Difference between the Laplacian and the sub-Laplacian of a Lie group

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.

• The sub-Laplacian depends on some extra structure, doesn't it? On the Heisenberg group, there might be a most natural choice, but what about other Lie groups? From your example, one might expect to obtain a Riemannian submersion $G\to G/H$ whose horizontal fields generate the Lie algebra of all vector fields, and the difference operator would be the fibrewise Laplacian. But this seems to use a nice correlation between the Riemannian metric on $G$ and the Lie algebra structure. – Sebastian Goette Apr 2 '16 at 11:01
• If you use two translation invariant Riemannian metrics on Euclidean space, you can already get quite a mess as the difference between their Laplace operators: elliptic, hyperbolic, or neither. So I think you need to make some special choice of which metric you use, and also which $G$-invariant subbundle to give your sub-Laplacian. This question would benefit from some thought about which operators you have in mind, since a Lie group does not have a single choice of Laplace operator or sub-Laplace operator. – Ben McKay Apr 2 '16 at 12:54

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. • @ Raziel: what we can also say for the spectrum$\sigma(\Delta)$of$\Delta$and of$\sigma(\Delta_{sub})\$ ? – Z. Alfata Apr 18 '16 at 18:06