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
5,355 answers , 22,793 comments
1,470 users with positive rep
820 active unimported users
More ...

  Questions on the $N=2$ superconformal algebra

+ 4 like - 0 dislike
1604 views

In my understanding, mirror symmetry in physics originates from representation of the $N=2$ superconformal algebra. Why do we need precisely 2 supersymmetries (why not 1 or 4)?

Moreover, a chiral (anti-chiral) field is defined as a state that is annihilated by $G^+_{-1/2}$ ($G^-_{-1/2}$), where $G^+_{-1/2}$ and $G^-_{-1/2}$ are coefficients of the Fourier mode expansion of some anti-commuting current $G^+(z)$ and $G^-(z)$ of conformal weight $3/2$. How should I understand this chiral (anti-chiral) field?

In $N=(2,2)$ superconformal algebra, there are four rings: $(c,c),(a,a),(a,c),(c,a)$. It is known that the first two are charge conjugate, but what does theism mean? Right and left-moving, and chiral and anti-chiral rings... these all confuse me.

This post imported from StackExchange Physics at 2014-04-02 13:01 (UCT), posted by SE-user Mathematician
asked Mar 1, 2014 in Theoretical Physics by mathematician (60 points) [ no revision ]

1 Answer

+ 1 like - 0 dislike
  1. It was shown by Zumino (Supersymmetry and Kahler Manifolds Phys.Lett. B87 (1979) 203 ) that the supersymmetric non-linear sigma model in four-dimensions (with target $M$) necessarily requires the manifold, $M$, to be Kahler. A dimensional reduction of this model leads to a two-dimensional nonlinear sigma model with $(2,2)$ supersymmetry. (See also: B. Zumino, “Supersymmetric sigma-models in two-dimensions,”)

  2. Consistency of string propagation on $M$ requires it to be Ricci-flat (this is a result due to Friedan). A six-dimensional compact manifold that is Kahler and Ricci-flat is a Calabi-Yau threefold.

These are the circle of ideas that eventually lead to mirror symmetry. Brian Greene's TASI lectures as well as Nick Warner's ICTP lectures on "N=2 Supersymmetric Integrable Models and Topological Field Theories"are two other references that might be of interest to you.

This post imported from StackExchange Physics at 2014-04-02 13:01 (UCT), posted by SE-user suresh
answered Mar 1, 2014 by suresh (1,545 points) [ no revision ]
Note one can define the A model even if the manifold is just Kahler.

This post imported from StackExchange Physics at 2014-04-02 13:01 (UCT), posted by SE-user Ryan Thorngren
@RyanThorngren I was discussing the untwisted theories. You are right that the topological A-model doesn't need point 2 mentioned above. There are many bells and whistles that can be added to the above story and I wanted to make my answer as simple as possible.

This post imported from StackExchange Physics at 2014-04-02 13:01 (UCT), posted by SE-user suresh
Is $N=2$ SCFT the same as $N=(2,2)$ field theory? I think the former give the latter.

This post imported from StackExchange Physics at 2014-04-02 13:01 (UCT), posted by SE-user Mathematician
$N=2$ SCFT usually refers to a chiral half. $(2,2)$ says that there are two left-moving and two right-moving supersymmetries. Thus $(2,2)$ has two copies of the $N=2$ SCFT.

This post imported from StackExchange Physics at 2014-04-02 13:01 (UCT), posted by SE-user suresh

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:
p$\hbar$ysic$\varnothing$Overflow
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
...