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Let M be a smooth manifold of dimension $n$. Given a Riemannian metric g on M, expressed in local coordinates $(x^1,...,x^n)$ as $$ g= g_{ij} dx^i \otimes dx^j $$ (unsing Einstein's convention) with $$ g_{ij} = \delta_{ij} + \frac{x^ix^j}{K^2- \Sigma_{i=1}^n(x^i)^2} .$$ Calculate $\Gamma^{\kappa}_{ij}, R_{ijkl}, R_{ij}$ and the scalar curvature R. K is a constant.

First, I tried using the expression for Christoffel symbols

$$ \Gamma^{\kappa}_{ij} = \frac{1}{2} g^{\kappa l}(\partial_{i}g_{jl} + \partial_{j}g_{il} - \partial_{l}g_{ij}) $$

for n=3, but it was getting a very long answer just for one specific symbol $\Gamma^{1}_{23}$ so it would be a nightmare to calculate then Riemann and Ricci tensors. I actually tried in mathematica but it got stucked on calculations. Searching around I found a chapter of some lecture notes where they say this is the spatial part of the FRW metric, they denote it as $\tilde{g}_{ij}$. In the FRW metric this is multiplied by a time depentend function $a(t)$ and they put the result:

$$ \Gamma^{i}_{j \kappa} = K \tilde{g}_{j \kappa}x^i $$

How could I get this result and move on to calculate the Riemann and Ricci tensor? I would aapreciate your help.

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It only changes constants in the computation, but the form of the spatial metric $\tilde g$ ((2.18) in the reference) is different from that in the question statement. The reference gives: $$\tilde g_{ij} = \delta_{ij} + K \frac{x^i x^j}{1 - K \sum_{\ell = 1}^n (x^\ell)^2} .$$ Hint Precomputing gives that $$\tilde g^{\ell m} = \delta^{\ell m} - K x^\ell x^m,$$ which is easier to differentiate than $\tilde g_{ij}$, so we can compute $$\partial_k \tilde g_{ij} = -\tilde g_{i \ell} (\partial_k \tilde g^{\ell m}) \tilde g_{m j} = -\tilde g_{i \ell} [-K (\delta^l{}_k x^m + \delta^m{}_k x^\ell)] \tilde g_{m j} = K (\tilde g_{i \ell} \tilde g_{j m} + \tilde g_{i m} \tilde g_{j \ell}) x^\ell .$$ Now, substitute in the coordinate formula for the Christoffel symbols, $$\tilde \Gamma_{ij}^k = \frac{1}{2} \tilde g^{k \ell}(\partial_i \tilde g_{\ell j} + \partial_j \tilde g_{i \ell} - \partial_\ell \tilde g_{ij}).$$

Once you've established the identity $\tilde\Gamma_{ij}^k = K \tilde g_{ij} x^k$, you can use that formula to compute the Riemann and then Ricci tensors efficiently, without using the coordinate formula for $\tilde g$: $$\tilde R_{ij}{}^k{}_\ell = \partial_i \tilde \Gamma^k_{j \ell} + \cdots .$$ For the first term, for example, we have $$\partial_i \tilde \Gamma_{j \ell}^k = \partial_i (K \tilde g_{j \ell} x^k) = K (x^k \partial_i \tilde g_{j \ell} + \delta^k{}_i \tilde g_{j \ell}),$$ and we can substitute our above formula for $\partial_i \tilde g_{j \ell}$.

We should find that the Riemann curvature is $\frac{K}{2}$ times the Kulkarni-Nomizu product of $\tilde g$ with itself, which in particular means that $\tilde g$ has constant sectional curvature. Contracting gives that $\widetilde{Ric} = K (n - 1) \tilde g$ and that the scalar curvature is $K (n - 1) n$.

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