Let $A_n=kQ_n$ be the path algebra of linear oriented Dynkin graph $Q_n$ of Dynkin type $\mathcal{A_n}$ (so $A_n$ is the unique hereditary Nakayama algebra given by quiver and relations).

The number of tilting $A_n$-modules equals the Catalan numbers $C_n$ when $n$ is the number of simple $A_n$-modules.

Question: Is there a natural bijection between Dyck paths of length $2n$ to the tilting modules of $A_n$?

I would think that this might be noted somewhere in the literature, but I did not find it.

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    $\begingroup$ The tilting modules correspond in a natural way to order ideals of the poset of positive roots, right? These order ideals are in obvious bijection with Dyck paths (because the poset is just a “triangle.”) $\endgroup$ – Sam Hopkins Apr 25 '18 at 21:18
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    $\begingroup$ Maybe Chapter 3 here explains this?: arxiv.org/abs/1502.06553v2 $\endgroup$ – Sam Hopkins Apr 25 '18 at 21:29
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    $\begingroup$ @SamHopkins Thanks. Can you make this an answer with a little more detail? So given a Dyck path, what is the corresponding tilting module in the Auslander-Reiten quiver of the algebra(which can be described as a set of $n$ dimension vectors)? When I draw the Auslander-Reiten quiver of the algebra for $n=3$ and put in the 5 tilting modules in the triangle, I do not get immediately Dyck paths. So Im note sure what the natural bijection is or should be. $\endgroup$ – Mare Apr 25 '18 at 21:33
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    $\begingroup$ In fact, section 4.7 of those notes suggest that finding a bijection to antichains of the poset of positive roots (i.e., to Dyck paths), may be an open problem... $\endgroup$ – Sam Hopkins Apr 25 '18 at 22:22
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    $\begingroup$ Catalan numbers also occur in the enumeration of binary trees. Lutz Hille gives a natural bijection between tilting modules of $A_n$ and rooted binary trees with $n+1$ leaves. This bijection is explained in Section 9 of the following article by L. Hille: On the volume of a tilting module, Abh. Math. Sem. Univ. Hamburg 76 (2006), 261--277. There more or less natural bijections between Dyck paths and rooted binary trees, and a composition gives a bijection between tilting modules and Dyck paths. $\endgroup$ – Philipp Lampe Apr 26 '18 at 15:10

This is more of an expansion of Sam's comments, but too long for a comment itself:

As pointed out by Sam in that Theorem, tilting modules of the linear type $A_n$ quiver are in natural correspondence to triangulations of a regular $(n+3)$-gon.

Short answer: Any bijection between triangulations and Dyck paths would work, and there are plenty. Everyone is "natural" in some respect and "unnatural" in others.

Longer answer: Triangulations of a polygon might be generalized to all finite type using either of the following:

  • noncrossing partitions $NC$
  • clusters $CL$
  • Coxeter-sortable elements $CS$

and there is a uniform type-independent bijection between them.

On the other hand, Dyck paths might be generalized to antichains in the root poset of a finite crystallographic type $AC$.

One now has $$|NC| = |CL| = |CS| = |AC|$$ and indeed many refinements of this identity hold (such as the one mentioned in Section 4.7).

The closest connection between the two that is known is (caution, self-reference!) the article A uniform bijection between nonnesting and noncrossing partitions jointly with Drew Armstrong and Hugh Thomas.

There is much more to say and more references to give, please look into (again caution!) Cataland: Why the Fuss? for all needed notions and references.


Start with a tilting module for the quiver $0\to 1\to 2\to ...\to n$.
Throw away the components with support at 0.
You get a support tilting module $T_1,...,T_k$ for $1\to 2\to ...\to n$.
Take, for each $T_i$, the part $\lambda_i$ equal to the top of $T_i$.
For example, every injective module gives $\lambda_i=1$.
Take the Young diagram of that partition.
The shadow of the Young diagram is the corresponding Dyck path of length $2n+2$.

Conversely, take a Dyck path of length $2n+2$. Then each down-step $D_i$ corresponds to a component $T_i$ of the corresponding tilting module.
The top of $T_i$ is at vertex $k$ where $k$ is the number of upsteps after $D_i$.
If $D_i$ goes from level $\ell+1$ to level $\ell$, the length of $T_i$ is equal to the number of upsteps since the last time the Dyck path went up to level $\ell$ (if none exists, $T_i$ is projective). For example, $T_{n+1}$ will be the projective injective $P_0$.


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