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

  • 3
    $\begingroup$ 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. $\endgroup$ – Sebastian Goette Apr 2 '16 at 11:01
  • $\begingroup$ 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. $\endgroup$ – 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.


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 $X_i^2(f) = X_i(X_i(f))$.


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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.