Editing Hodge star operator (section)

== Duality ==
Applying the Hodge star twice leaves a {{math|''k''}}-vector unchanged [[up to]] a sign: for <math>\eta\in {\textstyle\bigwedge}^k V</math> in an {{mvar|n}}-dimensional space {{math|''V''}}, one has
: <math>{\star} {\star} \eta = (-1)^{k(n-k)} s\, \eta ,</math>
where {{mvar|s}} is the parity of the [[metric signature|signature]] of the scalar product on {{math|''V''}}, that is, the sign of the [[determinant]] of the matrix of the scalar product with respect to any basis. For example, if {{math|''n'' {{=}} 4}} and the signature of the scalar product is either {{math|(+ − − −)}} or {{math|(− + + +)}} then {{math|''s'' {{=}} −1}}. For Riemannian manifolds (including Euclidean spaces), we always have {{math|''s'' {{=}} 1}}.

The above identity implies that the inverse of <math>{\star}</math> can be given as
: <math>\begin{align}
  {\star}^{-1}: ~ {\textstyle\bigwedge}^{\!k} V &\to {\textstyle\bigwedge}^{\!n-k} V \\
                                           \eta &\mapsto (-1)^{k(n-k)} \!s\, {\star} \eta
\end{align}</math>

If {{mvar|n}} is odd then {{math|''k''(''n'' − ''k'')}} is even for any {{mvar|k}}, whereas if {{mvar|n}} is even then {{math|''k''(''n'' − ''k'')}} has the parity of {{mvar|k}}. Therefore:
: <math>{\star}^{-1} = \begin{cases} s\, {\star} & n \text{ is odd} \\ (-1)^k s\, {\star} & n \text{ is even} \end{cases}</math>
where {{mvar|k}} is the degree of the element operated on.