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From my lecture notes, on a general Riemannian manifold $(M, g)$, the divergence operator $\operatorname{div}: T^\infty(TM)\rightarrow C^\infty(M)$ is defined as $\operatorname{div}X := \star^{-1}d(x\neg\operatorname{Vol})$ , where $\star$ is the Hodge star operator $C^\infty\rightarrow T^\infty(\bigwedge^nT^\ast M)$ defined as $f\rightarrow f\operatorname{Vol}_g$.

Now I've read this from other place that there is another characterized definition of divergence operator given by:

For $\vec{F}\in C^\infty(U)$ with an open subset $U\subset X$, the function $\operatorname{div}(\vec{F})\in C^\infty$, $$d(\star w(\vec{F})) = \operatorname{div}(\vec{F})dV_X|_U$$

The detail of this definition can be found here.

Now my question. How does the definition given in my lecture imply the characterization property of divergence operator?

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up vote 0 down vote accepted

If $ \overline{F} = (f_1,f_2,...,f_n) $ then you can identify it locally with the vector field $ X = \sum_i f_i\partial_i $ and thus you have definitions for both of them identical via this. There are several equivalent definitions of divergence, like defining $ div(X)dV = d(i_X dV) $, trace of Levi-Civita connection, etc.

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