## Explicit $p$-adic local Langlands for $p = 2$.

$\newcommand\Q{\mathbf{Q}}$ $\newcommand\Z{\mathbf{Z}}$ $\newcommand\Qbar{\overline{\Q}}$ $\newcommand\Gal{\mathrm{Gal}}$ $\newcommand\ss{\mathrm{ss}}$ $\newcommand\F{\mathbf{F}}$

Fix an integer $k \ge 2$, let $G = \Gal(\Qbar_2/\Q_2)$, and let $\chi: G \rightarrow \Qbar^{\times}_2$ be the $2$-adic cyclotomic character.

There are exactly two $2$-dimensional $2$-adic representations $\rho: G \rightarrow \mathrm{GL}(V)$ such that:

1. $\mathrm{det}(\rho) = \chi^{k-1}$,
2. $V$ is crystalline with Hodge--Tate weights $[0,k-1]$,
3. Frobenius does not act semisimply on $D_{\mathrm{cris}}(V)$.

(they differ by an unramified quadratic twist.) Choose either of these representations. Let $H$ denote the kernel of $\rho$. Then what is $(G/H)^{ab}$ in terms of $k$? More precisely, what is $(G/H)^{ab}$ as a quotient of $G^{ab} \simeq \widehat{\Q^{\times}_2}$ in terms of $k$? (The answer may depend on the twist, but it is easy to pass from the answer for one to the other.)

A related but slightly different question is to determine $(\overline{V})^{\ss}$; I am interested in both questions.

In principle, I could imagine trying to guess the answer in any particular case by fixing $k$ and assuming that $(G/H)^{ab}$ was locally constant as a function of the trace of Frobenius $a_2$, and then trying to find classical modular forms which were nearby. However, in practice, the valuations of $a_p$ are never that large. This suggests fixing $a_p$ and varying the weight. This leads to my second question: is $(G/H)^{ab}$ locally constant as a function of sufficiently large positive integers $k$ (with the topology coming from weight space)? Does Berger's argument (proving the analogous statement for $(\overline{V})^{\ss}$ when $a_p \ne 0$ and some other mild conditions) apply in this case?

When $k = 2$ the answer to the question is given by Fontaine--Laffaille theory; the quotient $(G/H)^{ab}$ corresponds to the projection $\widehat{\Q^{\times}_2} \rightarrow \Z^{\times}_2 \times \Z/2\Z$ sending $4$ to $1$ (for either twist).

EDIT: To respond to Kevin's comment, I think that $\overline{V}$ may well sometimes be reducible for some $k$. In general, we know that if $a_p$ is sufficiently small then $V$ is "close" to the corresponding representation with $a_p = 0$. On the other hand, if $a_p$ is small, then $-a_p$ is also small, and thus $V$ is "close" to its quadratic twist $V \otimes \eta$ where $\eta$ is unramified. Now an actual equality $V = V \otimes \eta$ would imply that $V$ is induced from the unramified quadratic extension of $\Q_p$, and this is indeed the case when $a_p = 0$. Yet it doesn't imply that $\overline{V}$ is irreducible, since the character one is inducing might be equal to its Galois conjugate modulo $p$. If $a_p = 0$, then $V$ is given by the induction of $\phi^{k-1}$ where $\phi \equiv \omega_2 \mod p$. Since $\omega^{p+1}_2$ is invariant under conjugation, it follows that $\overline{V}$ is reducible if and only if $p+1$ divides $k-1$. (When $p$ is odd and $k$ is even this doesn't happen very frequently.) Returning to small $a_p$, the fact that $\overline{V} = \overline{V} \otimes \eta$ is enough to say that there is always a surjection $$(G/H)^{ab} \rightarrow \Gal(\F_{p^2}/\F_p)$$ if $a_p$ is small enough. So one aspect of my question would be "is $a_2 = 2^{(k-1)/2}$ close enough to $a_2 = 0$ to deduce that $\overline{V}= \overline{V} \otimes \eta$? The result of Berger-Li-Zhu, even imagining that it applied with $p = 2$, would not be sufficient, because the bound there is something like $(k-2)/(p-1)$.

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I think the $k=2$ and $k>2$ situations are very different. When $k>2$ the $\phi$-stable subspace doesn't correspond to a sub-Galois representation, right? (I don't think it's weakly admissible). So I think the image of $\rho$ will be huge, e.g. for $k=4$ it might well be all of $GL(2,\mathbf{Z}_p)$ or some such thing (I am more hesitant to make an assertion about $k=3$ because there are more likely to be square roots of $p$ involved). Do you have any reason to believe any different? [surely for $k>2$ $\rho$ will be irred and now there's not too many choices left...]. That's my punt, anyway. – Kevin Buzzard Feb 4 2011 at 7:59
Aah---the point is beginning to dawn on me. Do you suspect that the mod $p$ representation is reducible in these cases? You know there's some recent preprint of Berger where he gives an algorithm to compute these things---does it apply in the $p=2$ non-ss case, and if so, what happens for $k=4$? – Kevin Buzzard Feb 4 2011 at 8:05
@Kevin : My algorithm does apply to the case $p=2$ (with a minor modification), but has not been implemented (for any $p$). – Laurent Berger Feb 4 2011 at 16:29
Laurent---your paper with Breuil proves that for $k\leq p+1$ or so, the associated mod $p$ representation is irreducible, right? The image lands in the normaliser of a non-split Cartan. So for $p=2$ and $k=3$ the image must already be quite big... – Kevin Buzzard Feb 4 2011 at 19:47
@Eggnog: you may find some of this : umpa.ens-lyon.fr/~mvienney/publis/reduction.pdf interesting. – Laurent Berger Feb 7 2011 at 12:13