Let $G$ be a Riemannian metric on a 3d set $\Omega\times (-h, h)\subset R^3$, where $\Omega$ is a disc in $R^2$ and $h$ is sufficiently small. We assume that $G(x\_1, x\_2, x\_3)=G(x\_1, x\_2)$, i.e. $G$ is independent of the third variable $x\_3$. 
By $G\_{2\times 2}$ we denote the $2\times 2$ principal minor of $G$ and treat it as a 2d metric on the 2d crossection $\Omega$. I would like to understand whether the following two conditions are equivalent:


(i) $G$ has an isometric immersion in $R^3$ (i.e. the Ricci tensor of $G$ is zero)


(ii) $G\_{2\times 2}$ has an isometric immersion in $R^3$ with the property that its second fundamental form $II$ equals: $II\_{ij} = \frac{1}{\sqrt{G^{33}}}\Gamma^3\_{ij}$. The Christoffel symbols are the symbols of the 3d metric $G$.


I can prove that (i) implies (ii), and it seems to me that the converse is also true. In particular, the Codazzi-Mainardi equations involving $II$ and $G\_{2\times 2}$ give precisely $R^3\_{121}=R^3\_{221}=0$. 


Have you seen such statement (that (i) is equivalent to (ii)) anywhere? Any hint or reference would be appreciated.



Question about immersability of a 3d and 2d Riemannian metrics
The answer is NO.


Take
$$g=
\left(\begin{matrix}
1&0&0\\
0&1&0\\
0&0&\lambda
\end{matrix}
\right)
$$
where $\lambda=\lambda(x\_1,x\_2)$.


In general the metric is not flat, but the second condition holds for the isometric flat embedding since $\Gamma^3\_{ij}\equiv0$ and $\mathrm{II}\_{ij}\equiv 0$.


