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 $\gamma$ is 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 $(\ell_1,\dots,\ell_n)\in (0,\infty)^n$ satisfying

\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…