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Let $Q$ always denote a Dynkin quiver.

Given a connected path algebra $A=kQ$ and a module $M$, is there a useful criterion on $M$ when $End_A(M)$ is again a connected quiver algebra?

Call a module $M$ hereditary in case $End_A(M)$ has this property (which is equivalent that it has global dimension at most 1 and is connected, which are exactly the connected hereditary algebras).

Recall that a module $M$ is called basic in case it has no direct summand of the form $N^2$ for non-zero $N$.

For a given Dynkin quiver $Q$ (with $s(Q)$ denoting the number of points of $Q$) define

$a_{Q,n}:= | \{ M \in mod-kQ | M$ is a hereditary basic module with $n$ indecomposable summands $\} |.$

Of special interest are the numbers

$b_{Q}:= | \{ M \in mod-kQ | M$ is a hereditary basic module with $s(Q)$ indecomposable summands $\} | .$

Can we calculate those numbers in general for Dynkin quivers $Q$?

Special cases are also interesting such as $Q$ being a linear oriented line.

For example we seem to have $a_{Q,2}=\frac{m(m+1)(m+2)(m+7)}{24}$ and $b_Q=2,6,23,114...$ when $Q$ is a linear oriented line with $m+1 \geq 2$ simple modules.

Maybe the question is also interesting when dropping the connecteness condition always.

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    $\begingroup$ Just out of curiosity: How does $b_Q$ continue, i.e. what are the next two numbers after 23? $\endgroup$
    – Julian Kuelshammer
    Commented Oct 15, 2018 at 9:24
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    $\begingroup$ For $a_{Q,2}$ I think you should be able to use the recursion $a_{Q,2}(m)=(m+1)m+a_{Q,2}(m-1)+(m-1)m(m+1)/6$ where $(m+1)m$ are all the pairs which contain a simple module, $a_{Q,2}(m-1)$ are all the sequences which come from the "upper subtriangle" in the Auslander-Reiten quiver and $(m-1)m(m+1)/6$ are the remaining ones that appear because in the "upper subtriangle" there is some interchange between zero-relations and commutativity relations. $\endgroup$
    – Julian Kuelshammer
    Commented Oct 15, 2018 at 11:31
  • $\begingroup$ Probably a better way to see this is to note that $a_{Q,2}(m)=\sum_{k=1}^{m}\frac{k(k-1)(k+4)}{6}$. What is summed over is the number of 2-element subsets which contain an injective module (on the $k$-th layer). There are $\frac{m(m+1)}{2}$ which contain two injectives and $\sum_{j=1}^{k-1}k(n-k)=\frac{k(k-1)(k+1)}{6}$ many which contain one injective and one other module. $\endgroup$
    – Julian Kuelshammer
    Commented Oct 15, 2018 at 12:06
  • $\begingroup$ @JulianKuelshammer Yes the $n=2$ case should not be hard since there can occur no relations. Ill calculate now one more term for $b_Q$ but it takes forever and the next term will probably the last one my computer can do in a reasonable time. $\endgroup$
    – Mare
    Commented Oct 15, 2018 at 12:49
  • $\begingroup$ @JulianKuelshammer The next term is 114. I added it. $\endgroup$
    – Mare
    Commented Oct 15, 2018 at 13:12

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