Closure of biased separable Brownian permutons under independent composition

Determine whether the product of two independent biased separable Brownian permutons can belong to the same one-parameter family; in particular, establish or refute the conjecture that the independent product family $\mathcal{S}\mathcal{S}$ is disjoint from $\mathcal{S}$, where $\mathcal{S}=\{\mu^p:p\in(0,1)\}$.

Background

The paper proves that the family of biased separable Brownian permutons is not closed under independent products by showing that at least one parameter value produces a product whose distribution lies outside the family. It then states the stronger conjectural possibility that every product of two members lies outside the family. The authors explain that proving this would require sufficient understanding of convolutions of the associated density functions αp\alpha_p.

References

It in fact seems plausible that $\mathcal{S} \mathcal{S}$ is disjoint from $\mathcal{S}$, although it is not as clear how to show this; sufficient understanding of convolutions of the functions $\alpha_p$ given in Theorem 1.7 of could achieve this.

— On the composition of permutons  (2610.12945 - Gorski, 9 Oct 2026) in Section 6, immediately before Section 7, and Section "Future work"