Principal bundles and Subriemannian Geometry

Question

In sub-Riemannian geometry, one considers manifolds $P$ equipped with a subbundle $\mathcal{H}$ of $TP$, the horizontal distribution. One then has a Riemannian metric only on this distribution $\mathcal{H}$, and is interested in horizontal curves $\gamma$, i.e. those with $\dot{\gamma}(t) \in \mathcal{H}$.

If one has a principal bundle $P\longrightarrow M$ with connection $\omega$, this defines a horizontal distribution $\mathcal{H}$ on $P$. If one has a Riemannian metric on $M$, one can pullback this metric to $\mathcal{H}$.

However, in the definition of a sub-Riemannian manifold, one usually has the requirement that $\mathcal{H}$ be bracket generating, i.e. the Lie hull of $\mathcal{H}$ is $TP$. In the principal bundle case, this is not always the case, e.g. when the connection $\omega$ is flat, because then $\mathcal{H}$ is integrable.

My Questions are:

1. Why does one pose this bracket-generating condition on $\mathcal{H}$ when doing sub-Riemannian geometry
2. Are there easy-to-formulate conditions on $\omega$ that tell my when the corresponding horizontal distribution satisfies the condition?

Some Background:

One can consider the horizontal Laplacian on $P$ defined by $$\Delta^{\mathrm{hor}} = -\sum_{i=1}^n \bigl((e_i^{\mathrm{hor}})^2 - (\nabla_i e_i)^{\mathrm{hor}}\bigr),$$ where $e_i$ is an orthonormal basis of $M$, $\nabla$ is the Levi-Civita connection and ${\mathrm{hor}}$ denotes the horizontal lift. If $X_t$ is a Brownian motion on $M$, then there is a horizontal lift $U_t$ of $X_t$ to $P$ (determined by the conditions $\pi U_t = X_t$ and that the Stratonovich integral $\int \omega(U_t) \delta U_t = 0$), and $\Delta^{\mathrm{hor}}$ is its generator.

I am interested in the transition functions (heat kernel) of this process, and in estimates on them. I learned that if $\mathcal{H}$ is bracket generating, then Hörmander's theorem applies and $\Delta^{\mathrm{hor}}$ is hypoelliptic. Otherwise, this might probably fail.

So my third question is probably

3. What does this mean for the transition functions and the process $U_t$?

Answer 1

Re question 2. The answer is provided by the Ambrose Singer theorem. You can see how it connects with your question in my book A Tour of sub-Riemannian Geometry. When the connection form is analytic then it is easy to state the Ambrose-Singer theorem. The curvature of the connection, together with all its covariant derivatives, are forms with values in the Lie algebra of the structure group G of the principal bundle. Take the Lie algebra generated by all of these values and then take the subgroup of the structure group having this Lie algebra. [This is the "holonomy group" of the bundle-with-connection.] The accessible set (points you can get to by horizontal paths) is a subbundle of the original bundle whose fiber is this (possibly) smaller connected Lie group.

Answer 2

I like to think of a sub-Riemannian manifold as a space you are trying to move around in, but you have the restriction that from point $x$ you may only move in the directions $\mathcal{H}_x \subset T_x P$. (The sub-Riemannian metric $g$ can be thought of saying that the (marginal) cost to move an infinitesimal distance in direction $v_x \in \mathcal{H}_x$ is $\sqrt{g(v_x, v_x)}$. For $w_x \in T_x P \setminus \mathcal{H}_x$, you should think $g(w_x, w_x) = \infty$; moving in the direction $w_x$ is infinitely expensive and hence impossible.)

You can define the so-called Carnot-Carathéodory distance $d(x,y)$ as the infimum of the length $\int_0^1 \sqrt{g(\dot{\gamma}(t), \dot{\gamma}(t))}\,dt$ over all horizontal paths connecting $x,y$; this is the minimum cost of getting from $x$ to $y$.

Consider this trivial example of a sub-Riemannian manifold that does not satisfy the bracket-generating condition: let $P = \mathbb{R}^3$ with its standard coordinates $(x,y,z)$, and set $\newcommand{\spanop}{\operatorname{span}}\mathcal{H} = \spanop\{\frac{\partial}{\partial x}, \frac{\partial}{\partial y}\}$. Since $[\frac{\partial}{\partial x}, \frac{\partial}{\partial y}] = 0$ the bracket-generating condition does not hold. If you start at $(0,0,0)$ and are only allowed to move in $\mathcal{H}$ directions, you can travel at will around the $xy$-plane, but you can never reach the point $(0,0,1)$. So the C-C distance $d$ is $$d((x,y,z), (x',y',z')) = \begin{cases} \sqrt{(x-x')^2 + (y-y')^2}, & z = z' \\ \infty, & z \ne z' \end{cases}$$ and the topology induced by $d$ is not the usual topology on $\mathbb{R}^3$, but rather is a disjoint union of uncountably many copies of $\mathbb{R}^2$.

One would think this is kind of a dumb space. If you are at $(0,0,0)$, for all intents and purposes you live in $\mathbb{R}^2$. The rest of $\mathbb{R}^3$ is inaccessible to you, so what's the point of even considering it? You might as well just work on $\mathbb{R}^2$.

Now, in an arbitrary sub-Riemannian manifold, suppose $X,Y$ are two horizontal vector fields (i.e. smooth sections of $\mathcal{H}$). Suppose you start at a point $x$ and then:

• flow for time $\epsilon$ along $Y$

• flow for time $\epsilon$ along $X$

• flow for time $\epsilon$ along $-Y$

• flow for time $\epsilon$ along $-X$.

The result is, mod lower order terms, the same as if you flowed for time $\epsilon^2$ along the vector field $[X,Y]$ (I may have the sign wrong). If $[X,Y](x) \notin \mathcal{H}_x$ then this is a big deal: starting at $x$ you were not allowed to move in the direction $[X,Y](x)$, but via a more elaborate sequence of "legal" moves you managed to make progress in the $[X,Y](x)$ direction anyway. (Of course, it was expensive: you paid roughly $4\epsilon$ dollars to make $\epsilon^2$ of progress, but that's better than nothing.) So you can "break free" of your restriction, in a sense. By induction, with more elaborate maneuvers, you can make progress in any direction $v_x \in T_x P$ that can be written as an iterated Lie bracket of horizontal vector fields. So if the bracket generating condition holds, you can really move away from $x$ in any direction you want.

Chow's theorem says you can exploit this local freedom to move globally. Namely, if $P$ is topologically connected and the bracket-generating condition holds, then any two points $x,y \in P$ can be joined by a horizontal curve, meaning that $d(x,y) < \infty$. Moreover the topology induced by $d$ equals the original topology on $P$. So even obeying your restrictions of only moving tangent to $\mathcal{H}$, you can travel throughout the space; there aren't useless unreachable components. Contrast this with the non-bracket-generating $\mathbb{R}^3$ example given above, where no maneuvers along horizontal vector fields can ever make progress in the $z$ direction.

As to question 3, the probabilistic implications of the bracket-generating condition are something like the following (again assume $P$ is topologically connected): for each $x \in P$ and each $t > 0$, the transition measure $P_t(x,\cdot)$ is a smooth measure whose support is all of $P$. ("Smooth measure" means that in each coordinate chart, the measure is absolutely continuous to Lebesgue measure, with a $C^\infty$ density.) So the Brownian motion $U_t$ is able to explore the entire manifold $P$ (it has positive probability to hit any open set).

Intuitively, this should make sense: by traveling along little loops in horizontal directions, you can make progress in all directions which are their brackets. Brownian motion is very wiggly and makes lots of little loops, so it should be able to make progress in all "bracket" directions. When the bracket-generating condition holds, Brownian motion is able to make progress in all directions.

If you work on the non-bracket-generating $P = \mathbb{R}^3$ example, you will see that the process $U_t$ started at, say, the origin 0, looks like a 2-D standard Brownian motion with the $z$ coordinate fixed at 0. The transition measure $P_t(0, \cdot)$ is 2-D Gaussian measure on $\mathbb{R}^2 \times 0$, which is not absolutely continuous to Lebesgue measure, and its support is not all of $\mathbb{R}^3$. This process is stuck in a submanifold of $P$ and will never explore the entire space.

(I have not addressed question 2 because I am less familiar with that setup, but maybe valeri's comment helps.)