- #1

gastkram

- 5

- 3

Start with the contracted Bianchi identity,

$$

\nabla_a \left( R^{ab}-\frac{1}{2}g^{ab} R\right)=0.

$$

Now rewrite with the metric so that we get the same factor R in both terms,

$$

\nabla_a \left( g^{ac}g^{bd}R_{cd}-\frac{1}{2}g^{ab}g^{cd} R_{cd}\right)=0,

$$

then factorize;

$$

\nabla_a \left( \left(g^{ac}g^{bd}-\frac{1}{2}g^{ab}g^{cd} \right)R_{cd}\right)=0.

$$

Now apply the product rule

$$

\nabla_a \left( g^{ac}g^{bd}-\frac{1}{2}g^{ab}g^{cd} \right) R_{cd} + \left(g^{ac}g^{bd}-\frac{1}{2}g^{ab}g^{cd} \right) \nabla_a R_{cd}=0.

$$

By metric compatibility the first term is zero, so we conclude that ##g^{ac}g^{bd}=\frac{1}{2} g^{ab}g^{cd}##

or the covariant derivative of the Ricci tensor is zero in general. The first option means that the metric is just zero and so makes the identity we started with trivial. The second option is just wrong.

What went wrong? I'm going nuts.