I am working through an exercise in the book 'Manifold Theory: An introduction for mathematical physicists' which asks me to show that path-independent parallel transport implies that the curvature tensor vanishes.
A hint given in the book says that I can use the fact that $$\tag{1} \partial_iX^k+\Gamma^k_{i\sigma}X^\sigma=0$$ is integrable. This equation comes from that fact that $X$ is undergoing parallel transport.
I believe I am almost there but I am unsure how exactly how I have used the fact that (1) is integrable/ path independent.
Ok so my solution went something like this;
Let $M$ be a differentiable manifold and let $X$ be smooth a vector field passing through a point $p \in M$. Let $\gamma:\mathbb{R}_+ \to M$ be a path for which $\gamma(0)=p$. Then by parallel transport we obtain a family of vector fields $S=\{X_{\gamma(t)}\}$ which have been obtained from $X$ by parallel transport. Accordingly we have that each vector in $S$ satisfies the parallel transport equations (1).
The Riemann curvature tensor in a coordinate expression is
$$R^\rho{}_{\sigma\mu\nu} = \partial_\mu\Gamma^\rho{}_{\nu\sigma} - \partial_\nu\Gamma^\rho{}_{\mu\sigma} + \Gamma^\rho{}_{\mu\lambda}\Gamma^\lambda{}_{\nu\sigma} - \Gamma^\rho{}_{\nu\lambda}\Gamma^\lambda{}_{\mu\sigma}$$ and hence we can compute the value of $R^\rho{}_{\sigma\mu\nu}X^\sigma_{\gamma(t)}$ at a particular point just by multiplying the above equation by $X^\sigma_{\gamma(t)}$. This is given by \begin{align} R^\rho{}_{\sigma\mu\nu} X^\sigma_{\gamma(t)}&= (\partial_\mu\Gamma^\rho{}_{\nu\sigma})X^\sigma_{\gamma(t)} - (\partial_\nu\Gamma^\rho{}_{\mu\sigma})X^\sigma_{\gamma(t)} + \Gamma^\rho{}_{\mu\lambda}(\Gamma^\lambda{}_{\nu\sigma}X^\sigma_{\gamma(t)}) - \Gamma^\rho{}_{\nu\lambda}(\Gamma^\lambda{}_{\mu\sigma}X^\sigma_{\gamma(t)})\\ &=(\partial_\mu\Gamma^\rho{}_{\nu\sigma})X^\sigma_{\gamma(t)} - (\partial_\nu\Gamma^\rho{}_{\mu\sigma})X^\sigma_{\gamma(t)} - \Gamma^\rho{}_{\mu\lambda}\left(\partial_\nu X^\lambda_{\gamma(t)}\right) + \Gamma^\rho{}_{\nu\lambda}\left(\partial_\mu X^\lambda_{\gamma(t)}\right)\\ &=(\partial_\mu\Gamma^\rho{}_{\nu\sigma})X^\sigma_{\gamma(t)} + \Gamma^\rho{}_{\nu\lambda}\left(\partial_\mu X^\lambda_{\gamma(t)}\right) - \left((\partial_\nu\Gamma^\rho{}_{\mu\sigma})X^\sigma_{\gamma(t)} + \Gamma^\rho{}_{\mu\lambda}\left(\partial_\nu X^\lambda_{\gamma(t)}\right) \right)\\ &=\partial_\mu\left(\Gamma^\rho{}_{\nu\sigma}X^\sigma_{\gamma(t)}\right)-\partial_\nu\left(\Gamma^\rho{}_{\mu\lambda}X^\lambda_{\gamma(t)}\right)\\ &=-\partial_\mu\partial_\nu(X^\rho_{\gamma(t)} )+\partial_\nu\partial_\mu(X^\rho_{\gamma(t)})\\ &=0 \end{align} Where I have used equation (1), some index gymnastics, the product rule, equation (1) again and equality of mixed partials, In that order.
Hence If we took the original vector field $X$ to be non-vanishing, it follows that $X_{\gamma(t)}$ is non vanishing, thus the curvature tensor is zero everywhere in which $X_{\gamma(t)}$ is defined. I believe now I must use the fact that I could of chosen a different path/ integrability of (1) to say that $X_{\gamma(t)}$ is defined everywhere and hence the curvature tensor vanishes. I am unsure on how to proceed.
Any hints of comments would be much appreciated.
If you are aware of the exponential map, one can apply Gauss Bonnet in the following way.
Given $p \in M$, let $U \subset M$ be a (totally) normal neighbourhood of $p$. That is, $exp_p: V \subset T_pM \to U$ is a diffeomorphism for some set $V$ contained in the tangent space at $p$.
Pick $q \in U$ and let $exp_p(y)=q$, then one can consider any plane $\sigma \subset T_pM$ containing the vector $y$. The image of $\sigma$ under $exp_p$ is a surface contained in our set $U$. The Gaussian curvature $K$ of this surface is the sectional curvature $K(\sigma)$.
One can proceed as in the question here to establish that this sectional curvature is $0$. (note the curves that one is transporting along must be contained in the surface).
This is true for any plane $\sigma$. The Riemann curvature tensor is determined entirely by its sectional curvatures and hence must be $0$.