I tryed resolve the follow problem, but I get not. Someone can help me ?
Let be $N^3$ a smooth Riemannian manifolds and let be $\Sigma^2$ a closed and embedded minimal surface of $ N $. Show that
$$K_\Sigma = K_N -2Ric^N - | A_\Sigma | ^ 2. $$
Where $ K $ is Gauss curvature and $A_\Sigma$ is second fundamental form of $\Sigma$. Hint: Took the trace in Gauss equation twice and useding that $\Sigma$ is minimal.
Your notation is confusing to me, but I think your question follows from this post. Basically, the following identity follows from the Gauss equation: $$ R_N - 2 \text{Ric}_N(n_\Sigma,n_\Sigma) - |A_\Sigma|^2 = 2K_\Sigma - H_\Sigma^2 $$ where $R_N$ is the scalar curvature, $n_\Sigma$ is a normal to $\Sigma$, $|A_\Sigma|$ is the Frobenius norm of the second fundamental form, and $K_\Sigma,H_\Sigma$ are the Gaussian and mean curvatures of the embedded hypersurface.
Since $\Sigma$ is minimal, $H_\Sigma = 0$. $$ R_\Sigma = R_N - 2 \text{Ric}_N(n_\Sigma,n_\Sigma) - |A_\Sigma|^2 $$ where we note that $R_\Sigma=2K_\Sigma$ because $\Sigma$ is 2D. Using $K_\Sigma$ instead makes the equation off by a coefficient of $2$, so I suspect you just mean this, or your notation is different.
