SUB-LAPLACIANS
On a given smooth manifold $M$, we consider a sub-Riemannian structure $(\mathcal{D},g)$, where $\mathcal{D} \subseteq TM$ is a vector distribution (a sub-bundle of the tangent bundle) and $g$ is a smooth metric defined on it. Furthermore, let $\mu$ be a smooth measure on $M$ (i.e. given by a smooth density).
Example: In the Riemannian case $\mathcal{D} = TM$ and $g$ is defined on the whole tangent space at any point. Moreover it is customary to choose the standard Riemannian measure in place of $\mu$ (that is $\mu = \mathrm{vol}_g = \sqrt{|g|}|dx^1\wedge\ldots \wedge dx^n|$ in local coordinates). In this case we obtain the standard Laplace-Beltrami.
Properties: The sub-Laplacian $\Delta_\mu$ is always symmetric on the space of smooth and compactly supported functions $C^\infty_c(M)$, with respect to the product of $L^2(M,\mu)$. If the distribution $\mathcal{D}$ is Lie-bracket generating (a standard assumption in this field, dating back to Hormander work on hypoelliptic operators), then $\Delta_\mu$ is hypoelliptic (and indeed subelliptic) for any choice of $\mu$. Moreover it is well known that if $M$ equipped with its sub-Riemannian distance is a complete metric space, then $\Delta_\mu$ is essentially self-adjoint on $C^\infty_c(M)$.
Lie groups: On Lie groups one can choose $\mu$ to be any left-invariant measure (any such a measure differs up to a constant rescaling, which does not change the divergence and thus the sub-Laplacian). Moreover it is natural to choose a left-invariant distribution $\mathcal{D}$. This gives you a left-invariant sub-Laplacian.
Local formula: In terms of a local (possibly left-invariant if you are on a Lie group) orthonormal frame $X_1,\ldots,X_k$ of $\mathcal{D}$ we have:
DIFFERENCE OPERATOR
The question you raised is well posed if you choose a Riemannian complement of the sub-Riemannian structure, that is a Riemannian metric $\hat{g}$ such that $\hat{g}|_{\mathcal{D}} = g$. In this case we define a "vertical distribution" $\mathcal{V}$ as the orthogonal complement to $\mathcal{D} w.r.t. $\hat{g}$, in such a way that
and $\hat{g}(\mathcal{D},\mathcal{V}) = 0$. In this caseNow you also have a well defined Laplace-Beltrami, the one of the Riemannian structure $\hat{g}$.
It is then a simple exercise to compute the difference between the sub-Laplacian $\Delta_\mu$ and the aforementioned Laplace-Beltrami pf the Riemannian structure. On Lie groups, where anyall left-invariant measure ismeasures are proportional, then the difference between the two operators is precisely the sub-Laplacian associated with the (possibly non-bracket generating) sub-Riemannian structure $(\mathcal{V},\hat{g}|_{\mathcal{V}})$. In particular ifMore explicitly, let $Z_1,\ldots,Z_{n-k}$ isbe a (left-invariant) local orthonormal frame for $\mathcal{V}$ (in, in such a way that $X_1,\ldots,X_k,Z_1,\ldots,Z_{n-k}$ is a frame for the RieamannianRiemannian metric $\hat{g}$) then. Then your difference operator is
$$ \sum_{i=1}^{n-k} Z_i^2 + \mathrm{div}_\mu(Z_i)Z_i $$$$ \Delta - \Delta_{sub} = \sum_{i=1}^{n-k} Z_i^2 + \mathrm{div}_\mu(Z_i)Z_i $$
In the Heisenberg casegroup $M = \mathbb{R}^3$, and following your notation, $(\mathcal{D},g)$ is generated by the orthonormal global left-invariant vector fields:
Notice that in this case theThe Lebesgue measure $\mu=dx dy dz$ is left-invariant (and also right-invariant) and the divergence term vanishes (but this is just a coincidence on unimodular groups, where the divergence of left-invariant fields vanishes). In particular, denotingDenoting with $\Delta_{sub}$ the sub-Laplacian associated with the standard sub-Riemannian structure and left-invariant measure $\mu = dxdydz$, we have:
$$\Delta_{sub} = X_1^2 + X_2^2 $$$$\Delta_{sub} = X_1^2 + X_2^2. $$
You recover your computation by choosing the "trivial" Riemannian extension $\hat{g}$ obtained by promoting $\partial_t$ to a global unit vector orthonormalorthogonal to the horizontal distribution$\mathcal{D} = \mathrm{span}\{X_1,X_2\}$.
