# Hodge star operator

In mathematics, the Hodge star operator or Hodge star is a linear map defined on the exterior algebra of a finite-dimensional oriented vector space endowed with a nondegenerate symmetric bilinear form. Applying the operator to an element of the algebra produces the Hodge dual of the element. This map was introduced by W. V. D. Hodge.

For example, in an oriented 3-dimensional Euclidean space, an oriented plane can be represented by the exterior product of two basis vectors, and its Hodge dual is the normal vector given by their cross product; conversely, any vector is dual to the oriented plane perpendicular to it, endowed with a suitable bivector. Generalizing this to an n-dimensional vector space, the Hodge star is a one-to-one mapping of k-vectors to (n – k)-vectors; the dimensions of these spaces are the binomial coefficients ${\displaystyle {\tbinom {n}{k}}={\tbinom {n}{n-k}}}$.

The naturalness of the star operator means it can play a role in differential geometry, when applied to the cotangent bundle of a pseudo-Riemannian manifold, and hence to differential k-forms. This allows the definition of the codifferential as the Hodge adjoint of the exterior derivative, leading to the Laplace–de Rham operator. This generalizes the case of 3-dimensional Euclidean space, in which divergence of a vector field may be realized as the codifferential opposite to the gradient operator, and the Laplace operator on a function is the divergence of its gradient. An important application is the Hodge decomposition of differential forms on a closed Riemannian manifold.

## Formal definition for k-vectors

Let V be an n-dimensional vector space with a nondegenerate symmetric bilinear form ${\displaystyle \langle \cdot ,\cdot \rangle }$ , referred to here as an inner product. This induces an inner product on k-vectors ${\displaystyle \alpha ,\beta \in {\textstyle \bigwedge }^{\!k}V}$ , for ${\displaystyle 0\leq k\leq n}$ , by defining it on decomposable k-vectors ${\displaystyle \alpha =\alpha _{1}\wedge \cdots \wedge \alpha _{k}}$  and ${\displaystyle \beta =\beta _{1}\wedge \cdots \wedge \beta _{k}}$  to equal the Gram determinant[1]:14

${\displaystyle \langle \alpha ,\beta \rangle =\det(\left\langle \alpha _{i},\beta _{j}\right\rangle )_{i,j=1}^{k}}$

extended to ${\displaystyle {\textstyle \bigwedge }^{\!k}V}$  through linearity.

The unit n-vector ${\displaystyle \omega \in {\textstyle \bigwedge }^{\!n}V}$  is defined in terms of an oriented orthonormal basis ${\displaystyle \{e_{1},\ldots ,e_{n}\}}$  of V as:

${\displaystyle \omega \ :=\ e_{1}\wedge \cdots \wedge e_{n}.}$

The Hodge star operator is a linear operator on the exterior algebra of V, mapping k-vectors to (nk)-vectors, for ${\displaystyle 0\leq k\leq n}$ . It has the following property, which defines it completely:[1]:15

${\displaystyle \alpha \wedge ({\star }\beta )=\langle \alpha ,\beta \rangle \,\omega }$  for every pair of k-vectors ${\displaystyle \alpha ,\beta \in {\textstyle \bigwedge }^{\!k}V.}$

Dually, in the space ${\displaystyle {\textstyle \bigwedge }^{\!n}V^{*}}$ of n-forms (alternating n-multilinear functions on ${\displaystyle V^{n}}$ ), the dual to ${\displaystyle \omega }$  is the volume form ${\displaystyle \det }$ , the function whose value on ${\displaystyle v_{1}\wedge \cdots \wedge v_{n}}$  is the determinant of the ${\displaystyle n\times n}$  matrix assembled from the column vectors of ${\displaystyle v_{i}}$  in ${\displaystyle e_{i}}$ -coordinates.

Applying ${\displaystyle \det }$  to the above equation, we obtain the dual definition:

${\displaystyle \det(\alpha \wedge {\star }\beta )=\langle \alpha ,\beta \rangle .}$

or equivalently, taking ${\displaystyle \alpha =\alpha _{1}\wedge \cdots \wedge \alpha _{k}}$ , ${\displaystyle \beta =\beta _{1}\wedge \cdots \wedge \beta _{k}}$ , and ${\displaystyle \star \beta =\beta _{1}^{\star }\wedge \cdots \wedge \beta _{n-k}^{\star }}$ :

${\displaystyle \det(\alpha _{1}\wedge \cdots \wedge \alpha _{k}\wedge \beta _{1}^{\star }\wedge \cdots \wedge \beta _{n-k}^{\star })\ =\ \det(\langle \alpha _{i},\beta _{j}\rangle ).}$

This means that, writing an orthonormal basis of k-vectors as ${\displaystyle e_{I}\ =\ e_{i_{1}}\wedge \cdots \wedge e_{i_{k}}}$  over all subsets ${\displaystyle I=\{i_{1}<\cdots   of ${\displaystyle [n]=\{1,\ldots ,n\}}$ , the Hodge dual is the (n – k)-vector corresponding to the complementary set ${\displaystyle {\bar {I}}=[n]\setminus I=\{{\bar {i}}_{1}<\cdots <{\bar {i}}_{n-k}\}}$ :

${\displaystyle {\star }e_{I}=(-1)^{\sigma (I)}e_{\bar {I}},}$

where ${\displaystyle (-1)^{\sigma (I)}}$  is the sign of the permutation ${\displaystyle \sigma (I)=i_{1}\cdots i_{k}{\bar {i}}_{1}\cdots {\bar {i}}_{n-k}}$ .

Since Hodge star takes an orthonormal basis to an orthonormal basis, it is an isometry on the exterior algebra ${\displaystyle {\textstyle \bigwedge }V}$ .

## Geometric explanation

The Hodge star is motivated by the correspondence between a subspace W of V and its orthogonal subspace (with respect to the inner product), where each space is endowed with an orientation and a numerical scaling factor. Specifically, a non-zero decomposable k-vector ${\displaystyle w_{1}\wedge \cdots \wedge w_{k}\in \textstyle \bigwedge ^{\!k}V}$  corresponds by the Plücker embedding to the subspace ${\displaystyle W}$  with oriented basis ${\displaystyle w_{1},\ldots ,w_{k}}$ , endowed with a scaling factor equal to the k-dimensional volume of the parallelopiped spanned by this basis (equal to the Gramian, the determinant of the matrix of inner products ${\displaystyle \langle w_{i},w_{j}\rangle }$ ). The Hodge star acting on a decomposable vector can be written as a decomposable (nk)-vector:

${\displaystyle \star (w_{1}\wedge \cdots \wedge w_{k})\,=\,u_{1}\wedge \cdots \wedge u_{n-k},}$

where ${\displaystyle u_{1},\ldots ,u_{n-k}}$  form an oriented basis of the orthogonal space ${\displaystyle U=W^{\perp }\!}$ . Furthermore, the (nk)-volume of the ${\displaystyle u_{i}}$ -parallelopiped must equal the k-volume of the ${\displaystyle w_{i}}$ -parallelopiped, and ${\displaystyle w_{1},\ldots ,w_{k},u_{1},\ldots ,u_{n-k}}$  must form an oriented basis of V.

A general k-vector is a linear combination of decomposable k-vectors, and the definition of the Hodge star is extended to general k-vectors by defining it as being linear.

## Examples

### Two dimensions

In two dimensions with the normalized Euclidean metric and orientation given by the ordering (x, y), the Hodge star on k-forms is given by

${\displaystyle {\star }\,1=dx\wedge dy}$
${\displaystyle {\star }\,dx=dy}$
${\displaystyle {\star }\,dy=-dx}$
${\displaystyle {\star }(dx\wedge dy)=1.}$

On the complex plane regarded as a real vector space with the standard sesquilinear form as the metric, the Hodge star has the remarkable property that it is invariant under holomorphic changes of coordinate. If z = x + iy is a holomorphic function of w = u + iv, then by the Cauchy–Riemann equations we have that x/u = y/v and y/u = –x/v. In the new coordinates

${\displaystyle \alpha \ =\ p\,dx+q\,dy\ =\ \left(p{\frac {\partial x}{\partial u}}+q{\frac {\partial y}{\partial u}}\right)\,du+\left(p{\frac {\partial x}{\partial v}}+q{\frac {\partial y}{\partial v}}\right)\,dv\ =\ p_{1}du+q_{1}\,dv,}$

so that

{\displaystyle {\begin{aligned}{\star }\alpha =-q_{1}\,du+p_{1}\,dv&=-\left(p{\frac {\partial x}{\partial v}}+q{\frac {\partial y}{\partial v}}\right)du+\left(p{\frac {\partial x}{\partial u}}+q{\frac {\partial y}{\partial u}}\right)dv\\[4pt]&=-q\left({\frac {\partial x}{\partial u}}du+{\frac {\partial x}{\partial v}}dv\right)+p\left({\frac {\partial y}{\partial u}}du+{\frac {\partial y}{\partial v}}dv\right)\\[4pt]&=-q\,dx+p\,dy,\end{aligned}}}

proving the claimed invariance.

### Three dimensions

A common example of the Hodge star operator is the case n = 3, when it can be taken as the correspondence between vectors and bivectors. Specifically, for Euclidean R3 with the basis ${\displaystyle dx,dy,dz}$  of one-forms often used in vector calculus, one finds that

{\displaystyle {\begin{aligned}{\star }\,dx&=dy\wedge dz\\{\star }\,dy&=dz\wedge dx\\{\star }\,dz&=dx\wedge dy.\end{aligned}}}

The Hodge star relates the exterior and cross product in three dimensions:[2]

${\displaystyle {\star }(\mathbf {u} \wedge \mathbf {v} )=\mathbf {u} \times \mathbf {v} \qquad {\star }(\mathbf {u} \times \mathbf {v} )=\mathbf {u} \wedge \mathbf {v} .}$

Applied to three dimensions, the Hodge star provides an isomorphism between axial vectors and bivectors, so each axial vector a is associated with a bivector A and vice versa, that is:[2] ${\displaystyle \mathbf {A} ={\star }\mathbf {a} ,\ \ \mathbf {a} ={\star }\mathbf {A} }$ . The Hodge star can also be interpreted as a form of the geometric correpondence between an axis and an infinitesimal rotation around the axis, with speed equal to the length of the axis vector. An inner product on a vector space ${\displaystyle V}$  gives an isomorphism ${\displaystyle V\cong V^{*}\!}$  identifying ${\displaystyle V}$  with its dual space, and the space of all linear operators ${\displaystyle L:V\to V}$  is naturally isomorphic to the tensor product ${\displaystyle V^{*}\!\!\otimes V\cong V\otimes V}$ . Thus for ${\displaystyle V=\mathbb {R} ^{3}}$ , the star mapping ${\displaystyle \textstyle \star :V\to \bigwedge ^{\!2}\!V\subset V\otimes V}$  takes each vector ${\displaystyle \mathbf {v} }$  to a bivector ${\displaystyle \star \mathbf {v} \in V\otimes V}$ , which corresponds to a linear operator ${\displaystyle L_{\mathbf {v} }:V\to V}$ . Specifically, ${\displaystyle L_{\mathbf {v} }}$  is a skew-symmetric operator, which corresponds to an infinitesimal rotation: that is, the macroscopic rotations around the axis ${\displaystyle \mathbb {v} }$  are given by the matrix exponential ${\displaystyle \exp(tL_{\mathbf {v} })}$ . With respect to the basis ${\displaystyle dx,dy,dz}$  of ${\displaystyle \mathbb {R} ^{3}}$ , the tensor ${\displaystyle dx\otimes dy}$  corresponds to a coordinate matrix with 1 in the ${\displaystyle dx}$  row and ${\displaystyle dy}$  column, etc., and the wedge ${\displaystyle dx\wedge dy\,=\,dx\otimes dy-dy\otimes dx}$  is the skew-symmetric matrix ${\displaystyle \scriptscriptstyle \left[{\begin{array}{rrr}\,0\!\!&\!\!1&\!\!\!\!0\!\!\!\!\!\!\\[-.5em]\,\!-1\!\!&\!\!0\!\!&\!\!\!\!0\!\!\!\!\!\!\\[-.5em]\,0\!\!&\!\!0\!\!&\!\!\!\!0\!\!\!\!\!\!\end{array}}\!\!\!\right]}$ , etc. That is, we may interpret the star operator as:

${\displaystyle \mathbf {v} =a\,dx+b\,dy+c\,dz\quad \longrightarrow \quad \star {\mathbf {v} }\ \cong \ L_{\mathbf {v} }\ =\left[{\begin{array}{rrr}0&c&-b\\-c&0&a\\b&-a&0\end{array}}\right].}$

Under this correspondence, cross product of vectors corresponds to the commutator Lie bracket of linear operators: ${\displaystyle L_{\mathbf {u} \times \mathbf {v} }=L_{\mathbf {u} }L_{\mathbf {v} }-L_{\mathbf {v} }L_{\mathbf {u} }}$ .

### Four dimensions

In case n = 4, the Hodge star acts as an endomorphism of the second exterior power (i.e. it maps two-forms to two-forms, since 4 − 2 = 2). If the signature of the metric tensor is all positive, i.e. on a Riemannian manifold, then the Hodge star is an involution; if the signature is mixed, then application twice will return the argument up to a sign – see § Duality below. For example, in Minkowski spacetime where n = 4 with metric signature (+ − − −) and coordinates (t, x, y, z) where (using ${\displaystyle \varepsilon _{0123}=1}$ ):

{\displaystyle {\begin{aligned}\star dt&=dx\wedge dy\wedge dz\\\star dx&=dt\wedge dy\wedge dz\\\star dy&=-dt\wedge dx\wedge dz\\\star dz&=dt\wedge dx\wedge dy\end{aligned}}}

for one-forms while

{\displaystyle {\begin{aligned}\star (dt\wedge dx)&=-dy\wedge dz\\\star (dt\wedge dy)&=dx\wedge dz\\\star (dt\wedge dz)&=-dx\wedge dy\\\star (dx\wedge dy)&=dt\wedge dz\\\star (dx\wedge dz)&=-dt\wedge dy\\\star (dy\wedge dz)&=dt\wedge dx\end{aligned}}}

for two-forms. Because their determinants are the same in both (+ − − −) and (− + + +), the signs of the Minkowski space two-form duals depend only on the chosen orientation.[verification needed]

An easy rule to remember for the above Hodge operations is that given a form ${\displaystyle \alpha }$ , its Hodge dual ${\displaystyle {\star }\alpha }$  may be obtained by writing the components not involved in ${\displaystyle \alpha }$  in an order such that ${\displaystyle \alpha \wedge (\star \alpha )=dt\wedge dx\wedge dy\wedge dz}$ .[verification needed] An extra minus sign will enter only if ${\displaystyle \alpha }$  does not contain ${\displaystyle dt}$ . (The latter convention stems from the choice (+ − − −) for the metric signature. For (− + + +), one puts in a minus sign only if ${\displaystyle \alpha }$  involves ${\displaystyle dt}$ .)

## Duality

Applying the Hodge star twice leaves a k-vector unchanged except for its sign: for ${\displaystyle \eta \in {\textstyle \bigwedge }^{k}V}$  in an n-dimensional space V, one has

${\displaystyle {\star }{\star }\eta =(-1)^{k(n-k)}s\eta ,}$

where s is the parity of the signature of the inner product on V, that is, the sign of the determinant of the matrix of the inner product with respect to any basis. For example, if n = 4 and the signature of the inner product is either (+ − − −) or (− + + +) then s = −1. For Riemannian manifolds (including Euclidean spaces), we always have s = 1.

The above identity implies that the inverse of ${\displaystyle \star }$  can be given as

{\displaystyle {\begin{aligned}{\star }^{-1}:~&{\textstyle \bigwedge }^{\!k}\to {\textstyle \bigwedge }^{\!n-k}\\&\eta \mapsto (-1)^{k(n-k)}\!s{\star }\eta \end{aligned}}}

If n is odd then k(nk) is even for any k, whereas if n is even then k(nk) has the parity of k. Therefore:

${\displaystyle {\star }^{-1}={\begin{cases}s{\star }&n{\text{ is odd}}\\(-1)^{k}s{\star }&n{\text{ is even}}\end{cases}}}$

where k is the degree of the element operated on.

## On manifolds

For an n-dimensional oriented pseudo-Riemannian manifold M, we apply the construction above to each cotangent space ${\displaystyle {\text{T}}_{p}^{*}M}$  and its exterior powers ${\displaystyle \wedge ^{k}{\text{T}}_{p}^{*}M}$ , and hence to the differential k-forms ${\displaystyle \zeta \in \Omega ^{k}(M)=\Gamma \left(\wedge ^{k}{\text{T}}^{*}\!M\right)}$ , the global sections of the bundle ${\displaystyle \wedge ^{k}\mathrm {T} ^{*}\!M\to M}$ . The Riemanninan metric induces an inner product on ${\displaystyle \wedge ^{k}{\text{T}}_{p}^{*}M}$  at each point ${\displaystyle p\in M}$ . We define the Hodge dual of a k-form ${\displaystyle \zeta }$ , defining ${\displaystyle {\star }\zeta }$  as the unique (nk)-form satisfying

${\displaystyle \eta \wedge {\star }\zeta \ =\ \langle \eta ,\zeta \rangle \,\omega }$

for every k-form ${\displaystyle \eta }$ , where ${\displaystyle \langle \eta ,\zeta \rangle }$  is a real-valued function on ${\displaystyle M}$ , and the volume form ${\displaystyle \omega }$  is induced by the Riemannian metric. Integrating this equation over ${\displaystyle M}$ , the right side beomes the ${\displaystyle L^{2}}$  inner product on k-forms, and we obtain:

${\displaystyle \int _{M}\eta \wedge {\star }\zeta \ =\ \langle \!\langle \eta ,\zeta \rangle \!\rangle .}$

More generally, if ${\displaystyle M}$  is non-oriented, one can define the Hodge star of a k-form as a (nk)-pseudo differential form; that is, a differential form with values in the canonical line bundle.

### Computation in index notation

We compute in terms of tensor index notation with respect to a (not necessarily orthonormal) basis ${\displaystyle \{{\tfrac {\partial }{\partial x_{1}}},\ldots ,{\tfrac {\partial }{\partial x_{n}}}\}}$  in a tangent space ${\displaystyle V=T_{p}M}$  and its dual basis ${\displaystyle \{dx_{1},\ldots ,dx_{n}\}}$  in ${\displaystyle V^{*}=T_{p}^{*}M}$ , having the metric matrix ${\displaystyle (g_{ij})\ =\ (\langle {\tfrac {\partial }{\partial x_{i}}},{\tfrac {\partial }{\partial x_{j}}}\rangle )}$  and its inverse matrix ${\displaystyle (g^{ij})\ =\ (\langle dx_{i},dx_{j}\rangle )}$ . The Hodge dual of a decomposable k-form is:

${\displaystyle \star \left(dx^{i_{1}}\wedge \dots \wedge dx^{i_{k}}\right)\ =\ {\frac {\sqrt {|\det[g_{ab}]|}}{(n-k)!}}g^{i_{1}j_{1}}\cdots g^{i_{k}j_{k}}\varepsilon _{j_{1}\dots j_{n}}dx^{j_{k+1}}\wedge \dots \wedge dx^{j_{n}}.}$

Here ${\displaystyle \varepsilon _{j_{1}\dots j_{n}}}$  is the Levi-Civita symbol with ${\displaystyle \varepsilon _{1\dots n}=1}$ , and we implicitly take the sum over all values of the repeated indices ${\displaystyle j_{1},\ldots ,j_{n}}$ . The factorial ${\displaystyle (n-k)!}$  accounts for double counting, and is not present if the summation indices are restricted so that ${\displaystyle j_{k+1}<\dots  . The absolute value of the determinant is necessary since it may be negative, as for tangent spaces to Lorentzian manifolds.

An arbitrary differential form can be written:

${\displaystyle \alpha \ =\ {\frac {1}{k!}}\alpha _{i_{1},\dots ,i_{k}}dx^{i_{1}}\wedge \dots \wedge dx^{i_{k}}\ =\ \sum _{i_{1}<\dots

The factorial ${\displaystyle k!}$  is again included to account for double counting when we allow non-increasing indices. We would like to define the dual of the component ${\displaystyle \alpha _{i_{1},\dots ,i_{k}}}$  so that the Hodge dual of the form is given by

${\displaystyle \star \alpha ={\frac {1}{(n-k)!}}(\star \alpha )_{i_{k+1},\dots ,i_{n}}dx^{i_{k+1}}\wedge \dots \wedge dx^{i_{n}}.}$

Using the above expression for the Hodge dual of ${\displaystyle dx^{i_{1}}\wedge \dots \wedge dx^{i_{k}}}$ , we find:[3]

${\displaystyle (\star \alpha )_{i_{k+1},\dots ,i_{n}}={\frac {\sqrt {|\det[g_{ab}]|}}{k!}}\alpha ^{i_{1},\dots ,i_{k}}\,\,\varepsilon _{i_{1},\dots ,i_{n}}.}$

Although one can apply this expression to any tensor ${\displaystyle \alpha }$ , the result is antisymmetric, since contraction with the completely anti-symmetric Levi-Civita symbol cancels all but the totally antisymmetric part of the tensor. It is thus equivalent to antisymmetrization followed by applying the Hodge star.

The unit volume form ${\displaystyle \omega =\star 1\in \wedge ^{n}V^{*}}$  is given by:

${\displaystyle \omega ={\sqrt {\left|\det[g_{ij}]\right|}}\;dx^{1}\wedge \cdots \wedge dx^{n}.}$

### Codifferential

The most important application of the Hodge star on manifolds is to define the codifferential ${\displaystyle \delta }$  on k-forms. Let

${\displaystyle \delta =(-1)^{n(k-1)+1}s\ {\star }d{\star }=(-1)^{k}\,{\star }^{-1}d{\star }}$

where ${\displaystyle d}$  is the exterior derivative or differential, and ${\displaystyle s=1}$  for Riemannian manifolds. Then

${\displaystyle d:\Omega ^{k}(M)\to \Omega ^{k+1}(M)}$

while

${\displaystyle \delta :\Omega ^{k}(M)\to \Omega ^{k-1}(M).}$

The codifferential is not an antiderivation on the exterior algebra, in contrast to the exterior derivative.

The codifferential is the adjoint of the exterior derivative with respect to the ${\displaystyle L^{2}}$  inner product:

${\displaystyle \langle \!\langle \eta ,\delta \zeta \rangle \!\rangle \ =\ \langle \!\langle d\eta ,\zeta \rangle \!\rangle ,}$

where ${\displaystyle \zeta }$  is a (k + 1)-form and ${\displaystyle \eta }$  a k-form. This identity follows from Stokes' theorem for smooth forms:

${\displaystyle 0\ =\ \int _{M}d(\eta \wedge {\star }\zeta )\ =\ \int _{M}\left(d\eta \wedge {\star }\zeta -\eta \wedge {\star }(-1)^{k+1}\,{\star }^{-1}d{\star }\zeta \right)\ =\ \langle \!\langle d\eta ,\zeta \rangle \!\rangle -\langle \!\langle \eta ,\delta \zeta \rangle \!\rangle ,}$

provided M has empty boundary, or ${\displaystyle \eta }$  or ${\displaystyle \star \zeta }$  has zero boundary values. (However, true adjointness follows after continuous continuation to the appropriate topological vector spaces as closures of the spaces of smooth forms.)

Since the differential satisfies ${\displaystyle d^{2}=0}$ , the codifferential has the corresponding property

${\displaystyle \delta ^{2}=s^{2}{\star }d{\star }{\star }d{\star }=(-1)^{k(n-k)}s^{3}{\star }d^{2}{\star }=0.}$

The Laplace–deRham operator is given by

${\displaystyle \Delta =(\delta +d)^{2}=\delta d+d\delta }$

and lies at the heart of Hodge theory. It is symmetric:

${\displaystyle \langle \!\langle \Delta \zeta ,\eta \rangle \!\rangle =\langle \!\langle \zeta ,\Delta \eta \rangle \!\rangle }$

and non-negative:

${\displaystyle \langle \!\langle \Delta \eta ,\eta \rangle \!\rangle \geq 0.}$

The Hodge star sends harmonic forms to harmonic forms. As a consequence of Hodge theory, the de Rham cohomology is naturally isomorphic to the space of harmonic k-forms, and so the Hodge star induces an isomorphism of cohomology groups

${\displaystyle {\star }:H_{\Delta }^{k}(M)\to H_{\Delta }^{n-k}(M),}$

which in turn gives canonical identifications via Poincaré duality of H k(M) with its dual space.

## Derivatives in three dimensions

The combination of the ${\displaystyle \star }$  operator and the exterior derivative d generates the classical operators grad, curl, and div on vector fields in three-dimensional Euclidean space. This works out as follows: 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 ${\displaystyle f=f(x,y,z)}$ , the first case written out in components gives:

${\displaystyle df={\frac {\partial f}{\partial x}}\,dx+{\frac {\partial f}{\partial y}}\,dy+{\frac {\partial f}{\partial z}}\,dz.}$

The inner product identifies 1-forms with vector fields as ${\displaystyle dx\mapsto (1,0,0)}$ , etc., so that ${\displaystyle df}$  becomes ${\displaystyle \mathrm {grad} \,f=({\tfrac {\partial f}{\partial x}},{\tfrac {\partial f}{\partial y}},{\tfrac {\partial f}{\partial z}})}$ .

In the second case, a vector field ${\displaystyle \mathbf {F} =(A,B,C)}$  corresponds to the 1-form ${\displaystyle \varphi =A\,dx+B\,dy+C\,dz}$ , which has exterior derivative:

${\displaystyle d\varphi =\left({\partial C \over \partial y}-{\partial B \over \partial z}\right)dy\wedge dz+\left({\partial C \over \partial x}-{\partial A \over \partial z}\right)dx\wedge dz+\left({\partial B \over \partial x}-{\partial A \over \partial y}\right)dx\wedge dy.}$

Applying the Hodge star gives the 1-form:

${\displaystyle \star d\varphi =\left({\partial C \over \partial y}-{\partial B \over \partial z}\right)\,dx-\left({\partial C \over \partial x}-{\partial A \over \partial z}\right)\,dy+\left({\partial B \over \partial x}-{\partial A \over \partial y}\right)\,dz,}$

which becomes the vector field ${\displaystyle \mathrm {curl} \,\mathbf {F} =({\tfrac {\partial C}{\partial y}}-{\tfrac {\partial B}{\partial z}},\,-{\tfrac {\partial C}{\partial x}}+{\tfrac {\partial A}{\partial z}},\,{\tfrac {\partial B}{\partial x}}-{\tfrac {\partial A}{\partial y}})}$ .

In the third case, ${\displaystyle \mathbf {F} =(A,B,C)}$  again corresponds to ${\displaystyle \varphi =A\,dx+B\,dy+C\,dz}$ . Applying Hodge star, exterior derivative, and Hodge star again:

{\displaystyle {\begin{aligned}\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}}=\mathrm {div} \,\mathbf {F} .\end{aligned}}}

One advantage of this expression is that the identity d2 = 0, which is true in all cases, sums up two others, namely that curl grad f = 0 and div curl F = 0. In particular, Maxwell's equations take on a particularly simple and elegant form, when expressed in terms of the exterior derivative and the Hodge star.

One can also obtain the Laplacian Δ f = div grad f in terms of the above operations:

${\displaystyle \Delta f=\star d{\star df}={\frac {\partial ^{2}f}{\partial x^{2}}}+{\frac {\partial ^{2}f}{\partial y^{2}}}+{\frac {\partial ^{2}f}{\partial z^{2}}}.}$

## Notes

1. ^ a b Harley Flanders (1963) Differential Forms with Applications to the Physical Sciences, Academic Press
2. ^ a b Pertti Lounesto (2001). "§3.6 The Hodge dual". Clifford Algebras and Spinors, Volume 286 of London Mathematical Society Lecture Note Series (2nd ed.). Cambridge University Press. p. 39. ISBN 0-521-00551-5.
3. ^ Frankel, T. (2012). The Geometry of Physics (3rd ed.). Cambridge University Press. ISBN 978-1-107-60260-1.