I got my copy of Computational Aspects of Modular Forms and Galois Representations in the mail yesterday. The goal of the book is "How one can compute in polynomial time the value of Ramanujan's tau at a prime", well, or any other modular form of level 1. It's all very thrilling!

The following fact is essential: for any modular form $f$ of level 1, and any prime $l$, the mod $l$ reduction of the semisimplification of the galois representation attached to $f$ by Deligne is a 2-dimensional subrepresentation of the galois representation of the $l$-torsion of the jacobian of the modular curve of level $l$.


From what I understand, this is somewhat equivalent (after Shimura and Deligne) to there being a modular form of weight 2 and level $l$ that is congruent to $f$ mod $l$, or something similar. Is this the right statement? Why is it true then?

Searching far and wide for an introduction to this topic yields very little.

  • 2
    $\begingroup$ I think Gross explains this in his Duke paper (A tameness criterion for Galois representations associated to modular forms (mod p), vol. 61, n.2, 1990). It's proposition 9.3 p. 478. (I'm not saying that it is an easy reference to read!) $\endgroup$ Commented Jul 8, 2011 at 11:52
  • $\begingroup$ [all the "editing" I did was fix the title's spelling of "congruent" - hope that's OK!] $\endgroup$ Commented Jul 8, 2011 at 15:06

2 Answers 2


By "level $\ell$" I assume you mean "level $\Gamma_1(\ell)$".

Here's a proof. By the Eichler-Shimura theorem, the system of eigenvalues associated to the modular form shows up in $H^1(SL(2,\mathbf{Z}),Symm^{k-2}(\mathbf{C}))$. Hence (by some easy commutative algebra) the mod $\ell$ reduction of the system of eigenvalues shows up in $H^1(SL(2,\mathbf{Z}),Symm^{k-2}(\mathbf{F}_\ell))$ and hence, by a standard diagram chase, in $H^1(SL(2,\mathbf{Z}),M)$ for $M$ an irreducible module for $GL(2,\mathbf{F}_\ell)$ (EDIT: here $M$ is a finite-dimensional vector space over $\mathbf{F}_\ell$, so it's just a twist of $Symm^n$ for some small $n$). But any such $M$ is a subquotient of $I:=Ind_{(* *;0 1)}^{GL(2,\mathbf{F}_\ell)}(1)$ so the system of eigenvalues shows up in $H^1(SL(2,\mathbf{Z}),I)$ and hence, by Shapiro, in $H^1(\Gamma_1(\ell),1)$. (EDIT: here $1$ means the trivial 1-d vector space over $\mathbf{F}_\ell$: one now deduces that the system of eigenvalues lifts to a system of evals showing up in $H^1(\Gamma_1(\ell),\mathbf{C})$).

Now using Eichler-Shimura again, this time at level $\ell$, shows that there's a weight 2 level $\Gamma_1(\ell)$ modular form giving rise to the same mod $\ell$ system of Hecke eigenvalues. This last statement is a little disingenuous because Eichler-Shimura only tells you about parabolic cohomology which isn't quite the same as group cohomology. But the extra stuff is all associated to reducible Galois representations so can be dealt with by hand using Eisenstein series.

You'll find these sorts of arguments in papers of Ribet from around 1987-1990. Another great place to look is papers of Ash and Stevens from slightly earlier -- I learnt the argument below from an Ash-Stevens paper.

  • $\begingroup$ @Kevin: Nice explanation. Your first $H^1(SL(2...$ is missing a right parenthesis after the Z, which makes it a bit confusing to read. $\endgroup$ Commented Jul 8, 2011 at 19:44
  • $\begingroup$ Thanks for the answer! Two small questions: 1. are you sure any such $M$ is a subquotient of $I$? Maybe you mean only cuspidal $M$? The subgroup taken is larger than the unipotent, and it is induction from the unipotent, i.e. Whitaker models, that contains all (except one-dimensional) irred. modules. 2. without giving two much away, can you hint at the players in the standard diagram chase mentioned? $\endgroup$ Commented Jul 8, 2011 at 21:04
  • 1
    $\begingroup$ @Dror: $M$ is a mod $\ell$ irreducible representation. I don't know what "cuspidal" means in this context. It's just a twist of $Symm^n$ for $n$ small (that's all there is!). One of the two Ash-Stevens papers from around the mid-80s explain why what I said is true via an explicit calculation. For the diagram chase, all one needs to know is: a system of eigenvalues shows up in a (finite-dimensional) module iff the module localized at the associated max ideal of the Hecke algebra is non-zero, and if $A\to B\to C$ is exact and $A_m$ and $C_m$ are zero, then $B_m$ is zero too (as lclzation is xct) $\endgroup$ Commented Jul 8, 2011 at 21:29
  • $\begingroup$ @Dror: I now realise that I did not make it clear that $M$ was supposed to be a mod $\ell$ rep in the answer -- I'll edit. $\endgroup$ Commented Jul 8, 2011 at 21:31
  • $\begingroup$ @Kevin: I think I'm beginning to understand this, thanks! I want to add a small question, but not sure it deserves its own thread: can we bound the degree of the number field of the coefficients of the congruent form, in terms of the of the original form? $\endgroup$ Commented Jul 9, 2011 at 19:08

If $N\geq 1$ is an integer not divisible by $p$, one can see that any system of Hecke eigenvalues $(a_\ell)$ arising from $S_k(\Gamma_1(N))$ is congruent mod $p$ to a system $(b_\ell)$ arising from $S_2(\Gamma_1(Np^n))$, for some $n$, using an interplay between a theorem of Serre (describing a purely mod $p$ Jacquet-Langlands correspondence), and the more classical, characteristic zero J-L between ${\rm GL}_2$ and the multiplicative group $G$ of the $\mathbf{Q}$-quaternion algebra ramified at $p$ and infinity.

I know that what I describe here is perhaps not the right way of proving the result you are asking, but it seems to me worth to mention.

In his '87 letter to Tate Serre proves:

${\rm Theorem:}$ Systems of mod $p$ Hecke eigenvalues arising from $M_k(\Gamma_1(N))$ are the same as those arising from locally constant function $f:G(A)\rightarrow\overline{F}_p$ that are left invariant under $G(\mathbf{Q})$ and right invariant under a certain open subgroup $K_N$.

Here $G(A)$ is the adelic group associated to $G$. Notice that the functions considered on the quaternion side are independent of the archimedean variable. Moreover, the double coset $G(\mathbf{Q})\backslash G(A)/K_N$ is finite and any mod $p$ system of eigenvalues arising from it can be lifted to characteristic zero.

Therefore applying the theorem and then lifting, we see that for any (char. zero) eigensystem $A=(a_\ell)$ arising from $M_k(\Gamma_1(N))$ there is a (char. zero) eigensystem $B=(b_\ell)$ arising from the space of locally constant functions $f:G(A)\rightarrow\mathbf{C}$ such that $A\equiv B$ mod $P$, where $P$ is a fixed prime of $\overline{\mathbf{Z}}$ lying over $p$.

Assuming that the automorphic form $\Pi_B$ on $G$ associated to $B$ is infinite dimensional, by the J-L correspondence we have that there is a cuspidal automorphic form $\Pi'_B$ on ${\rm GL}_2$ associated to the same eigensystem $B$. The type of $\Pi'_B$ at any finite place other than $p$ is the same as that of $\Pi_B$, while at infinity $\Pi'_B$ is the discrete series of lowest weight $2$. This basically says that there is a cusp form in $S_2(Np^n)$ whose associated system of eigenvalues is $B=(b_\ell)$.

We are only left with deciding when $\Pi_B$ is infinite dimensional, or can be chosen as such. This happens only for systems of eigenvalues of the form $B=(\chi(\ell)(1+\ell))_{\ell\nmid pN}$, where $\chi:\mathbf{Z}/p\rightarrow\mathbf{C}^*$ is any character (in order to show this one has to consider the particular shape of $K_N$, which I did not even define..). The reduction mod $P$ of such eigensystems are all of the form $(\ell^k+\ell^{k+1})_{\ell\nmid pN}$.

Concluding: Let $A=(a_\ell)$ be a sytstem of char. zero eigenvalues arising from $M_k(\Gamma_1(N))$, with $p\nmid N$. Assume that the mod $P$ reduction of $A$ is not of the form $(\ell^k+\ell^{k+1})_{\ell\nmid pN}$. Then, there exists a cusp form in $S_2(\Gamma_1(Np^n))$ such that its associated system of eigenvalues $B$ is congruent to $A$ mod $P$.

  • $\begingroup$ Remarks: (1) I gave the "etale cohomology" answer and your answer is the "coherent cohomology" answer -- i.e. it doesn't go via Eichler-Shimura. In fact there's another "coherent cohomology" approach to the question -- explained e.g. in Edixhoven's Inventiones paper from 1990 or so on Serre's conjecture. (2) Minor point: I am not so sure about your claims about when $\Pi_B$ is infinite-dimensional. You have a tame level $\Gamma_1(N)$ and so your character $\chi$ can I think be ramified at these primes too. But all these cases can be dealt with using Eisenstein series anyway without any truble. $\endgroup$ Commented Jul 9, 2011 at 12:55
  • $\begingroup$ Kevin thanks for your comments. Why is it the "coherent cohomology" answer? Concerning (2), let me add that the open subgroup $K_N$ that I am using is locally at $\ell\neq p$ given by invertible matrices that are congruent to $(* *; 0 1)$ mod $N$ (and it is max'l pro-p at p). If a function $f:G(A)\rightarrow \mathbf{C}$ factors through the reduced norm $G(A)\rightarrow A^*$, and it is invariant to the right by $K_N$, and to the left by $G(\mathbf{Q})$, then $f$ has to be invariant to the right under $K_1$. $\endgroup$ Commented Jul 9, 2011 at 13:41
  • $\begingroup$ (This is because the determinant is on the $\ell$-th component of $K_N$ surjective onto $\mathbf{Z}_\ell^*$) A consequence of Serre' thm. is that then for any mod p Eisestein eigensystem $B$ arising from level $\gamma_1(N)$, and different from $(\ell^k+\ell^{k+1})$, there is an infinte dimensional cusp form $\Pi$ on $G$ such that the associated eigensystem reduces mod $p$ to $B$. I know this looks strange (I think once we discussed this issue on MO). $\endgroup$ Commented Jul 9, 2011 at 13:41
  • $\begingroup$ @Tommaso: I thought that the way to relate forms to functions on the adelic quotient space was via considering a mod $p$ modular form as a section of the coherent sheaf $\omega^k$ on the mod $p$ modular curve, and then restrict these sections to the supersingular locus and analyse a la Edixhoven's paper. $\endgroup$ Commented Jul 10, 2011 at 11:34
  • $\begingroup$ @Kevin: thanks, I see now why you spoke of coherent cohomology, modular forms live in degree zero in this context. $\endgroup$ Commented Jul 10, 2011 at 12:43

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.