# Decomposition of $E_6$ into $SO(p,q)$

I've seen the following decomposition of the fundamental representation 27 of $E_6$ into

$$E_6 \rightarrow SU(2) \times SO(5,2) \times SO(1,1)$$

$$27 \rightarrow (1,1)(-4) + (1,7)(-2) + (2,8)(+1) + (3,1)(-2)$$

from the article hep-th/9812035 on the vacua of gauged ${\cal N}=8$ SUGRA for $D=5$. Is there a simple way to understand how to derive the commutant subgroup $SO(5,2) \times SO(1,1)$ of $SU(2)$ in $E_6$?

• Some quick points: a) There is certainly not a unique "commutant subgroup of $SU(2)$" - rather, you have to specify the particular $SU(2)$ within $E_6$. b) This decomposition involves the complexification of $E_6$. c) Ignoring the complexity issue, on the algebra level $so(5,2)"\sim"so(7)$, i.e. the $B_3$ algebra and $so(1,1)"\sim"so(2)\sim u(1)$. Maybe this helps in conjuction with the Dynkin diagram. – Toffomat May 15 '18 at 12:59

## 1 Answer

It turns out that a simple way (with no mathematical rigor) to see this decomposition is from the following chain of reasoning: $E_{6(6)}$ has as a maximal subgroup (among some other maximal subgroups) the group $SO(10)\times U(1)$. The non-compact form of this is $SO(5,5)\times SO(1,1)$. From here one can get the embedding relation

$E_{6(6)}\supset SO(5,5)\times SO(1,1) \supset SO(3)\times SO(5,2)\times SO(1,1)$

This is the same as

$E_{6(6)}\supset SU(2)\times SO(5,2)\times SO(1,1)$ since $SO(3)\cong SU(2)$.