Projectives Standard Filtrations and Bgg Reciprocity — Examples

By the action of [F1], for $n\ge 0$ the subspace $U={\sum }_{k\ge n+1}\mathbb{C}{v}_k$ is a submodule: it is $h$- and $f$-stable, and $e\cdot v_k=k(n-k+1)v_{k-1}$ lies in $U$ for $k\ge n+2$ while $e\cdot v_{n+1}=0$; the vector $v_{n+1}$ is a highest-weight vector of weight $-n-2$, so [F2] gives a nonzero map $M(-n-2)\to U$, which is surjective because the powers of $f$ on $v_{n+1}$ span $U$ and injective because $M(-n-2)$ is simple by [F3], hence an isomorphism; hence $U\cong L(-n-2)$. The quotient $M(n)\mathrm{/}U$ has basis the images $u_0,\dots ,u_n$ of $v_0,\dots ,v_n$, on which $e\cdot u_j=j(n-j+1)u_{j-1}\mathrm{\ne }0$ for $1\le j\le n$; any nonzero submodule contains some $u_j$, and applying $e^j$ with all factors $j!(n-j+1)\cdots n$ nonzero gives $u_0$, which generates the quotient, so $M(n)\mathrm{/}U$ is simple of highest weight $n$, that is $L(n)$. A splitting of $0\to L(-n-2)\to M(n)\to L(n)\to 0$ would exhibit a submodule of $M(n)$ isomorphic to $L(n)$, necessarily containing a nonzero vector of the weight-$n$ space $\mathbb{C}{v}_0$ and hence, since $v_0$ generates the infinite-dimensional $M(n)$, the whole of $M(n)$; so the sequence is nonsplit.

The dot orbit is $\{n,-n-2\}$ by [F4]. By [F5] and [F6], $\mathrm{\Delta }(n)$ is projective in its block. The Verma module is indecomposable, since its one-dimensional highest line lies in one summand and generates the whole module. Its unique simple quotient is $L(n)$ by step 1.1, so [F7] identifies $\mathrm{\Delta }(n)$ with $P(n)$. Its one-factor flag gives $(P(n):\mathrm{\Delta }(n))=1$ and $(P(n):\mathrm{\Delta }(-n-2))=0$.

By [F7] and [F8], the indecomposable cover $P(-n-2)$ is Verma-filtered with multiplicities $[\mathrm{\Delta }(\mu ):L(-n-2)]$. Step 1.1 gives these multiplicities as one for $\mu =n,-n-2$. No other label contributes: all Verma factors of a module in $C_n$ lie in $C_n$, whose only simple labels are $n,-n-2$, by [F5]. Thus the flag has exactly the factors $\mathrm{\Delta }(n)$ and $\mathrm{\Delta }(-n-2)$, once each.

The bottom flag factor cannot be $\mathrm{\Delta }(-n-2)$, since the resulting quotient $\mathrm{\Delta }(n)$ would give the simple quotient $L(n)$, contradicting the unique head $L(-n-2)$ of $P(-n-2)$. Hence the flag is the exact sequence $0\to \mathrm{\Delta }(n)\to P(-n-2)\to \mathrm{\Delta }(-n-2)\to 0$. It is nonsplit, since a splitting decomposes the cover into two nonzero summands.

The socle of $\mathrm{\Delta }(n)$ is $L(-n-2)$: it contains that simple submodule by step 1.1, and a simple submodule not contained there would map isomorphically to $L(n)$ and split that nonsplit sequence. Likewise, any simple submodule of $P(-n-2)$ outside $\mathrm{\Delta }(n)$ would map isomorphically to its quotient $L(-n-2)$ and split step 3.1. Thus $P(-n-2)$ has socle $L(-n-2)$, head $L(-n-2)$, and middle composition factor $L(n)$, since its three factors come from the flag and step 1.1.

Taking $n=0$ gives $P(0)=\mathrm{\Delta }(0)$ and the nonsplit sequence $0\to \mathrm{\Delta }(0)\to P(-2)\to \mathrm{\Delta }(-2)\to 0$, with head and socle $L(-2)$ and middle factor $L(0)$. The flag entries are $(P(0):\mathrm{\Delta }(0))=1$, $(P(0):\mathrm{\Delta }(-2))=0$, and $(P(-2):\mathrm{\Delta }(0))=(P(-2):\mathrm{\Delta }(-2))=1$; the standard composition entries are $[\mathrm{\Delta }(0):L(0)]=[\mathrm{\Delta }(0):L(-2)]=[\mathrm{\Delta }(-2):L(-2)]=1$ and $[\mathrm{\Delta }(-2):L(0)]=0$. They agree with BGG reciprocity by [F8].

In the order $(0,-2)$ the standard-composition matrix with entries $D_{\lambda \mu }=[\mathrm{\Delta }(\lambda ):L(\mu )]$ has rows indexed by the standards $\lambda =0,-2$ and columns indexed by the simples $\mu =0,-2$; by [F1] its entries are $D_{00}=D_{0,-2}=D_{-2,-2}=1$ and $D_{-2,0}=0$, that is, $D=\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)$.

In the same order the projective-flag matrix with entries $F_{\lambda \mu }=(P(\lambda ):\mathrm{\Delta }(\mu ))$ has, by [F2], $F_{00}=1$, $F_{0,-2}=0$, $F_{-2,0}=1$ and $F_{-2,-2}=1$, that is, $F=\left(\begin{array}{cc}1& 0\\ 1& 1\end{array}\right)$.

By [F3] each entry of $F$ equals the transposed entry of $D$: $F_{\lambda \mu }=[\mathrm{\Delta }(\mu ):L(\lambda )]=D_{\mu \lambda }$, so $F=D^{\mathsf{T}}$; this is the two-by-two instance of BGG reciprocity, as claimed.

By [F7] with $w=1$ and $w=s$, using [F2] to evaluate $s\cdot (-2)=0$ and $s\cdot (-1)=-1$, the translated wall functor satisfies $T_{-2}^{-1}\mathrm{\Delta }(-2)\cong \mathrm{\Delta }(-1)$ and $T_{-2}^{-1}\mathrm{\Delta }(0)\cong \mathrm{\Delta }(s\cdot (-1))=\mathrm{\Delta }(-1)$; by [F1] $T_0^{-1}=T_{-2}^{-1}$, so both standard objects of the regular block are sent to $\mathrm{\Delta }(-1)$.

By [F4] and [F5], $\mathrm{\Delta }(-1)$ is projective in ${\mathcal{O}}_{{\chi }_{-1}}={\mathcal{O}}_{\mathrm{\Gamma }}$; by [F8] its image $Q=T_{-1}^0\mathrm{\Delta }(-1)=T_{-1}^{-2}\mathrm{\Delta }(-1)$ under the left adjoint is projective in the regular block $C$, and by [F7] with $w=1$ it has a finite Verma flag with the two factors $\mathrm{\Delta }(-2)$ and $\mathrm{\Delta }(0)$, each once, so $(Q:\mathrm{\Delta }(0))=(Q:\mathrm{\Delta }(-2))=1$.

By [F8] the functor $T_0^{-1}$ is exact, so applying it to the nonsplit sequence $0\to \mathrm{\Delta }(-2)\to \mathrm{\Delta }(0)\to L(0)\to 0$ of [F6] yields the exact sequence $0\to T_0^{-1}\mathrm{\Delta }(-2)\to T_0^{-1}\mathrm{\Delta }(0)\to T_0^{-1}L(0)\to 0$; the first two terms are the simple $\mathrm{\Delta }(-1)=L(-1)$ by step 1.1 and [F3], and the first arrow is a monomorphism between these nonzero simple objects, hence an isomorphism. Thus the rightmost term is $T_0^{-1}L(0)=0$ and, using [F3], $T_0^{-1}L(-2)=T_0^{-1}\mathrm{\Delta }(-2)\cong \mathrm{\Delta }(-1)=L(-1)$.

By [F9] write $Q=Q_1\oplus \cdots \oplus Q_n$ with each $Q_i$ an indecomposable object; each $Q_i$ is projective because it is a direct summand of the projective $Q$, and its head is a simple object of $\mathcal{O}$, necessarily a composition factor of $Q$ and hence $L(0)$ or $L(-2)$ by [F6]; by [F6] and [F9] an indecomposable projective with head $L(0)$ is isomorphic to $P(0)=\mathrm{\Delta }(0)$ and one with head $L(-2)$ is isomorphic to $P(-2)$, so $Q\cong P(0{)}^a\oplus P(-2{)}^b$ for integers $a,b\ge 0$.

The multiplicities are additive over direct sums by [F10]; with the values of [F6] and step 1.2 this gives $1=(Q:\mathrm{\Delta }(0))=a(P(0):\mathrm{\Delta }(0))+b(P(-2):\mathrm{\Delta }(0))=a+b$ and $1=(Q:\mathrm{\Delta }(-2))=a(P(0):\mathrm{\Delta }(-2))+b(P(-2):\mathrm{\Delta }(-2))=b$, so $a=0$ and $b=1$.

Consequently $Q=T_{-1}^0\mathrm{\Delta }(-1)=T_{-1}^{0}L(-1)\cong P(-2)$, the nonsplit extension $0\to \mathrm{\Delta }(0)\to P(-2)\to \mathrm{\Delta }(-2)\to 0$ of [F6] with the two-step flag $\mathrm{\Delta }(0),\mathrm{\Delta }(-2)$; together with steps 1.1 and 2.1 this shows that translation to the wall sends $\mathrm{\Delta }(0)$ and $\mathrm{\Delta }(-2)=L(-2)$ to the wall standard $\mathrm{\Delta }(-1)=L(-1)$ and annihilates the finite-dimensional simple $L(0)$, while the reverse translation of the wall standard $L(-1)$ is the projective $P(-2)$, whose standard flag does not split.

By [F2] the maximal-label Verma $\mathrm{\Delta }(m)=M(m)$ is projective in ${\mathcal{O}}_{C_m}$ (and in $\mathcal{O}$). By [F1] the other Verma is simple, $\mathrm{\Delta }(-m-2)=M(-m-2)=L(-m-2)$, and the composition factors of $M(m)$ are $L(-m-2)$ and $L(m)$, each with multiplicity one.

By [F3] the cover $P(-m-2)$ is Verma-filtered and BGG reciprocity gives $(P(-m-2):\mathrm{\Delta }(\mu ))=[\mathrm{\Delta }(\mu ):L(-m-2)]$, which is $1$ for $\mu =-m-2$ and $\mu =m$ by [F1] and $0$ for all other $\mu$ because $L(-m-2)$ lies in the block $C_m$ and other linkage classes contribute no composition factors to it. Hence every Verma flag of $P(-m-2)$ has exactly the two factors $\mathrm{\Delta }(m)$ and $\mathrm{\Delta }(-m-2)$, each once.

In such a flag the bottom factor cannot be $\mathrm{\Delta }(-m-2)$: then $P(-m-2)$ would have $\mathrm{\Delta }(m)=M(m)$ as a quotient, and composing with $M(m)\twoheadrightarrow L(m)$ would exhibit $L(m)$ as a simple quotient, contradicting that $P(-m-2)$ has the unique simple quotient $L(-m-2)$ with $m\mathrm{\ne }-m-2$. So there is a subobject $\mathrm{\Delta }(m)\subseteq P(-m-2)$ with quotient $\mathrm{\Delta }(-m-2)$, giving the short exact sequence $0\to \mathrm{\Delta }(m)\to P(-m-2)\to \mathrm{\Delta }(-m-2)\to 0$; it is nonsplit because projective covers are indecomposable by [F3].

The Verma $\mathrm{\Delta }(-m-2)$ is not projective in the block $C_m$: if it were, the epimorphism $P(-m-2)\twoheadrightarrow \mathrm{\Delta }(-m-2)$ would split, contradicting the nonsplitness of step 2.1. Since $\mathrm{\Delta }(m)$ is projective by step 1.1, projectivity indeed depends on the position of the highest weight in the linkage poset, refuting the statement.

By [F1] the projective $P(-m-2)$ has the finite Verma flag $0\subseteq \mathrm{\Delta }(m)\subseteq P(-m-2)$ with quotient $\mathrm{\Delta }(-m-2)$, and the corresponding sequence is nonsplit. If the flag split, then $P(-m-2)\cong \mathrm{\Delta }(m)\oplus \mathrm{\Delta }(-m-2)$.

But $\mathrm{\Delta }(m)=M(m)$ has head $L(m)$, so the direct sum would have $L(m)$ as a simple quotient, in addition to the quotient $L(-m-2)$ from $\mathrm{\Delta }(-m-2)$; a projective cover has a unique simple quotient, its head, which is $L(-m-2)$ by [F1], and $m\mathrm{\ne }-m-2$. Hence no splitting exists.

For $m=0$ this is the module $P(-2)$ with head and socle $L(-2)$ and middle composition factor $L(0)$: the flag $0\subseteq \mathrm{\Delta }(0)\subseteq P(-2)$ with quotient $\mathrm{\Delta }(-2)=L(-2)$ does not split, so the existence of a finite Verma flag for a projective (from [F2]) is strictly weaker than a direct-sum decomposition into standards.

By [F1] the set $\mathrm{\Gamma }=\{-2\}$ is a finite downward-closed ideal of the linkage class $\{-2,0\}$ in which $-2$ is maximal, so [F2] makes $\mathrm{\Delta }(-2)$ projective in ${\mathcal{O}}_{\mathrm{\Gamma }}$.

In the full block $C=\{-2,0\}$, suppose $\mathrm{\Delta }(-2)$ were projective. Since $\mathrm{\Delta }(-2)=L(-2)$ is the head of $P(-2)$ and $P(-2)\twoheadrightarrow \mathrm{\Delta }(-2)$ is an epimorphism onto a projective object, it would split, contradicting the nonsplitness of the sequence in [F3]. Hence $\mathrm{\Delta }(-2)$ is not projective in ${\mathcal{O}}_C$.