Lie Algebra Representations, Enveloping Algebras, and PBW

Let $k$ be a field. A Lie algebra over $k$ is a $k$-vector space $\mathfrak{g}$ (Vector space over a field) together with a map

$[-,-]:\mathfrak{g}\times \mathfrak{g}⟶\mathfrak{g}$

that is linear in each variable (Linear map between vector spaces over the same field) and satisfies, for all $x,y,z\in \mathfrak{g}$,

$[x,x]=0$

and

$[x,[y,z]]+[y,[z,x]]+[z,[x,y]]=0.$

The first identity is alternation and the second is the Jacobi identity. No finite-dimensional hypothesis is imposed. Alternation implies $[x,y]=-[y,x]$ in every characteristic: expand $0=[x+y,x+y]=[x,y]+[y,x]$. Thus skew-symmetry is a consequence here, not a replacement for alternation in characteristic $2$.

The zero vector space, with its unique bracket, is a Lie algebra. A Lie algebra is abelian when $[x,y]=0$ for every $x,y$.

Let $\mathfrak{g}$ be a Lie algebra over $k$, and let $\mathfrak{h}$ be a linear subspace (Linear subspace of a vector space).

• It is a Lie subalgebra if $[\mathfrak{h},\mathfrak{h}]\subseteq \mathfrak{h}$. The restricted bracket then makes $\mathfrak{h}$ a Lie algebra: bilinearity, alternation, and Jacobi are inherited from $\mathfrak{g}$.
• It is an ideal, written $\mathfrak{i}⊴\mathfrak{g}$, if $[\mathfrak{g},\mathfrak{i}]\subseteq \mathfrak{i}$. Since the bracket is skew-symmetric, this is equivalent to $[\mathfrak{i},\mathfrak{g}]\subseteq \mathfrak{i}$.
• The center is $Z(\mathfrak{g})=\{z\in \mathfrak{g}:[z,x]=0\text{ for every }x\in \mathfrak{g}\}\mathrm{.}$

The center is an ideal: it is a linear subspace by bilinearity, and for $z\in Z(\mathfrak{g})$ and $x\in \mathfrak{g}$ one has $[x,z]=0$, which lies in $Z(\mathfrak{g})$. Every ideal is therefore a Lie subalgebra, but a Lie subalgebra need not be an ideal.

Let $\mathfrak{i}⊴\mathfrak{g}$. The rule

$[x+\mathfrak{i},y+\mathfrak{i}]=[x,y]+\mathfrak{i}$

is independent of representatives, is bilinear and alternating, and satisfies the Jacobi identity on the quotient vector space $\mathfrak{g}\mathrm{/}\mathfrak{i}$.

For an ideal $\mathfrak{i}⊴\mathfrak{g}$, the quotient vector space $\mathfrak{g}\mathrm{/}\mathfrak{i}$ equipped with

$[x+\mathfrak{i},y+\mathfrak{i}]=[x,y]+\mathfrak{i}$

is the quotient Lie algebra. The bracket is representative-independent and satisfies all Lie identities by The quotient Lie-algebra bracket is well-defined. The canonical projection $\pi :\mathfrak{g}\to \mathfrak{g}\mathrm{/}\mathfrak{i}$ is linear and satisfies $\pi ([x,y])=[\pi (x),\pi (y)]$.

Let $\mathfrak{g}$ and $\mathfrak{h}$ be Lie algebras over the same field $k$, with no dimension restriction. A Lie-algebra homomorphism $f:\mathfrak{g}\to \mathfrak{h}$ is a linear map (Linear map between vector spaces over the same field) satisfying

$f([x,y])=[f(x),f(y)]$

for all $x,y\in \mathfrak{g}$. A bijective Lie-algebra homomorphism is an isomorphism; its inverse preserves brackets because applying $f$ reduces that assertion to bracket preservation by $f$.

For a Lie-algebra homomorphism $f:\mathfrak{g}\to \mathfrak{h}$, the kernel is an ideal, the image is a Lie subalgebra, and

$\mathfrak{g}\mathrm{/}\ker f\cong \mathrm{im}f$

as Lie algebras via $x+\ker f\mapsto f(x)$.

For a family $({\mathfrak{g}}_i{)}_{i\in I}$ of Lie algebras over $k$, the Cartesian product ${\prod }_i{\mathfrak{g}}_i$ has componentwise vector operations and bracket

$[(x_i),(y_i)]=([x_i,y_i]{)}_{i\in I}\mathrm{.}$

Bilinearity, alternation, and Jacobi hold componentwise, so this is the direct product Lie algebra. Its algebraic direct sum

${⨁}_{i\in I}{\mathfrak{g}}_i$

is the finite-support subspace of the product (The direct sum of an indexed family of modules) with the restricted bracket. It is closed because $\mathrm{supp}([(x_i),(y_i)])\subseteq \mathrm{supp}(x)\cap \mathrm{supp}(y)$, a finite set when both inputs have finite support.

For finite $I$, product and direct sum are the same Lie algebra. For $I=\mathrm{\varnothing }$, both are the zero Lie algebra. For two algebras the notation is $\mathfrak{g}\oplus \mathfrak{h}$ and the bracket is $[(x,u),(y,v)]=([x,y],[u,v])$.

A derivation of a Lie algebra $\mathfrak{g}$ is a linear map $D:\mathfrak{g}\to \mathfrak{g}$ satisfying

$D[x,y]=[Dx,y]+[x,Dy]$

for all $x,y$. The vector space of derivations is denoted $\mathrm{Der}(\mathfrak{g})$. For $x\in \mathfrak{g}$, the linear map

${\mathrm{ad}}_x:\mathfrak{g}\to \mathfrak{g},{\mathrm{ad}}_x(y)=[x,y],$

is called an inner derivation once its derivation law is established by Jacobi in the following proposition.

$\mathrm{Der}(\mathfrak{g})$ is a Lie subalgebra of ${\mathrm{End}}_k(\mathfrak{g})$ under the commutator. The map $\mathrm{ad}:\mathfrak{g}\to \mathrm{Der}(\mathfrak{g})$ is a Lie-algebra homomorphism, its image is an ideal, and $\ker (\mathrm{ad})=Z(\mathfrak{g})$.

By Derivations form a Lie algebra and inner derivations an ideal, $\mathrm{Der}(\mathfrak{h})$ is a Lie algebra under the commutator. Let $\rho :\mathfrak{g}\to \mathrm{Der}(\mathfrak{h})$ be a Lie-algebra homomorphism. On $\mathfrak{g}\oplus \mathfrak{h}$ define the candidate bracket

$[(x,u),(y,v)]=([x,y],\rho (x)v-\rho (y)u+[u,v])\mathrm{.}$

The next lemma proves that this bracket satisfies Jacobi. The resulting Lie algebra is the semidirect product $\mathfrak{g}{⋉}_{\rho }\mathfrak{h}$. The subspace $0\oplus \mathfrak{h}$ is an ideal, the projection to $\mathfrak{g}$ is a Lie homomorphism, and the inclusion of $\mathfrak{g}$ is a section. When $\rho =0$, this is the direct sum Lie algebra.

The bracket defining $\mathfrak{g}{⋉}_{\rho }\mathfrak{h}$ is bilinear, alternating, and satisfies the Jacobi identity.

For a vector space $V$ over $k$, write

$\mathfrak{g}\mathfrak{l}(V)={\mathrm{End}}_k(V)$

with commutator bracket $[A,B]=AB-BA$. Expansion of triple composites shows that this is a Lie algebra in every characteristic.

A representation of a Lie algebra $\mathfrak{g}$ on $V$ is a Lie-algebra homomorphism

$\rho :\mathfrak{g}⟶\mathfrak{g}\mathfrak{l}(V)\mathrm{.}$

Writing $xv$ for $\rho (x)(v)$, this is equivalently a $k$-bilinear action $\mathfrak{g}\times V\to V$ satisfying

$[x,y]v=x(yv)-y(xv)$

for all $x,y\in \mathfrak{g}$ and $v\in V$. No finite-dimensional hypothesis is imposed on either $\mathfrak{g}$ or $V$. The zero vector space and the zero action are allowed.

Let $V$ be a representation of $\mathfrak{g}$.

A linear subspace $W\subseteq V$ is stable if $xw\in W$ for every $x\in \mathfrak{g}$ and $w\in W$. With the restricted action, such a $W$ is a subrepresentation.

For a stable subspace $W$, the quotient representation on $V\mathrm{/}W$ is

$x(v+W)=xv+W\mathrm{.}$

This is independent of the representative: replacing $v$ by $v+w$ changes $xv$ by $xw\in W$. The representation identity descends because both sides are the corresponding cosets of the identity in $V$.

For representations $V,V^{\mathrm{\prime }}$ of the same Lie algebra, an intertwiner is a linear map $T:V\to V^{\mathrm{\prime }}$ such that

$T(xv)=xT(v)$

for all $x\in \mathfrak{g}$ and $v\in V$.

A nonzero representation $V$ is irreducible if its only stable linear subspaces are $0$ and $V$.

A representation is completely reducible if it is isomorphic, as a representation, to an algebraic direct sum

${⨁}_{i\in I}{V}_i$

of irreducible representations, where elements have finite support. This is a definition, not an assertion that every representation has such a decomposition. The zero representation is the empty direct sum and is therefore completely reducible under this convention.

A representation $\rho :\mathfrak{g}\to \mathfrak{g}\mathfrak{l}(V)$ is faithful if $\rho$ is injective.

For a representation $\rho :\mathfrak{g}\to \mathfrak{g}\mathfrak{l}(V)$, the kernel is an ideal of $\mathfrak{g}$, and the representation is faithful if and only if this kernel is zero.

Let $(V_i{)}_{i\in I}$ be representations of $\mathfrak{g}$, and let $V,W$ be representations. Then the following formulas define representations:

$x(v_i{)}_{i\in I}=(x{v}_i{)}_{i\in I},$

$(x\lambda )(v)=-\lambda (xv),$

$(xT)(v)=xT(v)-T(xv),$

$x(v\otimes w)=xv\otimes w+v\otimes xw,$

on ${⨁}_i{V}_i$, $V^{\ast }$, ${\mathrm{Hom}}_k(V,W)$, and $V{\otimes }_{k}W$, respectively.

Let $V$ be a vector space over an arbitrary field $k$ and let $n\ge 0$. Permuting tensor factors gives an action of the symmetric group $S_n$ on $V^{\otimes n}$. Define

$S^n(V)=V^{\otimes n}\mathrm{/}{R}_n,$

where $R_n$ is the linear span of all $t-\pi t$ with $t\in V^{\otimes n}$ and $\pi \in S_n$. Thus $S^n(V)$ is the quotient by coinvariance relations, not the subspace of invariant tensors.

The exterior power ${\mathrm{\Lambda }}^n(V)$ is the repeated-vector quotient of The $k$th exterior power as the tensor-power quotient by repeated-vector relations: it is $V^{\otimes n}$ modulo the span of pure tensors in which two inputs are equal.

By convention $V^{\otimes 0}=k$, so $S^0(V)={\mathrm{\Lambda }}^0(V)=k$, while $S^1(V)={\mathrm{\Lambda }}^1(V)=V$. These definitions require no division by $n!$ and therefore apply in every characteristic.

If $V$ is a representation of $\mathfrak{g}$, the diagonal tensor action on $V^{\otimes n}$ preserves the symmetric and repeated-vector relation subspaces. It therefore descends to representations on $S^n(V)$ and ${\mathrm{\Lambda }}^n(V)$ for every $n\ge 0$ and over every field.

A representation of $\mathfrak{g}$ on $V$ is equivalently a $k$-bilinear action $(x,v)\mapsto xv$ satisfying

$[x,y]v=x(yv)-y(xv)\mathrm{.}$

In general this action does not, by itself, canonically make $\mathfrak{g}$ into an associative unital ring over which $V$ is a module.

For a vector space $V$ over $k$, set $V^{\otimes 0}=k$ and define the tensor algebra

$T(V)={⨁}_{n\ge 0}{V}^{\otimes n}\mathrm{.}$

After fixing the canonical tensor-product associators, multiplication on homogeneous pure tensors is concatenation:

$(v_1\otimes \cdots \otimes v_m)(w_1\otimes \cdots \otimes w_n)=v_1\otimes \cdots \otimes v_m\otimes w_1\otimes \cdots \otimes w_n\mathrm{.}$

Extend this bilinearly. Each element has finite degree support, so products are finite sums. Tensor associativity makes concatenation associative, and $1\in k=V^{\otimes 0}$ is its unit. Thus $T(V)$ is a graded unital associative $k$-algebra. The degree-one inclusion is denoted $j:V\to T(V)$.

If $A$ is a unital associative $k$-algebra, every linear map $f:V\to A$ extends uniquely to a unital $k$-algebra homomorphism $\widehat{f}:T(V)\to A$.

For a vector space $V$, let $J$ be the two-sided ideal of $T(V)$ generated by

$v\otimes w-w\otimes v(v,w\in V)\mathrm{.}$

The symmetric algebra of $V$ is the commutative graded algebra

$S(V)=T(V)\mathrm{/}J\mathrm{.}$

The quotient is commutative because its degree-one generators commute, and all elements are sums of products of those generators. Since the relators are homogeneous of degree two, $J$ is homogeneous. Its degree-$n$ relation space is the span of adjacent-transposition differences: an arbitrary permutation difference is a telescoping sum along adjacent transpositions, and each adjacent difference is a degree-$n$ multiple of a generator of $J$. Thus the degree-$n$ image is exactly the quotient $S^n(V)$ from Symmetric and exterior powers over an arbitrary field. The degree-zero unit is the image of $1\in k$.

If $A$ is a commutative unital $k$-algebra, every linear map $f:V\to A$ extends uniquely to a unital algebra homomorphism $\tilde{f}:S(V)\to A$.

Let $\mathfrak{g}$ be a Lie algebra over $k$. In $T(\mathfrak{g})$, regard $[x,y]\in \mathfrak{g}$ as an element of tensor degree one, and let $I$ be the two-sided ideal generated by

$x\otimes y-y\otimes x-[x,y](x,y\in \mathfrak{g})\mathrm{.}$

The universal enveloping algebra is

$U(\mathfrak{g})=T(\mathfrak{g})\mathrm{/}I\mathrm{.}$

Write

${\iota }_{\mathfrak{g}}:\mathfrak{g}⟶U(\mathfrak{g})$

for the composite of the degree-one inclusion with the quotient map. This definition makes $U(\mathfrak{g})$ a unital associative algebra and ${\iota }_{\mathfrak{g}}$ a linear map. It does not assert that ${\iota }_{\mathfrak{g}}$ is injective; that conclusion will come from PBW.

For every $x,y\in \mathfrak{g}$, the canonical map satisfies

${\iota }_{\mathfrak{g}}([x,y])={\iota }_{\mathfrak{g}}(x){\iota }_{\mathfrak{g}}(y)-{\iota }_{\mathfrak{g}}(y){\iota }_{\mathfrak{g}}(x)\mathrm{.}$

Consequently ${\iota }_{\mathfrak{g}}:\mathfrak{g}\to U(\mathfrak{g}{)}_{\mathrm{L}\mathrm{i}\mathrm{e}}$ is a Lie-algebra homomorphism into the commutator Lie algebra.

Let $A$ be a unital associative $k$-algebra, equipped with its commutator Lie bracket. Every Lie-algebra homomorphism $f:\mathfrak{g}\to A_{\mathrm{L}\mathrm{i}\mathrm{e}}$ extends uniquely to a unital algebra homomorphism $\bar{f}:U(\mathfrak{g})\to A$ satisfying $\bar{f}{\iota }_{\mathfrak{g}}=f$.

Restriction along ${\iota }_{\mathfrak{g}}:\mathfrak{g}\to U(\mathfrak{g})$ and extension by the universal property give mutually inverse correspondences between representations of $\mathfrak{g}$ and unital left $U(\mathfrak{g})$-module structures whose restriction along $k\to U(\mathfrak{g})$ is the given scalar action on $V$. A linear map is an intertwiner on one side exactly when it is a module homomorphism on the other.

A Lie-algebra homomorphism $f:\mathfrak{g}\to \mathfrak{h}$ induces a unique unital algebra homomorphism

$U(f):U(\mathfrak{g})⟶U(\mathfrak{h})$

such that $U(f){\iota }_{\mathfrak{g}}={\iota }_{\mathfrak{h}}f$. Moreover $U(\mathrm{id})=\mathrm{id}$ and $U(gf)=U(g)U(f)$.

Let $q:T(\mathfrak{g})\twoheadrightarrow U(\mathfrak{g})$ be the defining quotient map. For $n\ge 0$, define

$F_{n}U(\mathfrak{g})=q\left({⨁}_{0\le r\le n}{\mathfrak{g}}^{\otimes r}\right),$

and put $F_{-1}U(\mathfrak{g})=0$. This is the PBW filtration. It is increasing and exhaustive: every tensor-algebra element has finite degree support, hence every enveloping-algebra element belongs to some $F_n$.

Equivalently, $F_n$ is spanned by products of at most $n$ elements from the image ${\iota }_{\mathfrak{g}}(\mathfrak{g})$, with the empty product included. This definition does not assume that ${\iota }_{\mathfrak{g}}$ is injective.

Let $A$ be a unital algebra with an increasing multiplicative filtration

$0=F_{-1}A\subseteq F_{0}A\subseteq F_{1}A\subseteq \cdots ,$

where $1\in F_{0}A$ and $(F_{m}A)(F_{n}A)\subseteq F_{m+n}A$. Its associated graded algebra is

${\mathrm{gr}}_{F}A={⨁}_{n\ge 0}{F}_{n}A\mathrm{/}{F}_{n-1}A,$

with multiplication on homogeneous classes

$(a+F_{m-1}A)(b+F_{n-1}A)=ab+F_{m+n-1}A\mathrm{.}$

This product is well-defined. Indeed, if $a$ changes by $u\in F_{m-1}A$ and $b$ by $v\in F_{n-1}A$, the product changes by $ub+av+uv\in F_{m+n-1}A$. Associativity and the unit class descend from $A$.

For the PBW filtration,

$F_{m}U(\mathfrak{g})F_{n}U(\mathfrak{g})\subseteq F_{m+n}U(\mathfrak{g})\mathrm{.}$

Moreover $[F_m,F_n]\subseteq F_{m+n-1}$ when $m+n\ge 1$, and therefore $\mathrm{gr}U(\mathfrak{g})$ is commutative.