3
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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. $\endgroup$ – David Loeffler Jan 7 '15 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 Jan 8 '15 at 5:51
3
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.