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 radical $\{x\in \mathfrak{g}:B(x,\mathfrak{g})=0\}$ is an ideal because step 1.1 makes it stable under brackets. Cartan's semisimplicity criterion states that a finite-dimensional Lie algebra over characteristic $0$ is semisimple if and only if its Killing form is nondegenerate. Since $\mathfrak{g}$ is complex semisimple, this criterion makes the radical zero.

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 $U(\mathfrak{g})$.

Reductions on disjoint adjacent pairs commute. The only overlapping ambiguity occurs in a word $x_k{x}_j{x}_i$ with $k>j>i$. Reducing the left pair first and the right pair first gives expressions whose difference is $[x_k,[x_j,x_i]]+[x_j,[x_i,x_k]]+[x_i,[x_k,x_j]],$ which is zero by the Jacobi identity. Thus every overlap is resolvable, and termination from step 1.1 implies that every word has a unique ordered normal form.

If a linear combination of distinct ordered monomials represented zero in the tensor quotient, its unique normal form from step 1.2 would be that same nonzero combination and also the zero normal form, a contradiction. The ordered monomials are therefore linearly independent and hence a basis.

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 2.1 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.

Because $\mathfrak{h}$ is a Cartan subalgebra of a complex semisimple Lie algebra, the commuting operators $\mathrm{ad}(h)$ for $h\in \mathfrak{h}$ are simultaneously diagonalizable, so $\mathfrak{g}$ decomposes into common eigenspaces for the adjoint action of $\mathfrak{h}$.

The common eigenspace for the zero functional is the centralizer of $\mathfrak{h}$, which equals $\mathfrak{h}$ because $\mathfrak{h}$ is self-normalizing. Every other common eigenfunctional is nonzero and contributes a subspace of the displayed form ${\mathfrak{g}}_{\alpha }$.

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 Root-space decomposition relative to a Cartan subalgebra, step 1.1 is exactly the defining condition for $[x,y]$ to lie in the root space with weight $\alpha +\beta$. If that root space is zero, then $[x,y]=0$.

For any $h\in \mathfrak{h}$, invariance from The Killing form is invariant and nondegenerate on a complex semisimple Lie algebra 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, for $h_0\in \mathfrak{h}$, the same argument with $x=h_0$ and $y\in {\mathfrak{g}}_{\alpha }$ shows $B(h_0,y)=0$, 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 summand in Root-space decomposition relative to a Cartan subalgebra, hence to all of $\mathfrak{g}$. Nondegeneracy from The Killing form is invariant and nondegenerate on a complex semisimple Lie algebra gives $h_0=0$, so the restriction of $B$ to $\mathfrak{h}$ is nondegenerate.

By 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 The Killing form pairs only opposite root spaces forces $[x,y]=B(x,y)H_{\alpha }$.

Choosing $x$ and $y$ with $B(x,y)\mathrm{\ne }0$ shows that the image of $[{\mathfrak{g}}_{\alpha },{\mathfrak{g}}_{-\alpha }]$ is exactly the line $\mathbb{C}{H}_{\alpha }$.

The decomposition in Root-space decomposition relative to a Cartan subalgebra groups the positive, zero, and negative root spaces, so it immediately gives the direct-sum decomposition $\mathfrak{g}={\mathfrak{n}}^{-}\oplus \mathfrak{h}\oplus {\mathfrak{n}}^{+}$.

If $x\in {\mathfrak{g}}_{\alpha }$ and $y\in {\mathfrak{g}}_{\beta }$ with $\alpha ,\beta \in {\mathrm{\Phi }}^{+}$, then Brackets of root spaces add their roots shows $[x,y]\in {\mathfrak{g}}_{\alpha +\beta }$, which is again positive or zero. Because no sum of positive roots is zero, ${\mathfrak{n}}^{+}$ is a Lie subalgebra, and the same argument gives the same for ${\mathfrak{n}}^{-}$.

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 $x=x_0+{\sum }_{\alpha }{x}_{\alpha }$ according to Root-space decomposition relative to a Cartan subalgebra, 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.

In any fixed basis of $\mathfrak{g}$, the matrix entries of ${\mathrm{ad}}_x$ depend linearly on $x$. By Regular elements and rank for a complex semisimple Lie algebra, regularity is the condition that $\mathrm{rank}({\mathrm{ad}}_x)=n-r$, so the nonregular locus is cut out by the vanishing of all $(n-r)\times (n-r)$ minors and is therefore Zariski closed.

Standard structure theory for complex semisimple Lie algebras supplies at least one regular element, so the complement of the closed set from step 1.1 is a nonempty Zariski-open subset. Because a nonempty Zariski-open subset of the complex affine space underlying $\mathfrak{g}$ is dense, the regular set is dense.

The complement of a proper complex algebraic subset of a finite-dimensional complex vector space is connected, so the nonempty open regular set from step 2.1 is connected as well. Hence the regular elements form a connected dense open subset.

By [F1], choose maximal tori $T_1,T_2\le G_{\mathrm{a}\mathrm{d}}$ with $\mathrm{Lie}(T_i)={\mathfrak{h}}_i$.

By [F2], some $g\in G_{\mathrm{a}\mathrm{d}}$ satisfies $g{T}_2{g}^{-1}=T_1$. Taking Lie algebras gives $\mathrm{Ad}(g)({\mathfrak{h}}_2)={\mathfrak{h}}_1$.

The adjoint group is the identity component of the inner automorphism group of $\mathfrak{g}$, so $\mathrm{Ad}(g)$ is an automorphism in the identity component. This proves the claimed conjugacy.

Orthogonality of distinct root spaces and nondegeneracy of the Killing form imply that $B$ restricts to a nondegenerate pairing ${\mathfrak{g}}_{\alpha }\times {\mathfrak{g}}_{-\alpha }\to \mathbb{C}$. Choose $e_{\alpha },f_{\alpha }$ with $B(e_{\alpha },f_{\alpha })\mathrm{\ne }0$. Then Opposite root spaces bracket to the Killing-dual line gives $[e_{\alpha },f_{\alpha }]=B(e_{\alpha },f_{\alpha })H_{\alpha }$. The scalar $\alpha (H_{\alpha })$ is nonzero: otherwise the generated three-dimensional algebra would be solvable, so its adjoint action could be triangularized, making $\mathrm{ad}[e_{\alpha },f_{\alpha }]$ nilpotent; but this element lies in $\mathfrak{h}$ and acts semisimply, forcing it to be zero, a contradiction. After rescaling, $e_{\alpha },f_{\alpha }$ and $h_{\alpha }:=2H_{\alpha }\mathrm{/}\alpha (H_{\alpha })$ form an ${\mathfrak{s}\mathfrak{l}}_2$-triple.

The subspace $\mathbb{C}{H}_{\alpha }\oplus {⨁}_{k\mathrm{\ne }0}{\mathfrak{g}}_{k\alpha }$ is a finite-dimensional module for this ${\mathfrak{s}\mathfrak{l}}_2$. Its zero-weight space is the line $\mathbb{C}{H}_{\alpha }$, so complete reducibility of ${\mathfrak{s}\mathfrak{l}}_2$-modules leaves one irreducible summand with even weights. Consequently its weight-$2$ space ${\mathfrak{g}}_{\alpha }$ is one-dimensional. Since $[e_{\alpha },e_{\alpha }]=0$, the raising operator kills that weight-$2$ space, so it is the highest weight: no $k\alpha$ with $\mathrm{\mid }k\mathrm{\mid }\ge 2$ is a root.

For roots $\alpha ,\beta$, the root-string space $V_{\alpha ,\beta }:={⨁}_{k\in \mathbb{Z}}{\mathfrak{g}}_{\beta +k\alpha }$ is a finite-dimensional ${\mathfrak{s}\mathfrak{l}}_2$-module. Since its nonzero weight spaces are one-dimensional by step 2.1, the vector in ${\mathfrak{g}}_{\beta }$ lies in one irreducible summand. The $h_{\alpha }$-weight $\beta (h_{\alpha })=2(\beta ,\alpha )\mathrm{/}(\alpha ,\alpha )$ is therefore an integer, and symmetry of the weights in that summand supplies the nonzero opposite-weight space ${\mathfrak{g}}_{\beta -\beta (h_{\alpha })\alpha }$. Thus $s_{\alpha }(\beta )=\beta -\beta (h_{\alpha })\alpha$ is again a root.

The roots span ${\mathfrak{h}}^{\ast }$, since an element of $\mathfrak{h}$ annihilated by every root would commute with all of $\mathfrak{g}$ and hence be zero. On the real span of the coroots, the Killing form is positive definite: $B(h,h)=\mathrm{tr}({\mathrm{ad}}_h^2)$ is the sum of squares of the real root eigenvalues and is positive for $h\mathrm{\ne }0$. Finiteness comes from the finite root decomposition; step 2.1 gives reducedness; and step 3.1 gives crystallographic integrality and reflection stability. Hence $\mathrm{\Phi }$ is a finite reduced crystallographic root system, and all its root spaces are one-dimensional.

In any finite-dimensional module, one can choose a weight maximal with respect to the positive-root order. Its weight space contains a nonzero vector killed by ${\mathfrak{n}}^{+}$, so every finite-dimensional simple module is a cyclic highest-weight module in the sense of Highest-weight vectors and cyclic highest-weight modules.

Restricting that module to each simple-root ${\mathfrak{s}\mathfrak{l}}_2$-subalgebra shows that the highest weight pairs nonnegatively and integrally with every simple coroot. By Fundamental weights for a chosen simple root system, the highest weight is therefore a dominant integral combination of the fundamental weights.

Let $V$ be a finite-dimensional simple module. By step 1.1 it is an irreducible highest-weight module of some weight $\lambda$, and step 2.1 shows that $\lambda$ is dominant integral. By [F1], $V\cong L(\lambda )$. Conversely, if $\lambda$ is dominant integral, then [F2] gives a finite-dimensional module $L(\lambda )$, and [F1] makes it the unique simple module with highest weight $\lambda$. This is exactly the claimed 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.

The Killing form 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 $\mathfrak{g}\otimes \mathfrak{g}\to U(\mathfrak{g})$ sends that basis-independent tensor to the element ${\sum }_i{x}_i{x}^i$ from The quadratic Casimir element, so the Casimir element is 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.

By Central elements act by scalars on cyclic highest-weight modules and The quadratic Casimir element is central, it is enough to compute $Cv$ on the highest vector $v$. The root-system theorem The root set is a reduced crystallographic root system makes every root space one-dimensional, while nondegeneracy and root-space orthogonality make ${\mathfrak{g}}_{\alpha }$ and ${\mathfrak{g}}_{-\alpha }$ dual. Choose a basis of $\mathfrak{h}$ and root vectors $e_{\alpha }\in {\mathfrak{g}}_{\alpha }$, $f_{\alpha }\in {\mathfrak{g}}_{-\alpha }$ with $B(e_{\alpha },f_{\alpha })=1$; these are full dual bases, so the Casimir decomposes as $C={\sum }_j{h}_j{h}^j+{\sum }_{\alpha >0}(e_{\alpha }{f}_{\alpha }+f_{\alpha }{e}_{\alpha })$.

Since $e_{\alpha }v=0$ for every positive root, one has $f_{\alpha }{e}_{\alpha }v=0$ and $e_{\alpha }{f}_{\alpha }v=[e_{\alpha },f_{\alpha }]v$. By Opposite root spaces bracket to the Killing-dual line, that bracket is $H_{\alpha }$, so step 1.1 gives $Cv=({\sum }_j{h}_j{h}^j+{\sum }_{\alpha >0}{H}_{\alpha })v$.

The Cartan part acts on $v$ by $(\lambda ,\lambda )$, and the root contribution acts by ${\sum }_{\alpha >0}\lambda (H_{\alpha })=2(\lambda ,\rho )$. Therefore $Cv=(\lambda ,\lambda +2\rho )v$, so the Casimir scalar on $M$ is $(\lambda ,\lambda +2\rho )$.

If $z$ is central, then it commutes with every element of $\mathfrak{g}$, and in particular with every $h\in \mathfrak{h}$. Thus $[h,z]=0$ for all $h\in \mathfrak{h}$.

By definition of the adjoint weight decomposition that underlies The Harish-Chandra projection, the condition from step 1.1 is exactly the statement that $z$ has weight zero. Hence $z\in U(\mathfrak{g}{)}_0$.

By Central elements lie in the zero-weight subspace of $U(\mathfrak{g})$, both $z$ and $z^{\mathrm{\prime }}$ lie in $U(\mathfrak{g}{)}_0$. Write $z=\mathrm{pr}(z)+a$ and $z^{\mathrm{\prime }}=\mathrm{pr}(z^{\mathrm{\prime }})+a^{\mathrm{\prime }}$ with $a,a^{\mathrm{\prime }}\in {\mathfrak{n}}^{-}U(\mathfrak{g})+U(\mathfrak{g}){\mathfrak{n}}^{+}$.

Put $I:={\mathfrak{n}}^{-}U(\mathfrak{g})+U(\mathfrak{g}){\mathfrak{n}}^{+}$. Multiplication on either side by $U(\mathfrak{h})$ preserves $I$. Moreover, if $c$ is central, then $Ic\subseteq I$: for a term $xu$ with $x\in {\mathfrak{n}}^{-}$, one has $(xu)c=x(uc)$, while for a term $uy$ with $y\in {\mathfrak{n}}^{+}$, centrality gives $(uy)c=(uc)y$.

Using only the decomposition of $z$, write $z{z}^{\mathrm{\prime }}=\mathrm{pr}(z)z^{\mathrm{\prime }}+a{z}^{\mathrm{\prime }}=\mathrm{pr}(z)\mathrm{pr}(z^{\mathrm{\prime }})+\mathrm{pr}(z)a^{\mathrm{\prime }}+a{z}^{\mathrm{\prime }}\mathrm{.}$ The last two terms lie in $I$ by step 2.1; no assertion that $I$ is multiplicatively closed is needed.

Applying the projection from The Harish-Chandra projection to step 3.1 leaves exactly $\mathrm{pr}(z)\mathrm{pr}(z^{\mathrm{\prime }})$. Therefore $\mathrm{pr}(z{z}^{\mathrm{\prime }})=\mathrm{pr}(z)\mathrm{pr}(z^{\mathrm{\prime }})$.

By Central elements lie in the zero-weight subspace of $U(\mathfrak{g})$, the central element $z$ lies in $U(\mathfrak{g}{)}_0$, so write $z=\mathrm{pr}(z)+a$ using The Harish-Chandra projection, with $a\in {\mathfrak{n}}^{-}U(\mathfrak{g})+U(\mathfrak{g}){\mathfrak{n}}^{+}$.

Every summand of $a$ either has a factor from ${\mathfrak{n}}^{+}$ on the right, which kills the highest vector $v$, or has a nontrivial factor from ${\mathfrak{n}}^{-}$ on the left, which lowers the weight and therefore cannot contribute to the highest-weight line. Hence $av=0$.

The element $\mathrm{pr}(z)$ lies in $U(\mathfrak{h})$, so it acts on $v$ by the scalar obtained from the polynomial $\mathrm{pr}(z)$ by evaluation at the weight $\lambda$. Combining this with step 2.1 gives $zv=\mathrm{pr}(z)(\lambda )v$.

If $f$ is dot-invariant, then for every $w\in W$ and $\lambda \in {\mathfrak{h}}^{\ast }$ one has $(Tf)(w\lambda )=f(w\lambda -\rho )=f(w\cdot (\lambda -\rho ))=f(\lambda -\rho )=(Tf)(\lambda ),$ because [F1] gives $w\cdot (\lambda -\rho )=w((\lambda -\rho )+\rho )-\rho =w\lambda -\rho \mathrm{.}$