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  Does 4D N = 3 supersymmetry exist?

+ 18 like - 0 dislike

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?

This post has been migrated from (A51.SE)
asked Sep 16, 2011 in Theoretical Physics by Squark (1,725 points) [ no revision ]
retagged Mar 24, 2014 by dimension10
Although you do not say this explicitly, your question is about four-dimensional Poincaré supersymemtry. Certainly in three-dimensions, there are $N=3$ theories.

This post has been migrated from (A51.SE)
Of course. I changed the title to make it more precise

This post has been migrated from (A51.SE)
Has anybody read Chap. 12 of ["Harmonic Superspace" http://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

This post has been migrated from (A51.SE)
@Yuji--- It's N=4 on shell.

This post has been migrated from (A51.SE)

4 Answers

+ 18 like - 0 dislike

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.

This post has been migrated from (A51.SE)
answered Sep 16, 2011 by José Figueroa-O'Farrill (2,315 points) [ no revision ]
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

This post has been migrated from (A51.SE)
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.

This post has been migrated from (A51.SE)
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., http://relativity.livingreviews.org/Articles/lrr-2002-5/

This post has been migrated from (A51.SE)
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?

This post has been migrated from (A51.SE)
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).

This post has been migrated from (A51.SE)
Thanks, but actually, I think that the answer that should be accepted in Paul's answer below!

This post has been migrated from (A51.SE)
Well, Paul's answer is complementary but unfortunately I can't accept both! :)

This post has been migrated from (A51.SE)
+ 10 like - 0 dislike

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.

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answered Sep 25, 2011 by Paul_1 (340 points) [ no revision ]
+ 2 like - 0 dislike

The arguments given in the other answers for the fact that a $N=3$ 4d QFT is automatically $N=4$ only apply to theories with a Lagrangian description. They do not rule out the possibility of strongly coupled, non-Lagrangian strict (not $N=4$) $N=3$ QFT. In fact, such theories exist, see https://arxiv.org/abs/1512.06434 and follow-up (the construction uses F-theory).

answered May 8, 2017 by 40227 (5,140 points) [ revision history ]
+ 1 like - 0 dislike

Recently new N=3 4d theories have been found (without using an S-fold), to this regard have a look at 




answered Jun 5, 2019 by anonymous [ no revision ]

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