The Cartan-matrix monotonicity conjecture for τ-tilting counts

Let $Q$ be a quiver, and let $A=\mathbb{F}Q/I_1$ and $B=\mathbb{F}Q/I_2$ be algebras given by the same quiver with different admissible ideals. Write $a_s(A)$ for the number of pairwise non-isomorphic basic support $\tau$-tilting $A$-modules with $s$ indecomposable summands. If $P_i$ are the indecomposable projective modules, define the Cartan matrix by

$$c_{ij}^A=\mathsf{\dim}\operatorname{Hom}_A(P_i,P_j).$$

Assume that $B$ is $\tau$-tilting finite.

Cartan monotonicity conjecture. If

$$c_{ij}^A\leqslant c_{ij}^B$$

for every $i,j\in\{1,2,\dots,|Q_0|\}$, then

$$a_s(A)\leqslant a_s(B)$$

for $2\leqslant s\leqslant |Q_0|$.

The conjecture compares the numbers of support $\tau$-tilting modules through entrywise comparison of Cartan matrices. It is verified for the two-point algebras covered by the cited result of W. Two-point cases, but remains open in the generality stated here.