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. $\endgroup$ 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

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