$M$ as a totally geodesic submanifold of $TM$ Assume  that $M$ is a manifold.
What is a canonical Riemannian metric on $TM$  for which  the zero section (identified with $M$)  would be totally geodesic submanifold of $TM$?
What about if we fix in priory a metric on M and search for the metric on $TM$ with the above property with the extra condition that the restriction of the metric of $TM$ to zero section coincide the initial metric of $M$? In the  other  word,  how can we  extend  the  metric  on $M$  to  a  metric  on $TM$  such that $M$  would  be a  totally geodesic  submanifold  of  $TM$?
 A: Given a Riemannian metric $g$ on $M$, we have a natural splitting of $TTM$ into vertical and horizontal (as measured by the Levi-Civita connection) subbundles, each of which is isomorphic to $TM$; so we can think of an element of $TTM$ as a pair of vectors. With this in mind, there is an obvious extension of $g$ from the zero section to all of $TM$: just define $$\hat g((v,\xi),(w,\zeta)) = g(v,w) + g(\xi,\zeta).$$ This $\hat g$ is known as the Sasaki metric associated to $g$.
For any path $V(t) = (\gamma(t),v(t))$ in $TM$, its energy (as measured by $\hat g$) is $$E(V) = \frac12 \int_0^1 \left(|\gamma'(t)|^2  + |v'(t)|^2\right)dt = E(\gamma)+\frac12 \int_0^1|v'(t)|\,dt;$$ so the optimal path between any two points $(p,0)$ and $(q,0)$ in the zero section is always just the geodesic joining $p$ and $q$ in $M$. Thus the zero section is totally geodesic as desired.
This is not the unique metric with your desired properties (the most obvious counterexample is just distorting somewhere far away from the zero section), and clearly there's no canonical choice of such a metric in the case that you're just given a smooth manifold.
