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Bochner-Weitzenbock formula (-> Laplacian)

  1. Jan 2, 2013 #1
    Hi! I'm trying to understand a proof for the Bochner-Weitzenbock formular. 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.:

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

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

    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.
     
  2. jcsd
  3. Jan 3, 2013 #2
    I got it now :)
     
  4. Jan 3, 2013 #3

    dextercioby

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    You may try to post a solution/sketch of solution for the one interested. That would be nice of you.
     
  5. Feb 10, 2013 #4
    Sorry, i didn't notice the post. In case anyone ever finds this through google or the search function, here it is:

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