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Relative to a timelike unit normal $n$, the electric part is $E_{ab}=C_{acbd}{n}^c{n}^d$. It describes the tidal component of the Weyl tensor measured by that observer. Antisymmetry in each curvature pair makes it spatial in both slots; pair symmetry and vanishing traces make it symmetric and trace-free.

In four spacetime dimensions, relative to a timelike unit normal $n$, define $B_{ab}=\frac{1}{2}{ϵ}_{acde}{C}^{de}{}_{bf}{n}^c{n}^f$, with the chosen spacetime metric volume tensor. This is the dual tidal part of the Weyl tensor. Normal contractions vanish by curvature and volume-tensor antisymmetry. Its sign depends on the specified orientation and dualization convention.

A four-dimensional vacuum Einstein solution with the fixed cosmological constant $\Lambda$ has $R_{ab}=\Lambda g_{ab}$ and $R=4\Lambda$, constant. Thus every derivative of $f^{\prime }(R)$ vanishes andThe metric tensor is nondegenerate, so this is zero precisely whenThis is the necessary and sufficient Einstein metric condition in f(R) gravity for the specified $\Lambda$, and works for every such Einstein metric, even when its Weyl tensor is nonzero. There is no need to divide by $f^{\prime }(4\Lambda )$; the degenerate case where both $f$ and $f^{\prime }$ vanish at that curvature is included. At $\Lambda =0$, the condition is simply $f(0)=0$.

If the intention is to demand the inclusion for every real $\Lambda$ simultaneously, the stronger functional condition is $R{f}^{\prime }(R)=2f(R)$ for all $R$. On each nonzero half-line it integrates to $f(R)=C{R}^2$; smoothness across zero makes the constants equal. Thus the all-$\Lambda$ version gives $f(R)=C{R}^2$, including $C=0$. This stronger reading is distinct from fixing one cosmological constant.

Use the convention ${\Theta }^a{}_b=\frac{1}{2}{R}^a{}_{bcd}{e}^c\wedge e^d$ and $R_{ab}=R^c{}_{acb}$. Each independent two-form has only one coordinate-plane wedge, so the Ricci tensor is diagonal. Contracting the six coefficients givesBut $2B+F=2(1+q/2)+(1-q)=3$, soIn coordinate-independent form, $R_{\mu \nu }=-3g_{\mu \nu }$. Thus the vacuum Einstein field equations $G_{\mu \nu }+\Lambda g_{\mu \nu }=0$ hold withThe curvature radius is one in the metric's normalization. The parameter $\alpha$ changes the individual curvature 2-forms but cancels from their Ricci contraction. For $\alpha =0$ this is Anti-de Sitter spacetime; for nonzero $\alpha$ it is the planar Einstein metric with cubic radial function, with a nontrivial Weyl tensor rather than universally constant sectional curvature. The result is local on regular coordinate patches; a zero of $F$ invalidates this particular static coframe, not the tensor equation in a suitable regular extension.

For the Vacuum Einstein equations with zero cosmological constant, $R_{\alpha \beta }=0$, and its tensor contraction gives $R=0$. Every Ricci tensor and scalar curvature term in the Weyl tensor consequently disappears:Thus the Weyl tensor contains all vacuum curvature. Here vacuum means Ricci-flat; a nonzero cosmological constant would retain a constant-curvature contribution to the Riemann curvature tensor.

The electric part of the Weyl tensor has one normal in each antisymmetric pair of the Weyl tensor. ThereforeFor the magnetic part of the Weyl tensor, contracting the first slot gives $n^{\alpha }{n}^{\lambda }{ϵ}_{\alpha \lambda \mu \nu }=0$ by antisymmetry of the metric volume tensor; contracting the second gives $n^{\beta }{n}^{\rho }{C}^{\mu \nu }{}_{\beta \rho }=0$ by antisymmetry of the last curvature pair. Consequently both tensors are entirely spatial:These spatial projection tensor identities actually hold for any Weyl tensor; the vacuum assumption is needed subsequently to replace it by the Riemann curvature tensor.

Contract the defining formula in its first and third indices. In four dimensions,Thus

The Weyl tensor has the same algebraic symmetries as the Riemann curvature tensor: it is antisymmetric within each index pair, symmetric under exchanging the pairs, and obeys the first Bianchi identity. A contraction within an antisymmetric pair vanishes against the symmetric metric. Every contraction between the two pairs can be converted by the pair symmetries to the one just calculated, possibly with a sign. Hence every contraction vanishes, proving the Trace-free property of the Weyl tensor.

This metric has $R_{ab}=-3g_{ab}$ and cosmological constant $\Lambda =-3$ in unit curvature-radius normalization. Its regular static coframe is real when $F>0$ and $z\ne 0$. Setting $q=\alpha z^3$, its six independent curvature 2-form coefficients are $-(1-q)$ in the $03$ and $12$ planes and $-(1+q/2)$ in the other four planes. Their Ricci contractions cancel $q$. The Weyl tensor squared invariant is $12{\alpha }^2{z}^6$, so nonzero $\alpha$ does not give constant sectional curvature despite the Einstein equation.