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BochnerWeitzenbock formula (> Laplacian) 
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#1
Jan213, 11:30 AM

P: 48

Hi! I'm trying to understand a proof for the BochnerWeitzenbock 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.:
[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 LeviCivita identities, but I'm getting entangled in these formulas and don't get anywhere. Any help is appreciated. 


#2
Jan313, 09:11 AM

P: 48

I got it now :)



#3
Jan313, 10:01 AM

Sci Advisor
HW Helper
P: 11,955

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



#4
Feb1013, 08:32 AM

P: 48

BochnerWeitzenbock formula (> Laplacian)
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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