# 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?

• determinant of what? – Lord Shark the Unknown Jul 8 '18 at 10:05
• Determinant of the metric. – Higgsino Jul 8 '18 at 10:06
• I guess "vol" should be a coordinate independent object, right? – Higgsino Jul 8 '18 at 10:08
• Theres a good explanation of this in Baez & Munians Knots & Gravity - I was looking at it today. – Mozibur Ullah Jul 8 '18 at 12:33
• Look up the Gram determinant. The purpose of the Riemannian volume form is that if $X_1(p), X_2(p), \cdots, X_n(p)$ is a basis of orthonormal vectors in the inner product space $T_p M$ with inner product $g_p$, then $\text{vol}(X_1(p), \cdots, X_n(p)) = 1$. – Balarka Sen Jul 8 '18 at 13:06

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 $\big\{\frac{\partial}{\partial x^1},...,\frac{\partial}{\partial x^n}\big\}$.