First algebraic Bianchi identity of Riemann tensor (cyclic relation)

[binbagsss]

I am guessing that:

$R_{a[bcd]}=0$

can not be derived from the symmetries of

$R_{ab(cd)]}=R_{(ab)cd}=0$

$R_{[ab][cd]}=0$ ?Sorry when I search the proof for it I can not find much, it tends to come up with the covariant Bianchi instead.

I am guessing it will need one of the symmetries above (at least) and the second covariant Bianchi identity to prove it?

Thanks

 

[binbagsss]

* one of the symmetries above (at least) and either the second second Bianchi identity/ def of the Riemann tensor- basically will need involvement of the metric to prove? Thanks.

 

[strangerep]

$$R_{a[bcd]}=0$$ [...]
Sorry when I search the proof for it I can not find much, it tends to come up with the covariant Bianchi instead.

I am guessing it will need one of the symmetries above (at least) and the second covariant Bianchi identity to prove it?

It's easy to prove by brute force, using just the skewsymmetry on the last 2 indices, since the latter means that ##R_{a[bcd]}=0## can be re-written as $$R^a_{~bcd} + R^a_{~cdb} + R^a_{~dbc} = 0 ~.$$ Then just write out the LHS in full. It's not that hard. You'll find that everything cancels to zero, provided ##\Gamma## is symmetric in its lower indices.

 

[binbagsss]

It's easy to prove by brute force... provided ##\Gamma## is symmetric in its lower indices.

that's what I meant. So yeh it does need the definition of the tensor. that's all I was looking for, thanks.

 

[strangerep]

that's what I meant. So yeh it does need the definition of the tensor. that's all I was looking for, thanks.

What tensor? It doesn't need the definition of the metric tensor.

 

[binbagsss]

What tensor? It doesn't need the definition of the metric tensor.

the Riemann as a function of the metric/connection I meant.

 

[strangerep]

the Riemann as a function of the metric/connection I meant.

Oh, I see.

There is a more high-brow way to derive the identity. Start from the Jacobi identity $$[A,[B,C]] + [B,[C,A]] +[C,[A,B]] ~=~ 0 ~,$$and replace the A,B,C with ##\nabla## operators, e.g., ##\nabla_a##, etc. Then let the LHS act on an arbitrary scalar field ##\phi##. You'll get terms like a commutator acting on ##\nabla_? \phi##, and ##\nabla_? \phi## can be treated as a covector. So that gives you 3 terms involving the Riemann tensor. The other terms have ##[\nabla_a, \nabla_b] \phi## but these just involve the torsion, hence they vanish in a torsionless theory. That should give you the 1st Bianchi identity.

The 2nd Bianchi identity can be derived in a similar way, by acting with the Jacobi identity on an arbitrary vector field ##V^\alpha## instead the scalar field ##\phi##.

 

[binbagsss]

Oh, I see.

There is a more high-brow way to derive the identity. Start from the Jacobi identity $$[A,[B,C]] + [B,[C,A]] +[C,[A,B]] ~=~ 0 ~,$$and replace the A,B,C with ##\nabla## operators, e.g., ##\nabla_a##, etc. Then let the LHS act on an arbitrary scalar field ##\phi##.

which uses the definition of the Riemann tensor. my question was answered, as I said.