Physics Asked by user195583 on January 5, 2021

The scalar manifold of $D=5, mathcal N=8$ SUGRA is

$$mathcal M = frac{E_{6(6)}}{Usp(8)}$$

where $USp(8)$ is a maximal compact subgroup of $E_{6(6)}$

and the 42 scalars of the theory correspond to the noncompact generators of $E_{6(6)}$ orthogonal to those of $USp(8)$.

In this paper (Appendix B), the authors described a truncation of this scalar manifold from the above to the following manifold

$$mathcal M’ = [SO(1,1)times SO(1,1)] times frac{SO(4,4)}{SO(4)times SO(4)}$$

by keeping only those scalars invariant under a $mathbb Z_2^3 subset SO(6)times SL(2)$. These three $mathbb Z_2$ are generated by the following generators $P_1 Q, P_2 Q, P_3 Q$ where:

$$P_1 = text{diag}{-1,-1,1,1,1,1}subset SO(6)$$

$$P_2 = text{diag}{1,1,-1,-1,1,1}subset SO(6)$$

$$P_3 = text{diag}{1,1,1,1,-1,-1}subset SO(6)$$

$$Q = text{diag}{-1,-1} subset SL(2)$$

So my question is: Is there a way to derive, **group-theoretically**, the resulting scalar manifold, $mathcal M’$, from the action of projecting out $$mathbb Z_2^3subset SO(6)times SL(2)subset E_{6(6)}~?$$

Apparently there is no intuitive way to derive this via group-theoretical decomposition, starting from $E_{6(6)}$. The authors of this paper seemed to have derived the truncated scalar manifold by explicitly determining the invariant spectrum under the $mathbb Z_2^3$ truncation, and obtained the result that:

The resulting theory is an $mathcal N=2$ SUGRA because of 2 invariant gravitini left.

The moduli (scalar fields) that are left can be classified into the following field representation of $mathcal N=2$ SUGRA:

a. Two real scalars in 2 vector multiplets, each parameterising an $SO(1,1)$ factor.

b. Sixteen real scalars in 4 hypermultiplets, parameterising the factor $SO(4,4)/SO(4)times SO(4)$, which is a quaternionic manifold, as it should.

From there, the scalar manifold can be "intuitively" seen to be

$mathcal M = SO(1,1)^2 times frac{SO(4,4)}{SO(4)times SO(4)}$

Answered by user195583 on January 5, 2021

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