## Serre 1: Calegari 0

I just spent a week or so trying to determine whether Serre’s conjecture about the congruence subgroup property was false for a very specific class of S-arithmetic groups. The punch line, perhaps not surprisingly, was that I had made an error. I should note that I was pretty skeptical during the entire endeavour, so the final resolution was not a surpise, but there were still a few interesting twists along the way. (Thanks to Matt for some informative chats along the way.)

Let’s start by recalling Ribet’s proof of (what is one of many statements known as) Ihara’s lemma. Let $\Gamma$ be a congruence subgroup of $\mathrm{SL}_2(\mathbf{Z})$ of level prime to q. There is a congruence subgroup $\Gamma_0(q)$ defined in the usual way, where $q|c.$ However, there is also a second copy $\Gamma^0(q)$ of this group inside $\Gamma$ with $q|b.$ (Well, there are $q+1$ copies of this group, but let’s just consider these two for the moment.) The two groups are conjugate inside $\mathrm{GL}_2(\mathbf{Q}),$ but not inside $\Gamma$. An argument of Serre now shows that the amalgam of $\Gamma$ with itself along these groups (identified by conjugation by $[q,0;0,1])$ is the congruence subgroup $\Gamma[1/q]$ of $\mathrm{SL}_2(\mathbf{Z}[1/q]).$ That is, the congruence subgroup where the local conditions away from q are the same as $\Gamma.$ The Lyndon long exact sequence associated to an amalgam of groups shows that there is an exact sequence: $H_1(\Gamma_0(q),A) \rightarrow H_1(\Gamma,A)^2 \rightarrow H_1(\Gamma[1/q],A) \rightarrow 0,$

for any trivial coefficient system $A.$ Now the group $\Gamma[1/q]$ satisfies the congruence subgroup property, so the group on the right is easily seem to be finite and Eisenstein. By duality, there is also a map $H_1(\Gamma,A)^2 \rightarrow H_1(\Gamma_0(q),A),$

and the composition of this map with the projection above is a matrix with determinant $T^2_q - (1+q)^2.$ A bookkeeping argument now gives Ribet’s famous level raising theorem (taking coefficients $A = \mathbf{F}_p.$)

Fred Diamond and Richard Taylor generalized this theorem by replacing the modular curve with both definite and indefinite quaternion algebras. The actual theorem itself at this point is probably quite easily to prove by the K-W method, but that’s not relevant here. Instead, let’s think a little about the proof. The more difficult and interesting case is when $\Gamma$ comes from the norm one units in an indefinite quaternion algebra, which we consider from now on (the case of Shimura curves over $\mathbf{Q}.)$ Morally, the proof should be exactly the same. The only wrikle is that the corresponding group $\Gamma[1/q]$ is notoriously not known to satisfy the congruence subgroup property, although Serre conjectures that it does. Diamond and Taylor instead argued in the following way. (Let us specialize to the case of trivial weight, which is the only relevant case here.) Suppose that p is a prime greater than two and different from q. Then instead of working with Betti cohomology, one can instead, via a comparison theorem, use de Rham cohomology. The Hodge filtration consists of two pieces, one of which is $H^0(X,\Omega^1),$ and the other is $H^1(X,\mathcal{O}_X).$ They then investigate the kernel of the map: $H^1(X,\Omega^1)^2 \rightarrow H^1(X_0(q),\Omega^1)$

where everything is now over $\mathbf{F}_p.$ Here the two maps are the two pullbacks under the two projections $X_0(q) \rightarrow X.$ They now show that element in the kernel gives rise to a differential $\omega$ which vanishes at all the supersingular points or does not vanish at all. The first is impossible by a degree argument when $p > 3,$ and the second is always impossible. They conclude that, returning to etale cohomology, any kernel of the map $H^1(X,\mathbf{F}_p)^2 \rightarrow H^1(X_0(q),\mathbf{F}_p)$

must lie entirely in one filtered piece, from which they deduce it must be Eisenstein. But let’s look at this argument a little more closely. Even in Ribet’s case, the conclusion is really much stronger than level raising for non-Eisenstein primes; there is a very precise description of the kernel (or cokernel in homology) in terms of the homology of $\Gamma$ coming from congruence quotients, which one can compute quite explicitly. So Ribet’s theorem also gives level raising for Eisenstein representations in some contexts. In particular, for a suitable choice of congruence subgroup (with $p > 2)$ one can make the group $H_1(\Gamma[1/q],\mathbf{F}_p)$ vanish identically. Let’s now return to the argument of Diamond and Taylor when $p = 3.$ All the comparison theorems are still valid, so the only issue is that the map $H^1(X,\Omega^1)^2 \rightarrow H^1(X_0(q),\Omega^1)$

does have a kernel, namely, if one takes the “Hasse Invariant” $A$ which vanishes to degree one at all supersingular points, then the two pullbacks of $A$ to $X_0(q)$ coincide up to a scalar, and so the kernel is at least one dimensional. In fact, the argument of Diamond-Taylor shows that the kernel is at most one dimensional. But what does this mean in the proof of Ihara’s Lemma? It means that, assuming $X$ has good reduction at the prime $p = 3,$ the level raising map always has a kernel, and thus $H_1(\Gamma[1/q],\mathbf{F}_3)$ is always non-trivial.

This now seems suspicious: all we need to do is find a quaternion algebra which doesn’t have any congruence homology of degree $3.$ If the quaternion algebra $D/\mathbf{Q}$ is ramified at a prime $r,$ then the congruence homology coming from this prime (for $p \ne 2 r)$ is a subgroup of the norm one elements of $\mathbf{F}^{\times}_{r^2},$ which has order $r + 1.$ So it makes sense to take a quaternion algebra ramified at $7 \cdot 13,$ since these are the two smallest primes different from 3 which are congruent to 1.

Because this seemed to contradict Serre’s conjecture, I decided for fun to explicitly compute a presentation for the amalgam $\Gamma[1/2]$ to help work out what was going on. To first start, one needs a presentation for $\Gamma.$ John Voight (friend of the blog) has written a very nice magma package to do exactly this. (More precisely, it’s trivial to write down a presentation — $\Gamma$ is torsion free, and hence a surface group $\pi_1(\Sigma_g)$ for a genus $g$ one can compute via other means to be $g = 7;$ the point is that one also wants an explicit representation as well as an explicit identification with the norm one units of the correponding quaternion algebra.)

I then took an embarassingly long time to compute the subgroup $\Gamma_0(2).$ The main issue was finding a suitable element in $D$ to play the role of $\eta = [2,0;0,1]$ in $M_2(\mathbf{Q}).$ There certainly exists such a unit in $D \otimes \mathbf{Q}_2,$ so in real life one just has to find an actual norm 2 unit which is sufficiently close 2-adically to this. However, I am absolute rubbish at mathematica and so repeatedly made the following error: when you define suitable quaternions $i, j, k$ in $D \otimes_{\mathbf{Q}} E$ for some quadratic splitting field $E/\mathbf{Q},$ and then compute with the matrix $a + b i + c j + d k,$ mathematica helpfully interprets “ $a$” here as $[a,a;a,a]$ rather than a multiple of the identity, a programming decision which makes a lot of sense, said no one ever. I did this more times than I care to admit. Then, using John’s program, one can find the subgroup $\Gamma_0(2),$ and then write down a presentation for the amalgam by conjugating this subgroup by $\eta$ and identifying the correpsonding elements via a solution to the word problem as words in the original generators, and then substitute the names for these generators for the second copy of $\Gamma.$ The result is a group with $14 + 14$ generators and $2 + 38$ relations (corresponding to the 2 surface relations and the fact that $\Gamma_0(2)$ has $3(14 - 2) + 2 = 38$ generators.) Finally, one takes this group, plugs it into magma, and finds: $\mathtt{AbelianQuotientInvariants(G);}$ $\mathtt{> }$

There are known congruence factors coming from $7+1$ and $13 + 1,$ but here one sees that the factor of three survives!

And then, shortly after this point, I realized that $\mathrm{SL}_2(\mathbf{F}_3)$ has a quotient of order $3,$ because it is $A_4.$ So that degree three quotient is congruence after all… Oops! Still, it’s nice to see that mathematics is consistent.

However, at this point one might just ask why can’t one replace the quaternion algebra $D/\mathbf{Q}$ by (say) a real quadratic field in which $3$ is unramified and inert. Serre got away with it above because $\mathrm{SL}_2(\mathbf{F}_3)$ is solvable, but $\mathrm{SL}_2(\mathbf{F}_9)$ has the good manners not to have any such quotients. So why can’t one now run the same argument as above and disprove Serre’s conjecture? That’s a good question, and the entire argument works, up to the issue of defining the Hasse invariant. Quaternion algebras over fields other than $\mathbf{Q}$ are a bit of a disaster, because they don’t have nice moduli theoretic descriptions. That doesn’t mean they don’t have Hasse invariants, however. But now what happens, which at this point in the game I suspected but was confirmed and explained to be by George Boxer (Keerthi also suggested a computation which would lead to the same conclusion): the Hasse invariant is no longer a section of $\Omega^1 = \omega^{\otimes 2} = \omega^{p-1},$ but rather a section of $\omega^{p^2 - 1},$ and this has too large a degree to contribute to the cohomology of $\Omega^1.$ Since $2^2 - 1 > 2,$ it still has too large a degree when $p = 2,$ which is good, because otherwise working at this prime could have given rise to a counter-example to Serre’s conjecture because $\mathrm{SL}_2(\mathbf{F}_4) = A_5$ is perfect. (One would have to be slightly more careful with p=2 about comparison theorems, but at least one is dealing with curves.) So the conclusion is that Serre’s conjecture still stands, but only because various Hasse invariants in low weight are exactly accounted for by the solvability of $\mathrm{SL}_2(\mathbf{F})$ when $|\mathbf{F}| = 2,3.$

(Also, completely randomly and apropos of nothing, this link is now the top hit on the web to the search “Fred Diamond’s Beard.”)

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