## April 4th, 2021

### 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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### How to compute the volume of an arbitrary parallelotope embedded in R^n

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

linearly independent vectors in result in a -dimensional parallelotope (a generalization of parallelogram or parallelpiped in higher dimensions). We wish to determine its generalized volume with orientation in dimensions. The determinant of an matrix gives the (signed) volume of the parallelotope induced by the column vectors, the order of which affects the sign. (The proof of this is rather straightforward and will be left to the reader.) Independent of coordinates, a linear transformation between two -dimensional vector spaces the positively oriented orthogonal bases of which are respectively assigns each element in to a linear combination of the elements of , and the determinant is the volume of the parallelotope generated by elements of .

Let be the vectors for our -dimensional parallelotope in . Together they define a map from to that by the rank-nullity theorem is onto, with . Let be a positively oriented orthogonal basis of , the subspace of perpendicular to the kernel of with . We then define the same map on a restricted domain as

We then have

The coordinate free definition adjoint operator (or transpose) and its determinant

If is a linear map from and inner products are defined on -dimensional vector space with respect to their bases such that

Then the adjoint of , which we denote as is the map from such that

In the language of transposes, we have

Note how the left hand side of gives for every element in an ordered set of vectors in with respect to an orthonormal basis of , and then prescribes an ordered set of vectors in the coordinates of the th of which, with respect to an orthonormal basis of , is the th coordinate of the elements of , and vice versa.

We’ve essentially defined an matrix the elements of which are , and then the elements of transpose matrix . Applying the permutation based determinant formula gives us

Computing the volume of the parallelotope

The matrix formed by linearly independent column vectors in , , corresponds directly to a linear isomorphism from , the codomain of which is of course a dimensional subspace of . We showed in the previous section that the adjoint of , namely , has the same determinant. Since determinant is a multiplicative function, thus gives us the square of the volume of the parallelotope.

Per the rule of matrix multiplication, given a linear isomorphism which takes the elements of the orthonormal basis of its domain to vectors , regardless of basis or coordinates in the range, its composition with its adjoint is represented with respect to aforementioned basis by

wherein the invoked inner product is, of course, per the properties of inner product invariant with respect to coordinate transformations, which means is well-defined. This corresponds to the matrix formed from our vectors in ,

mutiplied by its transpose on the left, which gives us the result prescribed by , wherein the inner product is the Euclidean inner product in restricted to a dimensional subspace within it. The resulting Gramian matrix

is, as we’ve already explained, such that

or in words the square of the volume of the parallelotope generated by .

Decomposing a volume element into orthogonal components in -form space

There is also intimate connection here with differential forms, exterior products, and Hodge dual. In of [1], we defined an inner product on the th exterior product space with the Gramian determinant invoked in such that

Let be an orthonormal positively oriented basis for our vector space, which by definition results in the equivalence relation

as far as -dimensional volume is concerned, with which each of our s decomposes to . Substitution of this decomposition into yields for the coefficient of the determinant of the matrix assembled from the rows of our column vectors, with an appropriate sign adjustment. This coefficient is necessarily also an anti-symmetric th rank tensor.

Geometrically this is the oriented -dimensional volume obtained if only the components corresponding to indices are considered. The result in of [1], which was calculated via the Hodge star in a way that equates to the Gramian matrix definition of the inner product, yields for the value of the inner product given in the sum of squares of the coefficients with respect to our basis elements . Essentially there is a basis of of dimension volume elements of equivalent volume in space represented by -forms, which are mutually orthogonal with respect to the inner product we defined on the space of -forms, and we have projected our arbitrary dimensional volume element onto each of them. In this sense it is natural that the square of its norm or size would be the sum of the squares of its components.

I am dedicating this article to Seki Takakazu (1642-1708), who based on no more than 13th century Chinese algebra and arithmetic obtained results regarding determinants and resultants decades before the West and who discovered Bernoulli numbers (or Takakazu numbers) in connection to the closed formula for sum of the first th powers around the same time as did Jacob Bernoulli.

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