# 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 $I= < \alpha \beta \gamma >$

I calculated the following injective resolution for the representation $\begin{matrix} 2 &\\ 3 \end{matrix}$,

$0 \longrightarrow \begin{matrix} 2 &\\ 3 \end{matrix} \longrightarrow I(3)=\begin{matrix} 1 &\\ 2 &\\ 3 \end{matrix} \longrightarrow I(1)= 1 \longrightarrow 0$

applying the Inverse Nakayama functor $\nu^{-1}$ I get

$0 \longrightarrow \nu^{-1} \begin{matrix} 2 &\\ 3 \end{matrix} =0 \longrightarrow P(3) = \begin{matrix} 3 &\\ 4 \end{matrix} \longrightarrow P(1) =\begin{matrix} 1 &\\ 2 &\\ 3 \end{matrix} \longrightarrow \tau^{-1} \begin{matrix} 2 &\\ 3 \end{matrix} \longrightarrow 0$

BUT the sequence is not exact from $P(3) \longrightarrow P(1)$ since is not injective.

What is wrong here? I think that is because $\nu^{-1} \begin{matrix} 2 &\\ 3 \end{matrix} \neq 0$ but I thought the inverse nakayama functor is zero on non- injectives?

The inverse Nakayama functor is not zero on non-injective representations. Think about what this would imply: it would mean that for every non-injective representation $M$, the projective presentation of $\tau^{-1} M$ computed in your post would show that it has projective dimension at most $1$... which cannot be true for all representations of the form $\tau^{-1}M$.

So let's redo the computation. What is $\nu^{-1}\begin{matrix} 2 &\\ 3 \end{matrix}\!\!\!\!\!$? By definition, it is

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

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 $\nu^{-1}\begin{matrix} 2 &\\ 3 \end{matrix}\!\!\!\!\! = 4$.

Thus the exact sequence is

$$0 \longrightarrow \nu^{-1} \begin{matrix} 2 &\\ 3 \end{matrix}\!\!\!\!\! =4 \longrightarrow P(3) = \begin{matrix} 3 &\\ 4 \end{matrix} \longrightarrow P(1) =\begin{matrix} 1 &\\ 2 &\\ 3 \end{matrix} \longrightarrow \tau^{-1} \begin{matrix} 2 &\\ 3 \end{matrix} \longrightarrow 0,$$ which is more what we would expect. Thus $\tau^{-1} \begin{matrix} 2 &\\ 3 \end{matrix}\!\!\!\!\! = \begin{matrix} 1 &\\ 2 \end{matrix}\!\!\!\!\!.$

• I see, somehow I thought the behaviour of the nakayama functor (zero on representations with no projective sumands) was mirrored by the inverse functor (zero on representations with no injective sumands) – Eli Dec 19 '16 at 18:47
• So much thanks :) – Eli Dec 19 '16 at 18:48
• @Pierre-Guy Plamondon Hello, sir. I have a question to your answer: How do you get $\nu^{-1} \begin {matrix} 2 \\ 3\end {matrix} =4$ by $Hom_A(I(4), \begin {matrix} 2 \\ 3\end {matrix} )$ one-dimensional? Do we also need the form of $P(3)$ and $P(1)$? Or other ways? Thank you. – Xiaosong Peng Dec 21 '16 at 3:13
• @Penson That is because all the other $Hom_A(I(i), \begin{matrix} 2 \\ 3 \end{matrix})$ are zero. – Pierre-Guy Plamondon Dec 21 '16 at 10:08
• @Pierre-Guy Plamondon Yes, I know $Hom_A(I(i), \begin{matrix} 2 \\ 3 \end{matrix})=0$ for $i=1,2,3$ and $Hom_A(I(4), \begin{matrix} 2 \\ 3 \end{matrix})$ is one-dimensional. But I still don't know how to get $\nu ^{-1} \begin{matrix} 2 \\ 3 \end{matrix}=4$. Could you explain it more clear? – Xiaosong Peng Dec 21 '16 at 12:01