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.



Bounding the projective dimension of modules by the number of points and arrows
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**](https://doi.org/10.1007/BF02567802), Manuscr. Math. 88, No. 1, 25-31 (1995). [ZBL0851.16003](https://zbmath.org/?q=an:0851.16003).


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**](https://doi.org/10.1016/0021-8693(85)90069-9), J. Algebra 97, 14-16 (1985). [ZBL0571.16011](https://zbmath.org/?q=an:0571.16011).


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.


