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Consider the space $\mathbb{R}^n$ and let $x_1,\ldots, x_n$ be the coordinates. Fix the orientation $dx_1\wedge dx_2\ldots\wedge dx_n$. Let $E^p$ denote the space of smooth $p$ forms and let $d$ denote the exterior derivation map from $E^p$ to $E^{p+1}$. Using the orientation, we get the Hodge star operator

$$*:E^p\to E^{n-p}$$

The Laplace-Beltrami operator is defined as $$\Delta:=d\delta+\delta d=(-1)^{n(p+1)+1}d*d* + (-1)^{np+1}*d*d$$ It can be easily checked that for a 0 form (smooth function) $f$, we have $$\Delta(f)=-\sum_{i=1}^n\frac{\partial^2 f}{\partial x_i^2}$$ It is an exercise in Warner's book (Foundations of Differentiable Manifolds and Lie Groups), last Chapter, exercise 6 to show that $$\Delta(fdx_I)=\Delta(f)dx_I$$ Does someone know a clean/neat way to do this problem. Thanks in advance.

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