Suppose that it is given that the Riemann curvature tensor in a metric space of dimension $d\geq2$ can be written as $$R_{abcd}=k(x^a)(g_{ac}g_{bd}-g_{ad}g_{bc})$$ where $x^a$ is a vector in the space.
What condition can I impose on $d$ to guarantee that $k(x^a)$ is a constant? I have been told that such a condition exists.
Not knowing what to do, I computed the contractions of the curvature tensor and got $$R_{bd}=g^{ac}R_{abcd}=k(d-1)g_{bd}$$ and $$R=g^{bd}g^{ac}R_{abcd}=kd(d-1)$$ But it probably is of no use?
This is Schur's lemma. The statement is that in dimension at least 3, if the sectional curvature at each points does not depend on the tangent 2-plane taken (i.e. the sectional curvature only depends on the point), then the sectional curvature is constant.
Starting from $$R_{ijkl}=k(x)(g_{ik}g_{jl}-g_{il}g_{jk}),$$ this can be seen by substituting this into the second Bianchi identity $$\nabla_mR_{ijkl}+\nabla_iR_{jmkl}+\nabla_jR_{mikl}=0.$$ In particular you get $$\nabla_mk (g_{ik}g_{jl}-g_{il}g_{jk})+\nabla_ik(g_{jk}g_{ml}-g_{jl}g_{mk})+\nabla_jk(g_{mk}g_{il}-g_{ml}g_{ik})=0.$$ Since the dimension is at least 3, we can take $i=k,$ $j=l$ and $m$ all distinct and this gives $\nabla_mk=0$, so that $k$ is constnat.