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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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1 Answer 1

As usual, the trick is to use representation in local coordinates. Note first that using the linearity it is sufficient to show the statement holds for $\omega := f \mathrm d x^I$ for any increasing multi index $I$. Then we have to show that $$(\mathrm d\delta + \delta \mathrm d)\omega = (\Delta f) \mathrm dx^I.$$ We get

\begin{align*} \mathrm d \delta \omega &= \mathrm d \left( -\sum_{k=1}^p (-1)^{k-1} \partial_{jk} f \mathrm dx^{i_1} \wedge \dots \wedge \widehat{\mathrm dx^{i_1}} \wedge \dots \mathrm dx^{i_p} \right)\\ &= -\sum_{k=1}^p (-1)^{k-1} \sum_{l=1}^n \partial_l \partial_{jk} f \mathrm dx^{l} \wedge \mathrm dx^{i_1} \wedge \dots \wedge \widehat{\mathrm dx^{i_k}} \wedge \dots \wedge \mathrm dx^{i_p}\\ &= -\sum_{k=1}^p \partial_{jk}^2 f \mathrm dx^I - \sum_{k=1}^p (-1)^{k-1} \sum_{\substack{l=1\\ l \not\in \{i_1, \dots, i_p\}}}^n \partial_l \partial_{jk} f \mathrm dx^{l} \wedge \mathrm dx^{i_1} \wedge \dots \wedge \widehat{\mathrm dx^{i_k}} \wedge \dots \wedge \mathrm dx^{i_p} \end{align*} Analogously, but easier, one can show that \begin{align*} \delta\mathrm d\omega = -\sum_{k=1}^p \partial_{jk}^2 f \mathrm dx^I + \sum_{k=1}^p (-1)^{k-1} \sum_{\substack{l=1\\ l \not\in \{i_1, \dots, i_p\}}}^n \partial_l \partial_{jk} f \mathrm dx^{l} \wedge \mathrm dx^{i_1} \wedge \dots \wedge \widehat{\mathrm dx^{i_k}} \wedge \dots \wedge \mathrm dx^{i_p} \end{align*} Thus, we get $$ (\mathrm d \delta + \delta\mathrm d) \omega = \left(-\sum_k \partial_k^2 f\right)\mathrm dx^I = (-\Delta f) \mathrm dx^I.$$

PS: Maybe check if a don't mess up with the sign convention, since I usually use the definition $\delta := (-1)^{n(p+1)} * \mathrm d *$.

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