Let $A=KQ/I$ be a finite dimensional connected quiver algebra with admissible ideal $I$ with n points and m arrows and $M$ an $A$-module.
Question: Is the projective dimension of $M$ bounded by $n+m$ if it is finite?
Of course it would be surprising to see a proof, but maybe there is a simple counterexample.
A good strategy to find examples that break this bound is to use Xi's construction of the dual extension algebra, which keeps the number of vertices, doubles the number of arrows, and exactly doubles the global dimension:
Xi, Changchang, Global dimensions of dual extension algebras, Manuscr. Math. 88, No. 1, 25-31 (1995). ZBL0851.16003.
For instance if the quiver is an oriented cycle ($n=m$), then the maximal finite global dimension that can be attained is $2n-2$, see
Gustafson, William H., Global dimension in serial rings, J. Algebra 97, 14-16 (1985). ZBL0571.16011.
Since $\mathrm{gldim} (A)=2n -2 < 2n=n_A+m_A$, such an algebra $A$ with maximal finite global dimension is not itself a counterexample. But for its dual extension algebra $D$, we get $\mathrm{gldim} (D)=4n -4$ and $n_D+m_D=n + 2n = 3n$. So for $n \geq 5$, $$\mathrm{gldim} (D) >n_D+m_D,$$ and the bound does not hold.
A counterexample is the algebra $k\langle x,y| xyxy^2, xy^2xy^3, \dots ,xy^{k-1} xy^{k}\rangle $. It is connected ($n=1$) with $m=2$ generators and have global dimension $k$. So, the trivial module $M=k$ has projective dimension $k$.
Still, it seems unknown whether there exists a counterample among quadratic connected algebras.
