The longest increasing subsequence of Brownian separable permutons

We establish a scaling limit result for the length $\mathrm{LIS}({\sigma }_n)$ of the longest increasing subsequence of a permutation ${\sigma }_n$ of size $n$ sampled from the Brownian separable permuton ${\mathit{\mu }}_p$ of parameter $p\in (0,1)$, which is the universal limit of pattern-avoiding permutations. Specifically, we prove that $$\frac{\mathrm{LIS}({\sigma }_n)}{n^{\alpha }}X,$$ where $\alpha =\alpha (p)$ is the unique solution in the interval $(1\mathrm{/}2,1)$ to the equation $$\frac{1}{4^{\frac{1}{2\alpha }}\sqrt{\pi }}\frac{\mathrm{\Gamma }(\frac{1}{2}-\frac{1}{2\alpha })}{\mathrm{\Gamma }(1-\frac{1}{2\alpha })}=\frac{p}{p-1},$$ and $X=X(p)$ is a non-deterministic and a.s. positive and finite random variable, which is a measurable function of the Brownian separable permuton. Notably, the exponent $\alpha (p)$ is an increasing continuous function of $p$ with $\alpha (0^{+})=1\mathrm{/}2$, $\alpha (1^{-})=1$ and $\alpha (1\mathrm{/}2)\approx 0.815226$, which corresponds to the permuton limit of uniform separable permutations. We prove analogous results for the size of the largest clique of a graph sampled from the Brownian cographon of parameter $p\in (0,1)$.