Parallel Transport Examples

Differential Geometry · Axiom Academy

EXAMPLE Parallel Transport Examples Understanding curvature through vector transport on manifolds On a Euclidean plane, we transport a vector around a closed rectangular path. The vector maintains its direction throughout the transport. When it returns to the starting point, it points in the same direction as it started. This confirms the plane has zero curvature everywhere. Parallel transport around any closed loop produces no net rotation. Now we transport a vector around a spherical triangle with vertices at the north pole and two points on the equator separated by 90°. We start with vector pointing east at the north pole. We transport it along each edge of the triangle. Step 1 (N → A): Transport along a meridian from north pole to point A on equator. The vector remains perpendicular to the meridian, so at A it points south along the equator. Step 2 (A → B): Transport along the equator by 90°. The vector stays tangent to the equator, so at B it still points south (now along a meridian). Step 3 (B → N): Transport along a meridian from equator back to north pole. The vector stays perpendicular to the meridian, arriving at N pointing south . Initial vector at N: pointing east Final vector at N: pointing south The vector has rotated clockwise by: This 90° rotation is the holonomy angle for this closed loop. The Gauss-Bonnet theorem states that for a closed loop on a surface: where is the holonomy angle, is the Gaussian curvature, and is the area enclosed.

This is the written version of the interactive lesson above. See the full Differential Geometry course.