It is well known that for the algebra of $n \times n$-upper triangular matrices over a field the number of tilting modules is equal to the Catalan number $C_n$. This is just the (hereditary) Nakayama algebra with Kupisch series $[n,n-1,...,2,1]$ and can be viewed as the "mother" of all Nakayama algebras with a linear quiver because it has each such algebra as a quotient. The cyclic analogue of this algebra is the Nakayama algebra with Kupisch series $[n,2n-1,2n-2,....,n+2,n+1]$ for $n \geq 2$, which can be viewed as the "mother" of all Nakayama algebras with a cyclic quiver of finite global dimension because it has each such algebra as a quotient. Now let $A$ be this Nakayama algebra with Kupisch series $[n,2n-1,2n-2,....,n+2,n+1]$. This is an algebra with global dimension 2 (while the algebra of upper triangular matrices has global dimension 1).

I wondered what the tilting modules over this algebra are. The problem seems to contain the problem of classifying the tilting modules over the algebra of upper triangular matrices as a special case because the indecomposable modules with projective dimension one in $A$ just behave like modules over the algebra of upper triangular matrices and thus the number of 1-tilting modules of $A$ should be also equal to the Catalan numbers.

The number of tilting modules of $A$ starts with 1,3,10,35,126 and this suggests that the number of tilting modules equals $\binom{2n-1}{n}$ which are the monotone maps $\{1,...,n \} \rightarrow \{1,...,n \}$, see https://oeis.org/A001700.

This leads to the following guess:

There is a natural bijection from the set of monotone maps $\{1,...,n \} \rightarrow \{1,...,n \}$ to the set of tilting modules of $A$.

Note that the monotone maps $f$ with $f(i) \leq i$ are counted by the Catalan numbers (see for example exercise 78. in the book "Catalan numbers" by Richard Stanley) and thus the above bijection (if it exists) should restrict to a bijection between monotone maps $f$ with $f(i) \leq i$ and the 1-tilting modules of $A$ (at least if it is a nice bijection).

I wanted to ask whether there is a quick proof of this guess in case it is true using some advanced tools. I am able to translate the problem into a purely combinatorial problem but it looks very complicated at the moment and maybe there is an easy trick to obtain such a bijection or maybe this is even known.

The combinatorial translation gives the problem where $n$ points (corresponding to the indecomposable summands of the basic tilting module) are drawn into two triangles (whose points correspond to the 2-rigid indecomposable modules in the Auslander-Reiten quiver of the algebra). There is one bigger triangle with $\frac{n(n+1)}{2}$ points and one smaller triangle with $\frac{n(n-1)}{2}$ points so that both triangles have together $n^2$ points. Here the tilting modules for $n=3$ (maybe someone can see how they correspond to monotone sequences?): https://www.docdroid.net/YwBhi0k/monotonetilting.pdf

Here the configurations where the red market points only occur in the smaller triangle or on the leftmost boundary of the bigger triangle count the 1-tilting modules so that for n=3 we get 5 1-tilting modules.

  • $\begingroup$ I'm trying to unwind your notation, because I suspect that the answer will be that tilting modules for your A correspond to the facets of the "doubled Cambrian fan" from combinatorics.org/ojs/index.php/eljc/article/view/v22i4p46 in the same way that tilting modules modules for the hereditary Nakayama algebra correspond to the facets of the type A Cambrian fan arxiv.org/abs/math/0606201 (see arxiv.org/abs/1604.08401 for more explicit representation theory). $\endgroup$ – David E Speyer Jun 13 '18 at 13:57
  • $\begingroup$ Have I guessed correctly that $A$ is gotten by taking the cyclic quiver $1 \to 2 \to 3 \to \cdots \to n \to 1$ and quotienting by all paths with $n-1$ arrows? $\endgroup$ – David E Speyer Jun 13 '18 at 13:58
  • $\begingroup$ @DavidESpeyer No, that would make it a selfinjective algebra I think (when you mean quotienting out all paths with the same length n-1). The cyclic quiver you suggested is correct and Kupisch series $[c_0,c_1,...,c_{n-1}]=[n,2n-1,...,n+1]$ means that the indecomposable projective modules have vector space dimension $c_i$ (which characterises the algebra uniquely). $\endgroup$ – Mare Jun 13 '18 at 14:06
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    $\begingroup$ Okay, I take my guess back then. My example has cyclic symmetry which yours lacks. And I also see that I misremembered how many facets to expect: The ring I described should have number of support tilting modules equal to the type D Catalan number, which is $\tfrac{3n-2}{n} \binom{2n-2}{n-1}$, not $\binom{2n-1}{n}$. I'll think a bit about your ring. $\endgroup$ – David E Speyer Jun 13 '18 at 14:16
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    $\begingroup$ Yes, mine is self injective. (It is the Jacobian algebra for the potential which is the $n$-cycle.) $\endgroup$ – David E Speyer Jun 13 '18 at 14:17

I'm missing some details, but I think the answer is lurking in Buan-Krause (2004) and Adachi (2016).

Neither of these papers asks exactly the same question as Mare. Let Q be the oriented $n$-cycle, let $k[Q]$ be the path algebra and let $k[[Q]]$ be the completion of the path algebra. The Buan-Krause paper studies tilting modules for $k[[Q]]$; the Adachi paper studies Nakayama algebra $k[Q]/I$ where all the indecomposable projectives have length $\geq n$ (such as the OP's) but studies "basic $\tau$-tilting" and "basic proper support $\tau$-tilting" rather than "tilting". I'm not completely clear on the relation between these concepts, but I hope the OP is.

In any case, both these papers biject the modules they study to $n$-tuples $(a_1, \ldots, a_n)$ of nonnegative integers with $\sum a_i = n$. These correspond easily to monotone functions $f: [n] \to [n]$; put $a_i = \# f^{-1}(i)$.

These papers give bijections to the variants of tilting modules they study in Theorem D of Baun-Krause, and Theorems 2.16 and 2.19 in Adachi.

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  • $\begingroup$ Surprisingly the set of $\tau$-tilting modules and the set of tilting modules never coincide for this algebra, but their number seems to be equal (if a $\tau$-tilting module is a generalised tilting modules, it would be faithful and thus have projective dimension one. But for this algebra there are many generalised tilting modules of projective dimension 2). $\endgroup$ – Mare Jun 17 '18 at 12:48

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