Let $R_{\mu\nu}$ be the coordinates of the Ricci tensor. Using that $R^{\mu\alpha}R_{\alpha\nu}=\delta_\nu^\mu$ and $R_{\mu\nu}=R_{\nu\mu}$ it is possible to get the expressión of $R^{\mu\nu}$?
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What is your definition of $R^{\mu\alpha}$, is it really the inverse of $R_{\mu\alpha}$ (note it is in general not invertible) – Sep 18 '16 at 09:39
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Supposing that $R_{\mu\nu}$ is invertible, I define $R^{\mu\nu}$ as the inverse of $R_{\mu\nu}$. So I want to show that $R_{\mu\nu}=g^{\alpha\mu}g^{\beta\nu}R_{\mu\nu}$ – FUUNK1000 Sep 18 '16 at 09:54
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Do you mean $R^{\alpha\beta} = g^{\alpha\mu}g^{\beta\nu} R_{\mu\nu}$? – Sep 18 '16 at 10:28
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Yes, how can I prove it? – FUUNK1000 Sep 18 '16 at 10:41
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That's not true in general, if for example at a point we have $g_{ij} = \delta_{ij}$ and $R_{ij} = 2\delta_{ij}$, then the inverse of $R_{ij}$ is $1/2 \delta_{ij}$ while $R^{kl} = g^{ik} g^{jl} R_{ij} = 2\delta_{kl}$. – Sep 18 '16 at 10:49
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1You should note that the notation $R^{kl}$ you are using is an entity which, in Differential Geometry, has a defined special meaning (the one written down by John Ma). Therefore it's a very bad idea to use it as the inverse of $R_{kl}$. – Thomas Sep 18 '16 at 10:55
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First a comment on notation. In GR (and sometimes differential geometry more generally) the notation of moving a tensor's indices up is used to denote contraction with the metric. So $R^{\mu\nu}$ would refer to $g^{\mu\alpha}g^{\nu\beta}R_{\alpha\beta}$ rather than the inverse of $R_{\mu\nu}$. (A special case is the metric itself, where $g^{\mu\nu}$ represents both the inverse of $g_{\mu\nu}$ and what you get when you raise the two indices of $g_{\mu\nu}$. This is because we have chosen to define $g^{\mu\nu}$ in this way, so that the operation of raising an index will be the inverse to the operation of lowering it.)
So let's instead denote the inverse of $R_{\mu\nu}$ as $S^{\mu\nu}$. We want
$$R_{\mu\alpha}S^{\alpha\nu}=\delta_{\mu}^{\;\nu}\tag{$*$}$$
and
$$S^{\nu\alpha}R_{\alpha\mu}=\delta_{\;\mu}^{\nu}.\tag{$**$}$$
In fact any inverse can be calculated using the Levi-Civita symbol $\varepsilon^{\alpha\beta\gamma\delta}$. See here for more details. In particular we have
$$S^{\mu\nu}=4\frac{R_{\kappa\pi}R_{\lambda\rho}R_{\xi\sigma}\varepsilon^{\mu\kappa\lambda\xi}\varepsilon^{\nu\pi\rho\sigma}}{R_{\alpha\zeta}R_{\beta\eta}R_{\gamma\theta}R_{\delta\iota}\varepsilon^{\alpha\beta\gamma\delta}\varepsilon^{\zeta\eta\theta\iota}}\tag{$\dagger$}$$
Oscar Cunningham
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