Let $H_k$ be the vector space of degree $k$ homogeneous polynomials in two variables.I'm looking for a reference for the fact that $H^1(SL(2,\mathbb Z);H_k)=M^0(k+2)\oplus\overline{M^0(k+2)}\oplus E_{k+2}$, where

  1. $M^0(k+2)$ is the space of cuspidal modular forms of weight $k+2$.
  2. $\overline{M^0(k+2)}$ is its conjugate.
  3. $E_k$ is $1$-dimensional if $k\geq 4$ is even, and is zero otherwise.

I know how to get the $M^0(k+2)\oplus\overline{M^0(k+2)}$ piece. By the Eichler-Shimura isomorphism, this is the same as the cuspidal cohomology $H^1_{cusp}(SL(2,\mathbb Z);H_k)$, spanned by cocycles that vanish on the matrix $T$=[1 1 \ 0 1]. Serge Lang's book on Modular Forms has a detailed explanation of how to get this. I'd really like to know how to get the $E_{k+2}$ piece, which as far as I understand, corresponds somehow to the Eisenstein series, which is the "extra" modular form of weight $k+2$ that is not cuspidal.

It's not too hard to see that $H^1(SL(2,\mathbb Z);H_k)=H^1(PSL(2,\mathbb Z);H_k)$, with $PSL(2,\mathbb Z)$ having the nice presentation $\langle S,T\,|\, (ST)^3=S^2=I\rangle$, so one should be able to represent the $E_{k+2}$ cocycle by specifying its values on $S,T$.

So, to summarize, I'd be satisfied with any proof that $H^1(SL(2,\mathbb Z);H_k)=M^0(k+2)\oplus\overline{M^0(k+2)}\oplus E_{k+2}$, but I'd be happiest with an answer that lets me get my hands on the $E_{k+2}$ cocycle as explicitly as possible, perhaps even telling me its values on $S$ and $T$. Any comments that would help clarify this are appreciated.

Edit: I think I have a better understanding of what the "extra" Eisentstein cocycle is, based on Kevin's comments. It seems that the cuspidal cocycles vanish on $T$, whereas the Eisenstein cocycle vanishes on $S$, although I don't see how to show, for example, that there is only one dimension's worth of cocycles vanishing on $S$, up to coboundaries. (Edit: I'm not entirely sure about this.)

Edit: Shimura's Introduction to the Theory of Automorphic Forms only covers the cuspidal part of the above isomorphism.

  • $\begingroup$ Given the Eisenstein series $E(\tau)$ as a function on the upper half plane, isn't the cocycle you want simply the standard one: given $\gamma$ you integrate $\tau^n E(\tau)$ from $i$ to $\gamma i$ for $0\leq n\leq k$ and patch these $k+1$ numbers together into an element of $H_k$ in the standard way as in e.g. Shimura's book? $\endgroup$ Nov 5 '10 at 15:37
  • $\begingroup$ I know that for cuspidal forms $f(z)$ you integrate $f(z)z^s$ from $0$ to $\gamma 0$ to get a cocycle. I guess you're saying I should do a similar thing, but not use a boundary point. $\endgroup$
    – Jim Conant
    Nov 5 '10 at 15:57
  • $\begingroup$ @Kevin: which book of Shimura are you referring to? $\endgroup$
    – Jim Conant
    Nov 5 '10 at 16:04
  • $\begingroup$ You can integrate from $z$ to $\gamma z$ for any $z$ in the extended upper half plane to get a cocycle, and different choices will change the cocycle by a coboundary. I can't see where you are assuming that the form is cuspidal when doing this integration, at least if $z$ is in the upper half plane. As far as I know Shimura only wrote one book---his introduction to the theory of automorphic forms. But you seem to know the construction for cusp forms anyway. I don't understand what your question is really. Doesn't the same trick work for the Eisenstein series? If not why not? Sorrytobeconfused. $\endgroup$ Nov 5 '10 at 19:41
  • $\begingroup$ Shimura has written several books. There were at least five in our library and a cursory glance did not reveal anything to do with cocycles or cohomology. I agree that taking any $z$ in the upper half plane and integrating $z$ to $\gamma z$ will give a cocycle for any modular form, but you can't get enough of them this way. To get most of them, you have to start with a point on the boundary, such as $0$, but this construction only seems to be well-defined for cuspidal forms. $\endgroup$
    – Jim Conant
    Nov 6 '10 at 10:15

The result you're looking for is contained in the following article :

Haberland, Klaus. Perioden von Modulformen einer Variabler and Gruppencohomologie I (German) [Periods of modular forms of one variable and group cohomology I], Math. Nachr. 112 (1983), 245-282.

Let $S_k$ (resp. $M_k$) be the space of holomorphic cusp forms (resp. holomorphic modular forms) for $\Gamma = SL_2(\mathbf{Z})$. Let $\Gamma_{\infty}$ be the stabilizer of $\infty$ in $\Gamma$. Let $V_k$ be the space of polynomials of degree $\leq k-2$ with complex coefficients. Haberland proves an exact sequence

\begin{equation} (*) \qquad 0 \to S_k \oplus \overline{S_k} \to H^1(\Gamma,V_k) \to H^1(\Gamma_\infty,V_k) \to 0. \end{equation} Let $T = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \in \Gamma_{\infty}$. There is a natural map $V_{k+1} \to H^1(\Gamma_\infty,V_k)$ sending a polynomial $P$ to the cocycle $c_P$ determined by $c_P(T) = P(X+1)-P(X)$. It is easy to check that this map induces an isomorphism $\psi : V_{k+1}/V_k \cong H^1(\Gamma_\infty,V_k)$, so that the latter space is one-dimensional.

The "Eisenstein cocycle" you're looking for is a natural map $\delta : M_k \to H^1(\Gamma,V_k)$ which Haberland constructs the following way (actually I learnt this construction and many other properties of $\delta$ during Zagier's 2002-2003 lectures at the Collège de France).

Let $f \in M_k$. Let $\widetilde{f}$ be an Eichler integral of $f$, that is any holomorphic function on $\mathcal{H}$ such that

\begin{equation} \left(\frac{1}{2\pi i} \frac{d}{dz}\right)^{k-1} \widetilde{f}(z) = f(z). \end{equation} Note that $\widetilde{f}$ is unique up to adding some element of $V_k$. Since we integrate $k-1$ times, the function $\widetilde{f}$ should be thought of as a function of "weight" $k-2\cdot (k-1) = 2-k$ (of course this isn't true in the strict sense). Let us make this more precise.

For any $n \in \mathbf{Z}$, let $|_n$ denote the weight $n$ action of $SL_2(\mathbf{R})$ on the space of complex-valued functions on $\mathcal{H}$ (so that any $f \in M_k$ is a fixed vector of the weight $k$ action of $\Gamma$). Note also the weight $2-k$ action gives the usual action of $\Gamma$ on $V_k$. The crucial fact is that we have

\begin{equation} \widetilde{f} |_{2-k} (\gamma-1) \in V_k \qquad (\gamma \in \Gamma). \end{equation} This can be proved using Bol's identity

\begin{equation} \left(\frac{d}{dz} \right)^{k-1} (F |_{2-k} g) = \left(\frac{d^{k-1} F}{dz^{k-1}} \right) |_k g \end{equation} which holds for any holomorphic function $F$ on $\mathcal{H}$ and any $g \in SL_2(\mathbf{R})$.

Since $\gamma \mapsto \widetilde{f} |_{2-k} (\gamma-1)$ is obviously a coboundary in the space of functions on $\mathcal{H}$, it defines a cocycle in the space $V_k$. Therefore we get $\delta(f) \in H^1(\Gamma,V_k)$ and this element doesn't depend on the choice of $\widetilde{f}$. Thus we have constructed $\delta : M_k \to H^1(\Gamma,V_k)$.

It is not difficult to check that if $f =\sum_{n \geq 0} a_n e^{2i\pi nz}$ then the image of $\delta(f)$ in $H^1(\Gamma_\infty,V_k)$ is the image of the polynomial $\frac{a_0 \cdot (2\pi i)^{k-1}}{(k-1)!} \cdot X^{k-1} \in V_{k+1}$ under the isomorphism $\psi$ above. In particular $\delta$ is injective, and the exact sequence $(*)$ gives the isomorphism you want.

Note that there is a distinguished choice of $\widetilde{f}$, namely

\begin{equation} \widetilde{f} = \sum_{n \geq 1} \frac{a_n}{n^{k-1}} e^{2i\pi nz} + \frac{a_0 \cdot (2\pi i)^{k-1}}{(k-1)!} z^{k-1}. \end{equation} Let $c_f \in Z^1(\Gamma,V_k)$ be the cocycle associated to this choice of $\widetilde{f}$. Let us compute the value of $c_f$ on $T$ and $S= \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$. First as explained above, we have

\begin{equation} c_f(T)=\frac{a_0 \cdot (2\pi i)^{k-1}}{(k-1)!} ((X+1)^{k-1}-X^{k-1}). \end{equation} To compute $c_f(S)$, Haberland uses the natural integral representation of $\widetilde{f}$ in terms of $f-a_0$, and gets

\begin{equation} c_f(S) = \frac{(2\pi i)^{k-1}}{(k-2)!} \int_0^{\infty} \left(f(z)-\frac{a_0}{z^k}-a_0 \right) (z-X)^{k-2} dz \end{equation} (there is a similar but more complicated formula for $c_f(\gamma)$ for any $\gamma \in \Gamma$, see below). Then $c_f(S)$ can be expressed in terms of the special values of $L(f,s) := \sum_{n=1}^\infty a_n/n^s$ at integers $s = 1,\ldots,k-1$. It is then a good exercise to compute $c_f(S)$ when $f$ is the Eisenstein series $E_k$, in terms of Bernoulli numbers and of $\zeta(k-1)$ (this is Satz 3 in Haberland's article, Kapitel 1).

Please tell me if something isn't clear in my explanation.

EDIT : I found the following expression for $c_f(\gamma)$ where $\gamma \in \Gamma$. It is quite complicated (maybe it could be somewhat simplified) :

\begin{equation} \begin{aligned} \frac{(k-2)!}{(2\pi i)^{k-1}} c_f(\gamma) &= \int_{z_0}^{\infty} (f(z)-a_0)(z-X)^{k-2} dz + \int_{\gamma^{-1} \infty}^{z_0} \left(f(z) -\frac{a_0}{(cz+d)^k} \right) (z-X)^{k-2} dz \\ & + \frac{a_0}{k-1} \left((X-z_0)^{k-1}-(X-\gamma z_0)^{k-1} |_{2-k} \gamma + X^{k-1} |_{2-k} (\gamma-1) \right) \end{aligned} \end{equation} where $z_0 \in \mathcal{H}$ is arbitrary and $\gamma= \begin{pmatrix} a & b \\ c & d \end{pmatrix}$.

  • $\begingroup$ Sorry, it seems that my LaTeX code is wrong, and I don't know how to fix it... Doesn't "pmatrix" work here ? $\endgroup$ Nov 8 '10 at 18:25
  • $\begingroup$ I should also add that there is indeed an "integral representation" of the cocycle $r_f$, as Kevin Buzzard initially suggested. But I am not sure yet of the exact shape of the formula. So if I have time I will try to do the computations and then complete my answer. $\endgroup$ Nov 8 '10 at 22:14
  • $\begingroup$ There is a bug in the software that occasionally makes it choke on some code like pmatrix. You can work around the problem by enclosing the LaTeX in "accent grave"s. $\endgroup$
    – S. Carnahan
    Nov 9 '10 at 14:07
  • 2
    $\begingroup$ @Francois: I appreciate your detailed answer. Thank you. $\endgroup$
    – Jim Conant
    Nov 9 '10 at 14:10
  • $\begingroup$ Thank you very much for taking the time to edit my LaTeX code. Thanks a lot also for the tip, now I think I will manage to write matrices on MO! $\endgroup$ Nov 9 '10 at 14:13

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