Laplacian on a warped product. Let $(M, g)$, $(N, h)$ be complete Riemannian manifolds (not necessarily compact). Let $f : M \rightarrow (0, \infty)$ be a smooth function, and finally let
$$\overline{M} = M \times_f N$$
be the warped product of $M$ and $N$ with warping function $f$.
My question: is there a nice formula for $\Delta_{\overline{M}} : \Omega^p(\overline{M}) \rightarrow \Omega^p(\overline{M})$, the Laplacian operator of $\overline{M}$ on $p$-forms, in terms of $\Delta_M, \Delta_N$ and $f$?
 A: In the case $p=0$ (so this is the normal laplacian on smooth functions) there is the following formula, where $d_N$ is the dimension of $N$,
$$
\Delta_{\bar{M}}u =\Delta_M u+\frac{d_N}{f}\langle grad_{M_1}f,grad_{M_1}u\rangle+\frac{1}{f^2} \Delta_{N}u.
$$
For example if $M=\mathbb{R}^+, N=S^{d-1}, f=\rho$, so this is the usual spherical decomposition of Euclidean space, we have
$$
\Delta_{\mathbb{R}^d}u=\frac{\partial^2 u}{\partial \rho^2}+\frac{d-1}{\rho}\frac{\partial u}{\partial \rho}+\frac{1}{\rho^2}\Delta_{S^{d-1}}u.
$$
To prove the formula, one method is to choose local coordinates $x^i$ on $M$ and $y^i$ on $N$. With respect to these coordinates, the metric $\bar{g}$ on $\bar{M}$ has the block decomposition $\begin{pmatrix}g_{ij}(x) &0\\ 0 & f(x)^2h_{ij}(y)\end{pmatrix}$. If we then use the local coordinate expression for the Laplacian, we have that (using the shorthand $|m|:= \sqrt{\det m_{ij}}$ for matrices $m$), $|\bar{g}|=|g|\cdot |h|\cdot f^{2 d_N}$, and so
$$
\Delta u=\frac{1}{g(x)h(y)f^2(x)}\left(\partial_{x^i}(g^{ij}(x)|g(x)|\cdot |h(y)|\cdot f(x)^d \partial_{x^j}u)+\partial_{y^i}(\frac{h^{ij}(x)}{f(x)^2}|g(x)|\cdot |h(y)|\cdot f(x)^d \partial_{y^j}u)\right).
$$
By factoring out terms that only depend on $x$, the second term is clearly $\frac{1}{f(x)^2}\Delta_{N}u$, a similar computation works for the first term, but now one has to consider when $\partial_{x_i}$ falls on $f(x)^{d_N}$, which produces the rest of the expression.
