# Discrete Bending Energy

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}\$.