Strange result from Bianchi identity, me spot the error!

[gastkram]

I have accidentally derived a very wrong result from the contracted Bianchi identity and I can't see where the error is. I'm sure it's something obvious, but I need someone to point it out to me as I've gone blind. Thanks!

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.

 

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.

Hmm, your working up to here looks right to me at least but I think $g^{ac} g^{bd} - \frac{1}{2} g^{ab} g^{cd} = 0$ does not imply that the metric is zero, i.e. the indices don't match?

 

[gastkram]

Hmm, your working up to here looks right but I think $g^{ac} g^{bd} - \frac{1}{2} g^{ab} g^{cd} = 0$ does not imply that the metric is zero, i.e. the indices don't match?

Thank you for your help!

I mean that it is zero because we get
$$
g_{ac}g^{ac}g^{bd}=\frac{1}{2}g^{ab}g_{ac}g^{cd} \implies \delta_a^a g^{bd}=\frac{1}{2}g^{b}_{\phantom{b}c}g^{cd}\implies D g^{bd}=\frac{1}{2}g^{bd},
$$
but the dimension is not usually one half .

 

Maybe the problem is that the entire sum is zero,$$\left(g^{ac}g^{bd}-\frac{1}{2}g^{ab}g^{cd} \right) \nabla_a R_{cd}=0$$and you can't pull out the coefficients and set each to zero separately, $g^{ac}g^{bd}-\frac{1}{2}g^{ab}g^{cd} = 0$, because they're included in the summation [and three of the indices are dummy indices]? It's to say that each individual term in the sum is not necessarily zero.

 

[gastkram]

Maybe the problem is that the entire sum is zero,$$\left(g^{ac}g^{bd}-\frac{1}{2}g^{ab}g^{cd} \right) \nabla_a R_{cd}=0$$and you can't pull out the coefficients $g^{ac}g^{bd}-\frac{1}{2}g^{ab}g^{cd} = 0$ and set each to zero separately, because it's included in the summation [and three of the indices are dummy indices]?

Oh, that may be it. Let me think about that for a minute.

 

[gastkram]

Maybe the problem is that the entire sum is zero,$$\left(g^{ac}g^{bd}-\frac{1}{2}g^{ab}g^{cd} \right) \nabla_a R_{cd}=0$$and you can't pull out the coefficients and set each to zero separately, $g^{ac}g^{bd}-\frac{1}{2}g^{ab}g^{cd} = 0$, because they're included in the summation [and three of the indices are dummy indices]? It's to say that each individual term in the sum is not necessarily zero.

Yes, I guess I claimed that every coefficient has to be zero but that's not true. Mystery solved (I hope)! Thanks!

 

Yeah, that seems right!

 

[gastkram]

Yeah, that seems right!

Well, I was right that I had done something obviously wrong

 

haha, well contracted indices also sometimes make me feel like a ... dummy.

 

[PeterDonis]

Thank you for your help!

I mean that it is zero because we get
$$
g_{ac}g^{ac}g^{bd}=\frac{1}{2}g^{ab}g_{ac}g^{cd} \implies \delta_a^a g^{bd}=\frac{1}{2}g^{b}_{\phantom{b}c}g^{cd}\implies D g^{bd}=\frac{1}{2}g^{bd},
$$
but the dimension is not usually one half .

The operation you are doing here does not look correct. You can't contract the same index twice, but that is what you are doing by introducing the factor $g_{ac}$ on both sides. The $a$ and $c$ indexes are already contracted; they're not free. The only free index in the equation is $b$, so that's the only one available to contract anything with.

 

@PeterDonis I think that was just following from considering $g^{ac} g^{bd} = \frac{1}{2} g^{ab} g^{cd}$ as an equation in four free indices, which we eventually clocked in #4 that you can't do.

 

[PeterDonis]

I think that was just following from considering $g^{ac} g^{bd} = \frac{1}{2} g^{ab} g^{cd}$ as an equation in four free indices, which we eventually clocked in #4 that you can't do.

Yes, agreed. The "four free indices" part is the issue--only one of those indices, $b$, is actually free. The others, as you point out in #4, are dummy summation indices--or, as I put it, they are already contracted, so you can't contract them again.

 

[vanhees71]

Thank you for your help!

I mean that it is zero because we get
$$
g_{ac}g^{ac}g^{bd}=\frac{1}{2}g^{ab}g_{ac}g^{cd} \implies \delta_a^a g^{bd}=\frac{1}{2}g^{b}_{\phantom{b}c}g^{cd}\implies D g^{bd}=\frac{1}{2}g^{bd},
$$
but the dimension is not usually one half .

I don't understand your manipulations. On the left-hand side you simply have $g_{ac} g^{ac}=\delta_a^a=4$, i.e.,
$$g_{ac} g^{ac} g^{b d}=4 g^{b d}.$$
On the right-hand side you get
$$\frac{1}{2} g^{ab} g_{ac} g^{cd}=\frac{1}{2} \delta_c^b g^{cd}=\frac{1}{2} g^{bd}.$$
So why do you think both sides were equal?

 

[PeterDonis]

On the left-hand side you simply have $g_{ac} g^{ac}=\delta_a^a=4$

And, as already noted, this is already wrong, since both the $a$ and $c$ indexes on $g^{ac}$ were already contracted, so you can't contract them again with $g_{ac}$.