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?
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$\begingroup$ determinant of what? $\endgroup$– Angina SengJul 8, 2018 at 10:05
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$\begingroup$ Determinant of the metric. $\endgroup$– HiggsinoJul 8, 2018 at 10:06
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$\begingroup$ I guess "vol" should be a coordinate independent object, right? $\endgroup$– HiggsinoJul 8, 2018 at 10:08
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$\begingroup$ Theres a good explanation of this in Baez & Munians Knots & Gravity - I was looking at it today. $\endgroup$– Mozibur UllahJul 8, 2018 at 12:33
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$\begingroup$ 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 $\text{vol}(X_1(p), \cdots, X_n(p)) = 1$. $\endgroup$– Balarka SenJul 8, 2018 at 13:06
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1 Answer
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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 $\big\{\frac{\partial}{\partial x^1},...,\frac{\partial}{\partial x^n}\big\}$.
