[proofplan] We prove the assertion pointwise and then integrate. At a fixed point, choose a unitary coframe for the Hermitian metric and expand the two $(p,q)$-forms in the induced orthonormal wedge basis. Complex conjugation sends this basis to an [orthonormal basis](/page/Orthonormal%20Basis) of $(q,p)$-forms and conjugates the coefficients, so the pointwise Hermitian pairing reverses the two entries. Integrating the pointwise identity against the real Riemannian volume form gives the $L^2$ identity, and the norm equality follows by setting $\beta=\alpha$. [/proofplan]

[step:Compute the pointwise pairing in a unitary coframe]Fix a point $x\in X$, and let $n$ be the complex dimension of $X$. Choose a unitary coframe \begin{align*} \theta_1,\dots,\theta_n\in T_x^{1,0}X^* \end{align*} for the Hermitian metric $g_x$ on $T_x^{1,0}X$. For each $r\ge 0$, let $\mathcal I_r$ denote the set of strictly increasing $r$-tuples with entries in $\{1,\dots,n\}$. For $I=(i_1,\dots,i_p)\in\mathcal I_p$ and $J=(j_1,\dots,j_q)\in\mathcal I_q$, define \begin{align*} e_{I,J}:=\theta_{i_1}\wedge\cdots\wedge\theta_{i_p}\wedge\overline{\theta_{j_1}}\wedge\cdots\wedge\overline{\theta_{j_q}}\in \Lambda^{p,q}T_x^*X. \end{align*} By the induced Hermitian metric on exterior powers, the family $(e_{I,J})_{I,J}$ is an orthonormal basis of $\Lambda^{p,q}T_x^*X$. Write \begin{align*} \alpha_x=\sum_{I\in\mathcal I_p}\sum_{J\in\mathcal I_q} a_{I,J}e_{I,J} \end{align*} and \begin{align*} \beta_x=\sum_{I\in\mathcal I_p}\sum_{J\in\mathcal I_q} b_{I,J}e_{I,J}, \end{align*} where $a_{I,J},b_{I,J}\in\mathbb C$ are the coefficients of $\alpha_x$ and $\beta_x$ in this basis. Since the pointwise Hermitian pairing is linear in the first argument and conjugate-linear in the second, \begin{align*} (\beta_x,\alpha_x)_{g,x} = \sum_{I\in\mathcal I_p}\sum_{J\in\mathcal I_q} b_{I,J}\overline{a_{I,J}}. \end{align*} Complex conjugation sends $\Lambda^{p,q}T_x^*X$ to $\Lambda^{q,p}T_x^*X$. Define \begin{align*} f_{I,J}:=\overline{e_{I,J}}\in \Lambda^{q,p}T_x^*X. \end{align*} The family $(f_{I,J})_{I,J}$ is orthonormal: indeed, each $f_{I,J}$ differs from the usual ordered wedge basis \begin{align*} \theta_{j_1}\wedge\cdots\wedge\theta_{j_q}\wedge\overline{\theta_{i_1}}\wedge\cdots\wedge\overline{\theta_{i_p}} \end{align*} only by the sign $(-1)^{pq}$, and multiplying basis vectors by signs preserves orthonormality. Therefore \begin{align*} \overline{\alpha_x} = \sum_{I\in\mathcal I_p}\sum_{J\in\mathcal I_q} \overline{a_{I,J}}f_{I,J} \end{align*} and \begin{align*} \overline{\beta_x} = \sum_{I\in\mathcal I_p}\sum_{J\in\mathcal I_q} \overline{b_{I,J}}f_{I,J}. \end{align*} Using again the linear-in-first convention, \begin{align*} (\overline{\alpha_x},\overline{\beta_x})_{g,x} = \sum_{I\in\mathcal I_p}\sum_{J\in\mathcal I_q} \overline{a_{I,J}}\,\overline{\overline{b_{I,J}}} = \sum_{I\in\mathcal I_p}\sum_{J\in\mathcal I_q} b_{I,J}\overline{a_{I,J}}. \end{align*} Hence, for every $x\in X$, \begin{align*} (\overline{\alpha_x},\overline{\beta_x})_{g,x}=(\beta_x,\alpha_x)_{g,x}. \end{align*}[/step]

[guided]The pointwise computation is the whole content of the theorem. Fix $x\in X$, and choose a unitary coframe \begin{align*} \theta_1,\dots,\theta_n\in T_x^{1,0}X^* \end{align*} for the Hermitian metric at $x$. The point of using a unitary coframe is that the induced wedge basis is orthonormal, so the Hermitian pairing becomes the standard coordinate pairing on coefficients. Let $\mathcal I_r$ be the set of strictly increasing $r$-tuples in $\{1,\dots,n\}$. For $I=(i_1,\dots,i_p)\in\mathcal I_p$ and $J=(j_1,\dots,j_q)\in\mathcal I_q$, define \begin{align*} e_{I,J}:=\theta_{i_1}\wedge\cdots\wedge\theta_{i_p}\wedge\overline{\theta_{j_1}}\wedge\cdots\wedge\overline{\theta_{j_q}}. \end{align*} These forms constitute an orthonormal basis of $\Lambda^{p,q}T_x^*X$ for the Hermitian metric induced on exterior powers. Therefore there are unique coefficients $a_{I,J},b_{I,J}\in\mathbb C$ such that \begin{align*} \alpha_x=\sum_{I\in\mathcal I_p}\sum_{J\in\mathcal I_q} a_{I,J}e_{I,J} \end{align*} and \begin{align*} \beta_x=\sum_{I\in\mathcal I_p}\sum_{J\in\mathcal I_q} b_{I,J}e_{I,J}. \end{align*} Because the Hermitian pairing is linear in the first argument and conjugate-linear in the second, orthonormal coordinates give \begin{align*} (\beta_x,\alpha_x)_{g,x} = \sum_{I\in\mathcal I_p}\sum_{J\in\mathcal I_q} b_{I,J}\overline{a_{I,J}}. \end{align*} Now conjugate the basis. Complex conjugation sends $(1,0)$-covectors to $(0,1)$-covectors and sends $(0,1)$-covectors to $(1,0)$-covectors, so it sends $(p,q)$-forms to $(q,p)$-forms. Define \begin{align*} f_{I,J}:=\overline{e_{I,J}}\in \Lambda^{q,p}T_x^*X. \end{align*} The only small sign issue is the order of the wedge factors: $\overline{e_{I,J}}$ places the conjugates in the order inherited from $e_{I,J}$, while the usual ordered $(q,p)$-basis places the $(1,0)$-factors first. Moving the $q$ factors past the $p$ factors introduces the sign $(-1)^{pq}$. Thus $f_{I,J}$ is still an orthonormal basis vector, and the whole family $(f_{I,J})_{I,J}$ is orthonormal. Complex conjugation also conjugates the coefficients, so \begin{align*} \overline{\alpha_x} = \sum_{I\in\mathcal I_p}\sum_{J\in\mathcal I_q} \overline{a_{I,J}}f_{I,J} \end{align*} and \begin{align*} \overline{\beta_x} = \sum_{I\in\mathcal I_p}\sum_{J\in\mathcal I_q} \overline{b_{I,J}}f_{I,J}. \end{align*} Applying the same linear-in-first Hermitian pairing convention in this orthonormal basis gives \begin{align*} (\overline{\alpha_x},\overline{\beta_x})_{g,x} = \sum_{I\in\mathcal I_p}\sum_{J\in\mathcal I_q} \overline{a_{I,J}}\,\overline{\overline{b_{I,J}}}. \end{align*} Since $\overline{\overline{b_{I,J}}}=b_{I,J}$, this becomes \begin{align*} (\overline{\alpha_x},\overline{\beta_x})_{g,x} = \sum_{I\in\mathcal I_p}\sum_{J\in\mathcal I_q} b_{I,J}\overline{a_{I,J}}. \end{align*} Comparing with the earlier expression for $(\beta_x,\alpha_x)_{g,x}$ proves the pointwise identity \begin{align*} (\overline{\alpha_x},\overline{\beta_x})_{g,x}=(\beta_x,\alpha_x)_{g,x}. \end{align*}[/guided]

[step:Integrate the pointwise identity to obtain the $L^2$ identity] The previous step also gives \begin{align*} |\overline{\alpha_x}|_g^2=|\alpha_x|_g^2 \end{align*} and \begin{align*} |\overline{\beta_x}|_g^2=|\beta_x|_g^2 \end{align*} for every $x\in X$. Since $\alpha$ and $\beta$ are square-integrable with respect to $dV_g$, it follows that $\overline{\alpha}$ and $\overline{\beta}$ are square-integrable with respect to $dV_g$. By definition of the $L^2$ [inner product](/page/Inner%20Product) on complex-valued forms, \begin{align*} (\overline{\alpha},\overline{\beta})_{L^2} = \int_X (\overline{\alpha_x},\overline{\beta_x})_{g,x}\, dV_g(x). \end{align*} Using the pointwise identity from the previous step, \begin{align*} (\overline{\alpha},\overline{\beta})_{L^2} = \int_X (\beta_x,\alpha_x)_{g,x}\, dV_g(x). \end{align*} The right-hand side is exactly $(\beta,\alpha)_{L^2}$ by the same definition of the $L^2$ inner product. Hence \begin{align*} (\overline{\alpha},\overline{\beta})_{L^2}=(\beta,\alpha)_{L^2}. \end{align*} [/step]

[step:Set $\beta=\alpha$ to get preservation of the $L^2$ norm] Applying the identity already proved with $\beta=\alpha$ gives \begin{align*} (\overline{\alpha},\overline{\alpha})_{L^2}=(\alpha,\alpha)_{L^2}. \end{align*} By definition of the $L^2$ norm induced by the Hermitian inner product, \begin{align*} \|\overline{\alpha}\|_{L^2}^2=(\overline{\alpha},\overline{\alpha})_{L^2} \end{align*} and \begin{align*} \|\alpha\|_{L^2}^2=(\alpha,\alpha)_{L^2}. \end{align*} Therefore \begin{align*} \|\overline{\alpha}\|_{L^2}^2=\|\alpha\|_{L^2}^2. \end{align*} Both norms are nonnegative [real numbers](/page/Real%20Numbers), so taking the nonnegative square root gives \begin{align*} \|\overline{\alpha}\|_{L^2}=\|\alpha\|_{L^2}. \end{align*} This proves the antiunitarity of complex conjugation on these $L^2$ forms. [/step]