The oriented unit volume form

Let $V$ be an oriented finite-dimensional real inner product space with $\dim V=n\ge 1$ (Orientation of a finite-dimensional real vector space). Choose a positively oriented orthonormal basis $(b_1,\dots ,b_n)$, which exists by Every finite-dimensional real or complex inner product space has an orthonormal basis, and define the oriented unit volume form

$\omega :=b_1\wedge \cdots \wedge b_n\in {\mathrm{\Lambda }}^{n}V\mathrm{.}$

This is independent of the choice. Any other positively oriented orthonormal basis $(b_1^{\mathrm{\prime }},\dots ,b_n^{\mathrm{\prime }})$ is obtained from $(b_1,\dots ,b_n)$ by a linear map $Q$ that carries one orthonormal basis to another, hence is an orthogonal operator by For an endomorphism in finite dimension, preserving lengths, preserving inner products, carrying orthonormal bases to orthonormal bases, and $T^{\ast }T=I$ are equivalent; over the reals its determinant is therefore $\pm 1$ by Orthogonal and unitary operators form groups, and their determinants have modulus one. Positivity of the change of basis forces $\det Q=1$, and by On ${\mathrm{\Lambda }}^{n}V$, the induced map ${\mathrm{\Lambda }}^{n}T$ is multiplication by $\det T$,

$b_1^{\mathrm{\prime }}\wedge \cdots \wedge b_n^{\mathrm{\prime }}={\mathrm{\Lambda }}^{n}Q(b_1\wedge \cdots \wedge b_n)=\det Q\cdot (b_1\wedge \cdots \wedge b_n)=\omega \mathrm{.}$

The form is a unit. By the Gram pairing of The Gram formula gives a well-defined positive-definite inner product on exterior powers, and $\mathrm{\parallel }{v}_1\wedge \cdots \wedge v_k{\mathrm{\parallel }}^2$ is the Gram determinant, $⟨\omega ,\omega ⟩=\det (⟨b_i,b_j⟩)=\det I_n=1$. For $n=0$, set $\omega :=1\in {\mathrm{\Lambda }}^{0}V=\mathbb{R}$.

Reversing the orientation replaces $\omega$ by $-\omega$; the oriented unit volume form is where the orientation enters the Hodge-star data.