# Jordan-Hölder-like statements for modules with $\Delta$-filtrations over a quasihereditary algebra

## Definitions

Let $A$ be an Artin algebra (for instance, take $A$ to be a finite dimensional algebra over some field) and label the isomorphism classes of simple $A$-modules by the elements of a partially ordered set $(\Lambda, \leq)$.

Denote the simple $A$-modules (up to isomorphism) by $L_{\lambda}$, $\lambda \in \Lambda$, and let $P_{\lambda}$ be the projective cover of $L_{\lambda}$.

For every $\lambda \in \Lambda$, denote by $\Delta(\lambda)$ the largest factor module of $P_{\lambda}$ all of whose composition factors are of the form $L_{\mu}$, $\mu \leq \lambda$. Call this module the standard module with weight $\lambda$.

We say that $A$ is quasihereditary with respect to the poset $(\Lambda, \leq)$ if:

1. The poset $(\Lambda, \leq)$ is adapted to $B$: for every finitely generated module $M$ with simple socle $L_{\lambda}$ and with simple top $L_{\mu}$, where $\mu$ and $\lambda$ uncomparable in $(\Lambda, \leq)$, there is $\nu \in \Lambda$, $\nu > \mu$ or $\nu > \lambda$, such that $L_{\nu}$ is a composition factor of $M$,

2. $L_{\lambda}$ has multiplicity one in $\Delta(\lambda)$, for all $\lambda \in \Lambda$;

3. $P_{\lambda}$ has a $\Delta$-filtration (i.e. a filtration whose factors are standard modules), for all $\lambda \in \Lambda$.

Finally, call a module $\Delta$-semisimple if it is a direct sum of standard modules.

## Questions

Let $A$ be quasihereditary and let $M$ be some finitely genereted $A$-module with a $\Delta$-filtration. I can convince myself that the $\Delta$-filtrations of $M$ are essentially unique (à la Jordan-Holder), up to possible permutation of the factors (I could not find this in literature though -- does anyone know a reference?).

Let $N$ be a $\Delta$-semisimple submodule of $M$ such that $M/N$ still has a $\Delta$-filtration -- I have the feeling that $M$ has a unique maximal submodule $N$ with this property. I have not been able to prove this nor found a reference. Does anyone know if this is true? And would this be "real" uniqueness or just uniqueness up to isomorphism?

• Yes, $\Delta$-filtrations are essentially unique (more precisely, the filtration indexed by ideals in the poset $\Lambda$ is unique). I don't know a reference but the proof isn't difficult. I don't understand your second question. What does "with the property as above" mean? – Geordie Williamson Sep 5 '15 at 12:53
• @GeordieWilliamson - Thank you very much for your reply. By "the property above" I mean the property stated in the previous sentence. The second question is my main question. – 3 A's Sep 5 '15 at 13:12
• I just edited the second question. I hope it is more clear now. – 3 A's Sep 5 '15 at 14:10
• A reference for the first question is math.uni-bielefeld.de/~sek/select/BGG.pdf second paragraph, page 2. The standard modules form a basis for the Grothendieck group. – Dag Oskar Madsen Sep 5 '15 at 14:36

## 1 Answer

Let $M$ be a module with a $\Delta$-filtration. Let $N$ be a $\Delta$-semisimple submodule of M such that $M/N$ still has a $\Delta$-filtration -- I prove that there is a unique maximal module with this property (and hope that my justification is not immensely stupid).

This will be a consequence of the following Lemma.

Lemma Let $M_1$ and $M_2$ be submodules of $M$, with $M_i$ isomorphic to a standard module and such that $M/M_i$ has a $\Delta$-filtration, $i=1,2$. Then either $M_1=M_2$ or $M_1 \cap M_2 =0$.

Proof Suppose $M_1 \cap M_2 \neq 0$ and let $M_i \cong \Delta(x_i)$, for some $x_i \in \Phi$, $i=1,2$. Let $\iota_i$ be the inclusion map of $M_1 \cap M_2$ on $M_i$, and let $\tau_i$ be the inclusion of $M_i$ on $M$, $$\iota_i : M_1 \cap M_2 \longrightarrow M_i,$$ $$\tau_i : M_i \longrightarrow M.$$

Consider the (right exact) functor $F:=D\circ \operatorname{Hom}_A(-,T)$, where $D$ is the standard duality and $T$ the characteristic $A$-module. Denote $\operatorname{End}_A(T)^{op}$ by $\mathcal{R}(A)$ (the Ringel dual of $A$) -- the functor $F$ induces an equivalence between $\mathcal{F}(\Delta)$ and $\mathcal{F}(\nabla ')$ (where $\nabla '$ is the set of costandard modules over $\mathcal{R}(A)$).

The functor $F$ maps the monics $\tau_i$ in $\operatorname{mod}A$ to monics $F(\tau_i)$ in $\operatorname{mod} \mathcal{R}(A)$ -- this is because the cokernel of $\tau_i$ is in $\mathcal{F}(\Delta)$. I claim that the functor $F$ maps the monics $\iota_i$ in $\operatorname{mod}A$ to nonzero morphisms $F(\iota_i)$ in $\operatorname{mod} \mathcal{R}(A)$. Let $\varepsilon$ be the projective cover of $M_1 \cap M_2$. The morphisms $\varepsilon \circ \iota_i$ are nonzero, with domain and codomain in $\mathcal{F}(\Delta)$, thus $F(\varepsilon) \circ F(\iota_i)$ is nonzero, and hence $F(\iota_i) \neq 0$.

Now the image of $F(\iota_i)$ is contained in a module isomorphic to $\nabla '(x_i)$. Because $F(\tau_i)$ is monic, the image of $F(\tau_i) \circ F(\iota_i)$ has simple socle $L_{x_i}$. As $$F(\tau_1) \circ F(\iota_1)=F(\tau_1 \circ \iota_1)=F(\tau_2 \circ \iota_2) =F(\tau_2) \circ F(\iota_2),$$ we must have $x_1=x_2=:x$.

We want to show that $M_1$ and $M_2$ coincide. Suppose (by contradiction) that $M_1\neq M_2$. Then the (simple) top $L_x$ of $M_1$ is a composition factor of $M/ M_2$. Indeed, there must be some subquotient of $M/ M_2$, say $M_3$, satisfying: $M_3$ is part of a $\Delta$-filtration of $M/M_2$; $M_3$ is isomorphic to $\Delta(y)$ (for some $y \in \Phi$); $M_3$ contains the top of $M_1$ as a composition factor. Note that we must have $y \geq x$ in $\Phi$. In fact, by the definition of standard module and by the structure of the module $M$, the module $M_1$ (which is isomorphic to $\Delta(x)$) would have to be contained in $M_3\cong \Delta(y)$. On the other hand, we are assuming that $M_3$ is a subquotient of $M/ M_2$ and that $M_1\cap M_2\neq 0$.