The connection depends on the way we define the parallel transport the basis vector $\{e_\mu\}$: \begin{eqnarray} \triangledown_\nu e_\mu=\Gamma_{\nu\mu}^\lambda e_\lambda. \end{eqnarray} It is known that we have the degree of freedom to get another connection by a different parallel transport rule \begin{eqnarray} \tilde{\Gamma}_{\nu\mu}^\lambda={\Gamma}_{\nu\mu}^\lambda+t_{\nu\mu}^\lambda \end{eqnarray} where $t_{\nu\mu}^\lambda$ is a tensor.
However, the Riemann curvature \begin{eqnarray} R^\kappa_{\lambda\mu\nu}(\Gamma)=\partial_\mu\Gamma^\kappa_{\nu\lambda}-\partial_\nu\Gamma^\kappa_{\mu\lambda}+\Gamma^\eta_{\nu\lambda}\Gamma^\kappa_{\mu\eta}-\Gamma^\eta_{\mu\lambda}\Gamma^\kappa_{\nu\eta} \end{eqnarray} is not invariant under the above redefinition of parallel transport, or \begin{eqnarray} R^\kappa_{\lambda\mu\nu}(\tilde{\Gamma})\neq R^\kappa_{\lambda\mu\nu}(\Gamma). \end{eqnarray} My question is how to understand such non-invariance. If Riemann curvature indeed depends on the underlying parallel transport rule, does it mean that Riemann curvature is not so intrinsic a quantity? The same question can be asked on the torsion tensor as well.
Thanks very much in advance!