Given a Lie group $G$, what is the difference between the Laplacian $\Delta$ and the sub-Laplacian $\Delta\_{sub}$ of $G$. And what are the properties that we lose when going from sub-Laplace to Laplace and vice versa.


For example, what I know, for $G$ being the Heisenberg group $H^3= \mathbb C \times \mathbb R$, the difference between the Laplacian $\Delta$ and the sub-Laplacian $\Delta\_{sub}$ of $H^3$ is the standard Laplacian $\Delta\_{\mathbb R} = \frac{\partial^2}{\partial t^2} $ of $\mathbb R$, because 
$$ \Delta= \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + (x^2+y^2 ) \frac{\partial^2}{\partial t^2} + 2(x\frac{\partial}{\partial y} -y \frac{\partial}{\partial x} ) \frac{\partial}{\partial t} + \frac{\partial^2}{\partial t^2},$$
which can be rewritten in terms of the sub-Laplacian $\Delta\_{sub}$ as 
\begin{align}
\Delta &= \Delta\_{sub} + \Delta\_{\mathbb R},
\end{align}
and for the properties that we lose when going from sub-Laplace to Laplace are for example the ellipticity, because $\Delta\_{sub}$ is sub-elliptic but not elliptic, however $\Delta$ is elliptic.


Thank you in advance



Difference between the Laplacian and the sub-Laplacian of a Lie group
As Sebastian Goette explained in his comment, the sub-Laplacian $\Delta\_{sub}$ depends in general from an additional structure. And so does the Laplace-Beltrami $\Delta$ that you use to compute the difference. Let me elaborate.


**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).


We define the (horizontal) gradient of a smooth function $f$ as the unique vector field $\nabla f \in \Gamma(\mathcal{D})$ such that


$$ g(\nabla f, X) = df(X), \qquad \forall X \in \Gamma(\mathcal{D})$$


(here $\Gamma(\mathcal{D})$ denotes the space of smooth sections of the distribution, that is horizontal vector fields). Clearly this depends on the distribution and the metric.


Furthermore, we define the divergence of a smooth vector field $X \in TM$ as the smooth function $\mathrm{div}\_\mu(X)$ such that


$$ \mathcal{L}\_X \mu = \mathrm{div}\_\mu(X) \mu$$


where $\mathcal{L}\_X$ denotes the Lie derivative. Then the sub-Laplacian is


$$\Delta\_\mu f := \mathrm{div}\_\mu(\nabla f), \qquad f \in C^\infty(M)$$


Such an operator depends on the sub-Riemannian structure $(\mathcal{D},g)$ but also on the measure $\mu$.


**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:


$$\Delta\_\mu = \sum\_{i=1}^k X\_i^2 + \mathrm{div}\_\mu(X\_i) X\_i $$


where $X\_i^2$, when applied to functions, means that we apply it twice, that is $X\_i^2(f) = X\_i(X\_i(f))$.


**DIFFERENCE OPERATOR**


**Riemannian extensions:** 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


$$TM = \mathcal{D} \oplus \mathcal{V}$$


and $\hat{g}(\mathcal{D},\mathcal{V}) = 0$. Now you also have a well defined Laplace-Beltrami, the one of the Riemannian structure $\hat{g}$.


**Difference operator:** It is then a simple exercise to compute the difference between the sub-Laplacian $\Delta\_\mu$ and the Laplace-Beltrami pf the Riemannian structure. On Lie groups, where all left-invariant measures are proportional, then the difference between the two operators is precisely the sub-Laplacian associated with the (possibly non-bracket generating) sub-Riemannian structure $(\mathcal{V},\hat{g}|\_{\mathcal{V}})$. More explicitly, let $Z\_1,\ldots,Z\_{n-k}$ be a (left-invariant) local orthonormal frame for $\mathcal{V}$, in such a way that $X\_1,\ldots,X\_k,Z\_1,\ldots,Z\_{n-k}$ is a frame for the Riemannian metric $\hat{g}$. Then your difference operator is


$$ \Delta - \Delta\_{sub} = \sum\_{i=1}^{n-k} Z\_i^2 + \mathrm{div}\_\mu(Z\_i)Z\_i $$


**Example:** In the Heisenberg group $M = \mathbb{R}^3$, and following your notation, $(\mathcal{D},g)$ is generated by the left-invariant vector fields:


$$X\_1 = \partial\_x - y \partial\_t, \qquad X\_2 = \partial\_y +x\partial\_t$$


The 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). Denoting with $\Delta\_{sub}$ the sub-Laplacian associated with the standard sub-Riemannian structure and left-invariant measure $\mu = dxdydz$, we have:


$$\Delta\_{sub} = X\_1^2 + X\_2^2. $$


You recover your computation by choosing the "trivial" Riemannian extension $\hat{g}$ obtained by promoting $\partial\_t$ to a global unit vector orthogonal to $\mathcal{D} = \mathrm{span}\{X\_1,X\_2\}$.


**Remark:** In any case, the difference operator depends on the choice of a complementary Riemannian structure.


