Riemann tensor and the non-commutativity of parallel transports

Riemann curvature s denoted by $R^a_{bcd}$. If you try to attempt to parallel transport a tangent vector $\psi$ along an infinitesimal parallelogram given by two tangent vectors along non-parallel sides as $\eta$ and $\nu$, the following holds true. $$\psi^a_{;cd} - \psi^a_{;dc} = R^a_{cdb}\psi^b$$ where the $;$ indicates covariant differentiation. Here the $c$ and $d$ on the L.H.S. implies the two directions of the parallelogram. My professor said that the Riemann curvature tensor denotes the extent of the non commutativity of the covariant diferentiation along directions. In flat space, I would expect the Riemann Tensor to be zero. But does this imply commutativty of covariant differentiaton? Because covariant differentiation of a vector along a direction denotes how much the vector deviates during the parallel transport and one would not expect a deviation in flat space. So this should make the left hand side zero independent of the right hand side. What do you think?

• Note that it is not the commutativity of covariant derivatives, but, specifically, the commutativity of the 2nd-order covariant derivatives! – Dr. Ikjyot Singh Kohli Dec 30 '17 at 23:59

$\psi^a_{;cd} - \psi^a_{;dc} = R^a_{bdc}\psi^b$
• If the covariant derivative reduces to the partial derivative, the notion is still that of parallel transporting vector $\psi$. In this case, it might help to think of $\psi$ as a constant vector field $\psi(x)$ over the manifold. Partial differentiating a constant vector field at any point would give zero for each of the terms individually on the L.H.S. Would this be correct? – IanDsouza Dec 31 '17 at 22:42
The right hand side in that equation is identically $=0$ since the component of the vector are smooth and the covariant derivative reduces to partial derivative, in fact the $\Gamma$s are zero in a flat space.