# Ball-Box Theorem and Sequence of Distributions

Let $(e^k,g^k)$ be a sequence of 2d smooth distributions in $R^3$ (with Euclidean metric) s.t $e^k,g^k$ are orthogonal. Let $f^k$ normal direction to this distribution. Suppose $[e^k,g^k] \neq 0$ on some domain $D(x)$ around x. Moreover let $(e^k,g^k)$ converge to another distribution $(e,g)$

i). The ball-box theorem states that there exists constants $C_k,c_k$ and $r$ s.t

$exp_x(Box(c_kr)) \subset B^k_{sR}(r) \subset exp_x(Box(C_kr))$

where $B^k_{sR}(r)$ is the subriemannian ball of $k^{th}$ distribution. Now my first question is how does these constants depend on the distribution. From the proofs I have studies it seemes upper constant can be chosen to only depend on behaviour of $e^k,g^k$. The lower constant however depends on the map which exponentiates $e^k,g^k,[e^k,g^k]$. Thus it seems to me it would be uniform if for instance $(f^k,[e^k,g^k])$ is uniformly bigger than some constant and decreasing if this quantity goes to zero? However the magn. of $c_k$ depends on the implicit inner workings of inverse function theorem (in the proofs I have seen).

ii). Looking back at the ball box theorem if $(f^k,[e^k,g^k])$ goes to 0 then the boxes (or parallelopipeds) $exp_x(Box(c_kr))$ are becoming increasingly thin in the $f^k$ direction therefore the subRiemannian balls are converging to a 2 dimensional hypersurface? I am not sure how this reasoning could be true since it suggest that also the orbit of this distribution for time less than r also converges to a plane. But even if $(f^k,[e^k,g^k])$ goes to 0, further Lie brackets may be non-zero thus contributing to the orbit. Moreover even the limit distribution e,g may be non-involutive since the Lie brackets might not converge. These are the reasons why this reasoning seems faulty to me.

I am new to sR geometry so far have looked at book/notes of ivanov, agrachev and gromov. I would also be grateful if you have seen and could direct me to any resources where sequence of distributions are handled or studied.

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In what topology the distributions converge? If in $C^r$, $r\ge 1$, then the Lie bracket also converges (in $C^{r-1}$). If in some weaker topology like $C^0$, then indeed there may be a trouble. – Sergei Ivanov Jun 25 '13 at 14:11
In ii. they converge in $C^0$ and (possibly) have common Lipschitz constant and the Lie bracket of them in the orthogonal direction goes to zero. What I cant understand is that ball-box theorem seems to state that if the Lie bracket goes to zero in the orthogonal direction, the diameter of the boxes (for fixed r) also go to zero in that direction. So in the limit, somehow this seems to indicate that the orbit up to time r will have the same dimension as the distribution? The core of the ii. question is basically does $(f^k,[e^k,g^k])$ going to 0 tell anything about the sR balls or orbits? – Avicenna Jun 27 '13 at 9:21
by the way, in the question when I said that the orbit converges to a plane that was only a local, vague statement. What I meant was that the orbit becomes increasingly thin in the direction $f^k$. – Avicenna Jun 27 '13 at 9:31

I do not understand exactly what you mean by $\exp_x$, but I guess it is something like the diffeomorphism proposed by Sergei Ivanov. I want to stress out that the ball-box theorem holds only if the coordinates are privileged w.r.t. $(e_k,g_k)$, so you should check that the $exp_x$ you are using defines privileged coordinates for any $k$, in order to apply this theorem.

Regarding the question about $[e_k,g_k]\rightarrow [e,g]\in\text{span}(e,g)$, it is a quite delicate matter. Indeed, this is exactly what happens when we try to get uniform estimates for the sub-Riemannian balls on a sequence of points approaching a singular point (think for example of the Martinet distribution, which is of step 1 outside the plane $\{y=0\}$, where it is step 2, and take any sequence of points $(x_k,y_k,z_k)$ with $y_k>0$ but $y_k\rightarrow 0$). This problem has been treated in the paper

F. Jean, "Uniform Estimation of Sub-Riemannian Balls", Journal on Dynamical and Control Systems, vol. 7 (4), 2001 (http://www.ensta.fr/~fjean/jdcs03_07_01.pdf)

Also, in the notes on s-R geometry by Jean (http://arxiv.org/abs/1209.4387) you can find a somewhat more detailed discussion of privileged coordinates and their subtleties than in the Agrachev, Barilari, Boscain ones.

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I just had a chance to read the notes you have given. The second one has been tremendously helpful in understanding the points I have been having trouble with here and I actually was able to come up with some answers. Hopefully I will come up with more as I read. Didn`t look at the first one yet but will sure do soon. Thanks. – Avicenna Oct 10 '13 at 15:41

Dealing with such a low regularity is a tricky business. However in dimension 3 you can get away with your set of assumptions.

First, if the distributions are uniformly Lipschitz and converge in $C^0$, then for every fixed $r$ the $r$-balls converge in the Hausdorff metric. This is easy to see if you choose coordinates $(x,y,z)$ in $\mathbb R^3$ so that the planes of the distributions are separated away from the $z$-axis. In this case every path $t\mapsto(x(t),y(t))$ in the $xy$-plane starting at $(0,0)$ uniquely determines a path $t\mapsto(x(t),y(t),z(t))$ tangent to the distribution and starting at $(0,0,0)$. The coordinate $z(t)$ obtained as the solution of an O.D.E., and these solutions converge since the coefficients are uniformly Lipschitz and converge in $C^0$.

However, the convergence of balls doest not mean the the full orbits converge. If the orthogonal component of the Lie bracket does not vanish, then the orbit is the entire space no matter what. Yet the limit distribution may be integrable and hence have two-dimensional orbits. There is nothing mysterious in that. The orbit is the set of points where the sub-Riemannian distance to the origin is finite. It can be finite for every $k$ but go to infinity as $k\to\infty$, so the limit orbit can be smaller. You can see a similar effect if you consider just the images of linear maps $\mathbb R\to\mathbb R$: for a map $x\mapsto \frac1k x$ the image is the entire line, but the limit map is zero.

As for the "thickness" of balls, they are bounded above in terms of the maximum of $([e,g],f)$ over a neighborhood. And from below in terms of the minimum of $([e,g],f)$. So if the bracket goes to 0 uniformly, you have an expected upper bound for the thickness. But if you want to control the thickness more tightly, you probably need to assume something to prevent fast oscillation of $([e,g],f)$ between zero and the maximum value (or between positive an negative).

You can prove the above mentioned bounds "by hand" if you choose a suitable coordinate system. For example, the one given by a local diffeomorphism $$(x,y,z) \mapsto \exp(xe)\circ\exp(yg)\circ\exp(zf) .$$ Of course you may want to estimate the distortion of this coordinate system with respect to the original one. This is in some sense "inner working of the inverse function theorem", but you can do it directly by ODE analysis.

In these coordinates, one can move from $(x,y,z)$ to $(0,0,z)$ cheaply: just move time $-x$ along $e$, then time $-y$ along $g$. So you only need to control the changes of the $z$-coordinate as you move along the distribution. The vector field $e$ in these coordinates is the coordinate field $\partial/\partial x$, and $[e,g]=\partial g/\partial x$, so everything is indeed controlled by the 3rd coordinate og $[e,g]$.

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Thanks alot, reflecting on your answer clarified some misunderstandings I had in my mind. However I have some more questions:  1- Does the arguement rely on the fact that this 3 dimensional, it seems that the key point is that the distribution is codimension 1. Is it in the last part where 3d becomes important? – Avicenna Jun 28 '13 at 9:12
2- Is it vaguely true that the relation of the constants with maximum and minimum of $([e^k,g^k],f^k)$ somehow comes from the fact that (since e,g,f are orth) $([e^k,g^k],f^k) = df(e^k,g^k)$ and $df^k$ can be integrated over certain area pieces to get estimates about the length of the sR geodesics. 3- Suppose you only know that $([e,g],f)$ goes to 0 and that the surface $x\circ exp(te^k,sg^k)$ converges to a hypersurface for $t,s<T$. Then the ball box theorem states that $B_{sR}(x,T)$ and the orbit of x for all time less than T also converge in $C^0$ manner to that hypersurface? – Avicenna Jun 28 '13 at 9:31
4- Finally the part about trying to say whether if there exists a uniform constant c s.t ck>c for all k depends on the oscillation property of ([ek,gk],fk) right? So for instance as in my first question i.) if you assume it is uniformly bounded below but not going to zero as in ii.) then such a c exists? Thanks alot again. – Avicenna Jun 28 '13 at 9:54