In a Riemannian space, what is the value of the canonical volume form on a frame? In particular, say, an orthonormal frame. One does not usually need to know anything about the value of a differential form on vectors, since this value does not enter into the definition of the integral of a differential form. But it seems that the scalar factors do not come out right.

I am getting inconsistencies with the standard texts on the subject, which are inconsistent with each other and with themselves. (Due to the pandemic, and living out of a suitcase, I only have access to Choquet-Bruhat, Dirac, Einstein, Weyl, Nomizu, Spivak, and Fomenko et al., except that via the Internet I have been able to consult Kobayashi and Nomizu as well as Cartan, Burali-Forti, and Weil. But neither Dirac nor Einstein discuss wedge products or the Hodge operation at all, while Cartan and Burali-Forti are just unreadable.)

So consider the simplest example: Cartesian coordinates x,y in the Euclidean plane $V$. Orthonormal frame for tangent bundle $ \frac \partial {\partial x}, \frac \partial {\partial y}$. (Global.) Dual frame, also global and orthonormal, for cotangent bundle $dx, dy$. Since the Euclidean metric (quadratic form, inner product, whatever) is being used, there is no charge for raising or lowering indices.

Now the path divides. The first issue is, which tensor does the formula $dx\wedge dy$ represent? Many authors (e.g., Choquet-Bruhat, pp. 55--56, Nomizu, pp. 10, 11, Kobayashi and Nomizu, pp. 28, 35, and Varadarajan, pp. 37--39) normalise the anti-symmetrisation of a tensor, which yields

$$ dx \wedge dy = \frac 12 (dx\otimes dy - dy\otimes dx).$$ When this is applied to the orthonormal dual basis we get

$$\frac 12 (dx\otimes dy)(\frac \partial {\partial x}, \frac \partial {\partial y}) - \frac12 (dy\otimes dx) (\frac \partial {\partial x}, \frac \partial {\partial y}) = \frac 12 (1-0) = \frac 12.$$

But the same authors always say that the volume form is $dx \wedge dy$ (Choquet-Bruhat, p. 92, Kobayashi and Nomizu, p. 283, etc.) and if they go on to mention its value on a frame, any frame (not all of them bother to mention this evaluation at all, and not that much hangs on it, but it is mentioned by Cartan, Weil, and Kobayashi and Nomizu, loc. cit.), they always say that this value is the volume of the parallelipiped spanned by the basis vectors. But by my calculation, presented above, this is the volume of the triangle, and for dx^dy ^dz evaluated on $(\frac \partial {\partial x}, \frac \partial {\partial y}, \frac \partial {\partial z})$, it is 1/6, which is the volume of the tetrahedron. That is, the normalising factor used in the antisymmetrisation comes out. (It is still true that integrating this volume form on a region in Euclidean space gives the usual volume---because one does not evaluate the form on any vectors in order to calculate this volume, one simply uses a coordinate patch to pull back the formula to Euclidean space and performs a classical calculation of the integral.)

So what is wrong with my calculation?

Others make an equally reasonable definition, without the factorial:

$$ dx \wedge dy = dx\otimes dy - dy\otimes dx.$$ This includes Spivak Calculus on Manifolds and Fomenko et al. Geometrie Contemporaine, premier partie (pp. 165, 167). Then their assertions about the value of the volume form on any basis at all being the volume of the parallelipiped are true.

(This does not preserve them from further related mistakes, since Cartan's formula for the exterior derivative $d\omega$ of, e.g., a one-form $\omega$, depends on which convention one has picked for $dx \wedge dy$. Spivak does not treat this formula, but Fomenko et al. on p. 233 do, and wind up being inconsistent with themselves since they use the same form of Cartan's formula as does Nomizu, which is incompatible with their different normalisation of dx ^ dy:

$$ d\omega (X,Y) = \frac 12 ( X\cdot Y(\omega) -Y\cdot X(\omega) - \omega([X,Y])),$$ where $X$ and $Y$ are vector fields and $[X,Y]$ is their Lie bracket, i.e., their commutator, namely, the vector field $XY-YX$. That is, when the antisymmetrisation is normalised, one needs this factor of 1/(p+1). But when it is not normalised, then this $\frac 12$ must be omitted, since the wedge product is now double what it used to be, which means $d$ is doubled too---note that the left hand side of this formula depends on the definition of the wedge product, but the right hand side does not.)

Spivak (Calculus on Manifolds) makes no mistakes but does not treat the Hodge star operation or Cartan's formula for $d\omega$---nor does he ever take a contraction. Varadarajan, both a distinguished mathematician and a careful writer, makes no mistakes on pp. 37--39, but does not treat any kind of metric structure, so there is no discussion of the canonical volume form or the Hodge star operation, which is what is relevant to me.

I cannot help wondering if I am making some mistake, but on the other hand, the mistakes I have detected in quite reputable sources are worse than simple typos.

Wolfram's encyclopedia makes no mistakes by not asserting that the value of the volume form on a frame is the volume of the parallelipiped, which of course would be equal to the determinant of the coefficients of the coordinates of the basis vectors spanning the parallelipiped, but simply pointing out that under coordinate changes, the coefficient of the volume form transforms by the Jacobian. This is a more important property than its having one value or the other.

But Springer-Verlag's (well, formerly theirs, they have donated it to the European Mathematical Union) Exterior product. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Exterior_product&oldid=46889 makes the same mistakes. The article on the Exterior Product clearly defines dx ^ dy as the alternation of $dx \otimes dy$, with a linked reference to their own article on Alternation, which clearly includes the normalising factor 1/p!. (And both articles were written by the same author, and the revision history is just trivial.) But the article on the exterior product clearly displays an even more mistaken formula about the wedge produce of two forms, although one hopes it is a mere typo: the denominator should not be p!q! but (p+q)!.

This is relevant to the Hodge star operation as a contraction with the volume form, used in Maxwell's Equations and many other places.


2 Answers 2


I am not sure to understand well your question. However, a point is that there are two inequivalent definitions of the wedge product which give rise to corresponding (naturally isomorphic) structures of associative alternate algebra in the space of the forms.

In my view, the most natural is, for $\omega \in \Lambda^p(V^*)$, $\eta \in \Lambda^q(V^*)$

$$\omega \wedge \eta := A_{p+q}(\omega \otimes \eta)$$

where $A_{n}$ is the antisymmetrisation projector of covariant tensors of order $n$. The other definition is

$$\omega \wedge \eta := \frac{(p+q)!}{p!q!}A_{p+q}(\omega \otimes \eta)\:.$$

Each produces an associative alternating algebra that is equally indicated by $\Lambda(V^*)$ (the linear space structure is the same, but the algebra product is different). These structures are naturally isomorphic but not the same.

I naively thought that the associativity postulate of $\wedge$ led to the first definition. Instead it also works with the second one!

When applying the two constructions to the volume form the former produces the annoying factors you noticed: $$dx^1 \wedge \cdots \wedge dx^n\left(\frac{\partial}{\partial x^1}, \ldots, \frac{\partial}{\partial x^n}\right) = \frac{1}{n!}\:.$$ The latter instead fournishes $$dx^1 \wedge \cdots \wedge dx^n\left(\frac{\partial}{\partial x^1}, \ldots, \frac{\partial}{\partial x^n}\right) =1\:.$$ The exterior derivative is defined through the same procedure in both approaches and produces two different notions, however sharing the same properties if one does not check the action on vectors. In that case factors $\frac{1}{n+1}$ pops up and distinguishes the two setups.

The problematic fact is that in some textbooks the authors start with a definition and -- when changing subject-- they sometimes pass to use the other definition.

I suspected the existence of this mess when I was dealing with the symplectic formalism. It is stated everywhere that $$\omega(X_f,X_g) = -\{f,g\} \quad \mbox{(or $+\{f,g\}$ depending on the conventions.)}$$ where $$\{f,g\}:= \sum_{k=1}^n\frac{\partial f}{\partial q^k}\frac{\partial g}{\partial p_k}-\frac{\partial f}{\partial p_k}\frac{\partial g}{\partial q^k}\:,$$ and $$X_f := \sum_{k=1}^n\frac{\partial f}{\partial q^k}\frac{\partial }{\partial p_k}-\frac{\partial f}{\partial p_k}\frac{\partial }{\partial q^k}$$ and also $$\omega = d \sum_{k=1}^n p_k dq^k= \sum_{k=1}^n dp_k \wedge dq^k\:.$$ It is clear that these formulae are consistent if $$dq^k\wedge dp_h\left(\frac{\partial}{\partial q^a},\frac{\partial}{\partial p_b}\right) = \delta^k_a\delta_h^b$$ and not $$dq^k\wedge dp_h\left(\frac{\partial}{\partial q^a},\frac{\partial}{\partial p_b}\right) = \frac{1}{2}\delta^k_a\delta_h^b\:.$$ I went crezy trying to solve this puzzle (I used to adopt the former definition which produces the wrong factor $1/2$) untill I discovered the existence of the two inequivalent definitions of $\wedge$.

Warner's textbook on differential geometry and Lie groups discusses the two general definitions of $\wedge$ (and finally adopts the latter). As far as I understand it is the only author who does.

  • 1
    $\begingroup$ Yes, I felt like I was going crazy, too. Now I know I'm not, at least, not for that reason. Thanks. $\endgroup$ Commented Dec 7, 2020 at 10:39

The bottom line is that (multi-)covectors are geometric objects in themselves, defined by specific geometric properties and axioms. Their space is isomporphic to an appropriate subpace of completely antisymmetric tensors – but there is no canonical isomorphism. This is the origin of the different normalization factors. In fact an infinite number of other normalization factors can be used (they're discussed in Abraham, Marsden, Ratiu if I remember correctly). In principle we don't need to bring in tensor products to discuss and work with multivectors and multicovectors, so no normalization factors at all are needed. This also means that there's no a priori, canonical relation between a tensor like $\mathrm{d}x \otimes\mathrm{d}y$, or any operations done on it, and an area (in a 2D space). A primitive notion of area is represented by the primitive object $\mathrm{d}x \land \mathrm{d}y$, and a relation with the tensor product can be made only after a somewhat arbitrary relation between the tensor and the multicovector is established.

This point of view is presented and advocated in some works by Lindell:

  • Lindell: Differential Forms in Electromagnetics (IEEE & Wiley 2004).

and Deschamps:

Deschamps (1970), § III.4.1, p. 119 indeed writes: "The exterior algebra can be introduced independently of tensors and this result is some economy of notation".

This point of view is also supported by Burke

and Bossavit:

who present the spaces and calculus of multivectors and multicovectors in a geometric and visual way, somewhat independently of tensor products.

This point of view also connects with Grassmann/Peano spaces, where these objects appear (in an affine-space, rather than vector-space, setting) without any mention of tensor and tensor products. Peano's book is worth a reading:

as well as some remarks by

I personally found everything easier once I studied these spaces and calculi in their own right, without thinking in terms of tensor products. I'd recommend to start with Burke (who's fun to read), then Bossavit, then Lindell and Deschamps.

Feel also free to take a look at the "Some extra info" part in this answer.

  • $\begingroup$ Thx. But we do need to choose an identification in order to talk about the contraction of one form against another, such as the statement, often repeated, that the Hodge star operation is contraction against the volume form. And no matter which normalising factor I use, I cannot make this statement work out literally. Maybe it's just "a kinda contraction" or it's "like a contraction", but not formally a contraction. $\endgroup$ Commented Dec 11, 2020 at 0:36
  • $\begingroup$ @josephf.johnson There's a difference between a tensor contraction and an interior product of multicovectors and multivectors. Maybe that's what confuses you. The star operator involves the interior product, not the tensor contraction. See this answer and the analysis of the star operator in Burke or Bossavit, cited above. $\endgroup$
    – pglpm
    Commented Dec 11, 2020 at 5:11
  • $\begingroup$ @josephf.johnson I re-read your comment and realized I didn't reply to a crucial point: no we don't need an identification to talk about the interior product (also called "inner product" and "dot product") of multivectors and forms. The texts referenced in my answer give definitions that don't rely on tensor products and general tensor contraction. See for example the (recursive) definitions given in the answer to the other post. I fully understand your perplexities because I had the exact same ones, until I stumbled upon Deschamp's, Lindell's, and Peano's works. $\endgroup$
    – pglpm
    Commented Dec 11, 2020 at 13:54

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