# How to naturally construct and compute explicitly the Hodge dual of a differential form

Originally published at 狗和留美者不得入内. You can comment here or there.

There is antisymmetry in the wedge product of differential forms, i.e. . Take the decomposable -form in -dimensional space denoted by We can take and the corresponding -form as the Hodge dual with signs yet to be determined. As for how to reasonably prescribe the sign, we first note that a -form wedged producted with its Hodge dual gives The right hand side of this is equal to We let be the standard ordering of the basis of the underlying vector space, which we regard as positively oriented. Moreover, we let be the permutation such that It is reasonable to impose for symmetry that To satisfy this, we choose an ordering of remaining indices such that the sign of the corresponding permutation is positive. , the space of -forms over finite vector space , has dimension . The standard basis for it we represent as  has the same dimension . It is easy to see that the Hodge star operator maps the standard basis of to that of , and moreover that Imposing linearity on the Hodge star operator then gives us Now we wish to in Einstein notation in terms of indices represent the Hodge star of an arbitrary -form. In doing so, we will use the complete antisymmetric tensor of rank , which in addition to being zero when not all indices are different, satisfies In the basis above, we assumed , and the coefficients . If we impose no restriction on the ordering of the indices and for each combination pick its associated basis element arbitrarily, then for the sum of our linear combination of the basis elements to remain invariant must be a completely antisymmetric tensor. That is, when we swap adjacent elements in our wedge product which flips signs, we must flip the sign of the coefficient as well. Given this, we can also simply sum through all permutations of , the set of first integers, in which case the set is mapped to fixed combination in of the permutations. Thus, we have for any arbitrary -form, implicitly summing across all , Essentially, for each combination or basis element, we count it times, which we also divide by in the result to normalize. In calculating its Hodge dual, we note how In replacing the parenthesis with a subscript for , it is to emphasize that are different indices each of which are iterated across per Einstein notation. Complete antisymmetry means that a duplication of index results in . From , we get ### Some function isomorphisms between exterior product spaces

The Hodge star is a function of signature that is clearly also a linear isomorphism, which means that We also showed via the above calculation that wedge product is a bilinear function of signature We denote the volume corresponding to the wedge product of our basis vectors of real vector space in an order of positive orientation, , as and observe that every elements of is of the form , , or Moreover, tells us that plugging in a -form on the left results in a function of signature which is a functional over the -forms, which we denote with , and no different different -form inputs result in the same functional, which means we have induced an injective function of signature We now prove that it is also surjective. Every element in is, by definition of the vector space underlying the functional, uniquely defined by a collection of mappings of combinations to real numbers, or in other words, each in our basis is assigned to a real number . This corresponds uniquely to in our wedge product.

Bijectivity means ### Inducing an inner product on via the Hodge star

We now examine the function described by We now verify that it satisfies the properties of an inner product. Below, we per set and given this, by , We will for simplicity also omit the constant factor in subsequent calculations.

1. Linearity in the first argument is satisfied because the wedge product is bilinear.
2. The calculation below gives conjugate symmetry. 3. If in , we set in order to equate , we find that

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