Quantcast
  • 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.

News

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

Attributions

(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,082 questions , 2,232 unanswered
5,354 answers , 22,789 comments
1,470 users with positive rep
820 active unimported users
More ...

  Conformal invariance of $N = 4$ Supersymmetric Yang-Mills theory

+ 4 like - 0 dislike
3962 views

I will quote the following from the Wikipedia article on Supersymmetry Nonrenormalization theorems [1],

"In $N=4$ super Yang–Mills the $\beta$-function is zero for all couplings, meaning that the theory is conformal."

When we say that the $\beta$-function vanishes for a QFT, we conclude that scale invariance remains preserved at the quantum level.

However scale invariance doesn't necessarily imply conformal invariance, e.g. this paper has two such examples [2]. Hereby I am getting confused, this paper by Sohnius and West [3] originally shows why the $\beta$- function for $N=4$ super Yang–Mills vanishes. But then why is the claim for conformal invariance justified? 


  [1]: https://en.wikipedia.org/wiki/Supersymmetry_nonrenormalization_theorems#Nonrenormalization_in_supersymmetric_theories_and_holomorphy
  [2]: https://arxiv.org/abs/hep-th/0006037
  [3]: https://www.sciencedirect.com/science/article/pii/0370269381903269?via%3Dihub

asked May 21, 2018 in Theoretical Physics by Joyshaitan (85 points) [ no revision ]
recategorized May 21, 2018 by Dilaton

I had posted the same question on PSE, and got a good answer. https://physics.stackexchange.com/q/407265/50770

1 Answer

+ 2 like - 0 dislike

Isn't the claim of (super)conformal invariance justified by the fact that up to the trace anomaly the trace of the stress energy tensor is zero? This is exactly what you would expect for a CFT and this is exactly what happens at the origin of the moduli space of the $\mathcal{N}=4$ SYM theory.

P.S. In general, if specific conditions are satisfied, for a scale invariant theory one can make $T_{\mu \nu}$ such that $T_{\mu}^{\mu} =0$ which implies the conformal invariance. To understand what happens with the trace anomaly is a long story that need to be carefully studied. 

answered May 24, 2018 by conformal_gk (3,625 points) [ revision history ]
edited May 24, 2018 by conformal_gk

But why does this imply the conformal symmetry?

Sorry, misprint. Look again.

Is there a proof that under natural conditions, a vanishing trace implies conformal symmetry, or is it just something observed in typical examples?

No, I don't think there is a proof for that.

Then you should correct your claim ''which implies that''!

@conformal_gk I had a misunderstanding earlier that $T_{\mu}^{\mu} = 0$ implies only that the charge associated with dilatations is conserved, but it turns out that the charge associated with special conformal transformations is conserved too. So yeah, my problem is solved, for earlier I had the misconception that only scale invariance is preserved with this this condition.

It is not MY claim.

But it look like your claim. Your answer contains the statement that vanishing trace implies conformal invariance, without attribution. In the comment you write that you don't know of a proof. So you should qualify your answer accordingly. On the other hand, maybe the comment by Joyshaitan can be developed into a proof. Since I am not an expert on conformal  theory, I'd like to know the precise status of what is believed and what is actually proved.

The relation between scale invariance and conformal invariance is discussed  a bit in chapter 1 of https://arxiv.org/pdf/1601.05000.pdf.

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:
$\varnothing\hbar$ysicsOverflow
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
...