# Combinatorial problem on periodic dyck paths from homological algebra

edit: I added conjecture 2 that looks much more accessible.

Here is the elementary combinatorial translation of the problem (read below for the homological background):

Let $$n \geq 2$$. A Nakayama algebra is a list $$[c_0,c_1,...,c_{n-1}]$$ of $$n$$ integers with $$c_i \geq 2$$, $$c_{i+1} \geq c_i-1$$ and $$c_{n-1}=c_0+1$$ (it can be interpreted as a periodic Dyck path). Read the indices $$i$$ in the $$c_i$$ mod $$n$$ so that they are defined for any $$i \in \mathbb{Z}$$. A module is a tuple $$(i,k)$$ with $$0 \leq i \leq n-1$$ (the first entry $$i$$ is always viewed mod $$n$$) and $$1 \leq k \leq c_i$$. A proper module is a module with $$k < c_i$$. A proper module $$M=(i,k)$$ is called weird in case with $$k=s+tn$$ with $$0 \leq s we have $$n \leq k$$ and $$max(0,2k-c_i) \leq s+(t-1)n$$ (this should be equivalent to $$Ext_A^1(M,M) \neq 0$$ ). Define the $$i$$-th syzygy of a proper module $$(i,k)$$ as follows: $$\Omega^0(i,k)=(i,k)$$, $$\Omega^1(i,k)=(i+k,c_i-k)$$ and $$\Omega^l(i,k)=\Omega^1(\Omega^{l-1}(i,k))$$ for $$l \geq 2$$. A proper module $$M=(i,k)$$ is called periodic with period $$q$$ in case $$M = \Omega^q(M)$$ and it is called quasi-periodic in case $$\Omega^u(M)$$ is periodic for some $$u \geq 0$$.

The below guess can now be stated completely elementary (I state it here as a conjecture, but it is hard to check since there are infinitely many Nakayama algebras for a given $$n$$):

Conjecture: Any weird module is quasi-periodic.

Conjecture 2: In case $$M$$ is weird, also $$\Omega^1(M)$$ is weird.

Conjecture 2 implies conjecture 1 and looks much easier but I cant find a good argument yet.

Example: The Nakayama algebra [3,4] has the unique weird module (1,2) , which is periodic of period 1.

Homological background:

The strong no loop conjecture states that for a simple module $$S$$ over a finite dimensional algebra $$A$$ we have that $$S$$ has infinite projective dimension in case $$Ext_A^1(S,S) \neq 0$$. In general this is not true for arbitrary indecomposable modules instead of simple modules, but it seems that it might be true for Nakayama algebras.

Guess: In case $$A$$ is a Nakayama algebra with an indecomposable module $$M$$ with $$Ext_A^1(M,M) \neq 0$$ then $$M$$ has infinite projective dimension.

I can prove this for algebras with finite global dimension (which includes all Nakayama algebras with a linear quiver and thus we can focus on cyclic quivers). I should note that maybe there is an easy argument and Im too blind to see it at the moment.

• I think I might have found a homological proof. We have $Ext_A^1(M,M)= Hom(\Omega^1(M),M)$ and $Ext_A^1(M,M) \neq 0$ implies $Ext_A^1(\Omega^1(M),\Omega^1(M)) \neq 0$. Thus $Ext_A^1(\Omega^k(M),\Omega^k(M)) \neq 0$, which forces $\Omega^k(M)$ to be nonzero for all $k$ and $M$ has infinite projective dimension.
– Mare
Aug 14, 2019 at 9:43
• My proof had a gap. Does anyone see why $Hom(\Omega^1(M),M) \neq 0$ implies $Hom(\Omega^2(M), \Omega^1(M)) \neq 0$ for Nakayama algebras? It looks so elementary that I probably miss something easy. This statement corresponds exactly to conjecture 2.
– Mare
Aug 14, 2019 at 15:56
• These periodic Dyck paths, seem to be very much related with the following entry in OEIS: oeis.org/A194460 (with the circular area sequence definition). Is it possible to elaborate on this connection? Aug 14, 2019 at 19:14
– Mare
Aug 14, 2019 at 19:15
• ah, we obtain some counting results here, on circular Dyck paths: arxiv.org/abs/1903.01327 Aug 15, 2019 at 7:27

I think you have a typo in the definition of a Nakayama algebra list and it should read "$$c_{i+1}\geq c_i-1$$." If this is the case then conjecture 2 has a simple proof:
The condition for a module $$M=(i,k)$$ to be weird is that $$n\le k\le c_i-n$$. We have to show that this implies that $$\Omega^1(M)=(i+k,c_i-k)$$ is also weird. This means that we have to check that $$n\le c_i-k\le c_{i+k}-n.$$ Now, the inequality $$n\le c_i-k$$ follows immediately from the weirdness of $$M$$. The second inequality follows from observing that $$c_{i+k}-c_i=c_{i+k}-c_{i+n}=\sum_{j=n+1}^{k}(c_j-c_{j-1})\geq \sum_{j=n+1}^k (-1)=n-k.$$ After rearranging the terms we get $$c_i-k\le c_{i+k}-n$$ as desired.