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.

Note: This helps give an answer to my previous question (Eichler-Shimura Isomorphism) Proving c(f) 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.

  • $\begingroup$ You need to be a little careful to distinguish between $Y_\Gamma$ and its compactification $X_{\Gamma}$. It is $Y_\Gamma$ which is a $K(\Gamma, 1)$-space; but $Y_\Gamma$ is affine, so $H^0(Y_\Gamma, $anything) is likely to be infinite-dimensional. $\endgroup$ Commented Jan 7, 2015 at 7:53
  • $\begingroup$ I don't think it is true that $H^0(Y_\Gamma,\Omega^1 \otimes V^{k-2})$ is equal to $H^1(Y_\Gamma, V^{k-2})$. You need to mod out by the 1-forms that are exact (in general you only want to consider closed forms, but all holomorphic 1-forms are closed). Instead what you have that $H^1(Y_\Gamma,V^{k-2})$ is equal to the 1st deRham cohomology with twisted coefficients in $V^{k-2}$. By this I mean the homology of the complex $0\to \Gamma(Y_\Gamma,\Omega^0\otimes V^{k-2}) \to \Gamma(Y_\Gamma,\Omega^1\otimes V^{k-2}) \to ...$. Check out the appendix in Hida's book on Eisenstein series. $\endgroup$
    – jkramerm
    Commented Jan 8, 2015 at 5:51

1 Answer 1


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.


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