# Where does the matching condition for $U(1)$ subgroups come from in unified models?

The matching conditions for a breaking $G \rightarrow \prod_i G_i$ are

$$\omega_G-\frac{C_2(G)(\mu)}{12 \pi}=\omega_{G_i}-\frac{C_2(G_i)(\mu)}{12 \pi} ,$$

where $C_2(g)$ denotes the quadratic Casimir invariant for the adjoint representation of the group $g$. (See, for example, Eq. 7 in Implications of the CERN LEP results for SO(10) grand unification )

However, for example, for the braking chain $SU(4) \times SU(2)_L \times SU(2)_R \rightarrow SU(3_C) \times SU(2)_L \times U(1)_Y$ we have the matching condition

$$\omega_{U(1)_Y} = \frac{3}{5}\left( \omega_{SU(2)_R} - \frac{2}{12\pi} \right) + \frac{2}{5}\left( \omega_{SU(4)_C} - \frac{4}{12\pi} \right)$$

(See, for example, Eq. 8 in Implications of the CERN LEP results for SO(10) grand unification )

Where does this matching condition for $U(1)$ subgroups come from?

• What's $\omega_i$? Also, note that the Casimir of $\mathrm{U}(1)$ vanishes on the adjoint since the adjoint is the trivial representation. – ACuriousMind Apr 25 '16 at 10:39
• @ACuriousMind 1.) $\omega_i$ corresponds to the subgroup $i$. I changed it in the question to make it clearer. 2.) Yes, that's why there is no term $\propto \frac{1}{12\pi}$ on the left-hand side... – jak Apr 25 '16 at 10:43
• It was clear to me that $\omega_i$ refers to the $\omega$ belonging to $G_i$ - I don't know what object you denote by $\omega$ in the first place! Are these coupling constants? Also, you don't mean the tensor product $\otimes$, you mean the direct product $\times$ of groups there (many physicists strangely use the tensor product there). – ACuriousMind Apr 25 '16 at 10:46
• @ACuriousMind $\omega_i := \alpha_i^{-1} := \frac{4\pi}{g_i^2}$ – jak Apr 25 '16 at 10:47
• @ACuriousMind Presumably, the usage of $\otimes$ comes from the fact that one usually works with algebras in physics, instead of groups. – Danu Apr 25 '16 at 12:29

It results from the combination of two facts: $i$) the embedding of $U(1)_Y$ into $SU(2)_R \times U(1)_{B-L}$, and $ii$) the normalization of the $U(1)$ charges.

Let me take the simpler chain: $SU(2)_L \times SU(2)_R \times U(1)_{B-L} \to SU(2)_L \times U(1)_Y$.

$i$) At the scale of the left-right symmetry, the hypercharge gets merged into $SU(2)_R \times U(1)_{B-L}$, and we have the relation: $$\frac{Y}{2} = T_{3R} + \frac{B-L}{2}\,.$$ (the factors of 1/2 here are conventional). Here $T_{3R}$ is the generator of $SU(2)_R$.

$ii$) In order for the non-abelian generators to be normalized like their abelian counterparts, we define the new charges: $Y' = \sqrt{\frac{3}{5}} \left(\frac{Y}{2} \right)$ and $C' = \sqrt{\frac{3}{2}} \left(\frac{B-L}{2} \right)$. Therefore the equation above becomes:

$$Y' = \sqrt{\frac{3}{5}} T_{3R}+ \sqrt{\frac{2}{5}} C' \,.$$

Finally, when you use this expression with the coupling constants (squared) and do the matching, you get the relation which appears in your question. (note that: $SU(4)_C \to SU(3)_C\times U(1)_{B-L}$ .)

• Thanks for your answer. I guess my problem is then, where does $$\frac{Y}{2} = T_{3R} + \frac{B-L}{2}\,.$$ come from? – jak Apr 26 '16 at 7:53
• In general, we have: $Y/2= c_1 T_{3R} + c_2 {(B-L)/2}$, since the $U(1)_Y$ must be a linear combination of these generators. Then by matching with the (known) Standard Model hypercharges and the (known) $B-L$ charges which are fixed by the breaking, we obtain that $c_1=c_2=1$. – xi45 Apr 26 '16 at 11:34