How to prove Penrose "Bianchi symmetry" with non-zero torsion tensor using abstract indexing?

Question

I want to prove

$R_{[αβγ]}^{\ \ \ \ \ \ \ δ} + ∇_{[α}T_{βγ]}^{\ \ \ \ δ} + T_{[αβ}^{\ \ \ \ ρ}\ T_{γ]ρ}^{\ \ \ \ δ} = 0$

EDIT: A brief discussion of the solution found by Matt is at the bottom of this post.

The equation is called the "Bianchi symmetry" in Vol. 1 of "Spinors and space-time" by Penrose and Rindler, and is given there in equation (4.2.39) (with index $σ$ in place of the $δ$ here). The definitions of $T$, $R$ and all other symbols are taken from that same text, e.g:

$T_{αβ}^{\ \ \ \ γ}$ = Torsion tensor, satisfying $T_{αβ}^{\ \ \ \ γ} \ ∇_γf = ∇_α∇_βf - ∇_β∇_αf$, for scalars $f$
$R_{αβγ}^{\ \ \ \ \ \ \ δ}$ = Curvature tensor, satisfying $∇_α∇_βV^δ - ∇_β∇_αV^δ - T_{αβ}^{\ \ \ \ γ}\ ∇_γV^δ = R_{αβγ}^{\ \ \ \ \ \ \ δ}\ V^γ$, for vectors $V^γ$,
and for covariant vectors, $V_δ$:$\ \ \ \ \ \ \ \ \ \ \ \ \ ∇_α∇_βV_γ - ∇_β∇_αV_γ - T_{αβ}^{\ \ \ \ ρ}\ ∇_ρV_γ = -R_{αβγ}^{\ \ \ \ \ \ \ δ}\ V_δ$

I am trying to use the manner suggested in the text that is "along the same lines" as , but "more elaborate" than how the torsion-free version (4.2.37) is derived there, but also without using (4.2.52). The approach I have taken looks like this:

The first two (identical) equations below are shown as merely modest reformulations of others published (for other purposes) in the book as indicated (noting the fairly obvious book-typo in the first), so I am virtually certain the setup preceding the third equation here is correct.

What then follows are just some fairly straightforward transformations, so I am suspecting the tensor $T_{[αβ}^{\ \ \ \ δ}\ ∇_{γ]}∇_δf$ highlighted at the end must be zero, but am unable to show why, perhaps because it isn't.
(edit: In fact it isn't. See end of question.)

As in (4.2.40), but with f in place of $V^δ$ (not the $V^γ$-typo appearing in my edition):

$2∇_{[[α}∇_{β]}∇_{γ]} f = T_{[αβ}^{\ \ \ \ ρ}∇_{|ρ|}∇_{γ]}f − R_{[αβγ]}^{\ \ \ \ \ \ \ δ}\ ∇_δf$.

Equivalently, as in (4.2.35), but symmetrized in α, β, and γ and including T:

$−R_{[αβγ]}^{\ \ \ \ \ \ \ δ}∇_δ f = 2∇_{[[α}∇_{β]}∇_{γ]}f − T_{[αβ}^{\ \ \ \ ρ}\ ∇_{|ρ|}∇_{γ]}f$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 2∇_{[α}∇_{[β}∇_{γ]]}f − T_{[αβ}^{\ \ \ \ ρ}\ ∇_{|ρ|}∇_{γ]}f$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = ∇_{[α}Δ_{βγ]}f − T_{[αβ}^{\ \ \ \ ρ}\ ∇_{|ρ|}∇_{γ]}f\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $(with $Δ_{βγ} := 2∇_{[β}∇_{γ]}$)
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = ∇_{[α}T_{βγ]}^{\ \ \ \ δ}\ ∇_δf − T_{[αβ}^{\ \ \ \ ρ}\ ∇_{|ρ|}∇_{γ]}f\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $(4.2.22 applied to first term above)
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = ∇_{[α}T_{βγ]}^{\ \ \ \ ρ}\ ∇_δf + T_{[αβ}^{\ \ \ \ ρ}\ T_{γ]ρ}^{\ \ \ \ δ}\ ∇_δf − T_{[αβ}^{\ \ \ \ ρ}∇_{γ]} ∇_ρf\ \ $(4.2.22 applied to last term above)
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = ∇_{[α}T_{βγ]}^{\ \ \ \ δ}\ ∇_δf + T_{[αβ}^{\ \ \ \ ρ}\ T_{γ]ρ}^{\ \ \ \ δ}\ ∇_δf − T_{[αβ}^{\ \ \ \ δ}∇_{γ]}∇_δf$

Comparing to (4.2.39), $T_{[αβ}^{\ \ \ \ δ}\ ∇_{γ]}∇_δf$ must be zero. (edit: Not so. See discussion at end of question below.)

Hints about either why this might be true, or if not where I have been misled along the way will be greatly appreciated! (edit: See below for why not.)

P.S.: This book is a fabulous way for getting familiar with tensor indexing. I'm almost half way through reading it, and up until this point have figured out pretty much every single formula, minor aside and fottnote on my own (except for the curious bits on "irreducibility" at the end of 3.3) and savored every minute of it.

P.P.S.: This is my first post to the forum, and am very impressed by how easy it was to figure out everything I needed, especially using MathJax, a non-trivial bit of software that actually is implemented in an extremely self-explanatory way, unlike so much else that claims to be, but hardly ever is. I also find all the suggestions for making effective posts very sensible and helpful, something else that is very rare. I hope they are adequately reflected in this post!

Making this post has been truly a lot of fun, and I look forward to much more.

Cheers!

EDIT: Discussion of Solution due to Matt

As he points out, the Leibnitz expansion applied to the first term in the last line of my derivation cancels the last term on that line. The reason I missed this was failing to use parentheses in the relevant expression:

$\ ∇_{[α}(T_{βγ]}^{\ \ \ \ δ}\ ∇_δf) = (∇_{[α}T_{βγ]}^{\ \ \ \ δ})\ ∇_δf + T_{[αβ}^{\ \ \ \ δ}(\ ∇_{γ]}∇_δf)$

Penrose's abstract indexing is commutative and associative for the tensor objects themselves, which is one its greatest advantages. His covariant operators $∇_δ$ on the other hand are neither commutative (as he makes very clear) nor associative when followed by two or more tensors. While the latter is not specifically mentioned in the text, it becomes obvious with a small amount of thought, with this non-trivial example serving as a perfect illustration.

From this momentary stumbling block and the resulting great discussion and help on MSE, I have learned much. So now, because it is so much fun, I will show how to derive what Penrose calls the "Bianchi identity" that is actually a simpler formula than the "Bianchi symmetry", though more complicated to derive (it actually depends on the "symmetry" formula):

Derivation of "Bianchi identity"
This is equation (4.2.43) (as well as Fig. A-9) in Vol. 1 of "Spinors and space-time" by Penrose and Rindler, and uses the notation and other concepts and references from that text:

$2∇_{[[α}∇_{β]}∇_{γ]}V^δ = T_{[αβ}^{\ \ \ \ ρ}\ ∇_{|ρ|}∇_{γ]}V^δ + R_{[αβ|ρ|}^{\ \ \ \ \ \ \ \ \ δ}\ ∇_{γ]}V^ρ - R_{[αβγ]}^{\ \ \ \ \ \ \ \ ρ}\ ∇_ρV^δ\ \ \ \ \ \ \ $(4.2.40)
(LHS expanded using "generalized Ricci identity" as defined by Penrose)

$2∇_{[α}∇_{[β}∇_{γ]]}V^δ = ∇_{[α}(T_{βγ]}^{\ \ \ \ ρ}\ ∇_{ρ}V^δ) + R_{[αβ|ρ|}^{\ \ \ \ \ \ \ \ \ δ}\ ∇_{γ]}V^ρ + V^ρ∇_{[α}R_{βγ]ρ}^{\ \ \ \ \ \ \ δ}\ \ \ \ \ \ $(4.2.41)
(LHS expanded using definition of $R$ and Leibnitz law)

Subtracting the first from the second gives (since LHSs are equal):

$0 = V^ρ∇_{[α}R_{βγ]ρ}^{\ \ \ \ \ \ \ δ} - T_{[αβ}^{\ \ \ \ ρ}\ ∇_{|ρ|}∇_{γ]}V^δ$
$\ \ \ \ \ + ∇_{[α}(T_{βγ]}^{\ \ \ \ ρ}\ ∇_{ρ}V^δ) + R_{[αβγ]}^{\ \ \ \ \ \ \ \ ρ}\ ∇_ρV^δ$

$\ \ = V^ρ∇_{[α}R_{βγ]ρ}^{\ \ \ \ \ \ \ δ} - T_{[αβ}^{\ \ \ \ ρ}\ ∇_{|ρ|}∇_{γ]}V^δ$
$\ \ \ \ \ + (∇_{[α}\ ∇_{|ρ|}V^δ)T_{βγ]}^{\ \ \ \ ρ} + (∇_{[α}T_{βγ]}^{\ \ \ \ ρ})\ ∇_ρV^δ + R_{[αβγ]}^{\ \ \ \ \ \ \ \ ρ}\ ∇_ρV^δ$
$\ \ \ \ \ \ \ $(using Leibnitz law)

$\ \ = V^ρ∇_{[α}R_{βγ]ρ}^{\ \ \ \ \ \ \ δ} - T_{[αβ}^{\ \ \ \ ρ}\ ∇_{|ρ|}∇_{γ]}V^δ + T_{[αβ}^{\ \ \ \ ρ}\ ∇_{γ]}\ ∇_ρV^δ$
$\ \ \ \ \ - T_{[αβ}^{\ \ \ \ ρ}\ T_{γ]ρ}^{\ \ \ \ σ}\ ∇_σV^δ - R_{[αβγ]}^{\ \ \ \ \ \ \ \ ρ}\ ∇_ρV^δ + R_{[αβγ]}^{\ \ \ \ \ \ \ \ ρ}\ ∇_ρV^δ$
$\ \ \ \ \ \ \ $(using Bianchi symmetry)

$\ \ = V^ρ∇_{[α}R_{βγ]ρ}^{\ \ \ \ \ \ \ δ} + T_{[αβ}^{\ \ \ \ ρ}\ R_{γ]ρσ}^{\ \ \ \ \ \ \ δ}\ V^σ$
$\ \ \ \ \ \ \ $(using Ricci identity)

$\ \ =\ (∇_{[α}R_{βγ]ρ}^{\ \ \ \ \ \ \ δ}\ )V^ρ + T_{[αβ}^{\ \ \ \ σ}\ R_{γ]σρ}^{\ \ \ \ \ \ \ δ}\ V^ρ$

So we have arrived at the two Penrose formulas for

Bianchi symmetry: $\ \ -R_{[αβγ]}^{\ \ \ \ \ \ \ δ} = ∇_{[α}T_{βγ]}^{\ \ \ \ δ} + T_{[αβ}^{\ \ \ \ ρ}\ T_{γ]ρ}^{\ \ \ \ δ}$

Bianchi identity:$\ -∇_{[α}R_{βγ]ρ}^{\ \ \ \ \ \ \ δ} = T_{[αβ}^{\ \ \ \ σ}\ R_{γ]σρ}^{\ \ \ \ \ \ \ δ}$

Answer 1

First, I think you have some notation backwards. We typically use parentheses for symmetrization and square brackets for antisymmetrization, i.e, $$T_{(ij)}=\frac{1}{2}(T_{ij}+T_{ji}),\qquad T_{[ij]}=\frac{1}{2}(T_{ij}-T_{ji}).$$

Now, I think showing Bianchi's identity is easier to actually use vectors than index notation (especially when not dealing with a symmetric connection, which is your current setting). Also, I apologize for not using your conventions, as when I wrote this up, I forgot to check how you defined torsion and curvature, so I'll be explicit in my conventions.

Let $M$ be a smooth manifold with connection $\nabla$ on the tangent bundle $TM$. Then for smooth vector fields $X,Y,Z\in TM$, define the Riemannian curvature tensor $$R_{XY}Z=[\nabla_X,\nabla_Y]Z-\nabla_{[X,Y]}Z$$ and the torsion tensor $$T(X,Y)=\nabla_XY-\nabla_YX-[X,Y].$$ Let $\frak{S}$ denote the cyclic sum of $X,Y,Z$, e.g., $$\mathfrak{S}(R_{XY}Z)=R_{XY}Z+R_{YZ}X+R_{ZX}Y.$$

Expanding our curvature tensor and rearranging \begin{align*} \mathfrak{S}(R_{XY}Z)&=\nabla_X(\nabla_YZ-\nabla_ZY)+\nabla_Y(\nabla_ZX-\nabla_XZ)+\nabla_Z(\nabla_XY-\nabla_YX)-\nabla_{[X,Y]}Z-\nabla_{[Y,Z]}X-\nabla_{[Z,X]}Y\\ &=\mathfrak{S}(\nabla_X(\nabla_YZ-\nabla_ZY))-\mathfrak{S}(\nabla_{[X,Y]}Z)\\ &=\mathfrak{S}(\nabla_X(T(Y,Z))+[Y,Z])-\mathfrak{S}(\nabla_{[X,Y]}Z)\\ &=\mathfrak{S}(\nabla_X(T(Y,Z))+\mathfrak{S}(\nabla_X[Y,Z])-\mathfrak{S}(\nabla_{[X,Y]}Z). \end{align*} Now, noting that \begin{align*} \nabla_X(T(Y,Z))&=(\nabla_XT)(Y,Z)+T(\nabla_XY,Z)+T(Y,\nabla_XZ)\\ &=(\nabla_XT)(Y,Z)-T(Z,\nabla_XY)+T(Y,\nabla_XZ) \end{align*} by the fact that $T$ is antisymmetric. Hence we get the cyclic sum \begin{align*} \mathfrak{S}(\nabla_X(T(Y,Z))&=\mathfrak{S}((\nabla_XT)(Y,Z))-\mathfrak{S}(T(X,\nabla_YZ-\nabla_ZY))\\ &=\mathfrak{S}((\nabla_XT)(Y,Z))-\mathfrak{S}(T(X,T(Y,Z)+[Y,Z])\\ &=\mathfrak{S}((\nabla_XT)(Y,Z))-\mathfrak{S}(T(X,T(Y,Z))-\mathfrak{S}(T(X,[Y,Z])) \end{align*}

Now, for our other term in the expansion of curvature, we first note that $$\nabla_X[Y,Z]-\nabla_{[Y,Z]}X=T(X,[Y,Z])+[X,[Y,Z]],$$ so rearranging our cyclic sum as usual yields \begin{align*} \mathfrak{S}(\nabla_X[Y,Z])-\mathfrak{S}(\nabla_{[X,Y]}Z)&=\mathfrak{S}(T(X,[Y,Z]))+\mathfrak{S}([X,[Y,Z])\\ &=\mathfrak{S}(T(X,[Y,Z])) \end{align*} by Jacobi's identity for Lie brackets. Adding these final expression and cancelling the two like terms we see that \begin{align*} \mathfrak{S}(R_{XY}Z)&=\mathfrak{S}((\nabla_XT)(Y,Z))-\mathfrak{S}(T(X,T(Y,Z))\\ &=\mathfrak{S}((\nabla_XT)(Y,Z))+\mathfrak{S}(T(T(Y,Z),X)). \end{align*}

Edit and Correction:

I made a mistake in switching to index form, and stand corrected, the Bianchi identity is typically written using the antisymmetrization brackets (I'm not sure what I was thinking; I think I had a moment where I thought the cyclic sum was the non-index version of symmetrization, which is just wrong for more than $2$ indices).

Let consider the antisymmetrization \begin{align*} R_{[ijk]}^{\,\,\,\,\,\,l}&=\frac{1}{6}(R_{ijk}^{\,\,\,\,l}-R_{jik}^{\,\,\,\,l}+R_{jki}^{\,\,\,\,l}-R_{kji}^{\,\,\,\,l}+R_{kij}^{\,\,\,\,l}-R_{ikj}^{\,\,\,\,l})\\ &=\frac{1}{3}(R_{ijk}^{\,\,\,\,l}+R_{jki}^{\,\,\,\,l}+R_{kij}^l), \end{align*} where we used the fact that $R_{ijk}^{\,\,\,\,l}=-R_{jik}^{\,\,\,\,l}$. That is, $$3R_{[ijk]}^{\,\,\,\,\,\,l}X^iY^jZ^k=\mathfrak{S}(R_{XY}Z),$$ where $$X=X^i\partial_i,\qquad Y=Y^i\partial_i,\qquad Z=Z^i\partial_i.$$

Similarly, since the Torsion is antisymmetric $T_{ij}^{\,\,\,l}=-T_{ji}^{\,\,\,l}$, we get identical statements for the other two expressions (you can check this). Cancelling the $3$'s, we conclude thaat $$R_{[ijk]}^{\,\,\,\,\,\,l}=\nabla_{[i}T_{jk]}^{\,\,\,l}+T_{m[i}^{\,\,\,l}T_{jk]}^{\,\,\,\,m}.$$

Now this is exactly what you wanted to show (up to a sign difference in our definition of curvature).

Further Notes The author most likely calls this Bianchi's symmetry instead of Bianchi's identity because that name typically implies a torsion-free connection. That is, Bianchi's identity states $R_{[ijk]}^{\,\,\,\,\,l}=0$, since $T\equiv0$. So your case is more general, and I must admit I don't know of any introductory references which deal with the implications of a non-symmetric connection in detail.

Also, your other comment about the "harder" Bianchi identity is typically called the differential Bianchi identity (again you won't find many derivations with nonzero torsion).

Since you're fairly confident in index manipulation, I suggest writing out a few of the formal proofs you have in that notion again in "operator" form, and comparing the two. Both notations are useful depending on context, and it's good practice to be able to do both.