The Chevalley–Eilenberg Laplacian is scalar on weight components

(The metric data.) Realize the root system by a compact connected semisimple group. By complex semisimple classification and Cartan conjugacy [F2], identify its complexified Lie algebra and Cartan with $(\mathfrak{g},\mathfrak{h})$, transporting the compact form and conjugation $\tau$. Its simply connected compact form exists by [F2]. Integrate the restriction of the given representation $\pi$ to the compact real Lie algebra to this simply connected group by Lie's second theorem, and unitarize the resulting representation. This gives a positive Hermitian form invariant under the compact real form $\mathfrak{u}$. Unitarize also its adjoint representation by [F2]. For $X\in \mathfrak{u}$, $\mathrm{ad}X$ is skew-adjoint, so $B(X,X)=\mathrm{tr}((\mathrm{ad}X{)}^2)<0$ for $X\mathrm{\ne }0$: nondegeneracy of $B$ makes the center zero. For $x=X+iY$ with $X,Y\in \mathfrak{u}$, $-B(x,\tau x)=-B(X,X)-B(Y,Y)>0$ if $x\mathrm{\ne }0$. Transport the complexified maximal-torus Cartan in the same identification; its real root span is ${\mathfrak{h}}_{\mathbb{R}}=i\mathrm{Lie}(T)$, with $\tau (H)=-H$, so $\tau ({\mathfrak{g}}_{\alpha })={\mathfrak{g}}_{-\alpha }$ and each root vector can be scaled to $B(e_{\alpha },-\tau e_{\alpha })=1$. Invariance gives $\pi (X{)}^{\ast }=-\pi (X)$ on $\mathfrak{u}$; complex linearity gives $\pi (x{)}^{\ast }=-\pi (\tau x)$, hence $\pi (e_{\alpha }{)}^{\ast }=\pi (f_{\alpha })$ and $\pi (H{)}^{\ast }=\pi (H)$ on the real Cartan span. The ${\epsilon }_{\alpha }$ are orthonormal and the tensor form is positive Hermitian. The total Cartan operators $\mathrm{\Theta }(H)$ are self-adjoint on this span, so distinct weights are orthogonal.

(Hodge decomposition.) Since $C^q$ is finite dimensional and the form is positive definite, $⟨\mathrm{\square }c,c⟩=⟨d\delta c,c⟩+⟨\delta dc,c⟩=\mathrm{\parallel }\delta c{\mathrm{\parallel }}^2+\mathrm{\parallel }dc{\mathrm{\parallel }}^2\ge 0$, with equality precisely when $dc=\delta c=0$, which is precisely $\mathrm{\square }c=0$; moreover $\mathrm{im}d\perp \mathrm{im}\delta$ because $⟨d\alpha ,\delta \beta ⟩=⟨\alpha ,{\delta }^2\beta ⟩=0$, and $(\mathrm{im}d+\mathrm{im}\delta {)}^{\perp }=\ker \mathrm{\square }$, so $C^q=\mathrm{im}d\oplus \ker \mathrm{\square }\oplus \mathrm{im}\delta$. As $\ker d=\mathrm{im}d\oplus \ker \mathrm{\square }$ and $\ker \mathrm{\square }\cap \mathrm{im}d=0$, harmonic representatives give $H^q({\mathfrak{n}}^{+},V)\cong \ker \mathrm{\square }\cap C^q$, proving (i).

(Operator form and adjoints.) In the orthonormal basis, the differential of [F3] is $d=A+D$ with $A={\sum }_{\alpha }{\epsilon }_{\alpha }\pi (e_{\alpha })$ and $D=-\frac{1}{2}{\sum }_{\alpha ,\beta ,\gamma }{c}_{\alpha \beta }^{\gamma }{\epsilon }_{\alpha }{\epsilon }_{\beta }{\iota }_{\gamma }$: the first sum of the two-sum formula gives the action terms with the signs $(-1{)}^i$ after the exterior sign bookkeeping, and the second sum inserts the bracket into the first slot, which in the basis is the contraction expression displayed; the two sides are equal on the basis cochains ${\epsilon }_{i_1}\wedge \cdots \wedge {\epsilon }_{i_q}\otimes v$. For complex orthonormal exterior covectors, contraction deletes each occurrence with its alternating sign. On the orthonormal wedge basis this is the adjoint of wedge multiplication and satisfies the same anticommutation relations [F4], so the cited real-space formulas apply here by this basis computation. Taking adjoints with the linear-first Hermitian convention, using ${\epsilon }_{\alpha }^{\ast }={\iota }_{\alpha }$, $\pi (e_{\alpha }{)}^{\ast }=\pi (f_{\alpha })$ and the reversal of operator order with conjugated coefficients, gives $A^{\ast }={\sum }_{\alpha }\pi (f_{\alpha }){\iota }_{\alpha }$ and $D^{\ast }=-\frac{1}{2}{\sum }_{\alpha ,\beta ,\gamma }\overline{c_{\alpha \beta }^{\gamma }}{\epsilon }_{\gamma }{\iota }_{\beta }{\iota }_{\alpha }$.

(Mixed anticommutators.) Using $\{{\epsilon }_{\alpha },{\iota }_{\beta }\}={\delta }_{\alpha \beta }$, expansion of $A{A}^{\ast }+A^{\ast }A$ gives $\{A,A^{\ast }\}={\sum }_{\alpha }\pi (f_{\alpha })\pi (e_{\alpha })+{\sum }_{\alpha ,\beta }{\epsilon }_{\alpha }{\iota }_{\beta }\pi ([e_{\alpha },f_{\beta }])$, the double-sum term coming from ${\iota }_{\beta }{\epsilon }_{\alpha }={\delta }_{\alpha \beta }-{\epsilon }_{\alpha }{\iota }_{\beta }$ after moving the $\pi$-operators past the exterior operators. Likewise $A{D}^{\ast }+D^{\ast }A=-{\sum }_{\alpha ,\beta ,\gamma }\overline{c_{\alpha \beta }^{\gamma }}{\epsilon }_{\gamma }{\iota }_{\beta }\pi (e_{\alpha })$ and $D{A}^{\ast }+A^{\ast }D={\sum }_{\alpha ,\beta ,\gamma }{c}_{\alpha \beta }^{\gamma }{\epsilon }_{\alpha }{\iota }_{\gamma }\pi (f_{\beta })$: the degree-three exterior terms cancel between the two orders because ${\epsilon }_{\alpha }{\epsilon }_{\beta }$ is alternating while $c_{\alpha \beta }^{\gamma }$ is skew in $\alpha ,\beta$, and similarly for the degree-three contractions.

(Cancellation of root terms.) Under the normalization $B(e_{\alpha },f_{\alpha })=1$ and $f_{\alpha }=-\tau (e_{\alpha })$, the compact conjugation gives, whenever the stated differences are roots, $[e_{\alpha },f_{\beta }]=-\overline{c_{\beta \gamma }^{\alpha }}{e}_{\gamma }$ if $\alpha =\beta +\gamma$, $[e_{\alpha },f_{\beta }]=-c_{\alpha \gamma }^{\beta }{f}_{\gamma }$ if $\beta =\alpha +\gamma$, and $[e_{\alpha },f_{\alpha }]=H_{\alpha }$; hence $[e_{\beta },f_{\gamma }]={\mathbf{1}}_{\beta =\gamma }{H}_{\beta }+{\sum }_{s:\beta =s+\gamma }\overline{c_{s\gamma }^{\beta }}{e}_s-{\sum }_{s:\gamma =\beta +s}{c}_{\beta s}^{\gamma }{f}_s$. Substituting this into the double sum of $\{A,A^{\ast }\}$ term by term: every $e_s$-term cancels the corresponding term of $A{D}^{\ast }+D^{\ast }A$ and every $f_s$-term cancels the corresponding term of $D{A}^{\ast }+A^{\ast }D$, the index identifications being $\beta =s+\gamma$ against ${\gamma }^{\mathrm{\prime }}=\alpha +{\beta }^{\mathrm{\prime }}$ and $\gamma =\beta +s$ against ${\gamma }^{\mathrm{\prime }}={\alpha }^{\mathrm{\prime }}+{\beta }^{\mathrm{\prime }}$; the surviving diagonal is ${\sum }_{\alpha }{N}_{\alpha }\pi (H_{\alpha })=-{\sum }_j\pi (H_j)T(H_j)$, because ${\sum }_j\alpha (H_j)\pi (H_j)=\pi (H_{\alpha })$. Hence $\{A,A^{\ast }\}+\{A,D^{\ast }\}+\{D,A^{\ast }\}={\sum }_{\alpha }\pi (f_{\alpha })\pi (e_{\alpha })-{\sum }_j\pi (H_j)T(H_j)$, the second identity of (ii).

(Exterior anticommutator.) Normal-ordering $D{D}^{\ast }+D^{\ast }D$ with the relations of [F4], moving all contractions to the right, cancels all degree-six terms and yields $\{D,D^{\ast }\}=\frac{1}{2}{\sum }_{p,q,r}\mathrm{\mid }{c}_{pq}^r{\mathrm{\mid }}^2{N}_r+{\sum }_{a<b,c<d}{K}_{ab;cd}{\epsilon }_a{\epsilon }_b{\iota }_c{\iota }_d$ with $K_{ab;cd}=-{\sum }_s{c}_{ab}^s\overline{c_{cd}^s}-{\sum }_s\overline{c_{sc}^a}{c}_{sb}^d+{\sum }_s\overline{c_{sc}^b}{c}_{sa}^d+{\sum }_s\overline{c_{sd}^a}{c}_{sb}^c-{\sum }_s\overline{c_{sd}^b}{c}_{sa}^c$ for a fixed total order on the positive roots: the normal-ordering identities ${\iota }_r{\epsilon }_c={\delta }_{rc}-{\epsilon }_c{\iota }_r$ applied to the two orderings of the products produce, alongside the diagonal $\frac{1}{2}\sum \mathrm{\mid }c{\mathrm{\mid }}^2{N}_r$ read off the two $\delta \delta$ contractions, exactly the four ordered quartic monomials with the listed coefficient.

(Jacobi reduction of the quartic coefficient.) Write $X_{ac}=[e_a,f_c]=X_{ac}^{+}+X_{ac}^0+X_{ac}^{-}$ with $X_{ac}^{+}={\sum }_{s:a=s+c}\overline{c_{sc}^a}{e}_s$, $X_{ac}^0={\mathbf{1}}_{a=c}{H}_a$ and $X_{ac}^{-}=-{\sum }_{s:c=a+s}{c}_{as}^c{f}_s$, and put $S(X,Y)=B(X^{+},Y^{-})+B(Y^{+},X^{-})$. The four nonleading sums in $K$ combine with this notation as $-{\sum }_s\overline{c_{sc}^a}{c}_{sb}^d-{\sum }_s\overline{c_{sd}^b}{c}_{sa}^c+\cdots =-S(X_{ac},X_{bd})+S(X_{ad},X_{bc})$ after reindexing, and the leading sum is $B([e_a,e_b],[f_c,f_d])$ because $[f_c,f_d]=-{\sum }_s\overline{c_{cd}^s}{f}_s$ and $B(e_s,f_t)={\delta }_{st}$. Jacobi paired with $f_d$, using invariance $B([x,y],z)=B(y,[z,x])$, gives the identity $B([e_a,e_b],[f_c,f_d])=B(X_{ac},X_{bd})-B(X_{ad},X_{bc})$; since $B$ pairs only opposite root spaces, $B(X,Y)=S(X,Y)+B(X^0,Y^0)$, so the root terms cancel and $K_{ab;cd}=B(X_{ac}^0,X_{bd}^0)-B(X_{ad}^0,X_{bc}^0)=(a,b)({\mathbf{1}}_{a=c}{\mathbf{1}}_{b=d}-{\mathbf{1}}_{a=d}{\mathbf{1}}_{b=c})$. With the fixed order $a<b$, $c<d$ the quartic part is therefore $-{\sum }_{a<b}(a,b)N_a{N}_b$.

(Linear coefficient.) Fix a positive root $r$, and let $P:\mathfrak{g}\to {\mathfrak{n}}^{+}$ be the projection along ${\mathfrak{n}}^{-}\oplus \mathfrak{h}$. On ${\mathfrak{n}}^{+}$ put $E=\mathrm{ad}{e}_r$ and $F=P\mathrm{ad}{f}_r$; these are endomorphisms, so $\mathrm{tr}(EF-FE)=0$, by interchanging the indices in the finite matrix sums for their traces. Since $\mathrm{ad}{e}_r$ preserves ${\mathfrak{n}}^{+}$, one has $P\mathrm{ad}{H}_r{\mathrm{\mid }}_{{\mathfrak{n}}^{+}}=[E,F]+R$, with $R=P\mathrm{ad}{e}_r(1-P)\mathrm{ad}{f}_r{\mathrm{\mid }}_{{\mathfrak{n}}^{+}}$. Its diagonal is explicit: on $e_r$, $[f_r,e_r]=-H_r$ and $[e_r,-H_r]=\mathrm{\parallel }r{\mathrm{\parallel }}^2{e}_r$; for $\beta >0$ with $r=\beta +\gamma$, $[f_r,e_{\beta }]=c_{\beta \gamma }^r{f}_{\gamma }$ and $[e_r,f_{\gamma }]=\overline{c_{\beta \gamma }^r}{e}_{\beta }$, so $R{e}_{\beta }=\mathrm{\mid }{c}_{\beta \gamma }^r{\mathrm{\mid }}^2{e}_{\beta }$; all other root vectors give zero. Taking traces yields $2(\rho ,r)=\mathrm{\parallel }r{\mathrm{\parallel }}^2+{\sum }_{\beta +\gamma =r}\mathrm{\mid }{c}_{\beta \gamma }^r{\mathrm{\mid }}^2$, hence $\frac{1}{2}{\sum }_{a,b}\mathrm{\mid }{c}_{ab}^r{\mathrm{\mid }}^2=(\rho ,r)-\frac{1}{2}\mathrm{\parallel }r{\mathrm{\parallel }}^2$. With $N_r^2=N_r$, steps 1.6 and 1.7 therefore give $\{D,D^{\ast }\}={\sum }_r((\rho ,r)-\frac{1}{2}\mathrm{\parallel }r{\mathrm{\parallel }}^2)N_r-{\sum }_{a<b}(a,b)N_a{N}_b=-\frac{1}{2}{\sum }_{j}T(H_j{)}^2-T(H_{\rho })$, proving the first identity of (ii).

(Assembly and the scalar.) Since $\mathrm{\square }=d\delta +\delta d=\{d,d^{\ast }\}$ and $d=A+D$, steps 1.5 and 2.1 give $\mathrm{\square }={\sum }_{\alpha }\pi (f_{\alpha })\pi (e_{\alpha })-{\sum }_j\pi (H_j)T(H_j)-\frac{1}{2}{\sum }_{j}T(H_j{)}^2-T(H_{\rho })$; expanding ${\sum }_j\mathrm{\Theta }(H_j{)}^2+2\mathrm{\Theta }(H_{\rho })={\sum }_j\pi (H_j{)}^2+2{\sum }_j\pi (H_j)T(H_j)+{\sum }_{j}T(H_j{)}^2+2\pi (H_{\rho })+2T(H_{\rho })$ and using $C_{\mathfrak{g}}={\sum }_j{H}_j^2+2H_{\rho }+2{\sum }_{\alpha }{f}_{\alpha }{e}_{\alpha }$ from [F5] gives $1\otimes \pi (C_{\mathfrak{g}})-{\sum }_j\mathrm{\Theta }(H_j{)}^2-2\mathrm{\Theta }(H_{\rho })=2[{\sum }_{\alpha }\pi (f_{\alpha })\pi (e_{\alpha })-{\sum }_j\pi (H_j)T(H_j)-\frac{1}{2}{\sum }_{j}T(H_j{)}^2-T(H_{\rho })]$, hence $2\mathrm{\square }=1\otimes \pi (C_{\mathfrak{g}})-{\sum }_j\mathrm{\Theta }(H_j{)}^2-2\mathrm{\Theta }(H_{\rho })$, the first identity of (iii). On a cochain of weight $\mu$ the operator $\mathrm{\Theta }(H)$ acts by $\mu (H)$, and $\pi (C_{\mathfrak{g}})$ acts on $V=L(\lambda )$ by the scalar $\mathrm{\parallel }\lambda +\rho {\mathrm{\parallel }}^2-\mathrm{\parallel }\rho {\mathrm{\parallel }}^2$ by [F5], so $\mathrm{\square }$ acts on $C_{\mu }^{\bullet }$ by the scalar $\frac{1}{2}(\mathrm{\parallel }\lambda +\rho {\mathrm{\parallel }}^2-\mathrm{\parallel }\rho {\mathrm{\parallel }}^2-\mathrm{\parallel }\mu {\mathrm{\parallel }}^2-2(\mu ,\rho ))=\frac{1}{2}(\mathrm{\parallel }\lambda +\rho {\mathrm{\parallel }}^2-\mathrm{\parallel }\mu +\rho {\mathrm{\parallel }}^2)$; this is nonnegative, and it vanishes exactly when $\mathrm{\parallel }\mu +\rho \mathrm{\parallel }=\mathrm{\parallel }\lambda +\rho \mathrm{\parallel }$, which by the equality case of [F7] happens exactly for $\mu \in W\cdot \lambda$, with the corresponding cochain spaces one-dimensional in degree $\mathrm{\ell }(w)$. The normalization is common: $B$ fixes $H_j$, $C_{\mathfrak{g}}$ and the root-vector metric $B(e_{\alpha },f_{\alpha })=1$ simultaneously, and rescaling one of them independently changes the displayed scalar.