When applying a differential \(k\)-form \(\alpha(x) = \alpha_1(x) \wedge \cdots \wedge \alpha_k(x)\) to a \(k\)-vector field \(u(x) = u_1(x) \wedge \cdots \wedge u_k(x)\), swapping any pair of component differential \(1\)-forms in the \(k\)-form, i.e. \((\alpha_i(x), \alpha_j(x))\), or any pair of component vectors in the \(k\)-vector field, i.e. \((u_i(x), u_j(x))\), will change the sign of the resulted scalar value \(\alpha(u)\). This is because both the differential \(1\)-forms \(\{\alpha_1(x), \cdots, \alpha_k(x)\}\) in \(\alpha(x)\) and the vectors \(\{u_1(x), \cdots, u_k(x)\}\) in \(u(x)\) are ordered sets, the permutation of which defines an orientation.

The resulted value \(\alpha(u)\) is actually a determinant1

\[\alpha(u) = \begin{vmatrix} \alpha_1(u_1) & \cdots & \alpha_1(u_k) \\ \vdots & \vdots & \vdots \\ \alpha_k(u_1) & \cdots & \alpha_k(u_k) \end{vmatrix},\]

where each column is a projection of one component vector in \(u(x)\) to the differential \(k\)-form. Also the determinant itself is anti-symmetric, which is consistent with the rule for evaluation of \(\alpha(u)\). Therefore, the signed volume of the parallelogram spanned by the set of projected component vectors \(\left\{ (\alpha_1(u_i), \cdots, \alpha_k(u_i)) \right\}_{i=1}^{k}\) is the geometric meaning of determinant.

  1. The following display equation is wrapped around \begin{equation*} ... \end{equation*} for correct visualization in Emacs Org mode. ↩