Hodge Laplacian

Remarks

The sign in $\delta$ makes it the formal adjoint of $d$: $\int_M \ip{d\alpha}{\beta} \, \mathrm{vol} = \int_M \ip{\alpha}{\delta\beta} \, \mathrm{vol}$ when one of the forms has compact support. Changing the orientation changes $\star$ by a sign twice, so $\delta$ and $\Delta$ are defined on any Riemannian manifold. Some authors write $\Delta_d$ or $\Delta_H$; it is also called the Laplace–de Rham operator. It is a nonnegative, formally self-adjoint, elliptic operator of order $2$, and $\Delta = (d + \delta)^2$.

Example: on $\R^n$ with the flat metric, let $f$ be a function. Then $df = \sum_j \partial_j f \, dx_j$ and $\delta(\sum_j a_j \, dx_j) = -\sum_j \partial_j a_j$, so $\Delta f = \delta df = -\sum_j \partial_j^2 f$. So on functions $\Delta = -\operatorname{div} \operatorname{grad}$, the Laplace–Beltrami operator with the sign that makes it nonnegative. On $k$-forms of $\R^n$, $\Delta(\sum_I f_I \, dx_I) = \sum_I (-\sum_j \partial_j^2 f_I) \, dx_I$.

Non-example: the operator $d \delta$ alone is not the Hodge Laplacian; on functions it is zero, since $\delta f = 0$ for a $0$-form.

On a closed manifold, the Hodge theorem identifies $\ker \Delta$ on $k$-forms with the de Rham cohomology $H^k(M; \R)$. The Weitzenböck identity writes $\Delta$ as a rough Laplacian plus a curvature term.