# Why supergravity is extension of GR?

Why is it that supergravity is an extension to general relativity? In what ways? What I have read until now says the following:

GR got itself concerned with gravity after the revolution of special relativity. Then supersymmetry was a plausible new theory and after the investment of supersymmetry in general relativity, supergravity was born. Thus, this chain tells us that Sugra is an extension of GR.

If this is right, I don't think this is the only way to explain it, is it? How can I technically understand how this is true?

• $\uparrow$ Quote from where? Dec 25 '15 at 14:03
• @Qmechanic This is not a quote but rather a conclusion after reading many introductions and motivations on the subject. Dec 25 '15 at 14:04
• The introduction of the scalar field in Supergravity and Superstrings gives rise to a possible violation of the equivalence principle. What remains from GR ?
– user46925
Dec 25 '15 at 18:29
• Are you pointing out that gravity remains from relativity? @igael Dec 25 '15 at 18:32

OP's question (v2) is fairly broad. There exist various versions of SUGRA, such as e.g., linearized SUGRA, old mimimal SUGRA, new minimal SUGRA, non-minimal SUGRA, extended SUGRA, etc; with or without various matter multiplets such as e.g., chiral multiplet, vector multiplet, complex linear multiplet, etc; and in various spacetime dimensions. Here we will only try to indicate a road map to one pure SUGRA.

1. In the $D=4$ Einstein-Hilbert action of GR, the metric tensor $g_{mn}(x)$ is the dynamical variable. Here the so-called curved indices $m,n=0,1,2,3$ runs over spacetime dimensions. The spacetime is locally isomorphic to $\mathbb{R}^4$. The Euler-Lagrange (EL) equations are the EFE.

2. In order to couple fermionic matter to gravity, we need to use a vielbein $e^a{}_m(x)$ and a spin connection $\omega_{m,ab}(x)$ instead. Here $a,b=0,1,2,3$ are so-called flat indices. See e.g. my Phys.SE answer here. The EL eqs. are again the EFE.

3. The symmetries of GR in the vielbein formalism are:

• (i) reparametrization-invariance of spacetime $\delta x^m =\xi^m(x);$ and
• (ii) local Lorentz invariance $\delta e^a{}_m(x)=\lambda^a{}_b(x)~e^b{}_m(x)$ of the tangent frame.

4. Next it is convenient to use the ${\cal N}=1$ $D=4$ superfield formalism. Spacetime is replaced by a supermanifold, which is locally isomorphic to $\mathbb{R}^{4|4}$ with local coordinates $z^M=(x^m,\theta^{\mu},\bar{\theta}_{\dot{\mu}})$, where the new coordinates $\theta^{\mu}$ and $\bar{\theta}_{\dot{\mu}}$ are Grassmann-odd left- and right-handed Weyl spinors, respectively. The curved and flat indices are now replaced by $M=(m,\mu, \dot{\mu})$ and $A=(a,\alpha, \dot{\alpha})$, respectively.

5. The vielbein and spin connection are replaced by super-versions $E^A{}_M(z)$ and $\Omega_{M,AB}(z)$, respectively.

6. Similarly, the symmetries of SUGRA are super-versions of

• (i) reparametrization-invariance $\delta z^M =\Xi^M(z);$ and
• (ii) local Lorentz invariance $\delta E^a{}_M(z)=\Lambda^A{}_B(z)~E^B{}_M(z)$.
7. The original GR vielbein and spin connection, $e^a{}_m(x)$ and $\omega_{m,ab}(x)$, should be identified in pertinent Wess-Zumino-like gauge as a component sector of the super-versions $E^A{}_M(z)$ and $\Omega_{M,AB}(z)$, respectively. Also the EFE should appear in an appropriate sector of the SUGRA theory.

8. Finally, let us mention that the torsion tensor $T^{a}{}_{bc}(x)$ fits seamlessly into the vielbein formalism, cf. e.g. this Phys.SE post. Of course, all observations until now indicate that $T^{a}{}_{bc}(x)=0$ vanishes. Interestingly, it turns out that certain sectors of the corresponding super-version ${\cal T}^{A}{}_{BC}(z)\neq 0$ are not zero.

References:

1. J. Wess & J. Bagger, SUSY & SUGRA, 1992.

2. D.Z. Freedman & A. Van Proeyen, SUGRA, 2012.

• If we were i $N=2, D=4$ instead, what replaces the vielbein and the spin connection there? Dec 27 '15 at 21:38

Supergravity arises when we require that supersymmetry be a gauge symmetry. By this we mean that the Lagrangian that describes supergravity must be invariant if there is an infinitesimal supersymmetric tranformation that is an arbitrary function of position.

An infinitesimal supersymmetric operation produces a displacement in space. So for supersymmetry to be a local gauge symmetry we require the Lagrangian be invariant with respect to arbitrary infinitesimal displacements in space. But such displacements constitute a diffeomorphism, and the theory that is invariant under diffeomorphisms is general relativity. That's why any gauge theory that has supersymmetry as a local gauge transformation must have general relativity as a classical limit.

• Thanks a lot for your answer! Can you just elaborate on what you mean by "But such displacements constitute a diffeomorphism, and the theory that is invariant under diffeomorphisms is general relativity."? Dec 26 '15 at 16:17
• @Beyond-formulas: you need to explain what bit you want elaborated. Explaining what a diffeomorphism is and the role of diffeomorphism invariance in GR is a (lengthy) answer in itself. Have a look through the two Wikipedia articles I've linked and see if they help. Dec 26 '15 at 16:47
• Okay, I have read the articles, but let me rephrase my question above. 1) Firstly, when you said, "an infinitesimal supersymmetric operation produces a displacement in space", how is it that a supersymmetric operation produce that? Also does the word operation that you used in my quote involve an operation such as supersymmetric transformation? 2) Second of all, when you said "such displacements constitute a diffeomorphism" how is that? Maybe the difficulty that I'm facing here is to imagine how "displacements constitute a diffeomorphism" where I know that diffeomorphisms are mappings. Dec 26 '15 at 21:27