What am I actually doing when comparing tangent vectors of different fibres in the manifold $\mathbb R^n$?

Question

Suppose that $x, y \in \mathbb R^n$ are different points of the $n$-dimensional Euclidean manifold. Let $u \in T_x \mathbb R^n$ and $v \in T_y \mathbb R^n$ be tangent vectors in the two different fibres.

We usually treat them as members of the same vector space $\mathbb R^n$, so that we can take the sum $u+v$ and the difference $u-v$.

I would like to understand this from the differential geometry point of view, in particular with respect to "connections".

A connection over a smooth manifold defines, among other things, the transport of a tangential vector along a path. So when I calculate $u + v$, then I am actually transporting, say, $u$ along a path from $x$ to $y$ and compute the sum in the tangential fibre at $y$.

From the differential geometry point of view, what feature ensures that the transport is path-independent?

Is the transport of tangent vectors always reversible, that is, does transporting forward and then backward yield the original vector?

Answer 1

For the Levi-Civita connection on a Riemannian manifold, the parallel transport will be path-independent if the curvature vanishes. For surfaces this would be Gaussian curvature; for higher-dimensional manifolds, the sectional curvature. This is true if the manifold is simply connected; otherwise, require the paths to be homotopic (with fixed endpoints). Parallel transport along a given path is always reversible in the sense you mentioned.

Answer 2

A connection $\nabla$ allows one to define so-called parallel transports along curves. Given a curve $\gamma \colon [a,b] \to M$, and$u \in T_{\gamma(a)}M$, there exists a unique vector field $U \colon [a,b] \to \gamma^*(TM)$, called the parallel transport along $\gamma$, such that $U(a) = u$ and $\frac{\nabla}{dt} U = 0$ (where $\frac{\nabla}{dt}$ is the covariant derivative along $\gamma$). The parallel transport along $\gamma$ is defined as $T_{\gamma}(u) = U(b) \in T_{\gamma(b)}M$. One easily shows that $T_{\gamma} \colon T_{\gamma(a)}M \to T_{\gamma(b)}M$ is a linear isometry. In that sense, it is "reversible", as you said.

If $\gamma$ happens to be a closed loop and $\gamma(a)=p$, then $T_{\gamma} \in O(T_pM)$ is an orthogonal transformation. The set of all $T_{\gamma}$ is then a subgroup of $O(T_pM)$, called the holonomy group $\mathrm{Hol}_p(\nabla)$ of $\nabla$ at $p$. It is easy to show that if $M$ is connected, then all holonomy groups are isomorphic. One can therefore talk about the holonomy group of $\nabla$.

If now $M$ happens to be the Euclidean space $\Bbb R^n$, and if $\nabla$ is its Levi-Civita connection, then it turns out that the holonomy group is reduced to $\{\mathrm{Id}\}$ : this means that the parallel transport between two points is independent of the choice of a path connecting them. Hence, there is basically one unique way to identify all tangent spaces $T_p\Bbb R^n$ through parallel transport.

Luckily enough, this unique way to identify two different tangent spaces through the parallel transport in Euclidean spaces coincides with the identification we are used to make when writing $\gamma(t) = \gamma(0) + t \gamma'(0) + o(t)$ with $\gamma'(0) \in \Bbb R^n$.