Curvature Scalar in Riemannian Space

Question

Suppose that R_iklm=a(g_ilg_km-g_img_kl ) on some four dimensional Riemannian space and a is a constant. Question: Show that for the curvature scalar we have R=-12a.

What I know from calculating the Ricci curvature scalar is that it requires : 1. The Dual Metric Tensor [g^ij],which is the inverse of [g_ij],the metric tensor and 2. The Ricci Tensor(R_ij).

Also,I use R_iklm=sigma_i(g_i1Rⁱ_klm) when calculating the Gaussian curvature.

So,I'm stuck and don't know how to approach this.

Thanks.

Answer 1

I will do the computation for the Ricci tensor $R_{ij}$ and leave the computation of scalar curvature to you.

You are given

$$R_{iklm} = a(g_{il}g_{km} - g_{im}g_{kl})$$

Multiplying $g^{il}$ on both sides and using $R_{km} = g^{il}R_{iklm}$, we have

$$\begin{split} R_{km} = g^{il}R_{iklm} &= a g^{il}(g_{il}g_{km} - g_{im}g_{kl}) \\ &= a\big( (g^{il}g_{il}) g_{km} - (g^{il}g_{im})g_{kl}\big) \\ &= a\big( (g^{il}g_{il}) g_{km} - g_{km}\big) \end{split}$$

Note that repeated indices are summed over, so $$ g^{il}g_{il} = \sum_{i,l=1}^4 g^{il}g_{il} = \sum_{i=1}^4 \delta_{ii} = 4$$

since your manifold is of dimension four. Thus $$R_{km} = a (4g_{km} - g_{km} )= 3a g_{km}$$

(Unfortunately if you continue you will find $R = 12a$. Probably we are using different convention of the Ricci tensor)