1
$\begingroup$

We have a trinification model of $SU(3)_c\otimes SU(3)_L\otimes SU(3)_R$, where the first is the usual colour group, the second a left $SU(3)$ and the third a right $SU(3)$.

As usual, leptons and higgses are colour singlets, so under the left and right groups they both belong to the representations ($3$,$\bar{3}$). More explicitly:

$\psi_{lepton}=\pmatrix{(\mathcal{E}) & (E^c) & (\mathcal{L})\\ \mathcal{N}_1 & e^c & \mathcal{N}_2}$ and $\Phi_{higgs}=\pmatrix{(\phi_1) & (\phi_2) & (\phi_3)\\ S_1 & S_2 & S_3}$

where fields in parentheses are $SU(2)_L$ doublets (so they are actually $3\times3$ matrices). (obviously we also have extra fermions and higgses)

When taking the Yukawa interactions, we have:

$\frac{1}{2}\psi_{lepton}\psi_{lepton}\Phi_{higgs}= -(E^c \mathcal{N}_2-\mathcal{L}e^c)\phi_1 + (\mathcal{E}\mathcal{N}_2-\mathcal{L}\mathcal{N}_1)\phi_2 + (E^c \mathcal{N}_1-\mathcal{E} e^c)\phi_3 + E^c\mathcal{L} S_1-\mathcal{E}\mathcal{L}S_2-E^c\mathcal{E}S_3$

I don't understand how we obtain this result, in other words how do we isolate the singlets for the product $(3,\bar{3})\times(3,\bar{3})\times(3,\bar{3})$.

Since $3\times 3 \times 3=(6+\bar{3})\times 3=10+8+8+1$ and $\bar{3}\times \bar{3} \times \bar{3}=(\bar{6}+3)\times \bar{3}=\bar{10}+8+8+1$ I thought that taking the trace of the final $3\times 3$ matrix would give the singlets, but I get a different result. What am I doing wrong?

$\endgroup$

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.