Ngô Quốc Anh

December 5, 2009

R-G: Laplace-Beltrami operator

Filed under: Riemannian geometry — Ngô Quốc Anh @ 13:35

So far for a smooth function f : M \to \mathbb R, gradient of f is a vector field defined as follows

\displaystyle \nabla f = g^{kj} \dfrac{\partial f}{\partial x^j} \frac{\partial}{\partial x^k}.

Since the gradient of f is nothing but a vector field then it is reasonable to talk about the divergence of a vector field X. To be exact, we define

\displaystyle {\rm div} X = dx^i \left( \nabla_{\frac{\partial}{\partial x^i}} X\right)

where X is a vector field. All above was discussed in this topic.

In order to go further, I need to spend some time talking about {\rm div} X. Precisely, if f is a smooth function, then can we write down explicitly {\rm div}(\nabla f) in local coordinates?

Having the definition of divergence operator yields

\displaystyle {\rm div} X= \left\langle {{\nabla _{\frac{\partial }{{\partial {x^i}}}}}X,d{x^i}} \right\rangle

where \left\langle \cdot , \cdot \right\rangle presents a pairing between TM and its dual space T^\star M. If, in local coordinates, X=X^i\frac{\partial}{\partial x^i}, we then have

\displaystyle {\nabla _{\frac{\partial }{{\partial {x^i}}}}}X = {\nabla _{\frac{\partial }{{\partial {x^i}}}}}\left( {{X^j}\frac{\partial }{{\partial {x^j}}}} \right) = \frac{{\partial {X^j}}}{{\partial {x^i}}}\frac{\partial }{{\partial {x^j}}} + {X^j}{\nabla _{\frac{\partial }{{\partial {x^i}}}}}\frac{\partial }{{\partial {x^j}}} = \frac{{\partial {X^j}}}{{\partial {x^i}}}\frac{\partial }{{\partial {x^j}}} + {X^j}\Gamma _{ij}^k\frac{\partial }{{\partial {x^k}}}.


\displaystyle {\rm div} X = \left\langle {\frac{{\partial {X^j}}}{{\partial {x^i}}}\frac{\partial }{{\partial {x^j}}} + {X^j}\Gamma _{ij}^k\frac{\partial }{{\partial {x^k}}},d{x^i}} \right\rangle = \frac{{\partial {X^i}}}{{\partial {x^i}}} + {X^j}\Gamma _{ij}^i.

As a consequence,

\displaystyle {\rm div}(\nabla f)=\frac{\partial }{{\partial {x^i}}}\left( {{g^{ij}}\frac{{\partial f}}{{\partial {x^j}}}} \right) + {g^{jk}}\frac{{\partial f}}{{\partial {x^k}}}\Gamma _{ij}^i.

We now look at the Christoffel symbols. Clearly, by definition

\displaystyle\begin{gathered} \Gamma _{ij}^i = \frac{1}{2}{g^{ik}}\left( {\frac{{\partial {g_{ik}}}}{{\partial {x^j}}} + \frac{{\partial {g_{jk}}}}{{\partial {x^i}}} - \frac{{\partial {g_{ij}}}}{{\partial {x^k}}}} \right) = \frac{1}{2}{g^{ik}}\frac{{\partial {g_{ik}}}}{{\partial {x^j}}} \hfill \\ \qquad= \frac{1}{2}{\rm Trace}\left( {\left( {{g^{mn}}} \right)\frac{\partial }{{\partial {x^j}}}\left( {{g_{mn}}} \right)} \right) \hfill \\ \qquad= \frac{1}{2}{\rm Trace}\left( {{{\left( {{g_{mn}}} \right)}^{ - 1}}\frac{\partial }{{\partial {x^j}}}\left( {{g_{mn}}} \right)} \right) \hfill \\ \qquad = \frac{1}{2}\frac{{\dfrac{\partial }{{\partial {x^j}}}\det \left( {{g_{mn}}} \right)}}{{\det \left( {{g_{mn}}} \right)}} \hfill \\ \end{gathered}

where the main theorem posted here has been used. Thus,

\displaystyle \begin{gathered} {\rm div}X = \frac{{\partial {X^i}}}{{\partial {x^i}}} + {X^j}\Gamma _{ij}^i \hfill \\ \qquad= \frac{1}{{\sqrt {\det \left( {{g_{mn}}} \right)} }}\sqrt {\det \left( {{g_{mn}}} \right)} \frac{{\partial {X^i}}}{{\partial {x^i}}} + \frac{1}{{\sqrt {\det \left( {{g_{mn}}} \right)} }}\frac{1}{{2\sqrt {\det \left( {{g_{mn}}} \right)} }}\frac{{\partial \det \left( {{g_{mn}}} \right)}}{{\partial {x^j}}}{X^j} \hfill \\ \qquad= \frac{1}{{\sqrt {\det \left( {{g_{mn}}} \right)} }}\frac{\partial }{{\partial {x^j}}}\left( {\sqrt {\det \left( {{g_{mn}}} \right)} {X^i}} \right). \hfill \\ \end{gathered}

As a consequence,

\displaystyle\Delta f = \frac{1}{{\sqrt {\det \left( {{g_{mn}}} \right)} }}\frac{\partial }{{\partial {x^j}}}\left( {\sqrt {\det \left( {{g_{mn}}} \right)} {g^{ij}}\frac{{\partial f}}{{\partial {x^i}}}} \right).

The above \Delta acting on a smooth function f is called the Laplace-Beltrami operator.



  1. Thank you very much! This is a very helpful construction. I didn’t find anything as simple as this for the Laplace-Beltrami Operator. Excellent work!

    Comment by Samir — January 26, 2022 @ 2:09

  2. Thank you very much for this detailed proof! Excellent work.

    Comment by Samir — January 26, 2022 @ 3:30

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