• 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.


PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

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

205 submissions , 163 unreviewed
5,064 questions , 2,215 unanswered
5,347 answers , 22,731 comments
1,470 users with positive rep
818 active unimported users
More ...

  How to prove quantum N=4 Super-Yang-Mills is superconformal?

+ 5 like - 0 dislike

I'm especially interested in elegant illuminating proofs which don't involve a lot of straightforward technical computations

Also, does a non-perturbative proof exist?

This post has been migrated from (A51.SE)
asked Oct 26, 2011 in Theoretical Physics by Squark (1,725 points) [ no revision ]
retagged Apr 19, 2014 by dimension10

2 Answers

+ 8 like - 0 dislike

In any supersymmetric theory you can choose the gauge coupling to be the coefficient of W_\alpha^2 in the superpotential. This gauge coupling runs only at one-loop, which is a fundamental consequence of the non-renormalization theorems. The other possible running coefficients are the kinetic terms, Z(\mu)QQ^\dagger. These generally get renormalized to all orders in perturbation theory.

In N=4 the one-loop coefficient is zero. This is trivial (just counting the fields). Hence, the gauge coupling (as defined above) does not run. But N=4 relates the gauge particles with the chiral superfields (all the matter particles sit in one big representation of N=4) and so the latter cannot get renormalized either.

This is a slick and intuitive argument... Similar logic operates in many N=2 theories as well.

This post has been migrated from (A51.SE)
answered Nov 2, 2011 by Zohar Ko (650 points) [ no revision ]
Most voted comments show all comments
Yuji: I think you may need to resort to this argument anyway, even if you do Leigh-Strassler. This is because the latter only implies a one-dimensional line of CFTs, but it does not prove that this one-dimensional line coincides with the line on which N=4 sits. So at some point you have to invoke the higher symmetry.

This post has been migrated from (A51.SE)
Ok, so can you provide a ref to the nonrenormalization theorem you are using?

This post has been migrated from (A51.SE)
Page 96 in http://www.physics.uc.edu/~argyres/661/susy1996.pdf reviews the argument. In general you can prove that it is one-loop+non-perturbative contributions. In N=4 since there is no one-loop there is no \Lambda and so also the non-perturbative contributions are absent.

This post has been migrated from (A51.SE)
@Zohar I was just wondering if in reference to the original question one might refer to Seiberg's paper http://www.sciencedirect.com/science/article/pii/0370269388912658 Isn't this paper the first non-perturbative argument for the superconformality of N=4 SYM ? And also is the argument for superconformality in N=4 SYM conceptually different from the one needed in this one of my earlier questions - http://physics.stackexchange.com/questions/11438/argument-for-quantum-theoretic-conformality-of-caln-2-super-chern-simons-t

This post has been migrated from (A51.SE)
Yes that's right. The flavor of the argument there seems very similar to what I said. I assume a similar approach can be taken to theories in 3d, or more generally, any theory where the notion of holomorphy makes sense.

This post has been migrated from (A51.SE)
Most recent comments show all comments
Nice answer. Can you provide a ref pls? Do I understand correctly this argument is perturbative only? If so, is there a way to extend it to a nonperturbative one?

This post has been migrated from (A51.SE)
It is non-perturbative, because the gauge coupling does not run non-perturbatively. I have not seen it explicitly written anywhere, but I am sure I am (by far) not the first one who had this thought :)

This post has been migrated from (A51.SE)
+ 7 like - 0 dislike

Regard it as an N=1 super Yang-Mills theory with three adjoint chiral superfields, and apply the non-perturbative analysis of Leigh-Strassler.

This post has been migrated from (A51.SE)
answered Oct 27, 2011 by Yuji (1,395 points) [ no revision ]
It would be nice if you can add a short summary of this method

This post has been migrated from (A51.SE)

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).
Please complete the anti-spam verification

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

Your rights