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.