I have a very nice description of the Riemann tensor in the form $R_{\mu \nu \rho \sigma}$ only depending on the second derivatives on the metric.
Given $g^{a \mu}R_{a \nu \rho \sigma}=R^{\mu}_{\nu \rho \sigma} $ is it true that $R_{\nu\sigma}=R^{a}_{\nu a \sigma}$ verifies $R_{\nu\sigma}=g^{a b}R_{a \nu b \sigma}$ ?
I suspect this is wrong but I cannot find the good contraction.
We have, pluging in your definitions $$ R_{\nu\sigma} = R^a_{\nu a \sigma} = \delta_a^\rho R^a_{\nu \rho \sigma} = \delta_a^\rho g^{\mu a} R_{\mu\nu\rho\sigma} = g^{\mu\rho}R_{\mu\nu\rho\sigma} = g^{ab}R_{a\nu b \sigma}$$ So, the answer is yes.