# Length of non-horizontal curve

Let $M$ be a sub-Riemannian space. Consider a smooth curve $\gamma:[0,1]\to M$ such that $\dot\gamma(t)\not\in H_{\gamma(t)}$, where $H_{\gamma(t)}$ is the horizontal subbundle ( i.e. $\gamma$ is totally non-horizontal curve).

Is it obvious that the curve is not rectifiable or has infinite length?

• How do you define length for such a curve? I guess that you mean by $TM$ the subbundle of the tangent bundle on which the subRiemannian metric is defined. – Ben McKay Feb 16 '15 at 16:59
• I meant horizontal subbundle. Mistakenly written $TM$. – Nikita Evseev Feb 18 '15 at 4:52

Here is a rigorous proof that non-horizontal curves are not rectifiable.

First, recall that, given a metric space $(M,\delta)$ and a mapping $\gamma:[0,1]\to M$, the $\delta$-length of $\gamma$ is, by definition, $$\ell_\delta(\gamma) = \sup\left\{\delta\bigl(\gamma(0),\gamma(t_1)\bigr)+\delta\bigl(\gamma(t_1),\gamma(t_2)\bigr)+\cdots+\delta\bigl(\gamma(t_m),\gamma(1)\bigr)\ \bigl|\ 0< t_1 <t_2<\cdots<t_m<1 \right\}.$$ If $\ell_\delta(\gamma)<\infty$, one says that $\gamma$ is $\delta$-rectifiable.

Second, a basic result in Riemannian geometry is this: If $g:TM\to\mathbb{R}$ is a Riemannian metric on a connected manifold $M$ and $\delta_g:M\times M\to [0,\infty)$ is the associated distance metric, then, for any piecewise $C^1$-mapping $\gamma:[0,1]\to M$, one has $$\ell_{\delta_g}(\gamma) = \int_0^1 g\bigl(\gamma'(t)\bigr)^{1/2}\ \mathrm{d}t.$$

Now, suppose that $M$ is a manifold with a smooth plane field $H\subset TM$ with the property that each pair of points in $M$ can be joined by some $H$-curve, i.e., a piecewise $C^1$-curve $\gamma:[0,1]\to M$ such that $\gamma'(t)$ lies in $H_{\gamma(t)}$ for all $t\in[0,1]$. Suppose that $h:H\to\mathbb{R}$ is a smooth function that restricts to be a positive definite quadratic form on $H_x$ for each $x\in M$. Then one can define a metric $\delta:M\times M\to[0,\infty)$ by the formula $$\delta(x,y)=\inf\left\{ \int_0^1 h\bigl(\gamma'(t)\bigr)^{1/2}\ \mathrm{d}t\ \bigl|\ \gamma:[0,1]\to M\ \text{is an H-curve}, \gamma(0)=x,\gamma(1)=y\right\}.$$

Proposition: $\ell_\delta(\gamma) = \infty$ for any $\gamma:[0,1]\to M$ that is piecewise $C^1$ but not an $H$-curve.

Proof: Choose a smooth splitting $TM = K\oplus H$ where $K$ is a smooth plane field (necessarily of positive rank, or else all curves are $H$-curves and there is nothing to prove). Let $k:K\to \mathbb{R}$ be a smooth function that restricts to be a positive definite quadratic form on each $K_x$ for $x\in M$. Consider the family of Riemannian metrics $g_n = n\,k \oplus h$ on $M$, and let $\delta_n$ be the distance metric on $M$ associated to $g_n$. Then it follows directly from the definitions that $\delta(x,y)\ge \delta_n(x,y)$ for all $x,y\in M$. Consequently, it follows (again from the definitions) that $$\ell_\delta(\gamma) \ge \ell_{\delta_n}(\gamma)$$ for all maps $\gamma:[0,1]\to M$. Now suppose that $\gamma:[0,1]\to M$ is piecewise $C^1$ but not an $H$-curve. Writing $\gamma'(t) = a(t) + b(t)$, where $a(t)$ lies in $K_{\gamma(t)}$ and $b(t)$ lies in $H_{\gamma(t)}$, one has that $a(t)$ is non vanishing for $t$ in an open subset of $[0,1]$. Consequently, $\int_0^1 k\bigl(a(t)\bigr)^{1/2}\,\mathrm{d}t > 0$. But then, for all $n$, one has $$\ell_{\delta_n}(\gamma) = \int_0^1 \left(\,n\,k\bigl(a(t)\bigr) + h\bigl(b(t)\bigr)\,\right)^{1/2}\,\mathrm{d}t \ge \sqrt{n}\ \int_0^1 k\bigl(a(t)\bigr)^{1/2}\,\mathrm{d}t.$$ Thus, the inequality $\ell_\delta(\gamma) \ge \ell_{\delta_n}(\gamma)$ for all $n$ implies that $\ell_\delta(\gamma) = \infty$.

• In "Second", is $g$ a quadratic form (not metric) ? – Nikita Evseev Feb 18 '15 at 5:41
• @NikitaEvseev: Yes. It's an unfortunate convention in Riemannian geometry that the underlying quadratic form $g$ on $M$ is called a 'Riemannian metric', even though it is not a 'metric' in the sense of metric spaces. Of course, it does define a metric $\delta_g:M\times M\to [0,\infty)$, where $\delta_g(x,y)$ is the infimum of the $g$-lengths of piecewise $C^1$ curves joining $x$ to $y$ (when $M$ is connected). – Robert Bryant Feb 18 '15 at 9:46

Length is only defined for horizontal curves on a subRiemannian manifolds. This is due to the subRiemannian (or subFinsler) idea that you are only allowed move along a subbundle of the tangent bundle. For other curves it is natural to take the length to be infinite if you need to assign them a length.

One can approximate a subRiemannian metric by a sequence of Riemannian metrics which blow up in nonhorizontal directions. Now if you fix a curve and take the limit of its lengths with respect to these approximating Riemannian metrics, the limit length can only be finite if the curve is horizontal everywhere.

Since $M$ is a metric space, you can of course define the length of a smooth curve $\gamma:[0,1]\to M$ in the usual way as a supremum of sums of distance between points. If the curve is nonhorizontal, this supremum is infinite for the following reason. Let $\Delta\subset TM$ be the subbundle where the subRiemannian metric lives (I denote by $TM$ the whole tangent bundle). Iterated brackets of $\Delta$ generate the whole tangent bundle. If $\dot\gamma(t)\in[\Delta,\Delta]\setminus\Delta$, then $d(\gamma(t+\epsilon),\gamma(t))\approx\epsilon^{1/2}$ for small $\epsilon$. For more brackets you get smaller powers, so you should get $d(\gamma(t+\epsilon),\gamma(t))\gtrsim\epsilon^{1/2}$ when $\dot\gamma(t)\in TM\setminus\Delta$. This scaling leads to infinite length. I believe this argument is fairly obvious for those working with subRiemannian structures.