Shouldn't Torsion be eliminated on the basis of Equivalence Principle?

An affinely connected spacetime with a metric compatible connection can, in principle, have a non-vanishing anti-symmetric part; where the definition of the connection is given by defining parallel transport by

$$dA^{\mu} = - dx^{\sigma}\Gamma_{{\sigma}{\rho}}^{\mu}A^{\rho}$$

Now, in the light of Equivalence Principle, one can assert that at a point $\mathcal{P}$, one can always find a local inertial coordinate system with the coordinates of the point $\mathcal{P}$ being {$\xi^{\mu}$} and in this system, the parallel transported version of a vector $\vec{A}$ (whose coordinates at the point $\mathcal{P}$ are $\{A^\rho\}$) from $\mathcal{P}$ to a nearby point situated at coordinates $\{$$\xi^\mu+d\xi^\mu$$\}$ will also be $\{A^\rho\}$. Therefore, the components of the parallel transported version of $\vec{A}$ in a generic coordinate system with coordinates $\{x'^\mu\}$ will be

$\dfrac{\partial x'^{\mu}}{\partial \xi^{\rho}}\bigg|_{ \xi+d\xi}A^\rho$ and the components of the original vector $\vec{A}$ will be $\dfrac{\partial x'^{\mu}}{\partial \xi^{\rho}}\bigg|_{ \xi}A^\rho$. It is important to keep the distinction between coordinate transformation matrices of two different points because they will not generically be equal. Therefore, the difference in the coordinates of the parallel transported and the original vector in the generic coordinate system will become

\begin{aligned} dA'^{\mu} &= \bigg(\dfrac{\partial x'^{\mu}}{\partial \xi^{\rho}}\bigg|_{ \xi+d\xi} - \dfrac{\partial x'^{\mu}}{\partial \xi^{\rho}}\bigg|_{ \xi}\bigg) A^{\rho}\\ &= \dfrac{\partial^2x'^{\mu}}{\partial \xi^{\sigma}\partial\xi^{\rho}}d\xi^{\sigma}A^{\rho} = \dfrac{\partial^2x'^{\mu}}{\partial \xi^{\sigma}\partial\xi^{\rho}}\dfrac{\partial \xi^{\sigma}}{\partial x'^{\kappa}} dx'^{\kappa}A^{\rho} = \dfrac{\partial^2x'^{\mu}}{\partial \xi^{\sigma}\partial\xi^{\rho}}\dfrac{\partial \xi^{\sigma}}{\partial x'^{\kappa}} \dfrac{\partial \xi^{\rho}}{\partial x'^{\nu}}dx'^{\kappa}A'^{\nu} \end{aligned}

Thus, under the light of Equivalence Principle, one can conclude that

$$\Gamma_{{\kappa}{\nu}}^{{'}{\mu}} = - \dfrac{\partial^2x'^{\mu}}{\partial \xi^{\sigma}\partial\xi^{\rho}}\dfrac{\partial \xi^{\sigma}}{\partial x'^{\kappa}} \dfrac{\partial \xi^{\rho}}{\partial x'^{\nu}}$$

The expression obtained is manifestly symmetric in the lower two indices of the connection and thus, the anti-symmetric part of the connection is zero. Doesn't this argument suffice to conclude that although there is no apriori mathematical reason to believe that torsion ($T_{{\mu}{\nu}}^{\lambda}:=-2\Gamma_{[{\mu}{\nu}]}^{\lambda}$) must be zero, under the light of Equivalence Principle, it is proven that the torsion must always vanish? I know that in General Relativity, torsion is indeed zero but I have read at many places that torsion can be incorporated into a theory of gravity and it is just an assumption made in General Relativity that torsion vanishes. I find this inappropriate as the Equivalence Principle dictates the vanishing torsion and one doesn't need to assume that. Also, I wonder how a theory allowing non-vanishing torsion can possibly accommodate Equivalence Principle - without which I think the considered theory should be of no merit to a Physicist.

I don't believe the equivalence principle has anything to do with torsion. There are two common definitions of the equivalence principle, either:

1. The motion of objects under the force of gravity is independent of those objects' composition, or:

2. Experiments done in a local frame cannot distinguish whether that frame is "at rest" or in free fall.

For a theory to satisfy both versions of the principle, it is enough that:

1. Objects in free fall follow geodesics (specifically, autoparallels), and

2. Spacetime is locally Minkowskian.

The first of these conditions is just the standard postulate for the motion of test masses (I originally claimed that it follows from Einstein's equation, but apparently this is in dispute, see comments below), and the second is merely a restatement of the fact that spacetime is a manifold with a Lorentzian metric tensor. Neither condition interacts with the notion of torsion in any way.

Regarding the specifics of your question,

Now, in the light of Equivalence Principle, one can assert that if {$\xi^{\mu}$} represents the locally inertial coordinate system and {$x'^{\mu}$} represents the general coordinate system then parallel transport of a vector in the general coordinate system can be given as

$$dA'^{\mu}= \bigg(\dfrac{\partial x'^{\mu}}{\partial \xi^{\rho}}\bigg|_{ \xi+d\xi} - \dfrac{\partial x'^{\mu}}{\partial \xi^{\rho}}\bigg|_{ \xi}\bigg) A^{\rho}$$

I think you are actually assuming the consequent here. You can't hope to derive the connection coefficients by writing down some special coordinate transformation, because a connection is additional structure beyond the manifold structure which gives you the coordinates.

So, you are getting out the Levi-Civita connection at the end because you have somehow put it in at the beginning. It is hard for me to tell exactly how, because you've done some nonstandard things with your notation. But I can see you are attempting to compare the vector field $A^\mu$ with itself at two neighboring spacetime points, and this process requires input in the form of a differential structure (i.e., a connection). So there is some assumption you are making about how to relate $A^\mu$ at neighboring points where you have implicitly input the Levi-Civita connection.

Also, I wonder how a theory allowing non-vanishing torsion can possibly accommodate Equivalence Principle - without which I think the theory should be of no merit to a Physicist.

Generally if one wants to couple fermions to spacetime curvature, the torsion will be nonvanishing. In such theories, the torsion is not a propagating degree of freedom---it is related algebraically to the fermion field---but as a geometric quantity, it is there.

• There are various proofs in the literature that the geodesic motion of test masses actually follows from Einstein's equation. See this answer: physics.stackexchange.com/q/24359 – Ben Niehoff Mar 8 '17 at 22:12
• Fair enough. I know at least one of those papers takes a test-mass limit of an extended mass and obtains the result, but I guess that doesn't conclusively prove the postulate for test masses, because one has made assumptions about the composition. – Ben Niehoff Mar 8 '17 at 22:39
• Einstein claims that geodesic motion is required for test particles remains disputed. See en.wikipedia.org/wiki/…. Weinberg derives it from the Equivalence Principle which makes much more sense. For it to come from the EFE you have to work some magic like no other singularities, something on the stress energy tensor and limits, or other special assumptions. See philsci-archive.pitt.edu/9158/1/…. Maybe In an unknown quantum theory of gravity, – Bob Bee Mar 9 '17 at 4:16
• OK, I've edited the answer to avoid confusion. – Ben Niehoff Mar 9 '17 at 10:52
• @BenNiehoff By the way, for coupling to fermions, torsion is not necessary. For example, see: physics.stackexchange.com/a/339711/20427 – Dvij D.C. Jun 20 '17 at 12:41

This doesn't directly address your question, but I think you are overestimating the status of the equivalence principle.

For one thing it is notoriously difficult to state what the equivalence principle even is. Einstein's original formulation, grounded in Newtonian theory, was that a constant gravitational field is globally identical to a constantly accelerated frame. In GR a constant gravitational field is also diffeomorphic to Minkowski spacetime, so this doesn't make a lot of sense.

If the "equivalence principle" just means "spacetime is locally Minkowski", I would say that is on fairly solid footing, at least on scales where GR holds. That will, however, be true in any metrical theory of gravity.

The more familiar statement that "the paths of small objects are independent of their compositions" at least requires very careful consideration of what "composition" is supposed to mean and is arguably simply false, even theoretically within GR. In other words, GR predicts that point masses simply do not follow geodesics of their backgrounds.

This is because, as a consequence of GR being a field theory, point-particles deform the gravitational field around them and therefore have an effective finite size. Because the background field typically exerts tidal forces, and because of the motion of the particles, this deformation will generically be asymmetrical such that the point particle is pushed away from geodesic motion.

For example, stellar mass black holes are about a million to a billion times less massive than supermassive ones, so that a stellar mass orbiting a supermassive black hole is approximately a point mass. The spacetime around the supermassive black hole is very nearly the Kerr spacetime. But the stellar mass black holes actually radiate gravitationally, which drives them to inspiral into their supermassive partners. These paths are not geodesics of the Kerr spacetime, which contains no gravitational radiation.

The precise deformation will also certainly depend on the composition of the point mass. For example, a charged mass will also source an electromagnetic field, which will mutually interact with the spacetime curvature in a complicated way, so that its motion will differ from that of an uncharged one.

Accounting for this sort of thing requires a detailed quantitative study of the Einstein equations; i.e. the actual paths cannot be easily predicted from any sort of principle. There's a lot of wiggle room to account for them using theories that "break equivalence".