Is there an explanation of analogies between the cross-ratio and the Riemann curvature tensor?

Question

Define the cross-ratio of four real or complex numbers as follows: $$[a,b,c,d] = \frac{(a-c)(b-d)}{(a-d)(b-c)}.$$ Then its logarithm has the same symmetries as the curvature tensor: $$\log[a,b,c,d] = -\log[b,a,c,d] = -\log[a,b,d,c] = \log[c,d,a,b].$$ Moreover, if $[a,b,c,d] = \lambda$, then $[b,c,a,d] = 1 - \lambda^{-1}$ and $[c,a,b,d] = (1-\lambda)^{-1}$, which implies an analog of the algebraic Bianchi identity: $$\log[a,b,c,d] + \log[b,c,a,d] + \log[c,a,b,d] = \pi i.$$

Is there something behind these coincidences?

Answer 1

There is an indirect connection which goes via the representation theory of the symmetric group. The symmetries of the Riemann tensor are equivalent to saying that $R$ transforms according to the two-dimensional irreducible representation of $\mathbb S_4$ corresponding to the partition $[2,2]$. On the other hand let us consider the moduli space $M_{0,4}$ parametrizing four distinct ordered points on the Riemann sphere up to Möbius transformations. The cohomology group $H^1(M_{0,4},\mathbf C)$ is $2$-dimensional, and transforms according to the representation $[2,2]$ under its natural action of $\mathbb S_4$. But we may compute this cohomology group as the space of holomorphic $1$-forms on $M_{0,4}$ with at most logarithmic poles at infinity. Moreover, the cross-ratio $\chi$ may be considered as a holomorphic function on $M_{0,4}$ and its logarithmic derivative $d \log(\chi)$ is such a 1-form with log poles. (Note that differentiating your analogue of the Bianchi identity gets rid of the $\pi i$, so we really get something exactly like the usual Bianchi identity!)