differentiably extended to ${\Sigma}$. Then $$\begin{array}
{l} {\displaystyle A=\cosh\varphi\,I+\frac{\sinh\varphi}{\sqrt{H^2-K}}\,II'}\\[6mm]
{\displaystyle B=\sqrt{H^2-K}\sinh\varphi\,I+\cosh\varphi\,II'},
\end{array}$$ is a fundamental pair with mean curvature $H(A,B)=0$, extrinsic curvature $K(A,B)=-(H^2-K)$ and such that its Codazzi tensor $T_{\widetilde{S}}$ satisfies $$\label{TStilde}
T_{\widetilde{S}}(X,Y)=\omega(Y)\,X-\omega(X)\,Y,\qquad\omega=\frac{1}{2}(dH-\sqrt{H^2-K}d\varphi),$$ for all $X,Y\in{\mathfrak{X}}({\Sigma})$.

Let $(u,v)$ be doubly orthogonal parameters for the Codazzi pair $(I,II)$ such that $$I=E\,du^2+G\,dv^2,\qquad II=k_1\,E\,du^2+k_2\,G\,dv^2,$$ being $k_1\geq k_2$.

Then we can write $(A,B)$ as $$A=e^\varphi\,E\,du^2+e^{-\varphi}\,G\,dv^2,\qquad
B=\frac{k_1-k_2}{2}\,\left(e^\varphi\,E\,du^2-e^{-\varphi}\,G\,dv^2\right).$$

Hence, $A$ is a Riemannian metric on ${\Sigma}$ and the mean and extrinsic curvatures of the pair are given by $H(A,B)=0$ and $K(A,B)=-(H^2-K)$.

In addition, from (\[Tuv\]) and taking into account that $(I,II)$ is a Codazzi pair, we get $$\begin{aligned}
T_{\widetilde{S}}\left(\frac{\partial\ }{\partial u},\frac{\partial\ }{\partial
v}\right)&=&-\frac{1}{e^\varphi\,E}\left(\frac{k_1-k_2}{2}\,\left(e^\varphi\,E\right)\right)_v
\frac{\partial\ }{\partial
u}-\frac{1}{e^{-\varphi}\,G}\left(\frac{k_1-k_2}{2}\,\left(e^{-\varphi}\,G\right)\right)_u
\frac{\partial\ }{\partial v}\\
&=&-\frac{1}{2}\left((k_1)_v-(k_2)_v+(k_1-k_2)\varphi_v-2(k_1)_v)\right)\frac{\partial\
}{\partial u}\\
&&-\frac{1}{2}\left((k_1)_u-(k_2)_u-(k_1-k_2)\varphi_u+2(k_2)_u)\right)\frac{\partial\
}{\partial v}\\
&=&\omega\left(\frac{\partial\ }{\partial v}\right)\frac{\partial\ }{\partial
u}-\omega\left(\frac{\partial\ }{\partial u}\right)\frac{\partial\ }{\partial v}.\end{aligned}$$ Finally, by linearity we obtain (\[TStilde\]).

Under the assumptions of Lemma \[lemita\], if we take a conformal parameter $z$ for $A$ and put $$\label{elnuevopar}
A=2\lambda\,|dz|^2,\qquad B=Q\,dz^2+\overline{Q}\,d{\bar{z}}^2,$$ then from (\[primero1\]) and (\[TStilde\]) we get $$Q_{{\bar{z}}}=\lambda\,\omega\left(\frac{\partial\ }{\partial
z}\right)=\frac{\lambda}{2}\left(H_z-\sqrt{H^2-K}\,\varphi_z\right).$$

Moreover, we obtain from (\[TStilde\]) that the pair $(A,B)$ given by (\[elnuevopar\]) is a Codazzi pair if and only if $dH-\sqrt{H^2-K}d\varphi=0$, or equivalently $Q_{{\bar{z}}}=0$.

With all of this we have

\[corolario3\] Let $(I,II)$ be a Codazzi 