Let M be an n–dimensional manifold (n ≥ 1). An s-form on M (s ≥ 1) is a covariant field α_{b1…bs} that is anti-symmetric (i.e., anti-symmetric in each pair of indices). The case where s = n is of special interest.

Let α_{b1…bn} be an n-form on M. Further, let ξ^{i}^{b }(i = 1,…,n) be a basis for the tangent space at a point in M with dual basis η^{i}_{b }(i=1,…,n). Then α_{b1…bn} can be expressed there in the form

α_{b1…bn} = k n! η^{1}[b_{1}…η^{n}b_{n}] —– (1)

where

k = α_{b1…bn}ξ^{1b1}…ξ^{nbn}

(To see this, observe that the two sides of equation (1) have the same action on any collection of n vectors from the set {ξ^{1b}, . . . , ξ^{nb}}.) It follows that if α_{b1…bn} and β_{b1…bn} are any two smooth, non-vanishing n-forms on M, then

β_{b1…bn} = f α_{b1…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 (γ⃗_{1})a, . . . , (γ⃗_{n})a, and suppose η^{i}_{b}(i = 1, . . . , n) are dual fields. Then η^{1}[b_{1}…η^{n}b_{n}] 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 α_{ab} is any smooth two-form on M. We see that α_{ab} must vanish somewhere as follows.

*A 2-form α _{ab} 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 α_{ab} ≠ 0, and let ξ^{a} be any non-zero vector at p. Consider the number α_{ab} ξ^{a} ρ^{b} as ρ^{b} rotates though the vectors in M_{p}. If ρ^{b} = ±ξ^{b}, the number is zero. If ρ^{b} ≠ ±ξ^{b}, the number is non-zero. Therefore, as ρ^{b} rotates between ξ^{a} and −ξ^{a}, it is always positive or always negative. Thus α_{ab} determines a “positive direction of rotation” away from ξ^{a} on M_{p}. α^{ab} 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 g_{ab} of signature (n^{+}, n^{−}). We take a volume element on M (with respect to g_{ab}) to be a smooth n-form ε_{b1…bn} that satisfies the normalization condition

ε^{b1…bn} ε_{b1…bn} = (−1)^{n−}n! —– (2)

Suppose ε_{b1…bn} is a volume element on M, and ξ^{i 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),

ε_{b1…bn} = k n! ξ^{1}_{[b1} …ξ_{bn}] —– (3)

where

k = ε_{b1…bn }ξ^{1b1}…ξ^{nbn}

Hence, by the normalization condition (2),

(−1)^{n−}n! = (k n! ξ^{1}_{[b1} …ξ_{bn]}) (k n! ξ^{1}^{[b1} …ξ^{bn}])

= k^{2} n!^{2} 1/n! (ξ^{1}_{b1} ξ^{1}^{b1}) … (ξ^{n}_{bn} ξ^{n}^{bn}) = k^{2} (−1)^{n−}

So k^{2} = 1 and, therefore, equation (3) yields

ε_{b1…bn }ξ^{1b1}…ξ^{nbn} = ±1 —– (4)

Clearly, if ε_{b1…bn} is a volume element on M, then so is −ε_{b1…bn}. It follows from the normalization condition (4) that there cannot be any others. Thus, there are only two possibilities. Either (M, g_{ab}) 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 γ^{1a} , . . . , γ^{na} at a point, we can think of ε_{b1…bn} γ^{1}^{b1} … γ^{n}^{bn} as the volume of the (possibly degenerate) parallelepiped determined by the vectors. Notice that, up to sign, ε_{b1…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 g_{ab}). 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 ξ^{a}, ε_{b1…bn} ξ^{b1} ξ^{b2} = 0, then it must be the case that ε_{b1 …bn} is anti-symmetric in indices (b_{1}, b_{2}). 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 ε_{b1…bn} as a volume measure is that it respects orientation. If it assigns V to the ordered sequence γ^{1a} , . . . , γ^{na}, then it assigns (−V) to γ^{2a}, γ^{1a}, γ^{3a},…,γ^{na}, and so forth.