# How do objects with a (co)standard filtration behave under a duality?

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?

• 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)}. Jan 1 at 11:56
• 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! Jan 1 at 12:09
• 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 $\Lambda$. Jan 2 at 10:38
• @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. Jan 2 at 11:53
• 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? Jan 3 at 10:43

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 $$D_i/D_{i-1}\cong D(N_{n-i+1}/N_{n-i})\cong D(Q_{n-i+1}).$$