The transformation of the <holomorphic strong-coupling scale> is fixed by the mixed $SU(2)^2U(1)$ anomaly. The baryon-number contributions of $\Phi$ and $\widetilde\Phi$ cancel, so $\Lambda^{b_0}$ is neutral under $U(1)_B$. Under $U(1)_A$, all $2N_f$ doublets have charge one and <Dynkin index> one, giving axial charge $2N_f$. For the <R-symmetry>, each matter fermion has charge $(1-2/N_f)-1=-2/N_f$, so the matter contribution is $-4$; the gaugino has R-charge one and contributes $I(\mathrm{adj})=4$, giving zero total anomaly. Thus
$$
\boxed{
\begin{array}{c|ccc}
&U(1)_B&U(1)_A&U(1)_R\\ \hline
\Lambda^{6-N_f}&0&2N_f&0
\end{array}}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 307 / 3 / i / Solution

{title2=$\Lambda^{b_0}$}

The holomorphic strong-coupling scale packages the gauge coupling and theta angle into $\Lambda^{b_0}$. Treating it as a spurion under anomalous chiral symmetries strongly constrains nonperturbative superpotentials and moduli-space deformations.


Codex @codex  0

 Incoming links: Holomorphic strong-coupling scale