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Let $k$ be an algebraically closed field and $A$ be a symmetric algebra. I want to know how to compute almost split sequences ending at a non-projective indecomposable right $A$-module $X$.

Question: Let $P\xrightarrow{p} \Omega(M)$ be the projective cover of $\Omega(X)$, where $\Omega(X)$ is the first syzygy of $X$. Then, we have an exact sequence

$0\to \Omega ^2(X) \to P \xrightarrow{p} \Omega(X)\to 0$.

Now, let $\xi \in \underline{\mathsf{Hom}} _A(X,\Omega(X))$ such that $\xi$ generates the socle of $\underline{\mathsf{Hom}} _A(X,\Omega(X))$. Then, we can take pullback along $(p,\xi)$, and we have an exact sequence

$0\to \Omega^2(X)\to E \to X \to 0$.

My question is: Is the exact sequence $0\to \Omega^2(X)\to E \to X \to 0$ almost split?

Thank you for your consideration.

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  • $\begingroup$ Yes, this is correct. You can read about it in my Representations and Cohomology I, section 4.12. $\endgroup$
    – Dave Benson
    Commented Jun 8, 2023 at 7:36
  • $\begingroup$ Thank you for your help. I understood that the almost split sequence ending at $X$ is represented by a generator of simple socle of the stable endomorphism ring of $X$. But, why is the exact sequence from the pull-back along $(p,\xi)$ almost split? $\endgroup$
    – sola
    Commented Jun 8, 2023 at 13:06
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    $\begingroup$ No, it's the simple socle of the dual of the stable endomorphism ring. Use Tate duality, which is Auslander-Reiten duality in this case, to see that this is your stable homs. Then translating to Ext$^1$ is standard dimension shifting, and yours is one of the equivalent ways to do it. $\endgroup$
    – Dave Benson
    Commented Jun 8, 2023 at 13:14

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