8

(The conjecture is a homological algebra question, but question 2 is a pure combinatorics question given that the conjecture is true)

A Dyck path of length $n$ is a list of positive integers $[c_1,c_2,...,c_n]$ with $c_i -1 \leq c_{i+1}$ for all $i$ and $c_i \geq 2$ for $i \neq n$ and $c_n=1$. (One can show that those sequences really correspond to the classical Dyck paths via the area sequence and the number of Dyck paths of length $n$ is $C_{n-1}$ when $C_n$ denotes the Catalan numbers). Dyck paths can get naturally identified with the Nakayama algebra $A_D$ with a linear quiver having Kupisch series $[c_1,c_2,...,c_n]$, see for example https://arxiv.org/abs/1811.05846 .

Let $D=[c_1,c_2,...,c_n]$ be a Dyck path of length $n$. We define the Cartan matrix $C_D$ of $D$ as the $n \times n$ upper triangular matrix with entries 0 or 1 as follows: In the $i$-th row $C_D$ has entries equal to one in position $(i,i)$, $(i,i+1)$,...,$(i,i+c_i-1)$ and all other entries are zero. Define the Coxeter matrix $\phi_D$ as $-C_D^{-1} C_D^T$ and the coxeter polynomial $p_D$ as the characteristic polynomial of this matrix.

We say that a Dyck path is of Dynkin type $Q$ in case the corresponding Nakayama algebra $A_D$ is derived equivalent to $KQ$.

Conjecture: A Dyck path $D$ is of Dynkin type $Q$ if and only if the algebra $A_D$ has coxeter polynomial $p_Q$.

This is true for type $A_n$ as was proven in What are the periodic Dyck paths? by Gjergji Zaimi (together with the fact that bouncing Nakayama algebras are exactly those of Dynkin type $A_n$, which can be proved by using special tilting modules). With the help of a computer it is also true for all exceptional types $E_6, E_7$ and $E_8$ and true for $D_i$ for $i=4,5,6,7,8,9$ (which is why I made it a conjecture now), but maybe there is a nice uniform proof that works for all types.

In the theorem of page 23 in http://prospero.dmat.usherbrooke.ca/ibrahim/publications/Alg%C3%A8bres_pr%C3%A9inclin%C3%A9es_et_cat%C3%A9gories_d%C3%A9riv%C3%A9es.pdf one can find a homological characterisation when a (Nakayama) algebra/Dyck path is of Dynkin type $D_n$.

We have for the Dynkin types: $p_{A_n}=x^n+x^{n-1}+x^{n-2}+....+x^2+x+1$, $p_{D_n}=x^n+x^{n-1}+x+1$, $p_{E_6}=x_1^6+x_1^5-x_1^3+x_1+1$, $p_{E_7}=x_1^7+x_1^6-x_1^4-x_1^3+x_1+1$, $p_{E_8}=x_1^8+x_1^7-x_1^5-x_1^4-x_1^3+x_1+1$. See for example Table 1.1. in the book "Notes on Coxeter Transformations and the McKay Correspondence" by Rafael Stekolshchik.

The bouncing Dyck paths are exactly the Dyck paths of Dynkin type $A_n$ and there are $2^{n-2}$ many. A classification/enumeration seems not so easy in type $D_n$, which leads to the following question:

Question 2: How many Dyck paths are there for a given $n$ with Coxeter polynomial equal to $p_{D_n}$ (call this sequence $a_n$)? How many Dyck paths are there for a given $n$ that are of Dynkin type $D_n$ (call this sequence $b_n$)?

Note that in case the conjecture is true, we have $a_n=b_n$ and calculating $a_n$ is a purely elementary problem. The sequence $a_n$ starts for $n=4,5,6,7,8$ with 1,6,13,29,65 and does not appear in the oeis. Here are the Dyck paths with Coxeter polynomial equal to $p_{D_n}$ for $n=4,5,6,7$. Maybe someone sees a pattern what they might be:

$D_4$:

[ [ 3, 3, 2, 1 ] ]

$D_5$:

[ [ 3, 3, 2, 2, 1 ], [ 2, 3, 3, 2, 1 ], [ 3, 3, 3, 2, 1 ], [ 4, 3, 3, 2, 1 ], [ 3, 4, 3, 2, 1 ], [ 4, 4, 3, 2, 1 ] ]

$D_6$:

[ [ 3, 3, 2, 2, 2, 1 ], [ 3, 3, 3, 2, 2, 1 ], [ 3, 4, 3, 2, 2, 1 ], [ 3, 3, 2, 3, 2, 1 ], [ 2, 2, 3, 3, 2, 1 ], [ 3, 2, 3, 3, 2, 1 ], [ 2, 3, 3, 3, 2, 1 ], [ 4, 3, 3, 3, 2, 1 ], [ 2, 4, 3, 3, 2, 1 ], [ 5, 4, 3, 3, 2, 1 ], [ 3, 3, 4, 3, 2, 1 ], [ 3, 5, 4, 3, 2, 1 ], [ 5, 5, 4, 3, 2, 1 ] ]

$D_7$:

[ [ 3, 3, 2, 2, 2, 2, 1 ], [ 3, 3, 3, 2, 2, 2, 1 ], [ 3, 4, 3, 2, 2, 2, 1 ], [ 3, 3, 2, 3, 2, 2, 1 ], [ 2, 3, 3, 3, 2, 2, 1 ], [ 4, 3, 3, 3, 2, 2, 1 ], [ 3, 3, 4, 3, 2, 2, 1 ], [ 3, 5, 4, 3, 2, 2, 1 ], [ 3, 3, 2, 2, 3, 2, 1 ], [ 3, 3, 3, 2, 3, 2, 1 ], [ 3, 4, 3, 2, 3, 2, 1 ], [ 2, 2, 2, 3, 3, 2, 1 ], [ 3, 2, 2, 3, 3, 2, 1 ], [ 2, 3, 2, 3, 3, 2, 1 ], [ 4, 3, 2, 3, 3, 2, 1 ], [ 2, 2, 3, 3, 3, 2, 1 ], [ 3, 2, 3, 3, 3, 2, 1 ], [ 2, 4, 3, 3, 3, 2, 1 ], [ 5, 4, 3, 3, 3, 2, 1 ], [ 2, 2, 4, 3, 3, 2, 1 ], [ 3, 2, 4, 3, 3, 2, 1 ], [ 2, 5, 4, 3, 3, 2, 1 ], [ 6, 5, 4, 3, 3, 2, 1 ], [ 3, 3, 2, 4, 3, 2, 1 ], [ 2, 3, 3, 4, 3, 2, 1 ], [ 4, 3, 3, 4, 3, 2, 1 ], [ 3, 3, 5, 4, 3, 2, 1 ], [ 3, 6, 5, 4, 3, 2, 1 ], [ 6, 6, 5, 4, 3, 2, 1 ] ]

Here are the Dyck paths of Dynkin type $E_6$:

[ [ 2, 3, 3, 2, 2, 1 ], [ 4, 3, 3, 2, 2, 1 ], [ 4, 4, 3, 2, 2, 1 ], [ 3, 3, 3, 3, 2, 1 ], [ 4, 4, 3, 3, 2, 1 ], [ 2, 3, 4, 3, 2, 1 ], [ 4, 3, 4, 3, 2, 1 ], [ 2, 4, 4, 3, 2, 1 ], [ 3, 4, 4, 3, 2, 1 ], [ 4, 4, 4, 3, 2, 1 ], [ 5, 4, 4, 3, 2, 1 ], [ 4, 5, 4, 3, 2, 1 ] ]

(there are 12 Dyck paths with Coxeter polynomial $p_{E_6}$, 54 with Coxeter polynomial $p_{E_7}$ and 133 with Coxeter polynomial $p_{E_8}$.)

Note that it is in general not true that having the same Coxeter polynomial as $KQ$ implies that a finite dimensional algebra is derived to $KQ$( there even non-derived equivalent examples with the same Cartan matrix), so the conjecture might be special to Nakayama algebras in case it is true.

Mare
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  • Is this related to https://mathoverflow.net/q/366009 ? – Max Alekseyev Sep 01 '20 at 00:26
  • @MaxAlekseyev It is not really related. Also the arxiv paper is much older than this MO link and is also not directly related. – Mare Sep 01 '20 at 08:00
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    It seems to me that the values of the Coxeter polynomials for $D_n$ that you give do not match the product formula. For the given values, I have $p_{D_4} = \phi(6)\phi(2)^2$, $p_{D_5} = \phi(8)\phi(2)$, $p_{D_6} = \phi(10)\phi(2)^2$, $p_{D_7} = \phi(12)\phi(4)\phi(2)$, – Martin Rubey Sep 01 '20 at 08:07
  • @MartinRubey Thank you very much. It seems I forgot a factor $\Phi_2$ in the formula. I gave also a reference for the formula. – Mare Sep 01 '20 at 08:14
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    Yes, I saw that reference, but it still doesn't fit. In $p_{D_7}$, the factor $\phi(3)$ is missing, and $p_{D_5}$ and $p_{D_7}$ only have one factor $\phi(2)$. – Martin Rubey Sep 01 '20 at 08:16
  • @MartinRubey Thanks again. I deleted the cyclotomic factorisation as it is probably not needed here. Instead, I just wrote down the polynomials directly. – Mare Sep 01 '20 at 08:26
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    OK, then the correct expression would be the product over all divisors of $2(n-1)$ which are not divisors of $n-1$, times $\phi(2)$. – Martin Rubey Sep 01 '20 at 08:33
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    I very much hope that my code has a bug: the numbers of Dyck paths of type $D_n$ which only return to the axis with their final step are $1, 4, 5, 6, 7, 8, 11, 10$ and $1473$ for $n$ from $4$ to $12$. The numbers of all Dyck paths of type $D_n$ would be $1, 6, 13, 29, 65, 145, 323, 705, 3837$. – Martin Rubey Sep 01 '20 at 09:35
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    More data for $D_n$: the numbers of paths are $1, 6, 13, 29, 65, 145, 323, 705, 3837, 3329, 7169, 15361, 32769, 69633$, the numbers of paths which only return to the axis with their final step are $1, 4, 5, 6, 7, 8, 11, 10, 1473, 12, 13, 14, 15, 16$. I have no idea what makes $D_4$, $D_{10}$, $D_{12}$ special. – Martin Rubey Sep 02 '20 at 06:01
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    Put differently, the sequence would like to be $a_4=3, a_5=6$ and $a_{n+2} = 1 + 4(a_{n+1} - a_n)$, but isn't. – Martin Rubey Sep 02 '20 at 06:28
  • @MartinRubey Thanks, those numbers make it look like there is no really nice solution (at least not as nice in type $A_n$). – Mare Sep 02 '20 at 08:21
  • Also, maybe I am making an error calculating by hand, but are you sure that [3,3,3,2,1] is in $D_5$? – Gjergji Zaimi Sep 02 '20 at 20:42
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    @GjergjiZaimi Im pretty sure, I also send you an email. – Mare Sep 02 '20 at 21:03
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    @MartinRubey, empirical observations: $D_4$ is special because it has to be: there are $2^{n-2}$ paths of type $A_n$, so that only leaves one path which could be in $D_4$. For $D_{10}$ the two exceptional paths are $[3, 4, 5, 5, 4, 4, 3, 3, 2, 1]$ and $[5, 6, 5, 6, 5, 5, 4, 3, 2, 1]$; in general the $D_n$ paths which return to the axis only on the final step are $[n-1, n-1, n-2, \ldots, 2, 1]$; $[3, n-1, n-2, \ldots, 2, 1]$; and $[k, k-1, \ldots, 3, 3, n-k+1, n-k, \ldots, 2, 1]$ for $k \in [3, n-1]$ (and for $n=4$ they all coincide). – Peter Taylor Sep 28 '21 at 23:02

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