# Hodge dual

(Redirected from Hodge duality)

In mathematics, the Hodge star operator or Hodge dual is an important linear map introduced in general by W. V. D. Hodge. It is defined on the exterior algebra of a finite-dimensional oriented inner product space.

## Dimensionalities and algebra

Suppose that n is the dimensionality of the oriented inner product space and k is an integer such that 0 ≤ kn, then the Hodge star operator establishes a one-to-one mapping from the space of k-vectors to the space of (nk)-vectors. The image of a k-vector under this mapping is called the Hodge dual of the k-vector. The former space, of k-vectors, has dimensionality

${n \choose k}$

while the latter has dimensionality

${n \choose n - k},$

and by the symmetry of the binomial coefficients, these two dimensionalities are equal. Two vector spaces over the same field with the same dimensionality are always isomorphic; but not necessarily in a natural or canonical way. The Hodge duality, however, in this case exploits the inner product and orientation of the vector space. It singles out a unique isomorphism, that reflects therefore the pattern of the binomial coefficients in algebra. This in turn induces an inner product on the space of k-vectors. The 'natural' definition means that this duality relationship can play a geometrical role in theories.

The first interesting case is on three-dimensional Euclidean space V. In this context the relevant row of Pascal's triangle reads

1, 3, 3, 1

and the Hodge dual sets up an isomorphism between the two three-dimensional spaces, which are V itself and the space of wedge products of two vectors from V. See the Examples section for details. In this case the content is just that of the cross product of traditional vector calculus. While the properties of the cross product are special to three dimensions, the Hodge dual applies to all dimensionalities.

## Extensions

Since the space of alternating linear forms in k arguments on a vector space is naturally isomorphic to the dual of the space of k-vectors over that vector space, the Hodge dual can be defined for these spaces as well. As with most constructions from linear algebra, the Hodge dual can then be extended to a vector bundle. Thus a context in which the Hodge dual is very often seen is the exterior algebra of the cotangent bundle (i.e. the space of differential forms on a manifold) where it can be used to construct the codifferential from the exterior derivative, and thus the Laplace-de Rham operator, which leads to the Hodge decomposition of differential forms in the case of compact Riemannian manifolds.

## Formal definition of the Hodge star of k-vectors

The Hodge star operator on a vector space V with a nondegenerate symmetric bilinear form (herein referred to as the inner product) is a linear operator on the exterior algebra of V, mapping k-vectors to (nk)-vectors where n = dim V, for 0 ≤ kn. It has the following property, which defines it completely: given two k-vectors α, β

$\alpha \wedge (\star \beta) = \langle \alpha,\beta \rangle \omega$

where $\langle \cdot,\cdot \rangle$ denotes the inner product on k-vectors and ω is the preferred unit n-vector.

The inner product $\langle \cdot,\cdot \rangle$ on k-vectors is extended from that on V by requiring that

$\langle \alpha,\beta \rangle = \det \left (\left \langle \alpha_i,\beta_j \right \rangle \right )$

for any decomposable k-vectors $\alpha = \alpha_1 \wedge \dots \wedge \alpha_k$ and $\beta = \beta_1 \wedge \dots \wedge \beta_k$.

The unit n-vector ω is unique up to a sign. The preferred choice of ω defines an orientation on V.

## Explanation

Let W be a vector space, with an inner product $\langle\cdot, \cdot\rangle_W$. The Riesz representation theorem states that for every continuous (every in the finite-dimensional case) linear functional $f \in W^*$ there exists a unique vector v in W such that $f(w) = \langle w, v \rangle_W$ for all w in W. The map $W^* \to W$ given by $f \mapsto v$ is an isomorphism. This holds for all vector spaces with an inner product, and can be used to explain the Hodge dual.

Let V be an n-dimensional vector space with basis $\{e_1,\ldots,e_n\}$. For 0 ≤ kn, consider the exterior power spaces $\bigwedge^k V$ and $\bigwedge^{n-k} V$. For

$\lambda \in \bigwedge^k V, \quad \theta \in \bigwedge^{n-k} V,$

we have

$\lambda \wedge \theta \in \bigwedge^n V.$

There is, up to a scalar, only one n-vector, namely $e_1\wedge\ldots\wedge e_n$. In other words, $\lambda \wedge \theta$ must be a scalar multiple of $e_1\wedge\ldots\wedge e_n$ for all $\lambda \in \bigwedge^k V$ and $\theta \in \bigwedge^{n-k} V$.

Consider a fixed $\lambda \in \bigwedge^k V$. There exists a unique linear function

$f_{\lambda} \in \left(\bigwedge^{n-k} V\right)^{\! *}$

such that

$\forall \theta \in \bigwedge^{n-k} V: \qquad \lambda \wedge \theta = f_{\lambda}(\theta) \, (e_1\wedge\ldots\wedge e_n).$

This $f_{\lambda}(\theta)$ is the scalar multiple mentioned in the previous paragraph. If $\langle\cdot, \cdot\rangle$ denotes the inner product on (nk)-vectors, then there exists a unique (nk)-vector, say

$\star \lambda \in \bigwedge^{n-k} V,$

such that

$\forall \theta \in \bigwedge^{n-k} V: \qquad f_{\lambda}(\theta) = \langle \theta, \star \lambda\rangle.$

This (nk)-vector λ is the Hodge dual of λ, and is the image of the $f_{\lambda}$ under the isomorphism induced by the inner product,

$\left(\bigwedge^{n-k} V\right)^{\! *} \cong \bigwedge^{n-k} V.$

Thus,

$\star : \bigwedge^{k} V \to \bigwedge^{n-k} V$.

## Computation of the Hodge star

Given an orthonormal basis $(e_1,\cdots,e_n)$ ordered such that $\omega = e_1\wedge \cdots \wedge e_n$, we see that

$\star (e_{i_1} \wedge e_{i_2}\wedge \cdots \wedge e_{i_k})= e_{i_{k+1}} \wedge e_{i_{k+2}} \wedge \cdots \wedge e_{i_n},$

where $(i_1, i_2, \cdots, i_n)$ is an even permutation of {1, 2, ..., n}.

Of these $n! \over 2$ relations, only $n \choose k$ are independent. The first one in the usual lexicographical order reads

$\star (e_1\wedge e_2\wedge \cdots \wedge e_k)= e_{k+1}\wedge e_{k+2}\wedge \cdots \wedge e_n.$

## Index notation for the star operator

Using index notation, the Hodge dual is obtained by contracting the indices of a k-form with the n-dimensional completely antisymmetric Levi-Civita tensor. This differs from the Levi-Civita symbol by a factor of |det g|1/2, where g is an inner product (the metric tensor). The absolute value of the determinant is necessary if g is not positive-definite, e.g. for tangent spaces to Lorentzian manifolds.

Thus one writes[1]

$(\star \eta)_{i_1,i_2,\ldots,i_{n-k}} = \frac{1}{(k)!} \eta^{j_1,\ldots,j_k}\,\sqrt {|\det g|} \,\epsilon_{j_1,\ldots,j_k,i_1,\ldots,i_{n-k}}$

where η is an arbitrary antisymmetric tensor in k indices. It is understood that indices are raised and lowered using the same inner product g as in the definition of the Levi-Civita tensor. Although one can take the star of any tensor, the result is antisymmetric, since the symmetric components of the tensor completely cancel out when contracted with the completely anti-symmetric Levi-Civita symbol.

## Examples

A common example of the star operator is the case n = 3, when it can be taken as the correspondence between the vectors and the skew-symmetric matrices of that size. This is used implicitly in vector calculus, for example to create the cross product vector from the wedge product of two vectors. Specifically, for Euclidean R3, one easily finds that

$\star \mathrm{d}x=\mathrm{d}y\wedge \mathrm{d}z$
$\star \mathrm{d}y=\mathrm{d}z\wedge \mathrm{d}x$
$\star \mathrm{d}z=\mathrm{d}x\wedge \mathrm{d}y$

where dx, dy and dz are the standard orthonormal differential one-forms on R3. The Hodge dual in this case clearly relates the cross-product to the wedge product in three dimensions. A detailed presentation not restricted to differential geometry is provided next.

### Three-dimensional example

Applied to three dimensions, the Hodge dual 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]

$\mathbf{A} = \star \mathbf{a}\qquad\mathbf{a} = \star \mathbf{A}$

where indicates the dual operation. These dual relations can be implemented using multiplication by the unit pseudoscalar in C3(R),[3] i = e1e2e3 (the vectors {e} are an orthonormal basis in three dimensional Euclidean space) according to the relations:[4]

$\mathbf{A} = \mathbf{a}i\,,\quad\mathbf{a} = - \mathbf{A} i.$

The dual of a vector is obtained by multiplication by i, as established using the properties of the geometric product of the algebra as follows:

\begin{align} \mathbf{a}i &= \left(a_1 \mathbf{e_1} + a_2 \mathbf{e_2} +a_3 \mathbf {e_3}\right) \mathbf {e_1 e_2 e_3} \\ &= a_1 \mathbf{e_2 e_3} (\mathbf{e_1})^2 + a_2 \mathbf{e_3 e_1}(\mathbf{e_2})^2 +a_3 \mathbf{e_1 e_2}(\mathbf{e_3})^2 \\ &= a_1 \mathbf{e_2 e_3} +a_2 \mathbf{e_3 e_1} +a_3 \mathbf{e_1 e_2} \\ &= (\star \mathbf a ) \end{align}

and also, in the dual space spanned by {eem}:

\begin{align} \mathbf{A} i &= \left(A_1 \mathbf{e_2e_3} + A_2 \mathbf{e_3e_1} +A_3 \mathbf {e_1e_2}\right) \mathbf {e_1 e_2 e_3} \\ &= A_1 \mathbf{e_1} (\mathbf{e_2 e_3})^2 +A_2 \mathbf{e_2} (\mathbf{e_3 e_1})^2 +A_3 \mathbf{e_3}(\mathbf{e_1 e_2})^2 \\ &=-\left( A_1 \mathbf{e_1} + A_2 \mathbf{e_2} + A_3 \mathbf{e_3} \right) \\ &= - (\star \mathbf A ) \end{align}

In establishing these results, the identities are used:

$(\mathbf{e_1e_2})^2 =\mathbf{e_1e_2e_1e_2}= -\mathbf{e_1e_2e_2e_1} = -1$

and:

$\mathit{i}^2 =(\mathbf{e_1e_2e_3})^2 =\mathbf{e_1e_2e_3e_1e_2e_3}= \mathbf{e_1e_2e_3e_3e_1e_2} = \mathbf{e_1e_2e_1e_2} = -1.$

These relations between the dual and i apply to any vectors. Here they are applied to relate the axial vector created as the cross product a = u × v to the bivector-valued exterior product A = uv of two polar (that is, not axial) vectors u and v; the two products can be written as determinants expressed in the same way:

$\mathbf a = \mathbf{u} \times \mathbf{v} = \begin{vmatrix} \mathbf{e}_1 & \mathbf{e}_2 & \mathbf{e}_3\\u_1 & u_2 & u_3\\v_1 & v_2 & v_3 \end{vmatrix}\,,\quad\mathbf A = \mathbf{u} \wedge \mathbf{v} = \begin{vmatrix} \mathbf{e}_{23} & \mathbf{e}_{31} & \mathbf{e}_{12}\\u_1 & u_2 & u_3\\v_1 & v_2 & v_3 \end{vmatrix},$

using the notation em = eem. These expressions show these two types of vector are Hodge duals:[2]

$\star (\mathbf u \wedge \mathbf v )=\mathbf {u \times v}\,,\quad\star (\mathbf u \times \mathbf v ) = \mathbf u \wedge \mathbf v,$

as a result of the relations:

$\star \mathbf e_{\ell} = \mathbf e_{\ell} \mathit i =\mathbf e_{\ell} \mathbf{e_1e_2e_3} = \mathbf e_m \mathbf e_n \,,$

with ℓ, m, n cyclic,

and:

$\star ( \mathbf e_{\ell} \mathbf e_m ) =-( \mathbf e_{\ell} \mathbf e_m )\mathit{i} =-\left( \mathbf e_{\ell} \mathbf e_m \right)\mathbf{e_1e_2e_3} =\mathbf e_{n}$

also with ℓ, m, n cyclic.

Using the implementation of based upon i, the commonly used relations are:[5]

$\mathbf {u \times v} = -(\mathbf u \wedge \mathbf v ) i \,,\quad \mathbf u \wedge \mathbf v = (\mathbf {u \times v} ) i \ .$

### Four dimensions

In case n = 4, the Hodge dual acts as an endomorphism of the second exterior power (i.e. it maps two-forms to two-forms, since 4 − 2 = 2). It is an involution, so it splits it into self-dual and anti-self-dual subspaces, on which it acts respectively as +1 and −1.

Another useful example is n = 4 Minkowski spacetime with metric signature (+ − − −) and coordinates (t, x, y, z) where (using $\varepsilon_{0123} = 1$)

$\star \mathrm{d}t=\mathrm{d}x\wedge \mathrm{d}y \wedge\mathrm{d}z$
$\star \mathrm{d}x=\mathrm{d}t\wedge \mathrm{d}y \wedge\mathrm{d}z$
$\star \mathrm{d}y=\mathrm{d}t\wedge \mathrm{d}z \wedge\mathrm{d}x$
$\star \mathrm{d}z=\mathrm{d}t\wedge \mathrm{d}x \wedge\mathrm{d}y$

for one-forms while

$\star (\mathrm{d}t \wedge\mathrm{d}x) = - \mathrm{d}y\wedge \mathrm{d}z$
$\star (\mathrm{d}t \wedge\mathrm{d}y) = \mathrm{d}x\wedge \mathrm{d}z$
$\star (\mathrm{d}t \wedge\mathrm{d}z) = - \mathrm{d}x\wedge \mathrm{d}y$
$\star (\mathrm{d}x \wedge\mathrm{d}y) = \mathrm{d}t\wedge \mathrm{d}z$
$\star (\mathrm{d}x \wedge\mathrm{d}z) = - \mathrm{d}t\wedge \mathrm{d}y$
$\star (\mathrm{d}y \wedge\mathrm{d}z) = \mathrm{d}t\wedge \mathrm{d}x$

for two-forms.

## Inner product of k-vectors

The Hodge dual induces an inner product on the space of k-vectors, that is, on the exterior algebra of V. Given two k-vectors η and ζ, one has

$\zeta\wedge \star \eta = \langle\zeta, \eta \rangle\;\omega$

where ω is the normalised n-form (i.e. ω ∧ ★ω = ω). In the calculus of exterior differential forms on a pseudo-Riemannian manifold of dimension n, the normalised n-form is called the volume form and can be written as

$\omega=\sqrt{|\det [g_{ij}]|}\;\mathrm{d}x^1\wedge\cdots\wedge \mathrm{d}x^n$

where $[g_{ij}]$ is the matrix of components of the metric tensor on the manifold in the coordinate basis.

If an inner product is given on $\Lambda^k(V)$, then this equation can be regarded as an alternative definition of the Hodge dual.[6] The wedge products of elements of an orthonormal basis in V form an orthonormal basis of the exterior algebra of V.

## Duality

The Hodge star defines a dual in that when it is applied twice, the result is an identity on the exterior algebra, up to sign. Given a k-vector η in Λk(V) in an n-dimensional space V, one has

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

where s is related to the signature of the inner product on V. Specifically, s is the sign of the determinant of the inner product tensor. Thus, for example, if n = 4 and the signature of the inner product is either (+ − − −) or (− + + +) then s = −1. For ordinary Euclidean spaces, the signature is always positive, and so s = 1. When the Hodge star is extended to pseudo-Riemannian manifolds, then the above inner product is understood to be the metric in diagonal form.

Note that the above identity implies that the inverse of can be given as

$\begin{cases}\star^{-1}:\Lambda^k \to \Lambda^{n-k} \\ \eta \mapsto (-1)^{k(n-k)}s{\star \eta} \end{cases}$

Note that if n is odd k(nk) is even for any k whereas if n is even k(nk) has the parity of k. Therefore:

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

where k is the degree of the forms operated on.

## Hodge star on manifolds

One can repeat the construction above for each cotangent space of an n-dimensional oriented Riemannian or pseudo-Riemannian manifold, and get the Hodge dual (nk)-form, of a k-form. The Hodge star then induces an L2-norm inner product on the differential forms on the manifold. One writes

$(\eta,\zeta)=\int_M \eta\wedge \star \zeta = \int_M \langle \eta, \zeta \rangle \; \mathrm{d} \text{Vol}$

for the inner product of sections η and ζ of $\Lambda^k(T^*M)$. (The set of sections is frequently denoted as $\Omega^k(M)=\Gamma(\Lambda^k(T^*M))$. Elements of $\Omega^k(M)$ are called exterior k-forms).

More generally, in the non-oriented case, one can define the hodge star of a k-form as a (nk)-pseudo differential form; that is, a differential forms with values in the canonical line bundle.

### The codifferential

The most important application of the Hodge dual on manifolds is to define the codifferential δ on k-forms. Let

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

where d is the exterior derivative or differential, and s = 1 for Riemannian manifolds.

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

while

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

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, in that

$(\eta,\delta \zeta) = (\mathrm{d}\eta,\zeta).$

where ζ is a (k+1)-form and η a k-form. This identity follows from Stokes' theorem for smooth forms, when

$\int_M \mathrm{d}(\eta \wedge \star \zeta)=0 =\int_M (\mathrm{d}\eta \wedge \star \zeta - \eta\wedge \star (-1)^{k+1}\,{\star^{-1}\mathrm{d}{\star \zeta}}) =(\mathrm{d}\eta,\zeta) -(\eta,\delta\zeta)$

i.e. when M has empty boundary or when η or ζ has zero boundary values (of course, true adjointness follows after continuous continuation to the appropriate topological vector spaces as closures of the spaces of smooth forms).

Notice that since the differential satisfies d2 = 0, the codifferential has the corresponding property

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

The Laplace–deRham operator is given by

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

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

$(\Delta \zeta,\eta) = (\zeta,\Delta \eta)$

and non-negative:

$(\Delta\eta,\eta) \ge 0.$

The Hodge dual sends harmonic forms to harmonic forms. As a consequence of the 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

$\star : H^k_\Delta(M)\to H^{n-k}_\Delta(M),$

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 operator and the exterior derivative d generates the classical operators grad, curl, and div, in three dimensions. This works out as follows: d can take a 0-form (function) to a 1-form, a 1-form to a 2-form, and a 2-form to a 3-form (applied to a 3-form it just gives zero). For a 0-form, $\omega=f(x,y,z)$, the first case written out in components is identifiable as the grad operator:

$\mathrm{d}\omega=\frac{\partial f}{\partial x}\mathrm{d}x+\frac{\partial f}{\partial y}\mathrm{d}y+\frac{\partial f}{\partial z}\mathrm{d}z.$

The second case followed by is an operator on 1-forms ($\eta=A\,\mathrm{d}x+B\,\mathrm{d}y+C\,\mathrm{d}z$) that in components is the curl operator:

$\mathrm{d}\eta=\left({\partial C \over \partial y} - {\partial B \over \partial z}\right)\mathrm{d}y\wedge \mathrm{d}z + \left({\partial C \over \partial x} - {\partial A \over \partial z}\right)\mathrm{d}x\wedge \mathrm{d}z+\left({\partial B \over \partial x} - {\partial A \over \partial y}\right)\mathrm{d}x\wedge \mathrm{d}y.$

Applying the Hodge star gives:

$\star \mathrm{d}\eta=\left({\partial C \over \partial y} - {\partial B \over \partial z}\right)\mathrm{d}x - \left({\partial C \over \partial x} - {\partial A \over \partial z}\right)\mathrm{d}y+\left({\partial B \over \partial x} - {\partial A \over \partial y}\right)\mathrm{d}z.$

The final case prefaced and followed by , takes a 1-form ($\eta=A\,\mathrm{d}x+B\,\mathrm{d}y+C\,\mathrm{d}z$) to a 0-form (function); written out in components it is the divergence operator:

\begin{align} \star\eta &= A\,\mathrm{d}y\wedge \mathrm{d}z-B\,\mathrm{d}x\wedge \mathrm{d}z+C\,\mathrm{d}x\wedge \mathrm{d}y \\ \mathrm{d}{\star\eta} &= \left(\frac{\partial A}{\partial x}+\frac{\partial B}{\partial y}+\frac{\partial C}{\partial z}\right)\mathrm{d}x\wedge \mathrm{d}y\wedge \mathrm{d}z \\ \star \mathrm{d}{\star\eta} &= \frac{\partial A}{\partial x}+\frac{\partial B}{\partial y}+\frac{\partial C}{\partial z}. \end{align}

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. Using the information above and the fact that Δ f  = div grad f then for a 0-form, $\omega=f(x,y,z)$:

$\Delta \omega =\star \mathrm{d}{\star \mathrm{d}\omega}= \frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} + \frac{\partial^2 f}{\partial z^2}$

## Notes

1. ^ The Geometry of Physics (3rd edition), T. Frankel, Cambridge University Press, 2012, ISBN 978-1107-602601
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. ^ Venzo De Sabbata, Bidyut Kumar Datta (2007). "The pseudoscalar and imaginary unit". Geometric algebra and applications to physics. CRC Press. p. 53 ff. ISBN 1-58488-772-9.
4. ^ William E Baylis (2004). "Chapter 4: Applications of Clifford algebras in physics". In Rafal Ablamowicz, Garret Sobczyk. Lectures on Clifford (geometric) algebras and applications. Birkhäuser. p. 100 ff. ISBN 0-8176-3257-3.
5. ^ David Hestenes (1999). "The vector cross product". New foundations for classical mechanics: Fundamental Theories of Physics (2nd ed.). Springer. p. 60. ISBN 0-7923-5302-1.
6. ^ Darling, R. W. R. (1994). Differential forms and connections. Cambridge University Press.