[proofplan] We pass from Betti cohomology to complex de Rham cohomology and choose an auxiliary Hermitian metric on the compact Riemann surface $X_0(N)$. The [Hodge theorem](/theorems/3942) gives each cohomology class a unique harmonic representative. On a compact Riemann surface, harmonic complex-valued $1$-forms split uniquely into their $(1,0)$ and $(0,1)$ parts, which are respectively holomorphic differentials and conjugates of holomorphic differentials. Finally, the standard identification between weight $2$ cusp forms and holomorphic differentials on $X_0(N)$ transports this decomposition to the stated modular-form decomposition. [/proofplan] [step:Pass from Betti cohomology to harmonic complex-valued $1$-forms] Let $X := X_0(N)$, regarded as a compact Riemann surface. Let $H^1_{\mathrm{dR}}(X;\mathbb{C})$ denote the first cohomology group of the complex of smooth complex-valued differential forms on $X$, namely closed smooth complex-valued $1$-forms modulo exact smooth complex-valued $1$-forms. By the de Rham comparison isomorphism for compact smooth manifolds, there is a natural complex-linear isomorphism \begin{align*} \Phi_{\mathrm{dR}}: H^1_B(X,\mathbb{C}) &\longrightarrow H^1_{\mathrm{dR}}(X;\mathbb{C}). \end{align*} Choose a Hermitian metric $h$ on $X$. Let $\mathcal{A}^1(X;\mathbb{C})$ denote the complex [vector space](/page/Vector%20Space) of smooth complex-valued $1$-forms on $X$, and let \begin{align*} \mathcal{H}^1_h(X;\mathbb{C}) := \{\alpha \in \mathcal{A}^1(X;\mathbb{C}) : d\alpha = 0 \text{ and } d_h^*\alpha = 0\} \end{align*} denote the space of $h$-harmonic complex-valued $1$-forms, where $d_h^*$ is the codifferential determined by $h$. The Hermitian metric $h$ determines a smooth Riemannian metric on the underlying compact smooth surface $X$. The Hodge theorem for compact Riemannian manifolds applies to real-valued forms for this metric, and complexifying the real harmonic representative isomorphism gives that the map \begin{align*} \mathcal{H}^1_h(X;\mathbb{C}) &\longrightarrow H^1_{\mathrm{dR}}(X;\mathbb{C}) \\ \alpha &\longmapsto [\alpha] \end{align*} is a complex-linear isomorphism. Hence every class in $H^1_B(X,\mathbb{C})$ has a unique harmonic representative after applying $\Phi_{\mathrm{dR}}$. [guided] We first replace Betti cohomology by de Rham cohomology, because differential forms are the objects that carry the [Hodge decomposition](/theorems/2745). Let $X := X_0(N)$ be the compact Riemann surface in the statement. Let $H^1_{\mathrm{dR}}(X;\mathbb{C})$ denote the first cohomology group of the complex of smooth complex-valued differential forms on $X$, that is, closed smooth complex-valued $1$-forms modulo exact smooth complex-valued $1$-forms. The de Rham comparison theorem for compact smooth manifolds gives a natural complex-linear isomorphism \begin{align*} \Phi_{\mathrm{dR}}: H^1_B(X,\mathbb{C}) &\longrightarrow H^1_{\mathrm{dR}}(X;\mathbb{C}). \end{align*} This is the bridge between topological cohomology and differential forms. Now choose a Hermitian metric $h$ on $X$. Let $\mathcal{A}^1(X;\mathbb{C})$ be the complex vector space of smooth complex-valued $1$-forms on $X$. The metric $h$ defines a codifferential $d_h^*$, and we define \begin{align*} \mathcal{H}^1_h(X;\mathbb{C}) := \{\alpha \in \mathcal{A}^1(X;\mathbb{C}) : d\alpha = 0 \text{ and } d_h^*\alpha = 0\}. \end{align*} The Hermitian metric $h$ determines a smooth Riemannian metric on the underlying smooth surface $X$. Since $X$ is compact, the Hodge theorem for compact Riemannian manifolds applies to real-valued differential forms for this metric. Complex-valued forms are obtained by complexifying the real de Rham complex and the real harmonic representative isomorphism, so each complex de Rham cohomology class has a unique complex-valued harmonic representative. Therefore the map \begin{align*} \mathcal{H}^1_h(X;\mathbb{C}) &\longrightarrow H^1_{\mathrm{dR}}(X;\mathbb{C}) \\ \alpha &\longmapsto [\alpha] \end{align*} is an isomorphism. Thus proving the desired decomposition of $H^1_B(X,\mathbb{C})$ is equivalent to decomposing the harmonic complex-valued $1$-forms on $X$. [/guided] [/step] [step:Split harmonic $1$-forms into holomorphic and anti-holomorphic parts] Let $\mathcal{A}^{1,0}(X)$ denote the complex vector space of smooth $(1,0)$-forms on $X$, and let $\mathcal{A}^{0,1}(X)$ denote the complex vector space of smooth $(0,1)$-forms on $X$. Define the type components of the [exterior derivative](/theorems/1525) \begin{align*} \partial: \mathcal{A}^{p,q}(X) &\to \mathcal{A}^{p+1,q}(X), \\ \bar{\partial}: \mathcal{A}^{p,q}(X) &\to \mathcal{A}^{p,q+1}(X) \end{align*} by decomposing $d\beta = \partial\beta + \bar{\partial}\beta$ for each smooth $(p,q)$-form $\beta \in \mathcal{A}^{p,q}(X)$. Let \begin{align*} *_h: \mathcal{A}^1(X;\mathbb{C}) &\to \mathcal{A}^1(X;\mathbb{C}) \end{align*} denote the Hodge star operator determined by the Hermitian metric $h$, and recall that on $1$-forms the codifferential is $d_h^* = -*_h d *_h$. Every $\alpha \in \mathcal{A}^1(X;\mathbb{C})$ has a unique type decomposition \begin{align*} \alpha = \alpha^{1,0} + \alpha^{0,1}, \end{align*} with $\alpha^{1,0} \in \mathcal{A}^{1,0}(X)$ and $\alpha^{0,1} \in \mathcal{A}^{0,1}(X)$. Suppose $\alpha \in \mathcal{H}^1_h(X;\mathbb{C})$. Set \begin{align*} A := \bar{\partial}\alpha^{1,0} \in \mathcal{A}^{1,1}(X), \qquad B := \partial\alpha^{0,1} \in \mathcal{A}^{1,1}(X). \end{align*} Since the $(2,0)$ and $(0,2)$ components vanish identically in complex dimension $1$, decomposing $d\alpha=0$ by type gives \begin{align*} A + B = 0. \end{align*} The co-closed condition also gives $d(*_h\alpha)=0$, because $d_h^*\alpha=-*_h d *_h\alpha=0$ and $*_h$ is an isomorphism on $1$-forms. Since $*_h$ acts by $-i$ on $(1,0)$-forms and by $i$ on $(0,1)$-forms, we have \begin{align*} *_h\alpha = -i\alpha^{1,0} + i\alpha^{0,1}. \end{align*} Again using the vanishing of types $(2,0)$ and $(0,2)$ on a Riemann surface, the equation $d(*_h\alpha)=0$ becomes \begin{align*} -iA + iB = 0. \end{align*} The two equations $A+B=0$ and $-iA+iB=0$ imply $A=0$ and $B=0$. Hence $\bar{\partial}\alpha^{1,0}=0$, so $\alpha^{1,0}$ is a holomorphic differential, and $\partial\alpha^{0,1}=0$, so $\alpha^{0,1}$ is the complex conjugate of a holomorphic differential. Conversely, if $\omega \in H^0(X,\Omega^1)$, then $\omega$ is a smooth $(1,0)$-form satisfying $\bar{\partial}\omega = 0$. Since $\partial\omega$ has type $(2,0)$ and $X$ has complex dimension $1$, $\partial\omega = 0$, so $d\omega = 0$. The Hodge star operator satisfies $*_h\omega = -i\omega$ on $(1,0)$-forms, hence \begin{align*} d(*_h\omega) = -i\,d\omega = 0. \end{align*} Using $d_h^* = -*_h d *_h$ on $1$-forms, we obtain $d_h^*\omega = 0$. Thus $d\omega = 0$ and $d_h^*\omega = 0$, so $\omega$ is harmonic. The same argument applied to $\overline{\omega}$ gives that every element of $\overline{H^0(X,\Omega^1)}$ is harmonic. The sum is direct because a form of type $(1,0)$ and a form of type $(0,1)$ can be equal only if both forms are zero. Hence \begin{align*} \mathcal{H}^1_h(X;\mathbb{C}) = H^0(X,\Omega^1) \oplus \overline{H^0(X,\Omega^1)}. \end{align*} [guided] The complex structure on $X$ decomposes complex-valued $1$-forms by type. Let $\mathcal{A}^{1,0}(X)$ be the smooth $(1,0)$-forms and let $\mathcal{A}^{0,1}(X)$ be the smooth $(0,1)$-forms. For smooth $(p,q)$-forms, define the operators \begin{align*} \partial: \mathcal{A}^{p,q}(X) &\to \mathcal{A}^{p+1,q}(X), \\ \bar{\partial}: \mathcal{A}^{p,q}(X) &\to \mathcal{A}^{p,q+1}(X) \end{align*} by the type decomposition $d\beta = \partial\beta + \bar{\partial}\beta$ for each $\beta \in \mathcal{A}^{p,q}(X)$. Let \begin{align*} *_h: \mathcal{A}^1(X;\mathbb{C}) &\to \mathcal{A}^1(X;\mathbb{C}) \end{align*} be the Hodge star operator determined by $h$; on $1$-forms the corresponding codifferential is $d_h^* = -*_h d *_h$. For every smooth complex-valued $1$-form $\alpha \in \mathcal{A}^1(X;\mathbb{C})$, there are unique forms $\alpha^{1,0} \in \mathcal{A}^{1,0}(X)$ and $\alpha^{0,1} \in \mathcal{A}^{0,1}(X)$ such that \begin{align*} \alpha = \alpha^{1,0} + \alpha^{0,1}. \end{align*} Now assume $\alpha$ is harmonic. In particular, $d\alpha = 0$ and $d_h^*\alpha=0$. Define the two $(1,1)$-forms \begin{align*} A := \bar{\partial}\alpha^{1,0} \in \mathcal{A}^{1,1}(X), \qquad B := \partial\alpha^{0,1} \in \mathcal{A}^{1,1}(X). \end{align*} Since $X$ has complex dimension $1$, there are no nonzero smooth forms of type $(2,0)$ or $(0,2)$. Thus \begin{align*} \partial\alpha^{1,0}=0, \qquad \bar{\partial}\alpha^{0,1}=0. \end{align*} Decomposing $d\alpha=0$ by type therefore gives \begin{align*} 0=d\alpha=A+B. \end{align*} This equation alone is not enough to force $A$ and $B$ to vanish separately, because both have type $(1,1)$. The missing information is the co-closed condition. Since $d_h^*\alpha=-*_h d *_h\alpha=0$ and $*_h$ is an isomorphism on $1$-forms, we obtain \begin{align*} d(*_h\alpha)=0. \end{align*} The Hodge star satisfies $*_h\alpha^{1,0}=-i\alpha^{1,0}$ and $*_h\alpha^{0,1}=i\alpha^{0,1}$, so \begin{align*} *_h\alpha=-i\alpha^{1,0}+i\alpha^{0,1}. \end{align*} Using again that the $(2,0)$ and $(0,2)$ terms vanish in complex dimension $1$, the equation $d(*_h\alpha)=0$ becomes \begin{align*} 0=d(*_h\alpha)=-iA+iB. \end{align*} The system \begin{align*} A+B=0, \qquad -iA+iB=0 \end{align*} implies $A=0$ and $B=0$. Hence \begin{align*} \bar{\partial}\alpha^{1,0}=0, \qquad \partial\alpha^{0,1}=0. \end{align*} The first equation says exactly that $\alpha^{1,0}$ is a holomorphic differential. The second says that $\alpha^{0,1}$ is anti-holomorphic, equivalently that $\alpha^{0,1} = \overline{\omega}$ for a unique holomorphic differential $\omega \in H^0(X,\Omega^1)$. Conversely, let $\omega \in H^0(X,\Omega^1)$. Then $\omega$ is a smooth $(1,0)$-form satisfying $\bar{\partial}\omega = 0$. Since $X$ has complex dimension $1$, the form $\partial\omega$ has type $(2,0)$ and must vanish. Hence \begin{align*} d\omega = \partial\omega + \bar{\partial}\omega = 0. \end{align*} For a Hermitian metric on a Riemann surface, the Hodge star acts on $(1,0)$-forms by $*_h\omega = -i\omega$. Therefore \begin{align*} d(*_h\omega) = d(-i\omega) = -i\,d\omega = 0. \end{align*} The codifferential on $1$-forms is $d_h^* = -*_h d *_h$, so the last equality gives \begin{align*} d_h^*\omega = -*_h d(*_h\omega) = 0. \end{align*} Together with $d\omega = 0$, this proves that $\omega$ is harmonic. Complex conjugation gives the same conclusion for every $\overline{\omega} \in \overline{H^0(X,\Omega^1)}$. This proves \begin{align*} \mathcal{H}^1_h(X;\mathbb{C}) = H^0(X,\Omega^1) \oplus \overline{H^0(X,\Omega^1)}. \end{align*} The sum is direct because a form of type $(1,0)$ and a form of type $(0,1)$ can be equal only if both are zero. [/guided] [/step] [step:Transport the harmonic decomposition back to Betti cohomology] Composing the de Rham comparison isomorphism with the harmonic representative isomorphism gives \begin{align*} H^1_B(X,\mathbb{C}) \cong \mathcal{H}^1_h(X;\mathbb{C}) = H^0(X,\Omega^1) \oplus \overline{H^0(X,\Omega^1)}. \end{align*} The resulting subspaces are independent of the auxiliary metric $h$, because they are exactly the cohomology classes represented by holomorphic and anti-holomorphic differentials, which are defined by the complex structure of $X$ alone. Hence the decomposition is canonical. Finally, by the standard identification of weight $2$ cusp forms with holomorphic differentials on the compact modular curve, the map \begin{align*} S_2(\Gamma_0(N)) &\longrightarrow H^0(X,\Omega^1) \\ f &\longmapsto f(z)\,dz \end{align*} is a complex-linear isomorphism. Taking complex conjugates gives an isomorphism \begin{align*} \overline{S_2(\Gamma_0(N))} &\longrightarrow \overline{H^0(X,\Omega^1)}. \end{align*} Substituting these two isomorphisms into the [Hodge decomposition](/theorems/3941) yields \begin{align*} H^1_B(X_0(N),\mathbb{C}) \cong S_2(\Gamma_0(N)) \oplus \overline{S_2(\Gamma_0(N))}. \end{align*} This is the desired decomposition. [/step]