topological sphere and $(I,II_1)$, $(I,II_2)$ two Codazzi pairs with the same Riemanian metric $I$. If both pairs have the same mean curvature, then $II_1=II_2$.

In $\r^3$, this result says that two isometric immersions from a Riemannian sphere in $\r^3$ with the same mean curvature must coincide, up to an isometry of the ambient space.

Since $(I,II_1)$ and $(I,II_2)$ are Codazzi pairs, so is the new pair $(I,II_1-II_2)$. Besides, as $H(I,II_1)=H(I,II_2)$ the mean curvature of $(I,II_1-II_2)$ vanishes identically. In particular, by taking a local conformal parameter $z$, we can put (see (\[parholomorfo\])) $$I=2\lambda\,|dz|^2,\qquad II_1-II_2=Q\,dz^2+\overline{Q}\,d{\bar{z}}^2.$$ Thus, using Theorem \[elteorema\] for the pair $(I,II_1-II_2)$ and the functional $W(H,K)=H=0$, we get that $Q\equiv 0$, which finishes the proof.

For a fundamental pair $(I,II)$ with mean and extrinsic curvatures $H$ and $K$ respectively, the *third fundamental form* is given by $III=-K\,I+2H\,II$ (see, for instance, [@Mi]). In particular, given a surface isometrically immersed in a 3-dimensional manifold with first and second fundamental forms $I$ and $II$ respectively, $III$ is nothing but its classical third fundamental form. In other words, $III={\langle dN,dN \rangle }$ where $N$ is a unit normal vector field on the surface and ${\langle , \rangle }$ is the metric of the ambient space.

