# $SU(5)$ group theory: Contracting three adjoints to make a singlet

There's probably a trivial answer to my question, but I'm having trouble finding it.

In an $$SU(5)$$ GUT, we have a gauge field, $$A_{\mu} = A^{a}_{\mu}T^{a}$$ which lives in the $$24 (\text{Adjoint})$$ of $$SU(5)$$, as well as an adjoint scalar, $$\Sigma$$ which gets a vev that breaks $$SU(5) \rightarrow SU(3)\times SU(2)\times U(1).$$

Now in certain extensions of the minimal model, people consider adding higher dimensional operators to $$SU(5)$$, like:

$$\frac{1}{\Lambda}Tr(\Sigma GG)$$

where $$G_{\mu\nu}$$ is the field strength corresponding to the vector field $$A_{\mu}$$. But I don't understand the index structure of $$Tr(\Sigma GG)$$. $$A_{\mu}$$ (and therefore $$G_{\mu\nu}$$) carries one adjoint index, $$a$$, so $$\Sigma$$ should also carry only one index, so how do I contract these three fields to form a singlet?

I know $$24\otimes24$$ contains another 24, but with indices I don't know how it works.

• Have you tried composing the 3 respective Young tableaux, or contracting the three $q^i \bar q _j$ s? – Cosmas Zachos Apr 3 '19 at 18:54

It's not useful to think about "adjoint indices" for group theory. Instead denote fundamental representations with an upper index and antifundamentals with a lower index. Then an adjoint is a traceless rank $$2$$ tensor with mixed indices, as $$5 \times \bar{5} = 24 + 1.$$ Here, you start with something like $$\Sigma^a_b G^c_d G^e_f.$$ You can form an invariant by contracting up and down indices in pairs. You have to do this in a way that connects all three, because if you don't then you get traces, which vanish, so the only possibility is $$\Sigma^a_b G^b_c G^c_a$$ which can also be thought of as the trace $$\text{tr}(\Sigma GG)$$.