## The nearly ordinary deformation ring is (usually) torsion over weight space

Let $F/{\mathbf{Q}}$ be an arbitrary number field. Let $p$ be a prime which splits completely in $F$, and consider an absolutely irreducible representation:

$\rho: G_{F} \rightarrow {\mathrm{GL}}_2({\overline{\mathbf{Q}}}_p)$

which is unramified outside finitely many primes.

If one assumes that $\rho$ is geometric, then the Fontaine–Mazur conjecture predicts that $\rho$ should be motivic, and the Langlands reciprocity conjecture predicts that $\rho$ should be automorphic. This is probably difficult, so let’s make our lives easier by adding some hypotheses. For example, let us assume that:

• A: For all $v|p$, the representation $\rho|G_{v}$ is crystalline and nearly ordinary,
• B: The residual representation ${\overline{\rho}}$ has suitably big image (Taylor–Wiles type condition.)

Proving the modularity of $\rho$ under these hypotheses is still too ambitious — it still includes even icosahedral representations and Elliptic curves over arbitrary number fields. Natural further hypotheses to make include conditions on the Hodge–Tate weights and conditions on complex conjugation.

We prove the following:

Theorem I: Assume, in addition to conditions A and B+, that

• C: The Hodge-Tate weights $[a_v, b_v]$ at each $v|p$ are sufficiently generic,
• D: If $F$ is totally real, then there exists at least one infinite place such that $\rho$ is even.

Then $\rho$ does not exist.

The condition B+ (which will be defined during the proof) is more restrictive than the usual Taylor–Wiles condition — we shall see from the proof exactly what it entails. Condition C will also be explained — but let us note that, for any suitable method of counting, almost all choices of integers are generic, even after imposing some condition on the determinant (say $a_v + b_v$ is constant) to rule out stupidities.

One should think of this theorem as follows. If $F$ is totally real, then condition D should be sufficient to rule out the existence of any automorphic $\rho$ in regular weight, because (for motivic reasons) such representations should be totally odd. On the other hand, if $F$ is not totally real, then the weights of any motive (with coefficients) should satisfy a certain non-trivial symmetry property with respect to the action of complex conjugation. So, for example, if $F$ has signature $(1,2)$, then either condition C or D should be sufficient, but we will require both. In fact, even condition C is stronger than what should be necessary. In addition to assuming regularity at all primes, it amounts to (on the representation theoretic side) insisting that none of the $\mathrm{GL}_2(\mathbf{C})$ weights are fixed by any conjugate of complex conjugation, whereas a single such example should be enough for a contradiction.

Perhaps a useful way to think about Theorem I is to make the following comparison. Hida proves the following theorem:

Theorem [Hida]: The nearly ordinary Hida family for $\mathrm{SL}(2)/F$ is finite over weight space and has positive rank if and only if $F$ is totally real and the corresponding ${\overline{\rho}}$ is odd at all infinite places.

On the other hand, a consequence of Theorem I is:

Theorem II: The fixed determinant nearly ordinary deformation ring of a residual representation ${\overline{\rho}}$ satisfying condition B+ is finite over weight space and has positive rank if and only if $F$ is totally real and the corresponding ${\overline{\rho}}$ is odd at all infinite places.

In both cases, I am only considering the deformation rings up to twist — the deformation ring of the character is torsion over the corresponding weight space whenever $\mathcal{O}_F$ has infinitely many units. Also in both cases, it is of interest to determine the exact co-dimension of the ordinary family — this is a difficult problem, because strong enough results would allow you do deduce Leopoldt by considering induced representations.

OK, so what is the argument? If you have read some of my papers, you can probably guess.

Assume that $\rho$ exists. Let $U$ be the representation corresponding to $\rho$. Now replace $U$ by $V = {\mathrm{Sym}}^2(U)$. Now replace $V$ by the tensor induction:

$\displaystyle{ W = \bigotimes_{G_F/G_{{\mathbf{Q}}}} V}$

of dimension $3^{[F:{\mathbf{Q}}]}$. We now let C be the condition that $W$ has distinct Hodge–Tate weights. To see that this is generic, it really suffices to show that there is at least one choice of weights for which this is true. But one can let the weights of $U$ up to translation consist of the 2-uples $[0,1]$, $[0,3]$, $[0,9]$, etc. and then the weights of $W$ are, again up to translation, $[0,1,2,\ldots,3^{[F:{\mathbf{Q}}]} - 1].$ We now let B+ be the condition that the residual representation is absolutely irreducible, and that the prime $p > 2 \cdot 3^{[F:{\mathbf{Q}}]} + 1$. This is generically true, and amounts to saying that the conjugates of ${\overline{\rho}}$ under $G_{{\mathbf{Q}}}$ are sufficiently distinct. Since the dimension of $W$ is odd, and because it is essentially self-dual (exercise), orthogonal (obvious), nearly ordinary (by assumption), has distinct Hodge–Tate weights (by construction), satisfies the required sign condition (automatic in odd dimension), we deduce that it is potentially modular by [BLGGT]. In order to win, it suffices to show, by a theorem I made Richard prove, that the action of complex conjugation on $W$ has trace $\pm 1.$ However, this is equivalent to condition D (see below). QED.

One can relax condition B+ slightly by only inducing down to the largest totally real subfield of $F$. On the other hand, there are plenty of examples to which Theorem II applies. I think one can take any elliptic curve $E/F$ without CM and such that $j_E \in F$ does not lie in any subfield of $F$, and then take $p$ to be any sufficiently large ordinary prime which splits completely in $F$ (caveat emptor, I didn’t check this). Of course, the condition that $p$ splits isn’t really necessary either, I guess…

The second theorem follows along the exact same lines — the conditions are strong enough to ensure, using results of Thorne, that the nearly ordinary deformation ring of (the now residual) representation $W$ is finite over weight space, which translates back into finiteness of deformations of $U$ over weight space. The result is obvious if $F$ is totally real and ${\overline{\rho}}$ is odd. Otherwise, we choose a sufficiently generic point in weight space (in the sense of C), and then, by Theorem I, we see that the specialization of the nearly ordinary deformation ring at that point must be torsion.

It remains to compute the sign of $W$. This is an exercise in finite group theory, we only recall enough of the details for our purposes. Let $V$ be a representation of $H$ of dimension $d$. Consider the tensor induction:

$\displaystyle{\bigotimes_{G/H} \sigma V}.$

Let $T$ denote a set of representatives of right cosets of $H$ in $G$. Let $t g \in T$ denote the corresponding choice for the coset $Htg$. For $g \in G$, let $n(t)$ denote the size of the $\langle g \rangle$-orbit which contains $T$. If $g = c$ has order $2$, then either $n(t) = 1$ or $n(t) = 2$ Certainly

$t c^{n(t)} t^{-1} \in H, \quad t \in T.$

Let $T_0$ be a set of representatives for the $\langle g \rangle$ orbits on $T_0$. Then (proof omitted)

$\displaystyle{\phi^{\otimes G}(c) = \prod_{t \in T_0} \phi(t c^{n(t)} t^{-1})}.$

We observe that:

1. If $n(t) = 2$, then $\phi(t c^{n(t)} t^{-1}) = \phi(t t^{-1}) = \phi(1) = d$.
2. If $n(t) = 1$, then $tct^{-1} \in H$ and $\phi(t c t^{-1})$ is what it is. For example, it is $0,\pm 1$ if and only if $V$ is ${\mathrm{GL}}$-odd with respect to $tct^{-1}$.

Now suppose that $G = G_{{\mathbf{Q}}}$ and $H = G_{F}$. The elements $tct^{-1}$ are exactly the different complex conjugations of the representations of the conjugates of $H$. We deduce:

1. If $\dim(V)$ is even, then $W$ is ${\mathrm{GL}}$-odd if and only if there exists at least one real place of $F$ such that $V$ is ${\mathrm{GL}}$-odd.
2. If $\dim(V)$ is odd, then $W$ is ${\mathrm{GL}}$-odd if and only if $F$ is totally real and $V$ is ${\mathrm{GL}}$-odd at every real place.

Equivalently, a product of even integers can equal zero only if at least one of them is zero, and a product of odd integers can equal $\pm 1$ if and only if all of them are $\pm 1$.