6
$\begingroup$

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 $M$.

Improve this question
$\endgroup$
6
  • 1
    $\begingroup$ Why should it be true? $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Jan 1, 2018 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
    Commented Jan 1, 2018 at 20:48
  • 2
    $\begingroup$ 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
    Commented Jan 1, 2018 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
    Commented Jan 1, 2018 at 21:34
  • 2
    $\begingroup$ 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
    Commented Jan 1, 2018 at 21:57

2 Answers 2

Reset to default
5
$\begingroup$

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 $i=1,2$.

Improve this answer
$\endgroup$
4
  • $\begingroup$ 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
    Commented Jan 1, 2018 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
    Commented Jan 1, 2018 at 21:25
  • 1
    $\begingroup$ @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
    Commented Jan 1, 2018 at 21:30
  • $\begingroup$ Thanks, that is clever and makes it easy to calculate $Ext$. $\endgroup$
    – Mare
    Commented Jan 1, 2018 at 21:37
4
$\begingroup$

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).

Improve this answer
$\endgroup$
7
  • $\begingroup$ 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
    Commented Jan 1, 2018 at 19:54
  • $\begingroup$ @Mare: isn't Jeremy's answer a counter-example to that statement in non-commutative case? $\endgroup$
    – Hailong Dao
    Commented Jan 1, 2018 at 19:57
  • 1
    $\begingroup$ No, his example is not local. $\endgroup$
    – Mare
    Commented Jan 1, 2018 at 19:57
  • $\begingroup$ @Mare: interesting. Was that stated somewhere before your question? $\endgroup$
    – Hailong Dao
    Commented Jan 1, 2018 at 19:58
  • 2
    $\begingroup$ @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
    Commented Jan 1, 2018 at 20:07

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .