Lie Algebra Representations, Enveloping Algebras, and PBW — Examples

For the adjoint action, [L2] gives $[{\mathrm{ad}}_x,{\mathrm{ad}}_y]={\mathrm{ad}}_{[x,y]}$, exactly the identity in [L1].

For the trivial action, both $[0,0]$ and the operator assigned to $[x,y]$ are zero, so [L1] holds.

Hence both formulas define representations; the adjoint action is trivial precisely when $\mathfrak{g}$ is abelian.

For all endomorphisms $X,Y$ and $v\in V$, the commutator definition gives $[X,Y]v=(XY-YX)v=X(Yv)-Y(Xv)$.

Matrix multiplication is bilinear in the matrix and vector variables. Together with step 1.1, this verifies both conditions in the equivalence [L1], so restriction to each named bracket-closed matrix Lie algebra is a representation.

The zero bracket makes $V$ abelian, and every endomorphism of $V$ is then a derivation because both sides of the derivation identity are zero. Thus $\rho$ has the target required by [L1], and substituting the zero bracket on $V$ gives the displayed formula.

For $u,v\in V$, $[(0,u),(0,v)]=(0,0)$, while $[(x,w),(0,v)]=(0,\rho (x)v)$ lies in $0\oplus V$. Hence $V$ is an abelian ideal; it is also the kernel of the projection to $\mathfrak{g}$.

The Jacobi lemma in [L1] and steps 1.1–2.1 verify the claimed semidirect Lie algebra and its ideal.

Represent $(A,u)$ on $V\oplus k$ by $M(A,u)=\left(\begin{array}{cc}A& u\\ 0& 0\end{array}\right)$, where $u:k\to V$ sends $1$ to $u$. Multiplication gives $M(A,u)M(B,v)=\left(\begin{array}{cc}AB& Av\\ 0& 0\end{array}\right)$.

Subtracting the reversed product yields $[M(A,u),M(B,v)]=M([A,B],Av-Bu)$, exactly the bracket in [L1] because $V$ is abelian.

Thus the block realization identifies the affine Lie algebra with the stated semidirect product.

Applying two tensor operators to $v\otimes w$ produces the two unmixed commutator terms $[x,y]v\otimes w$ and $v\otimes [x,y]w$; the mixed terms $xv\otimes yw$ and $yv\otimes xw$ occur with opposite signs and cancel.

On the dual, two applications give $(x(y\lambda )-y(x\lambda ))(v)=\lambda (yxv-xyv)=-\lambda ([x,y]v)$, which is exactly the displayed dual action of $[x,y]$.

On Hom, expanding the commutator of $T\mapsto {\rho }_W(x)T-T{\rho }_V(x)$ and its $y$-analogue cancels the mixed composites and leaves ${\rho }_W([x,y])T-T{\rho }_V([x,y])$.

These computations verify the representation identity for all three formulas in [L1] and show why the dual minus sign and Hom subtraction are necessary.

The symmetric algebra $S(kx)$ has one basis monomial $x^n$ in every degree $n\ge 0$, and multiplication satisfies $x^m{x}^n=x^{m+n}$.

Sending $t^n\mapsto x^n$ therefore defines a bijective unital algebra map $k[t]\to S(kx)$. Composing with [L1] gives $k[t]\cong U(\mathfrak{g})$ and sends $t$ to ${\iota }_{\mathfrak{g}}(x)$.

A weakly increasing word in the order $x<y<z$ consists uniquely of $a$ copies of $x$, then $b$ copies of $y$, then $c$ copies of $z$, and is therefore $x^a{y}^b{z}^c$.

The enveloping relation is $yx=xy-z$, while centrality gives $zx=xz$ and $zy=yz$. These formulas concretely move every inversion toward the ordered form.

By [L1], the ordered forms identified in step 1.1 are linearly independent as well as spanning, so they are a basis; step 1.2 is the corresponding reordering rule.

The defining enveloping relations give $eh=(h-2)e$, $hf=f(h-2)$, and $ef=fe+h$. Each formula replaces an adjacent inversion for $f<h<e$ by an ordered pair plus a shorter term.

Every weakly increasing word has all $f$'s first, then all $h$'s, then all $e$'s, hence is uniquely $f^a{h}^b{e}^c$.

Step 1.1 rewrites every word into a linear combination of the forms in step 1.2, and [L1] makes those forms linearly independent, so the reordering result is unique.

The action is a representation because the acting Lie algebra is generated by one element and its chosen operator commutes with itself. Also $t(k{e}_1)=0$, so $k{e}_1$ is stable.

Every line complementary to $k{e}_1$ is spanned by $e_2+a{e}_1$ for some $a\in k$, but $t(e_2+a{e}_1)=e_1$ does not belong to that line. Hence no complementary line is stable.

The invariant short filtration $0\subset k{e}_1\subset V$ therefore does not split into subrepresentations, providing the claimed nonsplit example and, by [L1], a failure of complete reducibility.

In $U({\mathfrak{s}\mathfrak{l}}_2)$, $he-eh=2e$, so $\mathrm{sym}(eh)=\frac{1}{2}(eh+he)=eh+e$.

On degree-one factors, $\mathrm{sym}(e)\mathrm{sym}(h)=eh$. PBW injectivity from [L1] gives $e\mathrm{\ne }0$ in the enveloping algebra, so $eh+e\mathrm{\ne }eh$.

Therefore symmetrization does not preserve this product and is not an algebra homomorphism.

The line $k{v}_{+}$ is stable under $e$, since $e{v}_{+}=0\in k{v}_{+}$.

It is not stable under $f$, because $f{v}_{+}=v_{-}$ and $v_{-}$ is linearly independent from $v_{+}$. By [L1], $k{v}_{+}$ is therefore not a subrepresentation.

This line is stable under one named generator but not under the whole Lie algebra, refuting the statement.

Characteristic zero makes $2$ invertible, and [L1] therefore makes the displayed finite polynomial in $e,f,h$ a well-defined enveloping-algebra element.

Using [L2], it has the PBW-normal expression $\mathrm{\Omega }=2fe+h+\frac{1}{2}{h}^2$. This is an equality of elements, not a centrality computation.

Thus the stated Casimir expression and its normal form are justified; centrality is deliberately deferred to the later Casimir and central-character treatment.