2
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The general formula for the anomalous dimension can be found in Martin΄s review article (hep-ph/9709356), on page 62 relation (6.5.4). In the case of $SU(5)$ and especially in the paper of Kobayashi, Kubo, Mondragon and Zoupanos (hep-ph/9707425), which I am studying, I can deduce the first part of the relations (38), (page 15), which include the gauge coupling g.

However, I am extremely confused with the other part, which refers to the couplings $g^u$, $g^d$, $g^f$, $g^l$ in the superpotential $W$, relation (37). Is there anybody who can help me? The first two terms in the superpotential are: $$\frac{1}{2} \ g^{u} \ 10 \ 10 \ H +g^{d} \ 10 \ \bar5 \ \bar H,$$ where $10$, $\bar 5$, are the antisymmetric and the antifundamental representations of $SU(5)$ for the fermions respectively and $H$, $\bar H$,the Higgs quintets and antiquintets. According to the paper (hep-ph/9707425), the anomalous dimension for the $10$ is: $$\gamma_{10}=\frac{1}{16\pi^{2}} \ (\frac{-36}{5} g^{2}+ 3 (g^{u})^{2} + 2 (g^{d})^{2}).$$ The general formula for the anomalous dimension is: $$ \gamma_{i}^{j}=\frac{1}{2} Y_{ipq}Y^{jpq} -2 \delta_{i}^{j} C(i).$$ How can we deduce the factors in front of the gauge and Yukawa couplings?

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    $\begingroup$ It might be useful to include the relevant equations that are troubling you. $\endgroup$ – Kyle Kanos Nov 25 '14 at 17:59
  • $\begingroup$ The first two tems of the superpotential are: $\endgroup$ – ioannis Nov 28 '14 at 14:32
  • $\begingroup$ The first two terms of the superpotential are:W=1/2*g^u**10*10*H + g^d*10*5*5*H, where in the second term the fields are 5 bar, H bar ,(5 antifundamental, for the fermions ), and (H bar for the Higgs anti-quintents) where in the first term 10 and H stand for the antisymmetric SU(5), (for the fermions) and the Higgs quintents, resoectively.The anomalous dimension for the antisymmetric 10 is: γ_10=1/16π^2*(-36/5*g^2+3*(g^u)^2+2*(g^d)^2). The question which is troubling me is how we deduce the factors 3 and 2, in front of the Yukawa couplings.In general γ_i^j=1/2*Y_ipqY^jpq-2*δ_i^jg^2*C(i) $\endgroup$ – ioannis Nov 28 '14 at 15:12
  • $\begingroup$ Why not put them in your question (using the edit button). Note that MathJax (a LaTeX-type markup language) is enabled on this site, so you can have proper notation. $\endgroup$ – Kyle Kanos Nov 28 '14 at 15:13

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