[proofplan] Let $\omega_g$ be the fundamental form associated to the Hermitian metric $g$. Since a Riemann surface has real dimension $2$, the [exterior derivative](/theorems/1525) $d\omega_g$ is a smooth $3$-form on a real $2$-dimensional manifold, hence must vanish. Thus the fundamental form is closed, and the defining condition for a Hermitian metric to be Kähler is satisfied. [/proofplan] [step:Define the fundamental form associated to the Hermitian metric] Let $J:TX\to TX$ denote the complex structure on the real tangent bundle of $X$. Since $g$ is Hermitian, for every point $p\in X$ the [bilinear form](/page/Bilinear%20Form) \begin{align*} g_p:T_pX\times T_pX\to\mathbb R \end{align*} satisfies \begin{align*} g_p(J_pu,J_pv)=g_p(u,v) \end{align*} for all tangent vectors $u,v\in T_pX$. Define the fundamental form of $g$ to be the smooth real $2$-form \begin{align*} \omega_g\in \Omega^2(X) \end{align*} given pointwise by \begin{align*} (\omega_g)_p(u,v)=g_p(J_pu,v) \end{align*} for every $p\in X$ and every $u,v\in T_pX$. This is precisely the real $2$-form whose closedness is required in the definition of a Kähler metric. [/step] [step:Show that the fundamental form is automatically closed in real dimension two] Because $X$ is a Riemann surface, $X$ has complex dimension $1$ and therefore real dimension $2$. The exterior derivative of the fundamental form is \begin{align*} d\omega_g\in \Omega^3(X). \end{align*} For every point $p\in X$, the real [vector space](/page/Vector%20Space) $T_pX$ has dimension $2$, so \begin{align*} \Lambda^3T_p^*X=\{0\}. \end{align*} Hence $(d\omega_g)_p=0$ for every $p\in X$, and therefore \begin{align*} d\omega_g=0. \end{align*} [guided] The only possible obstruction to a Hermitian metric being Kähler is the closedness of its fundamental form. Here the dimension removes that obstruction. Since $X$ is a Riemann surface, each real tangent space $T_pX$ is a $2$-dimensional real vector space. A $3$-form at $p$ is an alternating trilinear map \begin{align*} \Lambda^3T_pX\to\mathbb R, \end{align*} or equivalently an element of $\Lambda^3T_p^*X$. But there are no nonzero alternating $3$-forms on a $2$-dimensional real vector space, because any three tangent vectors in $T_pX$ are linearly dependent. Therefore \begin{align*} \Lambda^3T_p^*X=\{0\}. \end{align*} Now $\omega_g$ is a smooth real $2$-form, so its exterior derivative is a smooth real $3$-form: \begin{align*} d\omega_g\in \Omega^3(X). \end{align*} Evaluating at any point $p\in X$, we get \begin{align*} (d\omega_g)_p\in \Lambda^3T_p^*X=\{0\}. \end{align*} Thus $(d\omega_g)_p=0$ for every $p\in X$, which is exactly the statement that \begin{align*} d\omega_g=0. \end{align*} [/guided] [/step] [step:Conclude that the Hermitian metric is Kähler] By definition, a Hermitian metric on a complex manifold is Kähler precisely when its associated fundamental form is closed. The metric $g$ is Hermitian by hypothesis, and the previous step proves that its fundamental form satisfies \begin{align*} d\omega_g=0. \end{align*} Therefore $g$ is a Kähler metric on $X$. [/step]