Harish Chandra Isomorphism Casimir and Central Characters

By the tensor-algebra universal property, $\rho$ extends uniquely to a unital algebra homomorphism $\widehat{\rho }\text{\hspace{0.17em}⁣}:T(\mathfrak{g})\to \mathrm{End}(V)$ with $\widehat{\rho }(x)=\rho (x)$ for $x\in \mathfrak{g}$.

For $x,y\in \mathfrak{g}$, the Lie-homomorphism identity gives $\widehat{\rho }(x\otimes y-y\otimes x-[x,y])=\rho (x)\rho (y)-\rho (y)\rho (x)-\rho ([x,y])=0$, so the defining ideal of The universal enveloping algebra as a tensor quotient lies in $\ker \widehat{\rho }$.

Therefore $\widehat{\rho }$ descends uniquely through the quotient $T(\mathfrak{g})\twoheadrightarrow U(\mathfrak{g})$ from The universal enveloping algebra as a tensor quotient, producing the required unital algebra homomorphism $\tilde{\rho }$.

Using ${\mathrm{ad}}_{[x,y]}=[{\mathrm{ad}}_x,{\mathrm{ad}}_y]$ and cyclicity of trace, one gets $B([x,y],z)=\mathrm{tr}([{\mathrm{ad}}_x,{\mathrm{ad}}_y]{\mathrm{ad}}_z)=\mathrm{tr}({\mathrm{ad}}_x[{\mathrm{ad}}_y,{\mathrm{ad}}_z])=B(x,[y,z])$.

The nondegeneracy clause of [L1] applies to $\mathfrak{g}$. It proves that the radical $\{x\in \mathfrak{g}:B(x,\mathfrak{g})=0\}$ is zero by an explicit finite trace-solvability argument, rather than leaving Cartan's criterion as an unproved appeal.

Hence $B$ is invariant and has zero radical, so it is nondegenerate.

For $x\in \mathfrak{g}$ and a monomial $y_1\cdots y_n$ of filtration degree $n$, repeatedly using $x{y}_i-y_{i}x=[x,y_i]$ from The PBW filtration by tensor degree on the enveloping algebra rewrites $x(y_1\cdots y_n)-(y_1\cdots y_n)x$ as a sum of terms with exactly one bracket and therefore filtration degree at most $n$.

Applying step 1.1 repeatedly to products of filtration degrees $m$ and $n$ shows that every commutator of elements of $F_{m}U(\mathfrak{g})$ and $F_{n}U(\mathfrak{g})$ drops by at least one filtration step, so it lies in $F_{m+n-1}U(\mathfrak{g})$.

Hence the degree-$m$ and degree-$n$ symbols commute in $\mathrm{gr}U(\mathfrak{g})$, and the associated graded algebra is commutative.

Orient [F1] as a rewriting rule whenever $j>i$. Order words first by length and then by their number of inverted index pairs. The swapped term has one fewer inversion and the bracket term has smaller length, so every sequence of reductions terminates in a linear combination of ordered words. Hence the ordered monomials span $U(\mathfrak{g})$.

Work first in the free tensor algebra, and extend each reduction linearly and inside arbitrary word contexts. Reductions at disjoint positions join by applying both rules in either order. The only overlapping positions are the two descents in $x_k{x}_j{x}_i$ with $k>j>i$. After applying the remaining swap in each branch, their difference is $[[x_j,x_i],x_k]+[x_j,[x_k,x_i]]+[[x_k,x_j],x_i],$ where each displayed Lie bracket is obtained by reducing the corresponding two-letter commutator (and expanding its Lie-algebra value in the $x_i$ basis). All these terms have length at most two after the first reduction, so their normal forms are already determined by induction on length. The displayed expression is zero by Jacobi. Hence both branches have the same normal form.

Here is the normal-form induction needed for linear reductions. Order words by (length, inversion count); below any fixed word there are finitely many words. Define its normal form recursively as the linear combination of normal forms produced by any one reduction, with an ordered word fixed by the recursion. Induct on this order to prove independence of the chosen reduction: for two choices, step 2.1 joins their immediate descendants (inside the same surrounding context), and the induction hypothesis gives the same normal form to both joined expressions. Linearity extends the resulting map $N$ to every tensor polynomial. Each one-step reduction of a polynomial preserves $N$, because it does so term by term. In particular, every contextual defining relation $u(x_j{x}_i-x_i{x}_j-[x_j,x_i])v$ has normal form zero; so does every element of the two-sided ideal generated by these relations.

If a linear combination of distinct ordered monomials represented zero in the tensor quotient, it would lie in the defining ideal. Its normal form would be zero by step 3.1, yet $N$ fixes every ordered word and thus fixes this nonzero combination. This contradiction proves linear independence; step 1.1 supplies spanning. [step 1.1, step 3.1].

In the associated graded algebra the bracket correction in [F1] has one lower tensor degree, so the symbol of an ordered monomial depends only on the corresponding commutative monomial. Step 4.1 shows that these symbols form a basis in each degree. Therefore the natural graded map $S(\mathfrak{g})\to \mathrm{gr}U(\mathfrak{g})$ is an isomorphism.

By [A1, F1], $\mathfrak{h}$ is maximal toral. Its adjoint operators therefore commute and are diagonalizable; [F2] supplies their simultaneous finite weight-space decomposition.

The zero weight space is the centralizer $C_{\mathfrak{g}}(\mathfrak{h})$, which equals $\mathfrak{h}$ by [F2]. The other nonzero weight spaces are exactly the displayed ${\mathfrak{g}}_{\alpha }$ by their simultaneous-eigenvector definition.

Collecting the finitely many nonzero weights gives the finite set $\mathrm{\Phi }$, and the simultaneous eigenspace decomposition from step 1.1 becomes the stated direct sum decomposition.

For every $h\in \mathfrak{h}$, the derivation property of $\mathrm{ad}(h)$ gives $[h,[x,y]]=[\mathrm{ad}(h)x,y]+[x,\mathrm{ad}(h)y]=\alpha (h)[x,y]+\beta (h)[x,y]=(\alpha +\beta )(h)[x,y]$.

By the simultaneous weight-space definition in the cited decomposition, step 1.1 places $[x,y]$ in the weight space for $\alpha +\beta$. By [F1] this space is $\mathfrak{h}$ when the weight is zero and is zero when the weight is neither zero nor a root. This gives both claims.

For any $h\in \mathfrak{h}$, invariance from [L2] gives $0=B([h,x],y)+B(x,[h,y])=(\alpha (h)+\beta (h))B(x,y)$.

If $\alpha +\beta \mathrm{\ne }0$, choose $h\in \mathfrak{h}$ with $(\alpha +\beta )(h)\mathrm{\ne }0$; then step 1.1 forces $B(x,y)=0$. Also, [L1] identifies $\mathfrak{h}$ as the zero-weight space. For $h_0\in \mathfrak{h}$, the same invariance argument with $x=h_0$ and $y\in {\mathfrak{g}}_{\alpha }$ gives $B(h_0,y)=0$, so $\mathfrak{h}$ is orthogonal to every nonzero root space.

If $h_0\in \mathfrak{h}$ is orthogonal to $\mathfrak{h}$, then step 2.1 makes it orthogonal to every nonzero root space. Under Choice, the complete direct sum in [L1] makes it orthogonal to all of $\mathfrak{g}$, and global nondegeneracy in [L2] gives $h_0=0$. Thus $B{\mathrm{\mid }}_{\mathfrak{h}}$ is nondegenerate.

By [L0] and Brackets of root spaces add their roots, the bracket $[x,y]$ lies in ${\mathfrak{g}}_0=\mathfrak{h}$. For every $h\in \mathfrak{h}$, invariance of the Killing form gives $B([x,y],h)=B(x,[y,h])=\alpha (h)B(x,y)=B(x,y)B(H_{\alpha },h)$.

Because both $[x,y]$ and $B(x,y)H_{\alpha }$ lie in $\mathfrak{h}$ and have the same Killing pairings with every $h\in \mathfrak{h}$, nondegeneracy of the restriction from Under Choice, the Killing form pairs only opposite root spaces forces $[x,y]=B(x,y)H_{\alpha }$.

The root space ${\mathfrak{g}}_{\alpha }$ is nonzero by the definition of a root. Choose $0\mathrm{\ne }x\in {\mathfrak{g}}_{\alpha }$; the perfect opposite-root pairing in [L1] supplies $y\in {\mathfrak{g}}_{-\alpha }$ with $B(x,y)\mathrm{\ne }0$. Also $H_{\alpha }\mathrm{\ne }0$, for otherwise its defining equation would make the nonzero root $\alpha$ the zero functional. Thus step 2.1 gives a nonzero scalar multiple of $H_{\alpha }$ in the bracket image, while every bracket lies on that line. Hence $[{\mathfrak{g}}_{\alpha },{\mathfrak{g}}_{-\alpha }]=\mathbb{C}{H}_{\alpha }$.

Group the positive, zero, and negative root spaces in the supplied decomposition. This gives the direct sum $\mathfrak{g}={\mathfrak{n}}^{-}\oplus \mathfrak{h}\oplus {\mathfrak{n}}^{+}$.

If $x\in {\mathfrak{g}}_{\alpha }$ and $y\in {\mathfrak{g}}_{\beta }$, Jacobi gives $[h,[x,y]]=[[h,x],y]+[x,[h,y]]=(\alpha +\beta )(h)[x,y]$ for every $h\in \mathfrak{h}$. Thus $[x,y]$ belongs to the supplied weight space for $\alpha +\beta$, and is zero if that nonzero weight is absent. A sum of two positive roots cannot be zero or negative, so ${\mathfrak{n}}^{+}$ is closed under brackets. The same argument applies to ${\mathfrak{n}}^{-}$.

Choose ordered bases of ${\mathfrak{n}}^{-}$, $\mathfrak{h}$, and ${\mathfrak{n}}^{+}$ and concatenate them in that order. The ordered PBW monomials from PBW gives an ordered monomial basis for the enveloping algebra are then exactly products of a monomial in $U({\mathfrak{n}}^{-})$, one in $U(\mathfrak{h})$, and one in $U({\mathfrak{n}}^{+})$, so multiplication gives the stated vector-space isomorphism.

Write $x=x_0+{\sum }_{\alpha }{x}_{\alpha }$ using the root-space decomposition supplied in the Statement, with $x_0\in \mathfrak{h}$ and $x_{\alpha }\in {\mathfrak{g}}_{\alpha }$. Then $[h,x]={\sum }_{\alpha }\alpha (h)x_{\alpha }$.

The directness of the root-space decomposition implies that $[h,x]=0$ holds exactly when $x_{\alpha }=0$ for every root with $\alpha (h)\mathrm{\ne }0$. Thus the centralizer consists precisely of $\mathfrak{h}$ together with the root spaces on which $h$ vanishes.

By Regular elements and rank for a complex semisimple Lie algebra, $r$ is the minimum of the dimensions of the centralizers of elements of $\mathfrak{g}$. These dimensions belong to the finite set $\{0,\dots ,n\}$, and $\mathfrak{g}$ is nonempty, so the minimum is attained. Thus at least one regular element exists.

In a fixed basis, the matrix entries of ${\mathrm{ad}}_x$ depend linearly on $x$. Rank-nullity gives $\dim C_{\mathfrak{g}}(x)=n-\mathrm{rank}({\mathrm{ad}}_x)$. Hence regularity is the condition that the rank equals its attained maximum $n-r$. For $n-r>0$, the nonregular locus is the common zero set of the $(n-r)\times (n-r)$ minors. If $n-r=0$, every element is regular. In either case the regular locus is a nonempty Zariski-open set. A nonzero minor at a regular point remains nonzero on a Euclidean neighborhood; a polynomial vanishing on the regular locus therefore vanishes identically. Thus the regular locus is Zariski dense.

Let $x,y$ be regular. On the complex affine line through them, choose a maximal minor nonzero at $x$ (when $n-r=0$, the whole space is regular). Its restriction is a nonzero one-variable polynomial, so the nonregular points on that line form a finite set. A complex line with finitely many points removed is path connected, giving a path of regular elements from $x$ to $y$. The zero-dimensional case is a singleton. Thus the regular locus is connected.

By [A1, F1], an element of the connected adjoint group carries ${\mathfrak{h}}_1$ to ${\mathfrak{h}}_2$ by its adjoint automorphism.

The continuous image of a connected group is connected and contains the identity. Therefore its adjoint automorphisms lie in the identity component of $\mathrm{Aut}(\mathfrak{g})$, proving the claim.

By [F0], the supplied Cartan subalgebra is maximal toral, so [F1] applies. Its Killing-form restriction is nondegenerate, and its real form is positive definite. Thus its induced bilinear form on the real span of roots agrees with the dual form used in [F1]; [F2] identifies the corresponding root vectors with the Killing-dual notation of this page.

The remaining assertions are exactly the conclusions of [F1]: $\mathrm{\Phi }$ is finite, spans the real dual, is reduced and crystallographic, and every ${\mathfrak{g}}_{\alpha }$ is one-dimensional. The empty root system when $\mathfrak{g}=0$ is included there.

Define $T(u\otimes c_{\lambda })=uv$. If $b=h+x\in \mathfrak{b}=\mathfrak{h}\oplus {\mathfrak{n}}^{+}$, then $bv=\lambda (h)v$, exactly the scalar by which $b$ acts on $c_{\lambda }$; hence $T(ub\otimes c_{\lambda })=T(u\otimes b{c}_{\lambda })$ and $T$ descends to the induced module.

The descended map is $\mathfrak{g}$-linear and sends $v_{\lambda }$ to $v$. Conversely a homomorphism is determined by $v_{\lambda }$, since that vector generates $M(\lambda )$; its image necessarily has the two displayed properties.

Triangular PBW from Triangular decomposition from a chosen positive root system identifies $U(\mathfrak{g})$ with $U({\mathfrak{n}}^{-})\otimes U(\mathfrak{b})$. Tensoring over $U(\mathfrak{b})$ with ${\mathbb{C}}_{\lambda }$ therefore gives $U({\mathfrak{n}}^{-})\otimes {\mathbb{C}}_{\lambda }$.

Under that identification, $u\otimes c_{\lambda }$ maps to $u{v}_{\lambda }$, so the map is bijective. Applying the ordered PBW basis in the negative factor proves the asserted basis statement.

The supplied root-space direct sum has zero weight space $\mathfrak{h}$. Thus every $\mathrm{ad}(h)$ is diagonalizable, $\mathfrak{h}$ is abelian, and its centralizer is its zero weight space $\mathfrak{h}$; hence it is maximal toral. Apply [L2] to choose nonzero $e_i\in {\mathfrak{g}}_{{\alpha }_i}$ and $f_i\in {\mathfrak{g}}_{-{\alpha }_i}$ with $[e_i,f_i]={\alpha }_i^{\vee }$. The resulting ${\mathfrak{s}\mathfrak{l}}_2$ relations give, by induction, $e_i{f}_i^r{v}_{\lambda }=r(⟨\lambda ,{\alpha }_i^{\vee }⟩-r+1)f_i^{r-1}{v}_{\lambda }\mathrm{.}$ At $r=m=⟨\lambda ,{\alpha }_i^{\vee }⟩+1$ this is zero, while [L1] shows $f_i^m{v}_{\lambda }\mathrm{\ne }0$.

For $j\mathrm{\ne }i$, $[e_j,f_i]=0$ because ${\alpha }_j-{\alpha }_i$ is not a root; hence every $e_j$ also kills $f_i^m{v}_{\lambda }$. Its weight is $\lambda -m{\alpha }_i=s_i(\lambda +\rho )-\rho$, so it is singular of the stated dot weight.

The finite PBW and highest-weight construction in [F1] proves that every finite-dimensional simple $\mathfrak{g}$-module is generated by a highest-weight vector of a unique dominant integral weight $\lambda$, and is isomorphic to $L(\lambda )$. This is a cyclic highest-weight module in the stated sense.

By [F2], the weight from step 1.1 is exactly $\lambda ={\sum }_i{m}_i{\omega }_i$ with $m_i\in {\mathbb{N}}_0$. The rank-one integrality and nonnegativity needed for this assertion are established in [F1].

Conversely, for each $\lambda$ of the form in step 2.1, [F1] constructs a finite-dimensional simple $L(\lambda )$ and proves uniqueness up to isomorphism. Together with step 1.1 this gives both directions of the classification.

The highest-weight line is one-dimensional, so $zv$ must equal $cv$ for a unique scalar $c\in \mathbb{C}$. Indeed, $z$ commutes with $\mathfrak{h}$ and ${\mathfrak{n}}^{+}$, so $zv$ has the same weight as $v$ and is again killed by ${\mathfrak{n}}^{+}$.

Every element of $M$ has the form $uv$ with $u\in U(\mathfrak{g})$, and centrality gives $z(uv)=u(zv)=u(cv)=cuv$. Thus $z$ acts as $c{\mathrm{id}}_M$.

Sending $z$ to the scalar $c$ from step 1.1 is an algebra homomorphism on the center, so $M$ has the central character promised in Central character of a Lie algebra module.

Nondegeneracy identifies $\mathfrak{g}$ with ${\mathfrak{g}}^{\ast }$, and under that identification the tensor ${\sum }_i{x}_i\otimes x^i$ is the image of the identity map on $\mathfrak{g}$. Therefore it depends only on the form, not on the chosen dual bases.

Multiplication $\mathfrak{g}\otimes \mathfrak{g}\to U(\mathfrak{g})$ (The universal enveloping algebra as a tensor quotient) sends that basis-independent tensor to ${\sum }_i{x}_i{x}^i$. Its image is therefore basis independent as well.

Choose dual bases $x_i,x^i$ for the Killing form. Then $[x,C]={\sum }_i[x,x_i]x^i+x_i[x,x^i]$.

The basis-independent tensor $\mathrm{\Omega }:={\sum }_i{x}_i\otimes x^i$ corresponds, through the Killing-form identification $\mathfrak{g}\cong {\mathfrak{g}}^{\ast }$, to the identity endomorphism of $\mathfrak{g}$. Invariance of the form says that this identity tensor is fixed by the diagonal adjoint action. Hence ${\sum }_i[x,x_i]\otimes x^i+x_i\otimes [x,x^i]=0.$ Multiplying the tensor factors and using step 1.1 gives $[x,C]=0$. The quadratic Casimir element is independent of the choice of dual bases

Since $[x,C]=0$ for every $x\in \mathfrak{g}$, the element $C$ commutes with the generators of $U(\mathfrak{g})$ and is therefore central.

The supplied root-space decomposition [F1] makes every ${\mathrm{ad}}_h$, $h\in \mathfrak{h}$, diagonalizable, and its zero weight space is $C_{\mathfrak{g}}(\mathfrak{h})=\mathfrak{h}$. The latter equality makes $\mathfrak{h}$ abelian and forces every toral enlargement of $\mathfrak{h}$ to equal $\mathfrak{h}$; thus $\mathfrak{h}$ is maximal toral. Apply [F2]: every root space is one-dimensional, $B{\mathrm{\mid }}_{\mathfrak{h}}$ is nondegenerate, and opposite root spaces are perfectly paired. Invariance makes distinct nonopposite weight spaces orthogonal. Choose a basis $h_j$ of $\mathfrak{h}$ with Killing-dual basis $h^j$, and for each positive root choose $e_{\alpha }\in {\mathfrak{g}}_{\alpha }$ and $f_{\alpha }\in {\mathfrak{g}}_{-\alpha }$ with $B(e_{\alpha },f_{\alpha })=1$. These are full dual bases, so $C={\sum }_j{h}_j{h}^j+{\sum }_{\alpha >0}(e_{\alpha }{f}_{\alpha }+f_{\alpha }{e}_{\alpha })$. All choices are finite.