Suppose we have the manifold $\mathbb{R}^3$ equipped with a Riemanian metric $g$ (not necessarily the Euclidean metric. And the induced metric on the $B\_1$ (the ball with radius $1$) is $\gamma$. 


Suppose we are given $Ric\_g$, $\gamma$, and $tr\_{\gamma} K$, where $K$ is the second fundamental form. Can we find $K$? (We don’t know $g$).


Here is my progress: 
All we need to find is the traceless part of $K$ which I denote by $\hat{K}$. Then $K = \hat{K} + \frac12 (tr\_{\gamma}K) \gamma$.


Choose a coordinate system $(x\_0, x\_1, x\_2)$ where $\frac{\partial}{\partial x\_0}$ is normal to $B\_1$ and is of unit length (so $g\_{00} = 1$). Then the following equations are true:


$$R\_g - 2Ric\_{00} = R\_{\gamma} - \frac12 (tr\_{\gamma}K)^2 + |\hat{K}|^2 $$
$$R\_{0i} = D^l \hat{K}\_{il} - \frac12 D\_i (tr\_{\gamma} K)$$
Where $D$ is the covariant derivative on $B\_1$ and $|\hat{K}|^2 = \hat{K}\_{ij} \hat{K}^{ij}.$


So in other words, we know $|\hat{K}|^2$ and $D^l \hat{K}\_{il}$ (since we know everything else in these equations). Does that uniquely determine $\hat{K}$?


Also notice that these are three functions but $\hat{K}$ is compromised of only 2 distinct functions. Is this overdetermined? More precisely, given any positive function $h$ and any 1-form $\omega\_i$, does there exist a unique symmetric traceless (0,2) tensor on $B\_1$ such that $h = |\hat{K}|^2$ and $\omega\_{i} = D^l \hat{K}\_{il}$ ?


Any help is appreciated. 



Given $Ric_g$ of 3-dim Riemannian manifold, induced metric $\gamma$ and mean curvature $tr_{\gamma}K$ on a hypersurface, do we have $K$?
The answer is 'no' even if $g$ is the standard flat metric. Indeed, it is well known that there are isometric *minimal* surfaces (the catenoid and the helicoid) such that the isometry between them does not align their second fundamental forms. Thus, knowing $Ric\_g=0$ and $\mathrm{tr}\_\gamma K = 0$ and knowing $\gamma$ on the surface is not (always) enough to determine $K$.


This is an unusual case though. The set of surface metrics that can be isometrically embedded into $\mathbb{R}^3$ in two ways so that they each induce the same mean curvature function is a very small set of all metrics (at least up to diffeomorphism).


(By the way, using $K$ to denote the second fundamental form is bound to lead to confusion, since $K$ is usually reserved for the Gauss curvature of the surface metric $\gamma$.)


