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Let $A$ be a selfinjective algebra and for an indecomposable module $M$ define $\psi_M:= \inf \{ i \geq 1 | Ext_A^i(M,M) \neq 0 \}$.

Questions:

  1. In case $A$ is symmetric, do we have $\psi_M \leq max \{ \psi_S | S $ is simple $\}$ for each indecomposable non-projective module $M$? This should be true in case $A$ is representation-finite.

  2. In case $A=kG$ is a group algebra over a field of characteristic $p$. Do we have even $\psi_M \leq \psi_K$ when $K$ is the trivial module and each indecomposable non-projective $M$ ? I can prove this for $p$-groups and in case $p$ does not divide the dimension of $M$.

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  • $\begingroup$ Why should it be true? $\endgroup$ – Mariano Suárez-Álvarez Jan 1 '18 at 20:45
  • $\begingroup$ @MarianoSuárez-Álvarez well it seems to be true in the local and the representation-finite case, so I think it is worth asking. $\endgroup$ – Mare Jan 1 '18 at 20:48
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    $\begingroup$ I am asking why it should be true. Explain why it should, as that can only improve your question — claims that something should or should not be unaccompanied with reasons are extremely vaporous in almost all contexts1. Also: it seems to be true in those cases or is it true? $\endgroup$ – Mariano Suárez-Álvarez Jan 1 '18 at 21:24
  • $\begingroup$ @MarianoSuárez-Álvarez 2. Is true for representation-finite and local group algebra. For 1. it is open whether it is true for local algebras. For me having some evidence for large classes of algebras is often enough to pose a question, together with the motivation that a postive answer would have some nice applications. I do not know what you expect as a good answer for why it should be true. $\endgroup$ – Mare Jan 1 '18 at 21:34
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    $\begingroup$ What I am unsuccessfully trying to tell you is to add that information to the body of your question. Instead of saying that something should be true or that it seems to be true, say why you think it is true, or why you expect it to be true, by mentioning the evidence you have and so on. Nothing has to be true. Newcomers to the subject, specially, will appreciate that — deontological claims only contribute to make them excluded of some inaccesible lore that would allow them to know, like you do, why things should be true. $\endgroup$ – Mariano Suárez-Álvarez Jan 1 '18 at 21:57
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I think this example answers both questions.

Let $k$ have characteristic $3$, and let $G=C_3\times S_3$.

Then $kG$ has two simple modules, both one-dimensional, and for each simple module $S$, $\text{Ext}^1(S,S)$ is one-dimensional.

But if $M=kC_3$, with $S_3$ acting trivially, then $\text{Ext}^i(M,M)=0$ for $i=1,2$.

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  • $\begingroup$ thanks, how to see that $Ext^i(M,M)=0$ for $i=1,2$ quickly? The quiver of $KG$ should have loops at both points so this shows that $Ext^1(S,S)$ should be nonzero for the simples. $\endgroup$ – Mare Jan 1 '18 at 19:41
  • $\begingroup$ @Mare A projective resolution for $M$ can be obtained by tensoring $M$, considered as a $kC_3$-module, with a projective $kS_3$-resolution of the trivial module, and then the calculation follows easily. $\endgroup$ – Jeremy Rickard Jan 1 '18 at 21:25
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    $\begingroup$ @Mare A more general example would be to take two algebras (over an algebraically closed field, say, to avoid non-separability issues): $B$ a non-semisimple local algebra, and $C$ an algebra such that $\text{Ext}^1(S,S)=0$ for every simple $C$-module. Then take $A=B\otimes_kC$ and $M=B\otimes_kS$ for $S$ a simple $C$-module. $\endgroup$ – Jeremy Rickard Jan 1 '18 at 21:30
  • $\begingroup$ Thanks, that is clever and makes it easy to calculate $Ext$. $\endgroup$ – Mare Jan 1 '18 at 21:37
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Suppose $A$ is commutative local (necessarily artinian) with the only simple $k\neq A$ then $\psi_k=1$, so your statement 1 will say that $Ext^1_A(M,M)=0$ implies $M$ is free. I stated it as a conjecture for complete intersections here (conjecture 9.1.3). Technically, it was stated as $Ext^1_A(M,N)=0$ implies $Ext^i_A(M,N)=0$ for all $i>0$, but when $M=N$ the latter condition is equivalent to $M$ being free.

One could also ask if $Ext^1_A(M,M)=0$ implies $M$ is free, still assuming that $A$ is Gorenstein. It was stated as a question (9.1.4) in the same survey.

As far as I know, both are open even for complete intersections unless $A$ is a hypersurface (which will be representation finite anyway).

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  • $\begingroup$ Thanks, as far as I know it is in general open whether $Ext_A^1(M,M) \neq 0$ for any non-projective indecomposable module over a finite dimensional local selfinjective algebra. (not necessarily commutative). $\endgroup$ – Mare Jan 1 '18 at 19:54
  • $\begingroup$ @Mare: isn't Jeremy's answer a counter-example to that statement in non-commutative case? $\endgroup$ – Hailong Dao Jan 1 '18 at 19:57
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    $\begingroup$ No, his example is not local. $\endgroup$ – Mare Jan 1 '18 at 19:57
  • $\begingroup$ @Mare: interesting. Was that stated somewhere before your question? $\endgroup$ – Hailong Dao Jan 1 '18 at 19:58
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    $\begingroup$ @Mare: yes, I saw that there were some issues with that paper, specified here, footnote on page 1: s.web.umkc.edu/segal/papers/AR.pdf $\endgroup$ – Hailong Dao Jan 1 '18 at 20:07

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