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 $0 \leq s <n$ 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.

  • $\begingroup$ 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. $\endgroup$
    – Mare
    Aug 14, 2019 at 9:43
  • $\begingroup$ 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. $\endgroup$
    – Mare
    Aug 14, 2019 at 15:56
  • $\begingroup$ 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? $\endgroup$ Aug 14, 2019 at 19:14
  • $\begingroup$ @PerAlexandersson See arxiv.org/abs/1811.05846 . $\endgroup$
    – Mare
    Aug 14, 2019 at 19:15
  • $\begingroup$ ah, we obtain some counting results here, on circular Dyck paths: arxiv.org/abs/1903.01327 $\endgroup$ Aug 15, 2019 at 7:27

1 Answer 1


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.

  • $\begingroup$ It is so simply ? I feel stupid now :( $\endgroup$
    – Mare
    Aug 14, 2019 at 17:12
  • $\begingroup$ Thanks, for noting the typo and the nice solution! $\endgroup$
    – Mare
    Aug 14, 2019 at 17:25

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