We know that the Riemannian curvature $(0,4)-$tensor may be decomposed as $$Rm=W\;+\;A *g$$ where $*$ is the Kulkarni-Nomizu product, and $A$ the Schouten tensor.
I am studying the proof of a theorem which uses that $W_{1212}+W_{1313}+W_{1414}=0.$ Why is this true? Can you please help me with this?
The Weyl curvature is totally tracefree, and in particular it satisfies $$g^{cd} W_{acbd} = 0.$$ In an orthonormal frame $(E_i)$, so that by definition $g^{cd} = \delta^{cd}$, expanding the implicit summation gives $$0 = W_{a1b1} + W_{a2b2} + \cdots + W_{anbn},$$ where $n$ is the dimension of the manifold. If we specialize to $a = 1$, the first term of the sum is $W_{11b1} = W(E_1, E_1, E_b, E_1)$, which is zero because $W$ (like the Riemann curvature tensor $R$) is skew-symmetric in its first two arguments. Thus, the sum simplifies to $$0 = W_{12b2} + \cdots + W_{1nbn}.$$ Specializing to $n = 4$ and taking $b = 1$ gives $$0 = W_{1212} + W_{1313} + W_{1414}$$ as desired.