NB: Basis one-forms and contravariant basis vectors (which, following Menzel, I am calling reciprocal) are the same thing. See, for example, the Mathematical Appendix to Gravitation and Inertia, by Wheeler and Ciufolini.

MTW, in Exercise 3.14. DUALS, we are told

[A previous and entirely distinct use of the word "dual" (section 2.7) called a set of basis one-forms $\left\{ \omega^{\alpha}\right\} $ dual to a set of basis vectors $\left\{ \mathbf{e}_{\alpha}\right\} $ if $\left\langle \omega^{\alpha},\mathbf{e}_{\beta}\right\rangle =\delta^{\alpha}{}_{\beta}.$ Fortunately there are no grounds for confusion between the two types of duality. One relates sets of vectors to sets of one-forms. The other relates antisymmetric tensors of rank $p$ to antisymmetric tensors of rank $4-p$.]

I'm not so sure these uses of "dual" actually are "entirely distinct". Consider the definition given by Menzel (Mathematical Physics) for the reciprocal (contravariant) basis for a 3-dimensional curvilinear coordinate system in Galilean 3-space.

Rectangular Cartesian coordinates are written $x^{i},$ with their primary basis vectors written $\hat{\mathfrak{e}}_{i},$ and the corresponding reciprocal basis vectors are written$\hat{\mathfrak{e}}^{i}$. For curvilinear coordinates $q^{\bar{i}},\mathfrak{e}_{\bar{i}}\equiv\hat{\mathfrak{e}}_{i}\frac{\partial x^{i}}{\partial q^{\bar{i}}},\mathfrak{e}^{\bar{i}}$ will be used. The symbol $\mathcal{E}_{\bar{i}\bar{j}\bar{k}}$ denotes a tensor density of weight $+1$, and is identical to the Levi-Civita tensor ($\varepsilon_{ijk}$) in orthonormal coordinates. Following Menzel we define the volume element $V$ and the reciprocal basis vectors $\mathfrak{e}^{\bar{i}}$ as follows:

$$ V\equiv\mathfrak{e}_{\bar{1}}\cdot\mathfrak{e}_{\bar{2}}\times\mathfrak{e}_{\bar{3}}, $$

$$ \mathfrak{e}^{\bar{i}}\equiv\frac{\mathfrak{e}_{\bar{2}}\times\mathfrak{e}_{\bar{3}}}{V}. $$

From standard tensor calculus, we may also express the volume element as the Jacobean determinant of the transformation matrix:

$$ \mathcal{E}_{ijk}\frac{\partial x^{i}}{\partial q^{\bar{i}}}\frac{\partial x^{j}}{\partial q^{\bar{j}}}\frac{\partial x^{k}}{\partial q^{\bar{k}}}=V\mathcal{E}_{\bar{i}\bar{j}\bar{k}}=\varepsilon_{\bar{i}\bar{j}\bar{k}}. $$

We now express the cross product of two barred basis vectors on the unbarred basis, and use the conventional transformation method to obtain the following:

$$ \mathfrak{e}_{\bar{2}}\times\mathfrak{e}_{\bar{3}}=\mathcal{E}_{ijk}\frac{\partial x^{j}}{\partial q^{\bar{2}}}\frac{\partial x^{k}}{\partial q^{\bar{3}}}\hat{\mathfrak{e}}^{i} $$

$$ =\mathcal{E}_{ijk}\frac{\partial x^{j}}{\partial q^{\bar{2}}}\frac{\partial x^{k}}{\partial q^{\bar{3}}}\frac{\partial x^{i}}{\partial q^{\bar{i}}}\mathfrak{e}^{\bar{i}} $$

$$ =V\mathcal{E}_{\bar{1}\bar{2}\bar{3}}\mathfrak{e}^{\bar{1}} $$

Which gives us back Menzel's definition. More generally we have

$$ \mathfrak{e}_{\bar{j}}\times\mathfrak{e}_{\bar{k}}=V\mathcal{E}_{\bar{i}\bar{j}\bar{k}}\mathfrak{e}^{\bar{i}}=\varepsilon_{\bar{i}\bar{j}\bar{k}}\mathfrak{e}^{\bar{i}}, $$

which is formally very similar to the first equation $\ast J_{\alpha\beta\gamma}=J^{\mu}\varepsilon_{\mu\alpha\beta\gamma}$ of (3.51) in MTW, Exercise 3.14.

If we replace the cross product with the wedge product, it seems that we could extend this definition of reciprocity to Minkowski 4-space. So, am I correct in concluding that the two uses of the term "dual" are not actually "entirely distinct," and in fact are closely related?

  • $\begingroup$ The Hodge dual is between forms and chains - or De Rham cohomology - see arxiv.org/abs/0807.4991. $\endgroup$ Commented Nov 2, 2019 at 20:30
  • $\begingroup$ I noticed that Menzel's book predates every source in the end note of that document. The only exception is Feynman's paper which predates the second edition. $\endgroup$ Commented Nov 2, 2019 at 21:02
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    $\begingroup$ Menzel's is using vector calculus which is useful in $3$ dimensions. If you're familiar with Stokes theorem, exterior derivative, closed and open differential forms, then De Rham cohomology is simply a generalization of vector calculus to any dimension - where chains are integral boundaries, e.g. $\oint_{\partial_{C}} \omega=\int_{C} d\omega$. $\endgroup$ Commented Nov 2, 2019 at 21:31
  • $\begingroup$ There are many different ways to arrive at the same results, such as the Riemann-Christoffel curvature tensor, and the generalized Stokes's theorem. Menzel's approach to extending his 3-D results to higher dimensions is: "Whenever $\times$ appears in any formula, we omit it, forming a tensor from the pair of base-vectors originally separated by the symbols." Page 127. But that's before he formally introduces the original form of Stokes's theorem. My point is that Menzel's definition of contravariant basis vectors looks a whole lot like a Hodge dual to me, contradicting MTW's assertion. $\endgroup$ Commented Nov 2, 2019 at 23:01
  • $\begingroup$ BTW, thanks for the link. It look well worth reading. $\endgroup$ Commented Nov 2, 2019 at 23:04

2 Answers 2


In three Euclidean dimensions basis duality and Hodge duality are identical. When a fourth dimension is introduced the identity bifurcates. One aspect of the duality becomes the identification of a volume 3-form (in 4-space) with a vector. The other is the condition $\mathbf{d}x^{\mu}\mathbf{e}_{\nu}=\delta^{\mu}_{\nu}.$

This is my rendering of Menzel's development for the case of Euclidean 3-space on rectangular Cartesian $x^i$ and general cruvilinear $q^i$ coordinates. The appearance of two basis vectors side-by-side is dyadic notation. Replace the cross with a wedge. The over-hat $\hat{\mathfrak{e}}_i$ indicates an orthonormal basis. All others are assumed general.

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  • 4
    $\begingroup$ This is absolutely not true. MTW are right; the use of "dual" as an adjective is distinct in both cases. "Dual" is "dual basis" refers to starting with a basis of a finite-dimensional vector space $V$, and arriving at a corresponding basis for $V^*:=\text{Hom}(V,\Bbb{F})$, where $\Bbb{F}$ is the underlying field (usually $\Bbb{R}$ in GR). The use of "dual" in "Hodge duality" is about using an inner product and orientation on a real vector space to provide isomorphisms $\bigwedge^k(V^*)\cong\bigwedge^{\dim V-k}(V^*)$. There's nothing in 3 dimensions which identifies these notions. $\endgroup$
    – peek-a-boo
    Commented Jul 2, 2022 at 11:03
  • $\begingroup$ See store.doverpublications.com/0486600564.html Part II Chapters 29 and 31. He introduces the dual basis of a three dimensional generalized coordinate system as reciprocal volume elements. The Hodge dual of a volume 3-form is a basis vector. $\endgroup$ Commented Jul 2, 2022 at 13:40
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    $\begingroup$ The Hodge dual of a volume 3-form is a number (if you're working on a single 3-dimensional vector space) or a smooth function (if you're working on the whole manifold). $\endgroup$
    – peek-a-boo
    Commented Jul 2, 2022 at 13:54
  • $\begingroup$ I was talking about a 3-form in 4-space. The volume element comes in when toggling indices. In 3-space $V=e_{1}\cdot e_{2}\wedge e_{3}$ and $e^{1}=\frac{e_{2}\wedge e_{3}}{V}.$ $\endgroup$ Commented Jul 3, 2022 at 9:52
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    $\begingroup$ Given a basis for $V$, one can construct the dual basis for $V^*$ with no additional input. On the other hand, you need both an inner product on $V$ and a choice of $\mathrm{dim}(V)$ form to serve as a volume form in order to define the Hodge duality between the elements of your basis for $V$ and the corresponding elements of $\bigwedge^{\mathrm{dim}(V)-1}V$. Is this not sufficient evidence that the two concepts are distinct? $\endgroup$
    – J. Murray
    Commented Jul 3, 2022 at 14:13

MTW (Gravitation by Misner, Thorne, and Wheeler) sec 2.7 makes no use of the volume element, while MTW sec 3.5 (Ex 3.14) does make use of the volume element.

The duality in MTW 2.7 is akin to the duality between points and [hyper]planes, column vectors and row vectors.

However, the Hodge-duality in MTW 3.5 (ex 3.14) is a different pairing using the volume element. It relates a column vector to an antisymmetric matrix, which have the same number of components only in 3 dimensions. The use of the volume element is associated with the notion of pseudo-quantities (like a pseudovector).

(I'm not going to comment about Menzel since the question in the title is about MTW and I would have to decode those sections in Menzel.)

Possibly useful resources:

  • $\begingroup$ I guess I will have to go spelunking through the first 15 Chapters of MTW to find the exact citations to support my argument. I hold Schouten's book in the highest esteem, though I haven't finished it. This really isn't that complicated. If you replace my $\mathfrak{e}^i\mapsto{dx^i}$ you will have things in the same form as they appear in MTW. I know what a Hodge dual is. $\star$ converts a representation on a p-form basis to an equivalent representation on the complementary (n-p)-polyvector basis; and vice-vera. $\endgroup$ Commented Jul 3, 2022 at 13:28
  • $\begingroup$ @StevenThomasHatton Schouten's book is good for indicating how two objects are somehow identified in the presence of additional structure (e.g. volume element, metric, etc). In 3-d Euclidean space, much of these identifications are blurred. So, in the MTW discussion of their two uses of duality, there are two different types of identifications. $\endgroup$
    – robphy
    Commented Jul 3, 2022 at 14:09

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