10
$\begingroup$

Call two Dyck paths $D_1$ and $D_2$ derived equivalent in case their corresponding Nakayama algebras are derived equivalent (The Dyck path of a Nakayama algebra with a linear quiver is just the top boundary of its Auslander-Reiten quiver, so we can identify Nakayama algebras with a linear quiver with Dyck paths). Derived categories of Nakayama algebras appeared in the literature (see for example https://www.sciencedirect.com/science/article/pii/S0001870813000182 ) but it seems to be wide open to classify Dyck paths with respect to derived equivalences (or?).

Call a Nakayama algebra (or the corresponding Dyck path) bouncing in case the Kupisch series is of the form $[a_1+1,a_1,...,3,2,a_2+1,a_2,...,2,...,a_r+1,a_r,...,3,2,1]$.

The truth of the conjecture in What are the periodic Dyck paths? would imply the following: In case a Dyck path $D_1$ is derived equivalent to a bouncing Dyck path $D_2$, then also $D_1$ is bouncing. Thus, it would give that being bouncing is a derived invariant.

Question 1:

Do we know whether it is true that being bouncing is a derived invariant? That is: In case a Nakayama algebra $D_1$ is derived equivalent to a bouncing Nakayama algebra $D_2$, then also $D_1$ is bouncing?

Question 2:

Is it known which bouncing Nakayama algebras with a fixed number of simple modules are derived equivalent to eachother? How many equivalence classes (up to derived equivalence) exist of bouncing Nakayama algebras?

Improve this question
$\endgroup$
6
  • 1
    $\begingroup$ In answer to the second part of the parenthesized Question 3, I think it's the case that the first Hochschild cohomology $HH^1(A,A)$ (which is an invariant of the derived category) is zero if $A$ is a Nakayama algebra with acyclic quiver, but non-zero if $A$ is a Nakayama algebra with non-acyclic quiver. $\endgroup$
    – Jeremy Rickard
    Sep 25, 2018 at 10:05
  • $\begingroup$ @JeremyRickard That is a good idea. That the Hochschild cohomology is zero for the linear quiver case is by a result of Happel. I cant remembers seeing the first Hochschild cohomology calculated for Nakayama algebras with a cyclic quiver but I guess you are right. $\endgroup$
    – Mare
    Sep 25, 2018 at 12:53
  • 1
    $\begingroup$ @JeremyRickard Here what my computer suggests: Let $A$ be a Nakayama algebra with a cyclic quiver and Kupisch series $[c_1,...,c_n]$ with minimal entry $c=s(n+1)+k \geq 2$ for $k \leq n$ and $s \geq 0$. Then the vector space dimension of the first Hochschild cohomology is equal to $s+1$. So the first Hochschild cohomology counts in a way how often you walk around the circle completely +1 it seems. $\endgroup$
    – Mare
    Sep 25, 2018 at 12:55
  • 2
    $\begingroup$ I think Happel showed, by producing an explicit non-inner derivation, that the path algebra of a non-acyclic quiver by an admissible and homogeneous (in terms of lengths of paths) ideal has non-zero $HH^1$. $\endgroup$
    – Jeremy Rickard
    Sep 25, 2018 at 13:01
  • 2
    $\begingroup$ I think, but am not sure, that I once knew of an example of derived equivalent algebras with acyclic/non-acyclic quivers. I’ll try to remember what it is or where I saw it. $\endgroup$
    – Jeremy Rickard
    Sep 25, 2018 at 13:04

1 Answer 1

Reset to default
6
$\begingroup$

Question 1 has a positive answer by the answer of Gjergji Zaimi in this thread: What are the periodic Dyck paths? .

Here a positive answer to question 1 and 2 using a complicated classification result that uses several other deep results from the literature (so more elementary proofs of Question 2 are welcome):

Question 1 and 2 have a positive answer by the classification of iterated tilted algebras of Dynkin type $\mathcal{A}$. See the main theorem of https://www.tandfonline.com/doi/abs/10.1080/00927878108822697 , which implies here that the bouncing Dyck paths correspond to Nakayama algebras with a linear quiver having only relations of length two.

(edit: I think I found an elementary answer to Question 2 using tilting modules. I might post it here soon)

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.