Csdrhrt

I know the Bianchi identity can be derived much more directly and simply (as is apparent in this post).

The point here is to follow this alternate path to it as proposed in the paragraph after equation (4.2.52) of Vol. 1 of "Spinors and space-time" by Penrose and Rindler:

For any pair of covariant derivatives $∇_ρ$ and $tilde ∇$, a unique tensor $Q_{αβ}^{ γ}$ exists such that

$ $
$tilde ∇_ρH_{λ...ν}^{α...γ} = ∇_ρH_{λ...ν}^{α...γ}
+ Q_{ρα_0}^{ α}H_{λ...ν}^{α_0...γ} +...+ Q_{ργ_0}^{ γ}H_{λ...ν}^{α...γ_0}$
$- Q_{ρλ}^{ λ_0}H_{λ_0...ν}^{α...γ} -...- Q_{ρν}^{ ν_0}H_{λ...ν_0}^{α...γ}.label{a}tag{4.2.48}$

Conversely, given $∇_ρ$ and any tensor $Q_{αβ}^{ γ}$, the operator $tilde ∇_ρ$ defined in (4.2.48) is a covariant derivative, whose torsion, $tilde T$, and curvature, $tilde R$, in terms of $Q$ and the original $T$, $R$ and $∇$ are given by

$tilde T_{αβ}^{ γ} = T_{αβ}^{ γ} - 2Q_{[αβ]}^{ γ}tag{4.2.50}$

$tilde R_{αβγ}^{ δ}
= R_{αβγ}^{ δ}
- T_{αβ}^{ ρ} Q_{ργ}^{ δ}
+2∇_{[α} Q_{β]γ}^{ δ}
+2Q_{[α|ρ|}^{ δ}Q_{β]γ}^{ ρ}.label{b}tag{4.2.51}$

By choosing $Q_{αβ}^{ γ}=frac{1}{2} T_{αβ}^{ γ}$, $tilde ∇_ρ$ is torsion-free, so the torsion-free Bianchi identity $tilde ∇_{[α} tilde R_{βγ]ρ}^{ σ} = 0$ applies.

The problem is to plug ref{b} into this, then expand it using ref{a} and somehow get the non-torsion-free Bianchi identity for $∇_ρ$ to pop out, i.e. reduce the maze of tensors in the resulting expansion to nothing but

$∇_{[α} R_{βγ]ρ}^{ δ}
+ T_{[αβ}^{ τ} R_{γ]τρ}^{ δ} = 0$.

To get the complete set of terms to vanish it is also necessary to apply one more nontrivial identity, namely
$ $
$(2∇_{[α}∇_{β]} - T_{αβ}^{ γ}∇_γ)H_{λ...ν}^{ρ...τ}
= R_{αβγρ_0}^{ ρ}H_{λ...ν}^{ρ_0...τ} +...+ R_{αβγτ_0}^{ τ}H_{λ...ν}^{ρ...τ_0}$
$- R_{αβγλ}^{ λ_0}H_{λ_0...ν}^{ρ...τ} -...- R_{αβγν}^{ ν_0}H_{λ...ν_0}^{ρ...τ},tag{4.2.33}$

called the "(generalized) Ricci identity" in Penrose and Rindler.

Good luck to all brave souls embarking on this hazard-strewn journey!

Will any one find a less tortuous way than I have yet been able to?

Although highlighting via bolding and color-coding helps give an overview of what's going on below, it is still very tedious to follow, and I see no way to reduce the maze of tensors that appear in the initial expansions. If anyone can point out easier way verify the result, that certainly would be interesting.

Expanding the alternate curvature before the differential:

$tilde ∇_{[α} tilde R_{βγ]ρ}^{ σ}
= color{red}{tilde ∇_{[α} R_{β γ]ρ}^{ σ}}
- color{magenta}{tilde ∇_{[α} (T_{βγ]}^{ δ} Q_{δρ}^{ σ})}
+color{green}{2tilde ∇_{[α}(∇_{[β} Q_{γ]]ρ}^{ σ})}
+color{blue}{2tilde ∇_{[α} (Q_{[β|δ|}^{ σ}Q_{γ]]ρ}^{ δ})}$

$color{red}{tilde ∇_{[α} R_{βγ]ρ}^{ σ}}
= ∇_{[α} R_{βγ]ρ}^{ σ}
+ Q_{[α|δ|}^{ σ} R_{βγ]ρ}^{ δ}
- Q_{[αβ}^{ δ} R_{|δ|γ]ρ}^{ σ}
- Q_{[αγ}^{ δ} R_{β]δρ}^{ σ}
- Q_{[α|ρ|}^{ δ} R_{βγ]δ}^{ σ}$

$- color{magenta}{tilde ∇_{[α} (T_{βγ]}^{ δ} Q_{δρ}^{ σ})}
= -(tilde ∇_{[α} T_{βγ]}^{ δ}) Q_{δρ}^{ σ}
- (tilde ∇_{[α} Q_{|δρ|}^{ σ})T_{βγ]}^{ δ}$
$
= -∇_{[α} (T_{βγ]}^{ δ}) Q_{δρ}^{ σ}
- Q_{[α|τ|}^{ δ} T_{βγ]}^{ τ} Q_{δρ}^{ σ}
+ Q_{[αβ}^{ τ} T_{|τ|γ}^{ δ} Q_{δρ}^{ σ}
+ Q_{[αγ}^{ τ} T_{β]τ}^{ δ} Q_{δρ}^{ σ}$
$
- (∇_{[α} Q_{|δρ|}^{ σ})T_{βγ]}^{ δ}
- Q_{[α|τ}^{ σ} Q_{δρ|}^{ τ} T_{βγ]}^{ δ}
+ Q_{[α|δ}^{ τ} Q_{τρ|}^{ σ} T_{βγ]}^{ δ} + Q_{[α|ρ}^{ τ} Q_{δτ|}^{ σ} T_{βγ]}^{ δ}$

$color{green}{2tilde ∇_{[α}(∇_{[β} Q_{γ]]ρ}^{ σ})}
= 2∇_{[α}(∇_{[β} Q_{γ]]ρ}^{ σ})
+ 2Q_{[α|τ|}^{ σ} ∇_β Q_{γ]ρ}^{ τ}
- 2Q_{[αβ}^{ τ} ∇_{[|τ|} Q_{γ]]ρ}^{ σ}$
$
- 2Q_{[αγ}^{ τ} ∇_{[β]} Q_{τ]]ρ}^{ σ}
- 2Q_{[α|ρ|}^{ τ} ∇_β Q_{γ]τ}^{ σ}$

$color{blue}{2tilde ∇_{[α} (Q_{[β|δ|}^{ σ}Q_{γ]]ρ}^{ δ})}
= 2(tilde ∇_{[α} Q_{[β|δ|}^{ σ})Q_{γ]]ρ}^{ δ}
- 2(tilde ∇_{[α} Q_{[β|ρ|}^{ δ})Q_{γ]]δ}^{ σ}$
$
= 2(∇_{[α} Q_{β|δ|}^{ σ})Q_{γ]ρ}^{ δ}
+ 2Q_{[α|τ|}^{ σ} Q_{[β|δ|}^{ τ} Q_{γ]ρ}^{ δ}
- 2Q_{[αβ}^{ τ} Q_{[|τ||δ|}^{ σ} Q_{γ]]ρ}^{ δ}
- 2Q_{[α|δ|}^{ τ} Q_{β|τ|}^{ σ} Q_{γ]ρ}^{ δ}$
$
- 2(∇_{[α} Q_{β|ρ|}^{ δ})Q_{γ]δ}^{ σ}
- 2Q_{[α|τ|}^{ δ} Q_{[β|ρ|}^{ τ} Q_{γ]δ}^{ σ}
+ 2Q_{[αβ}^{ τ} Q_{[|τ||ρ|}^{ δ} Q_{γ]]δ}^{ σ}
+ 2Q_{[α|ρ|}^{ τ} Q_{β|τ|}^{ δ} Q_{γ]δ}^{ σ}$
$ $

Substituting $frac{1}{2} T_{αβ}^{ γ}$ for $Q_{αβ}^{ γ} $on RHSs:

$color{red}{tilde ∇_{[α} R_{βγ]ρ}^{ σ}}
= ∇_{[α} R_{βγ]ρ}^{ σ}
+ frac{1}{2} T_{[α|δ|}^{ σ} R_{βγ]ρ}^{ δ}
- frac{1}{2} T_{[αβ}^{ δ} R_{|δ|γ]ρ}^{ σ}
- frac{1}{2} T_{[αγ}^{ δ} R_{β]δρ}^{ σ}
- frac{1}{2} T_{[α|ρ|}^{ δ} R_{βγ]δ}^{ σ}$
$
= ∇_{[α} R_{βγ]ρ}^{ σ}
+ T_{[αβ}^{ δ} R_{γ]δρ}^{ σ}
- ∇_{[α}(∇_β T_{γ]ρ}^{ σ})
{bf - frac{1}{2} R_{[αβγ]}^{ δ}T_{δρ}^{ σ}}
+ frac{1}{2} T_{[αβ}^{ τ} ∇_{|τ|} T_{γ]ρ}^{ σ}$

$- color{magenta}{tilde ∇_{[α} (T_{βγ]}^{ δ} Q_{δρ}^{ σ})}
= -(tilde ∇_{[α} T_{βγ]}^{ δ}) Q_{δρ}^{ σ}
- (tilde ∇_{[α} Q_{|δρ|}^{ σ})T_{βγ]}^{ δ}$
$
= -frac{1}{2} (∇_{[α} T_{βγ]}^{ δ}) T_{δρ}^{ σ}
- frac{1}{4} T_{[α|τ|}^{ δ} T_{βγ]}^{ τ} T_{δρ}^{ σ}
+ frac{1}{4} T_{[αβ}^{ τ} T_{|τ|γ]}^{ δ} T_{δρ}^{ σ}
+ frac{1}{4} T_{[αγ}^{ τ} T_{β]τ}^{ δ} T_{δρ}^{ σ}$
$
- frac{1}{2} (∇_{[α} T_{|δρ|}^{ σ})T_{βγ]}^{ δ}
- frac{1}{4} T_{[α|τ}^{ σ} T_{δρ|}^{ τ} T_{βγ]}^{ δ}
+ frac{1}{4} T_{[α|δ}^{ τ} T_{τρ|}^{ σ} T_{βγ]}^{ δ}
+ frac{1}{4} T_{[α|ρ}^{ τ} T_{δτ|}^{ σ} T_{βγ]}^{ δ}$

$
= {bf -frac{1}{2} (∇_{[α} T_{βγ]}^{ δ}) T_{δρ}^{ σ}
- frac{1}{2} T_{[αβ}^{ τ} T_{γ]τ}^{ δ} T_{δρ}^{ σ}}
+ color{purple}{frac{1}{4} T_{[αγ}^{ τ} T_{β]τ}^{ δ} T_{δρ}^{ σ}}$
$
- frac{1}{2} (∇_{[α} T_{|δρ|}^{ σ})T_{βγ]}^{ δ}
- frac{1}{4} T_{[α|τ}^{ σ} T_{δρ|}^{ τ} T_{βγ]}^{ δ}
+ color{purple}{frac{1}{4} T_{[α|δ}^{ τ} T_{τρ|}^{ σ} T_{βγ]}^{ δ}}
+ frac{1}{4} T_{[α|ρ}^{ τ} T_{δτ|}^{ σ} T_{βγ]}^{ δ}$

$color{green}{2tilde ∇_{[α}(∇_{[β} Q_{γ]]ρ}^{ σ})}
= ∇_{[α}(∇_β T_{γ]ρ}^{ σ})
+ color{cyan}{frac{1}{2} T_{[α|τ|}^{ σ} ∇_β T_{γ]ρ}^{ τ}}
- frac{1}{2} T_{[αβ}^{ τ} ∇_{[|τ|} T_{γ]]ρ}^{ σ}$
$
- frac{1}{2} T_{[αγ}^{ τ} ∇_{[β]} T_{τ]]ρ}^{ σ}
- color{violet}{frac{1}{2} T_{[α|ρ|}^{ τ} ∇_β T_{γ]τ}^{ σ}}$

$color{blue}{2tilde ∇_{[α} (Q_{[β|δ|}^{ σ}Q_{γ]]ρ}^{ δ})}
= 2(tilde ∇_{[α} Q_{[β|δ|}^{ σ})Q_{γ]]ρ}^{ δ}
- 2(tilde ∇_{[α} Q_{[β|ρ|}^{ δ})Q_{γ]]δ}^{ σ}$
$
= color{violet}{frac{1}{2} (∇_{[α} T_{β|δ|}^{ σ})T_{γ]ρ}^{ δ}}
+ color{orange}{frac{1}{4} T_{[α|τ|}^{ σ} T_{β|δ|}^{ τ} T_{γ]ρ}^{ δ}}
- frac{1}{4} T_{[αβ}^{ τ} T_{[|τ||δ|}^{ σ} T_{γ]]ρ}^{ δ}
- color{olive}{frac{1}{4} T_{[α|δ|}^{ τ} T_{β|τ|}^{ σ} T_{γ]ρ}^{ δ}}$
$
- color{cyan}{frac{1}{2} (∇_{[α} T_{β|ρ|}^{ δ})T_{γ]δ}^{ σ}}
- color{olive}{frac{1}{4} T_{[α|τ|}^{ δ} T_{β|ρ|}^{ τ} T_{γ]δ}^{ σ}}
+ frac{1}{4} T_{[αβ}^{ τ} T_{[|τ||ρ|}^{ δ} T_{γ]]δ}^{ σ}
+ color{orange}{frac{1}{4} T_{[α|ρ|}^{ τ} T_{β|τ|}^{ δ} T_{γ]δ}^{ σ}}$
$ $

All terms with three $T$s cancel:

The matching colored terms cancel directly. For the uncolored terms, expanding the second anti-symmetrization of the two terms with intersecting anti-symmetrizations reveals how they cancel in combination with the other two uncolored terms.

Except for the one in bold, all terms with two $T$s also cancel:

Again, matching colored terms cancel directly and expanding the intersecting symmetrizations reveals how the unbolded uncolored terms cancel.

The two terms with $∇_{[α}(∇_β T_{γ]ρ}^{ σ})$ of course also cancel.

Finally, all the three bold terms cancel because of the non-torsion-free Bianchi symmetry, leaving the full expansion of $tilde ∇_{[α} tilde R_{βγ]ρ}^{ σ}$ with only the first two terms of the final expansion of its first (red) term.

Having chosen $Q_{αβ}^{ γ} = frac{1}{2} T_{αβ}^{ γ} $, $tilde{T} $ is zero, so the torsion free version of Bianchi symmetry applies to $tilde{R} $, in other words $tilde ∇_{[α} tilde R_{βγ]ρ}^{ σ} = 0 $.

Therefore the first equation above reduces to

$0 = ∇_{[α} R_{βγ]ρ}^{ σ} + T_{[αβ}^{ δ} R_{γ]δρ}^{ σ} $,

yielding the non-torsion-free version of the Bianchi identity for $∇$, $T$ and $R$
$ $

Check by expanding the alternate differential first:

$tilde ∇_{[α} tilde R_{βγ]ρ}^{ σ}
= color{red}{∇_{[α} tilde R_{βγ]ρ}^{ σ}}
+ Q_{[α|δ|}^{ σ} tilde R_{βγ]ρ}^{ δ}
- Q_{[αβ}^{ δ} tilde R_{|δ|γ]ρ}^{ σ}
- Q_{[αγ}^{ δ} tilde R_{β]δρ}^{ σ}
- Q_{[α|ρ|}^{ δ} tilde R_{βγ]δ}^{ σ}$
$
= color{red}{∇_{[α} tilde R_{βγ]ρ}^{ σ}}
+ Q_{[α|δ|}^{ σ} R_{βγ]ρ}^{ δ}
- Q_{[α|δ|}^{ σ} T_{βγ]}^{ τ} Q_{τρ}^{ δ}
+ 2Q_{[α|δ|}^{ σ} ∇_β Q_{γ]ρ}^{ δ}
+ 2Q_{[α|δ|}^{ σ} Q_{β|τ|}^{ δ} Q_{γ]ρ}^{ τ}$
$
- Q_{[αβ}^{ δ} R_{|δ|γ]ρ}^{ σ}
+ Q_{[αβ}^{ δ} T_{|δ|γ]}^{ τ} Q_{τρ}^{ σ}
- 2Q_{[αβ}^{ δ} ∇_{[|δ|} Q_{γ]]ρ}^{ σ}
- 2Q_{[αβ}^{ δ} Q_{[|δ||τ|}^{ σ} Q_{γ]]ρ}^{ τ}$
$
- Q_{[αγ}^{ δ} R_{β]δρ}^{ σ}
+ Q_{[αγ}^{ δ} T_{β]δ}^{ τ} Q_{τρ}^{ σ}
- 2Q_{[αγ}^{ δ} ∇_{[β]} Q_{δ]ρ}^{ σ}
- 2Q_{[αγ}^{ δ} Q_{[β]|τ|}^{ σ} Q_{δ]ρ}^{ τ}$
$
- Q_{[α|ρ|}^{ δ} R_{βγ]δ}^{ σ}
+ Q_{[α|ρ|}^{ δ} T_{βγ]}^{ τ} Q_{τδ}^{ σ}
- 2Q_{[α|ρ|}^{ δ} ∇_β Q_{γ]δ}^{ σ}
- 2Q_{[α|ρ|}^{ δ} Q_{β|τ|}^{ σ} Q_{γ]δ}^{ τ}$
$ $

Substituting $frac{1}{2} T_{αβ}^{ γ}$ for $Q_{αβ}^{ γ} $:

$tilde ∇_{[α} tilde R_{βγ]ρ}^{ σ}
= color{red}{∇_{[α} tilde R_{βγ]ρ}^{ σ}}
+ frac{1}{2} T_{[α|δ|}^{ σ} R_{βγ]ρ}^{ δ}
- frac{1}{4} T_{[α|δ|}^{ σ} T_{βγ]}^{ τ} T_{τρ}^{ δ}
+ color{magenta}{frac{1}{2} T_{[α|δ|}^{ σ} ∇_β T_{γ]ρ}^{ δ}}
+ color{blue}{frac{1}{4} T_{[α|δ|}^{ σ} T_{β|τ|}^{ δ} T_{γ]ρ}^{ τ}}$
$
- frac{1}{2} T_{[αβ}^{ δ} R_{|δ|γ]ρ}^{ σ}
{bf + frac{1}{4} T_{[αβ}^{ δ} T_{|δ|γ]}^{ τ} T_{τρ}^{ σ}}
- frac{1}{2} T_{[αβ}^{ δ} ∇_{[|δ|} T_{γ]]ρ}^{ σ}
- frac{1}{4} T_{[αβ}^{ δ} T_{[|δ||τ|}^{ σ} T_{γ]]ρ}^{ τ}$
$
- frac{1}{2} T_{[αγ}^{ δ} R_{β]δρ}^{ σ}
{bf + frac{1}{4} T_{[αγ}^{ δ} T_{β]δ}^{ τ} T_{τρ}^{ σ}}
- frac{1}{2} T_{[αγ}^{ δ} ∇_{[β]} T_{δ]ρ}^{ σ}
- frac{1}{4} T_{[αγ}^{ δ} T_{[β]|τ|}^{ σ} T_{δ]ρ}^{ τ}$
$
- frac{1}{2} T_{[α|ρ|}^{ δ} R_{βγ]δ}^{ σ}
+ frac{1}{4} T_{[α|ρ|}^{ δ} T_{βγ]}^{ τ} T_{τδ}^{ σ}
- color{green}{frac{1}{2} T_{[α|ρ|}^{ δ} ∇_β T_{γ]δ}^{ σ}}
- color{blue}{frac{1}{4} T_{[α|ρ|}^{ δ} T_{β|τ|}^{ σ} T_{γ]δ}^{ τ}}$

$color{red}{∇_{[α} tilde R_{βγ]ρ}^{ σ}}
= ∇_{[α} R_{βγ]ρ}^{ σ}
- ∇_{[α} (T_{βγ]}^{ δ}Q_{δρ}^{ σ})
+ 2∇_{[α}∇_β Q_{γ]ρ}^{ σ}
+ 2∇_{[α}(Q_{β|δ|}^{ σ}Q_{γ]ρ}^{ δ})$
$
= ∇_{[α} R_{βγ]ρ}^{ σ}
- (∇_{[α} T_{βγ]}^{ δ})Q_{δρ}^{ σ}
- (∇_{[α} Q_{|δρ|}^{ σ})T_{βγ]}^{ δ}
+ 2∇_{[α}∇_β Q_{γ]ρ}^{ σ}$
$
+ 2(∇_{[α}Q_{β|δ|}^{ σ})Q_{γ]ρ}^{ δ}
- 2(∇_{[α}Q_{β|ρ|}^{ δ})Q_{γ]ρ}^{ σ}$
$ $

Again, substituting $frac{1}{2} T_{αβ}^{ γ}$ for $Q_{αβ}^{ γ} $:

$color{red}{∇_{[α} tilde R_{βγ]ρ}^{ σ}}
= ∇_{[α} R_{βγ]ρ}^{ σ}
- frac{1}{2} (∇_{[α} T_{βγ]}^{ δ})T_{δρ}^{ σ}
- frac{1}{2} (∇_{[α} T_{|δρ|}^{ σ})T_{βγ]}^{ δ}
+ ∇_{[α}∇_β T_{γ]ρ}^{ σ}$
$
+ frac{1}{2} (∇_{[α}T_{β|δ|}^{ σ})T_{γ]ρ}^{ δ}
- frac{1}{2} (∇_{[α}T_{β|ρ|}^{ δ})T_{γ]δ}^{ σ}$
$
= ∇_{[α} R_{βγ]ρ}^{ σ}
-{bf frac{1}{2} (∇_{[α} T_{βγ]}^{ δ})T_{δρ}^{ σ}}
- frac{1}{2} (∇_{[α} T_{|δρ|}^{ σ})T_{βγ]}^{ δ}
+ ∇_{[α}∇_β T_{γ]ρ}^{ σ}
- ∇_{[α}∇_β T_{γ]ρ}^{ σ}$
$
+ frac{1}{2} T_{[αβ}^{ τ} ∇_{|τ|} T_{γ]ρ}^{ σ}
+ frac{1}{2} R_{[αβ|δ|}^{ σ} T_{γ]ρ}^{ δ}
{bf - frac{1}{2} R_{[αβγ]}^{ δ}T_{δρ}^{ σ}}
- frac{1}{2} R_{[αβ|ρ|}^{ δ} T_{γ]δ}^{ σ}$
$
+ color{green}{frac{1}{2} (∇_{[α}T_{β|δ|}^{ σ})T_{γ]ρ}^{ δ}}
- color{magenta}{frac{1}{2} (∇_{[α}T_{β|ρ|}^{ δ})T_{γ]δ}^{ σ}}$

A reduced, slightly different set of terms with three $T$s cancel as above.

All others cancel the same, including those in bold, giving the desired result.