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.