Flatness implies path-independence on simply connected domains

On a simply connected region where the curvature vanishes, parallel transport depends only on the endpoints.
Flatness implies path-independence on simply connected domains

Let EME\to M be a vector bundle with a \nabla, and let FF_\nabla denote its .

Proposition (endpoint dependence on simply connected sets)

Let UMU\subset M be a connected, simply connected open set such that FU=0F_\nabla|_U=0. Then for any two points x,yUx,y\in U, the map

PTγ:ExEy \mathrm{PT}_\gamma : E_x \to E_y

depends only on the endpoints x,yx,y, not on the choice of piecewise smooth path γ\gamma in UU from xx to yy.

Equivalently, if γ0,γ1\gamma_0,\gamma_1 are two such paths with the same endpoints, then PTγ0=PTγ1\mathrm{PT}_{\gamma_0}=\mathrm{PT}_{\gamma_1}.

A parallel statement holds for a principal bundle with a flat principal connection: on a simply connected UU where curvature vanishes, the transport PxPyP_x\to P_y is independent of the path in UU.

Examples

  1. Euclidean space. On U=RnU=\mathbb R^n with the trivial bundle U×VU\times V and =d\nabla=d, curvature vanishes and parallel transport is the identity, hence depends only on endpoints.
  2. Restriction of a flat but globally nontrivial situation. A flat connection on a bundle over S1S^1 can have nontrivial holonomy around the circle, but on any simply connected arc US1U\subset S^1 the proposition applies, so transport inside UU is path-independent.
  3. Why simply connectedness matters. On S1S^1, take a flat U(1)U(1)-connection with holonomy eiθ1e^{i\theta}\neq 1. Curvature is still zero, but transport from a point to itself along the generator loop is multiplication by eiθe^{i\theta}, so dependence on the homotopy class of the path persists when the domain is not simply connected.