Let $M$ be a Riemannian manifold with Laplace de-Rham operator $\Delta = (d + \delta)^2$. If $g$ is a smooth $k$-form, and $f$ is a smooth function, is there a simple formula for $\Delta(fg)$ when $k > 0$?


Of course, this is a special case of $\Delta (f \wedge g)$. I would expect the formula to involve $\Delta f, \Delta g, \nabla f, \nabla g$ (like the case $k=0$) and something related to the curvature. I suspect that this is a very easy question to answer for people working with Riemannianian manifolds.



Product formula for Laplace de-Rham operator
Yes, and you can find it and its proof as a special case of Proposition 2.5 in the book of Berline-Getzler-Vergne: let $E$ be a hermitian vector bundle with unitary connection $\nabla,$ then the induced Laplacian $\Delta=\Delta^\nabla$ satisfies


$$\Delta (f s)=(\Delta f)s+f\Delta^\nabla s-2\langle\operatorname{grad} f,\nabla s\rangle.$$


Moreover, the Laplacian induced by the Levi-Civita-connection $\nabla$ on the exterior bundle is the standard Laplacian on the exterior bundle.


