# Confusion over the definition of a volume element

I'm fairly new to differential geometry (currently self-teaching) and I'm a bit confused over the definition of a volume form.

I've read that, given an $m$-dimensional manifold, a top-form $\omega\in\Omega^{m}(M)$ is an $m$-dimensional differential form that in a local coordinate chart $(U,\phi)$, $\phi (p)=x$, is given by $$\omega =f(x)dx^{1}\wedge\cdots\wedge dx^{m}$$ and according to some texts that I've read, this can be used as a volume form. This is the first point I'm unsure about; how is this related to the notion of a volume element in "ordinary" integral calculus in $\mathbb{R}^{n}$ (i.e. $dV=f(x^{1},\ldots,x^{n})dx^{1}\cdots dx^{n}$), is there simply some correspondence of the form $$f(x)dx^{1}\wedge\cdots\wedge dx^{m}\leftrightarrow f(x)dx^{1}\cdots dx^{m}$$ Does this also carry over for a change of coordinates in which the volume element changes by the absolute value of the Jacobian (determinant), $J$? That is, given two coordinate systems $(x^{1},\ldots,x^{n})$ and $(y^{1},\ldots, y^{n})$, it follows that in $\mathbb{R}^{n}$, $$f(x)dx^{1}\cdots dx^{m}=f(x(y))\lvert J\rvert dy^{1}\cdots dy^{n}$$ where $J=\text{det}\left(\frac{\partial x^{i}}{\partial y^{j}}\right)$. Is there a similar correspondence, $$f(x)dx^{1}\wedge\cdots\wedge dx^{m}=f(x(y))\lvert\text{det}\left(\frac{\partial x^{\mu}}{\partial y^{\nu}}\right)\rvert dy^{1}\wedge\cdots\wedge dy^{m}\leftrightarrow f(x(y))\lvert J\rvert dy^{1}\cdots dy^{n}$$

Secondly, I've also read that in Riemannian geometry the volume element can be expressed as $$dVol=\sqrt{\lvert g\rvert}dx^{1}\wedge\cdots\wedge dx^{m}$$ where $g$ is the determinant of the metric $g_{\mu\nu}$ defined on the manifold. In what I've read so far this is simply stated as the definition of a volume form, but is there an actually proof of why it has this particular form?

## 1 Answer

On ${\bf R}^n$ it has a volume form $\omega:= dx^1\cdots dx^n$ Simply volume form at a fixed point has a value $1$ on oriented orthonormal basis : If $e_i$ is orthonormal then $\omega(e_1,\cdots, e_n)=1$.

If $f$ is coordinate chart for $M$, then define a coordinate vector field $E_i=df\ e_i$. Then we have a coframe $E^i$ : $E^i(E_j)=\delta_{ij}$. If $g_{ij}:= g(E_i,E_j)$ then we have a volume form $$\omega:=\sqrt{{\rm det}\ g_{ij}} E^1\cdots E^n$$

Assume that $v_i:=A_{ij}E_j$ is orthonormal wrt $g$ so that $$\delta_{ij}=g(v_i,v_j)= A_{ik}A_{jm}g_{km}=(A gA^T)_{ij}$$ where $T$ is a transpose

Then $$\omega (v_1,\cdots, v_n) = \sqrt{{\rm det}\ g_{ij}} {\rm det} \ A_{ij}=1$$