# Why torsion vanishes in supergravity?

Why is it (more often than not) considered in supergravity that the torsion vanishes in Cartan's first structure equation? What does the vanishing of the torsion imply?

• A torsion term has, so far, not been observed in nature, so why add it to the equations? OTOH, since Einstein-Cartan theory seems to avoid singularities, maybe it wouldn't be such a bad idea, after all. – CuriousOne Jan 27 '16 at 17:13
• @CuriousOne But what does mean to solve it? I don't know if you're familiar with supergravity, but what happens while people aim to find susy solutions is that they solve for Killing Spinors equations. When they do they end up having some conditions involving the spin connection. Hence, they pack them all together and start plugging them in Cartan's first structure equation and hence they find some conditions resulting from solving that equation. I'm thinking now, there must be something that drove supergravity people to solve for the Cartan's equation with vanishing torsion, so what is it? – PhilosophicalPhysics Jan 27 '16 at 17:32
• All I am saying is that rational choices for theory have to be guided by empirical observations. Generally I don't think that any form of quantized gravity is even borderline rational, but if one wants to do it, anyway, then excluding torsion is rational based on empirical grounds. If supergravity does that, that's fine. Whether supergravity formally even works with torsion, that I can't tell you. Whether it does or does not, I would not take either possibility as much of a hint for anything, though. – CuriousOne Jan 27 '16 at 17:56
• More on torsion in gravity. – Qmechanic Jan 27 '16 at 18:00
• vanish in the title ? – user46925 Feb 2 '16 at 16:59

In SUGRA, torsion does NOT vanish. You are always left with fermionic torsion terms. The spin connection includes torsion terms in SUGRA that does not vanish

$\omega_\mu{}^{a b} (e, \psi_\mu) = \omega_\mu{}^{ab} (e) + \frac12 \bar\psi_\mu \gamma^{[a} \psi^{b]} + \frac14 \bar\psi^a \gamma_\mu \psi^b$

where $\omega_\mu{}^{ab}(e)$ is the torsion-free spin connection and $\psi_\mu$ is the gravitino. So, quite opposite to what you asked, in SUGRA, the torsion terms do not vanish.

• That's true. On the other hand, the condition that you write down means that bosonic components of the super-torsion do vanish. Still, in general also the full super-torsion does not vanish. But there are always strong constraints on the values that the torsion in supergravity may take, see ncatlab.org/nlab/show/torsion+constraints+in+supergravity – Urs Schreiber Aug 24 '16 at 18:27

Generally, the vanishing of torsion means that spacetime looks in each first order infinitesimal patch like the model spacetime (Minkowski spacetime). One may take this as a mathematical formulation of the principle of equivalence. A nice account of this is in section 3 of

John Lott, "The Geometry of Supergravity Torsion Constraints" Comm. Math. Phys. 133 (1990), 563–615, (exposition in arXiv:0108125

And least in 11d supergravity it is the bosonic part of the super-torsion that does vanish. In fact its vanishing here is equivalent to the equations of motion! Requiring the full super-torsion to vanish, in 11d, is equivalent to the EOM for purely bosonic solutions.

This is due to a remarkable result by Candiello-Lechner-Howe. For review and further pointers see the PhysicsForums-Insights article:

To add to what some of the others have said: In 10d supergravities at least, the 3-form field strength $H = dB$ can be re-cast as torsion. But it's usually easier just to say it's a 3-form field strength.

Nonzero fermions do require torsion, but usually when solving SUGRA, one is looking for classical solutions, on which the fermions vanish, and thus there is no torsion.

The two main motivations to not consider torsion are:

• Classical General Relativity (that has zero torsion) nicely fits the experimental data

• Computationally, it's (more) difficult to deal with a non torsion-free connection.

The two motivations are of course strongly related!

Moreover, a quote from Carroll's book could be relevant:

We could drop the demand that the connection be torsion-free, in which case the torsion tensor could lead to additional propagating degrees of freedom. Without going into details, the basic reason why such theories do not receive much attention is simply because the torsion is itself a tensor; there is nothing to distinguish it from other, "non-gravitational" tensor fields. Thus, we do not really lose any generality by considering theories of torsion-free connections (which lead to GR) plus any number of tensor fields, which we can name what we like.

EDIT (after the comments): In second order formalism of General Relativity, you deal only with a torsion free connection and torsion free spin connection. You can indeed show that the difference with the 1° order formalism (with torsion) is a quartic fermionic term. Just to mention, you can even go through the 1.5 formalism, that is a mix of these two. It could be helpful cap. 8 of Supergravity, Van Proeyen-Freedman

• Thanks for your answer. This answers part of my question. Do you know anything concerning the second part of my question? Please see my comment above to CuriousOne for detailed description of my question.. – PhilosophicalPhysics Jan 27 '16 at 17:37