Let $ f \in S\_k(\Gamma)$ be a weight k modular cusp form of level $\Gamma$, with modular curve $Y\_{\Gamma}$. Let $V^{k-2}$ be the homogenous polynomials in X and Y of degree k-2 with complex coefficients. Let $\Omega^1$ be the sheaf of differential 1 forms. We define the map:


$\phi : S\_k(\Gamma) \to H^0(Y\_{\Gamma}, \Omega^1 \otimes V^{k-2})$


$\phi(f) : = f(z)(X-zY)^{k-2}dz$


**Question: How do I show that this defines a global section of $Y\_{\Gamma}$?** There must be some argumentation about the coordinate charts of the modular curve.




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Note: This helps give an answer to my previous question ([Eichler-Shimura Isomorphism) Proving c(f) is not a boundary](https://mathoverflow.net/questions/188685/eichler-shimura-isomorphism-proving-cf-is-not-a-boundary)) via the following observations:


By Dolbeaut's theorem, $H^0(Y\_{\Gamma}, \Omega^1 \otimes V^{k-2} )\cong H^1(Y\_\Gamma, V^{k-2})$. Then, since $Y\_{\Gamma}$ is a $K(\Gamma, 1)$ surface, we know $H^1(Y\_{\Gamma}, V^{k-2}) \cong H^1(\Gamma, V^{k-2})$, which is group cohomology, and the subject of my previous post. By showing that this $f$ defines a global section, it is trivial to show it is not a boundary in cohomology, via the isomorphisms.



Proving the Eichler-Shimura Isomorphism defines a global section
What exactly is a global section of $\Omega^1 \otimes V^{k-2}$ on $Y\_\Gamma$? First consider $\Omega^1 \otimes V^{k-2}$ as a sheaf on $\mathbb{H}$. This sheaf is naturally endowed with a $\Gamma$ action through $\Gamma$'s action on $V^{k-2}$.


Cover $\mathbb{H}$ with open sets $\{U\_i\}$ such that $\gamma(U\_i)\cap U\_I=\varnothing$. For each $U\_i$ we choose a local section $f\_i\in\Gamma(U\_i, \Omega^1 \otimes V^{k-2})$ that is "compatible" under the action of $\Gamma$ in the following sense: For any open set $V\subset \gamma(U\_i) \cap U\_j$, the pullback of $f\_j|\_V$ along $\gamma$ gives $\gamma(f\_i)|\_{\gamma^{-1}(V)}$ (note that by $\gamma(f\_i)$ I mean the action of $\gamma$ on $f\_i$).


In our particular situation, set $\gamma =\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ and compute
\begin{eqnarray}
\gamma^\* (f(z)(X-zY)^{k-2}dz) &=& f(\gamma(z))(X-\gamma(z)Y)^{k-2}d\gamma(z) \\
&=&(cz+d)^kf(z)(X-\frac{az+b}{cz+d}Y)^{k-2}(cz+d)^{-2}dz \\
&=&f(z)\gamma((X-zY)^{k-2}) dz. 
\end{eqnarray}


When $k=2$, the $(cz+d)^{-2}$ that comes out of pulling back $dz$ is exactly what is needed to cancel out the factor of automorphy. When $k>2$, we need to add a "twisting" factor that will compensate. 


