# Help understanding Riemann volume form

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!

• – rych
Jun 21 at 11:32

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 $$v\in T_pM$$.
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 $$G_2 = P^tG_1P$$.
• Thank you for answering, but I don't understand your second sentence. Why does $g=A^T I A$?