and $\ti M_1$, there is only a mild dependence on $m_L$, which we have fixed as in mSUGRA for definiteness. Indeed, since the $A$-term amplitude in $d_i^{FC}$ arises from pure bino exchange, it does not involve [^9] $\mu$ nor $\tan \beta$. In mSUGRA there is an unphysical region in the plane $(\ti M_1, m_R)$ corresponding to $m_0^2 < 0$ and which has been indicated in light grey in the plots. Anyway, the dark grey region and below is also excluded because $m_R  \le \ti M_1$, in contrast with the requirement of neutrality for the LSP.

.5 cm

(-250,250)[Experiment]{} (-190,200)[$ r \ll 1 \downarrow $]{} (-15,118)[$ \swarrow r \approx 1  $]{} (-300,62)[ $ \left( \times \frac{d_e [{\rm e~cm}]}{10^{-27}} \right) $]{} (-86,62)[ $ \left( \times \frac{d_\mu [{\rm e~cm}]}{10^{-24}} \right) $ ]{}

To find an upper estimate for the seesaw induced $d_i^{FC}$, let us evaluate $| {\cal I}m (a_i) |/m_R$ from eq. (\[ss\_ae\]) by considering only its first term. This situation is representative because the terms proportional to $t_1$ are negligible when the lightest right-handed neutrino has smaller Yukawa couplings, as happens in many models and as one would guess from similarity with the charged fermion sectors. Anyway, the following discussion is trivially 