[proofplan] We fix a unitary coframe and use the Hodge star convention specified in the theorem statement, for which $dV$ is the positive volume form. The key point is the stated primitive Hodge star identity: for a primitive $(p,q)$-form $\alpha$ of degree $k\le n$, the star of $\overline{\alpha}$ is an explicit signed multiple of $L^{n-k}\overline{\alpha}$. Substituting this identity converts the primitive Hodge-Riemann expression into the Hermitian norm density $\alpha\wedge *\overline{\alpha}$. The general Lefschetz-summand statement then follows because the stated normalization is a positive scalar multiple of the primitive form, while mixed summands are orthogonal by definition of $Q_\ell$. [/proofplan]

[step:Fix the unitary coframe and the Hodge star convention]Let $V^{1,0*}$ denote the $(1,0)$-cotangent subspace of $V^*$ determined by the complex structure. Choose a unitary coframe \begin{align*} \theta_1,\dots,\theta_n\in V^{1,0*} \end{align*} for the Hermitian [vector space](/page/Vector%20Space) $(V,h)$. We use the convention \begin{align*} \omega=i\sum_{a=1}^n\theta_a\wedge\overline{\theta_a}. \end{align*} The positive volume form is \begin{align*} dV:=\frac{\omega^n}{n!}. \end{align*} This is the Hodge star convention from the theorem statement: the map \begin{align*} *: \Lambda^jV^*\to\Lambda^{2n-j}V^* \end{align*} is characterized by \begin{align*} \eta\wedge *\overline{\xi}=(\eta,\xi)_h dV \end{align*} for every $\eta,\xi\in\Lambda^jV^*$, where $(\cdot,\cdot)_h$ is the induced Hermitian [inner product](/page/Inner%20Product) on $\Lambda^jV^*$, linear in the first argument.[/step]

[guided]A sign-sensitive proof needs the orientation and star convention fixed before any computation. We choose a unitary coframe $\theta_1,\dots,\theta_n$ of $V^{1,0*}$, so the Kähler form is \begin{align*} \omega=i\sum_{a=1}^n\theta_a\wedge\overline{\theta_a}. \end{align*} This convention determines the positive volume form \begin{align*} dV:=\frac{\omega^n}{n!}. \end{align*} The Hodge star is the unique complex-[linear map](/page/Linear%20Map) \begin{align*} *: \Lambda^jV^*\to\Lambda^{2n-j}V^* \end{align*} such that, for every pair of $j$-forms $\eta,\xi\in\Lambda^jV^*$, \begin{align*} \eta\wedge *\overline{\xi}=(\eta,\xi)_h dV. \end{align*} The inner product is linear in $\eta$ and conjugate-linear in $\xi$. This is exactly the convention needed to identify a wedge product with a positive norm density, because substituting $\eta=\xi$ gives \begin{align*} \eta\wedge *\overline{\eta}=|\eta|_h^2dV. \end{align*}[/guided]

[step:Use the primitive Hodge star identity] Let $0\le p,q\le n$, let $k=p+q\le n$, and let $\alpha\in P^{p,q}$. By the primitive Hodge star identity assumed in the theorem statement, \begin{align*} *\overline{\alpha}=i^{p-q}(-1)^{k(k-1)/2}\frac{\omega^{n-k}}{(n-k)!}\wedge\overline{\alpha}. \end{align*} The hypothesis $\alpha\in P^{p,q}$ verifies the primitivity condition required for this identity. Thus, for every $\alpha,\beta\in P^{p,q}$, \begin{align*} \alpha\wedge *\overline{\beta}=i^{p-q}(-1)^{k(k-1)/2}\alpha\wedge\overline{\beta}\wedge\frac{\omega^{n-k}}{(n-k)!}. \end{align*} By the defining property of the Hodge star, \begin{align*} \alpha\wedge *\overline{\beta}=(\alpha,\beta)_h\frac{\omega^n}{n!}. \end{align*} Comparing the two displayed identities gives \begin{align*} Q_{p,q}(\alpha,\beta)=(\alpha,\beta)_h. \end{align*} Hence $Q_{p,q}$ is Hermitian because it equals the Hermitian inner product induced by $h$ on $P^{p,q}$. [/step]

[step:Deduce positivity on primitive forms] Let $\alpha\in P^{p,q}$ be nonzero. Since the Hermitian inner product induced by $h$ on $\Lambda^kV^*$ is positive definite, \begin{align*} (\alpha,\alpha)_h>0. \end{align*} From the previous step, \begin{align*} Q_{p,q}(\alpha,\alpha)=(\alpha,\alpha)_h. \end{align*} Therefore \begin{align*} Q_{p,q}(\alpha,\alpha)>0. \end{align*} This proves the primitive Hodge-Riemann positivity statement. [/step]

[step:Reduce the Lefschetz summand form to the primitive form]Fix $0\le\ell\le 2n$, an admissible integer $r$, and write \begin{align*} m=\ell-2r. \end{align*} Let $m=p+q$. By the injectivity hypothesis in the theorem statement, the restricted map $L^r:P^{p,q}\to L^rP^{p,q}$ gives each vector in $L^rP^{p,q}$ a unique primitive representative. Thus the rule defining $Q_\ell(L^r\alpha,L^r\beta)$ for $\alpha,\beta\in P^{p,q}$ is independent of representative choice and is a well-defined sesquilinear form on the summand. Let $\alpha,\beta\in P^{p,q}$. Since $\ell=m+2r$, we have \begin{align*} n-\ell+2r=n-m. \end{align*} The defining expression for $Q_\ell$ on $L^rP^{p,q}$ is therefore \begin{align*} i^{p-q}(-1)^{m(m-1)/2}\frac{r!}{(n-\ell+r)!}\alpha\wedge\overline{\beta}\wedge\omega^{n-m}. \end{align*} Because $n-\ell+r=n-m-r$, this equals \begin{align*} \frac{r!(n-m)!}{(n-m-r)!}\left(i^{p-q}(-1)^{m(m-1)/2}\alpha\wedge\overline{\beta}\wedge\frac{\omega^{n-m}}{(n-m)!}\right). \end{align*} By the definition of $Q_{p,q}$ in degree $m$, \begin{align*} Q_\ell(L^r\alpha,L^r\beta)=\frac{r!(n-m)!}{(n-m-r)!}Q_{p,q}(\alpha,\beta). \end{align*} The coefficient $r!(n-m)!/(n-m-r)!$ is a positive real number, so $Q_\ell$ is Hermitian on $L^rP^{p,q}$ because $Q_{p,q}$ is Hermitian.[/step]

[guided]The general form looks different from the primitive form only because it uses the Lefschetz summand $L^rP^{p,q}$ in total degree $\ell$. Before comparing constants, we must check that the displayed formula really defines a form on vectors in the image $L^rP^{p,q}$, not merely on chosen preimages. Set \begin{align*} m=\ell-2r. \end{align*} The theorem statement assumes that $L^r:P^{p,q}\to L^rP^{p,q}$ is injective. Hence if a vector $u\in L^rP^{p,q}$ is written as $u=L^r\alpha$, then the primitive representative $\alpha\in P^{p,q}$ is unique. The same applies to $v=L^r\beta$. Therefore the definition of $Q_\ell(u,v)$ using $\alpha$ and $\beta$ is independent of representative choice. Now let $\alpha,\beta\in P^{p,q}$. These forms have primitive degree $m=p+q$, while $L^r\alpha$ and $L^r\beta$ have total degree $\ell=m+2r$. The exponent of $\omega$ in the definition of $Q_\ell$ is \begin{align*} n-\ell+2r=n-(m+2r)+2r=n-m. \end{align*} The factorial in the denominator is \begin{align*} (n-\ell+r)!=(n-m-r)!. \end{align*} Thus the defining wedge expression for $Q_\ell(L^r\alpha,L^r\beta)$ is \begin{align*} i^{p-q}(-1)^{m(m-1)/2}\frac{r!}{(n-m-r)!}\alpha\wedge\overline{\beta}\wedge\omega^{n-m}. \end{align*} To compare this with the primitive form $Q_{p,q}$, insert and remove the factor $(n-m)!$: \begin{align*} i^{p-q}(-1)^{m(m-1)/2}\frac{r!}{(n-m-r)!}\alpha\wedge\overline{\beta}\wedge\omega^{n-m}=\frac{r!(n-m)!}{(n-m-r)!}\left(i^{p-q}(-1)^{m(m-1)/2}\alpha\wedge\overline{\beta}\wedge\frac{\omega^{n-m}}{(n-m)!}\right). \end{align*} The expression inside parentheses is exactly the wedge expression defining $Q_{p,q}(\alpha,\beta)$. Therefore the scalar multiplying $\omega^n/n!$ is \begin{align*} Q_\ell(L^r\alpha,L^r\beta)=\frac{r!(n-m)!}{(n-m-r)!}Q_{p,q}(\alpha,\beta). \end{align*} The coefficient is strictly positive because $r\ge0$, $n-m\ge0$, and admissibility gives $n-m-r\ge0$. Since this coefficient is real and positive, the displayed identity also shows that $Q_\ell$ is Hermitian on $L^rP^{p,q}$.[/guided]

[step:Conclude positivity and orthogonality on the Lefschetz decomposition] By the finite-dimensional Lefschetz decomposition assumed in the theorem statement, \begin{align*} \Lambda^\ell V^*=\bigoplus_{r,p,q}L^rP^{p,q},\qquad p+q=\ell-2r, \end{align*} with $r$ ranging over the admissible integers. Let $L^r\alpha\in L^rP^{p,q}$ be nonzero. Since the restricted map $L^r:P^{p,q}\to L^rP^{p,q}$ is injective by hypothesis, the assumption $L^r\alpha\ne0$ implies $\alpha\ne0$. Using the previous step and primitive positivity, \begin{align*} Q_\ell(L^r\alpha,L^r\alpha)=\frac{r!(n-m)!}{(n-m-r)!}Q_{p,q}(\alpha,\alpha)>0. \end{align*} Thus $Q_\ell$ is positive definite on every nonzero Lefschetz summand. Finally, if $(r,p,q)\ne(s,p',q')$, the theorem defines \begin{align*} Q_\ell(L^rP^{p,q},L^sP^{p',q'})=0. \end{align*} Hence the displayed Lefschetz decomposition is orthogonal for $Q_\ell$. This proves the full statement. [/step]