• 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.


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

174 submissions , 137 unreviewed
4,308 questions , 1,640 unanswered
5,089 answers , 21,602 comments
1,470 users with positive rep
635 active unimported users
More ...

  Why is there an Euler density in SCFT $T_{\mu}^{\mu}$?

+ 2 like - 0 dislike

The super conformal field theories are above all conformal. Conformal theories are defined on flat space-times. Despite that, if one looks at the stress tensor trace of a SCFT in 4d you get a contribution from the field strength of the gauge sector and the Euler density, i.e. $$T_{\mu}^{\mu} \backsim F_{\mu \nu }^2 - a(R_{\mu \nu \rho \sigma})^2 + \text{other terms}$$ Where does this $R_{\mu \nu \rho \sigma}$ come from in a super-conformal field theory which, from what I know, it is defined in flat space-time? Why we call this $a$ "central charge" despite it looks like some coupling?

This post imported from StackExchange Physics at 2014-12-23 15:08 (UTC), posted by SE-user Marion

asked Dec 23, 2014 in Theoretical Physics by Marion Edualdo (250 points) [ revision history ]
edited Dec 23, 2014 by conformal_gk

1 Answer

+ 3 like - 0 dislike

There is nothing wrong in putting a conformal field theory in a curved background as long as the conformal anomaly vanishes. A curved background is some Riemannian manifold $\mathcal{M}$ equipped with some metric $g$ from which one calculates both the Euler density and the Weyl tensor. Furthermore one can compute the trace Ward idenity which must vanish and finds that $\langle T_{\mu}^{\mu} \rangle \backsim aE + cW^2 +$ other terms. Check hep-th/9708042.

answered Dec 23, 2014 by conformal_gk (3,625 points) [ revision history ]

I wonder why nobody corrected me yet. Does the trace have to vanish? Hehe.

It s Christmas ... ;-) Hm I thought that the trace has to vanish for conformal invariance, whereas for just scale invariance to hold it is allowed to corrrspond to some kind of conserved current or something along these lines ...?
My impression is that the trace does not have to vanish as long as the background is not dynamical. If the trace vanishes then $a$ and $c$ are zero, right? Then does the $a$-theorem in 4d or the $c$-theorem in 2d hold? Of course if there is an anomaly the theory loses its conf. invariance.
Hm I have read somewhere that the quantities a or c correspond only in specific cases exactly to the anomaly would have to reconsider that ... Maybe @suresh knows?
$a$ and $c$ correspond to the central charges of the CFT (the corresponding QFT at a fixed point).

Please log in or register to answer this question.

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

Your rights