## Notes of Antonio Lerario’s lecture

Loop spaces in subRiemannian manifolds

with Agrachev and Gentile.

Let ${(M,\Delta)}$ be a subRiemannian manifold. How complicated is the space ${\Omega_{p_0,p_1}^E}$ of admissible curves joning ${p_0}$ and ${p_1}$ with energy ${\leq E}$ ?

By complicated, I mean topologically complicated, as measured by the sum of Betti numbers, ${b(X)}$.

Example 1 (Heisenberg group) Let ${(X_1,X_2)}$ be an orthonormal frame of the contact structure. Energy of a curve ${\gamma}$ such that ${\dot{\gamma}=u_1 X_1+u_2X_2}$ is ${J(\gamma)=\int u_1^2+u_2^2}$.

1. Morse Theory

It implies that the Betti sum ${b(\{f\leq E)\leq b(}$critical points with ${f\leq E)}$. In our case, critical points are geodesics joining ${p_0}$ and ${p_1}$.

In Heisenberg group, geodesics from origin to ${Z=(0,0,z)}$ come in one parameter families. For every integer ${N\geq 1}$, there is a circle of geodesics from origin to ${Z}$ with energy ${J=2\pi n|z|}$. So critical points with ${J\leq E}$ form a disjoint union of ${\lfloor \frac{E}{2\pi|z|}\rfloor}$ circles. This set has Betti sum ${b}$ which is linear in ${E}$. On the other hand, it turns out that ${\Omega_{e,Z}^E}$ is a sphere, its Betti sum equals 2.

2. Results

We shall see, more generally, for step 2 Carnot groups, that the Betti sum of the set of geodesics increases like ${E^\ell}$ where ${\ell}$ is the co-rank, whereas the Betti sum of the set of all admissible curves grows like ${E^{\ell-1}}$.

Theorem 1 Let ${M}$ be a 2-step Carnot group. For a generic final point, ${\Omega_{e,p}^{+\infty}}$ is a Hilbert manifold homeomorphic to the infinite dimensional sphere.

2.1. Structure of geodesics

Fact. Let ${M^{d+\ell}}$ be a 2-step Carnot group, with ${d<\ell}$ (should persist if ${d\geq \ell}$. If the final point ${p}$ is not vertical and generic, there are only a finite number of geodesics from ${e}$ to ${p}$.

Therefore, we shall concentrate on vertical final points.

A 2-step Carnot group is determined by a vectorspace ${W}$ of ${d\times d}$ skew-symmetric matrices. Denote by

$\displaystyle \begin{array}{rcl} \Lambda=\{A\in W\,;\,det(A-in1)=0\,, n\in{\mathbb N}\}. \end{array}$

This a countable collection of algebraic sets. View ${p\in {\mathbb R}^d \oplus W^*}$.

Theorem 2 Geodesics appear in families ${C_w}$. Each family is labelled by one of the points in ${\Lambda\cap}$sphere of radius ${E}$ where ${\Lambda}$ is orthogonal to the sphere…. The energy of the family is ${J(w)=w(p)}$. The family ${C_w}$ is a product of at most ${\ell}$ copies of ${S^1}$.

Corollary 3 ${b(crit(E))=O(E^{\ell})}$.

Theorem 4 ${b(\Omega_{e,p}^{E})=O(E^{\ell-1})}$. So Morse inequalities are not sharp.

Consider the end-point map in the Heisenberg case. Ending in the vertical direction is one equation (average zero). Morse theory (backwards) and infinite negative index implies one need only study curves with energy equal ${E}$. Endpoint map in quadratic. For a quadratic map on the spehere, real algebraic geometry tells us that complexity grows polynomially of the announced degree.

2.3. Case ${\ell=2}$

Theorem 5 ${b(\Omega_{e,p}^{E})=\tau(p)E+o(E)}$. Furthermore, ${\tau(p)}$ can be computed.

In that case, dim${(W)=2}$. ${p=\cos\theta A_1+\sin\theta A_2}$ where ${A_j}$ are skew-symmetric matrices. ${p}$ has eigenvalues ${i\alpha_j(\theta)}$… and ${\tau}$ is given by some integral.

2.4. Role of self-similarity

Applying dilations, we see that ${\Omega_{e,p}^{E}=\Omega_{e,\delta_{\epsilon}p}^{E/\epsilon^2}}$. Thus as ${p}$ tends to ${e}$ and energy bound decreases quadratically in ${\epsilon=d(e,p)}$, our formula gives the derivative of complexity as a function of ${\epsilon}$.

Note that when ${p=e}$, ${\Omega_{e,e}^{E}}$ has the same homotopy as ${\Omega_{e,e}^{+\infty}}$, i.e. trivial. And there is only one geodesic, the constant.