Question on block decomposition in Category O

Question

Let $V \in \mathcal{O}$. How can we show that if $V \in \mathcal{O}$, then $V$ admits a direct sum decomposition $$V = \bigoplus_{i=1}^{k}V_i$$ such that each $V_i$ belongs to a block while $V_i$ and $V_j$ do not belong to the same block for $i \neq j.$

A proof will be highly appreciated.

Answer 1

Let $Z=Z(\mathfrak{g})$ be the center of $U(\mathfrak{g})$. For a central character $\chi:Z\to \mathbb{C}$, define $$V_\chi=\{v\in V\mid (z-\chi(z))^Nv=0\mbox{ for all }z\in Z\mbox{ and }N>>0\}.$$ Then, $V_\chi$ is a $\mathfrak{g}$-invariant subspace since, for $z\in Z$, $x\in\mathfrak{g}$, and $N$ sufficiently large $$(z-\chi(z))^Nxv=x(z-\chi(z))^Nv=x0=0.$$

Moreover, $V_\chi\cap V_{\zeta}=0$ if $\chi\neq\zeta$. Indeed, choose $z$ such that $\chi(z)\neq \zeta(z)$. Assume $v\in V_\chi\cap V_{\zeta}$ is nonzero, then there exist $M,N>0$ such that \begin{align} (z-\chi(z))^{N-1}v&\neq 0\\ (z-\chi(z))^Nv&=0\\ (z-\zeta(z))^Mv&=0 \end{align} Now, set $v'=(z-\chi(z))^{N-1}v$. Then, $zv'=\chi(z)v'$ and $$(z-\zeta(z))^Mv'=(\chi(z)-\zeta(z))^Mv'\neq 0$$ since $\chi(z)\neq \zeta(z)$. But, \begin{align} (z-\zeta(z))^Mv'&=(z-\zeta(z))^M(z-\chi(z))^{N-1}v\\ &=(z-\chi(z))^{N-1}(z-\zeta(z))^Mv\\ &=0, \end{align} a contradiction. We now conclude that $V=\bigoplus_\chi V_\chi$, where the direct sum is over all central characters. Note that only finitely many summands can be nonzero since $V$ is finitely generated.

Answer 2

If $\mathcal{A}$ is an abelian category, then it has a block decomposition $\mathcal{A}=\prod_i\mathcal{A}_i$, i.e. there exist full subcategories $\mathcal{A}_i$ of $\mathcal{A}$ such that every $A\in\mathcal{A}$ decomposes uniquely (up to isomorphism of course) as $A=\bigoplus_i A_i$, where $A_i\in\mathcal{A}_i$, and that $i\neq j$ implies that $Hom_{\mathcal{A}}(A_i,A_j)=0$. This "clearly" always exists simply by splitting $\mathcal{A}$ step by step until you can't go further (this almost certainly encounters set theoretic problems, but let's not get sidetracked).

The point isn't that $\mathcal{O}$ has "a" block decomposition, it's that decomposition by central characters is the block decomposition of $ \mathcal{O}$. I can't point you to a reference off the top of my head I'm afraid.