Bochner-Weitzenbock formula (-> Laplacian)

[Sajet]

Hi! I'm trying to understand a proof for the Bochner-Weitzenbock formula. I'm sorry I have to bother you with such a basic question but I've worked at this for more than an hour now, but I just don't get the very first step, i.e.:

$-\frac{1}{2} \Delta |\nabla f|^2 = \frac{1}{2} \sum_{i}X_iX_i \langle \nabla f, \nabla f \rangle$

Where we are in a complete Riemannian manifold, $f \in C^\infty(M)$ at a point $p \in M$, with a local orthonormal frame $X_1, ..., X_n$ such that $\langle X_i, X_j \rangle = \delta_{ij}, D_{X_i}X_j(p) = 0$, and of course

$\langle \nabla f, X \rangle = X(f) = df(X)$
$\textrm{Hess }f(X, Y) = \langle D_X(\nabla f), Y \rangle$
$\Delta f = - \textrm{tr(Hess )}f$

I've tried to use the Levi-Civita identities, but I'm getting entangled in these formulas and don't get anywhere.

Any help is appreciated.

 

[Sajet]

I got it now :)

 

[dextercioby]

You may try to post a solution/sketch of solution for the one interested. That would be nice of you.

 

[Sajet]

Sorry, i didn't notice the post. In case anyone ever finds this through google or the search function, here it is:

$-\frac{1}{2} \Delta\|\nabla f\|^2 = \frac{1}{2}\text{tr}(\text{Hess}(\langle \nabla f, \nabla f \rangle ))$
$= \frac{1}{2}\sum_{i=1}^n \langle \nabla_{X_i} \text{grad}\langle \nabla f, \nabla f \rangle, X_i\rangle$ (<- these are the diagonal entries of the representation matrix)
$= \frac{1}{2}\sum_{i=1}^n X_i \langle \text{grad}\langle \nabla f, \nabla f\rangle, X_i\rangle - \langle \text{grad}\langle \nabla f, \nabla f\rangle, \nabla_{X_i} X_i\rangle$ (where the second summand is zero)
$= \frac{1}{2} \sum_{i=1}^n X_i X_i \langle \nabla f, \nabla f\rangle$