## A non-liftable weight one form modulo p^2

I once idly asked RLT (around 2004ish) whether one could use Buzzard-Taylor arguments to prove that any representation:

$\rho: \mathrm{Gal}(\overline{\mathbf{Q}}/\mathbf{Q}) \rightarrow \mathrm{GL}_2(\mathbf{Z}/p^2 \mathbf{Z})$

which was unramified at p and residually irreducible (and modular) was itself modular (in the Katz sense). Galois representations of this flavour are obviously something I’ve thought about (and worked on with David G) quite a lot since then. But I have never actually seen any examples of mod p^2 forms which didn’t lift to characteristic zero. I asked George Schaeffer about it once, but his computations were only set up to detect the primes for which non-liftable forms existed rather than to compute the precise structure of the torsion in H^1(X,omega). But just today I stumbled across an example in relation to a pairing I learned about from Akshay (which I will tell you all about some other time).

The particular form (or rather pair, since it comes with a twist by the nebentypus character) occurs at level $\Gamma_0(103) \cap \Gamma_1(3),$ and is defined over the ring $\mathbf{Z}/11^2 \mathbf{Z}.$ It doesn’t lift to a weight one form mod 11^3. The nebentypus character is the only one it could be at this level and weight: the odd quadratic character of conductor 3. When I looked again at Schaeffer’s thesis, he does indeed single out this particular level as a context where computations suggested their might exist a mod p^2 form. (Literally, he says that a computation “seems to imply the existence” of such a form.) I guess this remark was not in any previous versions of the document I had, so I hadn’t seen it. Here are the first few terms of the q-expansion(s):

$g = q + 16 q^2 + 20 q^3 + 15 q^4 + 58 q^5 + 78 q^6 + 22 q^7 + \ldots + 91 q^{11} + \ldots + 104 q^{103} + \ldots$

$f = q + 105 q^2 + 115 q^3 + 15 q^4 + 63 q^5 + 96 q^6 + 22 q^7 + \ldots$

Some remarks. Note that the coefficients of g and f satisfy $a(g,n) = \chi(n) a(f,n)$ for all $(n,3) = 1$ and where $\chi$ is the quadratic character of conductor 3 (the nebentypus character). On the other hand, at the prime 3, we have

$\rho_f = \left( \begin{matrix} \chi \psi^{-1} & 0 \\ 0 & \psi \end{matrix} \right), \qquad \rho_g = \chi \otimes \rho_f = \left( \begin{matrix} \chi \psi & 0 \\ 0 & \psi^{-1} \end{matrix} \right),$

and so the eigenvalue of U_3 is the image of Frobenius at 3 under $\psi$ or $\psi^{-1},$ and hence satisfies the equality

$a(g,3)a(f,3) = \psi(\mathrm{Frob}_3) \psi^{-1}(\mathrm{Frob}_3) = 20 \cdot 115 = 1 \mod 121.$

I was temporarily confused about the fact that $a_q = 1 + q$ for the Steinberg prime $q = 103$ rather than $a_q = \pm 1,$ and thought for a while I had made an error or mathematics was wrong. But then I realized this was weight one not weight two, and so one should have instead that $(a_q)^2 = q^{-1}$ (note that $\chi(103) = 1.$) And it just so happens that the equation

$(1+q)^2 = q^{-1}$

in a weird coincidence has a solution very close to 103 (this is a solution mod 11^3, in fact). It’s easy enough to see that the image of rho and its twist contains $\mathrm{SL}_2(\mathbf{Z}/11^2 \mathbf{Z})$ with index two, and so has degree 3513840. (At this level, the only real alternative is that the form is Eisenstein, which it isn’t.) The root discriminant is not particularly small, it is

$103^{1 - 1/11^2} \cdot 3^{1/2} = 171.6970 \ldots$

Finally, the Frobenius eigenvalues at the prime p = 11 are distinct, which is easy enough to see because otherwise the coefficient of q^11 would have to be twice the squareroot of chi(11) = -1, which isn’t even a square mod 11.

Perhaps there’s not too much more to say about this particular example, but I was happy to come across it, nonetheless. Well, perhaps I should also say that I computed this example in SAGE, as I slowly wean myself off magma dependency.

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