Q: Dependence of Riemann curvature on the connection 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! 
