Highest Weight Categories

Recall last time we end with the following theorem.

Proposition 1 (Humphreys). For any $\lambda\in\mathbb{X}^{\ast}$, the $G_ {1}T$-module $\hat{Q}(\lambda)$ admits a filtration with subquotients of the form $\hat{Z}(\mu)$ with $\mu\in \mathbb{X}$, and the multiplicity is equal to $[\hat{Z}(\mu):\hat{L}(\lambda)]$.

Today’s goal:

  1. Sketch a completion aof the argument for why $G_ {1}T$-modules and $\hat{Q}(\mu)$ help us connect simple and tilting objects.
  2. Why BGG reciprocity occures? Highest weight categories.

$G$-module structure on injective hulls

Definition 2.

  • We say a $G$-module $V$ is $p$-bounded if for any weight $\mu$ of $V$ and any dominant short root $\alpha$, we have that $$\langle \mu,\alpha^{\vee} \rangle \leq \langle (2p-1)\rho,\alpha^{\vee} \rangle.$$
  • We let $\mathrm{Rep}_ {b}(G)\subseteq \mathrm{Rep}(G)$ denote the Serre subcategory generated by $L(\mu)$ for $\mu$ $p$-bounded.
  • We write $\mathbb{X}_ {b}^{+}\subseteq \mathbb{X}$ the set of $p$-bounded dominant weights.
  • $\mathrm{Rep}_ {b}(G)$ has a highest weight category structure and each block has finitely many simples.

Definition 3. Let $R(\lambda)$ be the injective hull of $L(\lambda)$ in $\mathrm{Rep}_ {b}(G)$.

Theorem 4. Assume that $p\geq 2h-2$. For any $\lambda\in\mathbb{X}_ {\text{res}}^{+}$, we have an isomorphism of $G_ {1}$-modules $R(\lambda)|_ {G_ {1}}\cong Q(\lambda)$. In particular, $Q(\lambda)$ admits a structure of a $G$-module.

Corollary 5. Assume that $p\geq 2h-2$, for any $\lambda\in \mathbb{X}_ {\text{res}}^{+}$, there is an isomorphism of $G_ {1}T$-modules $R(\lambda)|_ {G_ {1}T}\cong \hat{Q}(\lambda)$.

Relation with tilting modules

For any $\lambda\in \mathbb{X}^{+}_ {b}$, since $R(\lambda)$ is injective in the highest weight category $\mathrm{Rep}(G)$, it admits a costandard filtration and satisfies the reciprocity formula $$(R(\lambda):N(\mu)) = [ M(\mu):L(\lambda) ]$$ for any $\mu\in\mathbb{X}_ {b}^{+}$.

Note that $\mathbb{X}_ {b}^{+}\subseteq \mathbb{X}^{+}$ is stable under the operation $\mu\mapsto -w_ {0}\mu$. Therefore, $\mathrm{Rep}_ {b}( G )$ is stable under duality $V\mapsto V^{\ast}$.

From now on, we assume that $p\geq 2h-2$.

Lemma 6. Let $M\in\mathrm{Rep}_ {b( G )}$ and $\mu\in\mathbb{X}^{+}_ {\text{res}}$ and assume that $M|_ {G_ {1}T}\cong \hat{Q}( \mu )$. Then $M\cong R( \mu )$ as $G$-modules.

Proof.

Suppose that $M|_ {G_ {1}T}\cong \hat{Q}( \mu )$. Then $$M|_ {G_ {1}}\cong Q( \mu ),$$ and hence $\text{soc}_ {G_ {1}}( M )$ is simple. Since $M$ is injective as a $G_ {1}$-module, the embedding $L( \mu )\hookrightarrow R( \mu )$ induces a surjection $$\mathrm{Hom}_ {G_ {1}}(R(\mu),M)\twoheadrightarrow \mathrm{Hom}_ {G_ {1}}(L(\mu),M),$$ and hence a surjection $$\mathrm{Hom}_ {G}(R(\mu),M)\twoheadrightarrow \mathrm{Hom}_ {G}(L(\mu),M),$$ since the socle of $M$ as a $G_ {1}T$-module is $\hat{L}( \mu )= L( \mu )|_ {G_ {1}T}$.

Therefore, there is a unique simple sub $G$-module isomorphic to $L( \mu )$. The embedding $L(\mu)\hookrightarrow M$ factors though $R( \mu )\rightarrow M$ since this morphism is injective on the unique simple submodule of $R(\mu)$, it must be injective. Since we understood $M|_ {G_ {1}T}$, we know $\text{dim}(M) =\text{dim}R(\mu)$. This injection is an isomorphism.

Corollary 7. For $\lambda\in\mathbb{X}^{+}_ {\text{res}}$, $R(\lambda)^{\ast}\cong R( -w_ {0}\lambda )$.

Proposition 8. For any $\lambda\in\mathbb{X}^{+}_ {\text{res}}$, the $G$-module $R( \lambda )$ is tilting and is isomorphic to $T( 2( p-1 )\rho+w_ {0}\lambda )$.

Proof.

Fix $\lambda\in \mathbb{X}_ {\text{res}}^{+}$, $R( \lambda )$ has a costandard filtration, but so does $R( \lambda)^{\ast}$, so $R( \lambda )$ must also have a standard filtration. What is its highest weight ?

All the weights $\mu$ of $R( \lambda )$ satisfy $\mu\leq 2( p-1 )\rho + w_ {0}\lambda$. Further, $R( \lambda )|_ {G_ {1}T}\cong \hat{Q}(\lambda)$. The baby Verma module $\hat{Z}( 2( p-1 )\rho+w_ {0}\lambda )$ admits $\hat{L}( -w_ {0}\lambda )^{\ast}\cong \hat{L}( \lambda )$ as a composition factor.

By reciprocity, we deduce that $\hat{Z}( 2( p-1 )\rho+w_ {0\lambda} )$ appears as quotient of $\hat{Q}( \lambda )$, and hence $2( p-1 )+w_ {0}\lambda$ is a $T$-weight of $\hat{Q}( \lambda )$. Therefore, $2( p-1 )+w_ {0}\lambda$ is a $T$-weight of $R( \lambda )$ and it is the highest weight.

Theorem 9 (Donkin’s tensor product theorem). For any $\lambda\in\mathbb{X}_ {\text{res}}^{+}$ and any $\mu\in \mathbb{X}^{\ast}(T^{(1)})$ dominant, the $G$-module $T((p-1)\rho+\lambda)\otimes \mathrm{Fr}^{\ast}_ {G}(T^{(1)}( \mu ))$ is tilting of highest weight $(p-1)\rho+\lambda+\mathrm{Fr}_ {T}^{\ast}(\mu)$. If
$T((p-1)\rho+\lambda)$ is indecomposable as a $G_ {1}$-module, then $T((p-1)\rho+\lambda)\otimes \mathrm{Fr}^{\ast}_ {G}(T^{(1)}( \mu ))\cong T((p-1)\rho+\lambda+\mathrm{Fr}_ {T}^{\ast}(\mu)).$

How do we actually compute simple characters from tilting ones?

Assume that we know the characters of $\text{ch}(T(2(p-1)\rho+w_ {0}\lambda))$ for any $\lambda\in\mathbb{X}_ {\text{res}}^{+}$. Then we know the characters of $\hat{Q}( \lambda )$ for any $\lambda\in\mathbb{X}_ {\text{res}}^{+}$ and hence for any $\lambda\in\mathbb{X}$. Since characters of baby Vermas are easy ([Riche, Excersice 4.8]), our problem is really understanding $[\hat{Z}(\lambda):\hat{L}( \mu )]$ for $\lambda,\mu\in\mathbb{X}$.

Now we claim that if we know all $[\hat{Z}(\lambda): \hat{L}(\lambda)]$, then we know characters of all $\hat{L}(\lambda)$. We can reduce to understanding characters of $\hat{L}(\lambda)$ for $\lambda\in \mathbb{X}^{+}_ {res}$. We want $\text{dim}\hat{L}(\lambda)_ {\mu}$ for $\mu\in\mathbb{X}$. For any $\nu\in\mathbb{X}$, weights of $\hat{Z}(\nu)$ are $\leq \nu$, so $[\hat{Z}(\nu):\hat{L}(\eta)]$ vanishes unless $\eta\leq \nu$ and $=1$ if $\eta=\nu$.

So we get $$\mathrm{ch}(\hat{L}(\lambda))=\sum_ {\nu\in \mathbb{X}_ {\lambda}}m_ {\nu}\text{ch}(\hat{Z}(\nu))+\sum_ {\nu\in\mathbb{X}_ {\lambda}}m_ {\nu}’\text{ch}(L(\nu)),$$ there is $n$ dominant $\mu$ such that $\mu\leq \nu$ for $\nu\in \mathbb{X}_ {\lambda}$. Therefore, $$\text{dim}(\hat{L}(\lambda)_ {\mu})=\sum_ {\nu\in\mathbb{X}_ {\lambda}}m_ {\nu}\text{dim}(\hat{Z}(\nu)_ {\mu}).$$

Highest weight categories essentials

Let

  • $\mathbb{k}$ be a field,
  • $\mathscr{A}$ a finite length $\mathbb{k}$-linear abelian category, such that $\mathrm{Hom}_ {\mathscr{A}}(M,N)$ is finite dimensional
  • Let $S$ be the set of isomorphism classes of irreducible objects in $\mathscr{A}$.
  • Assume that $(S,\leq)$ has a partial ordering.
  • Call $L_ {s}$ irreducible corresponding to $s\in S$.
  • Assume that for any $s$, we have objects $\nabla_ {s}$, $\triangle_ {s}$, such that $\triangle_ {s}\rightarrow L_ {s}$, and $L_ {s}\rightarrow \nabla_ {s}$.
  • For any $J\subseteq S$, we write $\mathscr{A}_ {J}$ for the Serre subcategory generated by $\{L_ {t}\}_ {t\in J}$.
  • For $s\in S$, we have $\mathscr{A}_ {\leq s}$ and $\mathscr{A}_ {<s}$.

Definition 10. We say $\mathscr{A}$ with this data is a highest weight category if

  1. for any $s\in S$, $\{t\in S: t\leq s\}$ is finite,
  2. for any $s\in S$, $\mathrm{Hom}(L_ {s},L_ {s})=\mathbb{k}$,
  3. for any $s\in S$, and any ideal $J\subseteq S$ such that $s\in J$ is max, then $\triangle_ {s}\rightarrow L_ {s}$ is a projective cover and $L_ {s}\rightarrow \nabla_ {s}$ is an injective envelope in $\mathscr{A}_ {J}$.
  4. The kernel of $\triangle_ {s}\rightarrow L_ {s}$ and cokernel of $L_ {s}\rightarrow \nabla_ {s}$ lie in $\mathscr{A}_ {<s}$.
  5. $\mathrm{Ext}^{2}(\triangle_ {s},\nabla_ {t})=0$ for $s,t\in S$.

We call $(S,\leq)$ the weight poset.

In particular, we have that $\triangle_ {s},\nabla_ {s}\in\mathscr{A}_ {\leq s}$ and $[\Delta_ {s}:L_ {s}]=[\nabla_ {s}:L_ {s}]=1$.

Theorem 11. Assume that $S$ is fintie. Then $\mathscr{A}$ has enough projectives and any projective object admits a $\triangle$-filtration. Furthermore, if $P_ {s}$ is the projective cover of $L$_ {s}, we have that $$[P_ {s}:\triangle_ {t}]=[\nabla_ {t}:L_ {s}]$$

Proof.

Both sides are equal to $\text{dim}(P_ {s}, \nabla_ {t})=\sum_ {i}\text{dim}\mathrm{Hom}(\triangle_ {i},\nabla_ {t})$.

Corollary 12. We have that $$ \mathrm{Ext}^{i}(\triangle_ {s},\nabla_ {t}) \begin{cases} \mathbb{k}& \text{if }s=t, i=0; \\ 0 & \text{otherwise} \end{cases} $$

Proposition 13. If $\mathscr{A}$ is a highest weight category, then the followings are equivalent:

  1. $M$ admits a $\nabla$-filtration,
  2. $\mathrm{Ext}^{i}(\triangle_ {s},M)=0$ for $i\in\mathbb{Z}_ {>0}$, for any $s$,
  3. $\mathrm{Ext}^{1}(\triangle_ {s},M)=0$ for any $s$.