Lie Algebra Representations, Enveloping Algebras, and PBW

If $x^{\mathrm{\prime }}=x+a$ and $y^{\mathrm{\prime }}=y+b$ with $a,b\in \mathfrak{i}$, bilinearity gives $[x^{\mathrm{\prime }},y^{\mathrm{\prime }}]-[x,y]=[a,y]+[x,b]+[a,b]$. Every term belongs to $\mathfrak{i}$ by [L1], so $[x^{\mathrm{\prime }},y^{\mathrm{\prime }}]+\mathfrak{i}=[x,y]+\mathfrak{i}$.

Bilinearity of the quotient bracket follows by choosing representatives, using bilinearity in $\mathfrak{g}$, and using [L2] for coset addition and scalar multiplication; step 1.1 makes the result independent of those choices.

For every $x$, one has $[x+\mathfrak{i},x+\mathfrak{i}]=[x,x]+\mathfrak{i}=\mathfrak{i}$, so the descended bracket is alternating.

For three cosets, their cyclic Jacobi sum is the coset of $[x,[y,z]]+[y,[z,x]]+[z,[x,y]]$, which is zero in $\mathfrak{g}$ and hence is the zero coset. Thus Jacobi descends.

Therefore the displayed rule has all Lie identities. For $\mathfrak{i}=0$ it is the original bracket, and for $\mathfrak{i}=\mathfrak{g}$ it is the unique bracket on the zero space; no choice of representatives is made in either case or in the argument above.

Linearity makes $\ker f$ and $\mathrm{im}f$ linear subspaces. If $a\in \ker f$ and $x\in \mathfrak{g}$, then $f([x,a])=[f(x),f(a)]=[f(x),0]=0$, so $[x,a]\in \ker f$; hence the kernel is an ideal.

If $u=f(x)$ and $v=f(y)$ are in the image, then $[u,v]=[f(x),f(y)]=f([x,y])$ is again in the image. Thus the image is a Lie subalgebra.

Let $\bar{f}$ be the vector-space isomorphism of [L2]. By [L3], $\bar{f}([x+\ker f,y+\ker f])=f([x,y])=[f(x),f(y)]=[\bar{f}(x+\ker f),\bar{f}(y+\ker f)]$, so it preserves brackets.

A bijective bracket-preserving linear map has bracket-preserving inverse: for $u=f(x),v=f(y)$, the inverse sends $[u,v]=f([x,y])$ to $[x,y]$. Hence $\bar{f}$ is a Lie-algebra isomorphism. The zero map gives $\mathfrak{g}\mathrm{/}\mathfrak{g}\cong 0$, while an injective map gives $\mathfrak{g}\mathrm{/}0\cong \mathrm{im}f$; these are included in the same proof.

Derivations form a linear subspace of ${\mathrm{End}}_k(\mathfrak{g})$, because the Leibniz identity is linear in $D$. For derivations $D,E$, expansion of $(DE-ED)[x,y]$ gives $[(DE-ED)x,y]+[x,(DE-ED)y]$: the two cross terms $[Dx,Ey]$ and $[Ex,Dy]$ occur once with each sign and cancel. Hence $[D,E]=DE-ED$ is a derivation.

Jacobi rewritten as $[x,[y,z]]-[y,[x,z]]=[[x,y],z]$ says $[{\mathrm{ad}}_x,{\mathrm{ad}}_y]={\mathrm{ad}}_{[x,y]}$. It also says ${\mathrm{ad}}_x[y,z]=[{\mathrm{ad}}_{x}y,z]+[y,{\mathrm{ad}}_{x}z]$, so every ${\mathrm{ad}}_x$ is a derivation and $\mathrm{ad}$ is a Lie homomorphism.

The endomorphism commutator is bilinear and alternating, and its Jacobi identity follows by expanding the six triple composites. Therefore the closed linear subspace in step 1.1 is a Lie subalgebra.

If $D$ is any derivation, then for every $y$, $[D,{\mathrm{ad}}_x](y)=D[x,y]-[x,Dy]=[Dx,y]$; hence $[D,{\mathrm{ad}}_x]={\mathrm{ad}}_{Dx}$. Thus the inner derivations form an ideal of $\mathrm{Der}(\mathfrak{g})$.

Finally, $x\in \ker (\mathrm{ad})$ exactly when $[x,y]=0$ for every $y$, which is exactly $x\in Z(\mathfrak{g})$ by [L3]. For an abelian algebra the inner ideal is zero; for the zero algebra all assertions remain valid.

Every term in the displayed bracket is bilinear. Substituting the same pair twice gives $([x,x],\rho (x)u-\rho (x)u+[u,u])=(0,0)$, so the bracket is alternating.

The first component of the cyclic Jacobi sum for $(x,u),(y,v),(z,w)$ is $[x,[y,z]]+[y,[z,x]]+[z,[x,y]]=0$ by Jacobi in $\mathfrak{g}$.

In the second component, the terms containing two elements of $\mathfrak{h}$ and one action are, cyclically, $\rho (x)[v,w]-[\rho (x)v,w]-[v,\rho (x)w]$; each vanishes by [L1]. The terms with three elements of $\mathfrak{h}$ form their Jacobi sum and vanish.

The remaining terms are $([\rho (x),\rho (y)]-\rho ([x,y]))w$ and its two cyclic analogues. They vanish by [L2]. Thus the entire second component is zero.

Both components of the Jacobi sum vanish, proving the claim. If one summand is zero, or if $\rho =0$, the calculation reduces to the componentwise Jacobi identity, so all boundary cases are already included and no choices are made.

Applying [L2] to the homomorphism in [L1] shows that $\ker \rho$ is an ideal.

A linear map is injective exactly when its kernel is zero; by [L3], this says that the representation is faithful exactly when $\ker \rho =0$.

Thus both assertions hold, including for $V=0$: in that case $\rho$ is faithful precisely when $\mathfrak{g}=0$.

The componentwise formula preserves finite support by [L3]. Its commutator is componentwise, so [L1] gives $[x,y](v_i)=(x(y{v}_i)-y(x{v}_i){)}_i$; hence it is a representation, including for an empty family.

For the dual formula, $(x(y\lambda )-y(x\lambda ))(v)=\lambda (yxv-xyv)=-\lambda ([x,y]v)=([x,y]\lambda )(v)$ by [L1]. Thus the minus sign gives the required bracket action.

On ${\mathrm{Hom}}_k(V,W)$, expand the commutator of $T\mapsto {\rho }_W(x)T-T{\rho }_V(x)$ and its $y$-analogue. The two mixed terms cancel, leaving ${\rho }_W([x,y])T-T{\rho }_V([x,y])$, which is the prescribed $[x,y]$-action by [L1].

For each fixed $x$, the displayed tensor formula is bilinear in $(v,w)$, so [L2] gives a linear endomorphism of $V\otimes W$. Applying the $x$- and $y$-operators successively to a pure tensor produces two mixed terms in each order; they cancel in the commutator, leaving $[x,y]v\otimes w+v\otimes [x,y]w$. Since pure tensors span, this is the representation identity everywhere.

Each construction is linear in $x$ and satisfies the bracket identity, so all four displayed actions are Lie-algebra representations without finite-dimensional assumptions.

Every diagonal operator $D_x$ commutes with every permutation of tensor positions, because permuting after applying $x$ in one position gives the same summand as applying $x$ in the permuted position. Hence $D_x(t-\pi t)=D_{x}t-\pi D_{x}t$ lies in the symmetric relation subspace.

Consider a pure tensor with equal entries $v$ in positions $p\mathrm{\ne }q$. Terms of $D_x$ differentiating another position still have equal entries in positions $p,q$. The sum of the two remaining terms, with $xv$ in position $p$ or $q$, equals the tensor having $v+xv$ in both positions minus the tensors having $v$ in both and $xv$ in both; it therefore belongs to the span of repeated-vector tensors in every characteristic. Thus the exterior relation subspace is stable.

Stability makes $D_x$ induce an endomorphism on each quotient by [L3]. Since $[D_x,D_y]=D_{[x,y]}$ on $V^{\otimes n}$, the same equality holds after passing to either quotient, and the induced actions are representations.

For $n=0$ the resulting action on $k$ is zero, and for $n=1$ it is the original action on $V$; the same construction proves the assertion for every $n$ without averaging or dividing by $n!$.

From a representation $\rho$, set $xv=\rho (x)v$. Linearity of $\rho$ and of each $\rho (x)$ makes the action bilinear, and bracket preservation expands to $[x,y]v=x(yv)-y(xv)$.

Conversely, a bilinear action defines a linear map $\rho (x)(v)=xv$ into ${\mathrm{End}}_k(V)$. The displayed identity says exactly that $\rho ([x,y])=\rho (x)\rho (y)-\rho (y)\rho (x)$, so $\rho$ is a representation.

The equivalence is therefore exact, but [L2] does not apply directly from the Lie-algebra data in general. The bracket need not be associative; if it is the zero bracket on a nonzero abelian Lie algebra, that multiplication has no unit. There is a genuine exceptional case: if $\mathfrak{g}=0$ and $V=0$, the zero bracket makes $\mathfrak{g}$ the permitted unital zero ring, and its unique action on $V$ is a unital module action. This exception does not give a general ring structure for Lie representations. The canonical associative-module formulation for arbitrary $\mathfrak{g},V$ uses $U(\mathfrak{g})$.

Applying the quotient map to the relator in [L1] gives ${\iota }_{\mathfrak{g}}(x){\iota }_{\mathfrak{g}}(y)-{\iota }_{\mathfrak{g}}(y){\iota }_{\mathfrak{g}}(x)-{\iota }_{\mathfrak{g}}([x,y])=0$, which is the displayed identity.

Since ${\iota }_{\mathfrak{g}}$ is linear by construction and step 1.1 is bracket preservation, it is a Lie-algebra homomorphism. No injectivity has been used.

By [L1], $f$ extends uniquely to a unital algebra homomorphism $F:T(\mathfrak{g})\to A$. Since $f$ preserves Lie brackets, $F(x\otimes y-y\otimes x-[x,y])=f(x)f(y)-f(y)f(x)-f([x,y])=0$ for every defining relator.

The kernel of $F$ is a two-sided ideal containing all defining relators, hence contains their generated ideal $I$. By [L2], $F$ factors uniquely as $T(\mathfrak{g})\twoheadrightarrow U(\mathfrak{g})\stackrel{\bar{f}}{\to }A$, and the factor satisfies $\bar{f}{\iota }_{\mathfrak{g}}=f$.

If $G:U(\mathfrak{g})\to A$ is another unital algebra map with $G{\iota }_{\mathfrak{g}}=f$, its composite with $T(\mathfrak{g})\twoheadrightarrow U(\mathfrak{g})$ is an algebra extension of $f$. It equals $F$ by [L1], and quotient-map surjectivity gives $G=\bar{f}$.

Thus the required extension exists uniquely; the argument uses only the quotient presentation and not injectivity of ${\iota }_{\mathfrak{g}}$.

A representation $\rho :\mathfrak{g}\to {\mathrm{End}}_k(V{)}_{\mathrm{L}\mathrm{i}\mathrm{e}}$ extends uniquely by [L2] to a unital algebra map $\bar{\rho }:U(\mathfrak{g})\to {\mathrm{End}}_k(V)$, hence to a unital left module structure by [L1].

Conversely, a scalar-compatible unital module gives a $k$-algebra map $\alpha :U(\mathfrak{g})\to {\mathrm{End}}_k(V)$. Its restriction $\alpha {\iota }_{\mathfrak{g}}$ preserves Lie brackets because both ${\iota }_{\mathfrak{g}}$ and every algebra map preserve commutators, so it is a representation. Starting with $\rho$ recovers $\rho$ by $\bar{\rho }{\iota }_{\mathfrak{g}}=\rho$; starting with $\alpha$ recovers $\alpha$ by uniqueness in [L2].

Let $T:V\to W$ be linear. If $T$ intertwines the $\mathfrak{g}$-actions, then it intertwines each $\rho (x)$, each finite product of these operators, and finite linear combinations of the products. By [L4], these include the actions of every element of $U(\mathfrak{g})$; the empty product acts as the identity on both modules. Thus $T$ is a module homomorphism. Conversely, restricting a module homomorphism to ${\iota }_{\mathfrak{g}}(\mathfrak{g})$ gives an intertwiner.

The object and morphism correspondences in steps 1.1–1.3 are mutually inverse, proving the claimed equivalence without any PBW or injectivity assumption.

The composite ${\iota }_{\mathfrak{h}}f:\mathfrak{g}\to U(\mathfrak{h}{)}_{\mathrm{L}\mathrm{i}\mathrm{e}}$ is a Lie map by [L1], so [L2] supplies the unique unital algebra map $U(f)$ with the stated generator equation.

Both $U({\mathrm{id}}_{\mathfrak{g}})$ and ${\mathrm{id}}_{U(\mathfrak{g})}$ compose with ${\iota }_{\mathfrak{g}}$ to ${\iota }_{\mathfrak{g}}$, so uniqueness in [L2] makes them equal.

For $\mathfrak{g}\stackrel{f}{\to }\mathfrak{h}\stackrel{g}{\to }\mathfrak{l}$, both $U(gf)$ and $U(g)U(f)$ send ${\iota }_{\mathfrak{g}}$ to ${\iota }_{\mathfrak{l}}gf$. Uniqueness in [L2] therefore gives $U(gf)=U(g)U(f)$.

The construction preserves identities and composition and is consequently functorial.

Concatenating a word of length at most $m$ with one of length at most $n$ gives length at most $m+n$. Taking spans and quotient images proves $F_m{F}_n\subseteq F_{m+n}$.

For a generator $x$ and a word $b_1\cdots b_s$, repeated use of $[x,ab]=[x,a]b+a[x,b]$ gives $[x,b_1\cdots b_s]={\sum }_j{b}_1\cdots [x,b_j]\cdots b_s$. By [L2], every $[x,b_j]$ is again the image of one element of $\mathfrak{g}$, so this commutator lies in $F_s$.

For words $a=a^{\mathrm{\prime }}x$ of length $r>0$ and $b$ of length $s>0$, the identity $[a^{\mathrm{\prime }}x,b]=a^{\mathrm{\prime }}[x,b]+[a^{\mathrm{\prime }},b]x$, together with step 1.2 and induction on $r$, puts both terms in $F_{r+s-1}$. The scalar boundary cases commute, and bilinearity therefore gives $[F_m,F_n]\subseteq F_{m+n-1}$.

If $a\in F_m$ and $b\in F_n$, step 2.1 says that $ab$ and $ba$ have the same class in $F_{m+n}\mathrm{/}{F}_{m+n-1}$. By [L3], all homogeneous elements of $\mathrm{gr}U(\mathfrak{g})$ commute, hence the whole associated graded algebra is commutative.

Let $P$ be the vector space freely spanned by the finite weakly increasing words in $B$, including the empty word. Multiply two basis words by sorting their concatenation. Totality of the supplied order makes the sorted word unique; this multiplication is commutative and associative, and the empty word is its unit.

By [L1], sending $b\in B$ to the one-letter word $b$ extends uniquely to a linear map $j:V\to P$. By [L2], it extends to a unital algebra map $\alpha :S(V)\to P$.

Send each ordered word $b_1\cdots b_n$ in the basis of $P$ to the product of the images of its letters in $S(V)$, and send the empty word to $1$. Extending linearly gives an algebra map $\beta :P\to S(V)$ because multiplication in $S(V)$ is commutative and the product of two word images is the image of their sorted concatenation.

The composite $\alpha \beta$ fixes every word-basis element of $P$, while $\beta \alpha$ fixes every generator from $V$ and hence all of $S(V)$ by [L2]. Thus $\alpha$ and $\beta$ are inverse isomorphisms.

Consequently the stated ordered words are a basis of $S(V)$. The construction uses only the supplied basis and order, and when $B$ is empty the empty word is the sole basis element.

Words of length zero or one are ordered, and every word with inversion number zero is weakly increasing. These are the base cases.

Assume every basis word with strictly smaller pair (length, inversion number) is a linear combination of ordered words.

A nonordered finite word has an adjacent inversion $xy$ with $x>y$. Replace it using [L2] by $yx+[x,y]$. The swapped word has the same length and one fewer inversion, while after the finite basis expansion in [L3] every bracket term has length one less. Step 1.2 therefore rewrites every resulting term as a linear combination of ordered words.

The lexicographic order on pairs of nonnegative integers is well-founded, so steps 1.1–2.1 prove that every basis word is in the ordered span. By [L1], that span is all of $U(\mathfrak{g})$; for the empty basis it consists only of $1$.