[proofplan] The proof is pointwise on each real tangent space $T_xX$. Starting from a smooth Hermitian metric $h$, the smooth projection $u\mapsto u^{1,0}$ transfers smoothness, positivity, and the unitary action of multiplication by $i$ to the real metric $g$. The identities for the fundamental form follow from $\omega(u,v)=g(Ju,v)$, the relation $J^2=-\operatorname{id}_{TX}$, and the $J$-invariance of $g$. Conversely, the map $u\mapsto u^{1,0}$ identifies the real tangent space with the $(1,0)$ tangent space, so the displayed formula defines $h$ uniquely, and the $J$-invariance of $g$ gives complex linearity, Hermitian symmetry, and positivity. [/proofplan]

[step:Work pointwise and record the type decomposition identities] Fix $x\in X$. Let $V=T_xX$ be the real tangent space, and let $J_x:V\to V$ be the complex structure at $x$. Its complex-linear extension to $V_{\mathbb C}=V\otimes_{\mathbb R}\mathbb C$ has the $i$-eigenspace $T_x^{1,0}X$. Define the real-[linear map](/page/Linear%20Map) $P:V\to T_x^{1,0}X$ by \begin{align*} P(u)=u^{1,0}=\frac{1}{2}(u-iJ_xu). \end{align*} Then \begin{align*} P(J_xu)=\frac{1}{2}(J_xu-iJ_x^2u)=\frac{1}{2}(J_xu+iu)=i\frac{1}{2}(u-iJ_xu)=iP(u). \end{align*} Thus \begin{align*} (J_xu)^{1,0}=i u^{1,0}. \end{align*} The map $P$ is a real-linear isomorphism. Indeed, if $P(u)=0$, then $u-iJ_xu=0$ in $V_{\mathbb C}$. Comparing real and imaginary parts gives $u=0$ and $J_xu=0$, so $P$ is injective. Since $\dim_{\mathbb R}V=2\dim_{\mathbb C}T_x^{1,0}X=\dim_{\mathbb R}T_x^{1,0}X$, injectivity implies bijectivity. [/step]

[step:Derive positivity and $J$-invariance of the real metric from the Hermitian metric]Assume first that $h$ is a Hermitian metric on $T^{1,0}X$, and define $g_x:V\times V\to\mathbb R$ by \begin{align*} g_x(u,v)=2\operatorname{Re} h_x(u^{1,0},v^{1,0}). \end{align*} Since $h_x$ is real-bilinear after taking real parts, the map $g_x$ is a real [bilinear form](/page/Bilinear%20Form). Since $h_x$ is Hermitian symmetric, $\operatorname{Re}h_x(Z,W)=\operatorname{Re}h_x(W,Z)$ for all $Z,W\in T_x^{1,0}X$, so $g_x$ is symmetric. For $u\in V$ with $u\neq 0$, the isomorphism $P$ gives $u^{1,0}\neq 0$. Positive definiteness of $h_x$ gives \begin{align*} g_x(u,u)=2h_x(u^{1,0},u^{1,0})>0. \end{align*} Thus $g_x$ is positive definite. It remains to check smoothness of the field $g$. Let $O\subset X$ be an [open set](/page/Open%20Set), and let $U,V:O\to TX$ be smooth real vector fields. Define smooth sections $U^{1,0},V^{1,0}:O\to T^{1,0}X$ by \begin{align*} U^{1,0}=\frac{1}{2}(U-iJU) \end{align*} and \begin{align*} V^{1,0}=\frac{1}{2}(V-iJV). \end{align*} These sections are smooth because $J:TX\to TX$ is smooth. Since $h$ is a smooth Hermitian metric, the function $p\mapsto h_p(U^{1,0}_p,V^{1,0}_p)$ is a smooth complex-valued function on $O$. Taking twice its real part shows that $p\mapsto g_p(U_p,V_p)$ is smooth on $O$. Hence $g$ is a smooth symmetric positive definite real bilinear form on $TX$, so $g$ is a Riemannian metric. For $u,v\in V$, using $(J_xu)^{1,0}=iu^{1,0}$ and the convention that $h_x$ is complex-linear in the first variable and conjugate-linear in the second, \begin{align*} g_x(J_xu,J_xv)=2\operatorname{Re}h_x(iu^{1,0},iv^{1,0})=2\operatorname{Re}\bigl(i\overline{i}h_x(u^{1,0},v^{1,0})\bigr)=g_x(u,v). \end{align*} Hence $J$ is an isometry for $g$.[/step]

[guided]We work at one point $x\in X$, because all the assertions are tensorial and therefore can be checked separately on each tangent space. Define $V=T_xX$ and define \begin{align*} g_x:V\times V\to\mathbb R \end{align*} by \begin{align*} g_x(u,v)=2\operatorname{Re}h_x(u^{1,0},v^{1,0}). \end{align*} First we check that this is a real [inner product](/page/Inner%20Product). Real bilinearity follows because $u\mapsto u^{1,0}$ is real-linear and because the real part of the Hermitian form is real-bilinear. Symmetry follows from Hermitian symmetry: for $Z,W\in T_x^{1,0}X$, the identity $h_x(W,Z)=\overline{h_x(Z,W)}$ implies \begin{align*} \operatorname{Re}h_x(W,Z)=\operatorname{Re}h_x(Z,W). \end{align*} Substituting $Z=u^{1,0}$ and $W=v^{1,0}$ gives \begin{align*} g_x(v,u)=g_x(u,v). \end{align*} Now we check positive definiteness. If $u\in V$ and $u\neq 0$, then $u^{1,0}\neq 0$ because the map $P:V\to T_x^{1,0}X$, $P(u)=u^{1,0}$, is injective. Since $h_x$ is a Hermitian metric, it is positive definite on $T_x^{1,0}X$, hence \begin{align*} h_x(u^{1,0},u^{1,0})>0. \end{align*} Therefore \begin{align*} g_x(u,u)=2h_x(u^{1,0},u^{1,0})>0. \end{align*} This proves that $g_x$ is an inner product on $T_xX$. We also have to check smoothness, because a Riemannian metric is a smooth field of inner products, not merely a pointwise inner product at each tangent space. Let $O\subset X$ be open, and let $U,V:O\to TX$ be smooth real vector fields. Define \begin{align*} U^{1,0}=\frac{1}{2}(U-iJU) \end{align*} and \begin{align*} V^{1,0}=\frac{1}{2}(V-iJV). \end{align*} The complex structure $J:TX\to TX$ is smooth, so $U^{1,0}$ and $V^{1,0}$ are smooth sections of $T^{1,0}X$ over $O$. Since $h$ is a smooth Hermitian metric, the function \begin{align*} p\mapsto h_p(U^{1,0}_p,V^{1,0}_p) \end{align*} is smooth and complex-valued on $O$. Therefore \begin{align*} p\mapsto g_p(U_p,V_p)=2\operatorname{Re}h_p(U^{1,0}_p,V^{1,0}_p) \end{align*} is a smooth real-valued function on $O$. Since $O$, $U$, and $V$ were arbitrary, $g$ is a smooth field of inner products, hence a Riemannian metric. Finally we verify that $J$ preserves this inner product. The key identity is \begin{align*} (J_xu)^{1,0}=iu^{1,0}. \end{align*} Using this identity for both arguments and using the convention that $h_x$ is complex-linear in the first argument and conjugate-linear in the second, we obtain \begin{align*} h_x((J_xu)^{1,0},(J_xv)^{1,0})=h_x(iu^{1,0},iv^{1,0})=i\overline{i}h_x(u^{1,0},v^{1,0})=h_x(u^{1,0},v^{1,0}). \end{align*} Taking twice the real part gives \begin{align*} g_x(J_xu,J_xv)=g_x(u,v). \end{align*} Thus $J$ is an isometry for $g$.[/guided]

[step:Obtain the invariance and skew-symmetry of the fundamental form] Define $\omega_x:V\times V\to\mathbb R$ by \begin{align*} \omega_x(u,v)=g_x(J_xu,v). \end{align*} Globally, $\omega$ is smooth because for every open set $O\subset X$ and smooth real vector fields $U,V:O\to TX$, the vector field $JU:O\to TX$ is smooth and the function $p\mapsto g_p(J_pU_p,V_p)$ is smooth by smoothness of $g$. For $u,v\in V$, $J$-invariance of $g_x$ gives \begin{align*} \omega_x(J_xu,J_xv)=g_x(J_x^2u,J_xv)=g_x(-u,J_xv)=-g_x(u,J_xv). \end{align*} Applying $J$-invariance of $g_x$ to the pair $(J_xu,v)$ gives \begin{align*} g_x(J_xu,J_xv)=g_x(u,v). \end{align*} Equivalently, applying it to the pair $(J_xu,J_x^2v)$ gives \begin{align*} g_x(J_xu,-v)=g_x(u,J_xv), \end{align*} so \begin{align*} -g_x(u,J_xv)=g_x(J_xu,v). \end{align*} Therefore \begin{align*} \omega_x(J_xu,J_xv)=g_x(J_xu,v)=\omega_x(u,v). \end{align*} The same identity gives skew-symmetry. Since $g_x$ is symmetric, \begin{align*} \omega_x(v,u)=g_x(J_xv,u)=g_x(u,J_xv)=-g_x(J_xu,v)=-\omega_x(u,v). \end{align*} Thus $\omega_x$ is $J$-invariant and skew-symmetric. Since $x$ was arbitrary, the stated identities hold on $TX$. [/step]

[step:Use the type decomposition to define the converse Hermitian form] Conversely, suppose $g$ is a Riemannian metric on $TX$ and satisfies \begin{align*} g_x(J_xu,J_xv)=g_x(u,v) \end{align*} for every $x\in X$ and all $u,v\in T_xX$. Fix $x\in X$ and set $V=T_xX$. Since $P:V\to T_x^{1,0}X$ is a real-linear isomorphism, every $Z\in T_x^{1,0}X$ has a unique real vector $u\in V$ such that $Z=u^{1,0}$. Define \begin{align*} h_x:T_x^{1,0}X\times T_x^{1,0}X\to\mathbb C \end{align*} by \begin{align*} h_x(u^{1,0},v^{1,0})=\frac{1}{2}\bigl(g_x(u,v)-i g_x(J_xu,v)\bigr). \end{align*} The uniqueness of $u$ and $v$ from the isomorphism $P$ makes this definition well-defined. Real bilinearity in $u$ and $v$ implies real bilinearity of $h_x$ in the two arguments before imposing complex scalar relations. [/step]

[step:Verify complex linearity and Hermitian symmetry of the converse form] Let $u,v\in V$. Since $i u^{1,0}=(J_xu)^{1,0}$, we compute \begin{align*} h_x(iu^{1,0},v^{1,0})=h_x((J_xu)^{1,0},v^{1,0})=\frac{1}{2}\bigl(g_x(J_xu,v)-i g_x(J_x^2u,v)\bigr). \end{align*} Using $J_x^2=-\operatorname{id}_V$, this becomes \begin{align*} h_x(iu^{1,0},v^{1,0})=\frac{1}{2}\bigl(g_x(J_xu,v)+i g_x(u,v)\bigr)=i h_x(u^{1,0},v^{1,0}). \end{align*} Hence $h_x$ is complex-linear in the first argument. Similarly, since $i v^{1,0}=(J_xv)^{1,0}$, \begin{align*} h_x(u^{1,0},iv^{1,0})=h_x(u^{1,0},(J_xv)^{1,0})=\frac{1}{2}\bigl(g_x(u,J_xv)-i g_x(J_xu,J_xv)\bigr). \end{align*} The $J$-invariance of $g_x$ gives $g_x(J_xu,J_xv)=g_x(u,v)$, and applying the same invariance to $(J_xu,J_x^2v)$ gives $g_x(u,J_xv)=-g_x(J_xu,v)$. Therefore \begin{align*} h_x(u^{1,0},iv^{1,0})=\frac{1}{2}\bigl(-g_x(J_xu,v)-i g_x(u,v)\bigr)=-i h_x(u^{1,0},v^{1,0}). \end{align*} Thus $h_x$ is conjugate-linear in the second argument. For Hermitian symmetry, symmetry of $g_x$ and the identity $g_x(J_xv,u)=-g_x(J_xu,v)$ give \begin{align*} h_x(v^{1,0},u^{1,0})=\frac{1}{2}\bigl(g_x(v,u)-i g_x(J_xv,u)\bigr)=\frac{1}{2}\bigl(g_x(u,v)+i g_x(J_xu,v)\bigr)=\overline{h_x(u^{1,0},v^{1,0})}. \end{align*} So $h_x$ is Hermitian symmetric. [/step]

[step:Verify positivity and recover the stated Hermitian metric] For $u\in V$, \begin{align*} h_x(u^{1,0},u^{1,0})=\frac{1}{2}\bigl(g_x(u,u)-i g_x(J_xu,u)\bigr). \end{align*} Using $g_x(u,J_xu)=-g_x(J_xu,u)$ and symmetry of $g_x$, we obtain \begin{align*} g_x(J_xu,u)=g_x(u,J_xu)=-g_x(J_xu,u). \end{align*} Hence $g_x(J_xu,u)=0$, and therefore \begin{align*} h_x(u^{1,0},u^{1,0})=\frac{1}{2}g_x(u,u). \end{align*} If $u^{1,0}\neq 0$, then $u\neq 0$ because $P$ is injective, and positive definiteness of the Riemannian metric $g_x$ gives $g_x(u,u)>0$. Thus \begin{align*} h_x(u^{1,0},u^{1,0})>0. \end{align*} Therefore $h_x$ is a positive definite Hermitian form on $T_x^{1,0}X$. It remains to verify that these pointwise Hermitian forms vary smoothly. Let $O\subset X$ be an open set, and let $Z,W:O\to T^{1,0}X$ be smooth sections. Since the bundle map $P:TX\to T^{1,0}X$, $P(u)=u^{1,0}$, is a smooth real-linear bundle isomorphism, there are unique smooth real vector fields $U,V:O\to TX$ such that $Z=U^{1,0}$ and $W=V^{1,0}$. The formula \begin{align*} p\mapsto h_p(Z_p,W_p)=\frac{1}{2}\bigl(g_p(U_p,V_p)-i g_p(J_pU_p,V_p)\bigr) \end{align*} defines a smooth complex-valued function on $O$, because $g$ and $J$ are smooth and $U,V$ are smooth. Hence the forms $h_x$ define a smooth positive definite Hermitian metric on $T^{1,0}X$. The construction uses the displayed formula for every $Z=u^{1,0}$ and $W=v^{1,0}$ in $T_x^{1,0}X$, so it proves exactly the converse assertion. This completes the proof. [/step]