Projectives Standard Filtrations and Bgg Reciprocity

Let $\nu$ be a weight of $X\in {\mathcal{O}}_{\mathrm{\Gamma }}$ with $\nu \ge \lambda$. By [F3] the weight $\nu$ occurs in some composition factor $L(\mu )$ of $X$, and $\mu \in \mathrm{\Gamma }$ because $X\in {\mathcal{O}}_{\mathrm{\Gamma }}$. By [F2], $\nu$ is then a weight of $M(\mu )$, so $\nu \le \mu$. From $\lambda \le \nu \le \mu$ and transitivity in [F4] we get $\lambda \le \mu$ with $\mu \in \mathrm{\Gamma }$, so maximality of $\lambda$ gives $\mu =\lambda$; then $\lambda \le \nu \le \lambda$ and antisymmetry give $\nu =\lambda$. Hence no object of ${\mathcal{O}}_{\mathrm{\Gamma }}$ has a weight $\lambda +\beta$ with $\beta \in Q^{+}\setminus \{0\}$.

The weight functor $X\mapsto X_{\lambda }$ is exact on $\mathfrak{h}$-semisimple $\mathfrak{g}$-modules: given a short exact sequence $0\to X^{\mathrm{\prime }}\stackrel{i}{\to }X\stackrel{p}{\to }{X}^{\mathrm{\prime }\mathrm{\prime }}\to 0$, injectivity of $i_{\lambda }$ is immediate, and if $x^{\mathrm{\prime }\mathrm{\prime }}\in X_{\lambda }^{\mathrm{\prime }\mathrm{\prime }}$ lifts to $x\in X$, then writing $x={\sum }_{\nu }{x}_{\nu }$ as a finite sum of weight vectors gives $x^{\mathrm{\prime }\mathrm{\prime }}=p(x)={\sum }_{\nu }p(x_{\nu })$ with $p(x_{\nu })$ of weight $\nu$; by the directness of the weight decomposition of $X^{\mathrm{\prime }\mathrm{\prime }}$ all terms with $\nu \mathrm{\ne }\lambda$ vanish and $x^{\mathrm{\prime }\mathrm{\prime }}=p(x_{\lambda })$, so $p_{\lambda }$ is surjective.

By step 1.1 no object of ${\mathcal{O}}_{\mathrm{\Gamma }}$ has a weight strictly above $\lambda$ in the sense of $\lambda +\beta$ with $\beta \in Q^{+}\setminus \{0\}$: if $v\in X_{\lambda }$ is nonzero and $x\in {\mathfrak{n}}^{+}$ has weight $\alpha >0$, then $xv\in X_{\lambda +\alpha }$ by [F4], and $\lambda +\alpha >\lambda$; if $xv\mathrm{\ne }0$ it would be a weight vector of weight $\lambda +\alpha$, contradicting step 1.1. Hence $xv=0$ for every $x\in {\mathfrak{n}}^{+}$ and $X_{\lambda }^{{\mathfrak{n}}^{+}}=X_{\lambda }$. Together with the exactness of the weight functor in step 1.2 this proves all three assertions.

For $X\in {\mathcal{O}}_{\mathrm{\Gamma }}$ the universal property [F2] identifies ${\mathrm{Hom}}_{{\mathcal{O}}_{\mathrm{\Gamma }}}(M(\lambda ),X)$ with the space of ${\mathfrak{n}}^{+}$-fixed vectors of weight $\lambda$ in $X$, naturally in $X$; by [F1] this space is $X_{\lambda }^{{\mathfrak{n}}^{+}}=X_{\lambda }$.

The functor $X\mapsto X_{\lambda }$ is exact on ${\mathcal{O}}_{\mathrm{\Gamma }}$ by [F1].

Combining steps 1.1 and 1.2, ${\mathrm{Hom}}_{{\mathcal{O}}_{\mathrm{\Gamma }}}(M(\lambda ),-)$ is naturally isomorphic to the exact functor $X\mapsto X_{\lambda }$, hence is exact; by the characterisation [F3] the Verma module $M(\lambda )$ is a projective object of ${\mathcal{O}}_{\mathrm{\Gamma }}$.

By [F2] the simple reflections generate $W$, so choose a reduced expression $w=s_{i_1}\cdots s_{i_k}$ with $k=\mathrm{\ell }(w)$ and put $w_j=s_{i_1}\cdots s_{i_j}$. Since $w_j=w_{j-1}{s}_{i_j}$, the telescoping sum gives $\zeta -w\zeta ={\sum }_{j=1}^k(w_{j-1}\zeta -w_j\zeta )={\sum }_{j=1}^k{w}_{j-1}(\zeta -s_{i_j}\zeta )={\sum }_{j=1}^k⟨\zeta ,{\alpha }_{i_j}^{\vee }⟩w_{j-1}{\alpha }_{i_j}$, using [F1] for the last equality.

Each coefficient is $⟨\zeta ,{\alpha }_{i_j}^{\vee }⟩\in {\mathbb{Z}}_{\ge 0}$ because ${\alpha }_{i_j}$ is simple and $\zeta$ is dominant integral, and each vector $w_{j-1}{\alpha }_{i_j}$ is a positive root: otherwise $\mathrm{\ell }(w_{j-1}{s}_{i_j})=\mathrm{\ell }(w_{j-1})-1$ by [F2], contradicting that the prefix $w_j$ of the reduced word $w$ is reduced and that $w=w_j{s}_{i_{j+1}}\cdots s_{i_k}$ has length $k$. Hence every term of the sum of step 1.1 lies in ${\mathbb{Z}}_{\ge 0}{\mathrm{\Phi }}^{+}\subseteq Q^{+}$, and $\zeta -w\zeta \in Q^{+}$ for every $w\in W$.

Since $\zeta -w\zeta \in Q^{+}$ means $w\zeta \le \zeta$, every Weyl conjugate of $\zeta$ lies below $\zeta$ in the root order, so $\zeta$ is the maximum of its Weyl orbit; and if in addition $\zeta \le w\zeta$, then $\zeta =w\zeta$ by antisymmetry of the partial order [F3].

For the dot-action statement let $\lambda$ be a weight with $\lambda +\rho$ dominant integral, and apply step 2.1 to $\zeta =\lambda +\rho$: then $(\lambda +\rho )-w(\lambda +\rho )\in Q^{+}$, that is, $\lambda -w\cdot \lambda \in Q^{+}$ for every $w\in W$; moreover $⟨\lambda +\rho ,{\alpha }_i^{\vee }⟩\in \mathbb{Z}$ for every simple root ${\alpha }_i$, so every simple reflection lies in $W_{\lambda }$ and $W_{\lambda }=W$ by [F2]; hence every element of the linkage class $W_{\lambda }\cdot \lambda$ is a Weyl conjugate of $\lambda$ and lies below $\lambda$.

Combining steps 3.1 and 3.2: for every $w\in W$ one has $\zeta -w\zeta \in Q^{+}$ and $\lambda -w\cdot \lambda \in Q^{+}$ in the dot setting, so $\zeta$ and $\lambda$ are the maxima of their orbits and linkage classes respectively, and $w\zeta \ge \zeta$ forces $w\zeta =\zeta$.

For every $X\in \mathcal{O}$ the tensor-Hom adjunction of [F1] gives a natural isomorphism ${\mathrm{Hom}}_{\mathcal{O}}(E\otimes P,X)\cong {\mathrm{Hom}}_{\mathcal{O}}(P,E^{\ast }\otimes X)$, and $E^{\ast }$ is finite-dimensional $\mathfrak{h}$-semisimple with $E^{\ast }\otimes -$ an exact endofunctor of $\mathcal{O}$.

If $I$ is injective, evaluation and coevaluation for the ordinary contragredient dual $E^{\ast }$ give ${\mathrm{Hom}}_{\mathcal{O}}(X,E\otimes I)\cong {\mathrm{Hom}}_{\mathcal{O}}(E^{\ast }\otimes X,I)$, naturally in $X$. Since $E^{\ast }\otimes -$ is exact by [F1] and $\mathrm{Hom}(-,I)$ is exact by [F3], their composite is exact. Thus $E\otimes I$ is injective.

Since $E\otimes P\in \mathcal{O}$ by [F1], the functor ${\mathrm{Hom}}_{\mathcal{O}}(E\otimes P,-)$ is naturally isomorphic to the composite of the exact functor $X\mapsto E^{\ast }\otimes X$ and the exact functor ${\mathrm{Hom}}_{\mathcal{O}}(P,-)$ of [F2]; composites of exact functors are exact, so ${\mathrm{Hom}}_{\mathcal{O}}(E\otimes P,-)$ is exact and [F2] makes $E\otimes P$ projective in $\mathcal{O}$.

Steps 2.1 and 1.2 prove that finite-dimensional tensoring preserves both projectives and injectives in $\mathcal{O}$.

By [F1] every object $X$ is ${⨁}_C{X}_C$ with finitely many nonzero terms, the decomposition is functorial, and morphisms between objects supported on disjoint collections of classes vanish. Hence a morphism $f:X\to Y$ between decomposed objects is block diagonal, $f={⨁}_C{f}_C$ with $f_C:X_C\to Y_C$.

By [F2] the projectivity of $P$ in $\mathcal{O}$ says exactly that ${\mathrm{Hom}}_{\mathcal{O}}(P,-)$ is exact on $\mathcal{O}$, and the hypothesis on $Q$ says that ${\mathrm{Hom}}_{{\mathcal{O}}_C}(Q,-)$ is exact on ${\mathcal{O}}_C$.

Apply [F3] to the block-diagonal differentials of step 1.1: a short exact sequence $0\to A\to E\to B\to 0$ in $\mathcal{O}$ decomposes into the short exact sequences $0\to A_C\to E_C\to B_C\to 0$ of its components, and conversely exactness of all components gives exactness of the sequence. Therefore ${\mathrm{pr}}_C$ is an exact functor $\mathcal{O}\to {\mathcal{O}}_C$, and $i_C$ is exact as the inclusion of a full subcategory closed under subobjects and quotients.

For $A\in {\mathcal{O}}_C$ and any $X\in \mathcal{O}$, decomposing $X$ and using the vanishing of morphisms from $A$ into components supported on other classes gives a natural isomorphism ${\mathrm{Hom}}_{\mathcal{O}}(A,X)\cong {\mathrm{Hom}}_{{\mathcal{O}}_C}(A,{\mathrm{pr}}_{C}X)$; here ${\mathrm{Hom}}_{\mathcal{O}}(i_{C}A,X)$ is written ${\mathrm{Hom}}_{\mathcal{O}}(A,X)$. Similarly, for $Y\in {\mathcal{O}}_C$, decomposing $P$ gives ${\mathrm{Hom}}_{\mathcal{O}}(P,{\mathrm{pr}}_{C}Y)\cong {\mathrm{Hom}}_{\mathcal{O}}({\mathrm{pr}}_{C}P,{\mathrm{pr}}_{C}Y)$, because all components of $P$ other than ${\mathrm{pr}}_{C}P$ map to zero into the object ${\mathrm{pr}}_{C}Y$ of ${\mathcal{O}}_C$.

Let $P\in \mathcal{O}$ be projective. Composing the isomorphisms of step 2.2, for every $X\in \mathcal{O}$ there is a natural isomorphism ${\mathrm{Hom}}_{\mathcal{O}}({\mathrm{pr}}_{C}P,X)\cong {\mathrm{Hom}}_{{\mathcal{O}}_C}({\mathrm{pr}}_{C}P,{\mathrm{pr}}_{C}X)\cong {\mathrm{Hom}}_{\mathcal{O}}(P,{\mathrm{pr}}_{C}X)$. Now $X\mapsto {\mathrm{pr}}_{C}X$ is exact by step 2.1 and ${\mathrm{Hom}}_{\mathcal{O}}(P,-)$ is exact by step 1.2, so the composite functor $X\mapsto {\mathrm{Hom}}_{\mathcal{O}}({\mathrm{pr}}_{C}P,X)$ is exact. By [F2] applied in $\mathcal{O}$, ${\mathrm{pr}}_{C}P$ is projective in $\mathcal{O}$.

Conversely let $Q\in {\mathcal{O}}_C$ be projective in ${\mathcal{O}}_C$. By step 2.2, for every $X\in \mathcal{O}$ there is a natural isomorphism ${\mathrm{Hom}}_{\mathcal{O}}(Q,X)\cong {\mathrm{Hom}}_{{\mathcal{O}}_C}(Q,{\mathrm{pr}}_{C}X)$. The first functor is the composite of the exact functor ${\mathrm{pr}}_C$ of step 2.1 with the exact functor ${\mathrm{Hom}}_{{\mathcal{O}}_C}(Q,-)$ of step 1.2, hence is exact; by [F2], $Q$ is projective in $\mathcal{O}$.

Choose the finite bound. Since a simple reflection reverses only ${\alpha }_i$ among the positive roots, their half-sum satisfies $\rho -s_i\rho ={\alpha }_i$, giving the simple coroot pairings one. Thus $\rho$ is regular dominant integral. By [F1], $d_w=\rho -w\rho \in Q^{+}\setminus \{0\}$ for $w\mathrm{\ne }1$; take the first positive coordinate in the fixed finite simple-root enumeration. The finitely many real numbers $-\mathrm{Re}(c_w{)}_{j(w)}\mathrm{/}(d_w{)}_{j(w)}$ have an upper bound, so choose an integer $N\ge 0$ strictly larger than all of them. For rank zero $W=\{1\}$ and there are no restrictions; use $N=0$. With $\lambda =\mu +N\rho$, the identity $\lambda -w\cdot \lambda =c_w+N{d}_w$ has positive real part at the selected coordinate for each $w\mathrm{\ne }1$. Therefore $w\cdot \lambda -\lambda$ cannot have all nonnegative integer simple-root coordinates; if that coordinate is nonreal it is not even in the real root lattice, and if real it is negative. No distinct dot conjugate lies above $\lambda$. By [F3] this makes $\lambda$ maximal in $C_{\lambda }$, without claiming it is a greatest weight.

The auxiliary tensor. The simple pairings $⟨N\rho ,{\alpha }_i^{\vee }⟩=N$ make $N\rho$ dominant integral, so $E=L(N\rho )$ is finite-dimensional and $\mathfrak{h}$-semisimple by [F2]. Since $w_0$ reverses the positive roots, $w_0\rho =-\rho$, and the dual-highest-weight formula gives $E^{\ast }\cong L(-w_{0}N\rho )=E$. This argument concerns $N\rho$ only and imposes no integrality on $\mu$ or $\lambda$.

Projectivity in the actual block. The Verma $M(\lambda )$ is indecomposable: in a direct-sum decomposition, its one-dimensional highest weight space lies in exactly one summand, and its highest vector generates the whole Verma, so all other summands vanish. The block decomposition in [F4] consequently places $M(\lambda )$ entirely in the block containing its simple quotient $L(\lambda )$, namely $C_{\lambda }$. This class is a finite ideal of itself by [F3], and step 1.1 makes $\lambda$ maximal there. Apply [F4] to obtain projectivity in ${\mathcal{O}}_{C_{\lambda }}$; the exact block-projection adjunction of [F4] makes it projective in $\mathcal{O}$. Step 2.1 and [F5] then make $E\otimes M(\lambda )$ projective in $\mathcal{O}$.

The product of a highest-weight vector $e\in E^{\ast }=E$ of weight $N\rho$ and a highest-weight vector $v\in L(\mu )$ of weight $\mu$ is nonzero of weight $N\rho +\mu =\lambda$ and is annihilated by ${\mathfrak{n}}^{+}$; by [F6] it is the image of a nonzero element of ${\mathrm{Hom}}_{\mathcal{O}}(E\otimes M(\lambda ),L(\mu ))$, so that Hom-space is nonzero.

By [F7] there is a natural isomorphism ${\mathrm{Hom}}_{\mathcal{O}}({\mathrm{pr}}_C(E\otimes M(\lambda )),L(\mu ))\cong {\mathrm{Hom}}_{\mathcal{O}}(E\otimes M(\lambda ),L(\mu ))\mathrm{\ne }0$, since $L(\mu )\in {\mathcal{O}}_C$. The object ${\mathrm{pr}}_C(E\otimes M(\lambda ))$ lies in ${\mathcal{O}}_C$ and is projective in $\mathcal{O}$ by step 3.1 and the first part of [F4]; a nonzero morphism from it to the simple object $L(\mu )$ is an epimorphism, so ${\mathcal{O}}_C$ contains a projective object mapping onto $L(\mu )$, as claimed.

A proper inclusion $A⊊B$ of subobjects of a finite-length object has $\mathrm{\ell }(A)<\mathrm{\ell }(B)$, because $B\mathrm{/}A\mathrm{\ne }0$ contributes at least one composition factor; consequently any ascending chain of subobjects stabilizes, and a nonzero object has a proper subobject of maximal length, hence a maximal proper subobject.

Every object $X$ is a finite direct sum of indecomposable objects, by induction on $\mathrm{\ell }(X)$: for $X=0$ take the empty sum; if $X\mathrm{\ne }0$ is indecomposable there is nothing to prove; otherwise $X\cong Y\oplus Z$ with $Y,Z\mathrm{\ne }0$, and $\mathrm{\ell }(Y),\mathrm{\ell }(Z)<\mathrm{\ell }(X)$, so the induction hypothesis applies to $Y$ and $Z$.

For an endomorphism $f:X\to X$, the image and kernel chains stabilize by [F1]. Choose $N$ with $\mathrm{im}{f}^N=\mathrm{im}{f}^{2N}$ and $\ker f^N=\ker f^{2N}$, and put $I=\mathrm{im}{f}^N$, $K=\ker f^N$. The restriction $f^N{\mathrm{\mid }}_I:I\to I$ is epic by image stabilization, hence is an isomorphism: length additivity makes its kernel zero. If $p:X\twoheadrightarrow I$ is the image factorization of $f^N$, then $(f^N{\mathrm{\mid }}_I{)}^{-1}p:X\to I$ retracts the inclusion $I\hookrightarrow X$ and has kernel $K$. The split exact sequence therefore gives $X\cong I\oplus K$.

If $X$ is indecomposable, step 2.1 forces $I=0$ or $K=0$. The first case gives $f^N=0$. In the second case $f^N$ is monic; length additivity makes its cokernel zero, so $f^N$ is an isomorphism. Since $f$ commutes with $f^N$ and its inverse, $f^{N-1}(f^N{)}^{-1}$ is an inverse of $f$. Thus every endomorphism of $X$ is either invertible or nilpotent.

Let $X$ be indecomposable and $f,g\in \mathrm{End}(X)$ with $f+g=u$ an isomorphism. If neither $f$ nor $g$ is an isomorphism, both are nilpotent by step 3.1; then $a:=u^{-1}f$ and $b:=u^{-1}g$ satisfy $a+b={\mathrm{id}}_X$ and are non-units, hence nilpotent, and $a={\mathrm{id}}_X-b$ gives $ab=b-b^2=ba$, so the commuting nilpotents $a,b$ have nilpotent sum: ${\mathrm{id}}_X=a+b$ is nilpotent, forcing ${\mathrm{id}}_X=0$ and $X=0$ by [F2], contrary to indecomposability. Hence $f$ or $g$ is an isomorphism.

More generally, if $f_1+\cdots +f_k={\mathrm{id}}_X$ with $X$ indecomposable, then some $f_i$ is an isomorphism: induct on $k$, the cases $k=1$ and $k=2$ being trivial and step 4.1; for $k\ge 3$ put $g=f_2+\cdots +f_k$, so that $f_1+g={\mathrm{id}}_X$, and if $f_1$ is not an isomorphism then $g$ is an isomorphism by step 4.1, and $g^{-1}{f}_2+\cdots +g^{-1}{f}_k={\mathrm{id}}_X$ has $k-1$ terms, so some $g^{-1}{f}_i$ is an isomorphism by induction and then $f_i=g(g^{-1}{f}_i)$ is an isomorphism. Consequently $\mathrm{End}(X)$ is local: in a ring $R$, locality is equivalent to the criterion that for every $a\in R$ either $a$ or $1-a$ is a unit, and here $a+(1-a)={\mathrm{id}}_X$ is the identity, so the two-term case applies. Also, a nonzero idempotent in a local ring is the identity.

Let $X=X_1\oplus A=Y_1\oplus \cdots \oplus Y_m$ with $X_1,Y_j$ indecomposable, and let $a_j:Y_j\to X_1$, $b_j:X_1\to Y_j$ be the composites of the inclusions and projections. Then ${\sum }_j{a}_j{b}_j={\mathrm{id}}_{X_1}$; by step 5.1 some $a_{j_0}{b}_{j_0}$ is an isomorphism, and after relabelling $j_0=1$ and setting $\beta :=a_1{b}_1$ we obtain $\epsilon :=b_1{\beta }^{-1}{a}_1\in \mathrm{End}(Y_1)$ with ${\epsilon }^2=b_1{\beta }^{-1}(a_1{b}_1){\beta }^{-1}{a}_1=\epsilon$, so $\epsilon$ is an idempotent; it is nonzero because ${\beta }^{-1}{a}_1$ is a left inverse of $b_1$ and $X_1\mathrm{\ne }0$, and its image is $b_1(X_1)$. By the last sentence of step 5.1, $\epsilon ={\mathrm{id}}_{Y_1}$; hence ${\beta }^{-1}{a}_1$ and $b_1$ are mutually inverse isomorphisms $X_1\cong Y_1$.