# Gaps in the projective dimensions of simple modules

Let $$A$$ be a finite dimensional algebra with finite global dimension $$g$$ and $$n$$ simple modules. Let $$d_1 be the sequence of projective dimension of simple $$A$$-modules in increasing order. Define $$\phi_A:= max \{ d_{i+1}-d_i | 1 \leq i \leq r-1 \}$$.

Question: Is there a class of examples where $$\phi_A$$ gets arbitrary large for a fixed $$n$$?

I did not even find an example with $$\phi_A >2$$ (feel free to post an example if you know one).

Here are two examples for $$\phi_A$$:

In Projective dimensions of simple modules in acyclic quiver algebras , Jeremy Rickard proved that $$\phi_A=1$$ for acyclic quiver algebras.

Dag Madsen showed me an argument, which shows that $$\phi_A \leq 2$$ for any Nakayama algebra (this gives also a very nice proof that the global dimension of a Nakayama algebra with $$n$$ simple modules is at most $$2n-2$$).

• What happens if you take a cyclic Nakayama algebra of finite global dimension such that the projective dimensions of all of the simple modules are very high, and then modify it by adding a new sink vertex (with one incoming arrow) somewhere? Then the new simple is projective, and it looks to me (but this might be wrong) that the simples at the other vertices will have the same projective dimension as they did in the original Nakayama algebra. – Matthew Pressland Oct 31 '19 at 16:14
• @MatthewPressland For Nakayama algebras with finitistic dimension $g$ (=global dimension in case this is finite), there exists a simple module of projective dimension $i \leq g$ for each odd i. So Nakayama algebras wont work. But your idea sounds good to construct such an algebra with $\phi_A > 2$. – Mare Oct 31 '19 at 16:40
• @MatthewPressland Your idea works taking the Fibonacci example algebras (see link.springer.com/article/10.1007/BF01189999 )of Happel and adding a new sink vertex. If you want, you can post this as an answer. – Mare Oct 31 '19 at 16:57
• Ah, I didn't realise that, although it does explain why I was having a hard time constructing a concrete example. I can post an answer based on the Fibonacci examples (but not immediately, so it is also ok with me if you want to do it). – Matthew Pressland Oct 31 '19 at 18:12

For $$n>0$$, let $$A_n=KQ_n/I_n$$ as follows: $$Q_n$$ is the quiver with vertex set $$\{1,2\}$$ and arrows $$\alpha_i\colon 1\to 2$$ and $$\beta_i\colon 2\to 1$$ for $$1\leq i\leq n$$, and $$I_n$$ is generated by $$\beta_j\alpha_i$$ for $$i\leq j$$ and $$\alpha_j\beta_i$$ for $$i\leq j-1$$ (reading composition from right-to-left). That is, to write down a path that is non-zero in $$A_n$$, an $$\alpha$$ may only be followed by a $$\beta$$ with a strictly lower index, and a $$\beta$$ may only be followed by an $$\alpha$$ with an equal or lower index.
These algebras were introduced by Green in Example 2.1 of Remarks on projective resolutions, who shows that the projective dimensions of $$S(1)$$ and $$S(2)$$ are $$2n$$ and $$2n-1$$ respectively. (Here I am taking only the even terms of Green's sequence since they are marginally easier to describe and still answer the question.)
Now let $$A_n'=KQ_n'/I_n'$$ as follows: $$Q'_n$$ is the quiver obtained from $$Q_n$$ by adding a vertex $$3$$ and a single arrow $$2\to 3$$ , and $$I'$$ is the ideal of $$KQ_n'$$ with the same generators as those given above for $$I$$. Then the simple module $$S(3)$$ for $$A_n'$$ is projective, i.e. has projective dimension zero, but one can check combinatorially that the simple modules $$S(1)$$ and $$S(2)$$ have the same projective dimension as the simple modules for $$A_n$$ of the same name. Thus $$\Phi_{A'_n}=2n-1$$ gets arbitrarily large.
The same construction will also work for the odd terms of Green's sequence, showing that $$\Phi_A$$ can take any value in $$\mathbb{N}$$.