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)
$= \frac{1}{2} \sum_{i=1}^n X_i X_i \langle \nabla f, \nabla f\rangle$

 

[blue_raver22]

Hello! The Bochner-Weitzenbock formula is a fundamental result in Riemannian geometry that relates the Laplacian operator to the curvature of a manifold. It is often used in differential geometry to study the behavior of functions on a manifold.

In this particular case, we are looking at the Laplacian of the function f at a point p, which is denoted by \Delta f. The Laplacian is defined as the trace of the Hessian matrix of f, which you have correctly stated in your question. The Hessian matrix is essentially a matrix of second derivatives of f, and the trace is the sum of the diagonal entries.

Now, the first step in the proof of the Bochner-Weitzenbock formula is to take the Laplacian of the square of the gradient of f, denoted by |\nabla f|^2. This is where the Levi-Civita identities come into play. By using these identities, we can rewrite the Laplacian of |\nabla f|^2 as a sum of terms involving the Riemann curvature tensor and the covariant derivatives of f.

Next, we use the fact that we are working on a complete Riemannian manifold with an orthonormal frame at the point p. This allows us to simplify some of the terms in the sum and ultimately arrive at the desired result, which is the Bochner-Weitzenbock formula.

I hope this helps clarify the first step in the proof. If you are still having trouble, I suggest consulting a textbook on Riemannian geometry or seeking help from a colleague who is familiar with this topic. Keep up the good work!