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I have read and heard that there are several theories of gravity and Quantum Gravity which treat the metric (defining rods and clocks) and connection (defining free fall equation of geodesic) of the manifold as independent quantities. I know that the Levi-Civita connections are given as, $$ \Gamma_{\mu\nu}^{\alpha}=\frac{1}{2}g^{\alpha\delta}(g_{\mu\delta,\nu}+g_{\nu\delta,\mu}-g_{\mu\nu,\delta}) $$ But in this case, the connection is uniquely determined by the metric tensor. However, as I see it, the converse is not true (that is the metric cannot be uniquely determined by a given Levi-Civita connection).

So, Given that the Levi-Civita connection can be uniquely determined by the metric tensor, does this fact ascribe certain general properties to the manifold? Hence, by violating these properties of the manifold, one can find other mathematical forms of connection which can be uniquely determined by the metric tensor.

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The Levi-Civita connection (or Christoffel symbols) are coming from the metric compatibility constraint :

$$\tag{1} \nabla_{\lambda} \, g_{\mu \nu} = 0,$$

plus the "desire" to get a symetrical connection : $\Gamma_{\mu \nu}^{\lambda} = \Gamma_{\nu \mu}^{\lambda}$. This last commandement is arbitrary, and is only justified to get the simplest theory possible (i.e classical General Relativity). If you remove the symetrical constraint but only imposes equ (1), you then get the Levi-Civita connection and a contorsion tensor :

$$\tag{2} \Gamma_{\mu \nu}^{\lambda} = \frac{1}{2} \, g^{\lambda \kappa} \, (\, \partial_{\mu} \, g_{\nu \kappa} + \partial_{\nu} \, g_{\mu \kappa} - \partial_{\kappa} \, g_{\mu \nu}) + K^{\lambda}_{\mu \nu}. $$ The antisymetrical part $T^{\lambda}_{\mu \nu} \equiv \Gamma_{\mu \nu}^{\lambda} - \Gamma_{\nu \mu}^{\lambda}$ is the torsion tensor, which is assumed to vanish in GR. This guy is not 0 in the Einstein-Cartan theory, which is the natural extension of GR (a bit more general than GR).

Some theories consider that the connection is an independant variable when they don't impose the symetric constraint, or even the compatibility (1) above.

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  • $\begingroup$ When they say they are independent does that mean that it is not possible to uniquely determine one from the other? For example if I relieve the symmetry requirement on Christoffel symbol, following the standard derivation of Christoffel symbols, it would not be possible to isolate it and write Christoffel symbols uniquely in terms of the metric tensor and its derivatives. In which case the equations determined from vanishing co-variant derivatives of the metric tensor would be a complicated mixture of various Christoffel components and the metric components and that is all that can be said. $\endgroup$
    – Shaz
    Jul 1, 2018 at 14:46
  • $\begingroup$ @Shaz, if you only imposes the compatibility constraint : $\nabla_{\lambda} \, g_{\mu \nu} = 0$, then the algebraic manipulations are almost the same as the standard procedure (in case of a symetric connection). You then get equ (2) above, with $K^{\lambda}_{\mu \nu}$ an arbitrary tensor which is antisymetric in tw0 of its indices. This is pretty standard and can be found in many papers. $\endgroup$
    – Cham
    Jul 1, 2018 at 15:18
  • $\begingroup$ Right. So, about the other part of the question: Do these conditions ascribe certain properties to the manifold they are defined on? Like a pseudo-Reimann manifold with vanishing covariant derivative and symmetric Christoffel symbols? I am thinking about other possibilities of the relation between metric and christoffel symbol. $\endgroup$
    – Shaz
    Jul 1, 2018 at 18:30
  • $\begingroup$ No, these conditions only determines the connection. You can have several manifolds with the same connection, and the reverse ; the same manifold with different connections. The connection is just another structure "superposed" on the manifold, like the metric itself. $\endgroup$
    – Cham
    Jul 1, 2018 at 18:43
  • $\begingroup$ Could you please shed some light on this post physics.stackexchange.com/q/414710. I could not add image here so I created a new question. $\endgroup$
    – Shaz
    Jul 2, 2018 at 6:03

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