Translation Functors and the Shapovalov Determinant

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

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 $\text{ad}:𝔤\times 𝔤\to 𝔤$ given by $$\text{ad}\left(x\right)y=[x,y]\text{.}$$ Suppose $𝔤$ is finite-dimensional. The Killing form $\kappa (,):𝔤\times 𝔤\to 𝔽$ is given by $$\kappa (x,y)=\text{Tr}\left(\text{ad}\left(x\right)\text{ad}\left(y\right)\right),$$ where Tr is the trace. The Killing form is invariant, that is, $\kappa ([x,y],z)=-\kappa (y,[x,z])\text{.}$

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 $$U\left(𝔤\right)=\underset{n\in {\mathbb{Z}}_{\ge 0}}{⨁}{g}^{\otimes n}/⟨x\otimes y-y\otimes x-[x,y]\hspace{0.17em}|\hspace{0.17em}x,y\in 𝔤⟩\text{.}$$ We can view $U\left(𝔤\right)$ as a Lie algebra $\text{Lie}\left(U\left(𝔤\right)\right)$ with bracket $[u,v]=uv-vu$ 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 $\left\{x_i\hspace{0.17em}\right|\hspace{0.17em}i\in I\},$ where $I$ is a totally ordered set. Then the following set is a basis for $U\left(𝔤\right)\text{:}$ $$\left\{x_{i_1}\cdots x_{i_n}\hspace{0.17em}\right|\hspace{0.17em}{i}_1\le \cdots \le i_n\}\text{.}$$

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 $\text{dim}\hspace{0.17em}𝔤>1$ 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 ${𝔤}^{\left(1\right)}=[𝔤,𝔤]$ and ${𝔤}^{\left(n\right)}=[{𝔤}^{(n-1)},{𝔤}^{(n-1)}]\text{.}$ The radical of $𝔤,$ $\text{Rad}\hspace{0.17em}𝔤,$ is the largest ideal of $𝔤$ such that ${\left(\text{Rad}\hspace{0.17em}𝔤\right)}^{\left(n\right)}=0$ 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 $\kappa (,)$ on $𝔤$ is nondegenerate;
•	$\text{Rad}\hspace{0.17em}𝔤=0\text{.}$

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 ${𝔥}^{*}={\text{Hom}}_{ℂ}(𝔥,ℂ)\text{.}$ Define $${𝔤}_{\alpha }=\{x\in 𝔤\hspace{0.17em}|\hspace{0.17em}\text{ad}\left(h\right)x=\alpha \left(h\right)x\}$$ 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 $\kappa (,)$ is nondegenerate on $𝔥\text{.}$ Therefore, we can identify an element $h\in 𝔥$ with an element $\alpha \in {𝔥}^{*}$ so that $\alpha \left(x\right)=\kappa (x,h)$ for all $x\in 𝔥\text{.}$ We use the Killing form to define a form $⟨,⟩:{𝔥}^{*}\times {𝔥}^{*}\to ℂ$ on ${𝔥}^{*}\text{.}$ For $\alpha ,\alpha \prime \in {𝔥}^{*},$ $$⟨\alpha ,\alpha \prime ⟩=\kappa (h,h\prime ),$$ where $\alpha \left(x\right)=\kappa (x,h)$ and $\alpha \prime \left(x\right)=\kappa (x,h\prime )$ 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 $$R^{+}=\{\beta \in R\hspace{0.17em}|\hspace{0.17em}\beta =\sum _i{k}_i{\alpha }_i,k_i\ge 0\}\text{.}$$ The height function $\text{ht}:R\to \mathbb{Z}$ is given by $$\text{ht}\left(\beta \right)=\sum _i{k}_i,\text{if}\hspace{0.17em}\alpha ={\sum }_i{k}_i{\alpha }_i\text{.}$$ 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 $$\rho =\frac{1}{2}\sum _{\beta \in R_{+}}\beta \text{.}$$ The dual Coxeter number $g$ is $$g=\frac{1}{2}⟨\theta ,\theta +2\rho ⟩\text{.}$$

Define ${\alpha }_i^{\vee }=\frac{2{\alpha }_i}{⟨{\alpha }_i,{\alpha }_i⟩}$ and let $h_i\in 𝔥$ so that ${\alpha }_i^{\vee }\left(x\right)=\kappa (x,h_i)$ 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)	$[h_i,h_j]=0\text{;}$
(2)	$[h_i,e_j]=a_{ij}{e}_j,$ $[h_i,f_j]=-a_{ij}{f}_j\text{;}$
(3)	$[e_i,f_j]={\delta }_{i,j}{h}_i\text{;}$
(4)	$\underset{\underset{1-a_{ij}\hspace{0.17em}\text{times}}{⏟}}{[e_i,\cdots [e_i},e_j]\cdots ]=0,$ $\underset{\underset{1-a_{ij}\hspace{0.17em}\text{times}}{⏟}}{[f_i,\cdots [f_i},f_j]\cdots ]=0,$ 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 $$\begin{array}{cc}{\displaystyle 𝔤={𝔤}_{>0}\otimes 𝔥\otimes {𝔤}_{<0}\text{.}}& \text{(1.1)}\end{array}$$ 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 $𝔤\text{:}$ $P=\{\lambda \in {𝔥}^{*}\hspace{0.17em}|\hspace{0.17em}⟨\lambda ,{\alpha }^{\vee }⟩\in \mathbb{Z}\}\text{;}$
•	the set of dominant integral weights of $𝔤\text{:}$ $P_{+}=\{\lambda \in {𝔥}^{*}\hspace{0.17em}|\hspace{0.17em}⟨\lambda ,{\alpha }^{\vee }⟩\in {\mathbb{Z}}_{\ge 0}\}\text{;}$
•	the root lattice of $𝔤\text{:}$ $$Q=\sum _{1\le i\le k}\mathbb{Z}{\alpha }_i\text{;}$$
•	the positive root lattice: $$Q_{+}=\sum _{1\le i\le k}{\mathbb{Z}}_{\ge 0}{\alpha }_i\text{.}$$

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 $$\begin{array}{cc}{\displaystyle ⟨{\alpha }_i,s_l{\alpha }_j⟩=⟨s_l{\alpha }_i,{\alpha }_j⟩}& \text{(1.2)}\end{array}$$ for $1\le i,j,l\le k\text{.}$ The dot action of $W$ is given by $s_i\circ \lambda =s_i(\lambda +\rho )-\rho \text{.}$

Recall $𝔤$ is finite-dimensional. Suppose $\text{dim}\hspace{0.17em}𝔤=n\text{.}$ Let $\left\{u_i\hspace{0.17em}\right|\hspace{0.17em}1\le i\le n\}$ be a basis of $𝔤,$ and let $\left\{u^i\hspace{0.17em}\right|\hspace{0.17em}1\le i\le n\}$ be a dual basis with respect to $\kappa (,),$ i.e. $\kappa (u_i,u^j)={\delta }_{i,j}\text{.}$ Define the Casimir element ${\mathrm{\Omega }}_0\in U\left(𝔤\right)$ by $${\mathrm{\Omega }}_0=\sum _{i=1}^n{u}_i{u}^i\text{.}$$

([Hum1978], 22.1-22.2). For $x\in 𝔤,$

•	$[{\mathrm{\Omega }}_0,x]=0\text{.}$
•	$\text{ad}\left({\mathrm{\Omega }}_0\right)x=\sum _{i=1}^n[u_i,[u^i,x]]=2gx,$ 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 $$\begin{array}{cc}{\displaystyle x=\sum _{i=1}^n\kappa (x,u_i)u^i\text{.}}& \text{(1.3)}\end{array}$$ 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 $$\begin{array}{cc}{\displaystyle \sum _{i=1}^n[u_i,u^i]=\sum _{i=1}^n{u}_i{u}^i-\sum _{i=1}^n{u}^i{u}_i=0\text{.}}& \text{(1.4)}\end{array}$$

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 $⟨,⟩:V\times W\to ℂ$ such that for all $v,v\prime \in V,$ $w,w\prime \in W$ and $\alpha \in ℂ,$ $$\begin{array}{rcl}{\displaystyle ⟨\alpha v+v\prime ,w⟩}& =& {\displaystyle \alpha ⟨v,w⟩+⟨v\prime ,w⟩}\\ {\displaystyle ⟨v,\alpha w+w\prime ⟩}& =& {\displaystyle \alpha ⟨v,w⟩+⟨v,w\prime ⟩\text{.}}\end{array}$$ A Hermitian bilinear form is a map $⟨,⟩:V\times W\to ℂ$ such that for all $v,v\prime \in V,$ $w,w\prime \in W$ and $\alpha \in ℂ,$ $$\begin{array}{rcl}{\displaystyle ⟨\alpha v+v\prime ,w⟩}& =& {\displaystyle \bar{\alpha }⟨v,w⟩+⟨v\prime ,w⟩,}\\ {\displaystyle ⟨v,\alpha w+w\prime ⟩}& =& {\displaystyle \alpha ⟨v,w⟩+⟨v,w\prime ⟩\text{.}}\end{array}$$ Here, $\bar{\alpha }$ is the complex conjugate of $\alpha \text{.}$

Let $q$ be an indeterminant. For vector spaces $V$ and $W$ over $\mathbb{Q}\left(q\right),$ we will also say a map $⟨,⟩:V\times W\to \mathbb{Q}\left(q\right)$ is a Hermitian bilinear form if for all $v,v\prime \in V,$ $w,w\prime \in W$ and $\alpha \in \mathbb{Q}\left(q\right),$ $$\begin{array}{rcl}{\displaystyle ⟨\alpha v+v\prime ,w⟩}& =& {\displaystyle \bar{\alpha }⟨v,w⟩+⟨v\prime ,w⟩\text{;}}\\ {\displaystyle ⟨v,\alpha w+w\prime ⟩}& =& {\displaystyle \alpha ⟨v,w⟩+⟨v,w\prime ⟩}\end{array}$$ where $\overline{·}:\mathbb{Q}\left(q\right)\to \mathbb{Q}\left(q\right)$ is the $\mathbb{Q}\text{-automorphism}$ 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 $x,y\in A,$ $$D\left(xy\right)=D\left(x\right)y+xD\left(y\right)\text{.}$$ The set of derivations $𝒟$ on $A$ naturally has a Lie algebra structure. That is, we can define a bracket $[,]:𝒟\times 𝒟\to 𝒟$ by $[D,D\prime ]=DD\prime -D\prime D\text{.}$

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 $$C=\left(\begin{array}{ccc}A{b}_{11}& \cdots & A{b}_{1m}\\ ⋮& \ddots & ⋮\\ A{b}_{m1}& \cdots & A{b}_{mm}\end{array}\right)\text{.}$$ Then, $$\text{det}\hspace{0.17em}A\otimes B={\left(\text{det}\hspace{0.17em}A\right)}^m{\left(\text{det}\hspace{0.17em}B\right)}^n\text{.}$$

		Proof.
	If $b_{1i}=0$ for all $1\le i\le m,$ then $\text{det}\hspace{0.17em}C=0$ and $\text{det}\hspace{0.17em}B=0\text{;}$ 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, $$\begin{array}{rcl}{\displaystyle \text{det}\hspace{0.17em}A\otimes B}& =& {\displaystyle \text{det}\left(\begin{array}{cccc}A{b}_{11}& A{b}_{12}& \cdots & A{b}_{1m}\\ 0& A(b_{22}-\frac{b_{12}{b}_{21}}{b_{11}})& \cdots & A(b_{2m}-\frac{b_{1m}{b}_{21}}{b_{11}})\\ ⋮& & & ⋮\\ 0& A(b_{m2}-\frac{b_{12}{b}_{m1}}{b_{11}})& \cdots & A(b_{mm}-\frac{b_{1m}{b}_{m1}}{b_{11}})\end{array}\right)}\\ & =& {\displaystyle {\left(b_{11}\right)}^n\text{det}\hspace{0.17em}A\times \text{det}\left(\begin{array}{ccc}A(b_{22}-\frac{b_{12}{b}_{21}}{b_{11}})& \cdots & A(b_{2m}-\frac{b_{1m}{b}_{21}}{b_{11}})\\ ⋮& & ⋮\\ A(b_{m2}-\frac{b_{12}{b}_{m1}}{b_{11}})& \cdots & A(b_{mm}-\frac{b_{1m}{b}_{m1}}{b_{11}})\end{array}\right)}\\ & =& {\displaystyle {\left(b_{11}\right)}^n\text{det}\hspace{0.17em}A\underset{\underset{\text{By the induction hypothesis}}{⏟}}{{\left(\text{det}\hspace{0.17em}A\right)}^{m-1}\text{det}{\left(\begin{array}{ccc}(b_{22}-\frac{b_{12}{b}_{21}}{b_{11}})& \cdots & (b_{2m}-\frac{b_{1m}{b}_{21}}{b_{11}})\\ ⋮& & ⋮\\ (b_{m2}-\frac{b_{12}{b}_{m1}}{b_{11}})& \cdots & (b_{mm}-\frac{b_{1m}{b}_{m1}}{b_{11}})\end{array}\right)}^n}}\\ & =& {\displaystyle {\left(\text{det}\hspace{0.17em}A\right)}^m{\left(b_{11}\text{det}\left(\begin{array}{ccc}(b_{22}-\frac{b_{12}{b}_{21}}{b_{11}})& \cdots & (b_{2m}-\frac{b_{1m}{b}_{21}}{b_{11}})\\ ⋮& & ⋮\\ (b_{m2}-\frac{b_{12}{b}_{m1}}{b_{11}})& \cdots & (b_{mm}-\frac{b_{1m}{b}_{m1}}{b_{11}})\end{array}\right)\right)}^n}\\ & =& {\displaystyle {\left(\text{det}\hspace{0.17em}A\right)}^m{\left(\text{det}\hspace{0.17em}B\right)}^n\text{.}}\end{array}$$ $\square$

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: $$\begin{array}{ccc}& & P\\ & \begin{array}{c} γ \end{array}& \downarrow \alpha \\ B& \stackrel{\beta }{⟶}& A& ⟶& 0\end{array}$$ 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 $$\begin{array}{ccc}& & P\\ & \begin{array}{c} γ \end{array}& \downarrow \\ \stackrel{\sim }{P}& ⟶& M& ⟶& 0\end{array}\begin{array}{ccc}& & \stackrel{\sim }{P}\\ & \begin{array}{c} γ∼ \end{array}& \downarrow \\ P& ⟶& M& ⟶& 0\end{array}$$ 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 $${\text{Hom}}_R(P,M/M\prime )={\text{Hom}}_R(P,M)/{\text{Hom}}_R(P,M\prime )$$ and $$\text{dim}\hspace{0.17em}{\text{Hom}}_R(P,M/M\prime )=\text{dim}\hspace{0.17em}{\text{Hom}}_R(P,M)-\text{dim}\hspace{0.17em}{\text{Hom}}_R(P,M\prime )\text{.}$$

		Proof.
	Clearly, we have a map $\overline{·}:{\text{Hom}}_R(P,M)\to {\text{Hom}}_R(P,M/M\prime ),$ where $\varphi :P\to M$ is sent to $\bar{\varphi }:P\to M\to M/M\prime \text{.}$ Then $\bar{\varphi }=0$ implies that $\varphi \left(P\right)\subset M\prime ,$ and so $\varphi$ is a map from $P$ to $M\prime \text{.}$ Thus, $\text{ker}\left(\overline{·}\right)\subset {\text{Hom}}_R(P,M\prime )\text{.}$ 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 $$0⟶B⟶E⟶A⟶0,$$ $$0⟶B⟶E\prime ⟶A⟶0,$$ are equivalent if there exists $\gamma :E\to E\prime$ such that the following diagram commutes. $$\begin{array}{ccccccccc}0& ⟶& B& ⟶& E& ⟶& A& ⟶& 0\\ & & \downarrow \text{id}& & \downarrow \gamma & & \downarrow \text{id}\\ 0& ⟶& B& ⟶& E\prime & ⟶& A& ⟶& 0\end{array}$$ Define ${\text{Ext}}^1(A,B)$ 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 $$0⟶B⟶B\prime ⟶B″⟶0$$ is exact. Then, $$0⟶{\text{Hom}}_R(A,B)⟶{\text{Hom}}_R(A,B\prime )⟶{\text{Hom}}_R{(A,B″)}^{\delta }⟶{\text{Ext}}^1(A,B)⟶{\text{Ext}}^1(A,B\prime )⟶{\text{Ext}}^1(A,B″)$$ is exact. The map $\delta :\text{Hom}(A,B″)\to {\text{Ext}}^1(A,B)$ 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. $$\begin{array}{ccccccccc}0& ⟶& B& ⟶& B\prime & ⟶& B″& ⟶& 0\\ & & \uparrow & & \uparrow & & f\uparrow \\ 0& ⟶& B& ⟶& E& ⟶& A& ⟶& 0\end{array}\text{.}$$ Also, $$0⟶{\text{Hom}}_R(B″,A)⟶{\text{Hom}}_R(B\prime ,A)⟶{\text{Hom}}_R{(B,A)}^{\delta \prime }⟶{\text{Ext}}^1(B″,A)⟶{\text{Ext}}^1(B\prime ,A\prime )⟶{\text{Ext}}^1(B,A)$$ is exact. The map $\delta \prime :\text{Hom}(B,A)\to {\text{Ext}}^1(B″,A)$ 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. $$\begin{array}{ccccccccc}0& ⟶& B& ⟶& B\prime & ⟶& B″& ⟶& 0\\ & & \downarrow f& & \downarrow & & \downarrow \\ 0& ⟶& A& ⟶& E\prime & ⟶& B″& ⟶& 0\end{array}\text{.}$$

Central Extensions

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

A covering (1.5) of $𝔤$ is a universal central extension if, for any central extension $$0⟶𝔠\prime \stackrel{i\prime }{⟶}\overline{𝔤}\prime \stackrel{\pi \prime }{⟶}𝔤⟶0,$$ the following diagram commutes: $$\begin{array}{ccccccccc}0& ⟶& 𝔠& \stackrel{i}{⟶}& \overline{𝔤}& \stackrel{\pi \prime }{⟶}& 𝔤& ⟶& 0\\ & & \downarrow & & \downarrow & & \downarrow \\ 0& ⟶& 𝔠\prime & \stackrel{i\prime }{⟶}& \overline{𝔤}\prime & \stackrel{\pi \prime }{⟶}& 𝔤& ⟶& 0\end{array}\text{.}$$ 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.

page history