Show that the Ricci curvature of a hypersurface $M\subset\mathbb{R}_v^{n+1}$ is given by:
$$Ric(V,W)=\epsilon[\langle SV,W\rangle]trace S-\langle SV,W\rangle$$
$S$ is the shape operator, defined as $Sv=-\nabla_V U$ where U is a unit normal vector.
$\langle Sv,u\rangle=\langle \Pi(v,w),u\rangle$
i tried to use the Gauss equation of curvature on the ricci curvature definition, but that does not yield $S$, at least I do not see how. This question does not seem like a straightforward calculation.
Question:
Can anyone provide me with some hints?
Thanks in advance.
I believe your formula has a mistake in its last term.
In general, we have the relation $$R(X,Y)Z = \langle SY, Z\rangle SX - \langle SX,Z\rangle SY$$between the curvature tensor $R$ of a hypersurface and the shape operator $S$ associated with a chosen unit normal field. See p. 16 on Section 1.4 of Dajczer & Tojeiro's Submanifold Theory.
We want to trace the above equation in the variable $X$ only. The left side gives ${\rm Ric}(Y,Z)$ by definition. In the right side, tracing $\langle SY,Z\rangle SX$ in $X$ obviously yields $\langle SY,Z\rangle \,{\rm tr}(S)$. As for the very last term, we have that ${\rm tr}(X\mapsto \langle SX,Z\rangle SY) = \langle S^2Y,Z\rangle = \langle SY,SZ\rangle$. Hence $${\rm Ric}(Y,Z) = \langle SY,Z\rangle\,{\rm tr}(S) - \langle SY,SZ\rangle.$$
Here, we're using the following linear algebra fact: whenever $V$ is a finite dimensional vector space, $v\in V$ and $f\in V^*$, and we define $T\colon V\to V$ by $T(x) = f(x)v$, then ${\rm tr}(T) = f(v)$. If $f=0$ this is trivial, and if $f\neq 0$ compute the trace using a basis $(e_1,\ldots, e_n)$ of $V$ with $f(e_1) = 1$ and $f(e_j) = 0$ for $j\geq 2$.