So I was studying the local shape (around the simple representation $S(3)$) of the Auslander Reiten quiver of this $Q$:

And so far I concluded that the neighborhood of the representation $S(3)$ looks like this:

enter image description here

Where the representations $S(3)$ and $\begin{matrix} 2 \\3\end{matrix}$ on the right and left side are identified. (so the AR quiver actually looks like a tube locally)

I wanted to prove that the three representations $N_1$, $N_2$ and $N_3$ are not isomorphic. Is this obvious from the graph, or how can I show it without messing too much with explicit calculations?

So far I proved that if two of them are isomorphic, all of them are isomorphic. And that implies (because of the uniqueness of the almost split sequences), that all the three representations in the second row, must have arrows (irreducible morphisms) into all three $N_i$'s. So there is one almost split sequence (one for each $N_i$ actually) $0 \rightarrow N \rightarrow \begin{matrix} 1 \\ 2 \ 4\end{matrix} \oplus \begin{matrix} 1 \ 3 \\ 4\end{matrix} \oplus \begin{matrix} 2 \\ 3 \end{matrix} \rightarrow N \rightarrow 0$.

But I don't see any contradiction. Is there any way to prove that they are not isomorphic by looking at the AR quiver or some abstract nonsense?

  • $\begingroup$ How did you arrive at this particular shape of the neighbourhood? $\endgroup$ – Julian Kuelshammer Jun 21 '17 at 11:07
  • 1
    $\begingroup$ Calculating the auslander reiten translation you can calculate the first row. Then to get $\begin{matrix} 2 \\ 3 \end{matrix}$ on the second row is because it is the only possible middle term in the almost split sequence between $S(3)$ and $S(2)$. Then I calculated the rest of the row using the auslander reiten translation. Finally I got the representation $N_1$ using an argument that it was the only possible term in the almost split sequence $0 \rightarrow \begin{matrix} 2 \\3 \end{matrix} \rightarrow S(2) \oplus ? \rightarrow \begin{matrix} 1 \\ 2 \ 4 \end{matrix} \rightarrow 0$ $\endgroup$ – Werner Germán Busch Jun 21 '17 at 21:14
  • $\begingroup$ And finally, I calculated the (inverse) auslander reiten translation to $N_1$ and got $N_2$, which has the same dimension vector. $\tau^{-1} N_1 = N_2$, $\tau^{-1} N_2 = N_3$ $\endgroup$ – Werner Germán Busch Jun 21 '17 at 21:18

I would use that the socle functor is left exact. This way you can see that $S(3)\in \operatorname{soc} N_1$ by applying the socle functor to the Auslander-Reiten sequence starting in $\begin{smallmatrix}2\\3\end{smallmatrix}$ and similarly, one can see that $S(3)\notin \operatorname{soc} N_2$ by applying the socle functor to the Auslander-Reiten sequence starting in $\begin{smallmatrix}&1\\2&&4\end{smallmatrix}$.

  • $\begingroup$ Yup, that looks very good! I would have never thought of using the socle functor. Thanks for teaching me that trick, thanks!. $\endgroup$ – Werner Germán Busch Jun 23 '17 at 20:47

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