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Rigid bricks imply brick-finiteness?

Determine whether, for any finite-dimensional algebra A over an algebraically closed field, the condition that every brick X satisfies Ext^1_A(X,X)=0 implies that A is brick-finite (i.e., has only finitely many bricks up to isomorphism).

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Background

Rigid modules have open orbits in their irreducible components, and the authors explain that the 2nd bBT conjecture would imply an affirmative answer to this question. The problem is therefore positioned as a consequence of the main conjecture in the survey.

The question probes whether homological rigidity of all bricks is strong enough to force finiteness of bricks, mirroring phenomena where rigidity controls orbit geometry and component structure.

References

For instance, it is known that the correctness of the 2nd bBT conjecture gives an affirmative answer to the following question, which (to our knowledge) is still an open problem. If every $X\in \brick(A)$ is rigid, then is algebra $A$ necessarily brick-finite?

On the bricks (Schur representations) of finite dimensional algebras (2508.11789 - Mousavand et al., 15 Aug 2025) in Question [Rigid bricks], Section 4.2