6
$\begingroup$

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 $C_l$.

$\endgroup$
  • 2
    $\begingroup$ 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 $A_{l-1}$. $\endgroup$ – Matthew Pressland Jul 24 '17 at 12:04
  • $\begingroup$ @MatthewPressland you are right, thanks. $\endgroup$ – Mare Jul 24 '17 at 12:29
  • 1
    $\begingroup$ 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. $\endgroup$ – Dag Oskar Madsen Sep 3 '17 at 18:37
  • 1
    $\begingroup$ This should work if you change the definition of $e$ a little. Do not include the "leftmost" primitive idempotent in the sum. $\endgroup$ – Dag Oskar Madsen Sep 3 '17 at 18:46
  • 1
    $\begingroup$ @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. $\endgroup$ – Mare Sep 4 '17 at 7:16
4
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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