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For a finitely generated algebra over $k$, choose a finite-dimensional generating subspace $V$ containing one. Its Gelfand–Kirillov dimension isChanging generators bounds the two filtrations by constant rescalings of the index, so the value is independent of $V$.

For the noncommutative Krull dimension, set the zero module's dimension to $-1$. A nonzero module has dimension zero exactly when it is an Artinian module. Recursively, dimension at most $d$ means that in every descending chain of submodules the successive factors eventually have dimension less than $d$; use the ordinal version of this definition if necessary. The left Krull dimension of $R$ is that of its left regular module. For commutative Noetherian rings this agrees with the usual prime-chain Krull dimension, but prime-chain counting alone is not its definition for a noncommutative ring.

Write $U=U({\mathfrak{sl}}_2(\mathbb{C}))$ with $[h,e]=2e$, $[h,f]=-2f$, $[e,f]=h$. By the Poincaré-Birkhoff-Witt theorem, $f^i{h}^j{e}^k$ form a basis. In the total-degree filtration there are $\left(\genfrac{}{}{0ex}{}{n+3}{3}\right)$ such monomials of degree at most $n$, soThe commutators lower degree, giving $\mathrm{gr}U\cong \mathbb{C}[f,h,e]$, which also proves that $U$ is left and right Noetherian by finite leading-term cancellation.

To determine the module-theoretic Krull dimension, use the generalized Weyl algebra description and verify the dimension criterion, rather than substituting the three-dimensional graded ring. PutCommutation with $h$ is immediate; with $e$, the contributions are $4he-2(eh+he)-4e=0$, using $he-eh=2e$. The corresponding calculation with $f$ also vanishes. Thus $z$ is a Casimir element. The subalgebra $D=\mathbb{C}[h,z]$ is a polynomial ring in two variables: for a nonzero polynomial in $z$ with coefficients in $\mathbb{C}[h]$, its highest $z$ power contributes a nonzero PBW term with that many copies of both $f$ and $e$, and no lower power can cancel it.

Take $\sigma (h)=h-2$, $\sigma (z)=z$ and $t=(z-h^2-2h)/4$. The relations becomeThey identify $U$ with $D(\sigma ,t)$. Surjectivity follows from its generators; injectivity follows from PBW: the graded pieces $D$, $D{e}^r$ and $D{f}^r$ for $r>0$ are linearly independent, using the difference between the $e$ and $f$ exponents and then the highest paired exponent. Every PBW monomial lies in one of these pieces by replacing paired $f,e$ powers with polynomials in $h,z$.

We use the following general Krull dimension criterion for rank-one generalized Weyl algebras: for a commutative Noetherian base $D$ of dimension $d$, the generalized Weyl algebra has Krull dimension $d$ provided no height-$d$ maximal ideal is periodic under $\sigma$ and none contains infinitely many translates ${\sigma }^n(t)$. This is Theorem 5.3, specialized to the commutative base, in Krull dimension of Generalized Weyl Algebras with non-commutative coefficients.

Here every maximal ideal is ${\mathfrak{m}}_{\lambda ,\mu }=(h-\lambda ,z-\mu )$ and has height two. Its translates have $h$ coordinate $\lambda +2n$, so no nonzero power of $\sigma$ fixes it. Further,This is a nonzero quadratic polynomial in $n$, so it vanishes for at most two integers. Both exceptional mechanisms are excluded for every maximal ideal, and the base has dimension two. The criterion consequently givesThe orbit and root calculations are needed for this upper bound on the entire left-ideal lattice; merely exhibiting a length-two chain of prime ideals would only prove a lower bound.

Let $B$ be the given invariant bilinear form on a Lie algebra. Choose a basis $x_1,\dots ,x_d$ and its $B$-dual basis $x^1,\dots ,x^d$, so $B(x_i,x^j)={\delta }_i^j$. The Casimir element isThe tensor ${\sum }_i{x}_i\otimes x^i$ is the inverse tensor of $B$, so it is independent of the chosen basis. Multiplication in the universal enveloping algebra therefore makes ${\Omega }_B$ independent of that choice as well.

For $z\in \mathfrak{g}$, write $[z,x_i]={\sum }_j{a}_{ji}{x}_j$. Invariance means $B([z,x],y)+B(x,[z,y])=0$, whence $[z,x^i]=-{\sum }_j{a}_{ij}{x}^j$. Thereforeby interchanging $i,j$ in the second sum. Since $\mathfrak{g}$ generates its universal enveloping algebra, ${\Omega }_B$ is central. This argument does not require adding a symmetry assumption to the given bilinear form.

For the sl2 Lie algebra, takeThe cyclic identity for the trace gives $B([Z,X],Y)+B(X,[Z,Y])=0$. The form is nondegenerate, since $B(e,f)=1$, $B(h,h)=2$, and all the other pairings except $B(f,e)$ are zero. The dual of $(e,f,h)$ is $(f,e,h/2)$, soBy the classification of finite-dimensional sl2 representations, an irreducible of dimension $n$ has a highest-weight vector $v$ with $ev=0$ and $hv=mv$, where $m=n-1$. Since $ef=fe+h$, one obtainsCentrality and the Schur lemma then give the answer on the entire representation:This Casimir eigenvalue for sl2 depends on normalization. Choosing the Killing form instead, $K=4B$, scales the dual basis and the Casimir element by $1/4$, giving $(n^2-1)I/8$.

Schur's lemma says that a nonzero intertwiner between irreducible representations is an isomorphism; over $\mathbb{C}$, every endomorphism of a finite-dimensional irreducible representation is scalar. To prove the Schur lemma, let $T:V\to W$ be a intertwining operator between irreducible representations. Its kernel and image of a linear map are invariant. If $T\ne 0$, irreducibility forces $\ker T=0$ and $\mathrm{im}T=W$. This proves the first assertion over any field. In particular the endomorphisms of an irreducible representation form a division ring.

When $V$ is finite-dimensional over an algebraically closed field, an endomorphism $T$ has an eigenvalue $\lambda$. The endomorphism $T-\lambda I$ has nonzero kernel. By the first assertion it must be zero, so $T=\lambda I$. Consequently, for complex finite-dimensional irreducible representations, the space of intertwining operators has dimension zero for nonisomorphic representations and dimension one for isomorphic representations.

Every finite-dimensional representation of a complex semisimple Lie algebra is completely reducible. We prove the Weyl complete reducibility theorem using the allowed Casimir operator facts, without assuming a splitting in advance. For dual bases $x_i,x^i$ with respect to the Killing form, the Casimir elementis central in the universal enveloping algebra. Hence its Casimir operator $C_M={\sum }_i{\rho }_M(x_i){\rho }_M(x^i)$ commutes with the action on every Lie algebra representation $M$ and is compatible with subrepresentations, quotient representations, and intertwining operators. On the trivial Lie algebra representation it acts by zero. On every nontrivial finite-dimensional Irreducible Lie algebra representation it acts by a nonzero scalar.

For clarity, the last fact can be expressed by the Casimir eigenvalue formula: on an irreducible with dominant integral weight $\lambda$, the scalar is $(\lambda ,\lambda +2\rho )$, with the inner product induced by the Killing form and $\rho$ the half-sum of positive roots. On the real span of the weights this inner product is positive definite, and the scalar is positive for $\lambda \ne 0$. For a semisimple Lie algebra with several simple factors the scalars add, so a nontrivial representation still gives a nonzero scalar. These are properties of the Casimir operator being used here.

First establish Casimir splitting of a trivial quotient. Supposeis a short exact sequence of finite-dimensional Lie algebra representations, with trivial quotient. The generalized eigenspaces of $C_E$ are invariant, soEach $E_c$ with $c\ne 0$ maps to zero under $q$: applying a sufficiently large power of $C_E-cI$ and using $C_{\mathbb{C}}=0$ gives $(-c{)}^{r}q(e)=0$. Thus $q(E_0)=\mathbb{C}$.

Take a composition series of a module of $E_0$. The Casimir operator is nilpotent on $E_0$, so its scalar on every irreducible composition factor is zero. The stated Casimir operator property makes every such factor trivial. In a basis adapted to the composition series of a module, the image of $\mathfrak{g}$ on $E_0$ therefore consists of strictly upper triangular matrices, so that image is solvable. The allowed fact that a semisimple Lie algebra acts trivially on every one-dimensional representation implies that $\mathfrak{g}$ is a perfect Lie algebra: otherwise a nonzero linear functional on $\mathfrak{g}/[\mathfrak{g},\mathfrak{g}]$ would define a nontrivial one-dimensional Lie algebra representation. Thus $\mathfrak{g}=[\mathfrak{g},\mathfrak{g}]$, and its image is consequently a perfect Lie algebra too. A perfect Lie algebra that is also a solvable Lie algebra is zero, since its derived series of a Lie algebra is constant until it vanishes. Hence $\mathfrak{g}$ acts trivially on $E_0$. Choose $e\in E_0$ with $q(e)=1$. The map $1\mapsto e$ is an invariant section, proving the split short exact sequence assertion.

Now let $W\subseteq V$ be any nonzero subrepresentation. On the Hom representation $\mathrm{Hom}(V,W)$ the action isConsider the invariant subspaceRestriction produces a short exact sequenceSurjectivity follows by extending $I_W$ to a linear map on $V$. The quotient is trivial, because $I_W$ commutes with the action on $W$. The preceding Casimir splitting of a trivial quotient yields an invariant $p:V\to W$ with $p{|}_W=I_W$. Thusand $\ker p$ is invariant. The zero subrepresentation also has a complement. Repeatedly splitting off an irreducible subrepresentation now gives a direct sum of irreducibles, proving the Weyl complete reducibility theorem.

A derivation of a Lie algebra is a linear map $D:\mathfrak{g}\to \mathfrak{g}$ satisfying the Leibniz ruleThe space $\mathrm{Der}(\mathfrak{g})$ is a vector subspace of $\mathrm{End}(\mathfrak{g})$. Equip it with the commutator $[D,E]=DE-ED$. Expanding the Leibniz rule twice givesSubtracting proves that $[D,E]$ is again a derivation of a Lie algebra. Antisymmetry and the Jacobi identity hold for the commutator in every associative endomorphism algebra, so this defines the derivation Lie algebra.

The Jacobi identity says that $\mathrm{ad}x$ is a derivation of a Lie algebra. Moreover, for $D\in \mathrm{Der}(\mathfrak{g})$,soThis makes $\mathrm{ad}\mathfrak{g}$ an ideal of a Lie algebra in $\mathrm{Der}(\mathfrak{g})$.

Finally suppose $\mathfrak{g}$ is semisimple. Let $\mathfrak{g}$ act on $\mathrm{Der}(\mathfrak{g})$ by $x\cdot D=[\mathrm{ad}x,D]$. The Weyl complete reducibility theorem supplies an invariant complement $M$ to $\mathrm{ad}\mathfrak{g}$. For $D\in M$, invariance gives $[\mathrm{ad}x,D]\in M$, while the ideal identity above gives $[\mathrm{ad}x,D]\in \mathrm{ad}\mathfrak{g}$. Their intersection is zero, so $\mathrm{ad}(Dx)=0$ for every $x$. The center of a Lie algebra of a semisimple Lie algebra is zero, hence $Dx=0$ for every $x$, and $M=0$. ThereforeEvery derivation of a Lie algebra is inner, and the element giving it is unique because the center of a Lie algebra vanishes.

Work away from $x=0$, where the inverse-square potential is singular. Variation of the phase-space action givesThe last equation is the constraint imposed by the Lagrange multiplier $e$. The canonical momentum conjugate to $t$ is $-E$, not $E$. Consequently the nonzero canonical Poisson brackets areIn particular,With the convention $\delta z=\alpha \{z,Q\}$, the Noether charge generating $\delta t=\alpha$ is $Q=-E$. Equivalently, the conserved physical energy is $E$; a convention that calls the energy the time-translation charge absorbs this minus sign into its parameter.

For the dilation charge,on the constraint surface. Its infinitesimal canonical transformations areFor constant $\alpha$, extending this by ${\delta }_{D}e=-\alpha e$ preserves the action: the kinetic terms are invariant and $E-H$ scales oppositely to $e$. Thus the dilation acts on time and position with their nonrelativistic relative scaling. Its Poisson bracket with the energy is

Put $C=E-H$. Direct variation of the Hamiltonian givessoThe supplied transformations are generated by the special conformal charge of inverse-square mechanicsChooseThen $\delta (eC)=0$ even for a time-dependent parameter. Expanding the kinetic-term variation, keeping the terms in $\dot{\beta }$, givesHenceFor a constant parameter only the boundary term remains, identifying $K$ as the Noether charge. Independently, the equations of motion giveUsing the same canonical Poisson bracket convention,Together, $\{D,E\}=-E$, $\{D,K\}=K$ and $\{E,K\}=2D$ form the sl2R Lie algebra of conformal mechanics. These are time translation, dilation and special conformal transformation. A useful normalization check is the Casimir elementon the constraint surface.

A finite-dimensional Lie algebra $L$ is semisimple when its solvable radical is zero, equivalently when it has no nonzero solvable ideals.

We prove the Weyl complete reducibility theorem. Induct on the dimension of a finite-dimensional $L$-module $V$. It is enough first to split a submodule $W$ of codimension one. The one-dimensional quotient is trivial because a semisimple Lie algebra is perfect. By induction, $W$ is a direct sum of irreducible modules. A Casimir element formed using the Killing form commutes with the $L$-action, acts as zero on every trivial summand, and acts by a nonzero scalar on every nontrivial irreducible summand. Its image is therefore the sum $W_1$ of the nontrivial summands, while its kernel contains the trivial summands $W_0$ and maps onto $V/W$. ThusInside $\ker \Omega$, choose a lift $v$ of a basis of $V/W$. For $x\in L$, $xv\in W_0$, and $L$ acts trivially on $W_0$. Hence $[x,y]v=0$ for all $x,y\in L$. Since $L=[L,L]$, actually $xv=0$ for every $x$, so $\mathbb{C}v$ is the required invariant complement.

This codimension-one case implies the general case. For an arbitrary submodule $W\subseteq V$, letwith the natural Hom representation. The maps vanishing on $W$ form an $L$-submodule $X_0$ of codimension one. Splitting $X_0$ supplies an $L$-equivariant $f$ with $f{|}_W=I_W$. Thenso every invariant subspace has an invariant complement and every finite-dimensional representation is completely reducible.

For the requested example, embed ${\mathfrak{sl}}_2$ as the upper-left $2\times 2$ block in the special linear Lie algebra ${\mathfrak{sl}}_3$. Under the restricted Adjoint representation,Here $V_n$ is the irreducible ${\mathfrak{sl}}_2$-module of highest weight $n$. The first summand is the three-dimensional adjoint module, the second is trivial, and the last two are the two-dimensional defining module and its dual, which are isomorphic. This explicit direct sum demonstrates complete reducibility.

A bilinear form $B$ on a Lie algebra is invariant whenequivalently $B([x,y],z)+B(y,[x,z])=0$.

Choose a basis $(x_i)$ of $\mathfrak{g}$ and its $B$-dual basis $(x^i)$. The Casimir element iswhere $U(\mathfrak{g})$ is the universal enveloping algebra. The tensor ${\sum }_i{x}_i\otimes x^i$ corresponds under $B:\mathfrak{g}\cong {\mathfrak{g}}^{\ast }$ to the identity endomorphism, so invariance of $B$ makes it fixed by the diagonal adjoint action. Applying multiplication $U(\mathfrak{g})\otimes U(\mathfrak{g})\to U(\mathfrak{g})$ givesfor every $x\in \mathfrak{g}$. Thus $\Omega$ lies in the center of an associative algebra of $U(\mathfrak{g})$.

Define the ladder operators $J_{\pm }=J_1\pm i{J}_2$. The SU(2) Lie algebra relations implyThus $d(J_{\pm })v_m$ has weight $m\pm 1$ when nonzero. Starting from a maximum-weight vector $v_I$, repeated lowering gives weightsThe norm formula derived from the Casimir element,shows that lowering stops precisely at $m=-I$. Therefore $2I$ is a nonnegative integer and the irreducible representation has dimension $2I+1$.

The one-dimensional quotient $V/W$ is trivial because every one-dimensional representation vanishes on the derived algebra $[{\mathfrak{sl}}_2,{\mathfrak{sl}}_2]={\mathfrak{sl}}_2$. Choose $v\in V$ mapping to $1\in V/W$. Thenis a $1$-cocycle:We show that it is a coboundary.

Decompose $W$ into generalized eigenspaces of its Casimir element $\Omega$. These are subrepresentations because $\Omega$ is central. On a generalized eigenspace with nonzero eigenvalue, $\Omega$ is invertible. If $(x_i)$ and $(x^i)$ are dual bases of ${\mathfrak{sl}}_2$ for the Killing form, putInvariance of the Killing form and the cocycle identity give the standard Casimir calculationThus on every nonzero generalized eigenspace, $c(x)=x({\Omega }^{-1}u)$.

On the zero generalized eigenspace, every irreducible composition factor has zero Casimir eigenvalue. By part i and the classification of finite-dimensional sl2 representations, each such factor is trivial. In a basis adapted to a composition series, the image of ${\mathfrak{sl}}_2$ is therefore strictly upper triangular and hence solvable. Since ${\mathfrak{sl}}_2$ is simple and non-solvable, that image is zero. The cocycle then vanishes because it kills $[{\mathfrak{sl}}_2,{\mathfrak{sl}}_2]$.

Combining the generalized eigenspaces gives $w\in W$ such that $c(x)=xw$ for every $x$. Hence $v-w$ is invariant, andis a decomposition into subrepresentations. This proves the codimension-one case of the Weyl complete reducibility theorem.

Write $e,h,f$ for the standard generators of the sl2 Lie algebra. For the representation $\varphi$, use the normalizationThis is the quadratic Casimir element, and by assumption it commutes with every $\varphi (x)$.

Schur lemma says that an endomorphism of a finite-dimensional irreducible complex representation which commutes with the representation is a scalar. Hence $\Omega =c{I}_V$ when $V$ is irreducible.

Let $v$ be a highest-weight vector of highest weight $m$, so $ev=0$ and $hv=mv$. Since $[e,f]=h$,ThereforeIt follows from scalarity thatThis is the Casimir eigenvalue for sl2 in the chosen normalization.

The antisymmetric $C_{\mu \nu }$ transforms as a Lorentz tensor, so its complete contractionis a Lorentz scalar. HenceParts ii and v give $[P_{\rho },{\tilde{C}}_2]=[Q_{\alpha },{\tilde{C}}_2]=[{\bar{Q}}_{\dot{\alpha }},{\tilde{C}}_2]=0$. It therefore commutes with every generator of the Super-Poincaré algebra and is a Casimir element, called the superspin Casimir.

The weights of $M_{\lambda }$ lie below $\lambda$ in the positive-root order, and every weight space is finite-dimensional. A nonzero submodule $V$ is stable under the Cartan subalgebra, so it is a direct sum of its weight spaces. Choose a maximal weight $\mu$ occurring in $V$. Every positive-root operator would raise its weight; maximality therefore makes it kill any nonzero $v\in V_{\mu }$. Thus $v$ is a singular vector.

The Casimir element is central and acts throughout $M_{\lambda }$ byThe same element acts on the highest-weight vector $v$ of weight $\mu$ byBoth are the action of one operator on the same module, so the scalars agree and