0
$\begingroup$

Let $\mathsf{C}$ be a locally finite abelian category. Assume that there exists a poset $\Lambda$ and a complete set $\{L(\lambda)\}_{\lambda \in \Lambda}$ of representatives of isomorphism classes of simple objects of $\mathsf{C}$. Choose such a set $\Lambda$ and a corresponding set $\{L(\lambda)\}_{\lambda \in \Lambda}$.

We call $\nabla(\lambda)$ a costandard object for $L(\lambda)$ if $\operatorname{Soc}\nabla(\lambda)\cong L(\lambda)$ and all composition factors $L(\mu)$ of $\nabla(\lambda)/\operatorname{Soc}\nabla(\lambda)$ satisfy $\mu < \lambda$. Here, $\operatorname{Soc}\nabla(\lambda)$ denotes the socle of $\nabla(\lambda)$.

We call $\Delta(\lambda)$ a standard object for $L(\lambda)$ if $\operatorname{Hd}\Delta(\lambda)\cong L(\lambda)$ and all composition factors $L(\mu)$ of $\operatorname{Rad}\Delta(\lambda)$ satisfy $\mu < \lambda$. Here, $\operatorname{Hd}\Delta(\lambda)$ denotes the head of $\Delta(\lambda)$ and $\operatorname{Rad}\Delta(\lambda)$ denotes the radical of $\Delta(\lambda)$.

We say that an object $N \in \mathsf{C}$ has a costandard filtration if there exists a filtration $$0=N_0 \subsetneq N_1 \subsetneq … \subsetneq N_{n-1} \subsetneq N_n=N$$ by subobjects $N_i$ of $N$ such that for each $i$ the quotient $N_i/N_{i-1}$ is isomorphic to a costandard object $\nabla(\lambda_i)$, for some $\lambda_i \in \Lambda$. Dually, we define a standard filtration.

Question
Assume that $N \in \mathsf{C}$ has a costandard filtration. Further assume that $\tilde{N} \in \mathsf{C}$ is a subobject of $N$ such that $N/\tilde{N}\cong \nabla (\lambda)$ for some $\lambda \in \Lambda$. Does there exist a costandard filtration of $N$ $$ 0=N_{0}\subsetneq \cdots\subsetneq N_{n-1}\subsetneq N_{n}=N$$ with $N_{n-1}\cong\tilde{N}$?

$\endgroup$

1 Answer 1

Reset to default
2
$\begingroup$

No.

For example, let $\mathsf{C}$ be the category of finite dimensional representations of the quiver $1\rightarrow2$. Then $\mathsf{C}$ has three indecomposable objects: the simple modules $S(1)$ and $S(2)$ and a two-dimensional module $X$ with $\operatorname{soc}X=S(2)$ and $\operatorname{head}X=S(1)$.

Take $\nabla(1)=S(1)$ and $\nabla(2)=X$. Then if $N=X$ and $\tilde{N}=\operatorname{soc}X$, both $N\cong\nabla(2)$ and $N/\tilde{N}\cong\nabla(1)$ certainly have costandard filtrations, but $\tilde{N}\cong S(2)$ does not.

$\endgroup$
3
  • $\begingroup$ Does anything change if we require $N$ to be an indecomposable tilting object? I am asking because in this paper arxiv.org/pdf/2107.07887.pdf the authors claim in Lemma 3.1 that $\operatorname{Ker}(\pi^\lambda)$ has a costandard filtration. I was trying to use Remark 2.15 somehow. If the claim of my question had been true, I could have defined $\pi^\lambda$ as the projection of the ultimate element of a certain costandard filtration (whose existence would have been guaranteed by the claim) onto the quotient with the penultimate element. How can I save the argument? $\endgroup$
    – Margaret
    Jan 5 at 10:42
  • $\begingroup$ That is, why does $\nabla(\lambda)$ occur at the top of a costandard filtration of $T(\lambda)$? Thank you for your helpful answer by the way. $\endgroup$
    – Margaret
    Jan 5 at 10:43
  • 1
    $\begingroup$ @Margaret In my example, $N$ is an indecomposable tilting module. ($\Delta(1)=S(1)$, $\Delta(2)=S(2)$, $T(1)=\nabla(1)$ and $T(2)=\nabla(2)$.) $\endgroup$
    – Jeremy Rickard
    Jan 6 at 9:35

You must log in to answer this question.

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