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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.

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  • $\begingroup$ Have you tried composing the 3 respective Young tableaux, or contracting the three $q^i \bar q _j$ s? $\endgroup$ – Cosmas Zachos Apr 3 at 18:54
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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 $\text{tr}(\Sigma GG)$.

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