# Number of tilting modules

Let $A=A_n$ be the algebra of upper triangular matrices over a field $K$ with $n$ simple modules. It is a nice result that there are $C_{n+1}=1,2,5,14,...$ (Catalan numbers for $n \geq 1$) tilting $A_n$-modules, where a tilting module $T$ is a module with $n$ indecomposable summands (we assume all modules are basic) and projective dimension 1 and $Ext^{1}(T,T)=0$. Let $J$ be the Jacobson radical of $A_n$ and $B_{n,l}:=A_n /J^l$ for some $n-1 \geq l \geq 2$.

Computer experiments with small n and l suggest the following generalisation:

The number of tilting $B_{n,l}$ modules equals $C_l$.

Is this true? If yes, there should be a simple reason, which I do not see at the moment.

If no, what is the correct number of $B_{n,l}-$tilting modules?

My guess goes as follows: Let $e$ be the idempotent of $B=B_{n,l}$ such that $eB$ is minimal faithful projective-injective. Then $eB$ is a summand of any $B$-tilting module. Thus any tilting module is of the form $T=eB \oplus X$ and $X$ corresponds to a tilting-module of $B/BeB$ (why?) which can be identified with $A_{l-1}$. Thus there are as many tilting $B$-modules as $A_{l-1}$ tilting modules which is $C_l$.

• You should ask that $l\leq n$, else the claim is certainly false (as $B_{n,n+k}=A_n$ for all $k\geq0$). I don't see why $B/BeB$ can be identified with $A_l$—it seems to be rather $A_{l-1}$. – Matthew Pressland Jul 24 '17 at 12:04
• @MatthewPressland you are right, thanks. – Mare Jul 24 '17 at 12:29
• It seems $B \to B/BeB$ is a homological epimorphism, and the essential image of ${\operatorname{mod}} (B/BeB) \to {\operatorname{mod}} B$ is the modules of projective dimension $\leq 1$. That should be the simple reason. – Dag Oskar Madsen Sep 3 '17 at 18:37
• This should work if you change the definition of $e$ a little. Do not include the "leftmost" primitive idempotent in the sum. – Dag Oskar Madsen Sep 3 '17 at 18:46
• @DagOskarMadsen you can find some data here findstat.org/StatisticsDatabase/St000949 , where the dyck path corresponds to the top of the Auslander-Reiten quiver of the Nakayama algebra with linear quiver. – Mare Sep 4 '17 at 7:16

## 1 Answer

Let $e$ be the idempotent in $B$ such that $Be$ is the direct sum of the $n-l$ indecomposable projective-injectives which do not have projective proper submodules.

Then the two-sided ideal $BeB=Be$ is projective as a left $B$-module, so $B \to B/BeB$ is a homological epimorphism, see for instance

Koenig, Steffen; Nagase, Hiroshi, Hochschild cohomology and stratifying ideals., J. Pure Appl. Algebra 213, No. 5, 886-891 (2009). ZBL1181.16009.

This means that $\operatorname{Ext}^*_{B/BeB}(M,N) \cong \operatorname{Ext}^*_B(M,N)$ for all $B/BeB$-modules $M$ and $N$. We also have an isomorphism of algebras $B/BeB \cong A_l$.

Claim: If $X$ is a tilting $B/BeB$-module, then $Be \oplus X$ is a tilting $B$-module.

Proof: Suppose $X$ is tiling $B/BeB$-module, and let $T=Be \oplus X$.

1) $X$ has $l$ indecomposable summands, so $T=Be \oplus X$ has $n$ indecomposable summands.

2) For $n \geq 2$, we have $$\operatorname{Ext}^n_B(T,B) \cong \operatorname{Ext}^n_B(X,B/BeB) \cong \operatorname{Ext}^n_{B/BeB}(X,B/BeB)=0,$$ so the projective dimension of $T$ is at most $1$.

3) $\operatorname{Ext}^1_B(T,T) \cong \operatorname{Ext}^1_B(X,X) \cong \operatorname{Ext}^1_{B/BeB}(X,X)=0.$

Claim: There are no other tilting $B$-modules.

Proof: Suppose $T=Be \oplus Y$ is a tilting $B$-module and $Y$ has an indecomposable direct summand $Y'$ with $(BeB)Y' \neq 0$. Let $f \colon P(Y') \to Y'$ be the projective cover. Then $P(Y')$ is an indecomposable direct summand of $Be$, so $\ker f$ is not projective. Hence the projective dimension of $Y'$ is at least $2$. Contradiction.

Conclusion: The number of tilting $B$-modules is equal to the number of tilting $A_l$-modules.