Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm following the proof of the local expression for the Laplacian on a compact manifold and I'm having problems understanding how the integral on a manifold translates into an integral in $R^n$, in particular in the following calculation:

\begin{align*} (\Delta f,\varphi)&=(d f, d \varphi)= \int_M \! df \wedge *d\varphi =\int_M \! <df,d\varphi> *(1) \\ &= \int \! \sum_{i,j} g^{ij} \frac{\partial f}{\partial x^i}\frac{\partial \varphi }{\partial x^j} \sqrt{g} dx^1 ... dx^n\\ &= - \sum_{i,j} \int \! \frac{1}{\sqrt{g}}\frac{\partial}{\partial x^j}\left( \sqrt{g} g^{ij} \frac{\partial f}{\partial x^i}\right) \varphi \sqrt{g} dx^1 ... dx^n \end{align*}

after the fourth equal symbol I think it should go $$ \sum_{\alpha} \int_{U_{\alpha}} \! \rho_\alpha\sum_{i,j} g^{ij} \frac{\partial f}{\partial x^i}\frac{\partial \varphi }{\partial x^j} \sqrt{g} dx^1 ... dx^n $$ where $\rho_\alpha$ are the local expression of a partition of unity, but then in the last step I would have $$ - \sum_{\alpha,i,j}\int_{U_{\alpha}} \! \frac{1}{\sqrt{g}}\frac{\partial}{\partial x^j}\left( \rho_{\alpha}\sqrt{g} g^{ij} \frac{\partial f}{\partial x^i} \right) \varphi \sqrt{g} dx^1 ... dx^n$$ and I don't know how to continue.

share|improve this question
    
Could you give the source of the proof? –  user31373 Sep 8 '12 at 0:38
    
@LVK "Riemannian Geometry and Geometric Analysis" Jost, page 88 –  inquisitor Sep 9 '12 at 9:00

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.