I am confused at the answer of the question: Why is the Auslander Reiten theory not working in this example?

Consider the quiver

$ 1 \xrightarrow{\alpha} 2 \xrightarrow{\beta} 3 \xrightarrow{\gamma} 4$

with relations $I= < \alpha \beta \gamma >$.

The answer talks about how to calculate $\nu^{-1} \begin{matrix} 2 \\ 3\end{matrix}$:

$$ \nu^{-1}\begin{matrix} 2 &\\ 3 \end{matrix} = Hom_A(DA,\begin{matrix} 2 &\\ 3 \end{matrix}\!\!\!\!\!) = Hom_A(I(1)\oplus I(2)\oplus I(3) \oplus I(4), \begin{matrix} 2 &\\ 3 \end{matrix}\!\!\!\!\!). $$

Since $Hom_A(I(1), \begin{matrix} 2 &\\ 3 \end{matrix}\!\!\!\!\!) = Hom_A(I(2), \begin{matrix} 2 &\\ 3 \end{matrix}\!\!\!\!\!) = Hom_A(I(3), \begin{matrix} 2 &\\ 3 \end{matrix}\!\!\!\!\!) = 0$ and $Hom_A(I(4), \begin{matrix} 2 &\\ 3 \end{matrix}\!\!\!\!\!)$ is one-dimensional, we get that $\nu^{-1}\begin{matrix} 2 &\\ 3 \end{matrix}\!\!\!\!\! = 4$.

Who can tell me how to get $\nu^{-1}\begin{matrix} 2 &\\ 3 \end{matrix}\!\!\!\!\! = 4$ just by these things? Thank you.


I find it confusing when I have to think of a Hom-space as itself being a module. In this case, though, $Hom(DA, {2 \atop 3})$ is one-dimensional by the given information, so it must be a simple module over some vertex, and the only question is which one. We can answer that question by analysing which $e_i$ doesn't annihilate it.

I am going to follow the conventions of Assem-Simson-Skrowroński, so I am assuming we are using right modules over $A$ which is the quotient of the path algebra. We get a right module structure on $Hom(DA,{2 \atop 3})$ by considering the left $A$-module structure on $DA$, and, defining $fa$ by $(fa)(x)=f(ax)$. The left $A$-module structure on $DA$ comes from the right $A$-module structure on $A$. $$Hom(DA,{2\atop 3})e_i=Hom(e_i(DA),{2\atop 3})= Hom(D(Ae_i),{2\atop 3})=Hom(I_i,{2\atop 3}).$$ It therefore follows that it is only $e_4$ which doesn't annihilate $Hom(DA,{2\atop 3})$, and thus $Hom(DA,{2\atop 3})$ as an $A$-module is isomorphic to the simple $S_4$.

  • $\begingroup$ Good answer, I think it also works even $Hom(DA, \begin{matrix} 2 \\ 3 \end{matrix})$ not one-dimensional. $\endgroup$ – Xiaosong Peng Dec 24 '16 at 2:51
  • $\begingroup$ If the Hom-space is not one-dimensional, then there's more to figuring out what its structure as an $A$-module is. The above analysis tells you about the effect of multiplying by $e_i$, but that's no longer enough information to be able to read off the full module structure in general. $\endgroup$ – Hugh Thomas Dec 24 '16 at 7:51
  • $\begingroup$ Sorry, I'm cofused again. For example, there is a quiver $Q: 1 \rightarrow 2$, what is the effect of multiplying by $e_i$? If $M$ is a module $k \rightarrow 0$, so $Me_1=k$ and $Me_2=0$ by your meaning? Then if $M=k \rightarrow k$, then what is $Me_i$? $\endgroup$ – Xiaosong Peng Dec 24 '16 at 13:17
  • $\begingroup$ If $M$ is a module for $kQ$, we can also think of it as a representation of $Q$, and vice versa. Then $Me_i$ is the vector space of $M$ corresponding to vertex $i$. See Assem-Simson-Skowroński Theorem III.1.6. (But this is really a different question from the one you started with, so if this quick answer is not sufficient, you might do better to try to formulate and post a question on this topic.) $\endgroup$ – Hugh Thomas Dec 24 '16 at 21:32
  • $\begingroup$ Ok, thank you. By the dimension of $Me_i$, we know $k^i$ at each point. But the maps between points may not clear. Is that right? $\endgroup$ – Xiaosong Peng Dec 26 '16 at 2:13

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