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$\mathcal{N} = 2^\star$ supersymmetry

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I have read in a paper (arXiv:1507.07709v3 [hep-th]) that the $(2,0)$ superconformal field theory (SCFT) in $d = 6$ when dimensionally reduced on $S^1$ yields a $d = 5$ $\mathcal{N} = 2^\star$ theory. In an effort to learn more about this $\star$, I figured out that the theory consists of the $\mathcal{N} = 2$ gauge vector multiplet along with an adjoint hypermultiplet of mass $m$. I have a number of questions about such a construction:

Question 1: In what sense is this mass deformation of $\mathcal{N} = 4$ Super Yang-Mills (SYM) theory?

The N=4 SYM multiplet consists of an N=2 vector multiplet and an N=2 hypermultiplet, and if you make the hypermultiplet infinitely massive, it is tantamount to integrating the hypermultiplet out, yielding an N=2 SYM. In the opposite limit (of taking the mass to zero), one gets back N=4 SYM. Is this all one means by a mass deformation?

The context of my question is the so called twisted compactification of (a stack of) M5-branes on $S^1$ which leads to this five-dimensional $\mathcal{N} = 2^\star$ theory.

Question 2: In this setting, what does the mass parameter m translate to in terms of the branes? What kind of "twist" relates to the mass?


This post imported from StackExchange Physics at 2016-08-11 19:19 (UTC), posted by SE-user leastaction

asked Aug 11, 2016 in Theoretical Physics by leastaction (425 points) [ revision history ]
edited Aug 12, 2016 by Dilaton

Could you please update the question from stackexchange? The question has been extensively edited and corrected for typos since it was first imported.

@leastaction done. Please have a look if everything is to your satisfaction.

Thank you @Dilaton!

1 Answer

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Remark: in the question, the four dimensional terminology for supersymmetry is used whereas the theories under consideration are five dimensional. I will do the same.

The ony field of $\mathcal{N}=4$ SYM is the $\mathcal{N}=4$ vector multiplet (massless). In the $\mathcal{N}=2$ language, it is the same than a $\mathcal{N}=2$ vector multiplet (massless) and a $\mathcal{N}=2$ hypermultiplet (massless) in the adjoint representation. Deforming this theory by giving a mass to the adjoint hypermultiplet gives the $\mathcal{N}=2^*$ theory.

For large mass, the adjoint hypermultiplet decouples and so one does not recover $\mathcal{N}=4$ SYM (which corresponds to the limit mass going to zero) but pure $\mathcal{N}=2$ SYM consisting only of one $\mathcal{N}=2$ vector multiplet.

Compactifying the 6d theory on a circle gives $\mathcal{N}=4$ SYM. Adding a twist around the circle corresponds to giving a mass to the adjoint hypermultiplet and one obtains $\mathcal{N}=2^*$ SYM.

answered Aug 12, 2016 by 40227 (4,660 points) [ no revision ]

Thanks @40227. The question was edited and corrected for a typo (the $\mathcal{N} = 2$) shortly after it was originally posted on SE. That being said, how does compactifying on a circle yield $\mathcal{N} = 4$ SYM, and how does the twisting work? Can you suggest a reference where I can read more about this? Also, where was this $\mathcal{N} = 2^\star$ terminology first introduced?

What are the details of this twisting and where can I find them?

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