Given a geodesically complete manifold M, can we define a global identification of tangent spaces by starting from a base point, and parallel transporting along smooth geodesics? For this to be consistent, we need the parallel transport along every geodesic loop to leave the tangent space invariant. Is there a simple condition on M that tells me whether this is possible?


(I am not a mathematician so I apologize if what I asked was not very precise.)



Identification of tangent spaces by parallel transport along geodesics
Ok, given the comments, what you are really asking for, is for a class of connected Riemannian manifolds for which the following construction (or the map $\Phi$) is a (smooth) trivialization of the tangent bundle of a Riemannian manifold $M$:


Fix a $p\in M$. For each $q\in M$ let $\gamma\_{qp}$ denote a unit speed geodesic connecting $q$ to $p$. Let $\Pi\_{qp}: T\_qM\to T\_pM$ denote the parallel transport along $\gamma\_{qp}$. Then take the map
$$
\Phi: TM\to M\times T\_pM, \quad \Phi(v)=(q, \Pi\_{qp}(v)), \quad v\in T\_qM.
$$


Then a **sufficient** (likely, also necessary, at the very least, you will need injectivity of the exponential map) condition for this to work is when $M$ has a **pole** at $p$, i.e. $\exp\_p: T\_pM\to M$ is a diffeomorphism. (This ensures that $\gamma\_{qp}$ exists, is unique and depends smoothly on $q$.) For instance, by the [Cartan-Hadamard theorem](https://en.wikipedia.org/wiki/Cartan%E2%80%93Hadamard_theorem), it suffices to assume that $M$ is complete, simply connected and has sectional curvature $\le 0$. An example which has positive curvature is a paraboloid of revolution in ${\mathbb R}^3$ (the point $p$ will be the tip of the paraboloid).


