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!
Best Answer
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 $v\in T_pM$.