= Solution
Contract the defining formula in its first and third indices. In four dimensions,
$$
\begin{aligned}
C^\alpha{}_{\beta\alpha\delta}
={}&R_{\beta\delta}\\
&-\frac12\left(
4R_{\beta\delta}
+g_{\beta\delta}R
-R_{\beta\delta}
-R_{\beta\delta}
\right)\\
&+\frac16(4g_{\beta\delta}-g_{\beta\delta})R\\
={}&0.
\end{aligned}
$$
Thus
$$
\boxed{C^\alpha{}_{\beta\alpha\delta}=0}.
$$
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>.
Back to article page
