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  Why is the symmetry group of the AN1 theory the Osp(2,6|4)?

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Why is the symmetry group of the AN1 theory the Osp(2,6|4)? And how do we extract the fact that the bosonic subgroup is Spin(2,6)×Spin(5)?

The AN1 theory is the resulting 6d SCFT made out of a stack of N coincident M5-branes. The Spin(5) is supposed to be the rotation group in the transverse space of the M5-branes and gives the R-symmetry but this is the double cover of SO(5) and I do not understand why the R-symmetry is not given by the SO(5). A reference is this one.

asked Apr 20, 2015 in Theoretical Physics by conformal_gk (3,625 points) [ no revision ]

1 Answer

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The AN1 theory is a N=(2,0) 6d superconformal theory. So its symmetry group is the N=(2,0) 6d superconformal group. Let G be this group and let us try to determine G. Let us first work at the level of Lie superalgebras. Let g be the Lie superalgebra of G. As in any superconformal algebra, the bosonic part of g is made of the ordinary conformal algebra and of the R-symmetry algebra.

It is well known that the conformal algebra of the d-dimensional Minkowski spacetime is so(2,d) (this result plays for example a key role in AdS/CFT because so(2,d) is naturally the Lie algebra of the isometry group of AdSd+1). In particular, in 6 dimensions, it is so(2,6). Remark that the 6d conformal algebra so(2,6) naturally contains the 6d Lorentz algebra so(1,5).

To go further, one has to understand the fermionic part. The 6d Lorentz algebra has Weyl spinors with 8 real components. The N=(2,0) supersymmetric algebra contains 16 real supercharges transforming in two copies of the same chirality of this Weyl spinor. Remark that there is an exceptional isomorphism of Lie algebras so(1,5)=sl(2,H), where H is the field of quaternions, and the 8 real spin representation of so(1,5) is simply the fundamental representation H2 of sl(2,H).

The ferminonic part of g contains not only these 16 real supercharges of the standard supersymmetry algebra, which square to a spacetime translation, but also conformal supercharges which square to special conformal transformations. Conformal supercharges form another copy of two Weyl spinors with respect to so(1,5). In conclusion, the full fermionic part of g has 16.2=32 real components and is the copy of two Weyl spinors of the conformal algebra so(2,4).

 R-symmetry rotates the N=1 superalgebra inside the N=2 superalgebra. As the supercharges are in quaternionic representations, the R-symmetry algebra is usp(4) i.e. the algebra of automorphisms of H2. There is an exceptional isomorphism of Lie algebras usp(4)=so(5) (this in clear at the level of Dynkin diagrams: B2 and C2 are the same).

Finally we have found that the bosonic part of g is so(2,6)usp(4)=so(2,6)so(5). This superalgebra is called osp(2,6|4): the "o(2,6)" notation signals the presence of so(2,6) in the bosonic part and the "sp(|4)" notation signals that the R-symmetry is usp(4).

To go from Lie algebras to Lie groups, one has to see what are the representations used. The superconformal algebra so(2,6) acts on the supercharges through the spinor representation so the relevant group is the full Spin(2,6). Similarly, the R-symmetry acts on the supercharges through the fundamental representation H2 of usp(4) which is a spinor representation of so(5) so the relevant group is the full Spin(5).

Much more information can be found in these notes

http://www.physics.rutgers.edu/~gmoore/FelixKleinLectureNotes.pdf

by Moore. The section 3 is particularly relevant.

answered Apr 21, 2015 by 40227 (5,140 points) [ revision history ]
edited Apr 21, 2015 by 40227

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