Consider now $M^n \subset N^{n+p}$ a Riemannian submanifold.
\begin{align} \Delta^{\perp}_M S = \sum\limits_{i=1}^n \nabla_{E_i}^{\perp} \nabla_{E_i}^{\perp} S - \sum\limits_{i=1}^n \nabla_{\nabla_{E_i} E_i}^{\perp} S \end{align} is the Laplacian on a normal bundle.
$$ \langle \Delta_M^{\perp} S,S \rangle = \frac{1}{2} \Delta_M |S|^2 - |\nabla_M^{\perp} S|^2, $$
where $\{ E_i \}_{i=1}^n$ are the coordinates basis of $M$, $S = S^{\alpha}\nu_{\alpha}$ and $\nabla_M^{\perp} S = E_i(S^{\alpha}) E_i \otimes \nu_{\alpha}$
I am trying to get the equality above from the definition of the Laplacian on a normal bundle, but I could not obtain this expression exactly. I did the following computation in normal coordinates:
\begin{align*} \Delta_M |S|^2 &= \nabla_{E_i} \nabla_{E_i} |S|^2\\ &= \nabla_{E_i} \nabla_{E_i} (S^{\alpha} S^{\alpha})\\ &= \nabla_{E_i} (2 S^{\alpha} \nabla_i S^{\alpha})\\ &= 2 \nabla_{E_i} S^{\alpha} \nabla_{E_i} S^{\alpha} + 2 S^{\alpha} \nabla_{E_i} \nabla_{E_i} S^{\alpha}. \end{align*}
The problem is that I got the equality for the tangential connection and not the normal connection. I would like to know how to got it for the normal connection.