## Translation Functors and the Shapovalov Determinant

Last updated: 10 February 2015

This is an excerpt from the PhD thesis Translation Functors and the Shapovalov Determinant by Emilie Wiesner, University of Wisconsin-Madison, 2005.

## Lie Algebras with Triangular Decomposition

Lie algebras with triangular decomposition generalize important properties of the decomposition of semisimple Lie algebras given in Equation 1.1. Our study of the Virasoro algebra (Chapter 3) relies on its Hermitian triangular decomposition.

The material in this chapter through Section 2.4 can be found in [MPi1995].

Let $𝔤$ be a Lie algebra over $ℂ\text{.}$ Then $𝔤$ has a triangular decomposition if

 • $𝔤={𝔤}_{<0}\oplus 𝔥\oplus {𝔤}_{>0},$ where $𝔤$ is abelian and ${𝔤}_{>0},$ ${𝔤}_{<0}\ne 0\text{;}$ • ${𝔤}_{>0}=\underset{\alpha \in {Q}_{+}}{⨁}{𝔤}_{\alpha }$ where ${Q}_{+}\subseteq {𝔥}^{*}={\text{Hom}}_{ℂ}\left(𝔥,ℂ\right)$ is a free additive semigroup and $𝔤α= { x∈𝔤 | [h,x]=α (h)x for all h∈𝔥 } ;$ • there is a basis $\left\{{\alpha }_{i} | i\in I\right\}$ of ${Q}_{+}$ of linearly independent elements of ${𝔥}^{*},$ i.e. $\gamma \in {Q}_{+}$ can be written $\gamma =\sum _{i\in I}{m}_{i}{\alpha }_{i}$ for ${m}_{i}\in {ℤ}_{\ge 0}\text{;}$ • there is a linear anti-involution $\sigma :𝔤\to 𝔤$ such that $\sigma \left({𝔤}_{>0}\right)={𝔤}_{<0}$ and $\sigma {|}_{𝔥}=\text{id}\text{.}$
We also wish to consider Lie algebras with maps $\sigma$ which are Hermitian (see section 1.3.1). Therefore, we say a Lie algebra $𝔤$ has an Hermitian triangular decomposition if there is a Hermitian anti-involution $\sigma :𝔤\to 𝔤$ and a Lie algebra $\stackrel{‾}{𝔤}$ over the real numbers $ℝ$ such that
 • $𝔤=\stackrel{‾}{𝔤}{\otimes }_{ℝ}ℂ$ • $\stackrel{‾}{𝔤}$ is a Lie algebra with triangular decomposition with anti-involution $\sigma {|}_{\stackrel{‾}{𝔤}}\text{.}$
In general, the results in this chapter for Lie algebras with a triangular decomposition hold for Lie algebras with an Hermitian triangular decomposition. The exception is in defining a contravariant form (section 2.4). Therefore, we will only distinguish between the two types of Lie algebras in that section.

Suppose $𝔤$ is a Lie algebra with a triangular decomposition. If $x\in {𝔤}_{\alpha }$ for some $\alpha \in {Q}_{+},$ then $\sigma \left(x\right)\in {𝔤}_{-\alpha }$ since $[h,σ(a)]= [σ(h),σ(x)]= σ([x,h])= σ(-α(h)x)=- α(h)σ(x).$ Because $\sigma \left({𝔤}_{>0}\right)={𝔤}_{<0},$ we have ${𝔤}_{<0}=\underset{\alpha \in {Q}_{+}}{⨁}{𝔤}_{-\alpha }\text{.}$

The roots of $𝔤$ are those $0\ne \alpha \in {𝔥}^{*}$ such that ${𝔤}_{\alpha }\ne 0\text{.}$ An element $0\ne x\in 𝔤$ is a root vector if $x\in {𝔤}_{\alpha }$ for some root $\alpha \text{.}$ Then ${𝔤}_{<0}$ has a basis of root vectors. Fix an index set $R$ for a basis of root vectors $\left\{{x}_{j} | j\in R\right\}$ of ${𝔤}_{<0}$ and a total order on $R\text{.}$ Let ${\beta }_{j}\in {Q}_{+}$ so that ${x}_{j}\in {𝔤}_{-{\beta }_{j}}\text{.}$

For a partition $\left(t\right):{t}_{1}\ge \cdots \ge {t}_{j}>0,$ write $x(t)=xtj⋯ xt1.$ Then Proposition 1.1.1 can be restated for $𝔤$ as follows.

(See Proposition 1.1.1). Let $𝔤$ be a Lie algebra with a triangular decomposition as described above. Then $U\left(𝔤\right)\cong U\left({𝔤}_{<0}\right)U\left(𝔥\right)U\left({𝔤}_{>0}\right),$ where

 • $U\left(𝔥\right)\cong S\left(𝔥\right),$ the symmetric algebra of $𝔥\text{;}$ • $U\left({𝔤}_{<0}\right)$ has a basis $\left\{{x}_{\left(t\right)} | \left(t\right):{t}_{1}\ge \cdots \ge {t}_{j}>0\right\}\text{;}$ • $U\left({𝔤}_{>0}\right)$ has a basis $\left\{\sigma \left({x}_{\left(t\right)}\right) | \left(t\right):{t}_{1}\ge \cdots \ge {t}_{j}>0\right\}\text{.}$

Let $Q=ℤ\text{-span}\left\{{\alpha }_{i} | i\in I\right\}\subseteq {𝔥}^{*}\text{.}$ Define the height function $\text{ht}:Q\to ℤ$ by $ht(∑i∈Imiαi) =∑i∈Imi.$

A triangular decomposition is regular if the weight spaces of ${𝔤}_{>0}$ are finite-dimensional. It is finite if the root lattice ${Q}_{+}$ is finitely generated. All highest weight modules will be in Category $𝒪$ only if $𝔤$ has a regular triangular decomposition. The proof that Category $𝒪$ decomposes into blocks will rely on $𝔤$ having a finite triangular decomposition.

### The Category $𝒪$

Let $𝔤$ be a Lie algebra with a regular triangular decomposition. Let $M$ be a $𝔤\text{-module}$ and $\mu \in {𝔥}^{*}\text{.}$ The $\mu \text{-weight}$ space of $M$ is $Mμ= {m∈M | hm=μ(h)m for all h∈𝔥}.$ If ${M}^{\mu }\ne 0,$ then $\mu$ is a weight of $M\text{.}$ The dominance order is the partial ordering on the set of weights ${𝔥}^{*}$ given by $λ≥μifλ-μ ∈Q+for λ,μ∈ 𝔥*.$ The Category $𝒪$ is the set of $𝔤\text{-modules}$ $M$ such that

 • $M$ is $𝔥\text{-diagonalizable:}$ $M=\underset{\mu \in {𝔥}^{*}}{⨁}{M}^{\mu }\text{;}$ • there exist ${\lambda }_{1},\dots ,{\lambda }_{n}\in {𝔥}^{*}$ such that ${M}^{\mu }\ne 0$ only if $\mu \le {\lambda }_{i}$ for some $1\le i\le n\text{;}$ • for all $\mu \in {𝔥}^{*},$ $\text{dim} {M}^{\mu }<\infty \text{.}$

([MPi1995] 2.6.2). Let $M$ and $N$ be modules in Category $𝒪\text{.}$ Then $M\oplus N$ and $M\otimes N$ are in Category $𝒪\text{.}$

Within Category $𝒪,$ we pay special attention to highest weight modules. A $𝔤\text{-module}$ $M$ is a highest weight module of highest weight $\lambda \in {𝔥}^{*}$ if there is $v\in {M}^{\lambda }$ such that $v$ generates $M,$ and ${𝔤}_{>0}v=0\text{.}$ Such a vector is a highest weight vector. If $M$ is a highest weight module of highest weight $\lambda ,$ then $M=\underset{\mu \le \lambda }{⨁}{M}^{\mu }\text{.}$

Among highest weight modules, Verma modules are of particular interest. Let $U\left(𝔤\right)$ be the universal enveloping algebra of $𝔤\text{.}$ The Verma module of weight $\lambda \in {𝔥}^{*}$ is the $𝔤\text{-module}$ $M(λ)=U(𝔤)/I$ where $I$ is the left $U\left(𝔤\right)\text{-ideal}$ generated by ${𝔤}_{>0}$ and $\left\{h-\lambda \left(h\right) | h\in 𝔥\right\}\text{.}$ Equivalently, we can define the Verma module $M\left(\lambda \right)$ as the induced module $M(λ)=U(𝔤) ⊗U(𝔤>0⊕𝔥) ℂλ,$ where ${ℂ}_{\lambda }$ is the $\left({𝔤}_{>0}\oplus 𝔤\right)\text{-module}$ $ℂ$ with action $xC=0$ and $hC=\lambda \left(h\right)C$ for $C\in ℂ,$ $x\in {𝔤}_{>0},$ and $h\in 𝔥\text{.}$ We will write ${v}^{+}$ for a choice of a highest weight vector of $M\left(\lambda \right)\text{.}$ (This choice is unique up to scalars.)

Let $\lambda \in {𝔥}^{*}\text{.}$ Then $M\left(\lambda \right)\in 𝒪,$ and $M\left(\lambda \right)=\underset{\mu \in {Q}_{+}}{⨁}M{\left(\lambda \right)}^{\lambda -\mu }\text{.}$ The weight space $M{\left(\lambda \right)}^{\lambda -\mu }$ has a basis $\left\{{x}_{\left(t\right)}{v}^{+} | \left(t\right):{t}_{1}\ge \cdots \ge {t}_{j}>0,\mu ={\beta }_{{t}_{1}}+\cdots +{\beta }_{{t}_{j}}\right\}\text{.}$ Proof. As $U\left({𝔤}_{<0}\right)\text{-modules,}$ $M\left(\lambda \right)\cong U\left({𝔤}_{<0}\right)\text{.}$ Therefore, Proposition 2.0.5 implies that $M\left(\lambda \right)=\underset{\mu \le \lambda }{⨁}M{\left(\lambda \right)}^{\mu }$ and $M{\left(\lambda \right)}^{\mu }$ has a basis $\left\{{x}_{\left(t\right)}{v}^{+} | \left(t\right):{t}_{1}\ge \cdots \ge {t}_{j}>0,\lambda -\mu ={\beta }_{{t}_{1}}+\cdots +{\beta }_{{t}_{j}}\right\}\text{.}$ Since $𝔤$ has a regular triangular decomposition, $\text{dim} M{\left(\lambda \right)}^{\mu }<\infty \text{.}$ Therefore, $M\left(\lambda \right)\in 𝒪\text{.}$ $\square$

Since Verma modules are free $U\left({𝔤}_{<0}\right)\text{-modules,}$ every highest weight module is the quotient of a Verma module. (If $M$ is a highest weight module generated by $v\in {M}^{\lambda },$ we can define a surjective homomorphism $M\left(\lambda \right)\to M$ by ${v}^{+}↦v\text{.)}$ Therefore, the previous result implies that all highest weight modules are in $𝒪\text{.}$

Let $M\in 𝒪$ be a finitely-generated $𝔤\text{-module.}$ Then $M$ has a maximal proper submodule. In particular, for $\lambda \in {𝔥}^{*},$ $M\left(\lambda \right)$ has a unique maximal proper submodule $J\left(\lambda \right)\text{.}$ Proof. Since every element of $M$ is a finite sum of weight vectors, $M$ is generated by a finite sum of weight vectors. Let $\left\{{m}_{1},\dots ,{m}_{n}\right\}$ be a minimal generating set of weight vectors for $M\text{.}$ Assume ${m}_{i}\in {M}^{{\gamma }_{i}}\text{.}$ Consider the set of submodules $𝒦={K⊆M | mn∉K}.$ This set is nonempty (since $\left\{0\right\}\in 𝒦\text{)}$ and partially ordered by inclusion. We claim that this set has a maximal element. To prove this claim, let ${K}_{1}\subset {K}_{2}\subset \cdots$ be an increasing chain of submodules in $𝒦,$ and define $K=\bigcup _{i}{K}_{i}\text{.}$ Then $K$ is a submodule of $M\text{.}$ Since $\text{dim} {M}^{{\gamma }_{n}}<\infty ,$ there is some $k\in {ℤ}_{>0}$ so that ${K}_{i}^{{\gamma }_{n}}={K}_{i+1}^{{\gamma }_{n}}$ for all $i\ge k\text{.}$ Since ${m}_{n}\notin {K}_{i}$ for all $i\in {ℤ}_{>0},$ this implies ${m}_{n}\notin K\text{.}$ Therefore, $K\in 𝒦,$ and so every increasing chain in $𝒦$ has a maximal element. Zorn's Lemma implies that $𝒦$ has a maximal element $\stackrel{‾}{K}\text{.}$ By construction, any submodule $M\prime$ of $M$ which properly contains $\stackrel{‾}{K}$ is not in $𝒦$ and so must contain ${m}_{n}\text{.}$ Because we have chosen the generating set $\left\{{m}_{1},\dots ,{m}_{n}\right\}$ to be minimal, the submodule generated by ${m}_{i}$ $\text{(}i\ne n\text{)}$ is contained in $𝒦,$ and so ${m}_{i}\in \stackrel{‾}{K}\subset M\prime$ for $1\le i\le n-1\text{.}$ This implies $M\prime =M\text{.}$ Therefore, if $\stackrel{‾}{K}$ is a maximal proper submodule. Any maximal proper submodule of $M$ must not contain at least one of the generators of $M\text{.}$ Since $M\left(\lambda \right)$ is generated by one element, this implies $M\left(\lambda \right)$ has a unique maximal proper submodule. $\square$

This result implies M(X) has a unique simple quotient $L(λ)=M(λ)/ J(λ).$

Up to isomorphism, the simple modules in Category $𝒪$ are $L\left(\lambda \right),$ $\lambda \in {𝔥}^{*}\text{.}$ Proof. Let $L\in 𝒪$ be a simple module. Since $L\in 𝒪,$ we can choose $\lambda \in {𝔥}^{*}$ so that ${L}^{\lambda }\ne 0$ and ${L}^{\mu }=0$ for all $\mu >\lambda \text{.}$ Let $0\ne v\in {L}^{\lambda }\text{.}$ Then $v$ is a highest weight vector, and so there is a map $\varphi :M\left(\lambda \right)\to L$ given by $\varphi :{v}^{+}↦v\text{.}$ Since $L$ is irreducible, $\text{ker}\left(\varphi \right)=J\left(\lambda \right)\text{.}$ To show that the simple modules $L\left(\lambda \right)$ are distinct, suppose $\lambda ,\mu \in {𝔥}^{*}$ with $\lambda \ne \mu \text{.}$ We may assume $\lambda \nleqq \mu \text{.}$ Therefore, $L{\left(\mu \right)}^{\lambda }=0$ and so $L\left(\mu \right)\ne L\left(\lambda \right)\text{.}$ $\square$

The following result is an analogue of the Krull-Schmidt theorem for groups (see [Isa1994], 10.9).

Let $P\in 𝒪$ be finitely generated. Then $P$ is a finite direct sum of indecomposable submodules which are also finitely generated. Proof. Let ${\left\{{v}_{i}\right\}}_{i=1}^{n}$ be a generating set for $P\text{.}$ Since $P\in 𝒪,$ every element of $P$ can be written as a finite sum of weight vectors. Therefore, we assume ${v}_{i}\in {P}^{{\nu }_{i}}$ for some ${\nu }_{i}\in {𝔥}^{*}\text{.}$ Suppose $P$ can be written as a finite direct sum of nonzero submodules, $P={P}_{1}\oplus \cdots \oplus {P}_{m}\text{.}$ Write ${v}_{i}={v}_{i}^{1}+\cdots +{v}_{i}^{m}$ where ${v}_{i}^{j}\in {P}_{j}\text{.}$ Then ${\left\{{v}_{i}^{j}\right\}}_{i=1}^{n}$ generates ${P}_{j}\text{.}$ This implies that for each $j$ there is at least one $1\le i\le n$ such that ${v}_{i}^{j}\ne 0\text{.}$ Therefore, $m≤∑i,jdim Pjνi=∑i dim Pνi<∞$ since $P\in 𝒪\text{.}$ Therefore, we can continue to decompose $P$ until the summands ${P}_{i}$ are indecomposable; since $m$ is bounded, the number of terms in this decomposition will be finite. $\square$

#### Characters

Let $ℤ\left[{𝔥}^{*}\right]=\left\{{a}_{1}{e}^{{\lambda }_{1}}+\cdots +{a}_{n}{e}^{{\lambda }_{n}} | {a}_{i}\in ℤ,{\lambda }_{i}\in {𝔥}^{*}\right\}$ be the group algebra of ${𝔥}^{*},$ and let $ℤ{\left[{𝔥}^{*}\right]}^{*}=\left\{\sum {a}_{\lambda }{e}^{\lambda } | \lambda \in {𝔥}^{*},{a}_{\lambda }\in ℤ\right\}$ be the set of formal infinite sums. For $M\in 𝒪,$ the character of $M,$ $\text{ch} M\in ℤ{\left[{𝔥}^{*}\right]}^{*},$ is $ch M=∑λ∈𝔥* dim Mλeλ. (2.1)$ For $M,N\in 𝒪,$ the product $\left(\text{ch} M\right)\left(\text{ch} N\right)$ is well-defined and $ch(M⊗N)= (ch M) (ch N).$

### Special Filtrations in Category $𝒪$

As we noted above, highest weight modules, and specifically Verma modules and irreducible highest weight modules, are special modules in Category $𝒪\text{.}$ Therefore, we would like to analyze other modules in Category $𝒪$ by means of these modules. The following filtrations provide a way to do this. The concepts of local and Verma composition series lead us to define the decomposition numbers $\left[M:L\left(\lambda \right)\right]$ and $\left[M:M\left(\lambda \right)\right]\text{.}$ Highest weight series provide an important tool in proving the decomposition of Category $𝒪$ into blocks (Theorem 2.3.12).

#### Local Composition Series

Let $M\in 𝒪\text{.}$ A composition series for $M$ is a finite sequence of submodules $M={M}_{n}\supseteq {M}_{n-1}\supseteq \cdots \supseteq {M}_{0}=0$ such that each factor ${M}_{i}/{M}_{i-1},$ $1\le i\le n,$ is irreducible. Not all modules in Category $𝒪$ have a composition series. Therefore, we use a more localized idea of a composition series. Let $M\in 𝒪$ and $\lambda \in {𝔥}^{*}\text{.}$ A local composition series for $M$ at $\lambda$ is a finite sequence of submodules $M={M}_{n}\supseteq {M}_{n-1}\supseteq \cdots \supseteq {M}_{0}=0$ and a subset $I\subseteq \left\{1,\dots ,n\right\}$ such that

 • for each $i\in I,$ there is ${\lambda }_{i}\ge \lambda$ so that ${M}_{i}/{M}_{i-1}\cong L\left({\lambda }_{i}\right)\text{;}$ • for $i\notin I,$ ${\left({M}_{i}/{M}_{i-1}\right)}^{\mu }=0$ for all $\mu \ge \lambda \text{.}$

Example. Consider the Verma module $M\left(0,1\right)$ for the Virasoro algebra. As we see in Chapter 3, this module has a unique chain of nested distinct submodules, which is infinite: $M(0,1)⊇ M(1,1)⊇⋯⊇ M(n2,1)⊇⋯.$ Therefore, $M\left(0,1\right)$ does not have a composition series. However, $M\left(0,1\right)$ does have a local composition series at $\lambda$ for any $\lambda \in {𝔥}^{*}\cong {ℂ}^{2}\text{.}$ For example,

 • For $\lambda =\left(h,c\right)$ with $c\ne 1$ or $h\le 0,$ the series $M\left(0,1\right)\supseteq 0$ is a local composition series at $\lambda \text{.}$ • For $\lambda =\left(h,1\right)$ with $h>0,$ the series $M\left(0,1\right)\supseteq M\left(1,1\right)\supseteq \cdots \supseteq M\left({n}^{2},1\right),$ where $n\in {ℤ}_{\ge 0}$ so that ${n}^{2}\ge h,$ is a local composition series at $\lambda \text{.}$

Let $M\in 𝒪$ and $\lambda \in {𝔥}^{*}\text{.}$ Then $M$ has a local composition series at $\lambda \text{.}$ Proof. We prove this claim by inducting on $\sum _{\mu \ge \lambda }\text{dim} {M}^{\mu }\text{.}$ (This sum is always finite since $\text{dim} {M}^{\mu }<\infty$ and there are a finite number of weights between any two weights.) If $\sum _{\mu \ge \lambda }\text{dim} {M}^{\mu }=0,$ then $M\supseteq 0$ is a local composition series at $\lambda \text{.}$ Suppose $\sum _{\mu \ge \lambda }\text{dim} {M}^{\mu }=n,$ and assume the result holds for all modules $M\prime \in 𝒪$ with $\sum _{\mu \ge \lambda }\text{dim}{\left(M\prime \right)}^{\mu } Choose $\mu \in {𝔥}^{*}$ maximal so that ${M}^{\mu }\ne 0$ and $\mu \ge \lambda \text{.}$ Let $0\ne v\in {M}^{\mu }\text{.}$ By the maximality of $\mu ,$ $v$ must be a highest weight vector. Let $V=U\left(𝔤\right)v$ and let $U$ be the maximal submodule of $V\text{.}$ Then $V/U\cong L\left(\mu \right)\text{.}$ We have $\sum _{\mu \ge \lambda }\text{dim}{\left(M/V\right)}^{\mu } and $\sum _{\mu \ge \lambda }\text{dim} {U}^{\mu }<\sum _{\mu \ge \lambda }\text{dim} {V}^{\mu }\le n\text{.}$ Therefore, by the inductive hypothesis, both $M/V$ and $U$ have local composition series at $\lambda \text{:}$ $M/V=Vk⊇⋯⊇V1 ⊇V0=0$ $U=Ul⊇⋯⊇U1⊇ U0=0.$ For each $0\le i\le k,$ let $V\subseteq {M}_{i}\subseteq M$ be a submodule of $M$ so that ${M}_{i}/V\cong {V}_{i}\text{.}$ Then, $M=Mk⊇⋯⊇M0= V⊇U=Ul⊇⋯⊇ U0=0$ is a local composition series for $M$ at $\lambda \text{.}$ $\square$

([MPi1995] 2.6.9). Let $M\in 𝒪$ and $\lambda ,\mu \in {𝔥}^{*}\text{.}$ Suppose that $M$ has a local composition series at $\lambda ,$ $M={M}_{k}\supseteq \cdots \supseteq {M}_{0},$ and at $\mu ,$ $M={\stackrel{\sim }{M}}_{l}\supseteq \cdots \supseteq {\stackrel{\sim }{M}}_{0}\text{.}$ Then there exist series $M={M}_{k}^{\prime }\supseteq \cdots \supseteq {M}_{0}^{\prime },$ a refinement of $\left\{{M}_{i}\right\},$ and $M={\stackrel{\sim }{M}}_{l}^{\prime }\supseteq \cdots \supseteq {\stackrel{\sim }{M}}_{0}^{\prime },$ a refinement of $\left\{{\stackrel{\sim }{M}}_{j}\right\},$ such that $k=l$ and the two series have isomorphic factors.

For $M\in 𝒪$ and $\mu \in {𝔥}^{*},$ define $[M:L(μ)]$ to be the number of occurrences of $L\left(\mu \right)$ in a local composition series for $M$ at $\mu \text{.}$

#### Verma Composition Series

Let $M\in 𝒪\text{.}$ A Verma composition series for $M$ is a finite sequence of submodules $M={M}_{n}\supseteq {M}_{n-1}\supseteq \cdots \supseteq {M}_{0}=0$ such that each factor ${M}_{i}/{M}_{i-1},$ $1\le i\le n,$ is a Verma module.

Let $M\in 𝒪\text{.}$ Then any two Verma composition series for $M$ have isomorphic factors. Proof. Let $M = Mn⊇Mn-1⊇ ⋯⊇M0=0 M = Mm′⊇ Mm-1′⊇ ⋯⊇ M0′=0$ be Verma composition series for $M$ with ${M}_{i}/{M}_{i-1}\cong M\left({\lambda }_{i}\right),$ ${\lambda }_{i}\in {𝔥}^{*}$ and ${M}_{i}^{\prime }/{M}_{i-1}^{\prime }\cong M\left({\mu }_{j}\right),$ ${\mu }_{j}\in {𝔥}^{*}\text{.}$ Then, $∑i=1n ch M(λi)= ch M= ∑j=1m ch M(μj). (2.2)$ Choose ${\lambda }_{k}$ maximal in $\left\{{\lambda }_{1},\dots ,{\lambda }_{m}\right\}\text{.}$ This implies the coefficient of ${e}^{{\lambda }_{k}}$ in (2.2) is $1$ and so ${\mu }_{l}={\lambda }_{i}$ for some $l\text{.}$ Therefore, $\sum _{i\ne k}\text{ch} M\left({\lambda }_{i}\right)=\sum _{j\ne l}\text{ch} M\left({\mu }_{j}\right)\text{.}$ Induct on $m$ to get the result. $\square$

Suppose $M\in 𝒪$ has a Verma composition series. For $\mu \in {𝔥}^{*},$ define $[M:M(μ)]$ to be the number of occurrences of $M\left(\mu \right)$ in a Verma composition series of $M\text{.}$

#### Highest Weight Series

Let $M\in 𝒪\text{.}$ Then a highest weight series for $M$ is an increasing filtration of submodules $0={M}^{0}\subseteq {M}^{1}\subseteq \cdots$ such that

 • $M=\bigcup _{i=0}^{\infty }{M}^{i}\text{;}$ • ${M}^{i}/{M}^{i-1}$ is a highest weight module.
Recall that a triangular decomposition for $𝔤$ is finite if ${Q}_{+}$ is finitely generated.

Let $𝔤$ be a Lie algebra with regular finite triangular decomposition. Then all modules in Category $𝒪$ have a highest weight series. Proof. Let $M\in 𝒪\text{.}$ Choose ${\lambda }_{1},\dots ,{\lambda }_{n}\in {𝔥}^{*}$ so that ${M}^{\mu }\ne 0$ only for $\mu \le {\lambda }_{i}\text{.}$ We may assume that any weight $\mu$ of $M$ is comparable to only one ${\lambda }_{i}\text{.}$ (If $\mu \le {\lambda }_{i}$ and $\mu \le {\lambda }_{j},$ then we can replace ${\lambda }_{i}$ and ${\lambda }_{j}$ by some $\lambda \in {𝔥}^{*}$ so that $\lambda >{\lambda }_{i},{\lambda }_{j}\text{.)}$ Let $\mu$ be a weight of $M$ and suppose $\mu \le {\lambda }_{i}\text{.}$ Then define $d\left(\mu \right)=\text{ht}\left({\lambda }_{i}-\mu \right)\text{.}$ Choose ${\mu }_{1}\in {𝔥}^{*}$ a weight of $M$ so that $d\left({\mu }_{1}\right)$ is minimal. This implies that any nonzero vector in ${M}^{{\mu }_{1}}$ is a highest weight vector. Therefore, choose $0\ne {v}_{1}\in {M}^{{\mu }_{1}}$ and let ${M}^{1}$ be the $𝔤\text{-submodule}$ of $M$ generated by ${v}_{1}\text{.}$ Now choose a weight ${\mu }_{2}$ of $M/{M}^{1}$ so that $d\left({\mu }_{2}\right)$ is minimal. Then we can choose $0\ne {v}_{2}\in {\left(M/{M}^{1}\right)}^{{\mu }_{2}}\text{.}$ Let ${M}^{2}\supseteq {M}^{1}$ be the submodule of $M$ so that ${M}^{2}/{M}^{1}$ is generated by ${v}_{2}\text{.}$ In this way we inductively define ${M}^{i}\text{.}$ The following picture illustrates this process of choosing the weights ${\mu }_{i}\text{:}$ ${\lambda }_{i} {\lambda }_{j} d\left({\mu }_{1}\right) d\left({\mu }_{2}\right) d\left({\mu }_{3}\right) {\mu }_{1} {\mu }_{2} {\mu }_{3}$ In this way, we construct a filtration $0={M}^{0}\subseteq {M}^{1}\subseteq {M}^{2}\subseteq \cdots$ such that each quotient is a highest weight module. Since $Q$ is finitely generated, there are only a finite number of weights $\mu$ of $M$ such that $d\left(\mu \right)$ has a given value. This, along with the fact that $\text{dim} {M}^{\mu }<\infty ,$ show that the construction outlined above yields a filtration so that $M=\bigcup _{i=1}^{\infty }{M}^{i}\text{.}$ $\square$

### Blocks

Let $𝔤$ be a Lie algebra with a regular, finite triangular decomposition. Recall, for $M\in 𝒪$ and $\lambda \in {𝔥}^{*},$ $\left[M:L\left(\lambda \right)\right]$ is the number of times $L\left(\lambda \right)$ appears in a local composition series for $M\text{.}$ Define a relation $\succ$ on ${𝔥}^{*}$ by $\lambda \succ \mu$ if $\left[M\left(\lambda \right):L\left(\mu \right)\right]>0\text{.}$ Extend this to an equivalence relation $\sim$ on ${𝔥}^{*}\text{.}$ The blocks of $𝔤$ are the equivalence classes of ${𝔥}^{*}$ under $\sim \text{.}$ For $\lambda \in {𝔥}^{*},$ let $\left[\lambda \right]$ represent the block containing $\lambda \text{.}$ Denote the set of blocks of $𝔤$ by $\left[{𝔥}^{*}\right]\text{.}$

For $\lambda \in {𝔥}^{*},$ define ${𝒪}^{\left[\lambda \right]}$ to be the full subcategory of $𝒪$ consisting of modules $M$ such that $\left[M:L\left(\mu \right)\right]>0$ only if $\mu \sim \lambda \text{.}$ As we will see in Theorem 2.3.12, the category $𝒪$ can be decomposed into these subcategories. This decomposition is the goal of this section, which follows Sections 2.10-2.12 of [MPi1995].

#### Projectives and BGG Reciprocity

In this section, we introduce a new class of $𝔤\text{-modules:}$ $P\left(\mu \right),$ the projective cover of the irreducible module $L\left(\mu \right)$ $\text{(}\mu \in {𝔥}^{*}\text{).}$ We then establish a reciprocity between these projective modules and Verma modules on the one hand and Verma modules and irreducibles on the other.

The irreducible modules $L\left(\mu \right)$ do not in general have a projective cover in the full category $𝒪\text{.}$ Therefore, we fix $\lambda \in {𝔥}^{*}$ and restrict to the subcategory $𝒪\left(\lambda \right)$ which consists of modules $M\in 𝒪$ such that ${M}^{\mu }\ne 0$ only if $\mu \le \lambda \text{.}$

Rather than defining $P\left(\mu \right)$ directly, we construct a larger projective module $\stackrel{\sim }{P}\left(\mu \right)$ in $𝒪\left(\lambda \right)\text{.}$ Let $\mu \le \lambda \text{.}$ We will view $U\left({𝔤}_{>0}\right)$ as a $𝔥\oplus {𝔤}_{>0}\text{-module}$ with the following action: $(h+x)·u≔ (ad(h)+(μ,h)) u+xu$ for $h\in 𝔥,$ $x\in {𝔤}_{>0},$ and $u\in U\left({𝔤}_{>0}\right)\text{.}$ Under this action, $U\left({𝔤}_{>0}\right)$ has a weight-space decomposition $U\left({𝔤}_{>0}\right)=\underset{\nu \ge \mu }{⨁}U{\left({𝔤}_{>0}\right)}^{\nu }\text{.}$ Then the subspace $\underset{\nu \nleqq \lambda }{⨁}U{\left({𝔤}_{>0}\right)}^{\nu }$ is a $𝔥\oplus {𝔤}_{>0}\text{-submodule}$ of $U\left({𝔤}_{>0}\right)\text{.}$ Define the $𝔥\oplus {𝔤}_{>0}\text{-module}$ $W(μ)=Wλ(μ) =U(𝔤>0)/ ⨁ν≰λ U(𝔤>0)ν.$ Then $\stackrel{\sim }{P}\left(\mu \right)$ is the induced $𝔤\text{-module}$ $P∼(μ)= P∼λ(μ) =U(𝔤) ⊗U(𝔥⊕𝔤>0) W(μ).$

$\stackrel{\sim }{P}\left(\mu \right)\in 𝒪\left(\lambda \right)\text{.}$ Proof. Let ${\left\{{w}_{i}\right\}}_{i=1}^{n}$ be a basis of weight vectors for $W\left(\mu \right)\text{.}$ (This set is finite since $\text{dim} W{\left(\mu \right)}^{\nu }\le \text{dim} U{\left({𝔤}_{>0}\right)}^{\nu }<\infty$ for all $\nu$ and $W{\left(\mu \right)}^{\nu }=0$ unless $\mu \le \nu \le \lambda \text{.)}$ As a $U\left({𝔤}_{<0}\right)\text{-module,}$ $\stackrel{\sim }{P}\left(\mu \right)$ is (freely) generated by $1\otimes {w}_{i}\text{.}$ Therefore, $\stackrel{\sim }{P}\left(\mu \right)$ is finitely generated as a $𝔤\text{-module}$ This implies $\stackrel{\sim }{P}{\left(\mu \right)}^{\nu }\ne 0$ only if $\nu \le \lambda$ and $\stackrel{\sim }{P}{\left(\mu \right)}^{\nu }<\infty \text{.}$ $\square$

$\stackrel{\sim }{P}\left(\mu \right)$ is projective in $𝒪\left(\lambda \right)\text{.}$ Proof. Let $M$ and $N$ be modules in $𝒪\left(\lambda \right)$ such that $P∼(μ) ↓ϕ N ⟶ψ M ⟶ 0.$ Note that elements of $\stackrel{\sim }{P}\left(\mu \right)$ are sums of elements of the form $x\otimes \stackrel{‾}{y},$ where $x\in U\left(𝔤\right)$ and $\stackrel{‾}{y}\in W\left(\mu \right)$ is the image of $y\in U\left({𝔤}_{>0}\right)\text{.}$ Suppose $\varphi \left(1\otimes \stackrel{‾}{1}\right)={m}_{0}\in M\text{.}$ Let ${n}_{0}\in N$ be such that $\psi \left({n}_{0}\right)={m}_{0}\text{.}$ Define a map $\varphi \prime :\stackrel{\sim }{P}\left(\mu \right)\to N$ by $ϕ′(x⊗y‾)= xyn0.$ If this map is well-defined, it is clearly a $𝔤\text{-module}$ homomorphism such that $\psi \otimes \varphi \prime =\varphi \text{.}$ Therefore we check that $\varphi \prime$ is well-defined. Suppose that $\stackrel{‾}{y}=\stackrel{‾}{y}\prime \text{.}$ Then $y=y\prime +u$ for some $u\in \underset{\nu \nleqq \lambda }{⨁}U{\left({𝔤}_{>0}\right)}^{\nu }\text{.}$ This implies $xyn0=xy′n0+ xun0.$ However, ${n}_{0}\in {N}^{\mu }$ and so $u{n}_{0}\in \underset{\nu \nleqq \lambda }{⨁}{N}^{\nu }=0\text{.}$ $\square$

$\stackrel{\sim }{P}\left(\mu \right)$ has a Verma composition series and $[P∼(μ):M(ν)] =dim Hom𝔤 (P∼(μ),M(ν)) =dim(M(ν)μ).$ Proof. Note that $\stackrel{\sim }{P}\left(\mu \right)$ is a free $U\left({𝔤}_{<0}\right)\text{-module.}$ Let ${\left\{{w}_{i}\right\}}_{i=1}^{n}$ be a basis of weight vectors for $W\left(\mu \right)\text{.}$ Assume that ${w}_{i}\in W{\left(\mu \right)}^{{\gamma }_{i}}$ and ${\gamma }_{i}>{\gamma }_{j}$ implies $i>j\text{.}$ Let ${M}_{i}$ be the submodule of $\stackrel{\sim }{P}\left(\mu \right)$ generated by $1\otimes {w}_{i}\text{.}$ Then the following is a Verma composition series for $\stackrel{\sim }{P}\left(\mu \right)\text{:}$ $0⊆Mn⊆Mn-1+ Mn⊆⋯⊆M1+⋯+ Mn=P∼ (μ).$ Also, for $\mu \le \nu \le \lambda ,$ $[P∼(μ):M(ν)] =dim W(μ)ν= dim M(ν)μ.$ Since $\stackrel{\sim }{P}\left(\mu \right)$ is projective, $dim M(ν)μ= dim Hom𝔤(P∼(μ),M(ν)).$ $\square$

Let $P\in 𝒪\left(\lambda \right)$ be a finitely generated indecomposable projective module. Then $P$ has a unique maximal submodule. Proof. Let $\mu \le \lambda \text{.}$ From Lemma 2.1.5, we have $\stackrel{\sim }{P}\left(\mu \right)=\underset{1\le i\le n}{⨁}{P}_{i},$ where each ${P}_{i}$ is an indecomposable projective module. Moreover, each ${P}_{i}$ is generated by one element, namely the projection of $1\otimes \stackrel{‾}{1}\in \stackrel{\sim }{P}\left(\mu \right)\text{.}$ Lemma 2.1.3 shows that ${P}_{i}$ has a maximal proper submodule. Since ${P}_{i}$ is generated by one element, this submodule is unique. Now consider any finitely generated indecomposable projective module $P\text{.}$ Let $\left\{{v}_{1},\dots ,{v}_{n}\right\}$ be a generating set for $P$ so that ${v}_{i}\in {P}^{{\nu }_{i}}\text{.}$ Then there is a surjective map $\varphi :\sum _{i}\stackrel{\sim }{P}\left({\nu }_{i}\right)\to P$ defined by $\varphi \left({v}_{{\nu }_{i}}\right)={v}_{i},$ where ${v}_{{\nu }_{i}}=1\otimes \stackrel{‾}{1}\in \stackrel{\sim }{P}\left({\nu }_{i}\right)\text{.}$ Since $P$ is projective and indecomposable, it must be isomorphic to an indecomposable summand of $\sum _{i}\stackrel{\sim }{P}\left({\nu }_{i}\right)\text{.}$ The previous paragraph implies that $P$ has a unique maximal submodule. $\square$

For each $\mu \le \lambda$ the irreducible module $L\left(\mu \right)$ has a projective cover $P\left(\mu \right)\in 𝒪\left(\lambda \right)\text{.}$ The modules $P\left(\mu \right),$ $\mu \le \lambda ,$ are all of the indecomposable projectives in $𝒪\text{.}$ Proof. Let $\mu \le \lambda \text{.}$ Then $L\left(\mu \right)$ is a quotient of $\stackrel{\sim }{P}\left(\mu \right)$ and therefore must be a quotient of a projective indecomposable summand $P$ of $\stackrel{\sim }{P}\left(\mu \right)\text{.}$ Then $P$ is a projective cover of $L\left(\mu \right),$ which we denote by $P\left(\mu \right)\text{.}$ The previous proposition shows that each finitely generated indecomposable module $P$ has a unique irreducible quotient, and so $P\cong P\left(\mu \right)$ for some $\mu \le \lambda \text{.}$ $\square$

([RCW1982]). Let $\mu ,\nu \in {𝔥}^{*}\text{.}$ Then, $[M(μ):L(ν)]= [P(ν):M(μ)].$ Proof. We begin by showing $\left[M\left(\mu \right):L\left(\nu \right)\right]=\text{dim}\left({\text{Hom}}_{𝔤}\left(P\left(\nu \right),M\left(\mu \right)\right)\right)\text{.}$ According to Lemma 2.2.1, $M\left(\mu \right)$ has a local composition series at $\nu ,$ $M\left(\mu \right)={M}_{1}\supseteq \cdots \supseteq {M}_{n}=0\text{.}$ Lemma 1.3.3 implies $dim(Hom𝔤(P(ν),M(ν))) =∑k=1n-1 dim(Hom𝔤(P(ν),Mk/Mk+1)).$ Consider $\text{dim}\left({\text{Hom}}_{𝔤}\left(P\left(\nu \right),{M}_{k}/{M}_{k+1}\right)\right)\text{.}$ By definition of a local composition series ${M}_{k}/{M}_{k+1}\cong L\left(\gamma \right)$ for some $\gamma \ge \nu$ or ${\left({M}_{k}/{M}_{k+1}\right)}^{\gamma }=0$ for all $\gamma \ge \nu \text{.}$ Now, $dim(Hom𝔤(P(ν),L(γ)))= { 0 if ν≠γ ℂ if ν=γ$ since $P\left(\nu \right)$ has a unique maximal submodule $L\left(\nu \right)\text{.}$ Suppose ${\left({M}_{k}/{M}_{k+1}\right)}^{\gamma }=0$ for all $\gamma \ge \nu \text{.}$ Since $L\left(\nu \right)$ is the unique irreducible quotient of $P\left(\nu \right)$ and $L{\left(\nu \right)}^{\nu }\ne 0,$ any homomorphism $\varphi :P\left(\nu \right)\to {M}_{k}/{M}_{k+1}$ must be trivial. Therefore, $\text{dim} {\text{Hom}}_{𝔤}\left(P\left(\nu \right),{M}_{k},{M}_{k+1}\right)=0\text{.}$ We now have $dim(Hom𝔤(P(ν),M(μ)))= ∑k=1ndim (Hom𝔤(P(ν),Mk/Mk+1)) =[M(μ):L(ν)].$ Next we show $\text{dim}\left({\text{Hom}}_{𝔤}\left(P\left(\nu \right),M\left(\mu \right)\right)\right)=\left[P\left(\nu \right),M\left(\mu \right)\right]\text{.}$ Recall $\text{dim}\left({\text{Hom}}_{𝔤}\left(\stackrel{\sim }{P}\left(\nu \right),M\left(\mu \right)\right)\right)=\left[\stackrel{\sim }{P}\left(\nu \right),M\left(\mu \right)\right]$ and $\stackrel{\sim }{P}\left(\nu \right)=\underset{\gamma }{⨁}{m}_{\nu }\left(\gamma \right)P\left(\gamma \right)$ for some ${m}_{\nu }\left(\gamma \right)\in {ℤ}_{\ge 0}\text{.}$ Therefore, $⨁γmν(γ) dim(Hom𝔤(P(γ),M(μ))) = dim(Hom𝔤(P∼(ν),M(μ))) = [P∼(ν):M(μ)] = ⨁γmν(γ) [P(γ):M(μ)] (2.3)$ Observe that $dim(Hom𝔤(P∼(ν),L(ω))) =⨁γmν(γ)dim (Hom𝔤(P(γ),L(ω))) =mν(ω).$ On the other hand, $\text{dim}\left({\text{Hom}}_{𝔤}\left(\stackrel{\sim }{P}\left(\nu \right),L\left(\omega \right)\right)\right)=\text{dim} L{\left(\omega \right)}^{\nu }$ since $\stackrel{\sim }{P}\left(\nu \right)$ acts like a free module in $𝒪\left(\lambda \right)\text{.}$ Therefore, ${m}_{\nu }\left(\nu \right)=1$ and ${m}_{\nu }\left(\gamma \right)=0$ unless $\nu \le \gamma \text{.}$ Using Equation 2.3, we can then induct on $\text{ht}\left(\lambda -\nu \right)$ to show $\left[P\left(\nu \right):M\left(\mu \right)\right]=\text{dim}\left({\text{Hom}}_{𝔤}\left(P\left(\nu \right):M\left(\mu \right)\right)\right)\text{.}$ $\square$

#### ${\text{Ext}}^{1}\left(·,·\right)$ and a Decomposition of Category $𝒪$

${\text{Ext}}^{1}\left(·,·\right)$ provides a useful tool for studying blocks, particularly in the proof of the decomposition of Category $𝒪$ by blocks (Theorem 2.3.12). For a definition of ${\text{Ext}}^{1}\left(·,·\right)$ see Section 1.3.4.

If ${\text{Ext}}^{1}\left(M\left(\mu \right),L\left(\lambda \right)\right)\ne 0,$ then $\left[M\left(\lambda \right):L\left(\mu \right)\right]>0\text{.}$ Proof. We first claim $\lambda \ngtr \mu$ implies ${\text{Ext}}^{1}\left(M\left(\mu \right),L\left(\lambda \right)\right)=0\text{.}$ To prove this, assume $\lambda \ngtr \mu$ and let $0⟶L(λ)⟶N ⟶ϕM(μ) ⟶0$ be an extension of $M\left(\mu \right)$ by $L\left(\lambda \right)\text{.}$ Let $v\in {N}^{\mu }$ so that $\varphi \left(v\right)={v}^{+}\in M\left(\mu \right)\text{.}$ If $\lambda \ngtr \mu ,$ then ${N}^{\gamma }=0$ for all $\gamma >\mu$ and so $v$ is a highest weight vector in $N\text{.}$ Therefore, we have a homomorphism $\psi :M\left(\mu \right)\to N$ given by $\psi \left({v}^{+}\right)=v\text{.}$ Then $\psi \circ \varphi ={\text{id}}_{M\left(\lambda \right)}$ and so the sequence splits. This implies ${\text{Ext}}^{1}\left(M\left(\mu \right),L\left(\lambda \right)\right)=0\text{.}$ We now assume ${\text{Ext}}^{1}\left(M\left(\mu \right),L\left(\lambda \right)\right)\ne 0\text{.}$ Since $\lambda >\mu ,$ $M\left(\mu \right)\in 𝒪\left(\lambda \right)\text{.}$ Using Theorem 2.3.6. we only need to show $\left[P\left(\mu \right),M\left(\lambda \right)\right]>0\text{.}$ Since $\stackrel{\sim }{P}\left(\mu \right)$ has a Verma composition series (Proposition 2.3.3) and $P\left(\mu \right)$ is a direct summand of $\stackrel{\sim }{P}\left(\mu \right),$ $P\left(\mu \right)$ has a Verma composition series $P\left(\mu \right)={P}_{0}\supseteq \cdots \supseteq {P}_{n}=0$ such that $P\left(\mu \right)/{P}_{1}\cong M\left(\mu \right)\text{.}$ We then have the following diagram $P(μ) ↓ϕ 0 ⟶ L(λ) ⟶ N ⟶ψ M(μ) ⟶ 0.$ Since $P\left(\mu \right)$ is projective, there is a homomorphism $\varphi \prime :P\left(\mu \right)\to N$ such that $\psi \circ \varphi \prime =\varphi \text{.}$ Then, $\varphi \prime \left({P}_{1}\right)\subseteq \text{Ker} \psi \cong L\left(\lambda \right)\text{.}$ We must have either $\varphi \prime \left({P}_{1}\right)=0$ or $\varphi \prime \left({P}_{1}\right)\cong L\left(\lambda \right)\text{.}$ If $\varphi \prime \left({P}_{1}\right)=0,$ $\varphi \prime$ induces a map $\varphi \prime :M\left(\mu \right)\to N$ and the sequence splits. Therefore, $\varphi \prime \left({P}_{1}\right)\cong L\left(\lambda \right)\text{.}$ Applying $\varphi \prime$ to the Verma composition series for $P\left(\mu \right),$ we have $ϕ′(P1)≅ L(λ)⊇⋯⊇ϕ′ (Pn)=0.$ Therefore, there is some $1\le k\le n-1$ such that $\varphi \prime \left({P}_{k}\right)\cong L\left(\lambda \right)$ and $\varphi \prime \left({P}_{k+1}\right)=0\text{.}$ Then ${P}_{k}/{P}_{k+1}\cong M\left(\lambda \right)\text{.}$ $\square$

Let $M\in 𝒪$ be a highest weight module of highest weight $\lambda \text{.}$ Suppose ${\text{Ext}}^{1}\left(M\left(\mu \right),M\right)\ne 0\text{.}$ Then there is some $\nu \in {𝔥}^{*}$ such that $\lambda \ge \nu \ge \mu$ and $\left[M\left(\lambda \right):L\left(\nu \right)\right],$ $\left[M\left(\nu \right):L\left(\mu \right)\right]>0\text{.}$ In particular, $\mu \in \left[\lambda \right]\text{.}$ Proof. We have $\lambda >\mu \text{.}$ Let $N$ be the maximal submodule of $M\text{.}$ Then we have a short exact sequence $0⟶N⟶M⟶L(λ)⟶0.$ Proposition 1.3.4 implies that ${\text{Ext}}^{1}\left(M\left(\mu \right),L\left(\lambda \right)\right)\ne 0$ or ${\text{Ext}}^{1}\left(M\left(\mu \right),N\right)>0\text{.}$ If ${\text{Ext}}^{1}\left(M\left(\mu \right),L\left(\lambda \right)\right)\ne 0,$ the proposition holds. Therefore assume that ${\text{Ext}}^{1}\left(M\left(\mu \right),N\right)\ne 0\text{.}$ Let $N=N0⊆N1⊆⋯$ be a local composition series for $N\text{.}$ Then, by Proposition 1.3.4, ${\text{Ext}}^{1}\left(M\left(\mu \right),{N}_{i}/{N}_{i+1}\right)\ne 0$ for some $i\text{.}$ Recall that ${N}_{i}/{N}_{i+1}\cong L\left(\nu \right)$ for some $\lambda \ge \nu \ge \mu$ or ${\left({N}_{i}/{N}_{i+1}\right)}^{\nu }=0$ for all $\nu \ge \mu \text{.}$ Suppose ${\left({N}_{i}/{N}_{i+1}\right)}^{\nu }=0$ for all $\nu \ge \mu \text{.}$ We claim this implies ${\text{Ext}}^{1}\left(M\left(\mu \right),{N}_{i}/{N}_{i+1}\right)=0,$ a contradiction. Let $0⟶Ni/Ni+1⟶ N′⟶M(μ)⟶0$ be a short exact sequence. Choose $v\in N\prime$ so that $v↦{v}^{+}\in M\left(\mu \right)\text{.}$ Then $v\in {\left(N\prime \right)}^{\mu }\text{.}$ Since ${\left({N}_{i}/{N}_{i+1}\right)}^{\nu }=0$ and $M{\left(\mu \right)}^{\nu }=0$ for $\nu >\mu ,$ ${\left(N\prime \right)}^{\nu }=0\text{.}$ This implies that $v$ is a highest weight vector. We then have a map $M\left(\mu \right)\to N$ given by ${v}^{+}↦v\text{.}$ Thus, the sequence splits, and so ${\text{Ext}}^{1}\left(M\left(\mu \right),{N}_{i}/{N}_{i+1}\right)=0\text{.}$ Therefore, ${N}_{i}/{N}_{i+1}\cong L\left(\nu \right)\text{.}$ By Lemma 2.3.7, $\left[M\left(\nu \right):L\left(\mu \right)\right]>0\text{.}$ Since ${N}_{i}/{N}_{i+1}$ is a subquotient of $M,$ and hence of $M\left(\lambda \right),$ $\left[M\left(\lambda \right):L\left(\nu \right)\right]>0\text{.}$ $\square$

Recall that the blocks of $𝔤$ are defined to be the equivalence classes of ${𝔥}^{*}$ where $\lambda \sim \mu$ if $\left[M\left(\lambda \right):L\left(\mu \right)\right]>0\text{.}$ The following proposition provides alternative definitions of this equivalence relation.

([Nei1984]). Let $\lambda ,\mu \in {𝔥}^{*}\text{.}$ Then the following are equivalent:

 • $\left[M\left(\lambda \right):L\left(\mu \right)\right]>0\text{.}$ • $\left[P\left(\mu \right):M\left(\lambda \right)\right]>0\text{.}$ • $M\left(\mu \right)↪M\left(\lambda \right)\text{.}$ Proof. Theorem 2.3.6 shows that the first two statements are equivalent. Also $M\left(\mu \right)↪M\left(\lambda \right)$ clearly implies $\left[M\left(\lambda \right):L\left(\mu \right)\right]>0\text{.}$ Therefore we only need to check that $\left[M\left(\lambda \right):L\left(\mu \right)\right]>0$ implies $M\left(\mu \right)↪M\left(\lambda \right)\text{.}$ Suppose $\left[M\left(\lambda \right):L\left(\mu \right)\right]>0\text{.}$ We will show that $M\left(\mu \right)↪M\left(\lambda \right)$ by inducting on $\text{ht}\left(\lambda -\mu \right)\text{.}$ If $\lambda =\mu ,$ the result clearly holds. Now suppose $\lambda >\mu \text{.}$ Let $J\left(\lambda \right)$ be the maximum proper submodule of $M\left(\lambda \right)\text{.}$ We may refine the series $0\subseteq J\left(\lambda \right)\subseteq M\left(\lambda \right)$ into a local composition series, which implies that $\left[J\left(\lambda \right):L\left(\mu \right)\right]>0\text{.}$ Let $0=J0⊆J1⊆⋯$ be a highest weight series for $J\left(\lambda \right),$ where ${J}^{i}/{J}^{i-1}$ is a highest weight module of highest weight ${\mu }_{i}\text{.}$ There is some $k$ such that ${\left({J}^{k}\right)}^{\mu }=J{\left(\lambda \right)}^{\mu }$ since the weight spaces of $J\left(\lambda \right)$ are finite-dimensional. This means $\left[J\left(\lambda \right)/{J}^{k}:L\left(\mu \right)\right]=0$ and so $\left[{J}^{k}:L\left(\mu \right)\right]>0\text{.}$ Therefore, $0={J}^{0}\subseteq {J}^{1}\subseteq \cdots \subseteq {J}^{k}$ can be refined to a local composition series with a factor of $L\left(\mu \right)\text{.}$ Choose $i$ minimal so that $\left[{J}^{i+1}/{J}^{i}:L\left(\mu \right)\right]>0\text{.}$ Then ${\mu }_{i+1}\ge \mu \text{.}$ Suppose $\mu ={\mu }_{i+1}\text{.}$ Then we have a nontrivial map $M\left(\mu \right)\to J\left(\lambda \right)/{J}^{i}\text{.}$ We may assume $i$ is minimal such that ${\text{Hom}}_{𝔤}\left(M\left(\mu \right),J\left(\lambda \right)/{J}^{i}\right)\ne 0\text{.}$ The sequence $0\to {J}^{i}/{J}^{i-1}\to J\left(\lambda \right)/{J}^{i-1}\to J\left(\lambda \right)/{J}^{i}\to 0$ is exact. Therefore, Proposition 1.3.4 implies that $Hom𝔤(M(μ),J(λ)/Ji-1)→ Hom𝔤(M(μ),J(λ)/Ji)→ Ext1(M(μ),Ji/Ji-1)$ is exact. The first term in this sequence is $0$ and the second is nonzero. Therefore, the third term is nonzero: ${\text{Ext}}^{1}\left(M\left(\mu \right),{J}^{i}/{J}^{i-1}\right)\ne 0\text{.}$ We use Lemma 2.3.8 to show there is some $\nu \in {𝔥}^{*}$ such that $\lambda >{\mu }_{i}\ge \nu \ge \mu$ and $\left[M\left({\mu }_{i}\right):L\left(\nu \right)\right]>0$ and $\left[M\left(\nu \right):L\left(\mu \right)\right]>0\text{.}$ Applying the inductive hypothesis, we have $M\left(\mu \right)↪M\left(\nu \right)↪M\left({\mu }_{i}\right)↪M\left(\lambda \right)\text{.}$ If $\mu \ne {\mu }_{i+1},$ we can use the induction hypothesis to conclude $M\left(\mu \right)↪M\left({\mu }_{i+1}\right)$ and $M\left({\mu }_{i+1}\right)↪M\left(\lambda \right)\text{.}$ $\square$

Recall that, for a block $\left[\lambda \right]\in \left[{𝔥}^{*}\right],$ ${𝒪}^{\left[\lambda \right]}$ is the full subcategory of $𝒪$ consisting of modules $M$ such that $L\left(\mu \right)$ is a local composition factor of $M$ only if $\mu \in \left[\lambda \right]\text{.}$

Let $\left[\lambda \right],\left[\mu \right]\in \left[{𝔥}^{*}\right]$ with $\left[\lambda \right]\ne \left[\mu \right]\text{.}$ Suppose $E\in {𝒪}^{\left[\lambda \right]}$ and $F\in {𝒪}^{\left[\mu \right]}$ are highest weight modules. Then, ${\text{Ext}}^{1}\left(E,F\right)=0\text{.}$ Proof. From Proposition 2.3.8, if ${\text{Ext}}^{1}\left(M\left(\mu \right),E\right)\ne 0,$ there exists $\nu \in {𝔥}^{*},$ $\lambda \ge \nu >\mu$ such that $\left[M\left(\lambda \right):L\left(\nu \right)\right]>0$ and $\left[M\left(\nu \right):L\left(\mu \right)\right]>0\text{.}$ This implies $\lambda \sim \nu \sim \mu ,$ a contradiction. Therefore, ${\text{Ext}}^{1}\left(M\left(\mu \right),E\right)=0\text{.}$ Since $F$ is a highest weight module of highest weight $\mu ,$ there is a submodule $N\subseteq M\left(\mu \right)$ such that $0\to N\to M\left(\mu \right)\to F\to 0$ is exact. Proposition 1.3.4 implies $Hom𝔤(N,E)⟶ Ext1(F,E)⟶ Ext1(M(μ),E)=0.$ If ${\text{Hom}}_{𝔤}\left(N,E\right)=0,$ then ${\text{Ext}}^{1}\left(F,E\right)=0\text{.}$ Therefore, we only need to check that ${\text{Hom}}_{𝔤}\left(N,E\right)=0\text{.}$ Let $\varphi \in {\text{Hom}}_{𝔤}\left(N,E\right)\text{.}$ Since $N\in {𝒪}^{\left[\mu \right]},$ $\varphi \left(N\right)\in {𝒪}^{\left[\mu \right]}\text{.}$ On the other hand, $\varphi \left(N\right)\in {𝒪}^{\left[\lambda \right]}\text{.}$ This implies $\varphi \left(N\right)=0\text{.}$ $\square$

([MPi1995], 2.12.3). Let $\left[\lambda \right],\left[\mu \right]\in \left[{𝔥}^{*}\right]$ with $\left[\lambda \right]\ne \left[\mu \right]\text{.}$ Suppose $E\in {𝒪}^{\left[\lambda \right]}$ and $F\in {𝒪}^{\left[\mu \right]},$ with $\left[\lambda \right]\ne \left[\mu \right],$ have a filtration by highest weight modules of finite length. Then, ${\text{Ext}}^{1}\left(E,F\right)=0\text{.}$

([MPi1995]). Let $𝔤$ be a Lie algebra with a regular, finite triangular decomposition. Then, the category $𝒪$ decomposes into subcategories: $𝒪=⨁[λ]∈[𝔥*] 𝒪[λ].$ This means that for $M\in 𝒪,$ $M=⨁[λ]∈[𝔥*] M[λ],$ where ${M}^{\left[\lambda \right]}\in {𝒪}^{\left[\lambda \right]}\text{.}$ Proof. Let $M\in 𝒪\text{.}$ Then $M$ has a filtration $0={M}_{0}\subseteq {M}_{1}\subseteq {M}_{2}\subseteq \cdots$ such that ${M}_{i}/{M}_{i-1}$ is a highest weight module. For $\left[\lambda \right]\in \left[{𝔥}^{*}\right],$ we construct ${M}^{\left[\lambda \right]}$ by constructing a highest weight series $0⊆M0[λ]⊆ M1[λ]⊆⋯$ such that $Mi = ⨁[λ]∈[𝔥*] Mi[λ]; Mi[λ]/ Mi-1[λ] = { Mi/Mi-1 if Mi/Mi-1 ∈𝒪[λ], 0 otherwise.$ We do this inductively. Suppose we have constructed these highest weight series for $j and suppose ${M}_{n}/{M}_{n-1}\in {𝒪}^{\left[\lambda \right]}\text{.}$ For $\left[\mu \right]\ne \left[\lambda \right],$ let ${M}_{n}^{\left[\mu \right]}={M}_{n-1}^{\left[\mu \right]}\text{.}$ To define ${M}_{n}^{\left[\lambda \right]},$ we use the short exact sequence $0⟶Mn-1/ Mn-1[λ]⟶ Mn/Mn-1[λ] ⟶Mn/Mn-1⟶0. (2.4)$ Since ${M}_{n-1}=\underset{\left[\mu \right]\in \left[{𝔥}^{*}\right]}{⨁}{M}_{n-1}^{\left[\mu \right]},$ ${M}_{n-1}/{M}_{n-1}^{\left[\lambda \right]}\cong \underset{\left[\mu \right]\ne \left[\lambda \right]}{⨁}{M}_{n-1}^{\left[\mu \right]}\in \underset{\left[\mu \right]\ne \left[\lambda \right]}{⨁}{𝒪}^{\left[\mu \right]}\text{.}$ Since ${M}_{n}/{M}_{n-1}\in {𝒪}^{\left[\lambda \right]},$ Lemma 2.3.11 implies ${\text{Ext}}^{1}\left({M}^{n}/{M}^{n-1},{M}^{n-1}/{M}_{\lambda }^{n-1}\right)=0,$ and so the sequence in (2.4) splits: $Mn/Mn-1[λ]≅ Mn-1/Mn-1[λ] ⊕Mn/Mn-1.$ Therefore there exists a module ${M}_{\lambda }^{n},$ ${M}_{\lambda }^{n-1}\subseteq {M}_{\lambda }^{n},$ so that $Mλn/ Mλn-1 ≅ Mn/Mn-1 Mn/Mλn-1 = Mλn/ Mλn-1⊕ Mn-1/ Mλn-1.$ To finish the induction step, we claim $Mn=⨁[μ]∈[𝔥*] Mn[μ]=Mn[λ] ⊕⨁[μ]≠[λ] Mn-1[μ].$ Clearly, ${M}_{n}={M}_{n}^{\left[\lambda \right]}+\underset{\left[\mu \right]\ne \left[\lambda \right]}{⨁}{M}_{n-1}^{\left[\mu \right]}\text{.}$ Also, ${M}_{n}^{\left[\lambda \right]}\cap {M}_{n-1}={M}_{n-1}^{\left[\lambda \right]},$ which shows $Mn[λ]∩ ⨁[μ]≠[λ] Mn-1[μ]=0.$ $\square$

### The Contravariant Form and the Shapovalov Determinant

Suppose $𝔤$ has a triangular decomposition. Proposition 2.0.5 implies the following decomposition, $U(𝔤)= U(𝔥)⊕ (𝔤<0U(𝔤)+U(𝔤)𝔤>0).$ Let $\tau :U\left(𝔤\right)\to U\left(𝔥\right)$ be the projection map. Recall that $\sigma :U\left(𝔤\right)\to U\left(𝔤\right)$ is the anti-involution such that $\sigma \left({𝔤}_{>0}\right)={𝔤}_{<0}$ and $\sigma {|}_{𝔥}=\text{id}\text{.}$ We can then define a bilinear form $⟨,⟩:U\left(𝔤\right)×U\left(𝔤\right)\to S\left(𝔥\right)$ on $U\left(𝔤\right)\text{.}$ For $x,y\in U\left(𝔤\right),$ define $⟨x,y⟩= τ(σ(x)y).$ This induces a bilinear form $⟨,⟩:M\left(\lambda \right)×M\left(\lambda \right)\to ℂ$ on $M\left(\lambda \right)\text{.}$ Fix a highest weight vector ${v}^{+}\in M\left(\lambda \right)$ of $M\left(\lambda \right)\text{.}$ For $x,y\in U\left(𝔤\right),$ define $⟨xv+,yv+⟩= λ(τ(σ(x)y)).$

If $𝔤$ has an Hermitian triangular decomposition, we make the following modifications to the definition of the contravariant form. In this case there is a real Lie algebra $\stackrel{‾}{𝔤}$ such that $𝔤=\stackrel{‾}{𝔤}{\otimes }_{ℝ}ℂ,$ and $\sigma$ is an Hermitian anti-involution such that $\sigma {|}_{\stackrel{‾}{𝔥}}=\text{id}\text{.}$ For $\stackrel{‾}{\lambda }\in {\stackrel{‾}{𝔥}}^{*},$ define $\stackrel{‾}{\lambda }\in {𝔥}^{*}$ by $\stackrel{‾}{\lambda }\left(\text{ch}\right)=c\stackrel{‾}{\lambda \left(h\right)}$ for $c\in ℂ$ and $h\in \stackrel{‾}{𝔥}\text{.}$ Then the Hermitian form $⟨,⟩:M\left(\stackrel{‾}{\lambda }\right)×M\left(\stackrel{‾}{\lambda }\right)\to ℂ$ is defined by $⟨x{v}^{+},y{v}^{+}⟩=\stackrel{‾}{\lambda }\left(\tau \left(\sigma \left(x\right)y\right)\right),$ where $x,y\in U\left(𝔤\right)$ and ${v}^{+}$ is a highest weight vector of $M\left(\stackrel{‾}{\lambda }\right)\text{.}$

For the remainder of this chapter we assume

 • $𝔤$ has a regular, finite triangular decomposition; OR • $𝔤$ has a regular, finite Hermitian triangular decomposition, and we restrict all choices of weights to $\lambda \in {𝔥}^{*}$ such that $\lambda =\stackrel{‾}{\lambda }$ for some $\stackrel{‾}{\lambda }\in {\stackrel{‾}{𝔥}}^{*}\text{.}$

For $\mu ,\nu \in {𝔥}^{*}$ with $\mu \ne \nu ,$ $M{\left(\lambda \right)}^{\mu }\perp M{\left(\lambda \right)}^{\nu }$ with respect to $⟨,⟩\text{.}$ Proof. Choose $h\in 𝔥$ so that $\mu \left(h\right)\ne \nu \left(h\right)\text{.}$ For $0\ne v\in M{\left(\lambda \right)}^{\mu }$ and $0\ne w\in M{\left(\lambda \right)}^{\nu },$ we have $μ(h)⟨v,w⟩= ⟨hv,w⟩= ⟨v,hw⟩= ν(h)⟨v,w⟩ ⇒⟨v,w⟩ =0.$ $\square$

The radical of $⟨,⟩$ is $Rad⟨,⟩= { m∈M(λ) | ⟨m,m′⟩=0 for all m′∈M(λ) } .$

Let $\lambda \in {𝔥}^{*}\text{.}$ Then $\text{Rad}⟨,⟩=J\left(\lambda \right),$ the unique maximal submodule of $M\left(\lambda \right)\text{.}$ Proof. We first show that $\text{Rad}⟨,⟩$ is a submodule of $M\left(\lambda \right)\text{.}$ Let $x\in 𝔤$ and $m\in \text{Rad}⟨,⟩\text{.}$ Then for any $m\prime \in M\left(\lambda \right),$ $⟨xm,m\prime ⟩=⟨m,\sigma \left(x\right)m\prime ⟩=0$ since $\sigma \left(x\right)m\prime \in M\left(\lambda \right)\text{.}$ Therefore, $xm\in \text{Rad}⟨,⟩\text{.}$ Clearly, linear combinations of elements in $\text{Rad}⟨,⟩$ are in $\text{Rad}⟨,⟩\text{.}$ Thus, $\text{Rad}⟨,⟩$ is a submodule of $M\left(\lambda \right)\text{.}$ Since ${v}^{+}\notin \text{Rad}⟨,⟩,$ $\text{Rad}⟨,⟩\subseteq J\left(\lambda \right)\text{.}$ We now show that $J\left(\lambda \right)\subseteq \text{Rad}⟨,⟩\text{.}$ Let $v\in J\left(\lambda \right)\text{.}$ Then there is no $x\in U\left(𝔤\right)$ such that $xv=c{v}^{+}$ for some $0\ne c\in ℂ\text{.}$ Therefore, for any $x\in U\left(𝔤\right),$ $⟨x{v}^{+},v⟩=⟨{v}^{+},\sigma \left(x\right)v⟩=0\text{.}$ Since any vector in $M\left(\lambda \right)$ is of the form $x{v}^{+}$ for some $x\in U\left(𝔤\right),$ this implies $v\in \text{Rad}⟨,⟩\text{.}$ $\square$

Let $M$ be any highest weight module of highest weight $\lambda \text{.}$ Since $M\cong M\left(\lambda \right)/M\prime$ for some submodule $M\prime \subseteq J\left(\lambda \right),$ the previous lemma implies that $⟨,⟩$ can be extended to $M\text{.}$

The weight space $M{\left(\lambda \right)}^{\lambda -\mu }$ has a basis $\left\{{x}_{\left(t\right)}{v}^{+} | \left(t\right):{t}_{1}\ge \cdots \ge {t}_{j}>0,{\beta }_{{t}_{1}}+\cdots +{\beta }_{{t}_{j}}=\mu \right\}\text{.}$ (Recall the ${\beta }_{{t}_{i}}\in {Q}_{+}$ are such that ${x}_{{t}_{i}}\in {𝔤}_{-{\beta }_{{t}_{i}}}\text{.)}$ We can associate to $⟨,⟩{|}_{M{\left(\lambda \right)}^{\lambda -\mu }}$ a matrix $A{\left(\lambda \right)}^{\lambda -\mu }$ defined by $A(λ)λ-μ= (⟨x(s)v+,x(t)v+⟩) βs1+⋯+βsk=μ =βt1+⋯+βtj .$ According to Lemmas 2.4.1 and 2.4.2, the matrix $A{\left(\lambda \right)}^{\lambda -\mu }$ is degenerate if and only if $J{\left(\lambda \right)}^{\lambda -\mu }\ne 0\text{.}$ Therefore we define $det M(λ)λ-μ= det A(λ)λ-μ.$ Then $\text{det} M{\left(\lambda \right)}^{\lambda -\mu }=0$ if and only if $J{\left(\lambda \right)}^{\lambda -\mu }\ne 0\text{.}$

Example. Consider ${sl}_{2}\left(ℂ\right),$ with the standard basis $\left\{h,e,f\right\}$ where $[h,e]=2e, [h,f]=-2f, [e,f]=h.$ The Cartan subalgebra is $𝔥=ℂh$ and let $\alpha \in {𝔥}^{*}$ so that $\alpha \left(h\right)=1\text{.}$ The anti-automorphism $\sigma :s{l}_{2}\left(ℂ\right)\to {sl}_{2}\left(ℂ\right)$ is given by $σ(h)=h, σ(e)=f, σ(f)=e.$ Let $\lambda \in {𝔥}^{*}$ and $n\in {ℤ}_{\ge 0}\text{.}$ Then $M{\left(\lambda \right)}^{\lambda -n\alpha }$ has a basis ${f}^{n}{v}^{+}\text{.}$ It is straightforward to check that $e{f}^{j}={f}^{j}e+j{f}^{j-1}\left(h-j+1\right),$ and so $det M(λ)μ = det(⟨fnv+,fnv+⟩) = ⟨v+,enfnv+⟩ = ∏j=1n j(λ(h)-j+1).$ More generally, Shapovalov [Sha1972] computed the determinant $\text{det} M{\left(\lambda \right)}^{\lambda -\mu }$ for any finite-dimensional semisimple Lie algebra $𝔤\text{:}$ $det M(λ)λ-μ= ∏β∈R+,r>0 (⟨ω+ρ,β⟩-rr) p(μ-rβ) ,$ where $p\left(\gamma \right)$ is the number of ways that $\gamma$ can be written as a sum of positive roots. As we will see, Kac also found an explicit formula for the determinant $M{\left(h,c\right)}^{\left(h+n,c\right)}$ for the Virasoro algebra (Theorem 3.4.3).

### Jantzen Filtrations

Let $t$ be an indeterminant. Define $𝔤\left[t\right]=ℂ\left[t\right]{\otimes }_{ℂ}𝔤,$ with the Lie bracket on $𝔤\left[t\right]$ given by $[p(t)⊗x,q(t)⊗y]= p(t)q(t)⊗[x,y].$ For $\lambda ,\gamma \in {𝔥}^{*},$ the Verma module $M\left(\lambda +t\gamma \right)$ is given by $M(λ+tγ)= U(𝔤[t])/I$ where $I$ is the left $U\left(𝔤\left[t\right]\right)\text{-ideal}$ generated by ${𝔤}_{>0}$ and $\left\{h-\left(\lambda +t\gamma \right)\left(h\right) | h\in 𝔥\right\}\text{.}$ Let $\epsilon :ℂ\left[t\right]\to ℂ$ be the $ℂ\text{-linear}$ map given by $t↦0\text{.}$ This can be extended to surjective homomorphisms $\epsilon :𝔤\left[t\right]\to 𝔤$ and $\epsilon :M\left(\lambda +t\delta \right)\to M\left(\lambda \right)\text{.}$

We can extend the Shapovalov form to a form ${⟨,⟩}_{\left(\lambda +t\gamma \right)}:M\left(\lambda +t\delta \right)×M\left(\lambda +t\delta \right)\to ℂ\left[t\right]\text{.}$ Define $M(λ+tγ)j= { v∈M(λ+tγ) | tj divides ⟨v,w⟩(λ+tγ) for all w∈M(λ+tγ) } .$ Then $M\left(\lambda +t\gamma \right)$ is a $𝔤\left[t\right]\text{-submodule}$ of $M\left(\lambda +t\gamma \right)\text{.}$ Define $M(λ)j=ε (M(λ+tγ)j)$ The Jantzen filtration of $M\left(\lambda \right)$ is $M(λ)=M (λ)0⊇ M(λ)1⊇ M(λ)2⊇⋯.$ For $x\in ℂ\left[t\right],$ define the order of $x$ by $\text{ord}\left(x\right)=k$ if ${t}^{k}|x$ and ${t}^{k+1}\nmid x\text{.}$

(See [MPi1995], Theorem 2.9.4) Suppose $\gamma \in {𝔥}^{*}$ is chosen so that ${⟨,⟩}_{\left(\lambda +t\gamma \right)}$ is nondegenerate. Then,

 • for $\mu \le \lambda ,$ $ord(det M(λ+tγ)(μ+tγ)) =∑j=1∞dim M(λ)jμ;$ • $M{\left(\lambda \right)}_{1}=J\left(\lambda \right),$ the maximal proper submodule of $M\left(\lambda \right)\text{;}$ • $\bigcap _{j=0}^{\infty }M{\left(\lambda \right)}_{j}=0\text{.}$

### Translation Functors

Let $𝔤$ be a Lie algebra with a regular, finite triangular decomposition. Recall that Category $𝒪$ is closed under tensor products. Theorem 2.3.12 then implies that $M\left(\lambda \right)\otimes L\left(\mu \right)$ decomposes by blocks. In this section, we consider the translation $M(λ)↦ (M(λ)⊗L(μ))[ν].$

Fix $\lambda ,\mu \in {𝔥}^{*}\text{.}$ For $\left[\nu \right]\in \left[{𝔥}^{*}\right],$ consider the set $[ν]∩ { γ∈𝔥* | λ+μ-γ∈ Q+,L (μ)γ-λ ≠0 } .$ If this set is nonempty, enumerate the elements ${\nu }_{1},{\nu }_{2},\dots$ so that ${\nu }_{i}\ngtr {\nu }_{j}$ for $i>j\text{.}$ The following proposition is a corollary of Theorem 2.3.12.

Let $\lambda ,\mu \in {𝔥}^{*}\text{.}$ Then, $M(λ)⊗L(μ)= ⨁[ν]∈[𝔥*] (M(λ)⊗L(μ))[ν].$ The module ${\left(M\left(\lambda \right)\otimes L\left(\mu \right)\right)}^{\left[\nu \right]}\in {𝒪}^{\left[\nu \right]}$ has a filtration $0=M0⊆M1⊆⋯$ such that ${\left(M\left(\lambda \right)\otimes L\left(\mu \right)\right)}^{\left[\nu \right]}=\bigcup {M}_{i}$ and ${M}_{i}/{M}_{i-1}\cong M{\left({\nu }_{i}\right)}^{\oplus \text{dim}L{\left(\mu \right)}^{{\nu }_{i}-\lambda }}\text{.}$ Proof. Let ${v}^{+}$ be a generator of $M\left(\lambda \right)\text{.}$ Enumerate the weights of $L\left(\mu \right)$ by ${\nu }_{1},{\nu }_{2},\dots$ so that ${\nu }_{i}\ngtr {\nu }_{j}$ for $i>j\text{.}$ For each weight ${\nu }_{i}$ of $L\left(\mu \right),$ let $\left\{{w}_{{\nu }_{i},j} | 1\le j\le \text{dim} L{\left(\mu \right)}^{{\nu }_{i}}\right\}$ be a basis for $L{\left(\mu \right)}^{{\nu }_{i}}\text{.}$ Define ${M}_{{\nu }_{i},j}$ to be the submodule of $M\left(\lambda \right)\otimes L\left(\mu \right)$ generated by ${v}^{+}\otimes {w}_{{\nu }_{i},j}\text{.}$ Then $0⊆ Mν1,1⊆ Mνi,2⋯⊆ Mν2,1⊆⋯$ gives a filtration by Verma modules. We then apply the proof of Theorem 2.3.12 to get the decomposition. $\square$

Recall that $𝔤$ is a Hopf algebra with coproduct $\mathrm{\Delta }\left(x\right)=x\otimes 1+1\otimes x\text{.}$ This allows us to define a contravariant form on the tensor product $M\left(\lambda \right)\otimes L\left(\mu \right)$ using the contravariant forms on $M\left(\lambda \right)$ and $L\left(\mu \right)\text{.}$ For $\nu ,\nu \prime \in M\left(\lambda \right)$ and $w,w\prime \in L\left(\mu \right),$ define $⟨,⟩:M\left(\lambda \right)\otimes L\left(\mu \right)×M\left(\lambda \right)\otimes L\left(\mu \right)\to ℂ$ by $⟨v⊗w,v′,w′⟩= ⟨v,v′⟩ ⟨w,w′⟩.$ This form will be contravariant: $⟨x(v⊗w),v′⊗w′⟩ = ⟨ xv⊗w+ v⊗xw, v′⊗w′ ⟩ = ⟨xv⊗w,v′⊗w′⟩+ ⟨v⊗xw,v′⊗w′⟩ = ⟨xv,v′⟩ ⟨w,w′⟩+ ⟨v,v′⟩ ⟨xw,w′⟩ = ⟨v,σ(x)v′⟩ ⟨w,w′⟩+ ⟨v,v′⟩ ⟨w,σ(x)w′⟩ = ⟨v⊗w,σ(x)v′⊗w′+v′⊗σ(x)w′⟩ = ⟨v⊗w,σ(x)(v′⊗w′)⟩$ for any $x\in 𝔤\text{.}$

Let $\lambda ,\mu \in {𝔥}^{*}\text{.}$ For $\nu \in {Q}_{+},$ let $\left\{{w}_{\nu ,i} | 1\le i\le \text{dim} L{\left(\mu \right)}^{\mu -\nu }\right\}$ be a basis for $L{\left(\mu \right)}^{\mu -\nu }\text{.}$ For $\gamma \in {Q}_{+},$ the following sets are bases for ${\left(M\left(\lambda \right)\otimes L\left(\mu \right)\right)}^{\lambda +\mu -\gamma }\text{:}$ $ℬ1= { x(t)v+⊗ wν,i | 1≤i≤dim L (μ)μ-ν, (t):t1≥⋯≥ tj>0,βt1 +⋯+βtj= γ-ν } ; (2.5)$ $ℬ2= { x(t) (v+⊗wν,i) | 1≤i≤dim L (μ)μ-ν, (t):t1≥⋯≥ tj>0,βt1 +⋯+βtj= γ-ν } . (2.6)$ Moreover, the transition matrix between these two bases has determinant $1\text{.}$ Proof. We have $x(t) (v+⊗wν,i)= x(t)v+⊗ wν,i+ ∑ζ>ν1≤j≤dim L(μ)μ-ζ(s)|βs1+βsk=γ-ζ x(s)v+⊗ wζ,j.$ By appropriately ordering the elements of (2.5) and (2.6), we see that the transition matrix taking (2.6) to (2.5) is upper triangular with ones on the diagonal. Therefore, this matrix is invertible with determinant 1. $\square$

Define $det L(μ)μ-ν =det (⟨wνi,wνj⟩) 1≤i,j≤dim L(μ)μ-ν .$ Note that $\text{det} L{\left(\mu \right)}^{\mu -\nu }\ne 0\text{.}$ Also, define $det(M(λ)⊗L(μ))λ+μ-γ= det (⟨x(s)⊗wν,i,x(t)v+⊗wζ,j⟩) x(s)v+⊗ wν,i,x(t) v+⊗wζ,j∈ ℬ1 .$

Let $\lambda ,\mu \in {𝔥}^{*}$ and $\gamma \in {Q}_{+}\text{.}$ Then $det(M(λ)⊗L(μ))λ+μ-γ= ∏ν+ζ=γ (det M(λ)λ-ν) dim L(μ)μ-ζ (det L(μ)μ-ζ)p(ν).$ Proof. Enumerate the weights between $0$ and $\gamma :$ ${\zeta }_{1},\dots ,{\zeta }_{n}\text{.}$ We have shown that $M{\left(\lambda \right)}^{\lambda -\nu }\perp M{\left(\lambda \right)}^{\lambda -\nu \prime }$ for $\nu \ne \nu \prime$ and $L{\left(\mu \right)}^{\mu -\zeta }\perp L{\left(\mu \right)}^{\mu -\zeta \prime }$ for $\zeta \ne \zeta \prime \text{.}$ Therefore, the matrix $\left(⟨{x}_{\left(s\right)}{v}^{+}\otimes {w}_{\nu ,i},{x}_{\left(t\right)}{v}^{+}\otimes {w}_{\zeta ,j}⟩\right)$ has the form $( Bζ1 0 0 0 ⋱ 0 0 0 Bζn ) ,$ where ${B}_{{\zeta }_{i}}$ is the matrix $(A(λ)λ-(γ-ζi)⟨wζi,j,wζi,k⟩) 1≤j,k≤dim L(μ)μ-ζi .$ Using Lemma 1.3.1, we have $\text{det}\left({B}_{{\zeta }_{i}}\right)={\left(\text{det}\left(A{\left(\lambda \right)}^{\lambda -\left(\gamma -{\zeta }_{i}\right)}\right)\right)}^{\text{dim} L{\left(\mu \right)}^{\mu -{\zeta }_{i}}}{\left(\text{det} L{\left(\mu \right)}^{\mu -{\zeta }_{i}}\right)}^{p\left(\nu \right)}\text{.}$ $\square$

Suppose $\left[\nu \right],\left[\gamma \right]\in \left[{𝔥}^{*}\right]$ with $\left[\nu \right]\ne \left[\gamma \right]\text{.}$ Then, ${\left(M\left(\lambda \right)\otimes L\left(\mu \right)\right)}^{\left[\nu \right]}\perp {\left(M\left(\lambda \right)\otimes L\left(\mu \right)\right)}^{\left[\gamma \right]}$ respect to the contravariant form. Proof. First we write down a basis for ${\left(M\left(\lambda \right)\otimes L\left(\mu \right)\right)}^{\left[\nu \right]}\text{.}$ Enumerate the set $[ν]∩ { γ∈𝔥* | λ+ μ-γ∈Q+,L (μ)γ-λ≠0 } .$ so that ${\nu }_{i}\ngtr {\nu }_{j}$ for $i>j\text{.}$ ${ νi∈Q+ | λ+μ-νi∈[ν] L(μ)μ-νi ≠0 } .$ so that ${\nu }_{i}\nless {\nu }_{j}$ for $i>j\text{.}$ Given the decomposition in Proposition 2.6.1, we can choose a set $\left\{{v}_{{\nu }_{i},j} | 1\le j\le \text{dim} L{\left(\mu \right)}^{\mu -{\nu }_{i}}\right\}\subseteq {\left(M{\left(\lambda ,\mu \right)}^{\left[\nu \right]}\right)}^{\lambda +\mu -{\nu }_{i}}$ so that for each weight space ${\left(M\left(\lambda \right)\otimes L\left(\mu \right)\right)}^{\lambda +\mu -\delta },$ the set $⋃νi<δ { x(s)vνi,j | (s):s1≥ ⋯≥sk>0,βs1+ ⋯+βsk=δ-νi ,1≤j≤dim L(μ)μ-νi }$ is a basis for ${\left({\left(M\left(\lambda \right)\otimes L\left(\mu \right)\right)}^{\left[\nu \right]}\right)}^{\lambda +\mu -\delta }\text{.}$ We will show $((M(λ)⊗L(μ))[ν]) λ+μ-δ ⊥ ((M(λ)⊗L(μ))[γ]) λ+μ-δ (2.7)$ by inducting on $\text{max}\left\{\text{ht}\left(\delta -{\nu }_{i}\right) | {\nu }_{i}\in \left[\nu \right]\right\}\text{.}$ To prove the base case, we need to show that (2.7) holds for all $\delta ={\nu }_{i}$ such that ${\nu }_{i}\ngtr {\nu }_{j}$ for any $j\text{.}$ If ${\left({\left(M\left(\lambda \right)\otimes L\left(\mu \right)\right)}^{\left[\gamma \right]}\right)}^{\lambda +\mu -{\nu }_{i}}=0,$ the result is obvious. Therefore, assume ${\left({\left(M\left(\lambda \right)\otimes L\left(\mu \right)\right)}^{\left[\gamma \right]}\right)}^{\lambda +\mu -{\nu }_{i}}\ne 0\text{.}$ Since $\lambda +\mu -{\nu }_{i}\notin \left[\gamma \right],$ ${\left({\left(M\left(\lambda \right)\otimes L\left(\mu \right)\right)}^{\left[\gamma \right]}\right)}^{\lambda +\mu -{\nu }_{i}}$ has a basis $\left\{{x}_{\left(s\right)}{v}_{{\gamma }_{k},l} | {\gamma }_{k}>{\nu }_{i},\left(s\right):{s}_{1}\ge \cdots \ge {s}_{l}>0,{\beta }_{{s}_{1}}+\cdots +{\beta }_{{s}_{l}}={\gamma }_{k}-{\nu }_{i}\right\}\text{.}$ Since ${\nu }_{i}\nless {\nu }_{j}$ for any $j,$ ${\left({\left(M\left(\lambda \right)\otimes L\left(\mu \right)\right)}^{\left[\nu \right]}\right)}^{\lambda +\mu -{\nu }_{i}}$ has a basis $\left\{{v}_{{\nu }_{i},j}\right\}\text{.}$ We then have $⟨x(s)vγk,l,vνi,j⟩= ⟨vγk,l,σ(x(s))vνi,j⟩ =0$ since $\sigma \left({x}_{\left(s\right)}\right){v}_{{\nu }_{i},j}\in {\left({\left(M\left(\lambda \right)\otimes L\left(\mu \right)\right)}^{\left[\nu \right]}\right)}^{\lambda +\mu -\left({\nu }_{i}-\left({\beta }_{{s}_{1}}+\cdots +{\beta }_{{s}_{k}}\right)\right)}=0\text{.}$ For general $\delta \in {Q}_{+},$ ${\left({\left(M\left(\lambda \right)\otimes L\left(\mu \right)\right)}^{\left[\nu \right]}\right)}^{\lambda +\mu -\delta }$ has a basis ${ x(s)vνi,j | νi<δ,(s) :s1≥⋯≥sk>0, βs1+⋯+βsk =δ-νi } ⏟S1 ∪ {vνi,j | νi=δ}⏟S2.$ For ${v}_{{\nu }_{i},j}\in {S}_{2},$ we have $⟨{x}_{\left(s\right)}{v}_{{\gamma }_{k},l},{v}_{{\nu }_{i},j}⟩=⟨{v}_{{\gamma }_{k},l},\sigma \left({x}_{\left(s\right)}{v}_{{\nu }_{i},j}\right)⟩\text{.}$ Since $\sigma \left({x}_{\left(s\right)}\right){v}_{{\nu }_{i},j}\in {\left({\left(M\left(\lambda \right)\otimes L\left(\mu \right)\right)}^{\left[\nu \right]}\right)}^{\lambda +\mu -\left({\nu }_{i}-\left({\beta }_{{s}_{1}}+\cdots +{\beta }_{{s}_{k}}\right)\right)}$ we can use the induction hypothesis to conclude $⟨{x}_{\left(s\right)}{v}_{{\gamma }_{k},l},{v}_{{\nu }_{i},j}⟩=0\text{.}$ For ${x}_{\left(s\right)}{v}_{{\nu }_{i},j}\in {S}_{1}$ and $v\in {\left({\left(M\left(\lambda \right)\otimes L\left(\mu \right)\right)}^{\left[\gamma \right]}\right)}^{\lambda +\mu -\delta },$ $⟨x(s)vνi,j,v⟩= ⟨vνi,j,σ(x(s))v⟩=0$ by the induction hypothesis. $\square$

## Notes and References

This is an excerpt from the PhD thesis Translation Functors and the Shapovalov Determinant by Emilie Wiesner, University of Wisconsin-Madison, 2005.