[proofplan] We choose a $g$-orthonormal real frame adapted to the complex structure and write the exterior algebra as a [tensor product](/page/Tensor%20Product) of $n$ two-dimensional exterior algebras. In each coordinate plane, wedging by the local Kähler form and contracting by the corresponding bivector form an elementary creation-annihilation pair. A direct computation shows that the local commutator is the identity minus the local degree operator. Summing these local identities gives $H\alpha=(n-k)\alpha$ on $k$-forms, and the commutator relations with $L$ and $\Lambda$ follow from the fact that $L$ raises degree by $2$ while $\Lambda$ lowers degree by $2$. [/proofplan]

[step:Choose an adapted orthonormal coframe and write the Kähler form] Choose a unitary basis $(z_1,\dots,z_n)$ of the complex [vector space](/page/Vector%20Space) $(V,h)$, and let $(e_1,\dots,e_n)$ be the same vectors regarded as real vectors in $V_{\mathbb R}$. Define $f_j:=Je_j$ for $j\in\{1,\dots,n\}$. Since $h$ is Hermitian and $g=\operatorname{Re}h$, the ordered real basis $(e_1,f_1,\dots,e_n,f_n)$ is $g$-orthonormal. Let $(x_1,y_1,\dots,x_n,y_n)$ be its dual real coframe, so $x_j(e_m)=\delta_{jm}$, $x_j(f_m)=0$, $y_j(e_m)=0$, and $y_j(f_m)=\delta_{jm}$. For $u,v\in V_{\mathbb R}$, the definition $\omega(u,v)=g(Ju,v)$ gives $\omega(e_j,f_j)=1$ and gives zero on all other ordered pairs of distinct basis vectors except by skew-symmetry. Hence \begin{align*} \omega=\sum_{j=1}^n x_j\wedge y_j \end{align*} [/step]

[step:Define the local wedge and contraction operators] For each $j\in\{1,\dots,n\}$, define linear maps \begin{align*} a_j:E\to E,\qquad a_j(\alpha)=x_j\wedge \alpha \end{align*} and \begin{align*} b_j:E\to E,\qquad b_j(\alpha)=y_j\wedge \alpha. \end{align*} Let $\iota_{e_j}:E\to E$ and $\iota_{f_j}:E\to E$ denote contraction by the real vectors $e_j$ and $f_j$, respectively, extended complex-linearly to $E$. Because the coframe is $g$-orthonormal and the Hermitian [inner product](/page/Inner%20Product) on $E$ is the induced exterior-algebra inner product, the Hermitian adjoints of $a_j$ and $b_j$ are \begin{align*} a_j^*=\iota_{e_j},\qquad b_j^*=\iota_{f_j}. \end{align*} Define \begin{align*} L_j:E\to E,\qquad L_j(\alpha)=x_j\wedge y_j\wedge \alpha \end{align*} and \begin{align*} \Lambda_j:E\to E,\qquad \Lambda_j(\alpha)=\iota_{f_j}\iota_{e_j}\alpha. \end{align*} Since $L_j=a_jb_j$, its adjoint is \begin{align*} L_j^*=(a_jb_j)^*=b_j^*a_j^*=\iota_{f_j}\iota_{e_j}=\Lambda_j. \end{align*} Using $\omega=\sum_{j=1}^n x_j\wedge y_j$, we have \begin{align*} L=\sum_{j=1}^n L_j. \end{align*} Taking adjoints and using linearity of the adjoint in finite dimension gives \begin{align*} \Lambda=L^*=\sum_{j=1}^n L_j^*=\sum_{j=1}^n \Lambda_j. \end{align*} [/step]

[step:Compute the local commutator as identity minus local degree]For each $j\in\{1,\dots,n\}$, define the local degree operator $N_j:E\to E$ as follows. On a basis monomial \begin{align*} \eta=\theta_1\wedge\cdots\wedge\theta_k \end{align*} with each $\theta_m$ one of the covectors $x_1,y_1,\dots,x_n,y_n$, set $N_j\eta=r_j\eta$, where $r_j$ is the number of factors among $\theta_1,\dots,\theta_k$ equal to $x_j$ or $y_j$, and extend complex-linearly. Thus $r_j\in\{0,1,2\}$. Let $I:E\to E$ denote the identity endomorphism of $E$. We claim that \begin{align*} [\Lambda_j,L_j]=I-N_j \end{align*} as endomorphisms of $E$. To prove the claim, it is enough to evaluate both sides on the exterior monomial basis determined by the coframe. If a monomial contains neither $x_j$ nor $y_j$, then $\Lambda_jL_j$ acts as the identity and $L_j\Lambda_j$ acts as zero, so the commutator acts as $1$. If a monomial contains exactly one of $x_j,y_j$, then both composites vanish. If a monomial contains both $x_j$ and $y_j$, write it as $\eta=\varepsilon\,x_j\wedge y_j\wedge\eta_0$, where $\varepsilon\in\{1,-1\}$ and $\eta_0$ contains neither $x_j$ nor $y_j$. Then $L_j\eta=0$, while $\Lambda_j\eta=\varepsilon\eta_0$ and hence $L_j\Lambda_j\eta=\varepsilon x_j\wedge y_j\wedge\eta_0=\eta$. Therefore the commutator has eigenvalues $1,0,-1$ according as the local degree is $0,1,2$, which is precisely $I-N_j$.[/step]

[guided]Fix $j\in\{1,\dots,n\}$. Let $I:E\to E$ denote the identity endomorphism of $E$. The point of introducing $N_j$ is to isolate the contribution of the two real covectors $x_j$ and $y_j$. Every exterior monomial contains neither of them, exactly one of them, or both of them. Let $\eta\in E$ be a basis monomial. If $\eta$ contains neither $x_j$ nor $y_j$, then $L_j\eta=x_j\wedge y_j\wedge\eta$, and contraction first by $e_j$ and then by $f_j$ removes precisely the newly inserted factors. With the chosen order $x_j\wedge y_j$, this gives \begin{align*} \Lambda_jL_j\eta=\eta. \end{align*} Also $\Lambda_j\eta=0$, because at least one of the contractions $\iota_{e_j}$ or $\iota_{f_j}$ has no matching local factor to remove. Therefore \begin{align*} [\Lambda_j,L_j]\eta=\eta. \end{align*} In this case $N_j\eta=0$, so $(I-N_j)\eta=\eta$. If $\eta$ contains exactly one of $x_j$ and $y_j$, then $L_j\eta=0$ because wedging by $x_j\wedge y_j$ repeats one of the exterior factors. Also $\Lambda_j\eta=0$, because two contractions cannot remove both $x_j$ and $y_j$ when only one of them is present. Hence \begin{align*} [\Lambda_j,L_j]\eta=0. \end{align*} In this case $N_j\eta=\eta$, so $(I-N_j)\eta=0$. If $\eta$ contains both $x_j$ and $y_j$, then $L_j\eta=0$ because wedging by $x_j\wedge y_j$ repeats both exterior factors. Write \begin{align*} \eta=\varepsilon\,x_j\wedge y_j\wedge \eta_0 \end{align*} where $\varepsilon\in\{1,-1\}$ and $\eta_0$ is an exterior monomial containing neither $x_j$ nor $y_j$. Since $\iota_{e_j}(x_j\wedge y_j\wedge\eta_0)=y_j\wedge\eta_0$ and $\iota_{f_j}(y_j\wedge\eta_0)=\eta_0$, we have \begin{align*} \Lambda_j\eta=\varepsilon\eta_0. \end{align*} Therefore \begin{align*} \Lambda_jL_j\eta=0,\qquad L_j\Lambda_j\eta=\varepsilon x_j\wedge y_j\wedge\eta_0=\eta. \end{align*} Thus \begin{align*} [\Lambda_j,L_j]\eta=-\eta. \end{align*} In this case $N_j\eta=2\eta$, so $(I-N_j)\eta=-\eta$. The exterior monomials in the coframe form a complex basis of $E$. Since the two endomorphisms $[\Lambda_j,L_j]$ and $I-N_j$ agree on every basis monomial, they are equal on all of $E$: \begin{align*} [\Lambda_j,L_j]=I-N_j. \end{align*}[/guided]

[step:Sum the local commutators to identify the action of $H$ on homogeneous forms] If $j\ne m$, then $L_m$ and $\Lambda_j$ act on disjoint pairs of exterior generators. Since both operators have even degree, their graded sign interaction is trivial, and the ordinary commutator vanishes: \begin{align*} [\Lambda_j,L_m]=0. \end{align*} Therefore \begin{align*} H=[\Lambda,L]=\sum_{j=1}^n[\Lambda_j,L_j]. \end{align*} Using the local identity from the previous step, this becomes \begin{align*} H=\sum_{j=1}^n(I-N_j)=nI-\sum_{j=1}^nN_j. \end{align*} Let $k\in\{0,\dots,2n\}$ and let $\alpha\in E_k$. The operator $\sum_{j=1}^nN_j$ counts the total number of one-form factors in a homogeneous $k$-form, so \begin{align*} \sum_{j=1}^nN_j\alpha=k\alpha. \end{align*} Hence \begin{align*} H\alpha=(n-k)\alpha. \end{align*} [/step]

[step:Use the degree shifts of $L$ and $\Lambda$ to obtain the commutation relations] Let $k\in\{0,\dots,2n\}$ and let $\alpha\in E_k$. Since $L$ is wedging by a two-form, $L\alpha\in E_{k+2}$, with the convention that $E_\ell=\{0\}$ for $\ell\notin\{0,\dots,2n\}$. Therefore the eigenvalue formula for $H$ gives \begin{align*} HL\alpha=(n-k-2)L\alpha \end{align*} and \begin{align*} LH\alpha=(n-k)L\alpha. \end{align*} Subtracting these two identities gives \begin{align*} [H,L]\alpha=-2L\alpha. \end{align*} Since $E$ is the direct sum of the homogeneous subspaces $E_k$, this proves \begin{align*} [H,L]=-2L. \end{align*} Similarly, $\Lambda$ lowers degree by $2$ because each $\Lambda_j=\iota_{f_j}\iota_{e_j}$ is a double contraction. Thus $\Lambda\alpha\in E_{k-2}$, again with $E_\ell=\{0\}$ outside the allowed range. The eigenvalue formula gives \begin{align*} H\Lambda\alpha=(n-k+2)\Lambda\alpha \end{align*} and \begin{align*} \Lambda H\alpha=(n-k)\Lambda\alpha. \end{align*} Subtracting gives \begin{align*} [H,\Lambda]\alpha=2\Lambda\alpha. \end{align*} Again by homogeneity and direct summation over degrees, \begin{align*} [H,\Lambda]=2\Lambda. \end{align*} The identity \begin{align*} [\Lambda,L]=H \end{align*} is the definition of $H$. Together with the eigenvalue formula already proved, this completes the proof. [/step]