Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Steven Weinberg's book "The Quantum Theory of Fields", volume 3, page 46 gives the following argument against N = 3 supersymmetry:

"For global N = 4 supersymmetry there is just one supermultiplet ... This is equivalent to the global supersymmetry theory with N = 3, which has two supermultiplets: 1 supermultiplet... and the other the CPT conjugate supermultiplet... Adding the numbers of particles of each helicity in these two N = 3 supermultiplets gives the same particle content as for N = 4 global supersymmetry"

However, this doesn't directly imply (as far as I can tell) that there is no N = 3 QFT. Such a QFT would have the particle content of N = 4 super-Yang-Mills but it wouldn't have the same symmetry. Is such a QFT known? If not, is it possible to prove it doesn't exist? I guess it might be possible to examine all possible Lagrangians that would give this particle content and show none of them has N = 3 (but not N = 4) supersymmetry. However, is it possible to give a more fundumental argument, relying only on general principles such as Lorentz invariance, cluster decomposition etc. that would rule out such a model?

share|cite|improve this question
Although you do not say this explicitly, your question is about four-dimensional Poincaré supersymemtry. Certainly in three-dimensions, there are $N=3$ theories. – José Figueroa-O'Farrill Sep 16 '11 at 18:45
Of course. I changed the title to make it more precise – Squark Sep 18 '11 at 18:38
Has anybody read Chap. 12 of ["Harmonic Superspace"], titled "N=3 super Yang-Mills theory"? I've never understood whether what they describe is N=3 or N=4 – Yuji Oct 5 '11 at 2:51
@Yuji--- It's N=4 on shell. – Ron Maimon Oct 7 '11 at 5:11
up vote 19 down vote accepted

Depending on what you mean by "exist", the answer to your question is Yes.

There is an $N=3$ Poincaré supersymmetry algebra, and there are field-theoretic realisations. In particular there is a four-dimensional $N=3$ supergravity theory. A good modern reference for the diverse flavours of supergravity theories is Toine Van Proeyen's Structure of Supergravity Theories.


Weinberg's argument is essentially the following observation. Take a massless unitary representation of the $N=3$ Poincaré superalgebra with helicity $|\lambda|\leq 1$. This representation is not stable under CPT, so the CPT theorem says that to realise that in a supersymmetric quantum field theory, you have to add the CPT-conjugate representation. Once you do that, though, the $\oplus$ representation admits in fact an action of the $N=4$ Poincaré superalgebra.

The reason the supergravity theory exists (and is different from $N=4$ supergravity) is that the $N=3$ gravity multiplet, which is a massless helicity $|\lambda|=2$ unitary representation, is already CPT-self-conjugate.

share|cite|improve this answer
OK, although I think that strictly speaking supergravity is not a QFT since a consistent quantization of a gravitational theory presumably requires something else than a QFT, namely superstring theory – Squark Sep 17 '11 at 18:07
Yes, I agree. But Weinberg's argument is purely kinematical. It's a property of the unitary representation theory of the $N=3$ Poincaré superalgebra with the additional requirement of CPT invariance. – José Figueroa-O'Farrill Sep 17 '11 at 18:26
Also, although $N=3$ supergravity is probably not renormalisable, this is not the case for all supergravity theories. There are many indications that $N=8$ supergravity is actually finite. See, e.g., – José Figueroa-O'Farrill Sep 17 '11 at 18:29
I feel that there are deep reasons that no QFT can be a theory of gravity (except in the holographic sense), but it is different subject. I still don't know the answer to my original question, namely is there an (honest, non-gravitational) QFT in 4D with N = 3 supersymmetry? – Squark Sep 18 '11 at 18:40
I'm accepting the answer since although quantum gravity is not a QFT, the existence of N=3 quantum gravity probably rules out the possibility for a no-go theorem along the lines of basic principles (since most of them aren't violated by gravity and the principle that is violated - locality - is only violated in a rather subtle way). – Squark Nov 12 '11 at 14:15

The discussion on pages 168-173 in Weinberg vol III looks to exclude rigid $N=3$ supersymmetric QFTs in 4d, at least those which are renormalisable and with a lagrangian description.

The first step is to note that, in order to identify the CPT-self-conjugate $N=4$ supermultiplet with the $N=3$ supermultiplet plus its CPT-conjugate, one must assume that all fields in both supermultiplets are valued in the adjoint representation of the gauge group. In $N=1$ language, the basic constituents in both supermultiplets are one gauge and three chiral supermultiplets, all adjoint-valued. The three chiral supermultiplets must transform as a triplet under the ${\mathfrak{su}}(3)$ part of the ${\mathfrak{u}}(3)$ R-symmetry of the $N=3$ superalgebra.

Any renormalisable lagrangian field theory in 4d that has a rigid $N \geq 2$ supersymmetry must take the form given by (27.9.33) in Weinberg. This just corresponds to the generic on-shell coupling of rigid $N=2$ vector and hyper multiplets, with renormalisable $N=2$ superpotential (27.9.29). For $N>2$, vector and hyper multiplets must both transform in the adjoint representation of the gauge group. ($N=2$ requires only that the hypermultiplet transforms in a real representation of the gauge group, i.e. a "non-chiral" representation in $N=1$ language.) Putting in this assumption, the $N>2$ case is easily deduced using Weinberg's analysis below (27.9.34). All terms except those in the last two lines of (27.9.33) assemble into precisely the $N=4$ supersymmetric Yang--Mills lagrangian. The remaining terms in the last two lines of (27.9.33) depend on a matrix $\mu$ which defines the quadratic term in the superpotential. As Weinberg argues, $N=4$ occurs only if these terms all vanish identically (e.g. if $\mu =0$). Whence $N=3$ can occur only if the terms in the last two lines of (27.9.33) are non-vanishing and $N=3$ supersymmetric on their own. This would require them to be invariant under the ${\mathfrak{u}}(3)$ R-symmetry of the $N=3$ superalgebra. However, only two of the three chiral superfields (coming from the hypermultiplet) appear in the $\mu$-dependent terms. Since the three chiral supermultiplets must transform as an ${\mathfrak{su}}(3)$ triplet under the R-symmetry, it is clearly impossible for the last two lines in (27.9.33) to be ${\mathfrak{u}}(3)$-invariant unless they vanish identically. Whence, $N>2$ implies $N=4$ in this context.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.