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I have my blank papers out and I am planning on showing that

$\displaystyle{\nabla^2 = \dfrac{1}{\sqrt{|g|}}\dfrac{\partial\left(\sqrt{|g|} g^{kl} \dfrac{\partial}{\partial u^l}\right)}{\partial u^k}}$

It would be nice if someone can sprinkle hints on how to go about this, or perharps just show this.

Ok, off to try and see if I can do this. Hoping for responses.

Sine of the Time
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Kevin Njokom
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  • Start with the Voss-Weyl formula, and then apply it to the special case where the vector field is the metric-gradient of a function. – peek-a-boo Nov 14 '22 at 02:58
  • This is only true on scalars. For example,$$\begin{align}\nabla^2V^a-\frac{\partial_b(\sqrt{|g|}g^{bc}\partial_cV_a)}{\sqrt{|g|}}&=g^{bc}(\partial_c\Gamma_{be}^a+\Gamma_{cd}^a\Gamma_{be}^d-\Gamma_{bc}^d\Gamma_{de}^a)V^e\&+(g^{bf}(\Gamma_{bg}^a-\delta_g^a\Gamma_{be}^e)-g^{bc}\delta_g^a\Gamma_{bc}^f-\delta_g^a\partial_bg^{bf})\partial_fV^g.\end{align}$$ – J.G. Nov 14 '22 at 09:19
  • What is $\nabla^2$ here? If $\nabla^2$ is the Laplacian $\text{div} \circ \text{grad}$, then you can get the formula by integrating by parts against a test function. – Mason Nov 15 '22 at 03:06

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