Given $Ricc_g$ of ambient space, and induced metric $\gamma$ and mean curvature $tr_{\gamma}K$ on hypersurface, do we have $K$? 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. 
 A: I assume you are talking about a surface $\Sigma$ in this $3$ manifold (diffeomorphic  to ${\mathbb R}^3$) with a metric $g$, and $K$ is the second fundamental form (usually you want to use the notation $\text{II}$ or $A$ for second fundamental form, $K$ is usually for curvature) on $\Sigma$, and $\gamma$ is the induced metric on $\Sigma$. 
Since the dimension is $3$, the Ricci curvature decides the full curvature tensor. For example, let $e_1, e_2, e_3$ be orthonormal, with $e_2, e_3$ tangent to 
$\Sigma$, then the sectional curvature in the $e_2, e_3$ plane (i.e. tangent plane of $\Sigma$ is $\frac 12 [Ric_g(e_2, e_2)+Ric_g(e_3, e_3)-Ric_g(e_1, e_1)]$. 
Now you know $\gamma$, you can compute its curvature; by the Gauss formula its difference from the sectional curvature in $e_2, e_3$ plane is the determinant of $K$. And you know the trace of $K$, you know the two eigenvalues of $K$. 
However it seems to me this cannot decide $K$ completely, namely it is not enough for us to know the eigenvectors of $K$. I mean that if you give me a coordinate system $(u, v)$, and tell me $Ric_g$, $\gamma$, and trace $K$ as functions of $u, v$, I cannot write out $K$ uniquely as a (matrix) function of $(u, v)$. 
For example consider the case $g$ is the flat metric, consider the surface $\Sigma_1$ defined by the equation $r(u, v)=(u, \cos v, \sin v)$ and the surface $\Sigma_2$ defined by the equation $r(u, v)=(\cos u, v, \sin u)$, both defined on the coordinate domain $(u, v)\in [-\frac {\pi}{2}, \frac {\pi}{2}]\times [-\frac {\pi}{2}, \frac {\pi}{2}]$, with identical data $Ric_g=0$, $\gamma$, trace $K$, yet $K$ on $\Sigma_1, \Sigma_2$ are different as functions of $(u, v)$. 
