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. [公式] . Take the decomposable [公式] -form in [公式] -dimensional space denoted by

[公式] We can take [公式] and the corresponding [公式] -form

[公式] as the Hodge dual with signs yet to be determined. As for how to reasonably prescribe the sign, we first note that a [公式] -form wedged producted with its Hodge dual gives

[公式] The right hand side of this is equal to

[公式] We let [公式] be the standard ordering of the basis of the underlying vector space, which we regard as positively oriented. Moreover, we let [公式] be the permutation such that

[公式] It is reasonable to impose for symmetry that

[公式] To satisfy this, we choose an ordering of remaining indices [公式] such that the sign of the corresponding permutation [公式] is positive.

[公式] , the space of [公式] -forms over finite vector space [公式], has dimension [公式] . The standard basis for it we represent as

[公式]

[公式] has the same dimension [公式] . It is easy to see that the Hodge star operator maps the standard basis of [公式] to that of [公式] , and moreover that

[公式] Imposing linearity on the Hodge star operator then gives us

[公式] Now we wish to in Einstein notation in terms of [公式] indices represent the Hodge star of an arbitrary [公式] -form. In doing so, we will use the complete antisymmetric tensor of rank [公式] , which in addition to being zero when not all indices are different, satisfies

[公式] In the basis above, we assumed [公式] , and the coefficients [公式] . 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 [公式]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 [公式] , the set of first [公式] integers, in which case the set [公式] is mapped to fixed combination [公式] in [公式] of the [公式] permutations. Thus, we have for any arbitrary [公式] -form, implicitly summing across all [公式] ,

[公式] Essentially, for each combination or basis element, we count it [公式] times, which we also divide by in the result to normalize. In calculating its Hodge dual, we note how

[公式] In replacing the parenthesis with a subscript for [公式], it is to emphasize that [公式] are different indices each of which are iterated across [公式] per Einstein notation. Complete antisymmetry means that a duplication of index results in [公式] . From [公式] , we get

[公式]

Some function isomorphisms between exterior product spaces

The Hodge star is a function of signature

[公式] that is clearly also a linear isomorphism, which means that

[公式] We also showed via the above calculation that wedge product is a bilinear function of signature

[公式]

We denote the volume corresponding to the wedge product of our [公式] basis vectors of real vector space [公式] in an order of positive orientation, [公式] , as [公式] and observe that every elements of [公式] is of the form [公式] , [公式] , or

[公式] Moreover, [公式] tells us that plugging in a [公式] -form on the left results in a function of signature

[公式] which is a functional over the [公式] -forms, which we denote with [公式] , and no different different [公式] -form inputs result in the same functional, which means we have induced an injective function of signature

[公式] We now prove that it is also surjective. Every element in [公式] is, by definition of the vector space underlying the functional, uniquely defined by a collection of [公式] mappings of combinations to real numbers, or in other words, each[公式] in our basis is assigned to a real number [公式] . This corresponds uniquely to [公式] in our wedge product.

Bijectivity means

[公式]

Inducing an inner product on [公式] via the Hodge star

We now examine the function described by

[公式]We now verify that it satisfies the properties of an inner product. Below, we per [公式] set

[公式]

and given this, by [公式] ,

[公式] 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.

[公式] 3. If in [公式] , we set [公式] in order to equate [公式] , we find that

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<p><small>Originally published at <a href="https://gmachine1729.wpcomstaging.com/2021/04/04/how-to-naturally-construct-and-compute-explicitly-the-hodge-dual-of-a-differential-form/">狗和留美者不得入内</a>. You can comment here or <a href="https://gmachine1729.wpcomstaging.com/2021/04/04/how-to-naturally-construct-and-compute-explicitly-the-hodge-dual-of-a-differential-form/#comments">there</a>.</small></p><div class="RichText ztext Post-RichText"> <p>There is antisymmetry in the wedge product of differential forms, i.e. <img src="https://www.zhihu.com/equation?tex=dx%5E%5Cmu+%5Cwedge+dx%5E%5Cnu+%3D+-+dx%5E%5Cnu+%5Cwedge+dx%5E%5Cmu+" alt="[公式]" eeimg="1" data-formula="dx^\mu \wedge dx^\nu = - dx^\nu \wedge dx^\mu "> . Take the decomposable <img src="https://www.zhihu.com/equation?tex=p" alt="[公式]" eeimg="1" data-formula="p"> -form in <img src="https://www.zhihu.com/equation?tex=n" alt="[公式]" eeimg="1" data-formula="n"> -dimensional space denoted by</p> <p><img src="https://www.zhihu.com/equation?tex=dx%5E%7B%5Cmu_1%7D%5Cwedge+dx%5E%7B%5Cmu_2%7D%5Cwedge+%5Cldots+%5Cwedge+dx%5E%7B%5Cmu_p%7D.%5C%5C" alt="[公式]" eeimg="1" data-formula="dx^{\mu_1}\wedge dx^{\mu_2}\wedge \ldots \wedge dx^{\mu_p}.\\"> We can take <img src="https://www.zhihu.com/equation?tex=%5C%7B%5Cnu_1%2C%5Cnu_2%2C%5Cldots%2C%5Cnu_%7Bn-p%7D%5C%7D+%3D+%5C%7B1%2C2%2C%5Cldots%2Cn%5C%7D+%5Csetminus+%5C%7B%5Cmu_1%2C%5Cmu_2%2C%5Cldots%2C%5Cmu_p%5C%7D" alt="[公式]" eeimg="1" data-formula="\{\nu_1,\nu_2,\ldots,\nu_{n-p}\} = \{1,2,\ldots,n\} \setminus \{\mu_1,\mu_2,\ldots,\mu_p\}"> and the corresponding <img src="https://www.zhihu.com/equation?tex=%28n-p%29" alt="[公式]" eeimg="1" data-formula="(n-p)"> -form</p> <p><img src="https://www.zhihu.com/equation?tex=%28dx%5E%7B%5Cmu_1%7D%5Cwedge+%5Cldots+%5Cwedge+dx%5E%7B%5Cmu_p%7D%29%5E%2A+%3D+%5Cpm+dx%5E%7B%5Cnu_1%7D%5Cwedge+%5Cldots+%5Cwedge+dx%5E%7B%5Cnu_%7Bn-p%7D%7D.%5C%5C" alt="[公式]" eeimg="1" data-formula="(dx^{\mu_1}\wedge \ldots \wedge dx^{\mu_p})^* = \pm dx^{\nu_1}\wedge \ldots \wedge dx^{\nu_{n-p}}.\\"> as the Hodge dual with signs yet to be determined. As for how to reasonably prescribe the sign, we first note that a <img src="https://www.zhihu.com/equation?tex=p" alt="[公式]" eeimg="1" data-formula="p"> -form wedged producted with its Hodge dual gives</p> <p><img src="https://www.zhihu.com/equation?tex=%28dx%5E%7B%5Cmu_1%7D%5Cwedge+%5Cldots+%5Cwedge+dx%5E%7B%5Cmu_p%7D%29%5Cwedge+%28dx%5E%7B%5Cmu_1%7D%5Cwedge+%5Cldots+%5Cwedge+dx%5E%7B%5Cmu_p%7D%29%5E%2A+%3D+%5Cpm+dx%5E%7B%5Cmu_1%7D%5Cwedge+dx%5E%7B%5Cmu_p%7D+%5Cwedge+dx%5E%7B%5Cnu_1%7D%5Cwedge+%5Cldots+%5Cwedge+dx%5E%7B%5Cnu_%7Bn-p%7D%7D.+%5Cqquad+%281%29%5C%5C" alt="[公式]" eeimg="1" data-formula="(dx^{\mu_1}\wedge \ldots \wedge dx^{\mu_p})\wedge (dx^{\mu_1}\wedge \ldots \wedge dx^{\mu_p})^* = \pm dx^{\mu_1}\wedge dx^{\mu_p} \wedge dx^{\nu_1}\wedge \ldots \wedge dx^{\nu_{n-p}}. \qquad (1)\\"> The right hand side of this is equal to</p> <p><img src="https://www.zhihu.com/equation?tex=%5Cpm+dx%5E%7B1%7D%5Cwedge+%5Cldots+%5Cwedge+dx%5E%7Bn%7D.%5C%5C" alt="[公式]" eeimg="1" data-formula="\pm dx^{1}\wedge \ldots \wedge dx^{n}.\\"> We let <img src="https://www.zhihu.com/equation?tex=dx%5E1%2Cdx%5E2%2C%5Cldots+dx%5En" alt="[公式]" eeimg="1" data-formula="dx^1,dx^2,\ldots dx^n"> be the standard ordering of the basis of the underlying vector space, which we regard as positively oriented. Moreover, we let <img src="https://www.zhihu.com/equation?tex=%5Csigma" alt="[公式]" eeimg="1" data-formula="\sigma"> be the permutation such that</p> <p><img src="https://www.zhihu.com/equation?tex=%5Csigma%281%2C2%2C%5Cldots%2C+n%29+%3D+%28%5Csigma%281%29%2C%5Csigma%282%29%2C%5Cldots%2C+%5Csigma%28n%29%29+%3D+%28%5Cmu_1%2C%5Cmu_2%2C%5Cldots%2C+%5Cmu_p%2C%5Cnu_1%2C%5Cldots%2C+%5Cnu_%7Bn-p%7D%29.%5C%5C" alt="[公式]" eeimg="1" data-formula="\sigma(1,2,\ldots, n) = (\sigma(1),\sigma(2),\ldots, \sigma(n)) = (\mu_1,\mu_2,\ldots, \mu_p,\nu_1,\ldots, \nu_{n-p}).\\"> It is reasonable to impose for symmetry that</p> <p><img src="https://www.zhihu.com/equation?tex=%28dx%5E%7B%5Cmu_1%7D%5Cwedge+%5Cldots+%5Cwedge+dx%5E%7B%5Cmu_p%7D%29%5Cwedge+%28dx%5E%7B%5Cmu_1%7D%5Cwedge+%5Cldots+%5Cwedge+dx%5E%7B%5Cmu_%7Bp%7D%7D%29%5E%2A+%3D+dx%5E1%5Cwedge+%5Cldots+%5Cwedge+dx%5En.%5C%5C" alt="[公式]" eeimg="1" data-formula="(dx^{\mu_1}\wedge \ldots \wedge dx^{\mu_p})\wedge (dx^{\mu_1}\wedge \ldots \wedge dx^{\mu_{p}})^* = dx^1\wedge \ldots \wedge dx^n.\\"> To satisfy this, we choose an ordering of remaining indices <img src="https://www.zhihu.com/equation?tex=%28%5Cnu_1%2C%5Cldots%2C+%5Cnu_%7Bn-p%7D%29" alt="[公式]" eeimg="1" data-formula="(\nu_1,\ldots, \nu_{n-p})"> such that the sign of the corresponding permutation <img src="https://www.zhihu.com/equation?tex=%5Csigma" alt="[公式]" eeimg="1" data-formula="\sigma"> is positive.</p> <p><img src="https://www.zhihu.com/equation?tex=%5Cbigwedge%5Ep%28V%29" alt="[公式]" eeimg="1" data-formula="\bigwedge^p(V)"> , the space of <img src="https://www.zhihu.com/equation?tex=p" alt="[公式]" eeimg="1" data-formula="p"> -forms over finite vector space <img src="https://www.zhihu.com/equation?tex=V" alt="[公式]" eeimg="1" data-formula="V">, has dimension <img src="https://www.zhihu.com/equation?tex=%5Cbinom%7Bn%7D%7Bp%7D" alt="[公式]" eeimg="1" data-formula="\binom{n}{p}"> . The standard basis for it we represent as</p> <p><img src="https://www.zhihu.com/equation?tex=%28e%5E1%2C+%5Cldots%2C+e%5E%7B%5Cbinom%7Bn%7D%7Bp%7D%7D%29+%3D+%5C%7Bdx%5E%7B%5Cmu_1%7D%5Cwedge+%5Cldots+%5Cwedge+dx%5E%7B%5Cmu_p%7D+%3A+1+%5Cleq+%5Cmu_1+%3C+%5Cmu_2+%3C+%5Cldots+%3C+%5Cmu_p+%5Cleq+n%5C%7D.%5C%5C" alt="[公式]" eeimg="1" data-formula="(e^1, \ldots, e^{\binom{n}{p}}) = \{dx^{\mu_1}\wedge \ldots \wedge dx^{\mu_p} : 1 \leq \mu_1 < \mu_2 < \ldots < \mu_p \leq n\}.\\"> </p> <p><img src="https://www.zhihu.com/equation?tex=%5Cbigwedge%5E%7Bn-p%7D%28V%29" alt="[公式]" eeimg="1" data-formula="\bigwedge^{n-p}(V)"> has the same dimension <img src="https://www.zhihu.com/equation?tex=%5Cbinom%7Bn%7D%7Bn-p%7D+%3D+%5Cbinom%7Bn%7D%7Bp%7D" alt="[公式]" eeimg="1" data-formula="\binom{n}{n-p} = \binom{n}{p}"> . It is easy to see that the Hodge star operator maps the standard basis of <img src="https://www.zhihu.com/equation?tex=%5CLambda%5Ep%28V%29" alt="[公式]" eeimg="1" data-formula="\Lambda^p(V)"> to that of <img src="https://www.zhihu.com/equation?tex=%5CLambda%5E%7Bn-p%7D%28V%29" alt="[公式]" eeimg="1" data-formula="\Lambda^{n-p}(V)"> , and moreover that</p> <p><img src="https://www.zhihu.com/equation?tex=e%5Ei+%5Cwedge+%28e%5Ej%29%5E%2A+%3D+%5Cdelta_i%5Ej+dx%5E1%5Cwedge+dx%5E2%5Cwedge+%5Cldots+%5Cwedge+dx%5En.%5C%5C" alt="[公式]" eeimg="1" data-formula="e^i \wedge (e^j)^* = \delta_i^j dx^1\wedge dx^2\wedge \ldots \wedge dx^n.\\"> Imposing linearity on the Hodge star operator then gives us</p> <p><img src="https://www.zhihu.com/equation?tex=%5Cbegin%7Beqnarray%7D+%28A_ie%5Ei%29%5Cwedge+%5B%28B_j+e%5Ej%29%5E%2A%5D+%26%3D%26+%28A_ie%5Ei%29%5Cwedge+%5BB_j+%28e%5Ej%29%5E%2A%5D%5C%5C+%26%3D%26+%28A_ie%5Ei%29%5Cwedge+%5BB_i+%28e%5Ei%29%5E%2A%5D%5C%5C+%26%3D%26+A_iB_i+dx_1+%5Cwedge+dx_2+%5Cldots+%5Cwedge+dx_n.+%5Cqquad+%282%29+%5Cend%7Beqnarray%7D%5C%5C" alt="[公式]" eeimg="1" data-formula="\begin{eqnarray} (A_ie^i)\wedge [(B_j e^j)^*] &amp;=&amp; (A_ie^i)\wedge [B_j (e^j)^*]\\ &amp;=&amp; (A_ie^i)\wedge [B_i (e^i)^*]\\ &amp;=&amp; A_iB_i dx_1 \wedge dx_2 \ldots \wedge dx_n. \qquad (2) \end{eqnarray}\\"> Now we wish to in Einstein notation in terms of <img src="https://www.zhihu.com/equation?tex=p" alt="[公式]" eeimg="1" data-formula="p"> indices represent the Hodge star of an arbitrary <img src="https://www.zhihu.com/equation?tex=p" alt="[公式]" eeimg="1" data-formula="p"> -form. In doing so, we will use the complete antisymmetric tensor of rank <img src="https://www.zhihu.com/equation?tex=n" alt="[公式]" eeimg="1" data-formula="n"> , which in addition to being zero when not all indices are different, satisfies</p> <p><img src="https://www.zhihu.com/equation?tex=e%5E%7B%5Csigma%281%29%5Csigma%282%29%5Cldots+%5Csigma%28n%29%7D+%3D+%5Cmathrm%7Bsgn%7D%28%5Csigma%29e%5E%7B12%5Cldots+n%7D+%3D+%5Cmathrm%7Bsgn%7D%28%5Csigma%29.%5C%5C" alt="[公式]" eeimg="1" data-formula="e^{\sigma(1)\sigma(2)\ldots \sigma(n)} = \mathrm{sgn}(\sigma)e^{12\ldots n} = \mathrm{sgn}(\sigma).\\"> In the basis above, we assumed <img src="https://www.zhihu.com/equation?tex=%5Cmu_1+%3C+%5Cldots+%3C+%5Cmu_n+" alt="[公式]" eeimg="1" data-formula="\mu_1 < \ldots < \mu_n "> , and the coefficients <img src="https://www.zhihu.com/equation?tex=A_i+%5Cequiv+A_%7B%5Cmu_1%5Cldots+%5Cmu_p%7D" alt="[公式]" eeimg="1" data-formula="A_i \equiv A_{\mu_1\ldots \mu_p}"> . 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 <img src="https://www.zhihu.com/equation?tex=A_%7B%5Cmu_1%5Cldots+%5Cmu_p%7D" alt="[公式]" eeimg="1" data-formula="A_{\mu_1\ldots \mu_p}">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 <img src="https://www.zhihu.com/equation?tex=%5Bn%5D" alt="[公式]" eeimg="1" data-formula="[n]"> , the set of first <img src="https://www.zhihu.com/equation?tex=n" alt="[公式]" eeimg="1" data-formula="n"> integers, in which case the set <img src="https://www.zhihu.com/equation?tex=%5C%7B1%2C+%5Cldots%2C+p%5C%7D" alt="[公式]" eeimg="1" data-formula="\{1, \ldots, p\}"> is mapped to fixed combination <img src="https://www.zhihu.com/equation?tex=%5C%7B%5Cmu_1%2C%5Cldots%2C+%5Cmu_p%5C%7D" alt="[公式]" eeimg="1" data-formula="\{\mu_1,\ldots, \mu_p\}"> in <img src="https://www.zhihu.com/equation?tex=p%21%28n-p%29%21" alt="[公式]" eeimg="1" data-formula="p!(n-p)!"> of the <img src="https://www.zhihu.com/equation?tex=n%21" alt="[公式]" eeimg="1" data-formula="n!"> permutations. Thus, we have for any arbitrary <img src="https://www.zhihu.com/equation?tex=p" alt="[公式]" eeimg="1" data-formula="p"> -form, implicitly summing across all <img src="https://www.zhihu.com/equation?tex=%5Csigma+%5Cin+S_n" alt="[公式]" eeimg="1" data-formula="\sigma \in S_n"> ,</p> <p><img src="https://www.zhihu.com/equation?tex=%5Comega+%5Cequiv+%5Cfrac%7B1%7D%7Bp%21%28n-p%29%21%7DA_%7B%5Csigma%281%29%5Cldots+%5Csigma%28p%29%7Ddx%5E%7B%5Csigma%281%29%7D%5Cwedge+%5Cldots+%5Cwedge+dx%5E%7B%5Csigma%28p%29%7D.+%5Cqquad+%283%29%5C%5C" alt="[公式]" eeimg="1" data-formula="\omega \equiv \frac{1}{p!(n-p)!}A_{\sigma(1)\ldots \sigma(p)}dx^{\sigma(1)}\wedge \ldots \wedge dx^{\sigma(p)}. \qquad (3)\\"> Essentially, for each combination or basis element, we count it <img src="https://www.zhihu.com/equation?tex=p%21%28n-p%29%21" alt="[公式]" eeimg="1" data-formula="p!(n-p)!"> times, which we also divide by in the result to normalize. In calculating its Hodge dual, we note how</p> <p><img src="https://www.zhihu.com/equation?tex=%5Cbegin%7Beqnarray%7D+%28dx%5E%7B%5Csigma%281%29%7D%5Cwedge+%5Cldots+%5Cwedge+dx%5E%7B%5Csigma%28p%29%7D%29%5E%2A+%26%3D%26+%5Cmathrm%7Bsgn%7D%28%5Csigma%29+dx%5E%7B%5Csigma%28p%2B1%29%7D%5Cwedge+%5Cldots+%5Cwedge+dx%5E%7B%5Csigma%28n%29%7D%5C%5C+%26%3D%26+e%5E%7B%5Csigma_1%5Csigma_2%5Cldots+%5Csigma_n%7Ddx%5E%7B%5Csigma_%7Bp%2B1%7D%7D%5Cwedge+%5Cldots+%5Cwedge+dx%5E%7B%5Csigma_n%7D.+%5Cqquad+%284%29+%5Cend%7Beqnarray%7D%5C%5C" alt="[公式]" eeimg="1" data-formula="\begin{eqnarray} (dx^{\sigma(1)}\wedge \ldots \wedge dx^{\sigma(p)})^* &amp;=&amp; \mathrm{sgn}(\sigma) dx^{\sigma(p+1)}\wedge \ldots \wedge dx^{\sigma(n)}\\ &amp;=&amp; e^{\sigma_1\sigma_2\ldots \sigma_n}dx^{\sigma_{p+1}}\wedge \ldots \wedge dx^{\sigma_n}. \qquad (4) \end{eqnarray}\\"> In replacing the parenthesis with a subscript for <img src="https://www.zhihu.com/equation?tex=%5Csigma" alt="[公式]" eeimg="1" data-formula="\sigma">, it is to emphasize that <img src="https://www.zhihu.com/equation?tex=%5Csigma_1%2C%5Csigma_2%2C%5Cldots+%5Csigma_n" alt="[公式]" eeimg="1" data-formula="\sigma_1,\sigma_2,\ldots \sigma_n"> are different indices each of which are iterated across <img src="https://www.zhihu.com/equation?tex=%5B1%5D" alt="[公式]" eeimg="1" data-formula="[1]"> per Einstein notation. Complete antisymmetry means that a duplication of index results in <img src="https://www.zhihu.com/equation?tex=e_%7B%5Csigma_1%5Csigma_2%5Cldots+%5Csigma_n%7D+%3D+0" alt="[公式]" eeimg="1" data-formula="e_{\sigma_1\sigma_2\ldots \sigma_n} = 0"> . From <img src="https://www.zhihu.com/equation?tex=%284%29" alt="[公式]" eeimg="1" data-formula="(4)"> , we get</p> <p><img src="https://www.zhihu.com/equation?tex=%5Cbegin%7Beqnarray%7D+%5Comega%5E%2A+%26%3D%26+%5Cfrac%7B1%7D%7Bp%21%28n-p%29%21%7DA_%7B%5Csigma_1%5Cldots+%5Csigma_p%7D%28dx%5E%7B%5Csigma_%7B1%7D%7D%5Cwedge+%5Cldots+%5Cwedge+dx%5E%7B%5Csigma_p%7D%29%5E%2A%5C%5C+%26%3D%26+%5Cfrac%7B1%7D%7Bp%21%28n-p%29%21%7DA_%7B%5Csigma_1%5Cldots+%5Csigma_p%7D+e%5E%7B%5Csigma_1%5Csigma_2%5Cldots+%5Csigma_n%7Ddx%5E%7B%5Csigma_%7Bp%2B1%7D%7D%5Cwedge+%5Cldots+%5Cwedge+dx%5E%7B%5Csigma_n%7D.+%5Cqquad+%285%29+%5Cend%7Beqnarray%7D+%5C%5C" alt="[公式]" eeimg="1" data-formula="\begin{eqnarray} \omega^* &amp;=&amp; \frac{1}{p!(n-p)!}A_{\sigma_1\ldots \sigma_p}(dx^{\sigma_{1}}\wedge \ldots \wedge dx^{\sigma_p})^*\\ &amp;=&amp; \frac{1}{p!(n-p)!}A_{\sigma_1\ldots \sigma_p} e^{\sigma_1\sigma_2\ldots \sigma_n}dx^{\sigma_{p+1}}\wedge \ldots \wedge dx^{\sigma_n}. \qquad (5) \end{eqnarray} \\"> </p> <h3>Some function isomorphisms between exterior product spaces</h3> <p>The Hodge star is a function of signature</p> <p><img src="https://www.zhihu.com/equation?tex=%5Cstar_p+%3A+%5Cbigwedge%5Ep%28V%29+%5Cto+%5Cbigwedge%5E%7Bn-p%7D%28V%29%5C%5C" alt="[公式]" eeimg="1" data-formula="\star_p : \bigwedge^p(V) \to \bigwedge^{n-p}(V)\\"> that is clearly also a linear isomorphism, which means that</p> <p><img src="https://www.zhihu.com/equation?tex=%5Cbigwedge%5Ep%28V%29+%5Csimeq+%5Cbigwedge%5E%7Bn-p%7D%28V%29.+%5Cqquad+%286%29%5C%5C" alt="[公式]" eeimg="1" data-formula="\bigwedge^p(V) \simeq \bigwedge^{n-p}(V). \qquad (6)\\"> We also showed via the above calculation that wedge product is a bilinear function of signature</p> <p><img src="https://www.zhihu.com/equation?tex=%5Cwedge+%3A+%5Cbigwedge%5Ep%28V%29+%5Ctimes+%5Cbigwedge%5E%7Bn-p%7D%28V%29+%5Cto+%5Cbigwedge%5E%7Bn%7D%28V%29.%5C%5C+" alt="[公式]" eeimg="1" data-formula="\wedge : \bigwedge^p(V) \times \bigwedge^{n-p}(V) \to \bigwedge^{n}(V).\\ "> </p> <p>We denote the volume corresponding to the wedge product of our <img src="https://www.zhihu.com/equation?tex=n" alt="[公式]" eeimg="1" data-formula="n"> basis vectors of real vector space <img src="https://www.zhihu.com/equation?tex=V" alt="[公式]" eeimg="1" data-formula="V"> in an order of positive orientation, <img src="https://www.zhihu.com/equation?tex=dx%5E1%5Cwedge+dx%5E2%5Cwedge+%5Cldots+%5Cwedge+dx%5En" alt="[公式]" eeimg="1" data-formula="dx^1\wedge dx^2\wedge \ldots \wedge dx^n"> , as <img src="https://www.zhihu.com/equation?tex=%5Cmathrm%7BVol%7D_%5Cmu" alt="[公式]" eeimg="1" data-formula="\mathrm{Vol}_\mu"> and observe that every elements of <img src="https://www.zhihu.com/equation?tex=%5Cbigwedge%5E%7Bn%7D%28V%29" alt="[公式]" eeimg="1" data-formula="\bigwedge^{n}(V)"> is of the form <img src="https://www.zhihu.com/equation?tex=c%5Ccdot+%5Cmathrm%7BVol%7D_%5Cmu" alt="[公式]" eeimg="1" data-formula="c\cdot \mathrm{Vol}_\mu"> , <img src="https://www.zhihu.com/equation?tex=c+%5Cin+%5Cmathbb%7BR%7D" alt="[公式]" eeimg="1" data-formula="c \in \mathbb{R}"> , or</p> <p><img src="https://www.zhihu.com/equation?tex=%5Cbigwedge%5E%7Bn%7D%28V%29+%5Csimeq+%5Cmathbb%7BR%7D%5Cmathrm%7BVol%7D_%5Cmu+%5Csimeq+%5Cmathbb%7BR%7D.+%5Cqquad+%287%29%5C%5C" alt="[公式]" eeimg="1" data-formula="\bigwedge^{n}(V) \simeq \mathbb{R}\mathrm{Vol}_\mu \simeq \mathbb{R}. \qquad (7)\\"> Moreover, <img src="https://www.zhihu.com/equation?tex=%282%29" alt="[公式]" eeimg="1" data-formula="(2)"> tells us that plugging in a <img src="https://www.zhihu.com/equation?tex=p" alt="[公式]" eeimg="1" data-formula="p"> -form on the left results in a function of signature</p> <p><img src="https://www.zhihu.com/equation?tex=%5Cbigwedge%5E%7Bn-p%7D%28V%29+%5Cto+%5Cleft%28%5Cbigwedge%5E%7Bn%7D%28V%29+%5Csimeq+%5Cmathbb%7BR%7D%5Cright%29%2C%5C%5C+" alt="[公式]" eeimg="1" data-formula="\bigwedge^{n-p}(V) \to \left(\bigwedge^{n}(V) \simeq \mathbb{R}\right),\\ "> which is a functional over the <img src="https://www.zhihu.com/equation?tex=%28n-p%29" alt="[公式]" eeimg="1" data-formula="(n-p)"> -forms, which we denote with <img src="https://www.zhihu.com/equation?tex=%5Cleft%28%5Cbigwedge%5E%7Bn-p%7D%28V%29%5Cright%29%5E%5Cvee" alt="[公式]" eeimg="1" data-formula="\left(\bigwedge^{n-p}(V)\right)^\vee"> , and no different different <img src="https://www.zhihu.com/equation?tex=p" alt="[公式]" eeimg="1" data-formula="p"> -form inputs result in the same functional, which means we have induced an injective function of signature</p> <p><img src="https://www.zhihu.com/equation?tex=%5Cbigwedge%5E%7Bp%7D%28V%29+%5Cto+%5Cleft%28%5Cbigwedge%5E%7Bn-p%7D%28V%29%5Cright%29%5E%5Cvee.%5C%5C" alt="[公式]" eeimg="1" data-formula="\bigwedge^{p}(V) \to \left(\bigwedge^{n-p}(V)\right)^\vee.\\"> We now prove that it is also surjective. Every element in <img src="https://www.zhihu.com/equation?tex=%5Cleft%28%5Cbigwedge%5E%7Bn-p%7D%28V%29%5Cright%29%5E%5Cvee" alt="[公式]" eeimg="1" data-formula="\left(\bigwedge^{n-p}(V)\right)^\vee"> is, by definition of the vector space underlying the functional, uniquely defined by a collection of <img src="https://www.zhihu.com/equation?tex=%5Cbinom%7Bn%7D%7Bn-p%7D" alt="[公式]" eeimg="1" data-formula="\binom{n}{n-p}"> mappings of combinations to real numbers, or in other words, each<img src="https://www.zhihu.com/equation?tex=dx%5E%7B%5Cnu_1%7D%5Cwedge+%5Cldots+%5Cwedge+dx%5E%7B%5Cnu_%7Bn-p%7D%7D" alt="[公式]" eeimg="1" data-formula="dx^{\nu_1}\wedge \ldots \wedge dx^{\nu_{n-p}}"> in our basis is assigned to a real number <img src="https://www.zhihu.com/equation?tex=C%5E%7B%5Cnu_1%5Cldots+%5Cnu_%7Bn-p%7D%7D" alt="[公式]" eeimg="1" data-formula="C^{\nu_1\ldots \nu_{n-p}}"> . This corresponds uniquely to <img src="https://www.zhihu.com/equation?tex=e%5E%7B%5Cmu_1%5Cldots+%5Cmu_p%5Cnu_1%5Cldots+%5Cnu_%7Bn-p%7D%7DC%5E%7B%5Cnu_1%5Cldots+%5Cnu_%7Bn-p%7D%7D+dx%5E%7B%5Cmu_1%7D%5Cwedge+%5Cldots+%5Cwedge+dx%5E%7B%5Cmu_p%7D+%5Cin+%5Cbigwedge%5Ep%28V%29" alt="[公式]" eeimg="1" data-formula="e^{\mu_1\ldots \mu_p\nu_1\ldots \nu_{n-p}}C^{\nu_1\ldots \nu_{n-p}} dx^{\mu_1}\wedge \ldots \wedge dx^{\mu_p} \in \bigwedge^p(V)"> in our wedge product.</p> <p>Bijectivity means</p> <p><img src="https://www.zhihu.com/equation?tex=%5Cbigwedge%5E%7Bp%7D%28V%29+%5Csimeq+%5Cleft%28%5Cbigwedge%5E%7Bn-p%7D%28V%29%5Cright%29%5E%5Cvee.+%5Cqquad+%288%29%5C%5C" alt="[公式]" eeimg="1" data-formula="\bigwedge^{p}(V) \simeq \left(\bigwedge^{n-p}(V)\right)^\vee. \qquad (8)\\"> </p> <h3>Inducing an inner product on <img src="https://www.zhihu.com/equation?tex=%5Cbigwedge%5Ep%28V%29" alt="[公式]" eeimg="1" data-formula="\bigwedge^p(V)"> via the Hodge star</h3> <p>We now examine the function described by</p> <p><img src="https://www.zhihu.com/equation?tex=%5Cbegin%7Balign%7D+%5Clangle%5Ccdot+%2C%5Ccdot+%5Crangle+%5Cbigwedge%5Ep%28V%29+%5Ctimes+%5Cbigwedge%5Ep%28V%29+%5Cto+%5Cmathbb%7BR%7D%2C%5C%5C+%5C%5C+%5Clangle+%5Comega%2C+%5Ceta+%5Crangle+%3D+%5Comega+%5Cwedge+%28%5Cstar_p+%5Ceta%29%2C%5C%5C%5C%5C++dx%5E1%5Cwedge+dx%5E2%5Cwedge+%5Cldots+%5Cwedge+dx%5En+%5Csim+%5Cmathrm%7BVol%7D_%5Cmu.+%5Cend%7Balign%7D%5Cqquad+%289%29%5C%5C" alt="[公式]" eeimg="1" data-formula="\begin{align} \langle\cdot ,\cdot \rangle \bigwedge^p(V) \times \bigwedge^p(V) \to \mathbb{R},\\ \\ \langle \omega, \eta \rangle = \omega \wedge (\star_p \eta),\\\\ dx^1\wedge dx^2\wedge \ldots \wedge dx^n \sim \mathrm{Vol}_\mu. \end{align}\qquad (9)\\">We now verify that it satisfies the properties of an inner product. Below, we per <img src="https://www.zhihu.com/equation?tex=%283%29" alt="[公式]" eeimg="1" data-formula="(3)"> set</p> <p><img src="https://www.zhihu.com/equation?tex=%5Cbegin%7Beqnarray%7D+%5Comega+%26%3D%26+%5Cfrac%7B1%7D%7Bp%21%28n-p%29%21%7DA_%7B%5Csigma_1%5Cldots+%5Csigma_p%7Ddx%5E%7B%5Csigma_1%7D%5Cwedge+%5Cldots+%5Cwedge+dx%5E%7B%5Csigma_p%7D%5C%5C+%5Ceta+%26%3D%26+%5Cfrac%7B1%7D%7Bp%21%28n-p%29%21%7DB_%7B%5Csigma_1%5Cldots+%5Csigma_p%7Ddx%5E%7B%5Csigma_1%7D%5Cwedge+%5Cldots+%5Cwedge+dx%5E%7B%5Csigma_p%7D%5C%5C+%5Cend%7Beqnarray%7D+%5Cqquad+%2810%29%5C%5C" alt="[公式]" eeimg="1" data-formula="\begin{eqnarray} \omega &amp;=&amp; \frac{1}{p!(n-p)!}A_{\sigma_1\ldots \sigma_p}dx^{\sigma_1}\wedge \ldots \wedge dx^{\sigma_p}\\ \eta &amp;=&amp; \frac{1}{p!(n-p)!}B_{\sigma_1\ldots \sigma_p}dx^{\sigma_1}\wedge \ldots \wedge dx^{\sigma_p}\\ \end{eqnarray} \qquad (10)\\"> </p> <p>and given this, by <img src="https://www.zhihu.com/equation?tex=%285%29" alt="[公式]" eeimg="1" data-formula="(5)"> ,</p> <p><img src="https://www.zhihu.com/equation?tex=%5Cbegin%7Beqnarray%7D+%5Cstar_p%5Comega+%26%3D%26+%5Cfrac%7B1%7D%7Bp%21%28n-p%29%21%7DA_%7B%5Csigma_1%5Cldots+%5Csigma_p%7D+e%5E%7B%5Csigma_1%5Csigma_2%5Cldots+%5Csigma_n%7Ddx%5E%7B%5Csigma_%7Bp%2B1%7D%7D%5Cwedge+%5Cldots+%5Cwedge+dx%5E%7B%5Csigma_n%7D%5C%5C+%5Cstar_p%5Ceta+%26%3D%26+%5Cfrac%7B1%7D%7Bp%21%28n-p%29%21%7DB_%7B%5Csigma_1%5Cldots+%5Csigma_p%7D+e%5E%7B%5Csigma_1%5Csigma_2%5Cldots+%5Csigma_n%7Ddx%5E%7B%5Csigma_%7Bp%2B1%7D%7D%5Cwedge+%5Cldots+%5Cwedge+dx%5E%7B%5Csigma_n%7D.%5C%5C+%5Cend%7Beqnarray%7D+%5Cqquad+%2811%29%5C%5C" alt="[公式]" eeimg="1" data-formula="\begin{eqnarray} \star_p\omega &amp;=&amp; \frac{1}{p!(n-p)!}A_{\sigma_1\ldots \sigma_p} e^{\sigma_1\sigma_2\ldots \sigma_n}dx^{\sigma_{p+1}}\wedge \ldots \wedge dx^{\sigma_n}\\ \star_p\eta &amp;=&amp; \frac{1}{p!(n-p)!}B_{\sigma_1\ldots \sigma_p} e^{\sigma_1\sigma_2\ldots \sigma_n}dx^{\sigma_{p+1}}\wedge \ldots \wedge dx^{\sigma_n}.\\ \end{eqnarray} \qquad (11)\\"> We will for simplicity also omit the constant factor in subsequent calculations.</p> <ol> <li>Linearity in the first argument is satisfied because the wedge product is bilinear.</li> <li>The calculation below gives conjugate symmetry.</li> </ol> <p><img src="https://www.zhihu.com/equation?tex=%5Cbegin%7Beqnarray%7D+%5Clangle+%5Comega%2C+%5Ceta+%5Crangle+%26%3D%26+%5Comega+%5Cwedge+%28%5Cstar_p+%5Ceta%29%5C%5C+%26%3D%26%28A_%7B%5Csigma_1%5Cldots+%5Csigma_p%7Ddx%5E%7B%5Csigma_1%7D%5Cwedge+%5Cldots+%5Cwedge+dx%5E%7B%5Csigma_p%7D%29%5Cwedge+%28B_%7B%5Csigma_1%5Cldots+%5Csigma_p%7D+e%5E%7B%5Csigma_1%5Csigma_2%5Cldots+%5Csigma_n%7Ddx%5E%7B%5Csigma_%7Bp%2B1%7D%7D%5Cwedge+%5Cldots+%5Cwedge+dx%5E%7B%5Csigma_n%7D%29%5C%5C+%26%3D%26+A_%7B%5Csigma_1%5Cldots+%5Csigma_p%7DB_%7B%5Csigma_1%5Cldots+%5Csigma_p%7De%5E%7B%5Csigma_1%5Csigma_2%5Cldots+%5Csigma_n%7Ddx%5E%7B%5Csigma_1%7D%5Cwedge+%5Cldots+%5Cwedge+dx%5E%7B%5Csigma_p%7D%5Cwedge+dx%5E%7B%5Csigma_%7Bp%2B1%7D%7D%5Cwedge+%5Cldots+%5Cwedge+dx%5E%7B%5Csigma_n%7D%5C%5C+%26%3D%26A_%7B%5Csigma_1%5Cldots+%5Csigma_p%7DB_%7B%5Csigma_1%5Cldots+%5Csigma_p%7Ddx%5E1%5Cwedge+dx%5E2%5Cldots+%5Cwedge+dx%5En%5C%5C+%26%3D%26%28B_%7B%5Csigma_1%5Cldots+%5Csigma_p%7Ddx%5E%7B%5Csigma_1%7D%5Cwedge+%5Cldots+%5Cwedge+dx%5E%7B%5Csigma_p%7D%29%5Cwedge+%28A_%7B%5Csigma_1%5Cldots+%5Csigma_p%7D+e%5E%7B%5Csigma_1%5Csigma_2%5Cldots+%5Csigma_n%7Ddx%5E%7B%5Csigma_%7Bp%2B1%7D%7D%5Cwedge+%5Cldots+%5Cwedge+dx%5E%7B%5Csigma_n%7D%29%5C%5C+%26%3D%26%5Ceta+%5Cwedge+%28%5Cstar_p+%5Comega%29%5C%5C+%26%3D%26%5Clangle+%5Ceta%2C+%5Comega+%5Crangle.+%5Cend%7Beqnarray%7D%5Cqquad+%2812%29%5C%5C+" alt="[公式]" eeimg="1" data-formula="\begin{eqnarray} \langle \omega, \eta \rangle &amp;=&amp; \omega \wedge (\star_p \eta)\\ &amp;=&amp;(A_{\sigma_1\ldots \sigma_p}dx^{\sigma_1}\wedge \ldots \wedge dx^{\sigma_p})\wedge (B_{\sigma_1\ldots \sigma_p} e^{\sigma_1\sigma_2\ldots \sigma_n}dx^{\sigma_{p+1}}\wedge \ldots \wedge dx^{\sigma_n})\\ &amp;=&amp; A_{\sigma_1\ldots \sigma_p}B_{\sigma_1\ldots \sigma_p}e^{\sigma_1\sigma_2\ldots \sigma_n}dx^{\sigma_1}\wedge \ldots \wedge dx^{\sigma_p}\wedge dx^{\sigma_{p+1}}\wedge \ldots \wedge dx^{\sigma_n}\\ &amp;=&amp;A_{\sigma_1\ldots \sigma_p}B_{\sigma_1\ldots \sigma_p}dx^1\wedge dx^2\ldots \wedge dx^n\\ &amp;=&amp;(B_{\sigma_1\ldots \sigma_p}dx^{\sigma_1}\wedge \ldots \wedge dx^{\sigma_p})\wedge (A_{\sigma_1\ldots \sigma_p} e^{\sigma_1\sigma_2\ldots \sigma_n}dx^{\sigma_{p+1}}\wedge \ldots \wedge dx^{\sigma_n})\\ &amp;=&amp;\eta \wedge (\star_p \omega)\\ &amp;=&amp;\langle \eta, \omega \rangle. \end{eqnarray}\qquad (12)\\ "> 3. If in <img src="https://www.zhihu.com/equation?tex=%2812%29" alt="[公式]" eeimg="1" data-formula="(12)"> , we set <img src="https://www.zhihu.com/equation?tex=A_%7B%5Csigma_1%5Cldots+%5Csigma_p%7D+%3D+B_%7B%5Csigma_1%5Cldots+%5Csigma_p%7D" alt="[公式]" eeimg="1" data-formula="A_{\sigma_1\ldots \sigma_p} = B_{\sigma_1\ldots \sigma_p}"> in order to equate <img src="https://www.zhihu.com/equation?tex=%5Comega+%3D+%5Ceta" alt="[公式]" eeimg="1" data-formula="\omega = \eta"> , we find that</p> <p><img src="https://www.zhihu.com/equation?tex=%5Clangle%5Comega%2C+%5Comega%5Crangle+%3D+A_%7B%5Csigma_1%5Cldots+%5Csigma_p%7DA_%7B%5Csigma_1%5Cldots+%5Csigma_p%7D%5Cmathrm%7BVol%7D_%5Cmu+%3E+0%2C%5C%5C" alt="[公式]" eeimg="1" data-formula="\langle\omega, \omega\rangle = A_{\sigma_1\ldots \sigma_p}A_{\sigma_1\ldots \sigma_p}\mathrm{Vol}_\mu > 0,\\&#8221;> being the product of a positive volume and a sum of squares holds iff only <img src="https://www.zhihu.com/equation?tex=%5Comega+%5Cneq+0" alt="[公式]" eeimg="1" data-formula="\omega \neq 0"> , which shows positive definiteness.</p> <p>We note how</p> <p><img src="https://www.zhihu.com/equation?tex=%5Clangle+dx%5E%7B%5Cmu_1%7D%5Cwedge+%5Cldots+%5Cwedge+dx%5E%7B%5Cmu_p%7D%2C+dx%5E%7B%5Cnu_1%7D%5Cwedge+%5Cldots+%5Cwedge+dx%5E%7B%5Cnu_p%7D+%5Crangle+%3D+%5Cpm+dx%5E1%5Cwedge+dx%5E2%5Cldots+%5Cwedge+dx%5En+%5C%5C" alt="[公式]" eeimg="1" data-formula="\langle dx^{\mu_1}\wedge \ldots \wedge dx^{\mu_p}, dx^{\nu_1}\wedge \ldots \wedge dx^{\nu_p} \rangle = \pm dx^1\wedge dx^2\ldots \wedge dx^n \\"> if and only if <img src="https://www.zhihu.com/equation?tex=%5C%7B%5Cmu_1%2C%5Cmu_2%5Cldots%2C+%5Cmu_p%5C%7D+%3D+%5C%7B%5Cnu_1%2C%5Cnu_2%5Cldots%2C+%5Cnu_p%5C%7D" alt="[公式]" eeimg="1" data-formula="\{\mu_1,\mu_2\ldots, \mu_p\} = \{\nu_1,\nu_2\ldots, \nu_p\}"> and that if such is not satisfied, the inner product is necessarily <img src="https://www.zhihu.com/equation?tex=0" alt="[公式]" eeimg="1" data-formula="0"> . Thus this inner product evaluates to <img src="https://www.zhihu.com/equation?tex=%5Cmathrm%7BVol%7D_%5Cmu" alt="[公式]" eeimg="1" data-formula="\mathrm{Vol}_\mu"> whenever both its inputs are the same basis vector. There is of course an inner product defined on the vector space spanned by <img src="https://www.zhihu.com/equation?tex=%5C%7Bdx%5E%5Cmu%5C%7D" alt="[公式]" eeimg="1" data-formula="\{dx^\mu\}"> , which is the inner product defined above for the case <img src="https://www.zhihu.com/equation?tex=p%3D1" alt="[公式]" eeimg="1" data-formula="p=1"> on<img src="https://www.zhihu.com/equation?tex=%5Cbigwedge%5E1%28V%29" alt="[公式]" eeimg="1" data-formula="\bigwedge^1(V)"> . We can use this to induce an inner product expressed in terms of the determinant for arbitrary <img src="https://www.zhihu.com/equation?tex=p" alt="[公式]" eeimg="1" data-formula="p"> . Treating <img src="https://www.zhihu.com/equation?tex=%5Cmathrm%7BVol%7D_%5Cmu" alt="[公式]" eeimg="1" data-formula="\mathrm{Vol}_\mu"> as a pre-set value representing an <img src="https://www.zhihu.com/equation?tex=n" alt="[公式]" eeimg="1" data-formula="n"> -dimensional volume corresponding to <img src="https://www.zhihu.com/equation?tex=n" alt="[公式]" eeimg="1" data-formula="n"> differentials. We can modify the inner product on <img src="https://www.zhihu.com/equation?tex=%5Cbigwedge%5E1%28V%29" alt="[公式]" eeimg="1" data-formula="\bigwedge^1(V)"> by changing to <img src="https://www.zhihu.com/equation?tex=dx%5E1%5Cwedge+dx%5E2%5Cwedge+%5Cldots+%5Cwedge+dx%5En+%5Csim+%28%5Cmathrm%7BVol%7D_%5Cmu%29%5E%7B1%2Fp%7D" alt="[公式]" eeimg="1" data-formula="dx^1\wedge dx^2\wedge \ldots \wedge dx^n \sim (\mathrm{Vol}_\mu)^{1/p}"> in that inner product, which we denote as <img src="https://www.zhihu.com/equation?tex=%5Clangle+%5Ccdot+%2C+%5Ccdot+%5Crangle_1" alt="[公式]" eeimg="1" data-formula="\langle \cdot , \cdot \rangle_1"> . Then, noting that the determinant of a diagonal matrix is the product of the diagonal entries, it is easy to observe that</p> <p><img src="https://www.zhihu.com/equation?tex=%5Clangle+dx%5E%7B%5Cmu_1%7D%5Cwedge+dx%5E%7B%5Cmu_2%7D%5Cwedge+%5Cldots+%5Cwedge+dx%5E%7B%5Cmu_p%7D%2C+dx%5E%7B%5Cnu_1%7D%5Cwedge+dx%5E%7B%5Cnu_2%7D%5Cwedge+%5Cldots+%5Cwedge+dx%5E%7B%5Cnu_p%7D%5Crangle+%3D+%5Cdet+%28%5Clangle+dx%5E%7B%5Cmu_i%7D%2C++dx%5E%7B%5Cnu_j%7D%5Crangle_1%29%2C+%5Cqquad+%2813%29%5C%5C" alt="[公式]" eeimg="1" data-formula="\langle dx^{\mu_1}\wedge dx^{\mu_2}\wedge \ldots \wedge dx^{\mu_p}, dx^{\nu_1}\wedge dx^{\nu_2}\wedge \ldots \wedge dx^{\nu_p}\rangle = \det (\langle dx^{\mu_i}, dx^{\nu_j}\rangle_1), \qquad (13)\\"> with <img src="https://www.zhihu.com/equation?tex=i%2Cj" alt="[公式]" eeimg="1" data-formula="i,j"> as the row and column indices of a <img src="https://www.zhihu.com/equation?tex=p%5Ctimes+p" alt="[公式]" eeimg="1" data-formula="p\times p"> matrix the elements of which are inner products. The determinant in <img src="https://www.zhihu.com/equation?tex=%2810%29" alt="[公式]" eeimg="1" data-formula="(10)"> is called the <i>Gram determinant</i> or <i>Gramian</i>. In this case <img src="https://www.zhihu.com/equation?tex=%5Clangle+dx%5Ei%2C+dx%5Ei+%5Crangle+%3D+%28%5Cmathrm%7BVol%7D_%5Cmu%29%5E%7B1%2Fp%7D" alt="[公式]" eeimg="1" data-formula="\langle dx^i, dx^i \rangle = (\mathrm{Vol}_\mu)^{1/p}"> for all <img src="https://www.zhihu.com/equation?tex=i" alt="[公式]" eeimg="1" data-formula="i"> , which results in</p> <p><img src="https://www.zhihu.com/equation?tex=%5Clangle+dx%5E%7B%5Cmu_1%7D%5Cwedge+dx%5E%7B%5Cmu_2%7D%5Cwedge+%5Cldots+%5Cwedge+dx%5E%7B%5Cmu_p%7D%2C+dx%5E%7B%5Cmu_1%7D%5Cwedge+dx%5E%7B%5Cmu_2%7D%5Cwedge+%5Cldots+%5Cwedge+dx%5E%7B%5Cmu_p%7D%5Crangle+%3D+%5Cmathrm%7BVol%7D_%5Cmu%2C+%5Cqquad+%2814%29%5C%5C" alt="[公式]" eeimg="1" data-formula="\langle dx^{\mu_1}\wedge dx^{\mu_2}\wedge \ldots \wedge dx^{\mu_p}, dx^{\mu_1}\wedge dx^{\mu_2}\wedge \ldots \wedge dx^{\mu_p}\rangle = \mathrm{Vol}_\mu, \qquad (14)\\"> which is consistent with <img src="https://www.zhihu.com/equation?tex=%2812%29" alt="[公式]" eeimg="1" data-formula="(12)"> . This inner product applied to pairs of basis vectors yield <img src="https://www.zhihu.com/equation?tex=0" alt="[公式]" eeimg="1" data-formula="0"> , when the two inputs are not the same. Linearity naturally extends this inner product to arbitrary vectors in <img src="https://www.zhihu.com/equation?tex=%5Cbigwedge%5Ep%28V%29" alt="[公式]" eeimg="1" data-formula="\bigwedge^p(V)"> . We will leave to the reader to check that <img src="https://www.zhihu.com/equation?tex=%2813%29" alt="[公式]" eeimg="1" data-formula="(13)"> is well-defined, that is, for <img src="https://www.zhihu.com/equation?tex=%5C%7Bu_1%2Cu_2%2C%5Cldots%2C+u_p%2C+v_1%2C+v_2%2C%5Cldots+v_p%5C%7D+%5Csubset+%5Cbigwedge%5E1%28V%29" alt="[公式]" eeimg="1" data-formula="\{u_1,u_2,\ldots, u_p, v_1, v_2,\ldots v_p\} \subset \bigwedge^1(V)"> ,</p> <p><img src="https://www.zhihu.com/equation?tex=%5Clangle+u_1%5Cwedge+u_2%5Cwedge+%5Cldots+%5Cwedge+u_p%2C+v_1%5Cwedge+v_2%5Cwedge+%5Cldots+%5Cwedge+v_p%5Crangle+%3D+%5Cdet+%28%5Clangle+u_i%2C++v_j%5Crangle_1%29.+%5Cqquad+%2815%29%5C%5C" alt="[公式]" eeimg="1" data-formula="\langle u_1\wedge u_2\wedge \ldots \wedge u_p, v_1\wedge v_2\wedge \ldots \wedge v_p\rangle = \det (\langle u_i, v_j\rangle_1). \qquad (15)\\"> </p> <hr> <p>The above was written after reading section 6 on four-vectors in [2] and then <a href="https://link.zhihu.com/?target=http%3A//math.stanford.edu/~conrad/diffgeomPage/handouts/star.pdf" class=" wrap external" target="_blank" rel="noreferrer noopener">[1]</a>. [2] gave an explicit formula for the Hodge dual in the case of two forms in 4 dimensions without explicitly mentioned &#8220;Hodge&#8221; using completely antisymmetric tensor (or pseudotensor) of rank 4. I wrote about this pseudotensor at <a href="https://zhuanlan.zhihu.com/p/361583379" class="internal" rel="noreferrer">[3]</a>. As for Brian Conrad&#8217;s notes, they did not present an explicit formula for Hodge dual though in the examples section, it was noted tht computation of them is mostly a matter of being careful with signs. I do remember that one of examples was under the Minkowski inner product in which case one would need to make an appropriate adjustment for the signs. Prof Conrad first gave the inner product induced by the Gramian determinant and he proceeded to in pure mathematician formalism to define the Hodge dual as the function satisfying <img src="https://www.zhihu.com/equation?tex=%289%29" alt="[公式]" eeimg="1" data-formula="(9)"> , the well-defined-ness of which is guaranteed implicitly by the chain of isomorphisms.</p> <p>I was quite pleased that while typing this up, I did not look up any sources directly, aside from a few peeks at Prof Conrad&#8217;s notes, which were entirely meant to ensure more notational consistency, which would be more convenient if one refers to both his notes and mine. I used to believe pessimistically that differential forms and the Hodge dual might actually be somewhat beyond my level of g (general intelligence factor, or IQ). When reading about them, I could always passively follow and feel like it all makes sense, but I never was able to comfortably write or talk about them in detail independent from another text. Now, after having written this without difficulty, I am much more confident about my prospect of learning differential geometry and general relativity and my mathematical and intellectual ability in general.</p> <p><b>References</b></p> <ul> <li>[1] Professor Brian Conrad&#8217;s notes on the Hodge star operator: <a href="https://link.zhihu.com/?target=http%3A//math.stanford.edu/~conrad/diffgeomPage/handouts/star.pdf" class=" external" target="_blank" rel="noreferrer noopener"><span class="invisible">http://</span><span class="visible">math.stanford.edu/~conr</span><span class="invisible">ad/diffgeomPage/handouts/star.pdf</span><span class="ellipsis"></span></a></li> <li>[2] Landau Lifshitz classical theory of fields (I downloaded it on libgenesis)</li> <li>[3] <a href="https://zhuanlan.zhihu.com/p/361583379" class="internal" rel="noreferrer">gmachine1729:On the completely antisymmetric unit rank 4 tensor (or pseudotensor) over spacetime coordinates</a></li> </ul> </div>

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.

[公式] linearly independent vectors in [公式] result in a [公式] -dimensional parallelotope (a generalization of parallelogram or parallelpiped in higher dimensions). We wish to determine its generalized volume with orientation in [公式] dimensions. The determinant of an [公式] 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 [公式] between two [公式] -dimensional vector spaces [公式] the positively oriented orthogonal bases of which are [公式] respectively assigns each element in [公式] to a linear combination of the elements of [公式] , and the determinant is the volume of the parallelotope generated by elements of [公式] [公式] .

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

[公式]We then have

[公式]

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

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

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

[公式] In the language of transposes, we have

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

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

[公式]

Computing the volume of the parallelotope

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

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

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

[公式]

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

[公式]

is, as we’ve already explained, such that

[公式] or in words the square of the volume of the parallelotope generated by [公式] .

Decomposing a [公式]volume element into orthogonal components in [公式] -form space

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

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

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

Geometrically this is the oriented [公式] -dimensional volume obtained if only the components corresponding to indices [公式] are considered. The result in [公式] 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 [公式] the sum of squares of the coefficients with respect to our [公式] basis elements [公式] . Essentially there is a basis of [公式] of dimension [公式] volume elements of equivalent volume in [公式] space represented by [公式] -forms, which are mutually orthogonal with respect to the inner product we defined on the space of [公式] -forms, and we have projected our arbitrary [公式] 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 [公式] th powers around the same time as did Jacob Bernoulli.

References