Help understanding Riemann volume form

Question

Let $(M, g)$ be an oriented Riemannian manifold, and let $p \in M$. Suppose $v_1, ..., v_n$ and $e_1, ..., e_n$ are positively oriented bases for $T_p M$, with $e_1, ..., e_n$ orthonormal. Then at $p$ the Riemann volume form is defined by

\begin{align} (dV)_p = e_1^* \wedge \cdots \wedge e_n^*. \end{align}

I am having trouble understanding why

\begin{align} (dV)_p = \sqrt{\det (g)} v_1^* \wedge \cdots \wedge v_n^*, \end{align}

where $g_{ij} = \langle v_i, v_j \rangle$. I understand that if $A$ is the change of basis matrix from $v_1, ..., v_n$ to $e_1, ..., e_n$, then

\begin{align} (dV)_p = \sqrt{\det (A^T A)} v_1^* \wedge \cdots \wedge v_n^*, \end{align}

so all that's left is showing that $A^T A = g$. I'm pretty sure we can write $A = [\langle e_i, v_j \rangle]$, which would suggest that

\begin{align} A^T A = \bigg[ \sum_{k=1}^n \langle e_j, v_k \rangle \langle e_i, v_k \rangle \bigg] \end{align}

So does this sum equal $g_{ij}$? Or have I made a mistake along the way. I fear I may have overcomplicated matters for myself... any help would be greatly appreciated!

Answer 1

The expression you wrote down for $A^TA$ is $AA^T$ instead. The correct expression for $A^TA$ is that its $ij$ entry is $\sum_k\langle e_k,v_i\rangle\langle e_k,v_j\rangle.$ This is equal to $\langle v_i,v_j\rangle$ since the $e_1,\dots,e_n$ basis is orthonormal so $v=\sum_k\langle e_k,v\rangle e_k$ for any $v\in T_pM$.

Answer 2

Take a point $p$, and let $A_p$ denote the transition matrix of the two above mentioned frames of the tangent space at $p$. Since the metric matrix respect an orthonormal basis is just the identity, we have by construction that $$(g_{ij})=A_p^t\cdot I\cdot A_p\implies \operatorname{det} A_p=\sqrt{\operatorname{det}(g_{ij})}.$$ In general, given an $n$-dimensional linear space $V$ endowed with inner product $(\alpha, \beta)$, suppose the Gramian matrices of $(\alpha, \beta)$ under two different bases $\{\alpha_1, \ldots, \alpha_n\}$ and $\{\beta_1, \ldots, \beta_n\}$ are $G_1$ and $G_2$, and $$(\beta_1, \ldots, \beta_n) = (\alpha_1, \ldots, \alpha_n)P \tag{1}$$ for some non-singular matrix $P$, then $G_2 = P^tG_1P$.