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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.

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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 $\mathrm{II}_{ij}\equiv 0$.

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  • $\begingroup$ Yes, this answers my question. Thank you very much. In fact, I have in the meantime proved that condition (ii) is equivalent to the vanishing of the following three Riemann curvatures of the 3d metric $G$: $$R^3_{221} = R^3_{112} = R_{1212} = 0.$$ Of course, this does not imply flatness of $G$, as your example shows. Let me nonetheless explain how I came up with condition (ii). Let $y$ be an isometric immersion of $G_{2\times 2}$ into $R^3$. Define now the vector field $\vec b:\Omega\to R^3$ with the property that it we write the matrix field with three columns: ... $\endgroup$
    – user41101
    Commented Oct 9, 2013 at 23:31
  • $\begingroup$ ... $Q=\left[\begin{array}{ccc}\partial_1y & \partial_2 y& \vec b\end{array}\right]\in R^{3\times 3}$, then: $$\det Q>0 \quad \mbox{ and } \quad Q^TQ=G$$ Of course one easily can compute $\vec b$ explicitly, but it is not important. The point is that the condition (ii) is equivalent to: (iii) $(\nabla y)^T (\nabla \vec b)$ is skew-symmetric (at each point $x'=(x_1, x_2)\in\Omega$). Now, attempting finding an isometric immersion of $G$, one writes: $$u(x',x_3)= y(x')+x_3\vec z(x') +\mbox{higher order terms}.$$ Then, the leading order terms in $(\nabla u)^T(\nabla u) - G$ vanish exactly when ... $\endgroup$
    – user41101
    Commented Oct 9, 2013 at 23:31
  • $\begingroup$ ... $z=\vec b$ and (iii). So, I thought that (ii) is a linearized condition for solving for the isometric immersion of G, and that for small $h$ it is also sufficient for it is existence. But evidently, it is not. Thanks again. $\endgroup$
    – user41101
    Commented Oct 9, 2013 at 23:31

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