is in correspondence. # Fixing the Edge Lengths In this post we will work with a fixed sequence of lengths$\ell_1,\dots,\ell_n>0$. This means we imagine polygons as composed of rigid edges connected by flexible joints at the vertices. Denote by$\mathcal{M}_m$the space of curvature functions$\kappa=(\kappa_1,\ldots,\kappa_n)$that correspond to closed regular polygons with$n$vertices and tangent winding number$m$. We can conveniently describe$\mathcal{M}_mas the zero set of the function \begin{align*}\qquad F:(-\pi,\pi)^n &\to \mathbb{R}^3 = \mathbb{R}\times \mathbb{C}\\\\ F(\kappa_1,\ldots,\kappa_n)&=\left(\begin{array}{c} \kappa_1+\ldots +\kappa_n -2 \pi n \\ \ell_0 e^{i\alpha_0}+\ldots +\ell_{n-1}\, e^{i\alpha_{n-1}}\end{array}\right).\end{align*} By the above calculation the derivative ofF$is given as $F'(\kappa_1,\ldots,\kappa_n)\left(\begin{array}{c}\dot{\kappa}_1\\ \vdots \\ \dot{\kappa}_n\end{array}\right)=\left(\begin{array}{c} \dot{\kappa}_1+\ldots +\dot{\kappa}_n \\ \dot{\kappa}_1 \gamma_1+\ldots + \dot{\kappa}_n \gamma_n \end{array}\right).$ Theorem: The rank of$F’$is three at all points of$\mathcal{M}_m$. Therefore, by the implicit function theorem,$\mathcal{M}_m$is a smooth manifold. Proof: At a point$(\kappa_1,\ldots,\kappa_n)$where the columns of$dF(\kappa_1,\ldots,\kappa_n)$were linearly dependent we would have$a,b,c\in \mathbb{R}$such that $a \mbox{Re}(\gamma_j)+b \mbox{Im}(\gamma_j)+c=0$ for all$j$. But this means that the whole polygon lies on a straight line, which is impossible for a regular polygon.$\square$Suppose we have a vector field tangent to$\mathcal{M}_m$that we want to follow numerically. This means that for any$\kappa=(\kappa_1,\ldots,\kappa_n)\in \mathcal{M}_m$we are given a certain tangent vector $X_\kappa \in T_\kappa \mathcal{M}_m=\mbox{ker}F'(\kappa).$ We want to find curves$t\mapsto \eta(t)\in \mathcal{M}_m$that satisfy for all$t$$\dot{\eta}(t)=X_\eta(t).$ If we already have constructed$\kappa=\eta(t)$we might attempt to find$\eta(t+\epsilon)$as $\tilde{\kappa}=\kappa+\epsilon \dot{\kappa}$ where$\dot{\kappa}=X_{\kappa(t)}$. Since$\dot{\kappa}$lies in the kernel of$F'(\kappa)$the condition $\tilde{\kappa}_1+\ldots+\tilde{\kappa}_n=2\pi m$ will be satisfied. The second condition $\ell_0 e^{i\tilde{\alpha}_0}+\ldots +\ell_{n-1} \,e^{i\tilde{\alpha}_{n-1}}$ however is non-linear and so we cannot be sure that$\tilde{\kappa}$corresponds to a closed regular polygon. We first have to project$\tilde{\kappa}$back to$\mathcal{M}_m$. To this end we add to$\tilde{\kappa}$an element of$(T_\kappa\mathcal{M}_m)^\perp$, the normal space to$\mathcal{M}_m$at the point$\kappa. This space is spanned by \begin{align*}N_0&=(1,\ldots,1)\\\\ N_1&=(x_1,\ldots,x_n) \\\\ N_2 &=(y_1,\ldots,y_n)\end{align*} Here\gamma_j=x_j+iy_j$and we assume that a translation has been applied to$\gamma$in order to achieve $\gamma_1+\ldots,+\gamma_n=0.$ Since the condition$\tilde{\kappa}_1+\ldots+\tilde{\kappa}_n=2\pi m$already has been dealt with there is no need to work with$N_1$. We therefore are looking for$\lambda_1,\lambda_2 \in \mathbb{R}$such that $f(\lambda_1,\lambda_2)=(0,0).$ Here $f(\lambda_1,\lambda_2)=G(\kappa+\lambda_1 N_1 + \lambda_2 N_2)$ where $G(\hat{\kappa}):=\ell_1e^{i\hat{\alpha}_1}+\ldots+\ell_{n-1}e^{i\hat{\alpha}_{n-1}}\,.$ Thus we are left with the problem of finding a zero of$f$. From the calculation at the beginning of the post we know the derivative of$f$, so a Newton method seems the right thing to implement. # The Discrete Whitney-Graustein Theorem Let us consider regular closed discrete plane curves$\gamma$with$n$vertices and tangent winding number$m$. We assume that the length of$\gamma$is normalized to some arbitrary (but henceforth fixed) constant$L$. Up to orientation-preserving rigid motions such a$\gammais uniquely determined by a point $(\ell_1,\dots,\ell_n,\kappa_1,\ldots,\kappa_n)\in (0,\infty)^n \times (-\pi,\pi)^n$ satisfying \begin{align*}\ell_1+\ldots+\ell_n&=L \\\\ \kappa_1+\ldots+\kappa_n &=2\pi m \\\\\quad \ell_1 e^{i\alpha_1}+\ldots+ \ell_n e^{i\alpha_n}&=0\end{align*} where $\alpha_j=\kappa_1+\ldots+\kappa_j.$ Proposition 1: Consider a fixed(\kappa_1,\ldots,\kappa_n) \in \times (-\pi,\pi)^n$satisfying$\kappa_1+\ldots+\kappa_n =2\pi m$for some$m\in \mathbb{Z}$and define$\alpha_1,\ldots,\alpha_n$as above. Then the set of$(\ell_1,\dots,\ell_n)\in (0,\infty)^nsatisfying \begin{align*}\ell_1+\ldots+\ell_n&=L \\\\ \quad \ell_1 e^{i\alpha_1}+\ldots+ \ell_n e^{i\alpha_n}&=0 \end{align*} is either empty or the interior of a compact convex polyhedron in an(n-3)$-dimensional affine subspace of$\mathbb{R}^n$. Proof: If the set is non-empty then there is a exists a regular closed polygon$\gamma$with angles$\kappa_1,\ldots, \kappa_n$. If$e^{i\alpha_1},\ldots, e^{i\alpha_n}$would be linearly dependent as vectors in$\mathbb{R}^2$the curve$\gamma$would be contained in a straight line, which is impossible. Therefore the two equations above define an$(n-3)$-dimensional affine subspace$A\subset \mathbb{R}^n$. The set in question is the intersection of$A$with the open convex cone$(0,\infty)^n$. This intersection clearly is bounded and is non-empty by assumption. Therefore inside of$A$the closure of this intersection is a compact convex polyhedron with non-empty interior.$\square$Proposition 2: Consider$(\kappa_1,\ldots,\kappa_n) \in \times (-\pi,\pi)^n$satisfying$\kappa_1+\ldots+\kappa_n =2\pi m$where$m \neq 0$. Then there exists a closed polygon with angles$\kappa_1,\ldots, \kappa_n$. Proof: If the unit vectors$e^{i\alpha_1},\ldots, e^{i\alpha_n}$were all contained in a closed half-circle the tangent winding number would be zero. Thus the origin must lie in the interior of the convex hull of$e^{i\alpha_1},\ldots, e^{i\alpha_n}$. This implies that there are$\ell_1,\ldots \ell_n>0$such that$\ell_1 e^{i\alpha_1}+\ldots+ \ell_n e^{i\alpha_n}=0$.$\square$Using this it is easy to prove the Whitney-Graustein theorem for polygons in the case of non-zero tangent winding number. A regular homotopy between two planar closed polygons$\hat{\gamma}$and$\tilde{\gamma}$is just a continuous path $[0,1]\mapsto \gamma(t)=(\gamma_1(t),\ldots ,\gamma_n(t)) \in \mathbb{R}^{2n}$ such that$\gamma(o)=\hat{\gamma}\,,\,\gamma(1)=\tilde{\gamma}$and all the$\gamma(t)$are regular closed polygons. As usual we call polygons regularly homotopic if there exists a regular homotopy between them. It is easy to see that regular homotopy defines an equivalence relation among regular closed polygons. Theorem: Let$\gamma,\tilde{\gamma}$be two regular cosed polygons in the plane with$n$vertices and the same tangent winding number$n\neq 0$. Then there exists a regular homotopy between$\hat{\gamma}$and$\tilde{\gamma}$. Proof: Without loss of generality we may assume that both curves have total length$L$, start at the origin of$\mathbb{C}$and have a first edge pointing in direction of the positive real axis. Then we can proceed by finding a path in the space of the invariants$(\ell_1,\ldots,\ell_n,\kappa_1,\ldots,\kappa_n)$. We begin by moving the initial lengths$(\hat{\ell}_1,\ldots,\hat{\ell}_n)$linearly to the center of mass of the polyhedron that corresponds to$\hat{\kappa}_1,\ldots,\hat{\kappa}_n$according to Proposition 1. Then we interpolate linearly between$\hat{\kappa}_1,\ldots,\hat{\kappa}_n$and$\tilde{\kappa}_1,\ldots,\tilde{\kappa}_n$. While doing this we keep the lengths at the center of the current polyhedron (which exists for all time because of Proposition 2). Finally we stick with the angles$\tilde{\kappa}_1,\ldots,\tilde{\kappa}_n$but interpolate the lengths linearly towards$(\tilde{\ell}_1,\ldots,\tilde{\ell}_n)$.$\square$I do not know how to adapt the above strategy of proof to the case of tangent winding number zero, but I believe that there should be a way… # The Schläfli Formula The moduli space (meaning congruent polygons are considered as equal) of open regular polygons$\gamma_0,\ldots,\gamma_n$in the plane can be parametrized by parameters $(\ell_0,\ldots,\ell_{n-1},\kappa_1,\ldots,\kappa_{n-1})\in (0,\infty)^{n-1} \times (-\pi,\pi)^{n-2}.$ If we we want to construct an actual polygon from such data we first define$\alpha_0,\ldots,\alpha_{n-1}$as $\alpha_j=\kappa_1+\ldots+\kappa_j.$ From these we define normalized edge vectors$T_j=e^{i\alpha_j}$and finally the polygon itself: For$j=0,\ldots,n$we set $\gamma_j=\ell_0 T_0+\ldots+\ell_{j-1} T_{j-1}.$ The direction of the edge joining$\gamma_j$to$\gamma_{j+1}$is given by$T_j=e^{i\alpha_j}$with . Then the position$\gamma_n$of the last vertex (equal to$\gamma_0$for a closed polygon) is given a $\gamma_n=\ell_0 e^{i\alpha_0}+\ldots +\ell_{n-1} e^{i\alpha_{n-1}}.$ Theorem (Schläfli Formula): From given parameters $(\ell_0,\dots,\ell_{n-1},\kappa_1,\ldots,\kappa_{n-1})\in (0,\infty)^{n-1} \times (-\pi,\pi)^{n-2}$ construct a polygon$\gamma_0,\ldots,\gamma_n$as described above. Consider a variation$(\dot{\ell}_0,\dots,\dot{\ell}_{n-1},\dot{\kappa}_1,\ldots,\dot{\kappa}_{n-1})$of these parameters and define $\dot{\kappa}_n:=-(\dot{\kappa}_1+\ldots+\kappa_{n-1}).$ Then the corresponding variation of$\gamma_n$will be given by $\dot{\gamma}_n=\sum_{j=0}^{n-1} \, \dot{\ell}_j T_j-i \sum_{j=1}^{n} \dot{\kappa}_j \gamma_j.$ Proof: For$j=0,\ldots,n$define$\dot{\alpha}_j:=\sum_{k=1}^j \,\dot{\kappa}_k$. Then$\dot{\alpha}_0=\dot{\alpha}_n=0and \begin{align*}\dot{\gamma}_n &=\sum_{j=0}^{n-1} \, (\dot{\ell}_j +i\ell_j \dot{\alpha}_j) e^{i\alpha_j}\\\\ &= \sum_{j=0}^{n-1} \, \dot{\ell}_j T_j +i \sum_{j=1}^{n-1} \dot{\alpha}_j (\gamma_{j+1}-\gamma_j) \\\\ &= \sum_{j=0}^{n-1} \, \dot{\ell}_j T_j+i \dot{\alpha}_{n-1}\gamma_n -\dot{\alpha}_1 \gamma_1-i \sum_{j=2}^{n-1} (\dot{\alpha}_j-\dot{\alpha}_{j-1}) \gamma_j\\\\ &=\sum_{j=0}^{n-1} \, \dot{\ell}_j T_j-i (\dot{\alpha}_n-\dot{\alpha}_{n-1})\gamma_n -i(\dot{\alpha}_1-\dot{\alpha_0}) \gamma_1-i \sum_{j=2}^{n-1} \dot{\kappa}_j \gamma_j\\\\ &=\sum_{j=0}^{n-1} \, \dot{\ell}_j T_j-i \sum_{j=1}^{n} \dot{\kappa}_j \gamma_j.\end{align*}\square$# Discrete Plane Curves. Let$(\gamma_0,\ldots,\gamma_n)$be a regular polygon in the plane$\mathbb{R}^2$. Then there are unique real numbers$\ell_0, \ldots ,\ell_{n-1}>0$and unit vectors$T_0,\ldots,T_{n-1}\in S^1$such that the edge vectors of$\gamma$have the form $\gamma_{j+1}-\gamma_j=\ell_j T_j.$ Furthermore, it never happens that$T_{j+1}=-T_j$and therefore there are unique real numbers$\kappa_1,\ldots,\kappa_{n-1}such that \begin{align*}-\pi &< \kappa_j < \pi \\\\ T_j &= e^{i\kappa_j}\,T_{j-1}.\end{align*} It is easy to see that we have $T_j=e^{i\alpha_j}\,T_0$ with $\alpha_j=\sum_{k=1}^j \,\kappa_j.$ So\kappa$determines$T$uniquely up to a multiplicative constant$a=T_0$of norm one. Together with the edge lengths$\ell_0,\ldots, \ell_{n-1}$the tangent directions$T$in turn determine$\gamma$up to an additive (translational) constant$b=\gamma_0$: $\gamma_j=\gamma(0)+\sum_{k=0}^{j-1}\, T_j.$ Thus for each collection of edge length$\ell_0,\ldots, \ell_{n-1}$and each curvature function$\kappa$there exists a discrete 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 discrete curve we must have$T_n=T_0$which means that there has to be an integer$m\in \mathbb{Z}$such that $\sum_{k=1}^n\, \kappa = \alpha_n=2\pi m.$$m$is called the tangent winding number of$\gamma$. # Smooth Plane Curves The euclidean plane$\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).$$J$is distinguished by the fact that it squares to minus the identity and for all$X,Y\in \mathbb{R}^2$we have $\langle J X,Y \rangle =\mbox{det}(X,Y) =-\langle X, JY\rangle.$ This unique feature of the two-dimensional case allows us to define for an arclength parametrization$\gamma: [0,L] \to \mathbb{R}^2$its curvature as follows: Because$T:=\gamma’$has unit length there must be a real-valued function$\kappa: [0,L] \to \mathbb{R}$such that $T’=\kappa JT.$ Define$\alpha: [0,L] \to \mathbb{R}$as $\alpha(s) = \int_0^s \kappa.$ It will be very convenient to identify$\mathbb{R}^2$with the complex plane$\mathbb{C}$. After this identification$J$just becomes multiplication with the imaginary unit$i=\sqrt{-1}. Define \begin{align*}\tilde{T}: [0,L] &\to \mathbb{C}\\ \tilde{T}&=e^{i\alpha}.\end{align*} Then also\tilde{T}’=i\kappa \tilde{T}$and from $\left(\frac{T}{\tilde{T}}\right)’=\frac{T’\tilde{T}-T\tilde{T}’}{\tilde{T}^2}=0$ we conclude that$T$is a constant multiple of$\tilde{T}$. Evaluating this at$s=0$reveals the constant to be$T(0)$and we obtain $T(s) = e^{i\alpha(s)}\,T(0).$ So$\kappa$determines$T$uniquely up to a multiplicative constant$a=T(0)$of norm one.$T$in turn determines$\gamma$up to an additive (translational) constant$b=\gamma(0)$: $\gamma(s)=\gamma(0)+\int_0^s T.$ Thus for each curvature function$\kappa$there exists a curve$\gamma$with curvature$\kappa$. Every other such curve$\tilde{\gamma}$differs from$\gamma$only by a euclidean motion: $\tilde{\gamma}=a \gamma+b$ with$|a|=1$. For a closed curve we must have$T(L)=T(0)$which means that there has to be an integer$m\in \mathbb{Z}$such that $\int_0^L \kappa = \alpha(L)=2\pi m.$$m$is called the tangent winding number of$\gamma$. # Discrete Curves A discrete curve in$\mathbb{R}^n$is just a finite sequence$\gamma=(\gamma_0,\ldots,\gamma_n)$of points in$\mathbb{R}^n$. Given a partition $a=t_0 < t_1 <t_2 <\ldots <t_n=b$ of a closed interval$[a,b]\subset \mathbb{R}$we can define a piecewise linear path$\hat{\gamma}: [a,b] \to \mathbb{R}^n$by setting for$t_{j-1}\leq t \leq t_j$$\hat{\gamma}(t) = \gamma_{j-1}+(t-t_{j-1}) (\gamma_j-\gamma_{j-1}).$ We call$\hat{\gamma}$a parametrization of$\gamma$. Any partitioning of another interval$[\tilde{a},\tilde{b}]$into$n$sub-intervals leads to another parametrization$\tilde{\gamma}$of$\gamma$.$\gamma$is called embedded if$\hat{\gamma}$is injective. It is called regular if$\hat{\gamma}$is at least locally injective, i.e. for each$t\in [a,b]$there is an$\epsilon >0$such that$\hat\gamma |_{(t-\epsilon,t+\epsilon)}$is injective. It is easy to see that these definitions do not depend on the choice of the parametrization$\hat{\gamma}$. The length of a discrete curve is defined as $L(\gamma_0,\ldots,\gamma_n)=\sum_{j=1}^n \,|\gamma_n-\gamma_{j-1}| .$ A parametrization$\hat{\gamma}$of$\gamma$is called a parametrization by arclength if each edge has the same length as the corresponding parameter interval: $|\gamma_n-\gamma_{j-1}|=t_j-t_{j-1}.$ Closed discrete curves are defined in a way similar to to the smooth case: They are pairs$(\gamma,n)$where$\gamma$is an$n$-periodic sequence of points in$\mathbb{R}^n$. For practical purposes they are the same thing as a discrete curve$(\gamma_1,\ldots,\gamma_n)$together with the convention that indices are to be counted modulo$n$. # Tractrices and Darboux Transforms Let us investigate the tracks traced out by the wheels of a moving bicycle during some time interval$[a,b]$. We denote the path of the (center of the) rear wheel by$\tau$and the path of the front wheel by$\gamma$. If$\ell$is the distance between the rear and the front wheel then at each time$t$the direction of the bicycle is given by a unit vector$T(t)$such that $\gamma=\tau+\ell\, T.$ In the picture below the front wheel$\gamma$moves on a straight line. The resulting path of the rear wheel is called the tractrix curve and has been investigated already in 1676 by Isaac Newton and in 1692 by Christian Huygens. Assuming that the rear wheel rolls without sliding it must move in the direction of$T$: There is a function$\rho: [a,b] \to \mathbb{R}$such that $\tau’=\rho T.$ Suppose the bicycle is prolonged in the rear direction by a rod. We assume that the tip of the rod has distance$\ell$to the center of the rear wheel (only projections to the horizontal plane are considered here). Then the position$\eta(t)$of the tip of the rod is given as $\eta=\tau-\ell T.$ The rotation speed$\omega(t)of the bicycle is defined via $T’=\omega JT.$ From \begin{align*}\gamma’ &= \rho’ T+\ell \omega JT \\\\ \eta’ &= \rho’ T-\ell \omega JT \end{align*} it is easy to see that the velocity vector\eta’$of the back end is equal in magnitude to the speed$\gamma’$of the front wheel: $|\eta’|^2=\rho’^2+\omega^2=|\gamma’|^2.$ Moreover, if we were to install another wheel at the rear end that also rolls without sliding it would have to make the same angle (though opposite in direction) with the bicycle as the front wheel. Such a bicycle (omitting the middle wheel) has in fact been built, even though I have to admit that the video does not look convincing. On the other hand, cars with four-wheel steering are manufactured in some variety. The motion of cars with four wheels is more difficult to analyze, but to some approximation we might assume also here that the center$\tau$of the rear axis moves towards the center$\gamma$of the front axis. The right picture below shows a DAF MC-139 prototype from 1940 (yes, it has two driver seats). The model in the left picture even has the middle wheels (showing the tractrix) and a video can be found here. Let us condense all this into a mathematical Definition: Let$\tau: [a,b]\to \mathbb{R}^n$be a path and suppose there are$T:[a,b] \to\mathbb{R}^n$of unit length$|T|=1$and$\rho: [a,b] \to \mathbb{R}$such that$\tau’=\rho T$. Then for every$\ell\in \mathbb{R}$the path$\tau$is called a tractrix of the path$\gamma:=\tau+\ell\,T$. Furthermore$\eta:=\tau-\ell\,T$is called a Darboux transform of$\gamma$. It is easy to see that in the above situation$\tau$is a tractrix also of$\eta$and$\gamma$also is a Darboux transform of$\eta$. The paths$\gamma$and$\eta$always have the same length. Note that if$\gamma$is regular then so is$\eta$and if$\gamma$is parametrized by arclength then the same holds for$\eta$. We can apply similar definitions in the case of closed curves by assuming that$\tau$,$T$and$\rho$are periodic functions. Theorem: Suppose$\eta$is a Darboux transform of a periodic path$\gamma$in$\mathbb{R}^n$that is parametrized by arc length. Then$\gamma$and$\eta$have the same length and the same total squared curvature $\oint |\gamma”|^2=\oint |\eta”|^2.$ If$n=3$then$\gamma$and$\eta$have the same area vector $\frac{1}{2}\oint \gamma \times \gamma’ = \frac{1}{2}\oint \eta \times \eta’.$ Proof: We have $\gamma’=\rho T + \ell T’$ and we may assume that$\gamma$and$\eta$are parametrized by arc length, i.e. $\rho^2+\ell^2|T’|^2=1.$ Then $\gamma”=\rho’T+\rho T’+\ell T”$ and $\oint \langle \gamma”,\gamma”\rangle =\oint \rho’^2 + 2\ell \langle \rho’ T +\rho T’,T”\rangle +\ell^2 |T”|^2.$ We want to show that this expression is unchanged if we replace$\ell$by$-\ell$, i.e. $\oint \langle \rho’ T +\rho T’,T”\rangle = \oint -\rho’ \langle T’,T’ \rangle – \frac{\rho}{2}\langle T’,T’\rangle’ =\oint \rho'(\rho^2-1)+\rho^2\rho’=0.$ Here we have used$0=\langle T,T’\rangle’=\langle T’,T’\rangle+\langle T,T”\rangle$, the equation$\rho^2+\langle T’,T’\rangle =1$and its derivative$2\rho \rho’+\langle T’,T’\rangle’=0$. Next we look at the area vector $\oint \gamma \times \gamma’ = \oint (\tau+\ell T)\times (\rho T+\ell T’).$ Again we only have to show that the coefficient of$\ell$vanishes, i.e. $\oint \tau \times \ell T’ = \oint (\tau \times \ell T)’=0.$$\square$The statement about the area vector implies for the case$n=2$that$\gamma$and$\eta\$ enclose the same area:

$\frac{1}{2}\oint \mbox{det}(\gamma, \gamma’) = \frac{1}{2}\oint \mbox{det}(\eta,\eta’).$