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Why is the volume element equal to the square root of the absolute value of the metric tensor determinant, i.e. $$dV=\sqrt{\left | g \right |} dx^0dx^1dx^2$$

BinaryBurst
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  • By definition? In general, you can choose a volume form completely independent from the metric. So if you want to "prove" the formula above, you need to have some additional premise, some externally imposed condition. You should probably tell us more about the situation, give us some context. However, if the formula in your question is the definition of the volume form, then it still remains to prove hat it doesn't depend on the choice of the local coordinate chart (at least when the choice is consistent with the chosen orientation). – Dan Shved Mar 25 '14 at 09:38
  • !? What do dot products between bases have to do with the volume element? – BinaryBurst Mar 25 '14 at 10:54
  • Yes, what do they have to do with each other? In an abstract setting, if you have a smooth oriented manifold, you can choose a metric and a volume form absolutely independent from each other. But maybe you're not in an abstract setting. Maybe you are asking about the standard space $\mathbb{R}^3$ with the standard metric and the standard volume, just non-standard coordinates. We need to know your setting to give an appropriate answer. – Dan Shved Mar 25 '14 at 11:41

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First, the definitions. Let $e_1,\ldots,e_n$ be an orthonormal basis, i.e. $\langle e_i,e_j\rangle=\delta_{ij}$, and $\theta_1,\ldots,\theta_n$ its dual, i.e. $\theta_i(e_j)=\delta_{ij}$. The volume element is defined by $vol:=\theta_1\wedge\ldots\wedge\theta_n$.

The rest is a simple calculation. If you change basis, say $e_i=\sum_j a_{ij} e'_j,$ with dual basis $\theta'_i$, then $\theta_i=\sum_j b_{ij}\theta'_j,$ where the change of basis matrices $A=(a_{ij}), B=(b_{ij})$, satisfy $AB^t=I$. Let $g_{ij}=\langle e'_i,e'_j\rangle.$ Then the matrix $G=(g_{ij})$ satisfies $I=AGA^t$. Now plug-in $\theta_i=\sum_j b_{ij}\theta'_{j} $ into $vol=\theta_1\wedge\ldots\wedge\theta_n$ and get $vol=\det(B)\theta'_1\wedge\ldots\wedge\theta'_n$. Taking determinants of the matrix equations $AB^t=AGA^t=I$, you get $g:=det(G)=det(B)^2.$

Note: there is a sign issue, having to do with orientation, but I guess this is not your concern here.

Gil Bor
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