• Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.


Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback


(propose a free ad)

Site Statistics

129 submissions , 109 unreviewed
3,743 questions , 1,307 unanswered
4,705 answers , 19,910 comments
1,470 users with positive rep
460 active unimported users
More ...

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 (1,705 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)

3 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 (2,135 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.

This post has been migrated from (A51.SE)
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 by 40227 (4,440 points) [ revision history ]

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
To avoid this verification in future, please log in or register.

user contributions licensed under cc by-sa 3.0 with attribution required

Your rights