# Confusion regarding Ricci Scalar

Source: Thomas Moore's A General Relativity Workbook

Equation 1: $$R= g^{\mu\nu}R_{\mu\nu} = R^\nu{}_\nu$$

Equation 2: $$R^{\mu\nu}=g^{\mu\beta}g^{\nu\sigma}R_\beta\sigma$$

Question: Does $$R$$ also equal $$g_{\mu\nu}R^{\mu\nu}$$?

The Ricci Scalar $$R$$ is simply the trace of the Ricci Tensor $$R^{\mu\nu}$$ with respect to the metric $$g_{\mu\nu}=g^{\mu\nu}$$, so I don't see why the Curvature Scalar $$R$$ cannot be written as $$g_{\mu\nu}R^{\mu\nu}$$.

However, I am struggling to use Equation 2 to turn $$R=g_{\mu\nu}R^{\mu\nu}=g_{\mu\nu}g^{\mu\beta}g^{\nu\sigma}R_\beta\sigma$$ into Equation 1.

I was working on transforming $$R^{\mu\nu}-(1/2)g^{\mu\nu}R+\Lambda g^{\mu\nu}=\kappa T^{\mu\nu}$$ into $$-R+\Lambda g^{\mu\nu}=\kappa T$$, and I ran into this question.

Using the equality $$g_{\mu\nu}g^{\nu\sigma}=\delta_{\mu}^\sigma$$ and the symmetry of the metric tensor, $$g^{\sigma\beta}=g^{\beta\sigma}$$, we have that \begin{align} g_{\mu\nu}R^{\mu\nu}&=g_{\mu\nu}g^{\mu\beta}g^{\nu\sigma}R_{\beta\sigma} \\ &=g^{\mu\beta}g_{\mu\nu}g^{\nu\sigma}R_{\beta\sigma} \\ &=g^{\mu\beta}\delta_{\mu}^\sigma R_{\beta\sigma} \\ &=g^{\sigma\beta} R_{\beta\sigma} \\ &=g^{\beta\sigma} R_{\beta\sigma}\\ &=R \end{align} Thus, $$R=g_{\mu\nu}R^{\mu\nu}$$
Yes, $$g_{\mu\nu} R^{\mu\nu} = g^{\mu\nu} R_{\mu\nu}$$. You should check that $$g^{\sigma\rho} = g_{\mu\nu} g^{\mu\sigma} g^{\nu\rho}$$.