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.

Assume that we are given a contravariant $K$-linear endofunctor $D$ on $\mathsf C$ and a natural isomorphism $\xi:\operatorname{id}_{\mathsf C}\rightarrow D^2$ of $K$-linear functors satisfying $\operatorname{id}_{D(X)}=D(\xi_X)\circ \xi_{D(X)}$. Further assume that $D$ exchanges standard and costandard objects (i.e. $D(\Delta(\lambda))\cong\nabla(\lambda)$).

I am trying to prove the following claim:

If $N\in \mathsf{C}$ has a costandard filtration, then $D(N)$ has a standard filtration.

This amounts to proving that $\operatorname{Ext}^1(\Delta(\lambda),N)=0$ for all $\lambda \in \Lambda$ implies that $\operatorname{Ext}^1(D(N),\nabla(\lambda))=0$ for all $\lambda \in \Lambda$.
Since an antiequivalence exchanges monomorphisms and epimorphisms, we know that a costandard filtration $$0=N_0 \subsetneq N_1 \subsetneq … \subsetneq N_{n-1} \subsetneq N_n=N$$ of $N$ induces a sequence of epimorphisms $$D(N)=D(N_n) \twoheadrightarrow D(N_{n-1}) \twoheadrightarrow… \twoheadrightarrow D(N_{1}) \twoheadrightarrow D(N_0)=D(0).$$ How to proceed from here?

  • 1
    $\begingroup$ Is $\nabla$ a function that takes the value $\nabla(\lambda)$ at $\lambda$, or is $\nabla(\lambda)$ just a suggestive but indivisible notation? (Similarly for $\Delta(\lambda)$.) \\ TeX note: \operatorname has some often-unexpected preceding space when used as the ‘denominator’ in a quotient. If you want to get rid of it, then you can turn it off with braces: $\nabla(\lambda)/\operatorname{Soc}(\lambda)$ \nabla(\lambda)/\operatorname{Soc}(\lambda) vs. $\nabla(\lambda)/{\operatorname{Soc}(\lambda)}$ \nabla(\lambda)/{\operatorname{Soc}(\lambda)}. $\endgroup$
    – LSpice
    Jan 1 at 11:56
  • $\begingroup$ Well, one can think of $\nabla$ as a function from $\Lambda$ into a subset of all objects of $\mathsf{C}$ (i.e. a subset of $\operatorname{Ob}(\mathsf{C})$ indexed by $\Lambda$). The image of this function then has to satisfy certain properties. I do not know the origin of this notation. I have seen it used in the theory of Verma modules as well. Thanks for the TeX-advice by the way! $\endgroup$
    – Margaret
    Jan 1 at 12:09
  • $\begingroup$ Are you missing something from your definition of costandard/standard objects? As you've written it, you could take $\nabla(\lambda)=L(\lambda)$ for all $\lambda$, in which case every object $N$ would have a costandard filtration, but its dual $D(N)$ might not have a standard filtration if you choose the standard objects to be more interesting. In contexts where I have seen costandard/standard objects there have been extra conditions that make the standard and costandard objects uniquely determined by the poset $\Lambda$. $\endgroup$ Jan 2 at 10:38
  • $\begingroup$ @JeremyRickard I am using Definition 2.6 from the following paper arXiv:2107.07887. The claim I am trying to verify is made in Lemma 3.13. Do you have an explicit example where $D(N)$ would not have a standard filtration? Isn‘t it true that the duality $D$ maps simple objects to simple objects? In our case — since $D$ exchanges standard and costandard objects — we even have $D(L(\lambda))\cong L(\lambda)$ for all $\lambda \in \Lambda$. Consequently, if we let $\nabla (\lambda)=L(\lambda)$, then $\Delta(\lambda)=L(\lambda)$ and every object has a standard filtration. $\endgroup$
    – Margaret
    Jan 2 at 11:53
  • $\begingroup$ OK. With that general a definition then $\Delta(\lambda)$ and $\nabla(\lambda)$ are not uniquely determined up to isomorphism. When you define a costandard filtration to have quotients isomorphic to costandard objects $\nabla(\lambda_i)$, I presume that you mean that you have fixed a choice of a costandard object for each $\lambda$ and the quotients are only allowed to be these particular costandard objects? $\endgroup$ Jan 3 at 10:43

1 Answer 1


Suppose $N$ has a filtration $$0=N_0 \subsetneq N_1 \subsetneq \dots \subsetneq N_{n-1} \subsetneq N_n=N$$ with quotients $Q_1,Q_2,\dots,Q_n$.

Applying any exact contravariant functor $D$, we get a sequence of epimorphisms $$D(N)=D(N_n) \twoheadrightarrow D(N_{n-1}) \twoheadrightarrow\dots\twoheadrightarrow D(N_{1}) \twoheadrightarrow D(N_0)=D(0),$$ from which we get a filtration $$0=D_0 \subsetneq D_1 \subsetneq \dots \subsetneq D_{n-1} \subsetneq D_n=D(N),$$ where $D_i=\ker\left(D(N)\twoheadrightarrow D(N_{n-i})\right)$, with quotients $$D_i/D_{i-1}\cong D(N_{n-i+1}/N_{n-i})\cong D(Q_{n-i+1}).$$


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