Ruminations on Philosophy of Science: A Case of Volume Measure Respecting Orientation

Let M be an n–dimensional manifold (n ≥ 1). An s-form on M (s ≥ 1) is a covariant field αb1…bs that is anti-symmetric (i.e., anti-symmetric in each pair of indices). The case where s = n is of special interest.

Let αb1…bn be an n-form on M. Further, let ξi(i = 1,…,n) be a basis for the tangent space at a point in M with dual basis ηi(i=1,…,n). Then αb1…bn can be expressed there in the form

αb1…bn = k n! η1[b1…ηnbn] —– (1)


k = αb1…bnξ1b1…ξnbn

(To see this, observe that the two sides of equation (1) have the same action on any collection of n vectors from the set {ξ1b, . . . , ξnb}.) It follows that if αb1…bn and βb1…bn are any two smooth, non-vanishing n-forms on M, then

βb1…bn = f αb1…bn

for some smooth non-vanishing scalar field f. Smooth, non-vanishing n-forms always exist locally on M. (Suppose (U, φ) is a chart with coordinate vector fields (γ⃗1)a, . . . , (γ⃗n)a, and suppose ηib(i = 1, . . . , n) are dual fields. Then η1[b1…ηnbn] qualifies as a smooth, non-vanishing n-form on U.) But they do not necessarily exist globally. Suppose, for example, that M is the two-dimensional Möbius strip, and αab is any smooth two-form on M. We see that αab must vanish somewhere as follows.


A 2-form αab on the Möbius strip determines a “positive direction of rotation” at every point where it is non-zero. So there cannot be a smooth, non-vanishing 2-form on the Möbius strip.

Let p be any point on M at which αab ≠ 0, and let ξa be any non-zero vector at p. Consider the number αab ξa ρb as ρb rotates though the vectors in Mp. If ρb = ±ξb, the number is zero. If ρb ≠ ±ξb, the number is non-zero. Therefore, as ρb rotates between ξa and −ξa, it is always positive or always negative. Thus αab determines a “positive direction of rotation” away from ξa on Mp. αab must vanish somewhere because one cannot continuously choose positive rotation directions over the entire Möbius strip.

M is said to be orientable if it admits a (globally defined) smooth, non- vanishing n-form. So far we have made no mention of metric structure. Suppose now that our manifold M is endowed with a metric gab of signature (n+, n). We take a volume element on M (with respect to gab) to be a smooth n-form εb1…bn that satisfies the normalization condition

εb1…bn εb1…bn = (−1)nn! —– (2)

Suppose εb1…bn is a volume element on M, and ξi b (i = 1,…,n) is an orthonormal basis for the tangent space at a point in M. Then at that point we have, by equation (1),

εb1…bn = k n! ξ1[b1 …ξbn] —– (3)


k = εb1…bn ξ1b1…ξnbn

Hence, by the normalization condition (2),

(−1)nn! = (k n! ξ1[b1 …ξbn]) (k n! ξ1[b1 …ξbn])

= k2 n!2 1/n! (ξ1b1 ξ1b1) … (ξnbn ξnbn) = k2 (−1)n

So k2 = 1 and, therefore, equation (3) yields

εb1…bn ξ1b1…ξnbn = ±1 —– (4)

Clearly, if εb1…bn is a volume element on M, then so is −εb1…bn. It follows from the normalization condition (4) that there cannot be any others. Thus, there are only two possibilities. Either (M, gab) admits no volume elements (at all) or it admits exactly two, and these agree up to sign.

Condition (4) also suggests where the term “volume element” comes from. Given arbitrary vectors γ1a , . . . , γna at a point, we can think of εb1…bn γ1b1 … γnbn as the volume of the (possibly degenerate) parallelepiped determined by the vectors. Notice that, up to sign, εb1…bn is characterized by three properties.

(VE1) It is linear in each index.

(VE2) It is anti-symmetric.

(VE3) It assigns a volume V with |V | = 1 to each orthonormal parallelepiped.

These are conditions we would demand of any would-be volume measure (with respect to gab). If the length of one edge of a parallelepiped is multiplied by a factor k, then its volume should increase by that factor. And if a parallelepiped is sliced into two parts, with the slice parallel to one face, then its volume should be equal to the sum of the volumes of the parts. This leads to (VE1). Furthermore, if any two edges of the parallelepiped are coalligned (i.e., if it is a degenerate parallelepiped), then its volume should be zero. This leads to (VE2). (If for all vectors ξa, εb1…bn ξb1 ξb2 = 0, then it must be the case that εb1 …bn is anti-symmetric in indices (b1, b2). And similarly for all other pairs of indices.) Finally, if the edges of a parallelepiped are orthogonal, then its volume should be equal to the product of the lengths of the edges. This leads to (VE3). The only unusual thing about εb1…bn as a volume measure is that it respects orientation. If it assigns V to the ordered sequence γ1a , . . . , γna, then it assigns (−V) to γ2a, γ1a, γ3a,…,γna, and so forth.