I've been working a lot with some PDE's on foliated manifolds and from times to times I realize basic stuff I've been missing. For instance, let $(M,g)$ be a closed Riemannian manifold with some (possible singular) Riemannian foliation $\mathcal F$. Given any smooth function $f : M \to \mathbb{R}$ one has the following $$\Delta_g f = \Delta^{\cal H}f + df(H),$$ where $\Delta^{\cal H}$ stands for the "Laplacian" considered only on horizontal directions, i.e, those orthogonal to the leaves and $H$ is the Mean Curvature Vector of the leaves.
The first thing here is that: the Lalplacian $\Delta_g$ is always a strongly elliptic operator. However, I am facing a little difficulty about understanding the operator $\Delta^{\cal H}$. It seems that it may not be elliptic, at least this is my impression. So I was wondering: in order to $\Delta^{\cal H}$ be elliptic, one could possible follow two ways: either $H\equiv 0$ or we are restricted to basic functions (those that are constant along the leaves). The precise statement here about $\Delta^{\cal H}$ being elliptic is what is called ''transversally elliptic'' in the sense p. 1337, Section 6, is this correct? The intuition is that on both cases, if we consider the submersion $M^{reg} \to M^{reg}/\cal F$, and suppose that it is Riemannian, where $reg$ denotes the regular part of the foliation, $\Delta^{\cal H}$ coincides with the Laplace operator on $M^{reg}/\cal F$.