# Number of hereditary modules of a hereditary algebra

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.

• Just out of curiosity: How does $b_Q$ continue, i.e. what are the next two numbers after 23? – Julian Kuelshammer Oct 15 '18 at 9:24
• 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. – Julian Kuelshammer Oct 15 '18 at 11:31
• 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. – Julian Kuelshammer Oct 15 '18 at 12:06
• @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. – Mare Oct 15 '18 at 12:49
• @JulianKuelshammer The next term is 114. I added it. – Mare Oct 15 '18 at 13:12