Are all algebras Igusa-Todorov?

Question

A finite dimensional algebra A is called (n-)Igusa-Todorov in case there exists a module V such that for any module M there is an exact sequence: $0 \rightarrow V_2 \rightarrow V_1 \rightarrow \Omega^{n}(M) \oplus P \rightarrow 0$, where $P$ is projective and $V_i \in \mathrm{add}(V)$. See this 2009 paper by Jiaqun Wei (ScienceDirect link).

In the same paper there is the problem, wheter any algebra is Igusa-Todorov. What is the status of this problem? How about selfinjective $A$? (Then the $n$ does not matter since $\Omega$ is an equivalence. )

The big motivation behind this concept is that if all algebras were Igusa-Todorov, the famous finitistic dimension conjecture would be true. I personally think that not all algebras are Igusa-Todorov and I see no reason why selfinjective algebras should be Igusa-Todorov in general.

Edit: I saw that a selfinjective algebra not being Igusa-Todorov was also noted here: https://teresaconde.xyz/files/thesis.pdf on page 96 (thesis memoir by Teresa Conde, Oxford).

Answer 1

If $A$ is self-injective, then I think that $A$ being Igusa-Todorov implies that the dimension (in the sense of Rouquier) of the stable module category $\textrm{stmod-}A$ is at most $1$.

In "Dimensions of triangulated categories", J. K-theory 1 (2008), no.2, 193-256 (link to arXiv), Rouquier defines the dimension of a triangulated category $\mathcal{T}$ as follows:

Given a subcategory $\mathcal{I}$, $\langle\mathcal{I}\rangle$ is the full subcategory containing all direct summands of finite direct sums of shifts of objects of $\mathcal{I}$, and inductively $\langle\mathcal{I}\rangle_1=\langle\mathcal{I}\rangle$ and $\langle\mathcal{I}\rangle_{n+1}=\langle\langle\mathcal{I}\rangle_n\ast\langle\mathcal{I}\rangle\rangle$, where $\langle\mathcal{I}\rangle_n\ast\langle\mathcal{I}\rangle$ is the full subcategory consisting of objects $M$ such that there is a distinguished triangle $$M_1\to M\to M_2\to\Sigma M_1$$ with $M_1$ in $\langle \mathcal{I}\rangle_n$ and $M_2$ in $\langle\mathcal{I}\rangle$. Then the dimension of $\mathcal{T}$ is defined to be the smallest $d$ such that there is some object $V$ with $\mathcal{T}=\langle V\rangle_{d+1}$.

So if $A$ is an Igusa-Todorov self-injective algebra, then for every module $M$ there is a distinguished triangle $$V_1\to M\to \Sigma V_2\to \Sigma V_1$$ in $\textrm{stmod-}A$, so $\textrm{stmod-}A=\langle V\rangle_2$, and so the dimension of $\textrm{stmod-}A$ is at most $1$.

So, for example, if $A$ is the exterior algebra of a $3$-dimensional vector space (for which Rouquier proves in Section 8.2 of the same paper that the dimension of $\textrm{stmod-}A$ is $2$), then $A$ can't be Igusa-Todorov.