Fix $N > 3$ and consider the modular curve $X(N)$ parametrizing elliptic curves with full level N structures. Let $\pi : E(N)\to X(N)$ be the universal elliptic curve. Then $V=R^1\pi_*\mathbf Q$ defines a local system on $X(N)$. Define $H^1_c = H^1_c(X(N),Sym^kV)$ and $H^1=H^1(X(N),Sym^kV)$ for $k>=0$. Parabolic cohomology is defined as $H^1_p = image H^1_c\to H^1$. Then there is a direct sum decomposition

$$ H^1 = H^1_p \oplus H^1_e$$

where $H^1_e$ is called the Eisenstein part of cohomology.

My question is: how does one prove that $H^1_p$ is in fact a direct summand of $H^1$? And is there any description of $H^1_e$ in terms of the geometric objects $X(N)$, $E(N)$?



1 Answer 1


The fact that $H^1_p$ is a direct summand in $H^1$ is obvious if we consider just those objects as $\mathbb{Q}$-vector space (since a sub vector space always has a suplementary). What you mean probably is "why is $H^1_p$ a direct summand of $H^1$ as a module over the Hecke operators?". Then the answer is still yes, but a proof is needed. The idea is that one can identify $H^1_p$, as a Hecke-module, to the sum of two copies of the space of cuspidal modular forms of weight $k+2$, while $H^1/H^1_p$ may be identified with the space of Eisenstein series of the same weight. Since the systems of Hecke-eigenvalues of Eisenstein series are not the same as the system of eigenvalues of cuspidal form (for example because the eigenvalues of $T_l$ goes to infinity as $l^{k+1}$ for an Eisenstein series, and in $O(l^{k/2+1})$ for a cusp form (Hecke estimates), $H^1_p$ has a unique Hecke-stable supplementary $H^1_e$ in $H^1$. Moreoever, this provides an interpretation of $H^1_e$ as Eisenstein series, which have a geometric interpretation as sections of certain sheaves of $X(N)$ satisfying some conditions at infinity.

All the above is known as the theory of the "Eichler-Shimura isomorphism". A complete reference for this is the book by Hida called "Elementary theory of L-function and Eisenstein series".

  • 3
    $\begingroup$ The use of "elementary" in Hida's title is, to put it mildly, a little disingenuous. $\endgroup$ Sep 26, 2011 at 15:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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