The volume element

The fact that $\operatorname{dim}\Lambda^n\left(\mathbb{R}^n\right)=1$ is probably not new to you, since $\operatorname{det}$ is often defined as the unique element $\omega\in\Lambda^n{\left(\mathbb{R}^n\right)}$ such that $\omega{\left(e_1,\ldots,e_n\right)}=1$. For a general vector space $V$ there is no extra criterion of this sort to distinguish a particular $\omega\in\Lambda^n{\left(\mathbb{R}^n\right)}$. Suppose, however, that an inner product $T$ for $V$ is given. If $v_1,\ldots,v_n$ and $w_1,\ldots, w_n$ are two bases which are orthonormal with respect to $T$, and the matrix $A=\left(a_{ij}\right)$ is defined by $w_i=\sum_{j=1}^n a_{ij}v_j$, then $$\delta_{ij}=T{\left(w_i,w_j\right)}= \sum_{k,l=1}^n a_{ik}a_{jl}\,T{\left(v_k,v_l\right)}= \sum_{k=1}^n a_{ik}a_{jk}.$$ \noindent In other words, if $A^T$ denotes the transpose of the matrix $A$, then we have $A\cdot A^T=I$, so $\operatorname{det}A=\pm 1$. It follows from Theorem 4-6 [see vignette `det.Rmd`] that if $\omega\in\Lambda^n(V)$ satisfies $\omega{\left(v_1,\ldots,v_n\right)}=\pm 1$, then $\omega{\left(w_1,\ldots,w_n\right)}=\pm 1$. If an orientation $\mu$ for $V$ has also been given, it follows that there is a unique $\omega\in\Lambda^n(V)$ such that $\omega\left(v_1,\ldots,v_n\right)=1$ whenever $v_1,\ldots,v_n$ is an orthornormal basis such that $\left[v_1,\ldots,v_n\right]=\mu$. This unique $\omega$ is called the *volume element* of $V$, determined by the inner product $T$ and orientation $\mu$. Note that $\operatorname{det}$ is the volume element of $\mathbb{R}^n$ determined by the usual inner product and usual orientation, and that $\left|\operatorname{det}\left(v_1,\ldots,v_n\right)\right|$ is the volume of the parallelepiped spanned by the line segments from $0$ to each of $v_1,\ldots,v_n$.

- Michael Spivak, 1969 (Calculus on Manifolds, Perseus books). Page 83