Ngô Quốc Anh

November 26, 2009

R-G: Hessian and Laplacian

Filed under: Riemannian geometry — Ngô Quốc Anh @ 1:17

For a given smooth function f on manifold M, the gradient of f is given by

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

Note that gradient of f is also a vector field on M. Thus, for each X \in TM, it is reasonable to talk about \nabla_X \nabla f.

Definition 1. Hessian of f, denoted by {\rm Hess}, is defined as the symmetric (0,2)-tensor

{\rm Hess} f (X,Y)=g(\nabla_X \nabla f, Y).

We also denote by f_{ij} the following

\displaystyle {\rm Hess} f \left(\frac{\partial}{\partial x^i}, \frac{\partial}{\partial x^j}\right).

Thus

\displaystyle\begin{gathered} {f_{ij}} = g\left( {{\nabla _{\frac{\partial }{{\partial {x^i}}}}}\nabla f,\frac{\partial }{{\partial {x^j}}}} \right) = g\left( {{\nabla _{\frac{\partial }{{\partial {x^i}}}}}\left( {{g^{kl}}\frac{{\partial f}}{{\partial {x^l}}}\frac{\partial }{{\partial {x^k}}}} \right),\frac{\partial }{{\partial {x^j}}}} \right) \hfill \\ \quad\; = g\left( {\frac{\partial }{{\partial {x^i}}}\left( {{g^{kl}}\frac{{\partial f}}{{\partial {x^l}}}} \right)\frac{\partial }{{\partial {x^k}}} + {g^{kl}}\frac{{\partial f}}{{\partial {x^l}}}{\nabla _{\frac{\partial }{{\partial {x^i}}}}}\frac{\partial }{{\partial {x^k}}},\frac{\partial }{{\partial {x^j}}}} \right) \hfill \\ \quad\; = \frac{\partial }{{\partial {x^i}}}\left( {{g^{kl}}\frac{{\partial f}}{{\partial {x^l}}}} \right)g\left( {\frac{\partial }{{\partial {x^k}}},\frac{\partial }{{\partial {x^j}}}} \right) + {g^{kl}}\frac{{\partial f}}{{\partial {x^l}}}g\left( {{\nabla _{\frac{\partial }{{\partial {x^i}}}}}\frac{\partial }{{\partial {x^k}}},\frac{\partial }{{\partial {x^j}}}} \right) \hfill \\\quad\; = \frac{\partial }{{\partial {x^i}}}\left( {{g^{kl}}\frac{{\partial f}}{{\partial {x^l}}}} \right){g_{kj}} + {g^{kl}}\frac{{\partial f}}{{\partial {x^l}}} \left[ \frac{1}{2}\left( {-\frac{{\partial {g_{ki}}}}{{\partial {x^j}}} + \frac{{\partial {g_{ij}}}}{{\partial {x^k}}} + \frac{{\partial {g_{kj}}}}{{\partial {x^i}}}} \right)\right]. \hfill\end{gathered}

Note that

\displaystyle \frac{\partial }{{\partial {x^i}}}\left( {{g^{kl}}\frac{{\partial f}}{{\partial {x^l}}}} \right){g_{kj}} = \frac{{\partial {g^{kl}}}}{{\partial {x^i}}}\frac{{\partial f}}{{\partial {x^l}}}{g_{kj}} + {g^{kl}}\frac{\partial }{{\partial {x^i}}}\left( {\frac{{\partial f}}{{\partial {x^l}}}} \right){g_{kj}} = \frac{{\partial {g^{kl}}}}{{\partial {x^i}}}\frac{{\partial f}}{{\partial {x^l}}}{g_{kj}} + \frac{{{\partial ^2}f}}{{\partial {x^i}\partial {x^j}}}.

Since 0=\frac{\partial}{\partial x^i}(g^{kl}g_{kj}) then

\displaystyle\frac{{\partial {g^{kl}}}}{{\partial {x^i}}}\frac{{\partial f}}{{\partial {x^l}}}{g_{kj}} = - \frac{{\partial {g_{kj}}}}{{\partial {x^i}}}\frac{{\partial f}}{{\partial {x^l}}}{g^{kl}}

which implies

\displaystyle f_{ij} =\frac{{{\partial ^2}f}}{{\partial {x^i}\partial {x^j}}} - \Gamma _{ij}^m\frac{{\partial f}}{{\partial {x^m}}}.

Definition 2. Laplacian of f, denoted by \Delta f, is defined as the trace of {\rm Hess} f.

Note that {\rm Hess} f is a (0,2)-tensor, then in local coordinates, one has

\displaystyle \Delta f = {g^{ij}}{f_{ij}}.

It is clear that \nabla X is a (1,1)-tensor field. To see this fact, one can assume X=X^i \frac{\partial}{\partial x^i} then from

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

one has

\displaystyle\nabla X = \left[ {\frac{{\partial {X^i}}}{{\partial {x^j}}}\frac{\partial }{{\partial {x^i}}} + {X^i}\Gamma _{ji}^l\frac{\partial }{{\partial {x^l}}}} \right] \otimes d{x^j}

since {\nabla _Y}X = \left\langle {Y,\nabla X} \right\rangle which is exactly an (1,1)-tensor. Then we can define divergence of a vector field X as following

Definition 3. Divergence of vector field X is given by

\displaystyle {\rm div} X = {\rm Trace}(\nabla X).

In coordinates, this is

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

and with respect to an orthornormal basis

\displaystyle {\rm div} X =g\left( {{\nabla _{\frac{\partial }{{\partial {x^i}}}}}X,\frac{\partial }{{\partial {x^i}}}} \right).

Thus \Delta f = {\rm Trace}(\nabla(\nabla f)) = {\rm div}(\nabla f).

NOTICE: To avoid any inconvenience caused, from now we denote gradient of f by {\rm grad}f instead of \nabla f. This is because \nabla f is covariant derivative of f, this is an (1,0)-tensor instead of a vector field as mentioned in this entry.

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