$V$, $W$ two vector fields on surface $M$, $S$ shape operator, show that $S(V) \cdot W=\nabla_{V} W \cdot U$ This is homework, and we are stuck.
Let $V$, $W$ be two vector fields on a surface $M$ (with assumed ambient space $\mathbb{R}^3$). Prove that if $S$ is the shape operator on $M$ corresponding to a given unit normal vector field $U$, then
$$
S(V)\cdot W = \nabla_{V} W \cdot U
$$
which I understand to be equivalent to, fixing arbitrary $p \in M$.
$$
S(V_p)\cdot W_p = \nabla_{V_p} W \cdot U_p
$$
What we did
We took $S(V)\cdot W$ and transformed it to $-\nabla_{V} U \cdot W$. Then the goal is equivalent to
\begin{align*}
-\nabla_{V} U \cdot W &= \nabla_{V} W \cdot U
\end{align*}
which is equivalent to,
$$
\nabla_{V} W \cdot U + \nabla_{V} U \cdot W = 0
$$
by some previous exercise we have that the goal is equivalent to
$$
V (W \cdot U) = V (p \mapsto W_p \cdot U_p) = 0
$$
Then we became stuck. We expanded the definition and simply could not reach anywhere. Any hints?
 A: Hint: $W$ is tangent and $U$ is normal, so their inproduct $W\cdot U$ is ...
A: The formula
$S(V) \cdot W = \nabla_V W \cdot U \tag 1$
where $V$ and $W$ are tangent vector fields to a surface
$M \subset \Bbb R^3, \tag 2$
and $U$ is a unit normal field to $M$, is difficult to prove because it is, in fact, false.  The correct formula is
$S(V) \cdot W = -\nabla_V W \cdot U, \tag 3$
i.e., $S(V) \cdot W$ is the equal to the negative of $\nabla_V W \cdot U$, not $\nabla_V W \cdot U$ itself.  This may easily be seen as follows:
by definiton,
$S(V) = \nabla_V U; \tag 4$
furthermore, since $V$ and $W$ are tangent to $M$ whilst $U$ is normal to it,  
$U \cdot W  = 0 = U \cdot V; \tag 5$
we take the covariant derivative of the left-hand equation with respect to $V$ and obtain
$\nabla_V U \cdot W + U \cdot \nabla_V W = \nabla_V (U \cdot W) = 0, \tag 6$
or
$\nabla_V U \cdot W = -U \cdot \nabla_V W, \tag 7$
which is, according to (4),
$S(V) \cdot W = -U \cdot \nabla_V W = -\nabla_V W \cdot U, \tag 8$
which shows that (3) is correct and (1) is erroneous.
The formula (3) and its evident compantion
$S(W) \cdot V = -\nabla_W V \cdot U, \tag 9$
whose derivation is essentially the same but based on $V \cdot U = 0$, may be exploited to show the symmetry of $S$:
$S(W) \cdot V = S(V) \cdot W, \tag{10}$
an important general result in surface theory.  Indeed, (3) and (9) yield
$S(V) \cdot W - S(W) \cdot V$
$= -\nabla_V W \cdot U - (-\nabla_W V \cdot U) = (\nabla_W V - \nabla_V W) \cdot U; \tag{11}$
we now recall that
$\nabla_W V - \nabla_V W = [W, V], \tag{12}$
which, being the Lie bracket of $W$ and $V$, is itself tangent to $M$; therefore
$S(V) \cdot W - S(W) \cdot V = [W, V] \cdot U = 0, \tag{13}$
establishing the symmetry of $S$.
