Compatibility with a metric, also referred to as metricity, means, I believe, that the covariant derivative of the metric is zero:


This is true identically for a symmetric metric and the associated Levi-Civita connection. If a torsion tensor $T$ was added to the connection, this becomes:


Now swap some indices and use the symmetry of g:


Subtracting these two equations gives:


So the antisymmetric component of the covariant derivative of $g_{ij}$ remains zero with the addition of a finite but otherwise arbitrary torsion $T$. Adding those two equations gives;


So it seems the symmetric component of the covariant derivative of $g_{ij}$ is not necessarily zero with arbitrary torsion $T$.

Here's the question, are there any nontrivial solutions for T that preserve metricity?

I have an idea of how to proceed for SR, where $g_{ij}=\eta_{ij}$, where this question seems simpler to approach, but I'm curious about the GR case of a Schwarzchild metric, and then more generally in the case of a Kerr metric, and I imagine and hope there's been some research already done on this subject.

  • $\begingroup$ In reading further here on the subject, I think the answer is no. If I understand the concepts, the symmetric component of the connection is modified from the Levi-Civita connection to maintain metricity with any torsion. In which case, wouldn't the torsion indirectly affect geodesics macroscopically, not just couple to fermion intrinsic spins? $\endgroup$
    – Ric
    Commented May 9 at 2:54

1 Answer 1


I think what's important to consider here is that the Levi-Civita connection is unique. Once you introduce torsion this is no longer true. Therefore, when you write $g_{ij;k}$ it isn't clear if you're considering the derivative with the Levi-Civita connection, or if you're considering a derivative operator with torsion. However, I think the answer to your question may be the following.

If you consider a derivative operator $\nabla_a$ such that

$$[\nabla_a, \nabla_b]f = -T^c_{\ ab} \nabla_c f,$$

where $T^c_{\ ab}$ is the torsion tensor, and for an arbitrary one-form $\omega_a$ you have

$$\nabla_a \omega_b = \partial_a \omega_b - C^c_{\ ab} \omega_c,$$

where $C^c_{\ ab}$ is the connection associated with our derivative operador $\nabla_a$; if you suppose $\nabla_a$ is metric compatible then you can show that

\begin{align} C^c_{\ ab} &= \frac{1}{2} g^{cd} (\partial_a g_{bd} + \partial_b g_{ad} - \partial_d g_{ab}) + \frac{1}{2} (T^c_{\ ab} + T^{\ c}_{a \ \ b} + T^{\ c}_{b \ \ a})\\ &= \Gamma^c_{\ ab} + K^c_{\ ab}, \end{align}

where $\Gamma^c_{\ ab}$ is the Christoffel symbol and $K^c_{\ ab}$ is the contorsion tensor. In fact, from this expresion you can see that if $D_a$ is the derivative operator associated with the Levi-Civita connection, then

$$(D_a - \nabla_a) \omega_b = K^c_{\ ab} \omega_c.$$

Thus, the contorsion tensor is the connection between the derivative operator associated with the Levi-Civita connection and the derivative operator with torsion.

Edit: I'd also like to give a bit of insight into the last part of your question. If you consider the Einstein-Cartan action with an additional matter action

$$S = S_{\text{EC}} [g^{ab}, K^c_{\ ab}] + S_{\text{M}} [g^{ab}, K^c_{\ ab}, \phi],$$

where $\phi$ is an arbitrary matter field, and

$$S_{\text{EC}} = \frac{1}{16 \pi} \int \mathcal{R} \sqrt{-g} \text{d}^4 x,$$

where $\mathcal{R}$ is the Ricci scalar associated with the derivative operator $\nabla_a$; the resultant equations of motion are

\begin{align} &G_{ab} + K^d_{\ (ab)} K^c_{\ cd} - K^d_{\ c(b} K^c_{\ a)d} - K^{d \ \ c}_{\ [c} K^e_{\ e]d} g_{ab} = 8 \pi \tau_{ab},\\ &g^{ab} K^d_{\ dc} + \delta^a_{\ c} K^{bd}_{\ \ \ d} - K^{ab}_{\ \ \ c} - K^{b \ \ a}_{\ \ c} = 16 \pi \sigma_c^{\ \ ab},\\ &\frac{\delta (\mathcal{L}_{\text{M}} \sqrt{-g})}{\delta \phi} = 0. \end{align}

Note that $G_{ab}$ is the Einstein tensor associated with the Levi-Civita derivative operator $D_a$. The stress-energy tensor $\tau_{ab}$ is defined as

$$\tau_{ab} := -\frac{2}{\sqrt{-g}} \frac{\delta (\mathcal{L}_{\text{M}} \sqrt{-g})}{\delta g^{ab}}.$$

The spin density $\sigma_c^{\ ab}$ is defined as

$$\sigma_c^{\ \ ab} := -\frac{1}{\sqrt{-g}} \frac{\delta (\mathcal{L}_{\text{M}} \sqrt{-g})}{\delta K^c_{\ ab}}.$$

What's important is that the equation of motion for the contorsion tensor is algebraic, there are no derivatives involved. It can also be shown that you can write the contorsion tensor in terms of the spin density. This means that if the spin density is zero, then the contorison tensor is zero and viceversa. Outside the region of a given concentration of matter, there's no spin density, which in turn means the contorsion tensor vanishes, and you are left with $G_{ab} = 8 \pi \tau_{ab}$; which is usual GR.

In the case of the Schwarzschild metric as well as the Kerr metric, you're working with a vacuum solution, which implies $\sigma^c_{\ ab} = 0$, which in turn implies $K^c_{\ ab} = 0$, and therefore you obtain the same dynamics that usual GR describes in the black hole's exterior region.

  • 1
    $\begingroup$ Thank you very much for your answer, it's very helpful to me. I need to take a step back and study what I've been reading. $\endgroup$
    – Ric
    Commented May 9 at 5:34
  • 1
    $\begingroup$ You're welcome. I added an edit that may help you with the last part of your question. $\endgroup$
    – R. M.
    Commented May 9 at 5:37
  • $\begingroup$ I seem to notice a cosmological term that depends on $K$. Is this a unique term, or can an additional cosmological term be added? With regard to zero spin density outside of matter, classically this suggests $K=0$, but what about with QM, where the probability density only asymptotically approaches zero, so there could be nonzero $K$ effectively everywhere, albeit small, except around very large concentrations of mass like galaxies. $\endgroup$
    – Ric
    Commented Jun 5 at 1:00
  • 1
    $\begingroup$ You can add an additional cosmological matter term since you can consider the geometric LaGrangian to be $\mathcal{L}_{\text{geo}} \sqrt{-g} = \frac{1}{16 \pi} (\mathcal{R} - 2 \Lambda) \sqrt{-g}$, where $\mathcal{R}$ is the Ricci scalar with torsion. I cannot comment on the latter part of your comment because I don't know, but I speculate a semiclassical analysis of this theory would give some insight. In particular, the contorsion tensor remains a classical field, but you would now consider the expectation value of the spin density tensor with Hadamard states. $\endgroup$
    – R. M.
    Commented Jun 5 at 1:27

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