In case of coupling of the Einstein's Field Equations to fermionic matter torsion is generated. However for a consistent set of EFEs it is required that the Einstein tensor

$$G_{ik} = R_{ik} -\frac{1}{2}g_{ik} R\quad \text{fulfills}\quad G_{i\, ;k}^k=0$$

which yields from the contraction of the second Bianchi identity. However, under non-zero torsion the second Bianchi identity changes its form to ($R^i_{jkl}$ are the components of the curvature tensor and $T^s_{km}$ the components of the torsion tensor):

$$\sum_{cyclic\, klm} R^i_{jkl;m} = \sum_{cyclic\, klm} T^s_{km} R^i_{jsl} \quad\quad\text{from N.Straumann: GR & relativistic Astrophysics}$$

With zero torsion the rhs of this equation would be zero and by using the definition of the Ricci tensor $R_{jl} = g^{ik}R_{ijkl}$ and the "usual" contracted Bianchi identity $G_{i\, ;k}^k=0$ would follow rather easily: by use of the definition and the symmetry of the curvature tensor we can write:

$$R_{j\,;m}^m = g^{ml} R_{jl;m} = g^{ml}g^{ik} R_{ijkl;m} = g^{ml}g^{ik} R_{klij;m}$$

followed by -- using the symmetry properties of the curvature tensor and the definition of the curvature scalar $R =g^{ik}R_{ik}$:

$$ R_{j\, ;m}^m = -g^{ml}g^{ik}(R_{kljm;i} + R_{klmi;j}) = - g^{ik} R_{kj;i} + g^{ik}R_{ki;j} = -R_{j\,;m}^m + R_{;j}$$

from which $G_{i\,\,;k}^k=0$ immediately follows. However, with non-zero torsion this equation would get the form:

$$ R_{j;m}^m = -g^{ml}g^{ik}(R_{kljm;i} + R_{klmi;j}) + g^{ml}g^{ik}\sum_{cyclic\, ijm} T^s_{im} R_{klsj}$$


I will expand the cyclic sum a bit further:

$$\text{cyclic sum}=g^{ml}g^{ik} ( T^s_{im} R_{klsj} + T^s_{mj} R_{klsi} + T^s_{ji} R_{klsm}) = T^{s\,\,l}_i R^i_{lsj} +T^{sl}_j R^i_{lsi} -T^{s\,\,k}_{j} R^m_{ksm}$$


$$\text{cyclic sum} = T^s_{il} R^{il}_{sj} -T^s_{lj} R^l_s +T^s_{jk} R^k_s =T^s_{il} R^{il}_{sj} -T^s_{lj} R^l_s +T^s_{jl} R^l_s $$

So if the torsion tensor were symmetric in its 2 last indices the sum would indeed vanish, however, it does not, otherwise the torsion form $\Theta^s$

$$\Theta^s = \frac{1}{2}T^s_{il} dx^i \wedge dx^l $$

would vanish too, but its non-zero-ness was actually the starting point of my question.

So how non-zero torsion is with the integrability condition $G_{i\,\,;k}^k=0$ compatibel ? Does the contracted torsion term eventually disappear or do we have to add a new contribution from the torsion to the energy-momentum tensor ?


For clarification I'd like to add the definition of cyclic sum used in this post:

$$\sum_{cyclic\,\, klm} R^i_{klm}:=R^i_{klm} + R^i_{lmk} + R^i_{mkl}$$

If this sum is set to zero it corresponds to the first Bianchi-identity (here without torsion). This how the above cited source (N.Straumann) also notes the first Bianchi-identity. So there cannot be an ambiguity with respect to the definition of the cyclic sum.

  • $\begingroup$ Won't that sum over cyclic KLM create a bunch of pairs of $T_{ab}$ + $T_{ba}$? $\endgroup$ Jun 19, 2021 at 20:09
  • $\begingroup$ @Jerry Schirmer what do you mean with $T_{ab}$ -- the torsion tensor has 3 indices. So $T_{ab}$ is the energy-momentum tensor ? $\endgroup$ Jun 19, 2021 at 22:44
  • $\begingroup$ Sorry, I typoed out the upper index. Irrespecitvely, you're symmetrizing over anti-symmetric indices. $\endgroup$ Jun 20, 2021 at 14:08
  • $\begingroup$ @Jerry Schirmer: At first I thought I understood your argument, but now with the indices right I no longer understand. $\endgroup$ Jun 24, 2021 at 12:47

1 Answer 1


This is only a partial answer, think of it more as an extended comment.

I have originally intended to work this out properly for Einstein-Cartan theory, but I wanted to go with the coordinate/metric formalism rather than the orthonormal tetrad formalism for certain reasons and I realized I am not sure how the degrees of freedom should be split properly in this formalism. Working it out would take too much of my time, so here is a much more general take on the issue.

In a modified/extended theory of gravity, eg. one involving torsion, the field equation is not (necessarily) the Einstein Field equation. What is referred to in the OP as an "integrability condition" is more of a Noether identity or "covariance identity" (see eg. Fatibene/Francaviglia: Natural and Gauge Natural Formalism in Classical Field Theory).

Suppose we have an action functional $S=S[g,\phi]$ depending on the metric tensor $g_{ij}$ and some additional field $\phi^A$ of unspecified type. We do assume that $\phi^A$ is Lie differentiable, so it is a geometric object (such as a tensor or a connection) rather than eg. a gauge field.

The action is assumed diffeomorphism invariant, and let $X$ be a compactly supported vector field that is otherwise arbitrary. The variation with respect to the 1-parameter family of diffeomorphisms induced by $X$ is the Lie derivative, thus we have $$ \delta_XS=\int\left(\frac{\delta S}{\delta g_{ij}}\mathscr L_X g_{ij}+\frac{\delta S}{\delta \phi^A}\mathscr L_X \phi^A\right)\mu_g, $$ where $\mu_g$ is the metric volume element. The Lie derivative of the metric is $$\mathscr L_X g_{ij}=\nabla_iX_j+\nabla_j X_i,$$ where $\nabla_i$ is the Levi-Civita covariant derivative. The Lie derivative of $\phi^A$ is unknown, but it will be a linear differential operator acting on $X$, i.e. of the form $$ \mathscr L_X\phi^A=Q^{Ai_1...i_k}_j\partial_{i_1...i_k}X^j+...+Q^{Ai}_j\partial_iX^j+Q^A_j X^j,$$ where the coefficients $Q$ may depend on $\phi^A$ and its derivatives. To simplify this madness, let us assume that $\phi^A$ is a tensor. Then the Lie derivative can be written in the form $$ \mathscr L_X \phi^A=Q^{Aij}\nabla_i X_j+Q^{Aj}X_j. $$

This is just a simplification, the general line of thought actually remains the same in the general case.

Since the vector field $X$ has compact support, we can integrate by parts freely, thus after integrations by parts we obtain $$ \delta_XS=\int\left(-2\nabla_iE^{ij}X_j-\nabla_i(F_AQ^{Aij})X_j+F_AQ^{Aj}X_j,\right)\mu_g $$ where $$ E^{ij}=\frac{\delta S}{\delta g_{ij}},\quad F_A=\frac{\delta S}{\delta\phi^A}. $$ Since the action is diffeomorphism-invariant, this must vanish, thus we get $$ \nabla_i E^{ij}=\frac{1}{2}\left(\nabla_i(F_AQ^{Aij})-F_AQ^{Aj}\right) $$ as an off-shell identity.

For Einstein gravity with $S$ being the Einstein-Hilbert action, $E^{ij}$ is proportional to the Einstein tensor, and $F_A$ is zero, we thus get back the $\nabla_i G^{ij}=0$ identity.

For some other theory of gravity, especially if it involves another field besides the metric, the required covariance identity will be of the above more complicated form, and thus I don't see the relevance of what happens to the Einstein tensor. The Einstein tensor will still satisfy its divergence identity provided that the Einstein tensor is calculated fully from the metric. The relevant identity as an "integrability condition" however is that I gave above, which just happens to reduce to the contracted Bianchi identity for the case of General Relativity.


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