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

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

145 submissions , 122 unreviewed
3,930 questions , 1,398 unanswered
4,853 answers , 20,624 comments
1,470 users with positive rep
501 active unimported users
More ...

Scalar field transformation behaviour of N=4 SYM after a topological twist

+ 5 like - 0 dislike
111 views

Why do scalars transform as 2-forms after a topological twist?

I'm currently looking at this paper https://arxiv.org/abs/1403.2530. There they analyze a topological twist on N=4 SYM theory on Kähler 2-fold.

After doing a topological twist with the $\mathcal{R}$-symmetry as well as an additional $U(1)$ symmetry in a $N=4$ SYM, the 6 real scalars of the multiplet are reorganized under the twisted holonomy group: $$ G = SU(2)_L \times U(1)_{J'} $$ with $U(1)_{J'} \subset SU(2)_R$. The important part is now that the 6 scalars can be reorganized in a way including two complex fields: $$ (\boldsymbol{1})_2 \oplus (\boldsymbol{1})_{-2} $$ It is now stated that these two fields can be identified with a (2,0) and a (0,2) form. My question is, how can this identification be reached exactly ?

It is of course obvious that after twisting and obtaining charge under the $U(1)_{J'} \subset U(2)$ of the holonomy group, these fields cannot transform simply as scalars anymore, but how can i conclude from their $U(1)$ charge on the exact transformation behaviour ?

This is probably connected to another question of mine, in another paper on this topic it is stated that the spinors in these theories could be understood as sections of $\boldsymbol{S}^{\pm} \otimes K^{1/2}$ with $\boldsymbol{S}^{\pm}$ denoting the spin bundle and $K^{1/2}$ the square root of the canonical bundle of the base manifold and that by twisting with the $\mathcal{R}$-symmetry we just untwist these twisted differential forms, but I don't get why spinors are not just sections of the spin bundle solely in the first place.

An answer for the first question only would be totally enough, just wanted to illuminate my confusion here a bit further.


This post imported from StackExchange Physics at 2016-12-25 09:32 (UTC), posted by SE-user Moe

asked Dec 23, 2016 in Theoretical Physics by Moe (25 points) [ revision history ]
edited Dec 31, 2016 by Dilaton

1 Answer

+ 3 like - 0 dislike

Spinors are globally defined on spin manifolds. In order to get spinors globally defined you need a $Spin$ structure on the manifold. Not all Kahler manifolds are $Spin$. Despite that, all Kahler manifolds are $Spin_c$ which requires twisting the Spin bundle with the "square-root" of a line bundle such that sign ambiguities are washed up. This should answer the second part of your question. 

*As for the first part, this is just a consequence of the embedding of the new rotation group to the old one. You just need to make sure agree. I will try to find some time to answer this question in more detail.

answered May 31 by conformal_gk (3,535 points) [ no revision ]

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$\hbar\varnothing$sicsOverflow
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
...