Why is it said that even if the baryon asymmetry existed as an initial condition, the asymmetry would have been destroyed by inflation. How does inflation get rid of initial Baryon asymmetry (if any)?
EDIT: For reference see the section 1.1 (last line of second paragraph) of this review.
Answer
The comment on the review you indicated can be explained in the following way. Inflation is a period of nearly exponential expansion of the Universe that is required to solve the flatness problem, the horizon problem, ecc.. These cosmological constraints, quite independently on the specific model of inflation (single field inflaton, modified gravity and so on), require for the Hubble function to remain almost constant for at least $N=50$ e-folds. In other words the scale factor at the end of inflation $a_f$ compared to the one at the beginning $a_i$ must be such that $$ \frac{a_f}{a_i} \simeq e^{50} $$
Now, lest define $\Delta = n_B - n_{\bar B}$ as the difference between the baryon and anti-baryon number densities. For our purposes we also assume for $\Delta$ to be different from zero at the start of inflation. Additionally, we assume that no baryon-violating processes happens during inflation. From this it follows that the total number of baryons and anti-baryons are conserved separately. Since the universe expands, then the number densities obey the differential equation: $$ \dot n_b + 3 H n_B = 0 $$ with $H\sim const$. The same equation holds also for $\Delta$. The solution is, approximately: $$ \Delta(t_f) = \Delta(t_i) e^{-3 H (t_f - t_i)} = \Delta(t_i) e^{-3 N} \approx \Delta(t_i) e^{-3 \times 50} $$
Therefore, in this picture of non-interacting baryons their asymmetry would be completely washed out at the end of inflation.
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