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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 $$\mathbb Z_2^3\subset SO(6)\times SL(2)\subset E_{6(6)}~?$$

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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:

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

  2. 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.

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

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

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