The Faraday Form of Conformal Products: Weyl Curvature and Four-Dimensional Einstein Rigidity
Let $(M^n,g)$ carry a conformal product structure whose adapted Weyl connection $D$ preserves orthogonal distributions of ranks $p,q\ge2$. We express the Faraday form $d\theta$ directly in terms of the Weyl curvature of $g$. If $S$ is the orthogonal involution determined by the splitting and \[ \mathcal K_W(X)=\sum_i W_{X,e_i}(Se_i) \] for any local orthonormal frame, then \[ (d\theta)^\sharp = \frac{n-2}{4(p-1)(q-1)}[\mathcal K_W,S]. \] No Ricci-curvature assumption is required. The same curvature calculation gives a companion formula for the symmetric part of $\nabla\theta$, \[ S\Theta^s-\Theta^sS+\theta^\sharp\wedge S\theta^\sharp = \frac{p-q}{n-2}(d\theta)^\sharp +\frac1{n-2}[S,\Ric^\sharp], \qquad \Theta=\nabla\theta. \] When the Ricci tensor is block diagonal with respect to the product splitting, these formulas lead to explicit identities for the two components of the Lee form. In dimension four, the rank-$(2,2)$ splitting determines an ambi-Hermitian pair. The classical ambi-Hermitian curvature identities imply local closedness when the Ricci tensor is invariant under both complex structures; we also recover this conclusion from the mixed Weyl trace above. Finally, if $(M^4,g)$ is compact and Einstein, no specialness assumption is needed: $D=\nabla^g$. Hence the universal cover is the product of two simply connected surfaces with the same constant Gaussian curvature.