## Notes of Andrei Agrachev’s lecture

Some models of constant curvature in sub-Riemannian geometry

This is a reaction on Luca Rizzi’s talk last time (october 22nd).

1. GO spaces

V. Berestovskii has classified a class of sub-Riemannian manifolds which I now describe.

Definition 1 Say a Sub-Riemannian manifold is GO if every geodesic is an orbit of a one-parameter group of isometries.

This happens for a rather large class of Riemannian homogeneous spaces. Taking limits of such Riemannian homogeneous spaces provides examples of GO sub-Riemannian homogeneous spaces. Berestovskii has found examples which do not arise in this way.

Why are we interested in GO spaces ? Luca’s comparison theorem involves a notion of curvature (asociated to a point and geodesic) which is pretty complicated to compute. In the GO case, such curvature is constant along each geodesic.

1.1. Examples

Tangent cones of GO spaces are GO. Many step 2 Carnot groups are GO (not the most interesting, since curvature vanishes for Carnot groups).

Here is a less trivial example discovered by Berestovskii. Let ${M}$ be a Lie group with a bi-invariant pseudo-Riemannian metric (e.g. ${M}$ is semi-simple). Fix a left-invariant distribution generated by ${\Delta=\mathfrak{k}^{\bot}}$, where ${\mathfrak{k}\subset\mathfrak{m}}$ is a subalgebra. If ${M}$ is simple, ${\Delta}$ is bracket-generating. Assume that metric is definite positive on ${\Delta}$. Then the corresponding sub-Riemannian metric is GO.

Question. Are these examples all 2-step or not ?

1.2. Proof that many 2-step Carnot groups are GO

Let ${X_1,\ldots,X_k}$ be a basis of the distribution ${\Delta}$. Each ${X_i}$ defines a function ${u_i(p,q)=p\cdot X_i(q)}$ on ${T^*M}$. Let

$\displaystyle \begin{array}{rcl} h=\frac{1}{2}\sum u_i^2. \end{array}$

denote the Hamiltonian. Set

$\displaystyle \begin{array}{rcl} u_{ij}=\{u_i,u_j\}=p\cdot[X_i,X_j] \end{array}$

be there Poisson brackets. Then ${u_{ij}}$‘s are constant along the motion. Indeed, there derivative is a Poisson bracket, which involves a third order Lie bracket, which vanishes, by assumption. The equations for geodesics are

$\displaystyle \begin{array}{rcl} \dot{q}=\sum u_i X_i,\quad \dot{u}_i=\sum u_{ji}u_j, \end{array}$

This can be readily integrated : ${u=e^{tA}v}$, then the horizontal projection

$\displaystyle \begin{array}{rcl} q^{hor}=\int \sum e^{tA}v_i X_i. \end{array}$

The matrix ${A=(u_{ji})_{ij}}$ is skew-symmetric. Therefore it is a good candidate for an isometric automorphism, provided it lifts to an isomorphism, which happens often.

Question. Classify 2-step Carnot groups which are GO.

1.3. Proof that examples attached to simple Lie groups are GO

For any ${a\in M}$ and ${b\in K}$, ${x\mapsto axb}$ is an isometry. On computes that every geodesic has the form ${t\mapsto e^{at}e^{-ut}}$ where ${a\in \mathfrak{m}}$, ${a=u+v}$ where ${v\in\mathfrak{k}}$ and ${u\in\Delta}$. So it is an orbit of a one-parameter group.

2. Structure of isometries

Capogna-Le Donne : sub-Riemannian isometries are smooth, they are determined by their derivative, they form compact Lie groups (on compact manifolds).

Question : is an isometry determined by its restriction to the distribution at one point ? Related question : on a Carnot group, is every isometry affine, i.e. translation times automorphism ?

2.1. Discussion

Breuillard : doesn’t this follow from Capogna and Le Donne’s results ?

Pansu : generic 2-step Carnot groups with dimension vectors ${(n,p)}$ such that ${3\leq p\leq \frac{n(n-1)}{2}-3}$ have no graded automorphisms but dilations, so no isometries fixing a point. Those cannot be GO.