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

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

However, this doesn't directly imply (as far as I can tell) that there is no ${\cal N} = 3$ QFT. Such a QFT would have the particle content of ${\cal N} = 4$ 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 ${\cal N} = 3$ (but not ${\cal N} = 4$) 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?

  • $\begingroup$ Although you do not say this explicitly, your question is about four-dimensional Poincaré supersymemtry. Certainly in three-dimensions, there are $N=3$ theories. $\endgroup$ Sep 16, 2011 at 18:45
  • $\begingroup$ Of course. I changed the title to make it more precise $\endgroup$
    – Squark
    Sep 18, 2011 at 18:38
  • $\begingroup$ Has anybody read Chap. 12 of ["Harmonic Superspace" ebooks.cambridge.org/ebook.jsf?bid=CBO9780511535109], titled "N=3 super Yang-Mills theory"? I've never understood whether what they describe is N=3 or N=4 $\endgroup$
    – Yuji
    Oct 5, 2011 at 2:51
  • $\begingroup$ @Yuji--- It's N=4 on shell. $\endgroup$
    – Ron Maimon
    Oct 7, 2011 at 5:11

3 Answers 3


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.

  • $\begingroup$ 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 $\endgroup$
    – Squark
    Sep 17, 2011 at 18:07
  • 3
    $\begingroup$ 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. $\endgroup$ Sep 17, 2011 at 18:26
  • $\begingroup$ 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., relativity.livingreviews.org/Articles/lrr-2002-5 $\endgroup$ Sep 17, 2011 at 18:29
  • $\begingroup$ 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? $\endgroup$
    – Squark
    Sep 18, 2011 at 18:40
  • $\begingroup$ 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). $\endgroup$
    – Squark
    Nov 12, 2011 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.


Weinberg's argument relies on the existence of Lagrangian/weak coupling regime. Genuine $\mathcal{N}=3$ theories were recently found in this paper. As expected, they are strongly coupled and believed to be non-Lagrangian.


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