I have a question about the proof of the following statement.
Let $M$ be an oriented hypersurface with constant mean curvature in $\mathbb{R}^{n+1}$ and with second fundamental form $B$. Let $v$ be the unit normal vector to $M$ then for all $a \in \mathbb{R}^{n+1}$ we have $$\Delta \langle a,v\rangle +|B|^2 \langle a,v\rangle =0.$$
Proof: Let $\nabla$ be the connection of $M$ and $\bar{\nabla}$ the connection of $\mathbb{R}^{n+1}$.
Choose orthonormal frame field $\{ e_i \}$ such that $\nabla_{e_j} e_i=0$ at the considered point.
Then $$\Delta \langle a,v \rangle =\nabla_{e_i} \nabla_{e_i} \langle a,v\rangle$$ $$=\nabla_{e_i} \langle a,\bar{\nabla}_{e_i} v\rangle$$
I'll stop the proof here because this is what my question is about.
Why is it the case that $\nabla_{e_i} \nabla_{e_i} \langle a,v\rangle =\nabla_{e_i} \langle a,\bar{\nabla}_{e_i} v \rangle$ shouldn't it be $\nabla_{e_i} \nabla_{e_i} \langle a,v\rangle =\nabla_{e_i} (\langle \nabla_{e_i} a,v \rangle +\langle a,\nabla_{e_i} v\rangle $?
Why is $\langle \nabla a ,v \rangle =0$?
Also how did $\bar{\nabla}$ appear? I know that $\bar{\nabla_X} Y =\nabla_X Y +B(X,Y)$ so shouldn't there be a $B(X,Y)$ term?
First of all, $a$ is a constant vector, so that $\bar{\nabla} a = 0$. One can think of it as a constant vector field on $\Bbb R^{n+1}$.
Second, notice that $\nabla v$ makes no sense: $v$ is not a vector field on $M$ (indeed, it is nowhere tangent to $M$). What makes sense is the following: the action of $\nabla$ on functions is by the differential. Hence, one has \begin{align} \nabla_{e_i} \langle a,v\rangle &= e_i \langle a,v \rangle & \text{by definition of the action of $\nabla$ on functions,}\\ &= \langle \bar{\nabla}_{e_i}a, v\rangle + \langle a, \bar{\nabla}_{e_i}v\rangle & \text{since $\bar{\nabla}$ is metric,} \\ &= \langle a , \bar{\nabla}_{e_i} v\rangle & \text{by $\bar{\nabla}a=0$.} \end{align}