## Notes of Nicola Garofalo’s lecture nr 3

1. Fundamental solutions

Exercise (related to the Hopf-Rinow): compute the sub-Riemannan metric associated to vectorfield ${X=(1+x^2)\partial_x}$. Observe that balls are non compact, i.e. metric is not complete.

2. Existence

Theorem 1 (Folland) On a Carnot group, all sub-Laplacians ${\Delta_H}$ have a unique fundamental solution, i.e. a smooth fonction ${\Gamma}$ on ${G\setminus\{e\}}$ such that

1. ${\Delta_H \Gamma=\delta}$, Dirac distribution at the origin,
2. ${\lim_{|g|\rightarrow\infty} \Gamma(g)=0}$.

It is homogeneous of degree ${2-Q}$ under dilations.

\proof of homogeneity. Consider ${v=\Gamma\circ\delta_\lambda-\lambda^{2-Q}}$. Then ${\Delta_H v=0}$. By hypoellipticity, ${v}$ is smooth and classically harmonic.

By Bony’s maximal principle, since ${v}$ teds to 0 at infinity, ${v=0}$. Alternatively, use Liouville’s theorem.

2.1. The case of groups of Heisenberg type

Charles Feffermann, studying several complex variables, suggested the form that the fundamental solution should take in the Heisenberg group. This was implemented by Folland and Kaplan.

Theorem 2 (Folland 1972, Kaplan 1981) Let ${G}$ be of Heiseberg type. The function

$\displaystyle \begin{array}{rcl} \Gamma(g)=\frac{C}{(|z|^4+16|t|^2)^{\frac{Q-2}{4}}} \end{array}$

is a fundamental solution of ${-\Delta_H}$. Here, ${C}$ is a suitable constant,

$\displaystyle \begin{array}{rcl} C^{-1}=m(Q-2)\int_{G}\frac{1}{((|z|^2+1)^2+16|t|^2)^{\frac{Q+2}{4}}}. \end{array}$

2.2. A Lemma

Lemma 3 If ${G}$ is of Heisenberg type,

1. ${\Delta_H(|t|^2)=\frac{k}{2}|z|^2}$,
2. ${|\nabla_H(|t|^2)|^2=|z|^2|t|^2}$,
3. ${\langle\nabla_H(|z|^2),\nabla_H(|t|^2)\rangle=0}$.

For this, use Baker-Campbell-Hausdorff to compute

$\displaystyle \begin{array}{rcl} z_j(g\exp(se_i))&=&z_j(g)+s\delta_{ij},\\ t_\ell(g\exp(se_i))&=&t_\ell(g)+\frac{s}{2}\langle[z,e_i],\epsilon_\ell\rangle. \end{array}$

Differentiating with respect to ${s}$ at ${s=0}$, this gives

$\displaystyle \begin{array}{rcl} X_i(z_j)(g)&=&\delta_{ij},\\ X_i(t_\ell)(g)&=&\frac{1}{2}\langle[z,e_i],\epsilon_\ell\rangle=\frac{1}{2}\langle J(\epsilon_\ell)z,e_i\rangle. \end{array}$

This leads rather easily to all 3 formulae.

2.3. Proof of the Folland-Kaplan Theorem

We see that ${\Gamma(g)=C\,\rho^{2-Q}}$ where ${\rho(g)=(|z|^4+16|t|^2)^{1/4}}$ is a gauge. Let us regularize it,

$\displaystyle \begin{array}{rcl} \rho_\epsilon(g)=((|z|^2+\epsilon^2)^2+16|t|^2)^{1/4}. \end{array}$

Then

$\displaystyle \begin{array}{rcl} |\nabla_H \rho_\epsilon|^2&=&\frac{|z|^2}{\rho_\epsilon^2},\\ \Delta_H\rho_\epsilon&=&\frac{Q-1}{\rho_\epsilon}|\nabla_H \rho_\epsilon|^2+\frac{m\epsilon^2}{\rho_\epsilon^3}. \end{array}$

Given an arbitrary function ${h:{\mathbb R}\rightarrow{\mathbb R}}$, differentiate ${v=h\circ\rho_\epsilon}$. Then apply it to ${h(t)=t^{2-Q}}$ and observe that this kills a term, yielding

$\displaystyle \begin{array}{rcl} \Delta_H v&=&\frac{m\epsilon^2}{\rho_\epsilon^3}h'(\rho_\epsilon)\\ &=&m(2-Q)\epsilon^2\rho_\epsilon^{-2-Q}\\ &=&-m(Q-2)\epsilon^2 v^{\frac{Q+2}{Q-2}}. \end{array}$

This equation is known as the CR Yamabe equation. This is the conformally invariant form of the sub-Laplacian. It indicates that ${v}$ is critical for the sub-Riemannian Sobolev inequality.

Observe that

$\displaystyle \begin{array}{rcl} \rho_\epsilon=\epsilon\delta_{\epsilon^{-1}}\circ\rho_1. \end{array}$

Thus

$\displaystyle \begin{array}{rcl} \Delta_H v&=&\epsilon^{-Q}\delta_{\epsilon^{-1}}\circ\Delta_H(\rho_1^{2-Q})\\ &=&-m(Q-2)\epsilon^{-Q}\delta_{\epsilon^{-1}}\circ v_1^{\frac{Q+2}{Q-2}}. \end{array}$

It turns out that ${v_1^{\frac{Q+2}{Q-2}}\in L^1(G)}$. So up to a multiplicative constant, ${\epsilon^{-Q}\delta_{\epsilon^{-1}}\circ v_1^{\frac{Q+2}{Q-2}}}$ converges to the Dirac distribution as ${\epsilon\rightarrow 0}$. Indeed, given a test function ${\phi}$,

$\displaystyle \begin{array}{rcl} \langle \rho^{2-Q},\Delta_H\phi\rangle&=&\langle v,\Delta_H\phi\rangle\\ &=&\lim_{\epsilon\rightarrow 0} \langle v,\Delta_H\phi\rangle\\ &=&\lim_{\epsilon\rightarrow 0} \langle\Delta_H v,\phi\rangle\\ &=&-m(Q-2)\lim_{\epsilon\rightarrow 0} \epsilon^{-Q}\langle\delta_{\epsilon^{-1}}\circ v_1^{\frac{Q+2}{Q-2}},\phi\rangle\\ &=&-m(Q-2)\lim_{\epsilon\rightarrow 0} \epsilon^{-Q}\langle v_1^{\frac{Q+2}{Q-2}},\phi\circ \delta_{\epsilon^{-1}}\rangle\\ &=&-m(Q-2)\phi(e)\int_G v_1^{\frac{Q+2}{Q-2}}. \end{array}$

2.4. The CR Yamabe problem

The problem: let ${M}$ be a compact strictly pseudoconvex CR manifold, find a choice of the contact form ${\theta}$, for which the Tanaka-Webster scalar curvature is constant.

This is a sub-Riemannian analogue of a problem posed in 1959 by Yamabe, and which has been solved (Yamabe, Trudinger, Aubin, Schoen).

Theorem 4 (Jerison-Lee 1990) The CR Yamabe problem is solvable when dim${(M)\geq 5}$ and ${M}$ is not locally CR equivalent to the round CR sphere.

The CR version

After a decade, Gamara and Yaccoub, two students of Abbas Bahri, solved the problem when ${M}$ is CR equivalent to the CR round sphere. The 3-dimensional case was later completed by Gamara.

These cases non treated by Jerison and Lee are analogues of the Riemannian cases where the positive mass conjecture in general relativity plays a role. There have been recent progress along similar lines in CR geometry recently. Attend the relevant workshop this fall!

2.5. The sub-Riemannian Sobolev embedding theorem

Observe that

$\displaystyle \begin{array}{rcl} \int_{G}|\nabla_H v|^2=-\int_{G}v\Delta_H v=\int_{G}v^{\frac{2Q}{Q-2}}. \end{array}$

This is an equality case in a Sobolev type inequality. The Euclidean Sobolev inequality reads

$\displaystyle \begin{array}{rcl} (\int_{{\mathbb R}^n}|u|^{q})^{1/q}\leq S(\int_{{\mathbb R}^n}|\nabla u|^p)^{1/p}. \end{array}$

The numerology ${\frac{1}{p}-\frac{1}{q}=\frac{1}{n}}$ is forced by dilaton invariance.

Theorem 5 (Folland-Stein 1975) In a Carnot group, let ${1. There exists a constant ${S_q(G)}$ such that, for all smooth compactly supported functions ${u}$,

$\displaystyle \begin{array}{rcl} (\int_{G}|u|^{q})^{1/q}\leq S_q(G)(\int_{G}|\nabla u|^p)^{1/p}, \end{array}$

provided ${\frac{1}{p}-\frac{1}{q}=\frac{1}{Q}}$.