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

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

Thus,

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

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.