Editing Hodge star operator (section)

=== Example: Derivatives in three dimensions===
The combination of the <math>{\star}</math> operator and the [[exterior derivative]] {{math|''d''}} generates the classical operators {{math|[[gradient|grad]]}}, {{math|[[Curl (mathematics)|curl]]}}, and {{math|[[divergence|div]]}} on [[vector field]]s in three-dimensional Euclidean space. This works out as follows: {{math|''d''}} takes a 0-form (a function) to a 1-form, a 1-form to a 2-form, and a 2-form to a 3-form (and takes a 3-form to zero). For a 0-form <math>f = f(x,y,z)</math>, the first case written out in components gives:
<math display="block">df = \frac{\partial f}{\partial x} \, dx + \frac{\partial f}{\partial y} \, dy + \frac{\partial f}{\partial z} \, dz.</math>

The scalar product [[Dual space#Bilinear products and dual spaces|identifies]] 1-forms with vector fields as <math>dx \mapsto (1,0,0)</math>, etc., so that <math>df</math> becomes <math display="inline">\operatorname{grad} f = \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}\right)</math>.

In the second case, a vector field <math>\mathbf F = (A,B,C)</math> corresponds to the 1-form <math>\varphi = A\,dx + B\,dy + C\,dz</math>, which has exterior derivative:
<math display="block">d\varphi =
\left(\frac{\partial C}{\partial y} - \frac{\partial B}{\partial z}\right) dy\wedge dz +
\left(\frac{\partial C}{\partial x} - \frac{\partial A}{\partial z}\right) dx\wedge dz +
\left({\partial B \over \partial x} - \frac{\partial A}{\partial y}\right) dx\wedge dy.</math>

Applying the Hodge star gives the 1-form:
<math display="block">{\star} d\varphi = \left({\partial C \over \partial y} - {\partial B \over \partial z} \right) \, dx  - \left({\partial C \over \partial x} - {\partial A \over \partial z} \right) \, dy + \left({\partial B \over \partial x} - {\partial A \over \partial y}\right) \, dz,</math>
which becomes the vector field <math display="inline">\operatorname{curl}\mathbf{F} = \left(
   \frac{\partial C}{\partial y} - \frac{\partial B}{\partial z},\,
  -\frac{\partial C}{\partial x} + \frac{\partial A}{\partial z},\,
   \frac{\partial B}{\partial x} - \frac{\partial A}{\partial y}
\right)</math>.

In the third case, <math>\mathbf F = (A,B,C)</math> again corresponds to <math>\varphi = A\,dx + B\,dy + C\,dz</math>. Applying Hodge star, exterior derivative, and Hodge star again:
<math display="block">\begin{align}
           {\star}\varphi &= A\,dy\wedge dz-B\,dx\wedge dz+C\,dx\wedge dy, \\
        d{\star\varphi} &= \left(\frac{\partial A}{\partial x}+\frac{\partial B}{\partial y}+\frac{\partial C}{\partial z}\right)dx\wedge dy\wedge dz, \\
  {\star} d{\star}\varphi &= \frac{\partial A}{\partial x}+\frac{\partial B}{\partial y}+\frac{\partial C}{\partial z}
= \operatorname{div}\mathbf{F}.
\end{align}</math>

One advantage of this expression is that the identity {{math|1=''d''{{i sup|2}} = 0}}, which is true in all cases, has as special cases two other identities: (1)  {{math|1=curl grad ''f'' = 0}}, and (2) {{math|1=div curl '''F''' = 0}}. In particular, [[Maxwell's equations#Relativistic formulations|Maxwell's equations]] take on a particularly simple and elegant form, when expressed in terms of the exterior derivative and the Hodge star. The expression <math>{\star}d{\star}</math> (multiplied by an appropriate power of −1) is called the ''codifferential''; it is defined in full generality, for any dimension, further in the article below.

One can also obtain the [[Laplacian]] {{math|1=Δ''f'' = div grad ''f''}}  in terms of the above operations:
<math display="block"> \Delta f = {\star}d{\star}d f = \frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} + \frac{\partial^2 f}{\partial z^2}.</math>

The Laplacian can also be seen as a special case of the more general [[Laplace–Beltrami operator|Laplace–deRham operator]] <math>\Delta = d\delta + \delta d</math> where in three dimensions, <math>\delta = (-1)^k {\star} d{\star}</math> is the codifferential for <math>k</math>-forms. Any function <math>f</math> is a 0-form, and <math>\delta f = 0</math> and so this reduces to the ordinary Laplacian. For the 1-form <math>\varphi</math> above, the codifferential is <math>\delta = - {\star} d{\star}</math> and after some straightforward calculations one obtains the Laplacian acting on <math>\varphi</math>.