I am stuck with a technical question concerning parabolic cohomology of modular groups and cup-products on them. Basically, I am trying to understand the appendix about cohomology of Hida's book "elementary theory of $L$-functions and Eisenstein Series" and related ideas, and though Hida's exposition is very detailed, I am not illuminated.

So, $\Gamma$ is a congruence subgroup of $SL_2(\mathbb Z)$ which acts freely on the Poincaré upper half-plane, for instance $\Gamma_1(\nu)$ for $\nu \geq 3$. We choose for each cusp of $\Gamma$ a generator of the subgroup fixing that cusp, and we define a subset $P$ of $\Gamma$ as the set of all conjugate of those elements. We let $M$ (and later $N$) be left $\Gamma$-modules.

So we have the cohomology groups $H^i(\Gamma,M)$ defined as the homology of the usual complex $C^0(\Gamma,M) \rightarrow C^1(\Gamma,M) \rightarrow C^2(\Gamma,M) \rightarrow \dots$ These groups have a geometric interpretation, namely there are canonical isomorphisms $H^i(\Gamma,M) \simeq H^i(Y_\Gamma,\tilde M)$ where $Y_\Gamma$ is the quotient of the upper half-plane by $\Gamma$ and $\tilde M$ the locally free sheaf on it defined by $M$, and the RHS $H^i$ is sheaf cohomology.

To define parabolic cohomology we consider a subcomplex $C^\bullet_P(\Gamma,M)$ of $C^\bullet(\Gamma,M)$ as follows: $C^i_P(\Gamma,M) = C^i(\Gamma,M)$ for $i=0,2,3,4,\dots$ and for $i=1$, we set $$C^1_P(\Gamma,M) = \{u \in C^1(\Gamma,M), u(\pi) \in (\pi-1) M \ \ \forall \pi \in P\}.$$ (To check that $C^\bullet_P$ is a subcomplex we just need to check that the differential sends $C^0$ into $C^1_P$ but this is trivial.) We define $H^i_P(\Gamma,M)$ as the homology of that complex. It is trivial that $H^i_P(\Gamma,M)=H^i(\Gamma,M)$ for all $i \neq 1,2$. Hida also gives a geometric interpretation of the new cohomology groups: there are canonical isomorphisms $$H^1_P(\Gamma,M) = H^1_! (Y_\Gamma, \tilde M) := Im \left(H^1_c(Y_\Gamma,\tilde M) \rightarrow H^1(Y_\Gamma,\tilde M)\right),$$ $$H^2_P(\Gamma,M) =H^2_c(Y_\Gamma,\tilde M),$$ where $H^i_c$ is the sheaf cohomology with compact support. (The second isomorphism is given in the last assertion of Prop. 2 of the appendix of Hida's book, and the first is Prop. 1 of his paper Congruences of cusp forms and special values of $L$-functions, invent. math 63 (1981).)

So far, so good. Now the cup-product $H^1_c(Y_\Gamma,\tilde M) \otimes H^1_c(Y_\Gamma,\tilde N) \rightarrow H^2_c(Y_\Gamma,\tilde M \otimes \tilde N)$ induces a cup-product $H^1_!(Y_\Gamma,\tilde M) \otimes H^1_!(Y_\Gamma,\tilde N) \rightarrow H^2_c(Y_\Gamma, \tilde M \otimes \tilde N)$ (as explained in Hida's paper, cf (2.2) and the first page of section 2), hence a cup-product $$H^1_P(\Gamma,M) \otimes H^1_P(\Gamma,N) \rightarrow H^2_P(\Gamma,M \otimes N).$$ Now my question:

How can this cup-product be described in terms of the definition of $H^\bullet_P$ as the homology of $C^\bullet_P$?

That is, how can this cup-product be described in purely group cohomological terms?

In theory, one should be able to answer this question by translating the definition of the cup-product in sheaf cohomology through all the isomorphisms involved, but this looks not very engaging. What worries me more is that I can't seem to be able to guess what the result could be, that is I am not able to define a cup-product $H^1_P \otimes H^1_P \rightarrow H^2_P$ in the only natural way I can think of, that is by defining it at the level of cochains. Indeed, the natural map $C^0(\Gamma,M) \otimes C^1_P(\Gamma,N) \rightarrow C^1(\Gamma,M \otimes N)$ does not seem to have image in $C^1_P$. So, I would be happy with any "natural" definition of the cup-product in terms of cochains, and quite ready to believe it corresponds to the one defined using geometric cup-product.

  • $\begingroup$ Of course, something in my exposition leading to my question may be wrong, and the cup-product I am trying to understood may actually not exist. Pointing out the possible mistake would be of course a perfectly correct answer. $\endgroup$
    – Joël
    Jun 2, 2017 at 20:10

2 Answers 2


It seems to me that the definition has been cooked up so that $H^0_P = H^0$, $H^1_P = \mathrm{Im}(H^1_c \to H^1)$, and $H^2_P = H^2_c$. This means that the parabolic cohomology is equal to the intersection cohomology. That is, if $j \colon Y_\Gamma \to X_\Gamma$ is the inclusion into the compact modular curve, then $H^\bullet_P(\Gamma,M) = I\!H^\bullet(X_\Gamma,\tilde M)$ (which also equals $H^\bullet(X_\Gamma,R^0 j_\ast\tilde M)$). But there is in general no natural cup product on intersection cohomology. This could explain why you don't see a natural multiplication.

Another remark is that the Manin-Drinfeld theorem provides a natural section of the map $H^1_c \to H^1_P$; the section is uniquely pinned down by the action of the Hecke algebra. Similarly the inclusion $H^1_P \to H^1$ splits, and $H^1_P$ is in a canonical way a direct summand of both $H^1_c$ and $H^1$. Hence the cup products on usual cohomology and compact support cohomology combine to induce cup products in parabolic cohomology. But perhaps you were after something rather explicit.

  • 1
    $\begingroup$ In this case, isn't there a pairing $j_*\tilde M\otimes j_*\tilde N\to j_*(\tilde M\otimes \tilde N)$, and wouldn't this give the cup product on $IH$? $\endgroup$ Jun 3, 2017 at 14:10
  • $\begingroup$ Thanks for the answer. But like Donu, I think than in this case there exists a canonical pairing $H^1_P \otimes H^1_P \rightarrow H^2_P$, defined without having to choose a section $H^1_P \rightarrow H^1_c$. Simply, let $\alpha$ and $\beta$ two elements of $H^1$ that actually belongs to $H^1_P$, so that they are the image by the natural maps of elements $\alpha_0$, $\beta_0$ of $H^1_c$. Then the element $\alpha_0 \cup \beta$ (from the pairing $H^1_c \otimes H^1 \rightarrow H^2_c$) is equal to $\alpha_0 \cup \beta_0$ (from the pairing $H^1_c \otimes H^1_c \rightarrow H^2_c$... $\endgroup$
    – Joël
    Jun 3, 2017 at 18:40
  • $\begingroup$ ) by the compatibility of those two product, and also to the element $\alpha \cup \beta_0$ (from the pairing $H^1 \otimes H^1_c \rightarrow H^2_c$ by the same reasoning symmetrically). Hence that element does not depend of the choice of $\alpha_0$ or $\beta_0$. We call it $\alpha \cup \beta$, that's our cup product $H^1_P \otimes H^1_P \rightarrow H^2_P$. $\endgroup$
    – Joël
    Jun 3, 2017 at 18:44
  • $\begingroup$ You're both right of course, oops. $\endgroup$ Jun 4, 2017 at 20:32

Since I seem to my stuck on something myself, let me take a crack at your problem. This is just a proposal, I haven't checked if this really works.

Let me start with the suggestion in Dan's answer and identify $H^i_P(\Gamma, M)= H^i(X, j_*M)$, here $X=X_\Gamma$ is a smooth compactification of $Y=Y_\Gamma$. I'm conflacting $M$ with $\tilde M$ and dropping various decorations for ease of writing. Then from the triangle $$j_*M\to \mathbb{R}j_*M\to R^1j_*M[-1]\to $$ we get $$H^1_P(\Gamma, M)\to H^1(Y, M)\to \bigoplus_p H^1(C_p,M)\to H_P^2(\Gamma,M)\to0$$ where $p$ (resp. $C_p$) are (resp. circles around) the cusps. This can be translated back to group cohomology $$H^1_P(\Gamma, M)\to H^1(\Gamma, M)\to \bigoplus_p H^1(P_p,M)\stackrel{\partial}{\to}H_P^2(\Gamma,M)\to 0$$ where now $P_p=\pi_1(C_p)$. Given classes $\alpha\in H^1_P(M),\beta\in H_P^1(N)$, their product, whatever it is, would have to be $\partial(?)$. Here is a guess as to what it might be.

Let us assume that the above sequence is induced by an exact sequence of suitable complexes $$0\to C_P^\bullet(M)\to C^\bullet(M)\to \bigoplus C^\bullet(P_p, M)\to 0$$ (Perhaps the ones specified in your question will work for this, I haven't checked.) Let $\alpha\in C^1_P(M),\beta\in C_P^1(N)$ denote cocycles representing classes above. Then $\alpha,\beta$ can be mapped to cocycles, and in fact coboundaries, $\alpha',\beta'$ in $\oplus C(P_p,-)$. Write $\alpha'=d \alpha''$. Try $\partial(\alpha''\cup\beta')$.


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