Rigid regular objects of path algebras of tame quivers

Question

In the paper On Maximal Green Sequences by Brustle, Dupont and Perotin the authors argued that in a path algebra $\Lambda=kQ$ of a tame quiver $Q$ with $n$ vertices each tilting module contains at most $n-2$ regular components. The same applies to silting objects between $\Lambda$ and $\Lambda[1]$. I think this is because for each homogeneous tube there can be no rigid indecomposable objects while for each nonhomogeneous tube $\mathbb{Z}\mathbb{A}_{\infty}/(\tau^k)$ there can be no rigid objects with $k$ or more summands without repeating summands.

Is there any known proof of the last statement I provided?

Answer 1

This question has been answered in my own paper Tame Quivers have finitely many $m$-Maximal Green Sequences. This statement can in fact be strengthened to the statement below:

Any pre-silting object in any standard stable tube of rank $k$ and its shifts can contain at most $k-1$ indecomposable summands.

Here I'm going to prove the one above while the proof of the general form is in my paper.

Assume that there exists a presilting object in the standard stable tube $\mathcal{T}$ of rank $k$ has a summand with composition series including all quasi-simples $M_1,M_2=\tau^{-1}M_1,\cdots,M_k=\tau M_1 $. We can assume that it contains an incomposable with quasi-socle $M_1$. Now we have to include some indecomposable with quasi-top $M_k$ which will cause the pre-silting object to have a self-extension which is impossible or some indecomposable with $M_k$ in the composition series but not as the quasi-top which will also cause the pre-silting object to have a self-extension. Hence we have reached a contradiction.