## There are non-liftable weight one forms modulo p for any p

Let $p$ be any prime. In this post, we show that there is an integer $N$ prime to $p$ such that $H^1(X_1(N),\omega_{\mathbf{Z}})$ has a torsion class of order $p$. Almost equivalently, there exists a Katz modular form of level $N$ and weight one over $\mathbf{F}_p$ which does not lift to characteristic zero. We shall give two different arguments. The first argument will have the virtue that the torsion class is non-trivial after localization at a maximal ideal $\mathfrak{m}$ which is new of level $N$. The second argument, in contrast, will produce torsion classes at fairly explicit levels. Neither proof, unfortunately, implies the existence of interesting Galois representations unramified at $p$ with image containing $\mathrm{SL}_2(\mathbf{F}_p)$. Rather, the classes will come from deformations of characteristic zero classes. (This post is an elaboration of my comment here.)

The first argument: Let $K/\mathbf{Q}$ be an imaginary cubic extension unramified outside $p$ with Galois closure $L/\mathbf{Q}$ with Galois group $S_3$. There is a corresponding Galois representation:

$\rho: G_{\mathbf{Q}} \rightarrow \mathrm{Gal}(L/\mathbf{Q}) = S_3 \rightarrow \mathrm{GL}_2(\mathbf{Q}_p).$

This representation is modular. Suppose for convenience that $p > 3$. Associated to $\rho$ is an absolutely irreducible residual representation $\overline{\rho}$. Let $R$ denote the corresponding universal unramified deformation. The only characteristic zero deformations are dihedral. Let $R^{\mathrm{dh}}$ denote the corresponding universal unramified dihedral deformation ring. It’s easy to identify this ring explicitly; it is

$R^{\mathrm{dh}} = \mathbf{Z}_p[C_E \otimes \mathbf{Z}_p],$

where $C_E$ is the class group of the imaginary quadratic subfield $E$ of $L$. The ring $R$ will fail to be $\mathbf{Z}_p$-flat exactly when $R \ne R^{\mathrm{dh}}$. Fortunately, this can be determined purely from the reduced tangent space of $R$. Note that

$\mathrm{ad}^0(\rho) \simeq \rho \oplus \eta,$

where $\eta$ is the quadratic character of $E/\mathbf{Q}$. The reduced tangent space of $R^{\mathrm{dh}}$ is the Bloch–Kato Selmer group $H^1_f(\mathbf{Q},\overline{\eta})$, where $H^1_f$ denotes the subring of cohomology classes which are unramified everywhere. So it all comes down to finding $K$ so that $H^1_f(\mathbf{Q},\overline{\rho})$ is non-zero. However, an elementary argument using inflation-restriction shows that this is equivalent to showing that the class number $h_K$ of $K$ is divisible by $p$. So we are done provided that we can find a suitable $K$ with class number divisible by $p$. (I should mention, of course, that we are using the theorem that $R = \mathbf{T}_{\mathfrak{m}}$ which was proved by David Geraghty and me.) The last step follows from the lemma below; the argument is essentially taken from this paper of Bilu–Luca.

Lemma: Fix a prime $p > 3$. There exists an imaginary cubic field $K/\mathbf{Q}$ of
discriminant prime to $p$ and class number divisible by $p$.

Proof: Consider the field $K = \mathbf{Q}(\theta)$, where

$(\theta^2 + 1)(\theta - t^p + 1) - 1 = 0$,

and $t$ is an element of $\mathbf{Q}$ to be chosen later. Note that $\theta^2 + 1$ is manifestly a unit in $K$. We may compute that

$(\theta^2 + \theta + 1) \theta = (1 + \theta^2) t^p.$

Since $(\theta,\theta^2 + \theta + 1) = (1)$ is trivial, it follows that $(\theta) = \mathfrak{a}^p$ for some ideal $\mathfrak{a}$. We shall show that, for a suitably chosen $t$, the element $\mathfrak{a}$ is non-trivial in the class group. If $\mathfrak{a}$ is trivial, then, up to a unit, $\theta$ is a $p$th power. On the other hand, the rank of the unit group of $K$ is one, and $\theta^2 + 1$ is a unit. Hence, it suffices to choose a $t$ such that:

1. $\theta^2 + 1$ generates a subgroup of $\mathcal{O}^{\times}_K$ of index prime to $p$. Equivalently, $\theta^2 + 1$ is not a perfect $p$th power in $K$.
2. None of the elements $\theta (\theta^2 + 1)^i$ for $i = 0,\ldots,p-1$ is a perfect $p$th power in $K$.
3. The polynomial defining $K$ is irreducible.
4. The discriminant of $K$ is not a square.
5. The discriminant of $K$ is prime to $p$.

By working over the function field $\mathbf{Q}(t)$ instead of $\mathbf{Q}$, one finds that the first four conditions hold for all $t \in \mathbf{Q}$ outside a thin set. (The discriminant $\Delta$ is always negative, so the signature of the field is always $(1,1)$.) On the other hand, the discriminant of the defining polynomial is $-3 \mod t$, so if one (for example) takes $t$ to be an integer divisible by $p$ then the discriminant will be prime to $p$. Note that the set of integers divisible by $p$ will contain elements not in any thin set, because the number of integral points of height at most $H$ in a thin set is $o(H)$.

Second Argument: Let $E = \mathbf{Q}(\sqrt{-23})$, and let $L = E[\theta]/(\theta^3 - \theta + 1)$ be the Hilbert class field of $E$. There is a weight one modular form of level $\Gamma_1(23)$ and quadratic character corresponding to the Galois representation:

$\rho: \mathrm{Gal}(L/\mathbf{Q}) \rightarrow \mathrm{GL}_2(\mathbf{Q}_p).$

Lemma: Let $p > 3$. Let $q = x^2 + 23 y^2$ be a prime such that $q \equiv 1 \mod p$. Equivalently, let $q$ be a prime which splits completely in $L(\zeta_p)$. Then

$\# H^1(X(\Gamma_1(23) \cap \Gamma_0(q)),\omega)^{\mathrm{tors}}$

is divisible by $p$. More generally, for any prime $q$, the quantity above is divisible by any prime divisor of $a^2_q - (1+q)^2$, and $a_{\ell} \in \{2,0,-1\}$ for a prime $\ell$ is the coefficient of $q^{\ell}$ in $q \prod (1-q^n) (1 - q^{23 n})$.

Proof: This follows from “level–raising” in characteristic $p$ for weight one forms. Under the hypothesis that $a^2_q - (1+q)^2$, we find that there is more cohomology (over $\mathbf{Z}_p$) in level $\Gamma_1(23) \cap \Gamma_0(q)$ than is accounted for by oldforms. Assuming that there is no torsion, this is inconsistent with the fact that there are no newforms in characteristic zero, because weight one forms cannot be Steinberg at any place. (The easiest way to see this is that the eigenvalue of $U_q$ would have to be non-integral — it also follows on the Galois side from local-global compatibility, but this is overkill.) Note that level–raising in this context does not follow from classical level–raising — for the details I refer to you my fifth lecture in Barbados on non-minimal modularity lifting theorems in weight one.

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### 2 Responses to There are non-liftable weight one forms modulo p for any p

1. AV says:

That is a lovely second argument indeed.

I still don’t understand why there is the experimental “suppression” of torsion classes here by characteristic zero classes. Hopefully the Arakelov version of analytic torsion will clarify the situation eventually – at first glance, the regulator seems negligible in this case, but something else must be going on.

• In a weak sense, the second argument is the same as the first, except one replaces class groups with *ray* class groups, and every field has a ray class group of order divisible by p if one is allowed to choose the conductor.

I do indeed look forward to seeing Arakelov analytic torsion!