TITLE: "Urban paths" in a Riemannian manifold
QUESTION [6 upvotes]: Fix a compact Riemannian manifold $M$, a base point $p\in M$, a basis $B$ for the tangent space $T_p$, and a destination $q$.  
The metric on $M$ determines its Levi-Civita connection.  
My goal consists of finding a short path from $p$ to $q$ of a special form: such a path must leave $p$ traveling along a geodesic lying in a direction contained in $B$ and, carrying $B$ along according to the connection, arrive at new point $p_1$; if $p_1\not=q$, then at $p_1$ the path must turn in a new direction belonging the the transported basis. The steps then repeat until finally we arrive at $q$ after a journey possessing a finite number of legs.  Call such a path urban (because we must stay on the "city streets" determined by $B$) unless the literature already has a name for this.
Clearly, with a bound on the number of legs, some urban path attains the
infinimum length of all urban paths with no more than that many legs.
My questions: without a bound on the number of legs, must some urban path still attain the infinimum?   If yes, can we get a bound, based solely on the dimension of $M$, for the number of legs in this ``urban geodesic''?

REPLY [3 votes]: Let us modify the problem a bit.
Consider all paths from $p$ to $q$.
For each path $\gamma$, transport the frame along  $\gamma$ and measure the $\ell^1$-length $\gamma$ --- the length of $\gamma$ in the $\ell^1$-norm provided by the frame.
Note that for urban paths the $\ell^1$-length coincide with the usual length
Now consider a path $\gamma_0$ from $p$ to $q$ which minimizes $\ell^1$-length.
Note that arbitrarily close to $\gamma_0$ there is an urban path with arbitrarily close length. 
So if $\gamma_0$ is a strict minimum and it is not an urban path then there's no chance of getting a minimum among the urban paths.
An example can be constructed the following way:
Take Euclidean plane and multiply its metric by conformal factor $1+(x-y)^2$;
the path $\gamma_0(t)=(t,t)$ satisfies the condition above.