In Spivak's Calculus on Manifolds, he develops the volume element in the following way:
The fact that $\dim \Lambda^n(\mathbb{R}^n) = 1$ is probably not new to you, since det is often defined as the unique element $\omega \in \Lambda^n(\mathbb{R}^n)$ such that $\omega (e_1, \ldots, e_n) = 1$. For a general vector space $V$ there is no extra criterion of this sort to distinguish a particular $\omega \in \Lambda^n(V)$. 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 $A = (a_{ij})$ is defined by $w_i = \sum_{j=1} ^n a_{ij}v_j,$ then $$\delta_{ij} = T(w_i, w_j) = \sum_{k,l = 1} ^n a_{ik}a_{jl}T(v_k,v_l) =\sum_{k=1} ^na_{ik}a_{jk}.$$ In other words, if $A^T$ denotes the transpose of the matrix $A$, then we have $A \cdot A^T = I$, so $\det A = \pm 1$. It follows from Theorem 4-6 (stated below) that if $\omega \in \Lambda^n(V)$ satisfies $\omega (v_1, \ldots, v_n) = \pm 1$, then $\omega (w_1, \ldots, w_n) = \pm 1$. If an orientation $\mu$ for $V$ has also been given, it follows that there is a unique $\omega \in \Lambda^n(V)$ such that $\omega (v_1,\ldots, v_n) = 1$ whenever $v_1, \ldots,v_n$ is an orthonormal basis such that $[v_1, \ldots, v_n] = \mu.$ This unique $\omega$ is called the volume element of $V$, determined by the inner product $T$ and orientation $\mu$.
Theorem 4-6: Let $v_1, \ldots, v_n$ be a basis for $V$, and let $\omega \in \Lambda^n (V)$. If $w_i = \sum_{j = 1}^na_{ij}v_j$ are $n$ vectors in $V$, then $$\omega (w_1, \ldots, w_n) = det (a_{ij}) \cdot \omega (v_1, \ldots, v_n)$$
Two questions:
- Why is $\omega$ unique? Is it because $\dim \Lambda^n(V) = 1$ so every other element is $c \cdot \omega$ for some $c$?
- What exactly is the role of the inner product in determining the volume element? Is it so that an orthonormal basis can be determined? Why are orthonormal bases important in this development?
Best Answer
Yes, exactly, and $(c\cdot\omega)(e_1,\ldots,e_n)=c$ for all $c\in\Bbb{R}$, making this $\omega$ unique.
Orthonormality requires an inner product; given an inner product $T$ on a vector space $V$, two vectors $u,v\in V$ are orthonormal with respect to $T$ if $T(u,v)=0$ and $T(u,u)=T(v,v)=1$.
In an inner product space, a basis transformation between two orthonormal bases has determinant $\pm1$. Hence there exists precisely two elements $\omega_1,\omega_2\in\bigwedge^n\Bbb{R}^n$ such that $\omega_i(v_1,\ldots,v_n)=\pm1$ for every orthonormal basis $\{v_1,\ldots,v_n\}$. These two elements differ only in sign, i.e. $\omega_1=-\omega_2$.
If the inner product space is also given an orientation, there is precisely one element $\omega\in\bigwedge^n\Bbb{R}^n$ such that $\omega(v_1,\ldots,v_n)=1$ for every orthonormal basis $(v_1,\ldots,v_n)$ with positive orientation.