Hermitian exterior Lefschetz decomposition and Weil star identity

Suppose $n=1$. Choose a unit vector $e_1$ and put $e_2=-J{e}_1$; then $g(e_1,e_2)=0$, $g(e_2,e_2)=1$, $J{e}_1=-e_2$, $J{e}_2=e_1$, and $\omega (e_1,e_2)=-g(J{e}_1,e_2)=g(e_2,e_2)=1$, so $\omega =e_1\wedge e_2$ is the positive unit volume form and $(e_1,e_2)$ is positively oriented. The basis $1,e_1,e_2,\omega$ of ${\bigwedge }^{\bullet }{V}_{\mathbb{C}}$ is orthonormal for $h$, and [F1] gives the star table $\star 1=\omega$, $\star e_1=e_2$, $\star e_2=-e_1$, $\star \omega =1$. Hence $L(1)=\omega$ and $L$ kills $e_1,e_2,\omega$, while the adjoint $\mathrm{\Lambda }$ satisfies $\mathrm{\Lambda }\omega =1$ and kills $1,e_1,e_2$. Therefore $[L,\mathrm{\Lambda }]=H$ is $-\mathrm{i}\mathrm{d}$, $0$, $0$, $\mathrm{i}\mathrm{d}$ on degrees $0,1,1,2$, that is $H=(k-1)\mathrm{i}\mathrm{d}$ on ${\bigwedge }^k$, and evaluating $HL-LH$ and $H\mathrm{\Lambda }-\mathrm{\Lambda }H$ on the four basis elements gives $[H,L]=2L$ and $[H,\mathrm{\Lambda }]=-2\mathrm{\Lambda }$. Evaluating the exponentials gives $W(1)=\omega$, $W(e_1)=e_1$, $W(e_2)=e_2$, $W(\omega )=-1$, while $J(1)=1$, $J{e}_1=-e_2$, $J{e}_2=e_1$, $J\omega =\omega$ give $ϵ(0)J\star 1=\omega$, $ϵ(1)J\star e_1=e_1$, $ϵ(1)J\star e_2=e_2$ and $ϵ(2)J\star \omega =-1$; since both sides are complex linear, $W=ϵ(k)J\star$ on ${\bigwedge }^k{V}_{\mathbb{C}}$ for $n=1$.

Let $V=V_1\oplus V_2$ with $V_1,V_2$ $g$-orthogonal and $J$-stable, oriented so that the positive volume form is ${\mathrm{vol}}_1\wedge {\mathrm{vol}}_2$; write ${\omega }_i=\omega {\mathrm{\mid }}_{V_i}$, so $\omega ={\omega }_1+{\omega }_2$, and let $L_i,{\mathrm{\Lambda }}_i,W_i,J_i$ be the corresponding operators on the factors. Under the linear isomorphism ${\bigwedge }^{\bullet }{V}_{1,\mathbb{C}}\otimes {\bigwedge }^{\bullet }{V}_{2,\mathbb{C}}\to {\bigwedge }^{\bullet }{V}_{\mathbb{C}}$, $a\otimes b\mapsto a\wedge b$, a check on orthonormal wedge bases gives the tensor rules $h(a\wedge b,a^{\mathrm{\prime }}\wedge b^{\mathrm{\prime }})=h_1(a,a^{\mathrm{\prime }})h_2(b,b^{\mathrm{\prime }})$, $\star (a\wedge b)=(-1{)}^{\mathrm{deg}a\mathrm{deg}b}({\star }_{1}a)\wedge ({\star }_{2}b)$ and $J(a\wedge b)=J_{1}a\wedge J_{2}b$ for homogeneous $a,a^{\mathrm{\prime }}$ and $b,b^{\mathrm{\prime }}$, the middle one being forced by the characterizing property [F1] and consistent with the isometry property [F2]. Consequently $L=L_1+L_2$, $\mathrm{\Lambda }={\mathrm{\Lambda }}_1+{\mathrm{\Lambda }}_2$ (the adjoint of a sum of operators acting on the two tensor factors is the sum of the adjoints), $H=H_1+H_2$ where $H_i=(\mathrm{deg}-n_i)\mathrm{i}\mathrm{d}$ on the $i$-th factor with $n_i=\dim V_i\mathrm{/}2$, and $W=W_1{W}_2$, $J=J_1{J}_2$, because operators on the two factors commute. If the three commutator relations and the identity $W=ϵ(k)J\star$ hold on both factors, then they hold on $V$: the commutator relations split termwise, and for homogeneous $a,b$ the identity gives $W(a\wedge b)=ϵ(\mathrm{deg}a)ϵ(\mathrm{deg}b)(J_1{\star }_{1}a)\wedge (J_2{\star }_{2}b)$, which equals $ϵ(\mathrm{deg}a+\mathrm{deg}b)J\star (a\wedge b)$ because $ϵ(m+\mathrm{\ell })=(-1{)}^{m\mathrm{\ell }}ϵ(m)ϵ(\mathrm{\ell })$ and the sign $(-1{)}^{\mathrm{deg}a\mathrm{deg}b}$ of the star rule cancels against $(-1{)}^{m\mathrm{\ell }}$. Splitting off a two-plane $W=\mathrm{s}\mathrm{p}\mathrm{a}\mathrm{n}(e_1,-J{e}_1)$ with $V_2=W^{\perp }$ and iterating the base case 1.1 proves the three relations of part 1 and the identity of part 4 for every $n\ge 0$: the degenerate factor $V_2=0$ has ${\bigwedge }^{\bullet }{V}_2=\mathbb{C}$ in degree zero, where all operators are the identity or zero and the relations are immediate, and for $n=0$ both sides of $W=ϵ(k)J\star$ are the identity on this one-dimensional space.

Let $u\mathrm{\ne }0$ be primitive of degree $t$, so $\mathrm{\Lambda }u=0$. From $[L,\mathrm{\Lambda }]=H$ and $H=(t+2r-n)\mathrm{i}\mathrm{d}$ on degree $t+2r$ we get, by induction on $r\ge 0$, the recurrence $\mathrm{\Lambda }{L}^{r}u=r(n-t-r+1)L^{r-1}u$: indeed $\mathrm{\Lambda }{L}^{r}u=L\mathrm{\Lambda }{L}^{r-1}u-(t+2(r-1)-n)L^{r-1}u$ and substituting the induction hypothesis for $r-1$ gives the claim. Iterating gives the displayed factorial formula. Let $N$ be the largest index with $L^{N}u\mathrm{\ne }0$, which exists because $L^{r}u=0$ once $t+2r>2n$. Then $L^{N+1}u=0$, so the recurrence with $r=N+1$ gives $0=\mathrm{\Lambda }{L}^{N+1}u=(N+1)(n-t-N)L^{N}u$; since $N+1\mathrm{\ne }0$ and $L^{N}u\mathrm{\ne }0$, we get $N=n-t$. Hence $n-t\ge 0$, so no nonzero primitive element has degree $t>n$; moreover $L^{n-t}u\mathrm{\ne }0$ and $L^{n-t+1}u=0$, and all coefficients in the factorial formula are nonzero for $0\le s\le r\le n-t$.

Let $\alpha \in {\bigwedge }^k{V}_{\mathbb{C}}$. Since $\mathrm{\Lambda }=L^{\ast }$ on the finite-dimensional space ${\bigwedge }^k{V}_{\mathbb{C}}$, the image of $L:{\bigwedge }^{k-2}{V}_{\mathbb{C}}\to {\bigwedge }^k{V}_{\mathbb{C}}$ is the orthogonal complement of $\ker \mathrm{\Lambda }$ on ${\bigwedge }^k{V}_{\mathbb{C}}$, so $\alpha =u_0+L{\alpha }_1$ with $\mathrm{\Lambda }{u}_0=0$ and ${\alpha }_1\in {\bigwedge }^{k-2}{V}_{\mathbb{C}}$. Inducting on $k$ (the cases $k=0$ and $k=1$ have no $L$-term) writes $\alpha ={\sum }_r{L}^r{u}_r$ with each $u_r$ primitive; terms with $\mathrm{deg}{u}_r>n$ vanish by step 3.1, and $L^r{u}_r=0$ when $r>n-\mathrm{deg}{u}_r$, equivalently $r<k-n$. Discarding these zero terms gives the stated range $r\ge \max (0,k-n)$. For uniqueness, suppose ${\sum }_{r=r_0}^s{L}^r{u}_r=0$ with each $u_r$ primitive and $u_s\mathrm{\ne }0$. Put $c_r=r!(n-k+2r)!\mathrm{/}(n-k+r)!$, which is a positive integer in this range. The factorial formula gives ${\mathrm{\Lambda }}^s{L}^s{u}_s=c_s{u}_s$, while for $r<s$ it gives ${\mathrm{\Lambda }}^s{L}^r{u}_r={\mathrm{\Lambda }}^{s-r}(c_r{u}_r)=0$ because $\mathrm{\Lambda }{u}_r=0$; applying ${\mathrm{\Lambda }}^s$ to the sum gives $c_s{u}_s=0$, a contradiction. For the projections start with the fixed index $s=\lfloor k\mathrm{/}2\rfloor$, allowing zero coefficients: $u_s=c_s^{-1}{\mathrm{\Lambda }}^s\alpha$. Subtract $L^s{u}_s$ and repeat with $s-1$ down to $\max (0,k-n)$. This expresses each $u_r$ by a fixed rational linear combination of compositions of $L,\mathrm{\Lambda }$, without choices.

Choose a basis $(u_{\alpha })$ of the direct sum $P={⨁}_j{P}^j$ of primitive spaces $P^j=\ker \mathrm{\Lambda }\cap {\bigwedge }^j{V}_{\mathbb{C}}$, a finite direct sum of finite-dimensional spaces. By step 4.1 every element of ${\bigwedge }^{\bullet }{V}_{\mathbb{C}}$ has a unique expression as a finite sum of string elements $L^r{u}_{\alpha }$ with $0\le r\le n-\mathrm{deg}{u}_{\alpha }$; hence these string elements form a basis, and the string of $u_{\alpha }$ occupies each degree $t_{\alpha },t_{\alpha }+2,\dots ,2n-t_{\alpha }$. Fix $k\le n$ and let $t_{\alpha }=\mathrm{deg}{u}_{\alpha }$. Multiplication by $L^{n-k}$ sends the string element at position $r$ to position $r+n-k$, which is a valid position of the same string because $r+n-k\le n-t_{\alpha }$ follows from $k=t_{\alpha }+2r$; conversely position $r^{\mathrm{\prime }}$ of a degree-$2n-k$ element of the string is the image of $r=r^{\mathrm{\prime }}-(n-k)$ with $k=t_{\alpha }+2r\ge 0$. On the direct sum of strings, $L^{n-k}$ is therefore a bijection from the degree-$k$ part onto the degree-$2n-k$ part, an isomorphism $L^{n-k}:{\bigwedge }^k{V}_{\mathbb{C}}\to {\bigwedge }^{2n-k}{V}_{\mathbb{C}}$. Similarly, $L^{n-k+1}{L}^r{u}_{\alpha }=L^{r+n-k+1}{u}_{\alpha }$ is zero exactly when $r=0$, and for $r\ge 1$ it is a nonzero string element; expanding $u={\sum }_{r\ge 0,\alpha }{c}_{r,\alpha }{L}^r{u}_{\alpha }$ in degree $k$ and using basis independence, $L^{n-k+1}u=0$ forces $c_{r,\alpha }=0$ for every $r\ge 1$, so $u={\sum }_{\alpha }{c}_{0,\alpha }{u}_{\alpha }$ is primitive. The converse is step 3.1.

Since $J$ acts as $i$ on $V^{1,0}$ and $-i$ on $V^{0,1}$ and is multiplicative, $J=i^{p-q}$ on ${\bigwedge }^{p,q}{V}_{\mathbb{C}}$. Let $u\mathrm{\ne }0$ be primitive of degree $k$ and type $(p,q)$, put $m=n-k\ge 0$, and apply $W=\exp L\exp (-\mathrm{\Lambda })\exp L$ to $\exp (-L)u$: since $\mathrm{\Lambda }u=0$ we get $W\exp (-L)u=\exp (L)\exp (-\mathrm{\Lambda })u=\exp (L)u$. Expanding $\exp (-L)u$ and comparing the components of degree $2n-k$, of which the left side has exactly $Wu$ (all other terms are mapped to lower degrees by $W$, which reverses the $H$-weight) and the right side exactly $L^{m}u\mathrm{/}m!$, gives $Wu=L^{m}u\mathrm{/}m!$. Substituting into the identity $Wu=ϵ(k)J\star u$ of step 2.1 and applying $J^{-1}$, which acts on $L^{m}u$ by $i^{-(p-q)}=i^{q-p}$ because $L^{m}u$ has type $(p+m,q+m)$, gives $\star u=ϵ(k)i^{q-p}{L}^{m}u\mathrm{/}m!$, which is the displayed formula. For $n=0$ both sides are the identity and the formula is $\ast 1=1$.