Proposition

If $w_1, \cdots, w_{n - 1} \in \mathbb{R}^n$, show that

\[\begin{align*} \abs{w_1 \times \cdots \times w_{n - 1}} = \sqrt{\det(g_{ij})}, \end{align*}\]

where $g_{ij} = \ev{w_i, w_j}$.

Solution

Suppose $w_1, \cdots, w_{n - 1}$ are linearly dependent. Then $w_1 \times \cdots \times w_{n - 1} = 0$ because $\ev{w_1 \times \cdots \times w_{n - 1}, v} = 0$ for any $v$.

Let $a_1, \cdots, a_{n - 1} \in \mathbb{R}$ be chosen such that they are not all 0 and $a_1w_1 + \cdots + a_{n - 1}w_{n - 1} = 0$. Then for each $i$,

\[\begin{align*} 0 &= \ev{a_1w_1 + \cdots + a_{n - 1}w_{n - 1}, w_i} \\ &= a_1\ev{w_1, w_i} + \cdots + a_{n - 1}\ev{w_{n - 1}, w_i} \\ &= a_1g_{1, i} + \cdots + a_{n - 1}g_{n - 1, i}. \end{align*}\]

Therefore,

\[\begin{align*} a_1\begin{bmatrix} g_{1,1} \\ \vdots \\ g_{1,n - 1} \end{bmatrix} + \cdots + a_{n - 1}\begin{bmatrix} g_{n - 1,1} \\ \vdots \\ g_{n - 1,n - 1} \end{bmatrix} = 0. \end{align*}\]

Thus $\det(g_{ij}) = 0$, so we are done.

Suppose $w_1, \cdots, w_{n - 1}$ are linearly independent. Let $V$ be the $(n - 1)$-dimensional vector space spanned by $w_1, \cdots, w_{n - 1}$. Let $w_n = \frac{w_1 \times \cdots \times w_{n - 1}}{\abs{w_1 \times \cdots \times w_{n - 1}}}$. Define

\[\begin{align*} \phi(x_1, \cdots, x_{n - 1}) = \det \begin{bmatrix} x_1 \\ \vdots \\ x_{n - 1} \\ w_n \end{bmatrix}. \end{align*}\]

Then $\phi$ is clearly multi-linear and alternating, so $\phi \in \Lambda^{n - 1}(V)$. Moreover,

\[\begin{align*} \phi(w_1, \cdots, w_{n - 1}) &= \det \begin{bmatrix} w_1 \\ \vdots \\ w_{n - 1} \\ w_n \end{bmatrix} \\ &= \ev{w_1 \times \cdots \times w_{n - 1}, w_n} \\ &= \frac{\ev{w_1 \times \cdots \times w_{n - 1}, w_1 \times \cdots \times w_{n - 1}}}{\abs{w_1 \times \cdots \times w_{n - 1}}} \\ &= \abs{w_1 \times \cdots \times w_{n - 1}}. \end{align*}\]

Thus, by Problem 4-3(Spivak), it suffices to show that $\phi$ is a volume element determined by $\ev{,}$ and some orientation $\mu$.

Since $V$ is an $(n - 1)$-dimensional vector space, there exists an orthonormal basis $u_1, \cdots, u_{n - 1}$ with respect to the usual inner product $\ev{,}$. Let $i$ be given. Then $\ev{u_i, w_n} = \ev{\sum_{j=1}^{n-1} a_jw_j, w_n} = \sum_{j=1}^{n-1} a_j\ev{w_j,w_n} = 0$. Moreover, $\ev{w_n, w_n} = 1$. Therefore, $\{ u_1, \cdots, u_{n - 1}, w_n \}$ is an orthonormal basis of $\mathbb{R}^n$ with respect to $\ev{,}$. This implies that $\phi(u_1, \cdots, u_{n - 1}, w_n) = \pm 1$ because $\begin{bmatrix} u_1 \\ \vdots \\ u_{n - 1} \\ w_n \end{bmatrix}$ is an orthogonal matrix, and the determinant of every orthogonal matrix is $\pm 1$.

If $\phi(u_1, \cdots, u_{n - 1}, w_n) = -1$, then set $u_1$ to be $-u_1$. Then $\phi(u_1, \cdots, u_{n - 1}, w_n) = 1$. We claim that $\phi$ is the volume element determined by $\ev{,}$ and $\mu = [u_1, \cdots, u_{n - 1}]$.

Let $v_1, \cdots, v_{n - 1}$ be an orthonormal basis of $V$ with $[v_1, \cdots, v_{n - 1}] = [u_1, \cdots, u_{n - 1}]$. Using the exact same argument as above, $\phi(v_1, \cdots, v_{n - 1}) = \pm 1$. By the discussion on PP.82-83 on Spivak, the sign of $\phi(v_1, \cdots, v_{n - 1})$ is equal to the sign of $\phi(u_1, \cdots, u_{n - 1})$. Therefore, $\phi(v_1, \cdots, v_{n - 1}) = 1$.

Hence, $\phi$ is indeed the volume element determined by $\ev{,}$ and $\mu$. By Problem 4-3(Spivak), $\abs{w_1 \times \cdots \times w_{n - 1}} = \sqrt{\det(g_{ij})}$.