Local form of $p$-Laplacian operator in Riemannian manifold

Question

Let $(M,g)$ be a connected oriented Riemannian manifold without boundary. The $p$-Laplacian of function $f:M\rightarrow\mathbb{R}$ is defined by $$\Delta_p f=\operatorname{div}\left(|\nabla f|^{p-2}\nabla f\right),$$ where $\nabla f$ is the gradient of $f$. I can not calculate the local form of $p$-Laplacian. I am trying to calculate it from the usual local form of laplacian operator but I am not getting any satisfactory form. Please help me.
Thank you.

Answer 1

The expression for the gradient vector field in local coordinates $(x_1 , . . . , x_n )$ is $$ \nabla_g \phi = g^{ij}\partial_i \phi, $$ where $\partial_i \phi= \frac{\partial \phi}{\partial x_i}$, and $g^{ij}$ the entries of the invese matrix of $g_{ij}$. This means $ \nabla_g \phi = \sum_{i,j=1}^n g^{ij}\partial_i \phi \, \partial_j, $ where $n=\dim(M)$. Then we have $$ |\nabla_g \phi|^2= g(\nabla_g \phi,\nabla_g \phi)=g_{ij}\, g^{hi}\, g^{kj} \, \partial_k\phi\, \partial_h \phi. $$ On the other hand, the expression for the divergence operator is $$ \text{div}(X)= \frac{1}{\sqrt{|\det g|}}\partial_i (\sqrt{|\det g|}\partial_i X). $$ Combining these expressions we have $$ \Delta_p\phi=\text{div}(|\nabla_g \phi|^{p-2}\nabla_g \phi)= $$ $$ \frac{1}{\sqrt{|\det g|}}\sum_m\partial_m\big\{\sqrt{|\det g|}\big(\sum_{i,j,k,h}g_{ij}\, g^{hi}\, g^{kj} \, \partial_k\phi\, \partial_h \phi\big)^{\frac{p-2}{2}}\sum_l g^{lm} \partial_l\phi\big\}. $$