operator $S$. We will call Codazzi function of $(I,II)$ to the map ${\cal T}_S:{\Sigma}\fl\r$ given by $$\label{funcioncodazzi}
I\left(T_S(v_1,v_2),T_S(v_1,v_2)\right)={\cal T}_S(p)\,
\left(I(v_1,v_1)\,I(v_2,v_2)-I(v_1,v_2)^2\right),$$ where $v_1,v_2\in T_p{\Sigma}$, $p\in{\Sigma}$.

Observe that ${\cal T}_S$ is a well-defined differentiable function since the Codazzi tensor is skew-symmetric. Besides, ${\cal T}_S$ vanishes identically if, and only if, $(I,II)$ is a Codazzi pair.

\[anterior\] Let $(I,II)$ be a fundamental pair on a surface ${\Sigma}$ with associated shape operator $S$, mean curvature $H$ and extrinsic curvature $K$. Let $z$ be a local conformal parameter for $I$ such that $$I=2\lambda\,|dz|^2,\qquad II=Q\,dz^2+2H\,\lambda\,|dz|^2+\overline{Q}\,d{\bar{z}}^2.$$ Then $$|Q_{{\bar{z}}}|^2=\frac{\lambda\,{\cal T}_{\widetilde{S}}}{2(H^2-K)}\,|Q|^2,$$ where $\widetilde{S}$ is the traceless operator $S-H\,Id$, $Id_p$ being the identity map on the tangent plane at $p\in {\Sigma}$.

Since the Levi-Civita connection of $I$ is given by (\[levicivita\]) and the shape operator $S$ by (\[endomorfismoweingarten\]), we have $$\begin{aligned}
 T_{\widetilde{S}}\left(\frac{\partial\ }{\partial z},\frac{\partial\ }{\partial
{\bar{z}}}\right)&=&\nabla_{\frac{\partial\ }{\partial z}}\widetilde{S}\frac{\partial\
}{\partial {\bar{z}}}-\nabla_{\frac{\partial\ }{\partial {\bar{z}}}}\widetilde{S}\frac{\partial\
}{\partial z}\ =\ \nabla_{\frac{\partial\ }{\partial
z}}\frac{\overline{Q}}{\lambda}\frac{\partial\ }{\partial z}-\nabla_{\frac{\partial\
}{\partial {\bar{z}}}}\frac{Q}{\lambda}\frac{\partial\ }{\partial {\bar{z}}}\nonumber\\
&=&\frac{1}{\lambda}\left(\overline{Q}_z\,\frac{\partial\ }{\partial z}-
Q_{{\bar{z}}}\,\frac{\partial\ }{\partial {\bar{z}}}\right).\label{primero1}\end{aligned}$$

Hence, using (\[funcioncodazzi\]) one gets $\frac{2}{\lambda}|Q_{{\bar{z}}}|^2={\cal T}_{\widetilde{S}}\,\lambda^2$. The proof finishes using (\[modulodeq\]).

Given a fundamental pair $(I,II)$ on a surface ${\Sigma}$, we will denote by ${\Sigma}_U\subseteq{\Sigma}$ the set of umbilical points of the pair. Then we have

\[thglover\] Let $(I,II)$ 