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Superconformal Multiplet Calculus in 6D

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A convenient method for dealing with off-shell formulations of supergravity theories is provided by the superconformal multiplet calculus. This calculus was originally constructed for 4d ${\cal N}=2$ supergravity by de Wit et al. For 5d supergravity, conformal supergravity approach was developed relatively recently by several groups of Bergshoeff et al, Ohashi et al,Hanaki et al.

http://xxx.lanl.gov/abs/hep-th/0104113 , http://xxx.lanl.gov/abs/hep-th/0104130 , http://xxx.lanl.gov/abs/hep-th/0611329 .

The main idea is that the Poincare algebra is extended to the superconformal algebra to obtain an off-shell version of Poincare supergravity. It turns out that extending conformal supergravity to a gauge theory of superconformal algebra provides an irreducible off-shell realization of the gravity and matter multiplets. Then, by imposing constraints, the gauge theory is identified as a gravity theory. Upon gauge fixing the extra superconformal symmetries, one obtains the Poincare supergravity. In this formalism, one has a systematic way to construct invariant actions since the transformation laws are simpler and completely fixed by the superconformal algebra. Following this approach, one gets an off-shell formulation of supergravity coupled to vector multiplets.

As far as I know, there is no off-shell formulation of 6d supergravity by superconformal multiplet calculas. Why is there no conformal supergravity in 6D? Is there any obstruction to the formulation?

This post has been migrated from (A51.SE)
asked Nov 30, 2011 in Theoretical Physics by Satoshi Nawata (335 points) [ no revision ]
retagged Apr 19, 2014 by dimension10

1 Answer

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Satoshi, I know of one paper which does this: "Nucl Phys B 264, Superconformal tensor calculus and matter couplings in six dimensions, Pages 653-686, E. Bergshoeff, E. Sezgin, A. Van Proeyen"

This post has been migrated from (A51.SE)
answered Dec 1, 2011 by Bindusar Sahoo (50 points) [ no revision ]
Thank you very much for the information. How about different supersymmetry? Namely ${\cal N}=(2,0)$ and ${\cal N}=(1,1)$?

This post has been migrated from (A51.SE)

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