If I look at the decomposition of the adjoint representation of the group $E_8$ under its maximal subgroup $E_6 \times SU(3)$, we have

$ \bf 248 \rightarrow (78, 1) \oplus (27, 3) \oplus (\bar{27}, \bar{3}) \oplus (1, 8) $

Let me focus on the ${\bf (27, 3)}$ representation.

Question: Is it true that I can express the associated (81?) broken generators of $E_8$ as

$T^{a,x} = k^x \otimes j^a $,

where $x$ is an $E_6$ index and $a$ is an $SU(3)$ index?

Doing this, the following identities seem to hold:

${\rm Tr}\, [T^{a,x} (T^{b,y})^{*} ] = \delta^a_b \delta^x_y \, , \qquad {\rm Tr}\, [T^{a,x} T^{b,y} T^{c,z} ] = \epsilon^{abc} d_{xyz} \,, $

where the trace is in the adjoint rep of $E_8$ and $\epsilon^{abc}$ is the $SU(3)$ invariant tensor, $d_{xyz}$ is the invariant tensor of $E_6$. See for example (apologies for the restricted access) the paper http://inspirehep.net/record/213124?ln=en , where the conventions are slightly different.

I guess that the ${\bf (\bar{27}, \bar{3})}$ representation has (81?) broken $E_8$ generators which can be labelled by $(T^{a,x})^* = T_{a,x}$.

Question: How do I evaluate the commutator $[T^{a,x}, (T^{b,y})^*]= [T^{a,x}, T_{b,y}]$?

My guess involves looking at the antisymmetric product $\sim ({\bf 27, 3}) \otimes_{Antisymmetric}({\bf \bar{27}, \bar{3}}) \sim \delta^a_b \delta^x_y$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.