Ngô Quốc Anh

May 6, 2011

Laplace-Beltrami vs. Laplacian

Filed under: Riemannian geometry — Ngô Quốc Anh @ 8:10

Let us do compare \Delta_g with respect to metric g and \Delta with respect to Euclidean metric. For simplicity, let us try with function u^{-\beta}. Obviously,

\displaystyle\begin{gathered} \Delta ({u^{ - \beta }}) = {\partial ^i}{\partial _i}({u^{ - \beta }}) \hfill \\ \qquad= {\partial ^i}( - \beta {u^{ - \beta - 1}}{\partial _i}u) \hfill \\ \qquad= \beta (\beta + 1){u^{ - \beta - 2}}({\partial ^i}u)({\partial _i}u) - \beta {u^{ - \beta - 1}}{\partial ^i}{\partial _i}u \hfill \\ \qquad= \beta (\beta + 1){u^{ - \beta - 2}}|\nabla u{|^2} - \beta {u^{ - \beta - 1}}\Delta u \hfill \\ \qquad= - \beta {u^{ - \beta - 1}}\left( {\Delta u - (\beta + 1)\frac{{|\nabla u{|^2}}}{u}} \right). \hfill \\ \end{gathered}

Next, by definition of \Delta_g we obtain

\displaystyle\begin{gathered} {\Delta _g}({u^{ - \beta }}) = \frac{1}{{\sqrt {\det g} }}{\partial _j}\left( {\sqrt {\det g} {g^{ij}}{\partial _i}({u^{ - \beta }})} \right) \hfill \\ \qquad= \frac{1}{{\sqrt {\det g} }}{\partial _j}\left( {( - \beta ){u^{ - \beta - 1}}\sqrt {\det g} {g^{ij}}{\partial _i}u} \right) \hfill \\ \qquad= \frac{1}{{\sqrt {\det g} }}\left[ {\beta (\beta + 1){u^{ - \beta - 2}}{\partial _ju}\sqrt {\det g} {g^{ij}}{\partial _i}u - \beta {u^{ - \beta - 1}}{\partial _j}\left( {\sqrt {\det g} {g^{ij}}{\partial _i}u} \right)} \right] \hfill \\ \qquad= \beta (\beta + 1){u^{ - \beta - 2}}|\nabla u{|_g^2} - \beta {u^{ - \beta - 1}}\Delta u_g \hfill \\ \qquad= - \beta {u^{ - \beta - 1}}\left( {\Delta_g u - (\beta + 1)\frac{{|\nabla u{|_g^2}}}{u}} \right). \hfill \\ \end{gathered}

Thus

\displaystyle {\Delta _g}({u^{ - \beta }}) = \Delta ({u^{ - \beta }}).

So the question is why do they equal?

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