inflation (although they have existed during inflation).

As a side remark, one may worry that the exponential form of the transfer function and $T_{{\cal S}{\cal S}}$ might enhance the isotropic fluctuations too much to conflict with the observations. However, if we scan its form more carefully, we will find that it is not the case. This is because that the factor $B$ will be quite small, namely only of order slow-roll parameters, while $N_\ast$ is nothing but the efolding number of the inflation, namely around the number $60$. As a special but explicit example, it has been numerically shown in Ref. [@Cai:2013caa] in which one of the two d.o.f.s act as curvaton, that the slow-roll parameters is around $10^{-3}$ to $10^{-6}$, making $T_{{\cal S}{\cal S}}\sim 0.9$. So no matter whether $B>0$ or $B<0$, the isocurvature perturbations will actually not expected to deviate much from the value at Hubble exit (although it does deviate). Moreover, since we can see that the most terms in $B$ are negative, there are large possibilities that $B$ is negative, meaning that the isocurvature modes are decaying. More exact identification of the sign of $B$ needs fitting at the whole parameter space, which is obviously beyond 