Laplacian on graph

Suppose {\Omega\subset \mathbb{R}^n} is a domain and {\Sigma=graph(F)\subset \mathbb{R}^{n+1}} is a hypersurface, where {F=F(u_1,\cdots,u_n)} is a function on {\Omega}. Define {f=f(u_1,\cdots,u_n)} on {\Omega}. Then {f} also can be considered as a function on {\Sigma}. How do we understand {\Delta_\Sigma f}?

Denote {\partial_i=\frac{\partial}{\partial{u_i}}} for short. If we pull the metric of {\mathbb{R}^{n+1}} back to {\Omega}, denote as {g}, then

\displaystyle g_{ij}=g(\partial_i,\partial_j)=\delta_{ij}+F_{u_i}F_{u_j},\quad g^{ij}=\delta_{ij}-\frac{F_{u_i}F_{u_j}}{W^2},\quad \det g=W^2

where {W=\sqrt{1+|\nabla F|^2}} and {F_{u_i}=\frac{\partial F}{\partial u_i}}. Then one can use the local coordinate to calculate {\Delta_\Sigma f}

\displaystyle \Delta_\Sigma f=\frac{1}{\sqrt{\det g}}\partial_{u_i}\left(\sqrt{\det g}\,g^{ij}f_{u_j}\right)

\displaystyle =g^{ij}f_{u_iu_j}+f_{u_j}\frac{\partial_{u_i}(\sqrt{\det g}\,g^{ij})}{\sqrt{\det g}}

also one can see from another definition of Laplacian

\displaystyle \Delta_\Sigma f=g^{ij}Hess(f)(\partial_i,\partial_j)=g^{ij}[\partial_j(\partial_i(f))-(\nabla_{\partial_j}\partial_i)f]

\displaystyle =g^{ij}f_{u_iu_j}-g^{ij}(\nabla_{\partial_j}\partial_i)f

By using the expression of {g^{ij}} stated above, we can calculate

\displaystyle \frac{\partial_{u_i}(\sqrt{\det g}\,g^{ij})}{\sqrt{\det g}}=\frac{-F_{u_j}F_{u_{i}u_{k}}g^{ik}}{W^2}

It follows from the definition of tangential derivative on {\Sigma}, see, that

\displaystyle \langle\nabla^\Sigma f, e_{n+1}\rangle=g^{ij}f_{u_{i}}F_{u_{j}}=f_{u_{i}}F_{u_{i}}-\frac{f_{u_i}F_{u_i}|\nabla F|^2}{W^2}=\frac{f_{u_{i}}F_{u_{i}}}{W^2}

then

\displaystyle f_{u_j}\frac{\partial_{u_i}(\sqrt{\det g}\,g^{ij})}{\sqrt{\det g}}=-\langle \nabla^\Sigma f,e_{n+1}\rangle F_{u_iu_k}g^{ik}=-\langle \nabla^\Sigma f,e_{n+1}\rangle HW

where {H} is the mean curvature of the {\Sigma}. Combining all the above calculations,

\displaystyle \Delta_\Sigma f=g^{ij}f_{u_iu_j}-\langle \nabla^\Sigma f,e_{n+1}\rangle HW

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