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?



Length of non-horizontal curve
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 $\ell\_\delta(\gamma) = \infty$.


