Parallel transport around a loop

[dEdt]

From the Wikipedia article on the Riemann curvature tensor:

Suppose that X and Y are a pair of commuting vector fields. Each of these fields generates a pair of one-parameter groups of diffeomorphisms in a neighborhood of x₀. Denote by τ_tX and τ_tY, respectively, the parallel transports along the flows of X and Y for time t. Parallel transport of a vector Z ∈ T_x0M around the quadrilateral with sides tY, sX, −tY, −sX is given by
\tau_{sX}^{-1}\tau_{tY}^{-1}\tau_{sX}\tau_{tY}Z.
This measures the failure of parallel transport to return Z to its original position in the tangent space T_x0M.

The last sentence assumes that $\tau_{sX}^{-1}\tau_{tY}^{-1}\tau_{sX}\tau_{tY}$ returns Z back to x₀. It isn't obvious to me that this should be true. My guess is that this is true because X and Y commute, but I can't think of a proof for this claim.

 

[Ben Niehoff]

X and Y being commuting fields is equivalent to the quadrilateral being closed (provided there is no torsion).

In the presence of torsion, quadrilaterals of commuting vectors fail to close, and the torsion tensor measures the amount by which they fail to close:

T(X,Y) \equiv \nabla_X Y - \nabla_Y X - [X,Y]
Here $Z = T(X,Y)$ is the vector that closes the gap left after taking into account the possibility that X and Y fail to commute.

 

[dEdt]

X and Y being commuting fields is equivalent to the quadrilateral being closed (provided there is no torsion).

That's what I suspected. Do you have a proof for this?