Volume form on a manifold. On a manifold M, we define the volume form to be
$vol = \sqrt{|det(g)|}dx^{1}\wedge...\wedge dx^{n}$. Where g is the metric.
But I don't really understand this definition. Why the square-root of the determinant?
 A: A Riemannian manifold has a canonically defined volume form, which is given by the above form $\text{vol}=\sqrt{\text{det}\,g}\,\text{d}x^1\wedge...\wedge\text{d}x^n$:
Given an arbitrary, positively oriented, chart $(U,x^1,...,x^n)$, apply the Gram-Schmidt process to the coordinate frame $\big\{\frac{\partial}{\partial x^1},...,\frac{\partial}{\partial x^n}\big\}$ to get an orthonormal frame, and then let $\{\theta^1,...,\theta^n\}$ be the frame dual to this orthonormal frame.  Then
$$
\omega=\theta^1\wedge...\wedge\theta^n
$$
is a nowhere vanishing $n$-form on $U$ which does not depend on the choice of positively oriented chart.  This means that if $(V,y^1,...,y^n)$ is another positively oriented, overlapping chart with coordinate frame $\big\{\frac{\partial}{\partial y^1},...,\frac{\partial}{\partial y^n}\big\}$, then by applying the Gram-Schmidt process to get an orthonormal frame, and letting $\{\alpha^1,...,\alpha^n\}$ be the frame dual to the orthonormal frame, we also have that $\omega=\alpha^1\wedge...\wedge\alpha^n$ on the intersection $U\cap V$.
We can then show that $\omega$ coincides with the volume form given above:
$$
\omega=\text{vol}=\sqrt{\text{det}\,g}\,\text{d}x^1\wedge...\wedge\text{d}x^n,
$$
where $\{\text{d}x^1,...,\text{d}x^n\}$ is the frame dual to $\big\{\frac{\partial}{\partial x^1},...,\frac{\partial}{\partial x^n}\big\}$.
