## 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.

## Basics and Background

### Some Definitions for Lie Algebras

Let $𝔤$ be a Lie algebra over a field $𝔽\text{.}$ We view $𝔤$ as a $𝔤\text{-module}$ via the adjoint action $\text{ad}:𝔤×𝔤\to 𝔤$ given by $ad(x)y= [x,y].$ Suppose $𝔤$ is finite-dimensional. The Killing form $\kappa \left(,\right):𝔤×𝔤\to 𝔽$ is given by $κ(x,y)=Tr (ad(x)ad(y)),$ where Tr is the trace. The Killing form is invariant, that is, $\kappa \left(\left[x,y\right],z\right)=-\kappa \left(y,\left[x,z\right]\right)\text{.}$

Fix a Lie algebra $𝔤\text{.}$ Write ${𝔤}^{\otimes n}=𝔤\otimes \cdots \otimes 𝔤$ $\text{(}n$ times). The universal enveloping algebra of $𝔤$ is the associative algebra $U(𝔤)=⨁n∈ℤ≥0 g⊗n/ ⟨x⊗y-y⊗x-[x,y] | x,y∈𝔤⟩.$ We can view $U\left(𝔤\right)$ as a Lie algebra $\text{Lie}\left(U\left(𝔤\right)\right)$ with bracket $\left[u,v\right]=uv-vu$ for $u,v\in U\left(𝔤\right)\text{.}$ Then the defining relation of $U\left(𝔤\right)$ implies that there is a Lie algebra homomorphism $𝔤\to \text{Lie}\left(U\left(𝔤\right)\right)\text{.}$

([Dix1996] 2.1.11). Let $𝔤$ be a Lie algebra with basis $\left\{{x}_{i} | i\in I\right\},$ where $I$ is a totally ordered set. Then the following set is a basis for $U\left(𝔤\right)\text{:}$ ${ xi1⋯ xin | i1≤⋯≤in } .$

Any Lie algebra $𝔤$ has a Hopf algebra structure given by

 coproduct: $\mathrm{\Delta }\left(x\right)=x\otimes 1+1\otimes x\text{;}$ counit: $\epsilon \left(x\right)=0\text{;}$ antipode: $s\left(x\right)=-x\text{.}$

### Semisimple Lie Algebras

In this section we give a brief introduction to semisimple Lie algebras. More on semisimple Lie algebras can be found in [Dix1996] and [Hum1978].

A Lie algebra $𝔤$ is simple if $\text{dim} 𝔤>1$ and the only ideals of $𝔤$ are 0 and $𝔤\text{.}$ A Lie algebra $𝔤$ is semisimple if $𝔤$ is the direct sum of simple Lie algebras.

Define ${𝔤}^{\left(1\right)}=\left[𝔤,𝔤\right]$ and ${𝔤}^{\left(n\right)}=\left[{𝔤}^{\left(n-1\right)},{𝔤}^{\left(n-1\right)}\right]\text{.}$ The radical of $𝔤,$ $\text{Rad} 𝔤,$ is the largest ideal of $𝔤$ such that ${\left(\text{Rad} 𝔤\right)}^{\left(n\right)}=0$ for some $n\text{.}$

([Dix1996], 1.5.2 and 1.5.12). Let $𝔤$ be a Lie algebra over $ℂ\text{.}$ The following are equivalent:

 • $𝔤$ is semisimple; • the Killing form $\kappa \left(,\right)$ on $𝔤$ is nondegenerate; • $\text{Rad} 𝔤=0\text{.}$

Let $𝔤$ be a finite-dimensional complex semisimple Lie algebra. Then $𝔤$ contains a unique maximal abelian subalgebra $𝔥$ (with respect to inclusion) such that the linear transformations $\text{ad}\left(h\right),$ $h\in 𝔥,$ are simultaneously diagonalizable on $𝔤\text{.}$ We call $𝔥$ the Cartan subalgebra of $𝔤\text{.}$

Let ${𝔥}^{*}={\text{Hom}}_{ℂ}\left(𝔥,ℂ\right)\text{.}$ Define $𝔤α= {x∈𝔤 | ad(h)x=α(h)x}$ and note that $𝔤={⨁}_{\alpha \in {𝔥}^{*}}{𝔤}_{\alpha }\text{.}$ A nonzero element $\alpha \in {𝔥}^{*}$ is a root of $𝔤$ if ${𝔤}_{\alpha }\ne 0\text{.}$ The set of roots of $𝔤$ is denoted $R\text{.}$

The Killing form $\kappa \left(,\right)$ is nondegenerate on $𝔥\text{.}$ Therefore, we can identify an element $h\in 𝔥$ with an element $\alpha \in {𝔥}^{*}$ so that $\alpha \left(x\right)=\kappa \left(x,h\right)$ for all $x\in 𝔥\text{.}$ We use the Killing form to define a form $⟨,⟩:{𝔥}^{*}×{𝔥}^{*}\to ℂ$ on ${𝔥}^{*}\text{.}$ For $\alpha ,\alpha \prime \in {𝔥}^{*},$ $⟨α,α′⟩= κ(h,h′),$ where $\alpha \left(x\right)=\kappa \left(x,h\right)$ and $\alpha \prime \left(x\right)=\kappa \left(x,h\prime \right)$ for all $x\in 𝔥\text{.}$

We choose a basis of ${𝔥}^{*}$ of simple roots ${\alpha }_{1},\dots ,{\alpha }_{k}\in R$ so that for each $\beta \in R$ we can write $\beta =\sum _{i}{k}_{i}{\alpha }_{i},$ where either ${k}_{i}\ge 0$ for all $i$ or ${k}_{i}\le 0$ for all $i\text{.}$ Given a choice of simple roots, the positive roots are $R+= { β∈R | β= ∑ikiαi, ki≥0 } .$ The height function $\text{ht}:R\to ℤ$ is given by $ht(β)=∑iki ,if α=∑i kiαi.$ The highest root is the unique element $\theta \in {R}^{+}$ such that $\text{ht}\left(\theta \right)\ge \text{ht}\left(\beta \right)$ for all $\beta \in R\text{.}$ Define $ρ=12∑β∈R+β.$ The dual Coxeter number $g$ is $g=12 ⟨θ,θ+2ρ⟩.$

Define ${\alpha }_{i}^{\vee }=\frac{2{\alpha }_{i}}{⟨{\alpha }_{i},{\alpha }_{i}⟩}$ and let ${h}_{i}\in 𝔥$ so that ${\alpha }_{i}^{\vee }\left(x\right)=\kappa \left(x,{h}_{i}\right)$ for all $x\in 𝔥\text{.}$

([Dix1996], 1.10.2). Let $𝔤$ be a finite-dimensional semisimple Lie algebra. Then, $𝔤$ is generated by elements ${e}_{i}\in {𝔤}_{{\alpha }_{i}},$ ${f}_{i}\in {𝔤}_{-{\alpha }_{i}},$ and ${h}_{i}\in 𝔥,$ $1\le i\le k,$ with relations

 (1) $\left[{h}_{i},{h}_{j}\right]=0\text{;}$ (2) $\left[{h}_{i},{e}_{j}\right]={a}_{ij}{e}_{j},$ $\left[{h}_{i},{f}_{j}\right]=-{a}_{ij}{f}_{j}\text{;}$ (3) $\left[{e}_{i},{f}_{j}\right]={\delta }_{i,j}{h}_{i}\text{;}$ (4) $\underset{\underset{1-{a}_{ij} \text{times}}{⏟}}{\left[{e}_{i},\cdots \left[{e}_{i}},{e}_{j}\right]\cdots \right]=0,$ $\underset{\underset{1-{a}_{ij} \text{times}}{⏟}}{\left[{f}_{i},\cdots \left[{f}_{i}},{f}_{j}\right]\cdots \right]=0,$ if $i\ne j$
where ${a}_{ij}=⟨{\alpha }_{j},{\alpha }_{i}^{\vee }⟩={\alpha }_{j}\left({h}_{i}\right)\text{.}$

The generators ${h}_{1},\dots ,{h}_{k}$ are a basis for $𝔥\text{.}$ Let ${𝔤}_{>0}$ be the subalgebra of $𝔤$ generated by ${e}_{1},\dots ,{e}_{k}$ and ${𝔤}_{<0}$ the subalgebra generated by ${f}_{1},\dots ,{f}_{k}\text{.}$ The previous proposition implies $𝔤=𝔤>0⊗𝔥⊗ 𝔤<0. (1.1)$ The Cartan matrix associated to $𝔤$ is $A={\left({a}_{ij}\right)}_{1\le i\le k}\text{.}$

We define the following subsets of ${𝔥}^{*}\text{:}$

 • the set of integral weights of $𝔤\text{:}$ $P=\left\{\lambda \in {𝔥}^{*} | ⟨\lambda ,{\alpha }^{\vee }⟩\in ℤ\right\}\text{;}$ • the set of dominant integral weights of $𝔤\text{:}$ ${P}_{+}=\left\{\lambda \in {𝔥}^{*} | ⟨\lambda ,{\alpha }^{\vee }⟩\in {ℤ}_{\ge 0}\right\}\text{;}$ • the root lattice of $𝔤\text{:}$ $Q=∑1≤i≤k ℤαi;$ • the positive root lattice: $Q+=∑1≤i≤k ℤ≥0αi.$
The dominance ordering on ${𝔥}^{*}$ is the partial ordering on ${𝔥}^{*}$ given by $\lambda \ge \mu$ if $\lambda -\mu \in {Q}_{+}\text{.}$

Associated to $𝔤$ is a Weyl group $W\text{.}$ The Weyl group is the group generated by reflections ${s}_{i},$ $1\le i\le k,$ which act on ${𝔥}^{*}$ by ${s}_{i}\left(\lambda \right)=\lambda -⟨\lambda ,{\alpha }_{i}^{\vee }⟩{\alpha }_{i}\text{.}$ It is straightforward to check that $⟨αi,slαj⟩= ⟨slαi,αj⟩ (1.2)$ for $1\le i,j,l\le k\text{.}$ The dot action of $W$ is given by ${s}_{i}\circ \lambda ={s}_{i}\left(\lambda +\rho \right)-\rho \text{.}$

Recall $𝔤$ is finite-dimensional. Suppose $\text{dim} 𝔤=n\text{.}$ Let $\left\{{u}_{i} | 1\le i\le n\right\}$ be a basis of $𝔤,$ and let $\left\{{u}^{i} | 1\le i\le n\right\}$ be a dual basis with respect to $\kappa \left(,\right),$ i.e. $\kappa \left({u}_{i},{u}^{j}\right)={\delta }_{i,j}\text{.}$ Define the Casimir element ${\mathrm{\Omega }}_{0}\in U\left(𝔤\right)$ by $Ω0=∑i=1n uiui.$

([Hum1978], 22.1-22.2). For $x\in 𝔤,$

 • $\left[{\mathrm{\Omega }}_{0},x\right]=0\text{.}$ • $\text{ad}\left({\mathrm{\Omega }}_{0}\right)x=\sum _{i=1}^{n}\left[{u}_{i},\left[{u}^{i},x\right]\right]=2gx,$ where $g=\frac{1}{2}⟨\theta ,\theta +2\rho ⟩$ is the dual Coxeter number of $𝔤\text{.}$

For any choice of dual bases $\left\{{u}_{i}\right\}$ and $\left\{{u}^{i}\right\},$ we can write $x\in 𝔤$ as $x=∑i=1nκ (x,ui)ui. (1.3)$ Using this, it is easy to check that ${\mathrm{\Omega }}_{0}$ is independent of the initial choice of basis $\left\{{u}_{i}\right\}\text{.}$ In particular, we can switch the roles of ${u}_{i}$ and ${u}^{i}$ to get $∑i=1n [ui,ui]= ∑i=1n uiui- ∑i=1n uiui=0. (1.4)$

#### Reductive Lie Algebras

A natural weakening of the definition of a semisimple Lie algebra is a reductive Lie algebra. A Lie algebra $𝔤$ is reductive if it is the direct sum of an abelian Lie algebra and a semisimple Lie algebra. Note that the Killing form is not necessarily nondegenerate on $𝔤\text{.}$ However, it is still possible to define a nondegenerate form on $𝔤$ ([Dix1996], 1.7.3).

### Odds and Ends

#### Bilinear Forms

Let $V$ and $W$ be $ℂ$ vector spaces. Then a bilinear form on $V$ and $W$ is a map $⟨,⟩:V×W\to ℂ$ such that for all $v,v\prime \in V,$ $w,w\prime \in W$ and $\alpha \in ℂ,$ $⟨αv+v′,w⟩ = α⟨v,w⟩+ ⟨v′,w⟩ ⟨v,αw+w′⟩ = α⟨v,w⟩+ ⟨v,w′⟩.$ A Hermitian bilinear form is a map $⟨,⟩:V×W\to ℂ$ such that for all $v,v\prime \in V,$ $w,w\prime \in W$ and $\alpha \in ℂ,$ $⟨αv+v′,w⟩ = α‾⟨v,w⟩ +⟨v′,w⟩, ⟨v,αw+w′⟩ = α⟨v,w⟩+ ⟨v,w′⟩.$ Here, $\stackrel{‾}{\alpha }$ is the complex conjugate of $\alpha \text{.}$

Let $q$ be an indeterminant. For vector spaces $V$ and $W$ over $ℚ\left(q\right),$ we will also say a map $⟨,⟩:V×W\to ℚ\left(q\right)$ is a Hermitian bilinear form if for all $v,v\prime \in V,$ $w,w\prime \in W$ and $\alpha \in ℚ\left(q\right),$ $⟨αv+v′,w⟩ = α‾⟨v,w⟩+ ⟨v′,w⟩; ⟨v,αw+w′⟩ = α⟨v,w⟩+ ⟨v,w′⟩$ where $\stackrel{‾}{·}:ℚ\left(q\right)\to ℚ\left(q\right)$ is the $ℚ\text{-automorphism}$ sending $q$ to ${q}^{-1}\text{.}$

#### Derivations

For an algebra $A,$ a derivation on $A$ is a linear map $D:A\to A$ such that for any $x,y\in A,$ $D(xy)= D(x)y+ xD(y).$ The set of derivations $𝒟$ on $A$ naturally has a Lie algebra structure. That is, we can define a bracket $\left[,\right]:𝒟×𝒟\to 𝒟$ by $\left[D,D\prime \right]=DD\prime -D\prime D\text{.}$

#### A Lemma for Determinants

Suppose $A\in {M}_{n}\left(ℂ\right)$ and $B\in {M}_{m}\left(ℂ\right)$ with $B=\left({b}_{kl}\right)\text{.}$ Define $C= ( Ab11 ⋯ Ab1m ⋮ ⋱ ⋮ Abm1 ⋯ Abmm ) .$ Then, $det A⊗B= (det A)m (det B)n.$

 Proof. If ${b}_{1i}=0$ for all $1\le i\le m,$ then $\text{det} C=0$ and $\text{det} B=0\text{;}$ then the lemma holds. Therefore, we may assume ${b}_{1i}\ne 0$ for at least one $1\le i\le m\text{.}$ Since a rearrangement of the columns of a matrix does not affect its determinant, we may assume ${b}_{11}\ne 0\text{.}$ We prove the result by inducting on $m\text{.}$ Now, $det A⊗B = det ( Ab11 Ab12 ⋯ Ab1m 0 A(b22-b12b21b11) ⋯ A(b2m-b1mb21b11) ⋮ ⋮ 0 A(bm2-b12bm1b11) ⋯ A(bmm-b1mbm1b11) ) = (b11)n det A×det ( A(b22-b12b21b11) ⋯ A(b2m-b1mb21b11) ⋮ ⋮ A(bm2-b12bm1b11) ⋯ A(bmm-b1mbm1b11) ) = (b11)n det A (det A)m-1det ( (b22-b12b21b11) ⋯ (b2m-b1mb21b11) ⋮ ⋮ (bm2-b12bm1b11) ⋯ (bmm-b1mbm1b11) ) n ⏟By the induction hypothesis = (det A)m ( b11det ( (b22-b12b21b11) ⋯ (b2m-b1mb21b11) ⋮ ⋮ (bm2-b12bm1b11) ⋯ (bmm-b1mbm1b11) ) ) n = (det A)m (det B)n.$ $\square$

#### Projectives and Extensions

Let $R$ be a ring and $𝒞$ an abelian category of left $R$ modules.

A module $P\in 𝒞$ is projective in the category $𝒞$ if whenever $A,B\in 𝒞$ with a map $\alpha :P\to A$ and a surjective map $\beta :B\to A,$ then there exists $\gamma :P\to B$ such that $\beta \circ \gamma =\alpha \text{.}$ In other words, we have the following diagram: $P {}^{\gamma } ↓α B ⟶β A ⟶ 0$ For $M\in 𝒞,$ a projective cover is a projective module $P\in 𝒞$ along with a surjective homomorphism $\varphi :P\to M$ such that $\varphi \left(P\prime \right)\ne M$ for any proper submodule $P\prime$ of $P\text{.}$

A projective cover $P$ of $M\in 𝒞$ is unique up to isomorphism.

 Proof. Suppose that $P$ and $\stackrel{\sim }{P}$ are both projective covers of $M\text{.}$ We then have $P {}^{\gamma } ↓ P∼ ⟶ M ⟶ 0 P∼ {}^{\stackrel{\sim }{\gamma }} ↓ P ⟶ M ⟶ 0$ By definition of a projective cover, both $\gamma$ and $\stackrel{\sim }{\gamma }$ must be surjective. Therefore, $\stackrel{\sim }{\gamma }\circ \gamma =\text{id}{|}_{P}$ and so $\gamma$ is an isomorphism. $\square$

Suppose $P,M,M\prime \in 𝒞$ with $P$ projective and $M\prime \subseteq M\text{.}$ Then $HomR(P,M/M′)= HomR(P,M)/ HomR(P,M′)$ and $dim HomR(P,M/M′)= dim HomR(P,M)- dim HomR(P,M′).$

 Proof. Clearly, we have a map $\stackrel{‾}{·}:{\text{Hom}}_{R}\left(P,M\right)\to {\text{Hom}}_{R}\left(P,M/M\prime \right),$ where $\varphi :P\to M$ is sent to $\stackrel{‾}{\varphi }:P\to M\to M/M\prime \text{.}$ Then $\stackrel{‾}{\varphi }=0$ implies that $\varphi \left(P\right)\subset M\prime ,$ and so $\varphi$ is a map from $P$ to $M\prime \text{.}$ Thus, $\text{ker}\left(\stackrel{‾}{·}\right)\subset {\text{Hom}}_{R}\left(P,M\prime \right)\text{.}$ For the other direction, note that a homomorphism from $P$ to $M/M\prime$ can be lifted to a homomorphism from $P$ to $M$ since $P$ is projective. $\square$

Let $A,B\in 𝒞\text{.}$ An extension of $A$ by $B$ is a short exact sequence $0⟶B⟶E⟶A⟶0,$ where $E\in 𝒪\text{.}$ Two extensions $0⟶B⟶E⟶A⟶0,$ $0⟶B⟶E′⟶A⟶0,$ are equivalent if there exists $\gamma :E\to E\prime$ such that the following diagram commutes. $0 ⟶ B ⟶ E ⟶ A ⟶ 0 ↓id ↓γ ↓id 0 ⟶ B ⟶ E′ ⟶ A ⟶ 0$ Define ${\text{Ext}}^{1}\left(A,B\right)$ to be the set of equivalence classes of extensions of $A$ by $B\text{.}$

([MPi1995], 1.12.4, 1.12.4'). Let $B,B\prime ,B″,A\in 𝒞,$ and suppose $0⟶B⟶B′⟶B″⟶0$ is exact. Then, $0⟶ HomR(A,B)⟶ HomR(A,B′)⟶ HomR(A,B″)δ⟶ Ext1(A,B)⟶ Ext1(A,B′)⟶ Ext1(A,B″)$ is exact. The map $\delta :\text{Hom}\left(A,B″\right)\to {\text{Ext}}^{1}\left(A,B\right)$ is given by $\delta \left(f\right)=ℰ,$ where $ℰ:0\to B\to E\to A\to 0$ is such that the following diagram commutes. $0 ⟶ B ⟶ B′ ⟶ B″ ⟶ 0 ↑ ↑ f↑ 0 ⟶ B ⟶ E ⟶ A ⟶ 0 .$ Also, $0⟶ HomR(B″,A)⟶ HomR(B′,A)⟶ HomR(B,A)δ′⟶ Ext1(B″,A)⟶ Ext1(B′,A′)⟶ Ext1(B,A)$ is exact. The map $\delta \prime :\text{Hom}\left(B,A\right)\to {\text{Ext}}^{1}\left(B″,A\right)$ is given by $\delta \prime \left(f\right)=ℰ\prime ,$ where $ℰ\prime :0\to A\to E\prime \to B″\to 0$ is such that the following diagram commutes. $0 ⟶ B ⟶ B′ ⟶ B″ ⟶ 0 ↓f ↓ ↓ 0 ⟶ A ⟶ E′ ⟶ B″ ⟶ 0 .$

#### Central Extensions

Let $𝔤$ and $𝔠$ be Lie algebras. A central extension of $𝔤$ by $𝔠$ is a short exact sequence of Lie algebras $0⟶𝔠⟶i 𝔤‾ ⟶π𝔤⟶0 (1.5)$ so that $𝔠$ is contained in the center of $\stackrel{‾}{𝔤}\text{.}$ We will also say that $\stackrel{‾}{𝔤}$ is a central extension of $𝔤,$ suppressing the choice of maps $i$ and $\pi \text{.}$ A Lie algebra $\stackrel{‾}{𝔤}$ is perfect if $\stackrel{‾}{𝔤}=\left[\stackrel{‾}{𝔤},\stackrel{‾}{𝔤}\right]\text{;}$ a central extension (1.5) so that $\stackrel{‾}{𝔤}$ is perfect is a covering of $𝔤\text{.}$

A covering (1.5) of $𝔤$ is a universal central extension if, for any central extension $0⟶𝔠′⟶i′ 𝔤‾′⟶π′ 𝔤⟶0,$ the following diagram commutes: $0 ⟶ 𝔠 ⟶i 𝔤‾ ⟶π′ 𝔤 ⟶ 0 ↓ ↓ ↓ 0 ⟶ 𝔠′ ⟶i′ 𝔤‾′ ⟶π′ 𝔤 ⟶ 0 .$ A Lie algebra $𝔤$ has a universal central extension if and only if $𝔤$ is perfect ([MPi1995] 1.9.2). A universal central extension will be unique up to an equivalence of extensions ([MPi1995], 1.9.1).

## 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.