Laplace-de Rham operator on $\mathbb{R}^n$ Let $\mathbb{R}^n$ have the standard orientation with volume element $dV = dx_1 \wedge...\wedge dx_n > 0$.  Show that $\Delta = - \sum_j \partial^2 / dx_j^2$ on 0-forms on $\mathbb{R}^n$.  Where $\Delta$ is the Laplace-de Rham operator.
Now, I understand we have: $\Delta f = (d\delta + \delta d)f = \delta d f$.  I don't really understand how we apply the $\delta$ operator.  In fact, I'm not sure I really understand that operator...  This seems like a simple problem, which is why any help would go a long way to aiding my understanding.
Thank you.
 A: By definition of $\delta$, we have 
$$\int_{\mathbb R^n} g(\delta \alpha) dx =  \int_{\mathbb R^n} \langle dg , \alpha \rangle dx$$
for any smooth one form $\alpha$, and smooth function $g$ with compact support. If we write 
$$\alpha = \sum_{i=1}^n \alpha_i dx^i, $$
then 
$$\langle dg, \alpha \rangle = \bigg\langle \sum_{j=1}^n \frac{\partial g}{\partial x^j}dx^j , \sum_{i=1}^n \alpha_i dx^i \bigg\rangle = \sum_{i=1}^n \frac{\partial g}{\partial x^i} \alpha_i . $$
The last equality holds as $\{dx^1, \cdots, dx^n\}$ are orthonormal in $\mathbb R^n$. So by integration by part, as $g$ are of compact support, 
$$ \int_{\mathbb R^n} \langle dg , \alpha \rangle dx = \sum_{i=1}^n \int_{\mathbb R^n} \frac{\partial g}{\partial x^i} \alpha_i dx = - \sum_{i=1}^n \int_{\mathbb R^n} g \frac{\partial \alpha_i }{\partial x^i } dx. $$
Thus we have 
$$\delta \alpha = -\sum_{i=1}^n \frac{\partial \alpha_i }{\partial x^i}$$
and (put $\alpha = df$)
$$ \Delta f = \delta df = -\sum_{i=1}^n \frac{\partial ^2 f}{\partial (x^i)^2}.$$
In general, $\delta$ in local coordinate would be of the form 
$$ \delta \alpha = - g^{ij} \nabla_i \alpha_j, $$
where $\nabla_i$ is the covariant differentiation on one form. So 
$$\Delta f = \delta df = - g^{ij} \nabla_i \nabla_j f.$$
