Editing Hodge star operator (section)

== Examples ==

=== Two dimensions ===
In two dimensions with the normalized Euclidean metric and orientation given by the ordering {{math|(''x'', ''y'')}}, the Hodge star on {{math|''k''}}-forms is given by
<math display="block">\begin{align}
              {\star} \, 1 &= dx \wedge dy \\
             {\star} \, dx &= dy \\
             {\star} \, dy &= -dx \\
  {\star} ( dx \wedge dy ) &= 1 .
\end{align}</math>

=== Three dimensions ===
A common example of the Hodge star operator is the case {{math|1=''n'' = 3}}, when it can be taken as the correspondence between vectors and  bivectors. Specifically, for [[Euclidean space|Euclidean]] '''R'''<sup>3</sup> with the basis <math>dx, dy, dz</math> of [[one-form]]s often used in [[vector calculus]], one finds that
<math display="block">\begin{align}
  {\star} \,dx &= dy \wedge dz \\
  {\star} \,dy &= dz \wedge dx \\
  {\star} \,dz &= dx \wedge dy.
\end{align}</math>

The Hodge star relates the exterior and cross product in three dimensions:<ref name="Lounesto" /> <math display="block">{\star} (\mathbf{u} \wedge \mathbf{v}) = \mathbf{u} \times \mathbf{v} \qquad {\star} (\mathbf{u} \times \mathbf {v}) = \mathbf{u} \wedge \mathbf{v} .</math> Applied to three dimensions, the Hodge star provides an [[isomorphism]] between [[axial vector]]s and [[bivector]]s, so each axial vector {{math|'''a'''}} is associated with a bivector {{math|'''A'''}} and vice versa, that is:<ref name="Lounesto">{{cite book |title=Clifford Algebras and Spinors, ''Volume 286 of London Mathematical Society Lecture Note Series'' |author=Pertti Lounesto |chapter-url = https://books.google.com/books?id=E_xvJuA4M7QC&pg=PA39 |page=39 |chapter=§3.6 The Hodge dual |isbn=0-521-00551-5 |year=2001 |edition=2nd |publisher=Cambridge University Press}}</ref> <math>\mathbf{A} = {\star} \mathbf{a}, \ \ \mathbf{a} = {\star} \mathbf{A}</math>.

The Hodge star can also be interpreted as a form of the geometric correspondence between an [[axis of rotation]] and an [[infinitesimal rotation]] (see also: [[3D rotation group#Lie algebra]]) around the axis, with speed equal to the length of the axis of rotation. A scalar product on a vector space <math>V</math> gives an [[Dual space#Bilinear products and dual spaces|isomorphism]] <math>V\cong V^*\!</math>  identifying <math>V</math> with its [[dual space]], and the vector space <math>L(V,V)</math> is naturally isomorphic to the [[tensor product]] <math>V^*\!\!\otimes V\cong V\otimes V</math>. Thus for <math>V = \mathbb{R}^3</math>, the star mapping <math display="inline">\textstyle {\star} : V\to\bigwedge^{\!2}\! V \subset V\otimes V</math> takes each vector <math>\mathbf{v}</math> to a bivector <math>{\star} \mathbf{v} \in V\otimes V</math>, which corresponds to a linear operator <math>L_{\mathbf{v}} : V\to V</math>. Specifically, <math>L_{\mathbf{v}}</math> is a [[Skew-symmetric matrix|skew-symmetric]] operator, which corresponds to an infinitesimal rotation: that is, the macroscopic rotations around the axis <math>\mathbb{v}</math> are given by the [[matrix exponential]] <math>\exp(t L_{\mathbf{v}})</math>. With respect to the basis <math>dx, dy, dz</math> of <math>\R^3</math>, the tensor <math>dx\otimes dy</math> corresponds to a coordinate matrix with 1 in the <math>dx</math> row and <math>dy</math> column, etc., and the wedge <math>dx\wedge dy \,=\, dx\otimes dy - dy\otimes dx</math> is the skew-symmetric matrix <math>\scriptscriptstyle\left[\begin{array}{rrr}
\,0\!\! & \!\!1 & \!\!\!\!0\!\!\!\!\!\! 
\\[-.5em]
\,\!-1\!\!&\!\!0\!\!&\!\!\!\!0\!\!\!\!\!\! 
\\[-.5em]
\,0\!\! & \!\!0\!\! & \!\!\!\!0\!\!\!\!\!\!
\end{array}\!\!\!\right]</math>, etc. That is, we may interpret the star operator as: <math display="block"> \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].</math> 
Under this correspondence, cross product of vectors corresponds to the commutator [[Lie algebra|Lie bracket]] of linear operators: <math>L_{\mathbf{u}\times\mathbf{v}} = L_{\mathbf{v}} L_{\mathbf{u}} - L_{\mathbf{u}} L_{\mathbf{v}}=-\left[L_{\mathbf{u}}, L_{\mathbf{v}}\right]</math>.
<!-- These dual relations can be implemented using multiplication by the [[Pseudoscalar (Clifford algebra)#Unit pseudoscalar|unit pseudoscalar]] in [[Clifford algebra#Examples: real and complex Clifford algebras|Cl<sub>3</sub>('''R''')]],<ref name=Datta>{{cite book |title=Geometric algebra and applications to physics |chapter=The pseudoscalar and imaginary unit | url=https://books.google.com/books?id=AXTQXnws8E8C&pg=PA53 |page=53 ''ff'' |author=Venzo De Sabbata, Bidyut Kumar Datta |isbn=1-58488-772-9 | publisher=CRC Press |year=2007}}</ref> {{math|1=''i'' = '''e'''<sub>1</sub>'''e'''<sub>2</sub>'''e'''<sub>3</sub>}} (the vectors {{math|{'''e'''<sub>ℓ</sub>} }} are an orthonormal basis in three-dimensional Euclidean space) according to the relations<ref name=Baylis>{{cite book |title=Lectures on Clifford (geometric) algebras and applications |editor=Rafal Ablamowicz, Garret Sobczyk |page=100 ''ff'' |chapter=Chapter 4: Applications of Clifford algebras in physics |author=William E Baylis  |isbn=0-8176-3257-3 |year=2004 | publisher=Birkhäuser | url=https://books.google.com/books?id=oaoLbMS3ErwC&pg=PA100}}</ref>
<math display="block">\mathbf{A} = \mathbf{a}i\,\quad\mathbf{a} = - \mathbf{A} i. </math>

The dual of a vector is obtained by multiplication by {{mvar|i}}, as established using the properties of the [[Geometric product#The geometric product|geometric product]] of the algebra as follows:
<math display="block">\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 \mathbf{e}_2 \mathbf{e}_3 \\
              &= a_1 \mathbf{e}_2 \mathbf{e}_3 (\mathbf{e}_1)^2 + a_2 \mathbf{e}_3 \mathbf{e}_1(\mathbf{e}_2)^2 + a_3  \mathbf{e}_1 \mathbf{e}_2(\mathbf{e}_3)^2 \\
              &= a_1 \mathbf{e}_2 \mathbf{e}_3 + a_2 \mathbf{e}_3 \mathbf{e}_1 + a_3 \mathbf{e}_1 \mathbf{e}_2 \\
              &= ({\star} \mathbf{a})
\end{align}</math>
and also, in the dual space spanned by {{math|{'''e'''<sub>ℓ</sub>'''e'''<sub>''m''</sub>}<nowiki/>}}:
<math display="block">\begin{align}
  \mathbf{A} i &= \left(A_1 \mathbf{e}_2\mathbf{e}_3 + A_2 \mathbf{e}_3\mathbf{e}_1 + A_3 \mathbf{e}_1\mathbf{e}_2\right) \mathbf{e}_1 \mathbf{e}_2 \mathbf{e}_3 \\
               &= A_1 \mathbf{e}_1 (\mathbf{e}_2 \mathbf{e}_3)^2 + A_2 \mathbf{e}_2 (\mathbf{e}_3 \mathbf{e}_1)^2 + A_3 \mathbf{e}_3(\mathbf{e}_1 \mathbf{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}</math>

In establishing these results, the identities are used:
<math display="block">(\mathbf{e}_1\mathbf{e}_2)^2 = \mathbf{e}_1\mathbf{e}_2\mathbf{e}_1\mathbf{e}_2= -\mathbf{e}_1\mathbf{e}_2\mathbf{e}_2\mathbf{e}_1 = -1</math>
and:
<math display="block">\mathit{i}^2 = (\mathbf{e}_1\mathbf{e}_2\mathbf{e}_3)^2 = \mathbf{e}_1\mathbf{e}_2\mathbf{e}_3\mathbf{e}_1\mathbf{e}_2\mathbf{e}_3 = \mathbf{e}_1\mathbf{e}_2\mathbf{e}_3\mathbf{e}_3\mathbf{e}_1\mathbf{e}_2 = \mathbf{e}_1\mathbf{e}_2\mathbf{e}_1\mathbf{e}_2 = -1.</math>

These relations between the dual <math>{\star}</math> and {{mvar|i}} apply to any vectors. Here they are applied to relate the axial vector created as the [[cross product]] {{math|1='''a''' = '''u''' × '''v'''}} to the bivector-valued [[exterior product]] {{math|1='''A''' = '''u''' ∧ '''v'''}} of two [[polar vectors|polar]] (that is, not axial) vectors {{math|'''u'''}} and {{math|'''v'''}}; the two products can be written as [[determinant]]s expressed in the same way:
<math display="block">\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}_{2}\mathbf{e}_{3} & \mathbf{e}_{3}\mathbf{e}_{1} & \mathbf{e}_{1}\mathbf{e}_{2} \\ u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \end{vmatrix}.</math>

These expressions show these two types of vector are Hodge duals:<ref name=Lounesto/>
<math display="block">{\star} (\mathbf{u} \wedge \mathbf{v}) = \mathbf{u \times v}\,,\quad{\star} (\mathbf{u} \times \mathbf {v}) = \mathbf{u} \wedge \mathbf{v},</math>
as a result of the relations:
<math display="block">{\star} \mathbf{e}_\ell = \mathbf{e}_\ell \mathit{i} = \mathbf{e}_\ell \mathbf{e}_1\mathbf{e}_2\mathbf{e}_3 = \mathbf{e}_m \mathbf{e}_n \,, </math>
with {{math|''ℓ'', ''m'', ''n''}} cyclic,
and:
<math display="block">{\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}_1\mathbf{e}_2\mathbf{e}_3 = \mathbf{e}_n </math>
also with {{math|''ℓ'', ''m'', ''n''}} cyclic.

Using the implementation of <math>{\star}</math> based upon {{mvar|i}}, the commonly used relations are:<ref name=Hestenes>{{cite book |title=New foundations for classical mechanics: Fundamental Theories of Physics  |isbn=0-7923-5302-1 |edition=2nd |year=1999 |publisher=Springer |chapter=The vector cross product |author-link = David Hestenes|author=David Hestenes |url=https://books.google.com/books?id=AlvTCEzSI5wC&pg=PA60 |page=60 }}</ref>
<math display="block"> \mathbf{u \times v} = -(\mathbf{u} \wedge \mathbf{v}) i  \,,\quad  \mathbf{u} \wedge \mathbf{v} = (\mathbf{u \times v} )  i \ . </math>
-->

=== Four dimensions ===
In case <math>n=4</math>, the Hodge star acts as an [[endomorphism]] of the second exterior power (i.e. it maps 2-forms to 2-forms, since {{math|1=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 (mathematics)|involution]]. If the signature is mixed, i.e., [[Pseudo-Riemannian_manifold|pseudo-Riemannian]], then applying the operator twice will return the argument up to a sign – see ''{{section link|#Duality}}'' below. This particular endomorphism property of 2-forms in four dimensions makes [[Yang–Mills equations#Anti-self-duality equations|self-dual and anti-self-dual two-forms]] natural geometric objects to study. That is, one can describe the space of 2-forms in four dimensions with a basis that "diagonalizes" the Hodge star operator with eigenvalues <math>\pm 1</math> (or <math>\pm i</math>, depending on the signature).

For concreteness, we discuss the Hodge star operator in Minkowski spacetime where <math>n=4</math> with metric signature {{math|(− + + +)}} and coordinates <math>(t,x,y,z)</math>. The [[volume form]] is oriented as <math>\varepsilon_{0123} = 1</math>. For [[one-form]]s,
<math display="block">\begin{align}
  {\star} dt &= -dx \wedge dy \wedge dz \,, \\
  {\star} dx &= -dt \wedge dy \wedge dz \,, \\
  {\star} dy &= -dt \wedge dz \wedge dx \,, \\
  {\star} dz &= -dt \wedge dx \wedge dy \,,
\end{align}</math>
while for [[2-form]]s,
<math display="block">\begin{align}
  {\star} (dt \wedge dx) &= - dy \wedge dz \,, \\
  {\star} (dt \wedge dy) &= - dz \wedge dx \,, \\
  {\star} (dt \wedge dz) &= - dx \wedge dy \,, \\
  {\star} (dx \wedge dy) &=   dt \wedge dz \,, \\
  {\star} (dz \wedge dx) &=   dt \wedge dy \,, \\
  {\star} (dy \wedge dz) &=   dt \wedge dx \,.
\end{align}</math>

These are summarized in the index notation as
<math display="block">\begin{align}
  {\star} (dx^\mu) &= \eta^{\mu\lambda} \varepsilon_{\lambda\nu\rho\sigma} \frac{1}{3!} dx^\nu \wedge dx^\rho 
\wedge dx^\sigma \,,\\
  {\star} (dx^\mu \wedge dx^\nu) &= \eta^{\mu\kappa} \eta^{\nu\lambda} \varepsilon_{\kappa\lambda\rho\sigma} \frac{1}{2!} dx^\rho \wedge dx^\sigma \,.
\end{align}</math>

Hodge dual of three- and four-forms can be easily deduced from the fact that, in the Lorentzian signature, <math>{\star}^2=1</math> for odd-rank forms and <math>{\star}^2=-1</math> for even-rank forms. An easy rule to remember for these Hodge operations is that given a form <math>\alpha</math>, its Hodge dual <math>{\star}\alpha</math> may be obtained by writing the components not involved in <math>\alpha</math> in an order such that <math>\alpha \wedge ({\star} \alpha) = dt \wedge dx \wedge dy \wedge dz </math>.{{verify source|date=September 2019}} An extra minus sign will enter only if <math>\alpha</math> contains <math>dt</math>. (For {{math|(+ − − −)}}, one puts in a minus sign only if <math>\alpha</math> involves an odd number of the space-associated forms <math>dx</math>, <math>dy</math> and <math>dz</math>.)

Note that the combinations
<math display="block"> (dx^\mu \wedge dx^\nu)^{\pm} := \frac{1}{2} \big( dx^\mu \wedge dx^\nu \mp i {\star} (dx^\mu \wedge dx^\nu) \big)</math>
take <math>\pm i</math> as the eigenvalue for Hodge star operator, i.e.,
<math display="block"> {\star} (dx^\mu \wedge dx^\nu)^{\pm} = \pm i (dx^\mu \wedge dx^\nu)^{\pm} , </math>
and hence deserve the name self-dual and anti-self-dual two-forms. Understanding the geometry, or kinematics, of Minkowski spacetime in self-dual and anti-self-dual sectors turns out to be insightful in both [[Low-dimensional topology#Four dimensions|mathematical]] and [[Chirality (physics)#Chirality and helicity|physical]] perspectives, making contacts to the use of the [[Weyl equation#Weyl spinors|two-spinor]] language in modern physics such as [[spinor-helicity formalism]] or [[twistor theory]].

=== Conformal invariance ===
The Hodge star is conformally invariant on {{math|''n''}}-forms on a {{math|2''n''}}-dimensional vector space <math> V </math>, i.e. if <math> g </math> is a metric on <math> V </math> and <math> \lambda > 0 </math>, then the induced Hodge stars 
<math display="block"> {\star}_g, {\star}_{\lambda g} : \Lambda^n V \to \Lambda^n V</math>
are the same. 

=== 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>.