decay channel of the resonance, substituting $\Gamma_{R,23}$ by $\Gamma_i$ of the particular decay channel. Since $\sum \Gamma_i = \Gamma_R$, the total width of the resonance, the sum of Eq. (\[Eq:5\]) over all decay channels, can be cast in terms of ${\rm Im}[1/(M_{\rm inv}^2 - M_R^2 + i M_R \Gamma_R)]$, which is a variant of the optical theorem.

The $K^* \bar K$ channel
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Let us look at the process depicted in Fig. \[Fig:1\] with 2, 3 being $K^* \bar K$. We shall investigate what happens when the resonance $R$ is the $f_1(1285)$. We can use directly Eq. (\[Eq:5\]) replacing the coupling $g_{R,23}$ in Eq. (\[Eq:4\]) by the $f_1, K^* \bar K$ coupling $$\label{Eq:6}
g_{R,23} \to g_{f_1, K^* \bar K} \; \vec{\epsilon}_{f_1}  \cdot  \vec{\epsilon}_{K^*} \, ,$$ and summing and averaging over polarizations, and we use $g_{f_1, K^* \bar K} = 7555$ MeV, from Ref. [@Roca:2005nm], ignoring for the moment the isospin and $C$-parity structure. The $\Gamma_{f_1, K^* \bar K}$ width is then given by $$\label{Eq:7}
  \Gamma_{f_1, K^* \bar K} = \frac{1}{8\pi}\frac{1}{M_{\rm inv}^2} g_{f_1, K^* \bar K}^2 \, \widetilde{p}_{\bar K} \, \theta(M_{\rm inv} - m_{K^*} -m_K).$$

The $K^* \bar K$ production will begin at a threshold of 1383 MeV, 100 MeV above 