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Let $K$ be a field. Let $\mathsf{C}$ be an essentially small abelian category in which every hom-space is a finite-dimensional $K$-vector space.

For an object $X\in \mathsf C$ denote by $\operatorname{Sub}(X)$ the set of subobjects of $X$. Take the posetal reflection of the usual preorder on $\operatorname{Sub}(X)$. One can show that $\operatorname{Sub}(X)$ is a complete lattice. Hence, we can define $$\operatorname{Rad}(X)\ \colon = \bigcap_{M \in \operatorname{Sub}(X) \\ M \text{ is maximal}} M$$ $$\operatorname{Soc}(X)\ \colon =\bigcup_{N \in \operatorname{Sub}(X) \\ N \text{ is minimal}} N$$ $$\operatorname{Hd}(X)\ \colon = X / \operatorname{Rad}(X)$$

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 $\operatorname{id}_{D(X)}=D(\xi_X)\circ \xi_{D(X)}$. I am trying to prove that the following two objects are isomorphic for any object $X \in \mathsf C$: $$D(\operatorname{Soc}(X))\cong \operatorname{Hd}(D(X)).$$

Is this claim even true? In categories of modules with usual duality it holds, but I am struggling to generalise. How do I show from this that for any simple object $S\in \mathsf C$ we have $D(S)\cong S$?

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Groundwork and observations

Subobjects and quotient objects in arbitrary categories

In any object $X$ in a category $\newcommand{\cat}{\mathsf}\cat{C}$ we can not only consider the partially ordered set $\newcommand{\Sub}{\operatorname{Sub}}\Sub(X)$ of subobjects of $X$, but also the partially ordered set $\newcommand{\Quot}{\operatorname{Quot}}\Quot(X)$ of quotient objects of $X$.

In general, we don’t have a correspondence between the two partially ordered sets $\Sub(X)$ and $\Quot(X)$. For example, if $X$ is an object of $\cat{Set}$ with $n$ elements, then $\Sub(X)$ has $2^n$ elements, whereas the number of elements of $\Quot(X)$ is given by the $n$-th Bell number.

Subobjects and quotient objects in abelian categories

However, in an abelian category $\cat{A}$ we do have a one-to-one correspondence between $\Sub(X)$ and $\Quot(X)$: to every subobject $S \to X$ we associate its cokernel $X \to C$, and to every quotient object $X \to Q$ we associate its kernel $K \to X$. These two assignments are mutually inverse because in an abelian category every monomorphism is the kernel of its cokernel and every epimorphism is the cokernel of its kernel.

We get in this way an anti-isomorphism between $\Sub(X)$ and $\Quot(X)$. Let us denote this anti-isomorphism by $Φ_X \colon \Sub(X) \to \Quot(X)$.

Subobjects and quotient objects under dualities

Suppose that $D \colon \cat{C} \to \cat{D}$ is a contravariant equivalence of categories (i.e., a duality). Then $D$ induces isomorphisms (not anti-isomorphisms!) of partially ordered sets $$ \Sub(X) \to \Quot(D(X)) \,, \quad \Quot(X) \to \Sub(D(X)) \,, $$ which we shall again denote by $D$.

Suppose that $D \colon \cat{A} \to \cat{B}$ is a duality between abelian categories. Then $D$ interchanges kernels and cokernels. We thus find that the following diagram commutes: \begin{equation} \require{AMScd} \begin{CD} \Quot(X) @>{D}>> \Sub(D(X)) \\ @V{Φ^{-1}_X}VV @VV{Φ_{D(X)}}V \\ \Sub(X) @>>{D}> \Quot(D(X)) \end{CD} \tag{$\ast$} \end{equation}

Radicals and socles from an order-theoretic perspective

Given a partially ordered set $P$ let us make the following auxiliary definitions:

  • An element of $P$ is proper if it is not the greatest element of $P$. An element of $P$ is maximal if it is maximal among all proper elements of $P$. The radical of $P$ is the infimum of all proper elements of $P$, and it is denoted by $\newcommand{\Rad}{\operatorname{Rad}} \Rad(P)$.

  • An element of $P$ is non-trivial if it is not the least element of $P$. An element of $P$ is minimal if it is minimal among all non-trivial elements of $P$. The socle of $P$ is the supremum of all minimal elements of $P$, and it is denoted by $\newcommand{\Soc}{\operatorname{Soc}} \Soc(X)$.

With these auxiliary definitions, we have in any abelian category $\cat{A}$ the equalities of subobjects of $X$, $$ \Rad(X) = \Rad(\Sub(X)) \,, \quad \Soc(X) = \Soc(\Sub(X)) \,. $$

Radials and socles under (anti-)isomorphisms

An isomorphism of partially ordered sets preserves radicals and socles.

An anti-isomorphism, on the other hand,

  • interchanges the greatest element with the least element;
  • consequently interchanges proper elements with non-trivial elements;
  • consequently interchanges maximal elements with minimal elements;
  • interchanges infima with suprema;
  • consequently, interchanges the socle with the radical.

Simple objects

An object $X$ of a category $\cat{C}$ is called simple if $\Quot(X)$ consists of precisely two objects. In an abelian category, this is equivalent to saying that $\Sub(X)$ has precisely two objects (because of the anti-isomorphism between $\Sub(X)$ and $\Quot(X)$).

Consequences

Let $D \colon \cat{A} \to \cat{B}$ be a duality between abelian categories,

I am trying to prove that the following two objects are isomorphic for any object $\newcommand{\Hd}{\operatorname{Hd}}X ∈ \cat{C}$: $$ D(\Soc(X)) ≅ \Hd(D(X)) \,. $$

Let us now consider the commutative diagram $(\ast)$ of partially ordered sets. The horizontal arrows are isomorphisms, whereas the vertical arrows are anti-isomorphisms. We have therefore the following equalities of quotient objects of $D(X)$: \begin{align*} \Hd(D(X)) &= D(X) / \Rad(D(X)) \\ &= Φ_{D(X)}( \Rad(D(X)) ) \\ &= Φ_{D(X)}( \Rad(\Sub(D(X))) ) \\ &= Φ_{D(X)}( D(\Rad(\Quot(X))) ) \\ &= D( Φ^{-1}_X( \Rad(\Quot(X)) ) ) \\ &= D( \Soc(\Sub(X)) )\\ &= D( \Soc(X) ) \,. \end{align*}

How do I show from this that for any simple object $S ∈ \cat{C}$ we have $D(S) ≅ S$?

An object $S$ of $\cat{A}$ is simple in $\cat{A}$ if and only if $D(S)$ is simple in $\cat{B}$, because of the isomorphism $\Sub(D(X)) ≅ \Quot(D)$. So if $S$ is simple, then $D(S)$ is again simple.

But I don’t think that we can show $S ≅ D(S)$ for $D \colon \cat{A} \to \cat{A}$, even if $ξ \colon D^2 ≅ \mathrm{Id}_{\cat{A}}$ with $\mathrm{id}_{D(X)} = D(ξ_X) ∘ ξ_{D(X)}$.

To see this, consider the group $G = ℤ/n$ for $n ≥ 3$ and let $\cat{A}$ be the category of finite-dimensional complex representations of $G$. The category $\cat{A}$ is essentially small and $ℂ$-linear with finite-dimensional $\mathrm{Hom}$-spaces. (And according to Maschke’s theorem, $\cat{A}$ is semisimple.) The category $\cat{A}$ admits the auto-equivalence $D ≔ (-)^*$ with $D^2 ≅ \mathrm{Id}_{\cat{A}}$. More explicitly, we have the natural isomorphisms $$ ξ_V \colon V \longrightarrow V^{**} $$ given by $$ ξ_V(v)(v^*) = v^*(v) $$ for all $v ∈ V$, $v^* ∈ V^*$. These isomorphisms satisfy the equation $\mathrm{id}_{V^*} = (ξ_V)^* ∘ ξ_{V^*}$ because $$ (ξ_V)^*\Bigl( ξ_{V^*}( v^* ) \Bigr)(v) = \Bigl( ξ_{V^*}( v^* ) ∘ ξ_V \Bigr)(v) = ξ_{V^*}( v^* )( ξ_V(v) ) = ξ_V(v)(v^*) = v^*(v) $$ for all $v ∈ V$, $v^* ∈ V^*$.

We have for every $n$-th root of unity $λ$ a one-dimensional representation $V_λ$ of $G$, on which the cyclic generator $[1]$ of $G$ acts by multiplication with $λ$. These representations are pairwise non-isomorphic, every irreducible complex representations of $G$ is of this form, and we have $(V_λ)^* ≅ V_{\overline{λ}}$ for every $n$-th root of unity $λ$. Therefore, $D(V_λ) ≅ V_λ$ if and only if $λ = \overline{λ}$, which happens only for $λ = 1$ (the trivial representation) and $λ = -1$ (the alternating representation, existing only if $n$ is even.) So, since $n ≥ 2$, we have simple objects $S$ in $\cat{A}$ with $D(S) ≇ S$.

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  • $\begingroup$ Thanks a lot! Is there a way to prove that $D$ exchanges heads and socles via the Freyd-Mitchell embedding theorem? $\endgroup$
    – Margaret
    Commented Dec 31, 2022 at 8:17
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    $\begingroup$ @Margaret Freyd–Mitchell allows us to realize $\mathsf{A}$ as a full exact subcategory of $\mathrm{Mod}(R)$ for some ring $R$. But at this point, $D$ is still just an abstract functor $D \colon \mathsf{A} \to \mathsf{A}$ without any connection to the newly introduced module structures on the objects of $\mathsf{A}$. So I don’t see how Freyd–Mitchell would help us here. (But I also can’t rule out that there is a way.) $\endgroup$ Commented Jan 1, 2023 at 3:08
  • $\begingroup$ I did not see a way either. Your reasoning makes sense — I was simply wondering whether one can generalize from the module-case. $\endgroup$
    – Margaret
    Commented Jan 1, 2023 at 14:58

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