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Hi,

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 $X(N)$, $E(N)$?

Thanks!

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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".

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The use of "elementary" in Hida's title is, to put it mildly, a little disingenuous. –  David Hansen Sep 26 '11 at 15:49
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