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.


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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