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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.

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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 $\gamma$.)

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