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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):

$$\sum_{cyclic\, klm} R^i_{jkl;m} = \sum_{cyclic\, klm} T^s_{km} R^i_{jsl}$$

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:

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

EDIT:

I will expand the cyclic sum a bit further:

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

Moreover:

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

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 ?

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:

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

EDIT:

I will expand the cyclic sum a bit further:

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

Moreover:

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

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 ?

EDIT 2

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.

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):

$$\sum_{cyclic\, klm} R^i_{jkl;m} = \sum_{cyclic\, klm} T^s_{km} R^i_{jsl}$$

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:

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

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 ?

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):

$$\sum_{cyclic\, klm} R^i_{jkl;m} = \sum_{cyclic\, klm} T^s_{km} R^i_{jsl}$$

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:

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

EDIT:

I will expand the cyclic sum a bit further:

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

Moreover:

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

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 ?

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):

$$\sum_{cyclic\, klm} R^i_{jkl;m} = \sum_{cyclic\, klm} T^s_{km} R^i_{jsl}$$

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:

$$ R_{j;m}^m = -g^{ml}g^{ik}(R_{kljm;i} + R_{klmi;j}) + g^{ml}g^{ik}\sum_{cyclic\, klm} T^s_{km} R^i_{jsl}$$

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 ?

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):

$$\sum_{cyclic\, klm} R^i_{jkl;m} = \sum_{cyclic\, klm} T^s_{km} R^i_{jsl}$$

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:

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

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 ?