gradient

Let $(X,g)$ be a Riemannian manifold and $f \in C^\infty(X)$ a function.

The **gradient** of $f$ is the vector field

$\nabla f := g^{-1} d_{dR} f \in \Gamma(T X)
\,,$

where $d_{dR} : C^\infty(X) \to \Omega^1(X)$ is the de Rham differential.

This is the unique vector field $\nabla f$ such that

$d_{dR} f = g(-,\nabla f)$

or equivalently, if the manifold is oriented, this is the unique vector field such that

$d_{dR} f = \star_g \iota_{\nabla f} vol_g
\,,$

where $vol_g$ is the volume form and $\star_g$ is the Hodge star operator induced by $g$. (The result is independent of orientation, which can be made explicit by interpreting both $vol$ and $\star$ as valued in pseudoforms.)

Alternatively, the gradient of a scalar field $A$ in some point $x\in M$ is calculated (or alternatively defined) by the integral formula

$grad A = lim_{vol D\to 0} \frac{1}{vol D} \oint_{\partial D} \vec{n} A d S$

where $D$ runs over the domains with smooth boundary $\partial D$ containing point $x$ and $\vec{n}$ is the unit vector of outer normal to the surface $S$. The formula does not depend on the shape of boundaries taken in limiting process, so one can typically take a coordinate chart and balls with decreasing radius in this particular coordinate chart.

If $(M,g)$ is the Cartesian space $\mathbb{R}^n$ endowed with the standard Euclidean metric, then

$\nabla f= \sum_{i=1}^n\frac{\partial f}{\partial x^i}\partial_i
.$

This is the classical gradient from vector analysis?.

In many classical applications of the gradient in vector analysis?, the Riemannian structure is actually irrelevant, and the gradient can be replaced with the differential 1-form.

Revised on September 5, 2011 18:11:45
by Toby Bartels
(75.88.82.16)